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If 0 lights -> turn on corrupted If 1 light -> All lights off if corrupted is in 1st position, 1,2 on= 2nd light corrupted 1,3 on= 3rd light corrupted 1,4 on= 4th light corrupted 2,3 on= 4th light corrupted 2,4 on= 3rd light corrupted 3,4 on = 2nd light corrupted If 2 lights -> Make the corrupted light either be the only one on or the only one off If 3 lights -> All lights on if corrupted is in 1st position. The rest is the same as “if 1 light” 4 lights on -> turn off corrupted I think I’ve found an alternate solution.
@@-ZH suppose that the 2nd light is corrupted and the third light is on. According to your answer, we would turn on both the 1st and 2nd light and also make sure to close the 3rd one. That means doing three different tasks when you can only do one.
It's neat to see the ingenious answer, but sometimes these riddles' rules are so specific that it feels like they started with the solution and worked their way back to the question.
I mean, the answers actually pretty simple if you don’t think about it in binary or whatever. You can always either make one light on or one light off be the corrupted one unless three are on or off to begin with. In that case you can make four distinct patterns that you can assign whatever meaning you want to. It’s not that complicated. I still had to think about it for a while before I could solve it though.
More like, the very specific way they solve it is so far removed from what most people would even consider that it feels ridiculous? It seems like there are other viable solutions that don’t involve the binary rules they showed, and that most people presented with this riddle would just not use binary. Yeah, TEDED is also trying to teach people new things about logic or mathematics with these riddles, but for some of them the shown solution is kind of a reach.
@@christianboi7690 They definitely started with the solution. Cuz most people would probably come up with some kind of answers like yours. It's simpler and more intuitive. I know I did too, I had a vague idea they might try to do something with binary, but I didn't explore further because your line of thinking is just much simpler and works just as well
I don't think I've ever solved one of these riddles on my own. Heaven knows I've tried. But I keep watching them anyways, hoping one day.... one day I'll be able to solve one.
XOR is actually so cool. Like addition, it's commutative and associative, and has the identity element 0. But the coolest part is that the xor inverse of a number is always itself, so if you want to add a specific number X to the xor sum of the lighted lamps, for example, if lamp X is off, then you simply turn it on (and commutativity and associativity does the work for you), but if the lamp is already on, then you can simply turn it off, and everything works the same way.
Me : *Explaining this to my team* My team : Okay Me : *Goes inside and sacrifice* My team : *Goes in for the attack* My team : Wait, do we number the disk from the left side or the right side?
Ok. I'm at the problem-pause screen. Here's my solution. If all of them are off or on, I can indicate which one by having an "odd-one-out." I can also create the same scenario if two are off and two are on. If there are three one way and one the other way, I can flip one to make them all match. Or I can flip one to make a 2-2 split down the middle. Or I can flip one to make every other light match. Or I can flip one to make a 1-2-1 pattern, where the two in the center match each other. So my instructions to my allied agents are these. 1. If there's three lights matching, the corrupted disk is the one that doesn't match. 2. Failing that, see the chart below All matching - disk 0 2-2 pattern - disk 1 Alternating pattern - disk 2 1-2-1 pattern - disk 3 Edit: I think I used something that works because of what was explained? I think? I just didn't use the binary math to explain it?
I thought it was something about feeling the heat of the lightbulbs. If the corrupted disk light was on, you would turn it off and the squad could feel the heat from a bulb that was currently off but still emitting heat from when it was on. If the corrupted disk was off, you would turn it on and the squad notices that one bulb isn’t as warm as the other bulbs that are on. Assuming the bulbs that are off are cool and the bulbs that are on are supposed to be hot. It probably doesn’t work in some cases though, what do I know.
I did a different solution too: Step One: If all lights are on or off, the rightmost disk is corrupted. Step Two: If exactly one light is on or off, that disk is corrupted. Step Three: If exactly two lights are on, ignore the rightmost disk and repeat Step Two. If you start with all lights on or off, swap the corrupted disk. If you start with exactly two on or off; swap the other on off if the corrupted is on or the other off on if the corrupted is off. If you start with exactly one on or off; if said different disk is the corrupted disk, turn the rightmost disk on/off; if it's the rightmost disk turn the corrupted disk on/off, if it's one of the other two swap the other one so that only the corrupted disk and the last disk are on/off.
Instead of using binary numbers, i thought you could directly signal which light was corrupted to your friends. If the number of lights on when you get there is an even number (0. 2 or 4), then it is straightforward because you can turn one light switch to make it such that the odd light out (the only one turned off or on) is the corrupted light. It is a bit more complicated if the number of lights on is an odd number (1 or 3), but you could arrange a code with your friends, one example which I have listed below. 1. To signal that the 1st light is the corrupted one, turn the light switch that makes it such that all 4 are on, or all 4 are off (in other words: on-on-on-on or off-off-off-off) 2. To signal that the 2nd light is the corrupted one, turn the light switch that makes it such that the lights 1 & 2 share the same configuration, while lights 3 & 4 share the opposite configuration (in other words on-on-off-off or off-off-on-on) 3. To signal that the 3rd light is the corrupted one, turn the light switch that makes it such that lights 1 & 3 share the same configuration, while lights 2 & 4 share the opposite configuration (in other words on-off-on-off or off-on-off-on) 4. To signal that the 4th light is the corrupted one, turn the light switch that makes it such that lights 1 & 4 share the same configuration, while lights 2 & 3 share the opposite configuration (in other words on-off-off-on or off-on-on-off) No matter what the starting configuration of lights is, you should always be able to signal each of the 4 lights depending on which one is revealed to be corrupted. When you expand the problem and number of lights on a much greater scale, the binary solution definitely makes a lot more sense, but in the case of just having 4 lights, I believe this is a more straightforward solution.
Unluckely this does not work because you can't turn more than 1 light, so, if for example you find in front of you on on off off, you will not be able to signal that the 1 light is infected since you established that the signal should be either off off off off or on on on on. I know cause I ran into the same problem thinking of the solution, but I managed to solve it. Next comment i explain
@@regularhuman2887 1+10=110 would work if you're using concatenation as the symbol of addition. Otherwise no, the video said : 01 + 10 + 11 = 110 and if you converted that from binary to decimal it would be : 1 + 2 + 3 = 6, which is 100% correct.. Don't mock binary if you don't understand it
I've watched enough Ted-Ed riddles to know that the solution involves parity but the binary is what I didn't think of. Hats off Ted-Ed, thanks for the entertaining educational riddle!
There is a much easier way. Take the first 3 bits, ignore the 4th when trying to figure out which bit is corrupted. Rule 1: If the corrupted bit is 1, 2 or 3 then switch one bit within them to make the corrupted bit different. Rule 2: If the corrupted bit is 4 then switch one bit within the first 3 to make them all same. Rule 3: If the original state of the first 3 bits are already what you wanted then switch bit 4 since it will be ignored.
I came up with a very similar solution, albeit a lot more complicated. It was basically halfway using binary. The first two disks represented 0s when off and 1s when on. If I had a light situation that had both of them in the wrong position, disk 3 would be on to flip what they meant. An example would be if the first 3 were on and the fourth was corrupted, I would need the first two to be on, or the first two off with the 3rd on. In that scenario I turn the 3rd off and it works. If it was the first one, I need both off so with 3 on, I just turn on the 4th one since it can be ignored and the 3rd switch flips the output. As I said, more complicated lol, but it works
@@bill5922 Uhh that was long time ago but let me see if I still remember. First reason is because the chosen agent must flip one switch. So in case you don't want to switch 1, 2 or 3, you can switch the 4th which is irrelevant. The second reason is that 3 bits are enough to show which one of the 4 discs is corrupted: 100 or 011 -> disc 1 010 or 101 -> disc 2 001 or 110 -> disc 3 000 or 111 -> disc 4 And if you try it out, you will see that you can always (at any state) set the 3 bits to the disc you want with just one switch or no switch at all.
My solution was to tell my team something like: "Use the binary number indicated by these two lights to find the answer, but reverse the digits if this third light is on. Ignore the fourth." It's pretty similar to what's in the video, I suppose, but I didn't think about using addition.
This is actually equivalent Though in your solution disk 4 acts like disk 1 in their solution, disk 3 acts like disk 4, disk 2 acts like disk 2 and disk 1 acts like disk 3
The Ted-Ed riddle playlist is just growing 1 by 1, and I love binge watching/listening to all the riddles in the background while I play games, there’s just something relaxing about that combination and it makes the Ted-Ed Riddle Playlist my favourite playlist on UA-cam for that reason. I really hope there are a lot more riddles to come 😁🧐
This was a fun puzzle, and I have found an alternate solution. If each light represents a bit with OFF meaning 0, and ON meaning 1, then the four lights will represent a 4-bit number which in decimal represents a number between 0 to 15. We only need 2-bits of information to represent the corrupted disk, so this means, we can use 4 numbers from the 16 numbers to represent each disk. After working it out (and easy to check) let us make the following definition: 0, 1, 14, 15 represents the first disk 2, 3, 12, 13 represents the second disk 4, 5, 10, 11 represents the third disk 6, 7, 8, 9 represents the fourth disk Starting from any given initial configuration of lights (i.e. any number from 0 to 15), it is easy to check that you can always flip just one light to get at least one of the four numbers to represent the correct corrupted disk. As an example, if initially the lights are 0 1 1 0 (=6 in decimal) you can flip bit#1 to get 0 1 1 1 (=7 or disk 4), or flip bit#2 to get 0 1 0 0 (=4 or disk 3), or flip bit#3 to get 0 0 1 0 (=2 or disk 2) or flip bit#4 to get 1 1 1 0 (=14 or disk 1). This works for all possible cases. I like this solution better. :)
For those who doesn't like math, I've created a solution that only uses patterns. There's 4 types of pattern 0 = on X = off Full (0-0-0-0 or X-X-X-X) Side (X-X-0-0 or 0-0-X-X) Center (X-0-0-X or 0-0-X-X) Split (0-X-0-X or X-0-X-0) These patterns are named as it will be useful for Scenario 3 Let's call the corrupted disk as "c-disk" Scenario 1: If all lights are on, turn off the c-disk. If all lights are off, turn on the c-disk Scenario 2: If there are 2 lights that are on, flip one light that is in the same state as the c-disk, so if the light of c-disk is on, turn off the other light that is on and if c-disk is off, turn off the other light that is also turned off. Scenario 3: Only one light is on or off Now this is the tricky part, but Its actually easy, what we need to do is create a pattern and each pattern will represent which c-disk is corrupted. If the c-disk is bulb no.1, we'll use the "full" pattern, so all we have to do is flip the odd light. If the c-disk is bulb no.2, we'll use the "side pattern, so basically we'll just have to flip the other light that is on the same side as the odd light. If the c-disk is bulb no.3, we'll create the "center" pattern that will look like a sandwich, basically the one in the center is in the same state, and the one on both sides are the same If c-disk is on bulb no.4 we'll create the "split" pattern, where we just have to create an alternating pattern like on-off-on-off or off-on-off-on Now try it yourself 😉
Another solution that should work for any amount of disks (although difficult to find a pattern that works), is to assign four 4-bit configurations to indicate which disk is the corrupt one such that by flipping one bit on any random configuration, you can always change the configuration to indicate which disk is the corrupt one. (Sorry for long and convoluted sentence). In this example you can use the following: "Disk light configuration" = "which disk is the corrupt one" 0000 = 1 0001 = 1 0010 = 2 0011 = 2 0100 = 3 0101 = 3 0110 = 4 0111 = 4 1000 = 4 1001 = 4 1010 = 3 1011 = 3 1100 = 2 1101 = 2 1110 = 1 1111 = 1 With this you can always change any configuration to indicate the correct disk, for example. If agent A comes in and is told that disk number 3 is the corrupt one, and the current configuration is 0110, he can simply change it to be 0100. I'm pretty sure there's a boolean theorem describing this and gives a method to find such a table for x amount of disks, but I can't remember what it's called ¯\_(ツ)_/¯
I ended up doing the same thing and found that for a number of disks D and infected drive X, there is always an ending configuration C that is exactly one binary digit away, with the relationship being described by the function C = (D-.5)-abs(floor(X/2)-(D-.5)), assuming the disks are numbered from 0 to D-1.
Fantastic puzzle with a great explanation! I figured out another way to solve this particular scenario though: Take the first 3 disks as bits: off -> 0, on ->1, the 4th disk is a 'throw-away'. Encode the position of the corrupted disk as follows: disk 1: 000 or 111, disk 2: 100 or 011, disk 3: 010 or 101, disk 4: 001 or 110 With this encoding it is ensured that one can indicate the position of the corrupt disk via a single toggle (if the code is already correct, just toggle the 4th disk): 1 2 3 4 000 100, 010, 001 111 011, 101, 110 2 1 3 4 100 000, 101, 110 011 111, 010, 001 3 1 2 4 010 000, 011, 110 101 111, 100, 001 4 1 2 3 001 000, 011, 101 110 111, 100, 010
Ooh, very nice solution! I had a different solution (which Dillon Teo also found and explained in this comment: ua-cam.com/video/i0WH4SFpeB8/v-deo.html&lc=UgxZ1dRebvvZQ5Tzojh4AaABAg ), but this one is much more elegant, with fewer "if-thens".
Yeah, this is the solution I came up with. I thought it made a lot of sense to only use as many bits as you need and save the last one in case they're all already in the correct position.
I had a similar solution. Focus on the first three to figure out the answer. If the first three are all on or off then it means it’s the fourth one. If it’s not the forth one then it’s the odd ball from the first three.
I personally came up with a different solution, though I could be wrong. If they are all the same value, then flip the corrupted one. If there is two of each, flip the one with the same value of the one that is corrupted. So, if the team enters to only one switch having a distinct value, they will know that one is corrupted. However, if there is a 3-1 split, we will have to change the strategy a little. If the first switch is corrupted, then change the odd one out to give everything the same value. If the second one is, then create a clean split down the middle (aka 1100, 0011). If the third one is, choose the option that has a one switch gap in between (so 1010, 0101). If the last one is corrupted, then mirror it across the center (1001, 0110).
If anyone is interested, I came to a different solution that works perfectly as well and I think is easy to understand. Suppose we assign each disk/light a value. The first disk = 0, the second disk = 1, the third disk = 2 and the last disk = -3. Notice the last one is -3, not 3! The trick is now to flip a switch, so that the sum of these numbers equals the disk number (0, 1, 2, 3). If you end up in negatives, take the absolute value of the number. It works for every configuration, because you can add or substract any number of 0, 1, 2 or 3 from any configuration. It works so beautifully because we allow ourselves to go into the negatives. The sum of these disks can lie between -3 and 3. (-3, -2, -1, 0, 1, 2, 3). If the sum is at any of these points, you can always add or substract in such a way to go to any of the other values (in absolute sense). Examples with lights: - Suppose all the lights are off. Take the sum of these numbers, which equals 0. This means the first disk ( = 0 ) has the virus. You can flip any switch to indicate that disk has the virus (e.g. flip last light, sum of disks = (0+0+0-3) = -3 --> 3rd disk has virus). - Suppose all the lights are on. Sum of disks = 0 +1 + 2 -3 = 0. If the second disk has that virus, You can flip that disk to make the sum of disks = -2 --> 2. - Suppose only the second light is on (sum = 1). You can go to disk 0 by turning it off, stay at disk 1 by switching the first switch, go to disk 2 by adding -3 (1 - 3 = -2), and go to disk 3 by adding 2. Numerical examples: From -3 you can go to 0 by substracting -3. You can go to 1 by adding 2 (= -1 -> disk 1 has virus). To 2 by adding 1 (-3 + 1 = -2 --> disk 2). To 3 by adding 0. From -2 (= 1 + -3, these lights are on) you can go to 0 by adding 2, to 1 by substracting -3 (-2 - -3 = disk1), to 2 by adding 0 (-2 --> disk 2) and to 3 by removing 1 (-2 + -1 = -3 --> disk 3). From -1 (= 2 + - 3) you can go to 0 by adding 1, to 1 by adding 0, to 2 by substracting -3, to 3 by removing 2. From 0 you can go to 1 by adding or substracting 1, to 2 by adding or substracting 2, and to 3 by adding or substracting -3. From 1 you can go to 0 by substracting 1, to 1 by adding 0, to 2 by adding -3 and to 3 by adding 2. From 2 you can go to 0 by substracting 2, to 1 by adding -3, to 2 by adding 0 and to 3 by adding 1. From 3 (= 2 +1) you can go to 0 by adding -3, to 1 by substracting 2, to 2 by substracting 1 and to 3 by adding 0. In the numerical example we have explored every possible solution. Not sure if my explanation is clear, but it seems very elegant to me. Very easy to understand/use, just sum disk values and flip switch to get to the virus. I think the principle is the same as the solution in the video, just applied differently.
The way I solved this was to look at what information the others have. When the others come in, all they know is what the lights show, so they have 4 ordered bits. This gives 2^4 = 16 possible configurations, and from that they must be able to find which is corrupted - in other words, we need to sort these 16 possible configurations into 4 groups, so that of the configuration when they enter is in (let's say) group 1, then disk 1 is corrupted, etc. Furthermore, on my end, it must be possible for me to switch any configuration into any of the 4 groups, in order to indicate which disk is corrupted. And thus for any initial state when I enter, there must be at least one element in each group that is exactly 1 bit flip away from this state, so that I can put the configuration into the correct group. Thus for any configuration, each of its 4 bit flip "neighbours" must be in different groups. If you think about it, every solution must follow this rule actually, and necessarily fits into this scheme. It's just that other solutions have rules for what configuration corresponds to what disk, while this method allows for all possible configuration groups. In total, ignoring which group corresponds to which disk, there are 42 possible configuration groupings over all 4 groups. It's actually not that hard to find a working grouping. Keep picking configurations until all initial states are accounted for, i.e. any configuration is exactly 1 bit flip away from at least one of the elements in the group. Each configuration accounts for 4 initial states, so we need at least 4 in each group to account for all 16, which means we must split them into 4 groups of 4 - they just barely cover all possibilities, there can be no overlap, so any configuration is actually 1 bit flip away from exactly one of the elements in each group - its 4 bit flip neighbours. For the first group, pick any configuration, it does not matter. This configuration is one bit flip away from 4 other configurations, and so those initial states are now accounted for. Crucially, this does not account for the configuration we picked, since we must flip exactly 1 bit. Thus the second configuration of the group must be exactly 1 bit flip away from our first configuration, with a total of 4 choices. These choices will never overlap with the first. We now have half the group filled, and half the initial states accounted for. For the second half, we can use a trick: If we take our current two configurations, and flip all 4 of their bits, then we get two more configurations that also work. This is because these are 4 bit flips away from their counterparts, and so they are 3 bit flips away from the other. Thus there is no overlap, and because they don't overlap each other either they must account for 8 confurations - the other half of the set. As it turns out, this is actually the only way to complete the group for this case of 4 bits. For the remaining groups, follow the same rules, while avoiding any configurations that have already been chosen. This should be easy. Note that this also explains why the number of disks must be a power of 2. Let n be the number of disks. Each configuration in the group accounts for n initial states, and so we need at least 2^n/n configurations in each group. This must be an integer, so we have to round up. The only times this divides cleanly is when n is a power of two, as otherwise there will be a prime factor other than 2 in n. There are n groups, and so we need ceil(2^n/n)*n total configurations. But we only have 2^n, so this is only possible if n is a power of 2.
There is a different solution that works without binary. If zero, two, or four lights are on, you can always flip a switch that leaves the corrupted disk in a different state than the other three (only one on or only one off). If there are exactly one or three lights on, you can always configure the lights into one of four patterns: AAAA, AABB, ABAB, or ABBA. Where A and B refer to lights on/off or vice versa. Then just assign a disk number to each pattern.
Yeah brother! I worked it out exactly the same way. My solution is the same for when we get in and we see 0,2,4 lights on and we make the corrupted one unique so that the team comes in they destroy the unique light. However, with when we arrive and 1 or 3 lights are on. It's a bit different for me and we'll do the following and team will recognize that as follows: Zero or All On = Corrupted disk is the left most one Two Lights are on with 1 Off Light between them = Corrupted disk is the right most one If the two center ones OR the left most & right most lights are on = Corrupted disk is 2nd from the left most. If the either two corner ones are ones together (1100,0011) = Corrupted disk is 3rd from the left most. I checked it out. It's legit. Works every time. Took me a while to figure it out but I'm so glad I made a workable strategy for this one.
I have a different solution. Let the three first discs represent the corrupted disc.100 and 011 are 1, 010 and 101 are 2, 001 and 110 are 3, 000 and 111 are 4. That way you no matter which lights are on you can communicate which one of the disc is corrupted with 1 step, and if the lights are showing the correct answer just flip the fourth one.
This is equivalent to the solution I thought of! I figured that since you HAVE to flip one switch, one light would have to be designated as meaningless. From there it seemed to me that no matter what state the lights were in, I could make any one of the remaining three lights the odd one out, or make them all the same. Be nice if I could figure out a mathematical proof, though...
I was thinking something along the lines of (since you can communicate with your team) setting up a bunch of scenarios like “if all off or all on but one, that’s it.” Etc…. But creating all those rules essentially breaks out to this math. This is freaking cool
I made a mapping table from all possible states to modified states. Then I assigned each modified state a number from 1 to 4 and make sure that you can reach all possible numbers (1 to 4) from every starting state.
That's what I did. If the 4 lights are considered 4-bit binary number, then if the lights indicate 0,1,14,15 - Disk A is corrupted; 2,3,12,13 - B; 4,5,10,11 - C; and 6,7,8,9 - D. All possible starting states can go to exactly one of the 4 disks by changing one bit.
This is, of course, just one of many possible solutions. More generally, there are 16 possible combinations of lights. For any given combination, there are 4 other combinations which can be reached by toggling a single switch. You need to assign each combination of lights to mean a particular disk is the corrupted, and do so in such a way that, for any given starting combination, there's one reachable combination to indicate each possible corrupt disk.
Hey Ted Ed , this document is about the rich and beautiful of Vietnamese (Rich and Beautiful Vietnamese - VIETNAM IS Identities Vietnamese is the common language of Vietnamese people - a language that is growing more and more in line with the development of the country towards international integration. Vietnamese is not only diverse by the system of tones and 29 letters, but also rich by the combination of local languages of the above regions and the territory of Vietnam.More specifically, going from South to North on this S-shaped piece of land, we will clearly see the difference in production and living methods of people in different regions. Similarly, each region also uses different languages called dialects. In the South the word "go back" will often be pronounced "go back". In the Central region, people will not be able to distinguish the first consonants "D", "Gi" and "N", but they will only pronounce it as "Gi". For example, old, old and house will all be read as old by the people here. As for people living in the North, they often confuse the pronunciations between the two letters "l" and "n" such as: "profit and loss" will be read as "effort". Not only are they different in pronunciation, but the local language also has a difference in terms of vocabulary: Corn - Corn, Snakehead - Banana fish, Going to work - Going to mam ... And also thanks to the above differences. The system of vocabulary as well as pronunciation in Vietnamese becomes extremely diverse and rich. Not only that, this also creates an interesting, it can be said to be a bit "difficult" for foreigners when they learn Vietnamese and then travel to Vietnam.Why is it called "challenging"? Because in Vietnam, each locality has different names such as: Bu (Thai Binh), Bam (Bac Ninh), U (Ha Nam), Ma (Hue) and Ma are commonly used in the Southern region. With just one word "mother", Vietnamese people have created many different ways of calling it. This is both a "brain hack" game for foreigners when learning Vietnamese, and also an extremely interesting thing that makes them want to learn more about the Vietnamese language as well as the country and people of Vietnam. Thanks to the presence of local languages, the Vietnamese vocabulary and phonetic system becomes richer and more diverse. Each region has its own voice, expressing the daily life of our people from different regions. It is the interesting things from the local language that have contributed to creating a culture imbued with the national identity of the Vietnamese people) . It's not perfect and also have a lots of wrong letter so please help me fix the wrong and install it into your UA-cam channel, please!
So, whatever the state of total parity maybe at some point, like EO. You have choices to either flip both parities(first and second place), or only first parity, or only second parity. 00 corresponds to no flips, 01 flips second, 10 flips first, 11 flips both. By fliping to the parities that corresponds to corrupt light, we can always reach the right answer. This isnt the complete explanation of all thats happeing, but rest is along these lines.
Alternative solution: Since you're allowed to plan ahead with your team before going in, you can assign number values 1 to 4 to certain configurations. You will need to create 2 cheat sheets, one for odd configurations and one for even configurations. With that sorted out before going in, you will always be able to set a configuration that will tell your team which disk is corrupted by flipping exactly one switch
I absolutely brute forced my way to a solution: If my team finds only one light on or only one light off, destroy that one. If they find any combination of two offs and two ons or all ons and all offs, i made a cheat sheet with each of the 8 combinations equal to 1, 2, 3 or 4. Not quite elegant but it would work
YES! Thank you! This was my solution as well. Since you're allowed to plan ahead with your team before going in, you can assign number values 1 to 4 to certain configurations. You will need to create 2 cheat sheets, one for odd configurations and one for even configurations. With that sorted out before going in, you will always be able to set a configuration that will tell your team which disk is corrupted by flipping exactly one switch.
Honestly, there is an inspiration to come up with this problem. The idea is that when you are given the corrupt light (or whatever it was), you are given one of 4 possible pieces of information, and then you have the power to show your team one of 4 possible lighting arrangements. So if there is a solution, it has to be precisely created, and that is exactly what is shown in the video.
This is interesting. I've come to another solution though and it seems to work as well. I knew binary code was the answer and we need such an arrangement that with a single switch we should be able to give code for the corrupted disk. Out of the 16 possible values, I've assigned 4 for each disk such that with a single switch we can change it to a value belonging to the corrupted disk. [0000,0001,1111,1110] - values for disk 1 (group 1) [0010,0011,1101,1100]- values for disk 2 (group 2) [0100,0101,1011,1010]- values for disk 3 (group 3) [0110,1000,1001,0111]- values for disk 4 (group 4) For example - if the initial setting of the disks are 1010 (belonging to group 3) and our corrupted disk is 2, We can switch off the 4th place converting 1010 to 0010 (belonging to group 2) which will tell us the corrupted disk is 2. If our corrupted disk is 1, switch on the third place from 1010 to 1110. If the corrupted disk is 4, switch off the second place from 1010 to 1000. For corrupted disk three, although the existing values belong to group 3, since we must make a change, if we switch on the first place from 1010 to 1011, it would remain in group 3, signaling the corrupted disk is 3.
There's a much simpler answer. Only work with the first 3 and ignore the fourth. No matter what combination of lights there are there will always be a way to make one of the first three lights the odd one out if it's corrupted. Example: 1, 3 and 4 are on, corrupted is 1. You hit 3 so 1 is on and 2 and 3 are off which signals it to be 1. If the corrupted light is already the odd one out just flip 4. Also, if the corrupted is 4, you make it so that the first 3 are either all on or all off which works for any combination. If they're already all on or off just flip 4 since its on or off is meaningless.
I actually managed to solve this one! I didn't translate each light into binary, but I created my own ruleset that I would communicate with my team that would work for any solution. It's a bit complicated, but it essentially follows the same rules as the one shown in the video
Wow, I was so happy with my solution and it felt so TED-Edy, but then it was totally different. I basically only use the middle switches for the code: 00 for disk 1, 01 for 2, 10 for 3 and 11 for 4. But what if you have to flip two switches? Then you can establish the first light is a true/false one. If it's off, you read the middle slots as they are. If it's on, you read the opposite, so 10 means 01 and so on. Since the virus makes you flip a switch, if everything is already in its proper position when you arrive you just flip switch 4, which means nothing.
I really like the simplicity of your solution. I did something very similar, but without the use of a true/false light. First, number the lights starting at 0, so you can directly translate the binary instead of shifting everything by 1. So instead of numbering lights 1234, they are 0123. Then I divided the last three lights into two groups of two with lights 1 and 2 together in a group, and lights 2 and 3 together in a group. I then said that if a group had an even number of lights on (0 or 2) then that grouping was equal to 1. If only 1 light was on in a group, then it was equal to 0. So for example 1100 would translate to 01. So that way you can reach any binary number from 00, 01, 10, 11 with just one flip without having to remember if you need to reverse your answer based on a true/false light. If the left digit needs to change, flip light 1, if the right digit needs to change, flip light 3. If both digits need to change, flip light 2. If nothing needs to change, flip light 0.
Alternate answer 1: Tell your team which disk was corrupted and if the virus stops you just tell him the hardest Ted Ed riddle you know which should stump him and buy you enough time to break into the mainframe and destroy the corrupted disk
I make the valiant sacrifice, and my team rushes in... They have no idea what they I did, we didn't agree on anything before hand. They destroy all the disks. Destroying the corruption and freeing the system.
I solved it, but with a much simpler method 1) if your friends see that there is an odd light out, it’s the corrupted one. - This takes care of any setup where you start with two of each (two on or two off), and any set up where you start with all of one (all on or all off) because with both set ups you can manipulate them so that any one light is the odd one out. (Now the tricky part is dealing with if you are giving a set up where three lights are the same and one light is different. That’s where you can discuss with your friends different patterns that indicate a different corrupted light) 2) a) if your friends see that all the lights are matching (all off or all on) then the first one is corrupted. b) If your friends see that it’s two pairs next to each other (double on then double off, or double off then double on) the the corrupted light is the second one. c) If your friends see that it’s alternating (on off on off, or off on off on) then the corrupted light is the third one. d) If your friends see that there’s a pair in the center with a pair split on the outside (on off off on, or off on on off) then the corrupted light is the fourth light. This works for any scenario and all this pretty much gives a “cheat sheet” for your friends to know which one is corrupted.
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YAY! I finally solved one of these riddles - and not using the same method to reach it as TED-ED! Basically there are 16 different configurations of how the lights can be illuminated. Each configuration can (and indeed must) be switched to one of 4 other configurations. Therefore you can assign a number from 1-4 to each of the configurations, and agree in advance with your team what configuration refers to what. For example, in my solution: All lights off = 3. Left hand light only = 1, Middle left only = 2, Middle right only =3, far right only = 4. You can basically map it out so that no matter what the starting combo is, you can always change it to be on a configuration that you've agreed in advance is one of the numbers 1-4.
For anyone interested, here is my solution (where “X” is a light that’s switched off, and “0” is a light that’s switched on): XXXX = 3 0XXX = 1, X0XX = 2, XX0X = 3, XXX0 = 4 00XX = 1, 0X0X = 2, X00X = 4, XOXO = 2, XX00 = 1, 0XX0 = 4 00X0 = 3, 000X = 4, X000 = 1, 0X00 = 2 0000 = 3 I’d give my team that list above, and go in, knowing that whichever combination is showing when I arrive, I can switch it to be the number that I need, out of the above list. Hope that makes sense, please ask what I mean if not 😊
And before anyone says it - no my answer isn't as elegant as TED-ED's, and it can't be used to solve bigger versions of the puzzle - but then again the riddle never asked for that, so :P
Before watching the answer: My brain inmediatly thought of binary. With more brainstorm, i came up with "What if i assign to 4 values between 0 and 15 the tag [Corrupt Disk = 1], to other ones [Corrupt Disk= 2]..." And then i had to find those values. And i come up with: Binary value: [00] [01] [02] [03] [04] [05] [06] [07] [08] [09] [10] [11] [12] [13] [14] [15] Corrupt Disk: [d1] [d2] [d3] [d4] [d1] [d2] [d3] [d4] [d4] [d3] [d2] [d1] [d4] [d3] [d2] [d1] The thing is to make the binary lights output to be: - 0, 4, 11 or 15 if the corrupt disk is 1 (That is: 0000, 0100, 1011, 1111). - 1, 5, 10 or 14 if the corrupt disk is 2 (That is: 0001, 0101, 1010, 1110). - 2, 6, 9 or 13 if the corrupt disk is 3 (That is: 0010, 0110, 1001, 1101). - 3, 7, 8 or 12 if the corrupt disk is 4 (That is: 0011, 0111, 1000, 1100). This is because in each case, those values can be obtained with a single change of a light in a 4 lights array. EX for Corrupt Disk is 1: 0110 (6) -> 0100 (4); 1110 (14) -> 1111 (15); 0101 (5) -> 0100 (4). About to comment down in this same comment to react to the TEd-ed answer.
I am keeping my strat, but the Ted-ed answer is not only more easy to understand to follow and elegant, but easy to scale up. I still cannot explain how i come up with *that* tag assigment to each value. I just started from the extremes and, when I realized that it was a reflection, I just went to fill the gaps with the most obvious pattern. And worked lol.
@@sirnikkel6746 I did the same method, but I used a Karnaugh map to obtain the assignments. It's a table that is characterised by a switch of 1 digit when you move from a box to an adjacent one.
I was about to do a lengthy reply, and then I saw this comment. I did the EXACT same thing you did. I agree the Ted-ed answer is much more elegant. I'm curious why this solution works though.
I managed to solve this one, but I came up with a much more complicated and less intuitive and generalizable answer. Basically, if you go into the mainframe and there are 0 or 4 active lights, flip the corrupted disk. If there are 2 on and 2 off, find the good disk that is the same setting as the corrupted one and flip it. In these cases, your team will discover the lights split 1-3, and can destroy the odd man out. If there are 1 or 3 lights on at the start, things get more complicated. If the corrupted disk is the first one, flip the odd man out. If it isn’t the first disk, then if it has a different setting from the first disk, flip whichever of the first and corrupted disks isn’t the odd man out. If the corrupted disk isn’t the first one but has the same setting, then flip whichever of the other two disks has the same setting as the first disk. Then, when your team goes in, they’ll either discover 2 on and 2 off (in which case they’ll check the setting of the first disk, find the other disk with the same setting, and destroy it), or all 4 the same setting (in which case they’ll destroy the first disk). For example, if the lights are initially set 0010, you could indicate disks 1, 2, 3, and 4 with 0000, 0011, 1010, and 0110 respectively.
so, if i understand correctly: use disc 11 as a NOT operation, discs 01 and 10 to point to a different disc, and disc 00 to not change anything if its already correct
I would have preferred this explanation as well. The "break the rules regard 2 as 0" thing seemed like an unnecessary complication, but I am speaking from a programming background.
I came up with a different solution. I will use 0 for off and 1 for on. 2^4 = 16 possible starting points which are as follows: --- 2 options of all the same number (0000/1111) -> then flip the faulty one --- 6 options of two on and two off (eg. 1010) -> flip in a way that so that the faulty one is always the odd one (eg cont. 1st faulty: 1000, 2nd: 1011, 3rd: 0010, 4th: 1110) --- 8 options of only one odd either on/off (eg.0100, 1110) -> there are 4 non-symmetric patterns that can be created by flipping only one light. The patterns are discussed and mapped with the team before the sacrifice. For eg. : 1100/0011 -> 1st faulty 1010/0101 -> 2nd 1001/0110 -> 3rd 0000/1111 -> 4th Hence, when the team comes in together to attack and they see that: - only 1 light is on and all the rest are off (or vice versa), then the odd one is the faulty one. - two lights are on and the other two are off, then they follow the mapping done above
So I broke this down into 5 cases (which is really 3, but whatever), based on how many lights are on when you go in. If all the lights are on, or all off, you can swap just the corrupted light and your team will know to hit the different one. If 2 are on and 2 are off, ensure that the corrupted disk is showing a different light from the other 3. So then you have just the case where 1 or 3 lights are on at the start. In this case we will end up in one of 8 different end cases: all lights on, all off, or 6 patterns of 2 on 2 off. By assigning opposite cases to the same disk, you assign each pattern to one of the disks. So 1111 and 0000 would mean, perhaps, the right most disk, while 1010 and 0101 would mean, perhaps, the 2nd from right. In any case where 1 or 3 are on you can get to one of the settings for whichever disk is corrupted. It doesn't scale as well as your answer though, I didn't get to the idea of assigning the disks themselves values and adding them.
Sorry dude it's totally wrong 3 people have made the same MISTAKE IN THE CHATS AND I HAVE HAD TO REPPY 3 TIMES !! LETS ASSUME THE TWO LIGHT ON AND 2 OFF SCENARIO ,LETS SAY LIGHT NO 1 AND 2 ARE ONE BUT NOT CORRUPT , NO 3 IS JUST LIKE 1 AND 2,meaning not corrupt , BUT ITS SWITHCED OFF AND SO THAT LEAVES LIGHT NO 4 AS CORRUPT AND ALSO SWITCHED OFF , THEN ACCORDING TO YOU THE LIGHT WHICH IS CORRUPTED MUST BE MADE ODD ONE OUT ISNT IT ?? WHICH IN THIS SCENARIO IS LIGHT NO 4 . AND THE ONLY POSSIBLE WAY TO MAKE LIGHT NO4 ODD ONE IS TO CHANGE THE STATUS OF LIGHT 1 AND 2 WHICH ARE BOTH ONE INTO OFF AND THEN LEAVE LIGHT 3 AS IT IS AND THEN TURN ON LIGHT NO 4 AS ON bcoz that's the corrupt one to be made an odd one out so that's 3 flips dude
Okay, I have another solution. I haven't wrote every possiblity down to check if that would work, but I am curious to have your opinion on this : What we need is to communicate a number : 1, 2, 3 or 4. If we think in binary, we only need 2 bits (lights) to write those numbers : they usually allow us to write numbers between 0 and 3, but let's say that we want our values between 1 and 4 so we'll say that 00 is 1, 01 is 2, 10 is 3 and 11 is 4. We will therefore use the first 2 lights to communicate the number of the corrupted disk. We have several possibilities now : let's say that when I come in I am told which disk is corrupted and realize that I need to switch exactly one of the first 2 lights to get the correct number : that's what I do, great. But now imagine that I need to switch both lights (for example, the corrupted disk is the 4th one and both lights are off). Our solution is to invert our code : if we decide that lights off don't mean 0 anymore but 1 and lights on mean 0, then it's the same as switching both lights. Now how do I tell my team what means 0 and what means 1 ? Third light is here to help. The situation of this light will indicate what configuration means 1. So in our example, if I want to indicate the number 4 with both my first 2 lights off, I need my 3rd light off too. If it is on when I come in, I switch if off and problem solved. But what if it is already off ? Everything is how I want it, I don't need to change anything. Well, let's switch 4th light. This is the useless light, doesn't provide any information. I could just leave everything the way it is and my team would be able to determine the solution. But the rules say I have to switch a light, the 4th one will serve in that case. So to synthetize : if, when I come in, and considering the version of the code the 3rd light is currently telling me : - I need to turn exactly one of the first 2 lights for them to give me the correct number : I switch it. - I need to switch both of them : I switch the 3rd one, this is equivalent. - I don't need to switch anything, I switch the 4th one. What do you think ?
You can do it in a less abstract way by making a code: If 0, 2 or 4 lights are on you can make the one different from the others; If it's 3 and 1 (or 1 and 3, same thing) you can make 4 patterns: alternative, grouped, middle/edges and all the same, which would correspond to 4 predetermined positions positions; It's nice how it doesn't matter whether lights are on or off, just their relation to the other lights. The given solution cooler and much more practical, but at this scale you can find unintended solutions still, which makes the riddle more fun as you have less to go on.
You can just number the lights 0,1,2,3 then work out the sum of them mod 4 and it works in exactly the same way. More than that it will work for any number of lights, not just powers of 2.
I have a way simpler solution. You can tell your crew the plan... If the lights are 1,1,1,1 then you can switch one off to leave an odd one out. Likewise, if the lights are 0,0,0,0 you can switch one on to leave an odd one out. If the lights have 2 on and 2 off, then you can ALWAYS switch one light in a way to leave the special light as the odd one out (try a mental experiment to prove this example if it's 1,0,0,1 and the special light is in position 2, you can simply switch position three to make position 2 the odd one out). The tricky part is when there is 3 of one kind and 1 of another when you first get there. If this happens, you can create one of four patterns that you've agreed with your team will correspond to a specific light. For example, if you switch the odd light, you will make uniform lights, and that could correspond to spot one. The other three combinations are alternating (ex 1,0,1,0), same at either end (ex 1,1,0,0) or same at The opposite end (ex 1,0,0,1). You can create those four combinations for any sequence in which you have three of the lights the same and one of the lights different (run a mental experiment to prove this). I hope that was helpful!
Alternative answer: If there are 0, 2 or 4 lights on, you can flip one so the odd one out (the only one which is on/off) is corrupted. If there are 1 or 3 on, you can ignore the first light, and make the odd one out corrupted (if the corrupted is already odd one out, swap the first light) If there are 1 or 3 on, and the first one is corrupted, you can make all of the lights the same (all on or all off) For the decoder: If there is 1 or 3 on, the odd one out is corrupted. If there are 2 on, the one matching the first light is corrupted. If there are 0 or 4 on, the first light is corrupted
I arrived at the same answer through a different thought process, feels nice to see divergent thinking converge to the same answer 4 bits: Useless bit: flip if what u want is already done The "negative" bit: take the reverse of the following bits Last 2 bits: 2 bit number
Taking the reverse here means that u treat 0 as 1 and 1 as 0 if the second bit is a 1, else treat them as normal. This is just a rudimentary thinking of parity flipping ig
I used patterns to solve it. If you come in and 0, 2 or 4 lights are on, just flip one so that the corrupted disc is the odd one out. Now, if you come in and 1 or 3 lights are on it's a bit trickier, but the math still works out. You have to disregard the exact values and instead use patterns. All these can be got from any 1/3 lights on: 0000 and 1111 -> disc 1 0011 and 1100 -> disc 2 0110 and 1001 -> disc 3 0101 and 1010 -> disc 4 You just have to agree which disc corresponds to which pattern beforehand.
Your solution was much neater and more extendable than mine, but I think mine still works: If you see all four lights on or all four lights off, flip the corrupted switch (so that it becomes the odd one out) If you see two lights on and two lights off, flip whichever is the same as the corrupted switch (so that the corrupted switch again becomes the odd one out) If you see three lights on and one light off, or vice versa, there is always one switch you can pull to get the lights into each of the following four patterns: - All lights the same (all on or all off) - Lights alternate between on and off - Left two lights on and right two lights off, or vice versa - Middle two lights on and edge two lights off, or vice versa So you can assign each of those four patterns to one of the disks. If your team comes in and sees one light is the odd one out, they know that's the right one. Otherwise they can identify the correct pattern and act accordingly.
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I would not let myself watch the solution until I figured it out. This is way more elegant than what I came up with. I treated the 4 mainframes as a bit in a number from 0 to 15 represented as binary. When you walk in the 4 disks can represent any one of those 16 numbers. If the team comes in and sees the number 8 (1000) it has to mean the same thing regardless of how you found it or which switch you flipped to make it that way. So 16 possible numbers need to be grouped into 4 groups of 4 numbers. And each group must contain a possible outcome for any scenario you walk into. The possibilities look like this START | Switch 1 , Switch 2, Switch 3, Switch 4 0 | 8,4,2,1 1 | 9,5,3,0 2 | 10,6,0,3 3 | 11,7,1,2 4 | 12,0,6,5 5 | 13,1,7,4 6 | 14,2,4,7 7 | 15,3,5,6 8 | 0,12,10,9 9 | 1,13,11,8 10 | 2,14,8,11 11 | 3,15,9,10 12 | 4,8,14,13 13 | 5,9,15,12 14 | 6,10,12,15 15 | 7,11,13,14 The logic works in reverse as well. For example, the only starting numbers that would let you get to 15 by flipping a switch are 7,11,13,14 A pattern emerges where the 16 numbers fall into groups that are exclusive, meaning only one number from the group is possible in any starting position. Group A: 1,2,15,16 Group B: 2,3,12,13 Group C: 4,5,10,11 Group D: 6,7,8,9 You tell the team that if the corrupted server is the first you will leave a number from group A, 2nd Group B and so on. So now, regardless of what the lights are when you walk in, you can flip a switch that shows the "Group" translating to the corrupted server. I realize after watching the solution that this is a very convoluted way to get the same binary math, but I wanted to do it on my own...
If you need help solving this riddle, a good idea is to simplify the problem down to its basics - two switches. A switch can either be on or off. It's a binary choice. We can represent off with a zero and on with a one. Now we can represent each configuration of lights as binary. 00, 01, 10 and 11. 00 is when both are off, 11 is when both are on. And with these binary numbers, we can also convert them into decimal to make it easier for us to understand. 0, 1, 2 and 3. As for the disks, we'll label them A and B, to keep them separate from all the numbers. Let's say we enter the mainframe and see 00 (0 in decimal). If we flip one switch, we get 01 (1) and if we flip the other switch, we get 10 (2). Now, let's say if disk A is broken, we turn 00 into 01. Therefore, if disk B is broken, we MUST turn 00 into 10 instead. Otherwise, we have NO way to communicate if disk B is broken. If we enter the mainframe and see 11 (3), we can flip a switch to get 01 or flip a switch to get 10, just like before. However, if we see 01, we can't make 01 or 10. Instead, we can make 00 (0) or 11 (3) so we must assign one of these to disk A and the other one MUST be with disk B. It doesn't matter if you assign 00 to A or B, as long as you set its partner, 11, to the opposite disk. There are two unique plans. You can assign 00 (0) and 01 (1) to A, while 10 (2) and 11 (3) to B. Here you will notice that A is 2 or =2. Or you can assign 00 (0) and 10 (2) to A, while 01 (1) and 11 (3) to B. Here you will notice that A is even while B is odd. Going back to binary, the right-most digit is the one that determines if the number is even or odd. Likewise, the left-most digit is the one that determines if the number is less than 2. In decimal, we have the ones, tens, hundreds, thousands digits. In binary, this is ones, twos, fours, eights. Instead of getting 10 times bigger, it gets 2 times bigger, as each digit can only be one of two things instead of the whole 0-9.
knew this one for once cause of error correcting codes. surprised they didnt mention that, this same concept is what makes the internet and your computers work today, and the guy who invented them won a turing award for this.
There are actually 567 (=24*24) strategies that work. Here is a simple strategy I came up with: The corrupted disk is the odd one out of the first 3; if the first 3 lights are all the same, the last disk is corrupted. Example: If the virus shows you 0101 (second and fourth lights on) and tells you the first disk is corrupted, you can make it the odd one out of the first 3 by flipping the third light: 0111. Example 2: Same lights, but now the fourth disk is corrupted. The first 3 lights must be the same, so now we flip the second light: 0001. Each possible strategy comes down to distributing the 16 possible light combinations into 4 groups: (for example) light 1 corrupted = 1000 / 0111 / 0110 / 1001 light 2 corrupted = 0100 / 1011 / 0101 / 1010 light 3 corrupted = 0010 / 1101 / 0011 / 1100 light 4 corrupted = 0001 / 1110 / 0000 / 1111 All groups must be reachable from any starting combination, therefore if two combinations are 2 flips apart, they must be in different groups! A consequence of this rule is that a combination and its complement (i.e. all lights flipped) have to be in the same group. Another consequence is that each group must have 2 even combinations (i.e. even number of 1's and 0's) and two uneven combinations. Consider these uneven combinations: 1000, 0100, 0100, 0001. They are all separated by two flips, so they must all be in different groups. You can assign them in 24 ways (4 possible choices for the first one, 3 for the second, and 2 for the third; 4*3*2 = 24). Assign the complements of each combination to the correct group, which is forced. This takes care of all the uneven combinations. Now for the even combinations: 0000, 0011, 0101, 1001. For the same reason as above, these can only be assigned in 24 ways. The complements are again forced. Multiply the 24 choices from the uneven combinations with the 24 choices from the even combinations to get the total number of possible groupings: 567. Finally, you can try to come up with a more intuitive method based on the grouping (like I did above).
Used a different solution: based on the limited configurations, you can use parity to create two different rule sets, one for even number of lights on and off (or just all on or off), and 3/1 lights on/off (and vice versa). So, say you show up and it’s a 2/2; play odd-man out, and flick the flight that makes the corrupt one not match the other three lights. If you show up and it’s a 4/0, then same rule. But for 3/1, we need a slight variation: it does not matter what the last light is. Do odd man out on the first three. If the last light is the corrupt, make all 3 others match. Ex: you find 1001. team shows up to 0001. It’s odd parity, meaning it came from a 2/2 or a 0/4; finding the odd light out it’s the last light. Ex2: you find 0010. The first is corrupt. You flip to make 0110. Your team shows up. Seeing a 2/2, they ignore the last light, and see that the first 0 is the odd light out.
Lots of possible solutions for this one. Here was mine: I need to deliver two binary pieces of information, which could give four possible outcomes in total, enough to communicate which disk. Now, consider two pairs of lights. I can't control the initial state, but if we consider the output of a pair not based on which lights are on, but on whether both lights in that pair are the same or different. But how to manipulate two pairs with one action? Overlap the pairs, with one unit in common. Now, with one flip, I can change the sameness of both pairs (flip the common unit), either (flip the one belonging to that pair) or neither (flip the last unit.) So, instructions left with the team: if the two units on the left are both on, or both off, the corrupted unit is on the left side, and if they're different, it's on the right. If the two units in the middle are both on or off, the bad drive is toward the inside of the row, otherwise on an end. From any initial arrangement, I can see which of those pairs is already appropriately same/different and which need to change to indicate the right drive, and can change either pair, both, or neither with one flip.
Found a different solution before looking at the original one: Condition 1: If all are off or all are on/two are on. Then flip one bulb so only the corrupted one remains (1 off or 1 on). The team will recognize that 1 of the lights is unique so they know that is the one. Condition two: 3 are on/1 is on. This one is a bit more difficult, but still easily solvable. Solution: if it’s 4, turn all of them on or off (depending on wether 3 lights are on or off). If it’s 3, flip one bulb until the split is 1-2/3-4 (if 1-2 is on, 3-4 should be off and vice versa). If it’s 2, flip one bulb until the split is 1-3/2-4. If it’s 1, split should be 1-4/2-3. The team will recognize that 2 lights are on and will destroy the corrupted one based on the split, or if all are on/off. Should work :)
I had a solution that only works with the four disks that is different. I'll represent the lights that are ON with an O and the lights that are OFF with an I. The corrupted disk will be in brackets. Let's say this was the arrangement numbering them 1 to 4 from left to right: 1 2 3 4 I O I (O) I would indicate to my team to look at the centre two disks (2 and 3) and see whether the corrupted disk is on the left(1, 2) or on the right (3, 4). If they are both ON/OFF then the corrupted disk is on the right(3, 4). If one is ON and one is OFF the corrupted disk is on the left(1, 2). Once we figure out the side the corrupted disk is on (left/right), we can indicate between the two odd and even numbers which one it is. If both lights on the side the corrupted disk is on are ON/OFF the even numbered disk is corrupted (2 or 4). If one is ON and the other is OFF, it's the odd number (1 or 3). Applying those rules to the light arrangement looking at the two in the middle, they indicate the light is on the left (1 or 2) currently. To change that we need to either turn OFF 2, or turn ON 3. The corrupted disk is an even number so 3 and 4 BOTH must be ON/OFF. For this scenario turning ON 3 indicates the corrupted disk is on the RIGHT and is EVEN. So then it has to be 4. If this was the new arrangement: 1 2 3 4 I O O (O) You would switch ON 1 because it wouldn't matter. If this doesn't work for an arrangement of lights let me know. I tested it out for a couple and they all seemed to work.
Hi was hoping if you could make a video on how we could try solving these . I solved one of your riddles ( the catching the rare fish Einstein problem one ) it took me quite a while but once I figured out how to solve it I pieced the puzzle also your videos are extremely engaging and keep my brain on its toes , hunting for answers constantly . Thank you for the amazing content . Have a great day !
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Brilliant.
Hey Ted Ed. You should do a video on the Collatz Conjecture.
I'm sure that would be interesting.
If 0 lights -> turn on corrupted
If 1 light ->
All lights off if corrupted is in 1st position,
1,2 on= 2nd light corrupted
1,3 on= 3rd light corrupted
1,4 on= 4th light corrupted
2,3 on= 4th light corrupted
2,4 on= 3rd light corrupted
3,4 on = 2nd light corrupted
If 2 lights ->
Make the corrupted light either be the only one on or the only one off
If 3 lights ->
All lights on if corrupted is in 1st position.
The rest is the same as “if 1 light”
4 lights on -> turn off corrupted
I think I’ve found an alternate solution.
@@-ZH suppose that the 2nd light is corrupted and the third light is on.
According to your answer, we would turn on both the 1st and 2nd light and also make sure to close the 3rd one. That means doing three different tasks when you can only do one.
It's neat to see the ingenious answer, but sometimes these riddles' rules are so specific that it feels like they started with the solution and worked their way back to the question.
Most likely they did ;-).
I mean, the answers actually pretty simple if you don’t think about it in binary or whatever. You can always either make one light on or one light off be the corrupted one unless three are on or off to begin with. In that case you can make four distinct patterns that you can assign whatever meaning you want to. It’s not that complicated. I still had to think about it for a while before I could solve it though.
That's how riddles usually work...
More like, the very specific way they solve it is so far removed from what most people would even consider that it feels ridiculous? It seems like there are other viable solutions that don’t involve the binary rules they showed, and that most people presented with this riddle would just not use binary.
Yeah, TEDED is also trying to teach people new things about logic or mathematics with these riddles, but for some of them the shown solution is kind of a reach.
@@christianboi7690 They definitely started with the solution. Cuz most people would probably come up with some kind of answers like yours. It's simpler and more intuitive. I know I did too, I had a vague idea they might try to do something with binary, but I didn't explore further because your line of thinking is just much simpler and works just as well
I don't think I've ever solved one of these riddles on my own. Heaven knows I've tried. But I keep watching them anyways, hoping one day.... one day I'll be able to solve one.
"Heaven knows I've tried"
Heard that phrase from somewhere about 8 years ago… 😳
Try Einstein's riddle by Dan Van Vieren
(It's actually not his)
This is the only one i could solve so far Took me a whole day
@@ReadTheBible001 I solved this too!
Why this is so relatable
@@sb3picman619, Frozen?
“Don’t use “beef stew” as a computer password. Its not stroganoff.”
That right there, made my day ❤️
That's the only thing I understood
Still have no idea what this means.
Edit: ohh wait it's meant to say "strong enough" 🤦♀️
www.youtubeuis.com/watch?v=deRHlcbwuucQ
@@mice2188 yep and stroganoff is a Russian meal mainly containing beef with sour cream
I love how the it also says “-unknown”
I can’t solve these riddles but the storyline is interesting so I’ll still watch
Same😞
🙋♀️ Me too! lol
I'm just happy when I understand the solution...which was not this time...
And this one especially is so beyond me 😊
That is exactly me!!!
I guess we are going to look over the fact that we are still vaporized after defeating the malware.
Sometimes a sacrifice needed to be made to save your homies
Bokurano
You live to die. It is your choice as to whether to take out the rest of your team, however.
But as software, you don't care either way!😁
Hopefully they backed up the file of you!
@@Thaplayer1209 HAHAHA NO WAY YOU SAID THAT
0:01 Unknown has really good quotes. He posts them everywhere
Sure Jan
No the quotes by anonymous are better.
@@MarkWatney
Agreed. Better yet by [DELETED USER]
They're not trying to be fake deep they're just funny.
To anyone who’s interested, that operation is called an XOR, or exclusive or.
That’s absolutely true, though it’s often explained by either add but don’t carry(like here) or one or the other but not both
XOR is actually so cool. Like addition, it's commutative and associative, and has the identity element 0. But the coolest part is that the xor inverse of a number is always itself, so if you want to add a specific number X to the xor sum of the lighted lamps, for example, if lamp X is off, then you simply turn it on (and commutativity and associativity does the work for you), but if the lamp is already on, then you can simply turn it off, and everything works the same way.
nimsum :D
I’m learning about those operators in calc 2 right now
*laughs in Scrap Mechanic*
Me : *Explaining this to my team*
My team : Okay
Me : *Goes inside and sacrifice*
My team : *Goes in for the attack*
My team : Wait, do we number the disk from the left side or the right side?
Your ghost: My team are idiots -_-
Left from the front or left from the back?
Me: ok, this is the plan
Team after I die: wait what's the plan again?
ugh. group projects
My team: *does the math wrong*
Ok. I'm at the problem-pause screen.
Here's my solution.
If all of them are off or on, I can indicate which one by having an "odd-one-out."
I can also create the same scenario if two are off and two are on.
If there are three one way and one the other way, I can flip one to make them all match. Or I can flip one to make a 2-2 split down the middle. Or I can flip one to make every other light match. Or I can flip one to make a 1-2-1 pattern, where the two in the center match each other.
So my instructions to my allied agents are these.
1. If there's three lights matching, the corrupted disk is the one that doesn't match.
2. Failing that, see the chart below
All matching - disk 0
2-2 pattern - disk 1
Alternating pattern - disk 2
1-2-1 pattern - disk 3
Edit: I think I used something that works because of what was explained? I think? I just didn't use the binary math to explain it?
Exactly
That’s the solution I came to! No need for the crew to do any math when they get there
I was so proud of myself when I came to this solution. The real answer bummed me out lol
@@shawman25 any answer that works is a real answer
I got exactly the same answer, even down to which pattern correlates to which disc.
I thought it was something about feeling the heat of the lightbulbs.
If the corrupted disk light was on, you would turn it off and the squad could feel the heat from a bulb that was currently off but still emitting heat from when it was on.
If the corrupted disk was off, you would turn it on and the squad notices that one bulb isn’t as warm as the other bulbs that are on.
Assuming the bulbs that are off are cool and the bulbs that are on are supposed to be hot. It probably doesn’t work in some cases though, what do I know.
This is a digital environment. Bulbs don't get hot.
That was my first thought too.
That is the solution to a different light puzzle though, so you did do well!
nice idea, but these look like LEDs rather than old timey bulbs
I did a different solution too:
Step One: If all lights are on or off, the rightmost disk is corrupted.
Step Two: If exactly one light is on or off, that disk is corrupted.
Step Three: If exactly two lights are on, ignore the rightmost disk and repeat Step Two.
If you start with all lights on or off, swap the corrupted disk.
If you start with exactly two on or off; swap the other on off if the corrupted is on or the other off on if the corrupted is off.
If you start with exactly one on or off; if said different disk is the corrupted disk, turn the rightmost disk on/off; if it's the rightmost disk turn the corrupted disk on/off, if it's one of the other two swap the other one so that only the corrupted disk and the last disk are on/off.
Instead of using binary numbers, i thought you could directly signal which light was corrupted to your friends.
If the number of lights on when you get there is an even number (0. 2 or 4), then it is straightforward because you can turn one light switch to make it such that the odd light out (the only one turned off or on) is the corrupted light.
It is a bit more complicated if the number of lights on is an odd number (1 or 3), but you could arrange a code with your friends, one example which I have listed below.
1. To signal that the 1st light is the corrupted one, turn the light switch that makes it such that all 4 are on, or all 4 are off (in other words: on-on-on-on or off-off-off-off)
2. To signal that the 2nd light is the corrupted one, turn the light switch that makes it such that the lights 1 & 2 share the same configuration, while lights 3 & 4 share the opposite configuration (in other words on-on-off-off or off-off-on-on)
3. To signal that the 3rd light is the corrupted one, turn the light switch that makes it such that lights 1 & 3 share the same configuration, while lights 2 & 4 share the opposite configuration (in other words on-off-on-off or off-on-off-on)
4. To signal that the 4th light is the corrupted one, turn the light switch that makes it such that lights 1 & 4 share the same configuration, while lights 2 & 3 share the opposite configuration (in other words on-off-off-on or off-on-on-off)
No matter what the starting configuration of lights is, you should always be able to signal each of the 4 lights depending on which one is revealed to be corrupted.
When you expand the problem and number of lights on a much greater scale, the binary solution definitely makes a lot more sense, but in the case of just having 4 lights, I believe this is a more straightforward solution.
Good
Brilliant work.
Wow, nice unique solution. Works!
That’s the solution I came up with as well! And there’s no need for your crew to do any math once they get in!
Unluckely this does not work because you can't turn more than 1 light, so, if for example you find in front of you on on off off, you will not be able to signal that the 1 light is infected since you established that the signal should be either off off off off or on on on on.
I know cause I ran into the same problem thinking of the solution, but I managed to solve it.
Next comment i explain
Huh, this is the first time you actually die in a riddle.
Gladly he didn't said me and the squad are perfect logicians, now I don't feel bad for the fact we failed miserably
He also didn't say they had good memories or were good at math, so we'll 100% fail and let the virus destroy everything.
Ted-Ed : Asks seemingly *impossible to solve* riddles
Also Ted-Ed : Gives the best answer and teaches us perfect logic
1+10=110
@@regularhuman2887 1+10=110 would work if you're using concatenation as the symbol of addition. Otherwise no, the video said : 01 + 10 + 11 = 110
and if you converted that from binary to decimal it would be : 1 + 2 + 3 = 6, which is 100% correct.. Don't mock binary if you don't understand it
Not the best answer
I solved it with an "easy to explain" answer
Definitely not the best answer, I've already seen 2 better solutions in the comments not including my own.
@@johndoberman4970 Oh, I actually didn't try to solve it any other way so, I thought that's the best answer. Sorry and thank you
this was a missed opportunity to name it "the risky disky riddle"
The Simplest solution:
Confirm you have green eyes,
Ask the malware to leave.
this joke is getting old...
@@kevin-ec9hf IKR...
@@kevin-ec9hf the solution:
Confirm that you have green eyes
Ask the joke to not be old
@@kevin-ec9hf But what if you _don't_ have green eyes?
Okay I'll leave.
I dont understand
I've watched enough Ted-Ed riddles to know that the solution involves parity but the binary is what I didn't think of. Hats off Ted-Ed, thanks for the entertaining educational riddle!
Tbf, the answer is literally just parity if there were only two lights instead of four.
Meanwhile I picked up on the solution involving binary but not the parity part /lh
There is a much easier way. Take the first 3 bits, ignore the 4th when trying to figure out which bit is corrupted.
Rule 1: If the corrupted bit is 1, 2 or 3 then switch one bit within them to make the corrupted bit different.
Rule 2: If the corrupted bit is 4 then switch one bit within the first 3 to make them all same.
Rule 3: If the original state of the first 3 bits are already what you wanted then switch bit 4 since it will be ignored.
I came up with a very similar solution, albeit a lot more complicated. It was basically halfway using binary. The first two disks represented 0s when off and 1s when on. If I had a light situation that had both of them in the wrong position, disk 3 would be on to flip what they meant. An example would be if the first 3 were on and the fourth was corrupted, I would need the first two to be on, or the first two off with the 3rd on. In that scenario I turn the 3rd off and it works. If it was the first one, I need both off so with 3 on, I just turn on the 4th one since it can be ignored and the 3rd switch flips the output. As I said, more complicated lol, but it works
That's really good!
How did you think to ignore the 4th? What was your thought process
@@bill5922 Uhh that was long time ago but let me see if I still remember.
First reason is because the chosen agent must flip one switch. So in case you don't want to switch 1, 2 or 3, you can switch the 4th which is irrelevant.
The second reason is that 3 bits are enough to show which one of the 4 discs is corrupted:
100 or 011 -> disc 1
010 or 101 -> disc 2
001 or 110 -> disc 3
000 or 111 -> disc 4
And if you try it out, you will see that you can always (at any state) set the 3 bits to the disc you want with just one switch or no switch at all.
@@Yusso thanks for answering
My solution was to tell my team something like:
"Use the binary number indicated by these two lights to find the answer, but reverse the digits if this third light is on. Ignore the fourth."
It's pretty similar to what's in the video, I suppose, but I didn't think about using addition.
This is actually equivalent
Though in your solution disk 4 acts like disk 1 in their solution, disk 3 acts like disk 4, disk 2 acts like disk 2 and disk 1 acts like disk 3
This UA-cam channel was recommended to me by my professor and I slowly liking it as I watch your vids. It helps a lot!
It helps a lot… with what?
I just found this channel one day, never stopped watching
The narrators's voice is so sooothing.
Addison Anderson
The Ted-Ed riddle playlist is just growing 1 by 1, and I love binge watching/listening to all the riddles in the background while I play games, there’s just something relaxing about that combination and it makes the Ted-Ed Riddle Playlist my favourite playlist on UA-cam for that reason.
I really hope there are a lot more riddles to come 😁🧐
This was a fun puzzle, and I have found an alternate solution.
If each light represents a bit with OFF meaning 0, and ON meaning 1, then the four lights will represent a 4-bit number which in decimal represents a number between 0 to 15. We only need 2-bits of information to represent the corrupted disk, so this means, we can use 4 numbers from the 16 numbers to represent each disk. After working it out (and easy to check) let us make the following definition:
0, 1, 14, 15 represents the first disk
2, 3, 12, 13 represents the second disk
4, 5, 10, 11 represents the third disk
6, 7, 8, 9 represents the fourth disk
Starting from any given initial configuration of lights (i.e. any number from 0 to 15), it is easy to check that you can always flip just one light to get at least one of the four numbers to represent the correct corrupted disk.
As an example, if initially the lights are 0 1 1 0 (=6 in decimal) you can flip bit#1 to get 0 1 1 1 (=7 or disk 4), or flip bit#2 to get 0 1 0 0 (=4 or disk 3), or flip bit#3 to get 0 0 1 0 (=2 or disk 2) or flip bit#4 to get 1 1 1 0 (=14 or disk 1).
This works for all possible cases. I like this solution better. :)
yeah, that is the same solution i found
same bro nice
For those who doesn't like math, I've created a solution that only uses patterns.
There's 4 types of pattern
0 = on X = off
Full (0-0-0-0 or X-X-X-X)
Side (X-X-0-0 or 0-0-X-X)
Center (X-0-0-X or 0-0-X-X)
Split (0-X-0-X or X-0-X-0)
These patterns are named as it will be useful for Scenario 3
Let's call the corrupted disk as "c-disk"
Scenario 1:
If all lights are on, turn off the c-disk.
If all lights are off, turn on the c-disk
Scenario 2:
If there are 2 lights that are on, flip one light that is in the same state as the c-disk, so if the light of c-disk is on, turn off the other light that is on and if c-disk is off, turn off the other light that is also turned off.
Scenario 3: Only one light is on or off
Now this is the tricky part, but Its actually easy, what we need to do is create a pattern and each pattern will represent which c-disk is corrupted.
If the c-disk is bulb no.1, we'll use the "full" pattern, so all we have to do is flip the odd light.
If the c-disk is bulb no.2, we'll use the "side pattern, so basically we'll just have to flip the other light that is on the same side as the odd light.
If the c-disk is bulb no.3, we'll create the "center" pattern that will look like a sandwich, basically the one in the center is in the same state, and the one on both sides are the same
If c-disk is on bulb no.4 we'll create the "split" pattern, where we just have to create an alternating pattern like on-off-on-off or off-on-off-on
Now try it yourself 😉
Another solution that should work for any amount of disks (although difficult to find a pattern that works), is to assign four 4-bit configurations to indicate which disk is the corrupt one such that by flipping one bit on any random configuration, you can always change the configuration to indicate which disk is the corrupt one. (Sorry for long and convoluted sentence).
In this example you can use the following:
"Disk light configuration" = "which disk is the corrupt one"
0000 = 1
0001 = 1
0010 = 2
0011 = 2
0100 = 3
0101 = 3
0110 = 4
0111 = 4
1000 = 4
1001 = 4
1010 = 3
1011 = 3
1100 = 2
1101 = 2
1110 = 1
1111 = 1
With this you can always change any configuration to indicate the correct disk, for example. If agent A comes in and is told that disk number 3 is the corrupt one, and the current configuration is 0110, he can simply change it to be 0100.
I'm pretty sure there's a boolean theorem describing this and gives a method to find such a table for x amount of disks, but I can't remember what it's called ¯\_(ツ)_/¯
This is exactly what I did!
I ended up doing the same thing and found that for a number of disks D and infected drive X, there is always an ending configuration C that is exactly one binary digit away, with the relationship being described by the function C = (D-.5)-abs(floor(X/2)-(D-.5)), assuming the disks are numbered from 0 to D-1.
Fantastic puzzle with a great explanation!
I figured out another way to solve this particular scenario though:
Take the first 3 disks as bits: off -> 0, on ->1, the 4th disk is a 'throw-away'.
Encode the position of the corrupted disk as follows:
disk 1: 000 or 111, disk 2: 100 or 011, disk 3: 010 or 101, disk 4: 001 or 110
With this encoding it is ensured that one can indicate the position of the corrupt disk via a single toggle (if the code is already correct, just toggle the 4th disk):
1 2 3 4
000 100, 010, 001
111 011, 101, 110
2 1 3 4
100 000, 101, 110
011 111, 010, 001
3 1 2 4
010 000, 011, 110
101 111, 100, 001
4 1 2 3
001 000, 011, 101
110 111, 100, 010
Ooh, very nice solution!
I had a different solution (which Dillon Teo also found and explained in this comment: ua-cam.com/video/i0WH4SFpeB8/v-deo.html&lc=UgxZ1dRebvvZQ5Tzojh4AaABAg ), but this one is much more elegant, with fewer "if-thens".
Yeah, this is the solution I came up with. I thought it made a lot of sense to only use as many bits as you need and save the last one in case they're all already in the correct position.
You can frame this as:
1st bit = "whether to negate 2nd and 3rd bits"
2nd & 3rd bits = the corrupted disk
4th bit = throwaway
I had a similar solution. Focus on the first three to figure out the answer. If the first three are all on or off then it means it’s the fourth one. If it’s not the forth one then it’s the odd ball from the first three.
I personally came up with a different solution, though I could be wrong. If they are all the same value, then flip the corrupted one. If there is two of each, flip the one with the same value of the one that is corrupted. So, if the team enters to only one switch having a distinct value, they will know that one is corrupted. However, if there is a 3-1 split, we will have to change the strategy a little. If the first switch is corrupted, then change the odd one out to give everything the same value. If the second one is, then create a clean split down the middle (aka 1100, 0011). If the third one is, choose the option that has a one switch gap in between (so 1010, 0101). If the last one is corrupted, then mirror it across the center (1001, 0110).
That's basically what I came up with
That’s actually identical to the given solution, just with a different explanation.
Came up with the same thing down to what pattern to assign to each value on my own too.
No need for binary
I came up with the same idea!
If anyone is interested, I came to a different solution that works perfectly as well and I think is easy to understand.
Suppose we assign each disk/light a value. The first disk = 0, the second disk = 1, the third disk = 2 and the last disk = -3.
Notice the last one is -3, not 3!
The trick is now to flip a switch, so that the sum of these numbers equals the disk number (0, 1, 2, 3). If you end up in negatives, take the absolute value of the number.
It works for every configuration, because you can add or substract any number of 0, 1, 2 or 3 from any configuration. It works so beautifully because we allow ourselves to go into the negatives.
The sum of these disks can lie between -3 and 3. (-3, -2, -1, 0, 1, 2, 3). If the sum is at any of these points, you can always add or substract in such a way to go to any of the other values (in absolute sense).
Examples with lights:
- Suppose all the lights are off. Take the sum of these numbers, which equals 0. This means the first disk ( = 0 ) has the virus. You can flip any switch to indicate that disk has the virus (e.g. flip last light, sum of disks = (0+0+0-3) = -3 --> 3rd disk has virus).
- Suppose all the lights are on. Sum of disks = 0 +1 + 2 -3 = 0. If the second disk has that virus, You can flip that disk to make the sum of disks = -2 --> 2.
- Suppose only the second light is on (sum = 1). You can go to disk 0 by turning it off, stay at disk 1 by switching the first switch, go to disk 2 by adding -3 (1 - 3 = -2), and go to disk 3 by adding 2.
Numerical examples:
From -3 you can go to 0 by substracting -3. You can go to 1 by adding 2 (= -1 -> disk 1 has virus). To 2 by adding 1 (-3 + 1 = -2 --> disk 2). To 3 by adding 0.
From -2 (= 1 + -3, these lights are on) you can go to 0 by adding 2, to 1 by substracting -3 (-2 - -3 = disk1), to 2 by adding 0 (-2 --> disk 2) and to 3 by removing 1 (-2 + -1 = -3 --> disk 3).
From -1 (= 2 + - 3) you can go to 0 by adding 1, to 1 by adding 0, to 2 by substracting -3, to 3 by removing 2.
From 0 you can go to 1 by adding or substracting 1, to 2 by adding or substracting 2, and to 3 by adding or substracting -3.
From 1 you can go to 0 by substracting 1, to 1 by adding 0, to 2 by adding -3 and to 3 by adding 2.
From 2 you can go to 0 by substracting 2, to 1 by adding -3, to 2 by adding 0 and to 3 by adding 1.
From 3 (= 2 +1) you can go to 0 by adding -3, to 1 by substracting 2, to 2 by substracting 1 and to 3 by adding 0.
In the numerical example we have explored every possible solution.
Not sure if my explanation is clear, but it seems very elegant to me. Very easy to understand/use, just sum disk values and flip switch to get to the virus. I think the principle is the same as the solution in the video, just applied differently.
The way I solved this was to look at what information the others have. When the others come in, all they know is what the lights show, so they have 4 ordered bits. This gives 2^4 = 16 possible configurations, and from that they must be able to find which is corrupted - in other words, we need to sort these 16 possible configurations into 4 groups, so that of the configuration when they enter is in (let's say) group 1, then disk 1 is corrupted, etc.
Furthermore, on my end, it must be possible for me to switch any configuration into any of the 4 groups, in order to indicate which disk is corrupted. And thus for any initial state when I enter, there must be at least one element in each group that is exactly 1 bit flip away from this state, so that I can put the configuration into the correct group. Thus for any configuration, each of its 4 bit flip "neighbours" must be in different groups.
If you think about it, every solution must follow this rule actually, and necessarily fits into this scheme. It's just that other solutions have rules for what configuration corresponds to what disk, while this method allows for all possible configuration groups. In total, ignoring which group corresponds to which disk, there are 42 possible configuration groupings over all 4 groups.
It's actually not that hard to find a working grouping. Keep picking configurations until all initial states are accounted for, i.e. any configuration is exactly 1 bit flip away from at least one of the elements in the group. Each configuration accounts for 4 initial states, so we need at least 4 in each group to account for all 16, which means we must split them into 4 groups of 4 - they just barely cover all possibilities, there can be no overlap, so any configuration is actually 1 bit flip away from exactly one of the elements in each group - its 4 bit flip neighbours.
For the first group, pick any configuration, it does not matter. This configuration is one bit flip away from 4 other configurations, and so those initial states are now accounted for. Crucially, this does not account for the configuration we picked, since we must flip exactly 1 bit. Thus the second configuration of the group must be exactly 1 bit flip away from our first configuration, with a total of 4 choices. These choices will never overlap with the first. We now have half the group filled, and half the initial states accounted for. For the second half, we can use a trick: If we take our current two configurations, and flip all 4 of their bits, then we get two more configurations that also work. This is because these are 4 bit flips away from their counterparts, and so they are 3 bit flips away from the other. Thus there is no overlap, and because they don't overlap each other either they must account for 8 confurations - the other half of the set. As it turns out, this is actually the only way to complete the group for this case of 4 bits.
For the remaining groups, follow the same rules, while avoiding any configurations that have already been chosen. This should be easy.
Note that this also explains why the number of disks must be a power of 2. Let n be the number of disks. Each configuration in the group accounts for n initial states, and so we need at least 2^n/n configurations in each group. This must be an integer, so we have to round up. The only times this divides cleanly is when n is a power of two, as otherwise there will be a prime factor other than 2 in n. There are n groups, and so we need ceil(2^n/n)*n total configurations. But we only have 2^n, so this is only possible if n is a power of 2.
There is a different solution that works without binary.
If zero, two, or four lights are on, you can always flip a switch that leaves the corrupted disk in a different state than the other three (only one on or only one off).
If there are exactly one or three lights on, you can always configure the lights into one of four patterns: AAAA, AABB, ABAB, or ABBA. Where A and B refer to lights on/off or vice versa. Then just assign a disk number to each pattern.
Pretty much solved it the same way. Glad to see this solution on here too!
Yeah brother! I worked it out exactly the same way. My solution is the same for when we get in and we see 0,2,4 lights on and we make the corrupted one unique so that the team comes in they destroy the unique light. However, with when we arrive and 1 or 3 lights are on. It's a bit different for me and we'll do the following and team will recognize that as follows:
Zero or All On = Corrupted disk is the left most one
Two Lights are on with 1 Off Light between them = Corrupted disk is the right most one
If the two center ones OR the left most & right most lights are on = Corrupted disk is 2nd from the left most.
If the either two corner ones are ones together (1100,0011) = Corrupted disk is 3rd from the left most.
I checked it out. It's legit. Works every time. Took me a while to figure it out but I'm so glad I made a workable strategy for this one.
I have a different solution. Let the three first discs represent the corrupted disc.100 and 011 are 1, 010 and 101 are 2, 001 and 110 are 3, 000 and 111 are 4. That way you no matter which lights are on you can communicate which one of the disc is corrupted with 1 step, and if the lights are showing the correct answer just flip the fourth one.
Well done! Wow this is way more simple than what I did!
This is equivalent to the solution I thought of!
I figured that since you HAVE to flip one switch, one light would have to be designated as meaningless. From there it seemed to me that no matter what state the lights were in, I could make any one of the remaining three lights the odd one out, or make them all the same.
Be nice if I could figure out a mathematical proof, though...
Good solution.
I was thinking something along the lines of (since you can communicate with your team) setting up a bunch of scenarios like “if all off or all on but one, that’s it.” Etc…. But creating all those rules essentially breaks out to this math. This is freaking cool
I made a mapping table from all possible states to modified states. Then I assigned each modified state a number from 1 to 4 and make sure that you can reach all possible numbers (1 to 4) from every starting state.
That's what I did. If the 4 lights are considered 4-bit binary number, then if the lights indicate 0,1,14,15 - Disk A is corrupted; 2,3,12,13 - B; 4,5,10,11 - C; and 6,7,8,9 - D. All possible starting states can go to exactly one of the 4 disks by changing one bit.
Same!
Virus : "The corrupted disk is #3 muahahahah"
Me : "Yo guys its #3"
My Team : "01101000 01101001"
This is, of course, just one of many possible solutions. More generally, there are 16 possible combinations of lights. For any given combination, there are 4 other combinations which can be reached by toggling a single switch. You need to assign each combination of lights to mean a particular disk is the corrupted, and do so in such a way that, for any given starting combination, there's one reachable combination to indicate each possible corrupt disk.
Hey Ted Ed , this document is about the rich and beautiful of Vietnamese (Rich and Beautiful Vietnamese - VIETNAM IS Identities Vietnamese is the common language of Vietnamese people - a language that is growing more and more in line with the development of the country towards international integration. Vietnamese is not only diverse by the system of tones and 29 letters, but also rich by the combination of local languages of the above regions and the territory of Vietnam.More specifically, going from South to North on this S-shaped piece of land, we will clearly see the difference in production and living methods of people in different regions. Similarly, each region also uses different languages called dialects. In the South the word "go back" will often be pronounced "go back". In the Central region, people will not be able to distinguish the first consonants "D", "Gi" and "N", but they will only pronounce it as "Gi". For example, old, old and house will all be read as old by the people here. As for people living in the North, they often confuse the pronunciations between the two letters "l" and "n" such as: "profit and loss" will be read as "effort". Not only are they different in pronunciation, but the local language also has a difference in terms of vocabulary: Corn - Corn, Snakehead - Banana fish, Going to work - Going to mam ... And also thanks to the above differences. The system of vocabulary as well as pronunciation in Vietnamese becomes extremely diverse and rich. Not only that, this also creates an interesting, it can be said to be a bit "difficult" for foreigners when they learn Vietnamese and then travel to Vietnam.Why is it called "challenging"? Because in Vietnam, each locality has different names such as: Bu (Thai Binh), Bam (Bac Ninh), U (Ha Nam), Ma (Hue) and Ma are commonly used in the Southern region. With just one word "mother", Vietnamese people have created many different ways of calling it. This is both a "brain hack" game for foreigners when learning Vietnamese, and also an extremely interesting thing that makes them want to learn more about the Vietnamese language as well as the country and people of Vietnam. Thanks to the presence of local languages, the Vietnamese vocabulary and phonetic system becomes richer and more diverse. Each region has its own voice, expressing the daily life of our people from different regions. It is the interesting things from the local language that have contributed to creating a culture imbued with the national identity of the Vietnamese people) . It's not perfect and also have a lots of wrong letter so please help me fix the wrong and install it into your UA-cam channel, please!
Please help me!!!!!!!!!
Step 1: Find out if you have green eyes.
Step 2: Ask the virus to leave.
So, whatever the state of total parity maybe at some point, like EO. You have choices to either flip both parities(first and second place), or only first parity, or only second parity. 00 corresponds to no flips, 01 flips second, 10 flips first, 11 flips both. By fliping to the parities that corresponds to corrupt light, we can always reach the right answer. This isnt the complete explanation of all thats happeing, but rest is along these lines.
Ted-Ed: "Can you solve..."
Me: "You know goddamn well that I cannot"
Fittingly, this is the 64th TED-Ed riddle video in the playlist.
Alternative solution: Since you're allowed to plan ahead with your team before going in, you can assign number values 1 to 4 to certain configurations. You will need to create 2 cheat sheets, one for odd configurations and one for even configurations. With that sorted out before going in, you will always be able to set a configuration that will tell your team which disk is corrupted by flipping exactly one switch
0:03 best TED-Ed quote I have seen yet.
I absolutely brute forced my way to a solution:
If my team finds only one light on or only one light off, destroy that one.
If they find any combination of two offs and two ons or all ons and all offs, i made a cheat sheet with each of the 8 combinations equal to 1, 2, 3 or 4.
Not quite elegant but it would work
YES! Thank you! This was my solution as well. Since you're allowed to plan ahead with your team before going in, you can assign number values 1 to 4 to certain configurations. You will need to create 2 cheat sheets, one for odd configurations and one for even configurations. With that sorted out before going in, you will always be able to set a configuration that will tell your team which disk is corrupted by flipping exactly one switch.
There are more than 8 combinations with the lights
Can we just appreciate how smooth and on-flowing that gradient is?
Fun fact - TED-ed didn't come up with a question and made the answer, they came up with an answer and worked all the way to the question.
Honestly, there is an inspiration to come up with this problem. The idea is that when you are given the corrupt light (or whatever it was), you are given one of 4 possible pieces of information, and then you have the power to show your team one of 4 possible lighting arrangements. So if there is a solution, it has to be precisely created, and that is exactly what is shown in the video.
This is interesting. I've come to another solution though and it seems to work as well. I knew binary code was the answer and we need such an arrangement that with a single switch we should be able to give code for the corrupted disk. Out of the 16 possible values, I've assigned 4 for each disk such that with a single switch we can change it to a value belonging to the corrupted disk.
[0000,0001,1111,1110] - values for disk 1 (group 1)
[0010,0011,1101,1100]- values for disk 2 (group 2)
[0100,0101,1011,1010]- values for disk 3 (group 3)
[0110,1000,1001,0111]- values for disk 4 (group 4)
For example - if the initial setting of the disks are 1010 (belonging to group 3) and our corrupted disk is 2, We can switch off the 4th place converting 1010 to 0010 (belonging to group 2) which will tell us the corrupted disk is 2. If our corrupted disk is 1, switch on the third place from 1010 to 1110. If the corrupted disk is 4, switch off the second place from 1010 to 1000. For corrupted disk three, although the existing values belong to group 3, since we must make a change, if we switch on the first place from 1010 to 1011, it would remain in group 3, signaling the corrupted disk is 3.
There's a much simpler answer. Only work with the first 3 and ignore the fourth. No matter what combination of lights there are there will always be a way to make one of the first three lights the odd one out if it's corrupted. Example: 1, 3 and 4 are on, corrupted is 1. You hit 3 so 1 is on and 2 and 3 are off which signals it to be 1. If the corrupted light is already the odd one out just flip 4. Also, if the corrupted is 4, you make it so that the first 3 are either all on or all off which works for any combination. If they're already all on or off just flip 4 since its on or off is meaningless.
I actually managed to solve this one! I didn't translate each light into binary, but I created my own ruleset that I would communicate with my team that would work for any solution. It's a bit complicated, but it essentially follows the same rules as the one shown in the video
Yeah yeah yeah
It's always a great day when there's a new Ted-ed riddle
1:18 pause here *'if'* you want to figure it out for yourself.... Are you poking me TedEd🤨
We are loving your riddles😍 please keep Posting more riddles 👍🏻
Wow, I was so happy with my solution and it felt so TED-Edy, but then it was totally different. I basically only use the middle switches for the code: 00 for disk 1, 01 for 2, 10 for 3 and 11 for 4. But what if you have to flip two switches? Then you can establish the first light is a true/false one. If it's off, you read the middle slots as they are. If it's on, you read the opposite, so 10 means 01 and so on.
Since the virus makes you flip a switch, if everything is already in its proper position when you arrive you just flip switch 4, which means nothing.
It would be easier if you have to flip two switches because there are 6 ways to flip two switches but there are 4 ways to flip only one switch.
I actually really like this solution. It's extremely simple and unique. It's also similar to the first idea I had, except yours actually works.
I really like the simplicity of your solution. I did something very similar, but without the use of a true/false light.
First, number the lights starting at 0, so you can directly translate the binary instead of shifting everything by 1. So instead of numbering lights 1234, they are 0123.
Then I divided the last three lights into two groups of two with lights 1 and 2 together in a group, and lights 2 and 3 together in a group.
I then said that if a group had an even number of lights on (0 or 2) then that grouping was equal to 1. If only 1 light was on in a group, then it was equal to 0. So for example 1100 would translate to 01.
So that way you can reach any binary number from 00, 01, 10, 11 with just one flip without having to remember if you need to reverse your answer based on a true/false light. If the left digit needs to change, flip light 1, if the right digit needs to change, flip light 3. If both digits need to change, flip light 2. If nothing needs to change, flip light 0.
Wow intersting riddle from TED - Ed with coding Decoding of riddles
Thank you ted ed. Now I am more knowledgeable than ever.
Alternate answer 1: Tell your team which disk was corrupted and if the virus stops you just tell him the hardest Ted Ed riddle you know which should stump him and buy you enough time to break into the mainframe and destroy the corrupted disk
I make the valiant sacrifice, and my team rushes in... They have no idea what they I did, we didn't agree on anything before hand. They destroy all the disks. Destroying the corruption and freeing the system.
I solved it, but with a much simpler method
1) if your friends see that there is an odd light out, it’s the corrupted one.
- This takes care of any setup where you start with two of each (two on or two off), and any set up where you start with all of one (all on or all off) because with both set ups you can manipulate them so that any one light is the odd one out.
(Now the tricky part is dealing with if you are giving a set up where three lights are the same and one light is different. That’s where you can discuss with your friends different patterns that indicate a different corrupted light)
2)
a) if your friends see that all the lights are matching (all off or all on) then the first one is corrupted.
b) If your friends see that it’s two pairs next to each other (double on then double off, or double off then double on) the the corrupted light is the second one.
c) If your friends see that it’s alternating (on off on off, or off on off on) then the corrupted light is the third one.
d) If your friends see that there’s a pair in the center with a pair split on the outside (on off off on, or off on on off) then the corrupted light is the fourth light.
This works for any scenario and all this pretty much gives a “cheat sheet” for your friends to know which one is corrupted.
I would totally replace History class for a problem solving class.
But I sadly can't.
:*(
Check out Hardcore History by Dan Carlin. He makes history far more interesting than a traditional class which has to force many topics with few details in a limited time space.
Boo! History is the best class, there is wars and empires, food, religion, politics, and so on.
@@Seth9809 you're right!
I would add a problem solving class without leaving out history class!
In my point of view the goal of history knowledge is that we learn from the mistakes the last generations made.
Well, you do have a problem solving class: math.
YAY! I finally solved one of these riddles - and not using the same method to reach it as TED-ED! Basically there are 16 different configurations of how the lights can be illuminated. Each configuration can (and indeed must) be switched to one of 4 other configurations. Therefore you can assign a number from 1-4 to each of the configurations, and agree in advance with your team what configuration refers to what. For example, in my solution: All lights off = 3. Left hand light only = 1, Middle left only = 2, Middle right only =3, far right only = 4. You can basically map it out so that no matter what the starting combo is, you can always change it to be on a configuration that you've agreed in advance is one of the numbers 1-4.
For anyone interested, here is my solution (where “X” is a light that’s switched off, and “0” is a light that’s switched on):
XXXX = 3
0XXX = 1, X0XX = 2, XX0X = 3, XXX0 = 4
00XX = 1, 0X0X = 2, X00X = 4, XOXO = 2, XX00 = 1, 0XX0 = 4
00X0 = 3, 000X = 4, X000 = 1, 0X00 = 2
0000 = 3
I’d give my team that list above, and go in, knowing that whichever combination is showing when I arrive, I can switch it to be the number that I need, out of the above list.
Hope that makes sense, please ask what I mean if not 😊
And before anyone says it - no my answer isn't as elegant as TED-ED's, and it can't be used to solve bigger versions of the puzzle - but then again the riddle never asked for that, so :P
Before watching the answer: My brain inmediatly thought of binary. With more brainstorm, i came up with "What if i assign to 4 values between 0 and 15 the tag [Corrupt Disk = 1], to other ones [Corrupt Disk= 2]..." And then i had to find those values. And i come up with:
Binary value: [00] [01] [02] [03] [04] [05] [06] [07] [08] [09] [10] [11] [12] [13] [14] [15]
Corrupt Disk: [d1] [d2] [d3] [d4] [d1] [d2] [d3] [d4] [d4] [d3] [d2] [d1] [d4] [d3] [d2] [d1]
The thing is to make the binary lights output to be:
- 0, 4, 11 or 15 if the corrupt disk is 1 (That is: 0000, 0100, 1011, 1111).
- 1, 5, 10 or 14 if the corrupt disk is 2 (That is: 0001, 0101, 1010, 1110).
- 2, 6, 9 or 13 if the corrupt disk is 3 (That is: 0010, 0110, 1001, 1101).
- 3, 7, 8 or 12 if the corrupt disk is 4 (That is: 0011, 0111, 1000, 1100).
This is because in each case, those values can be obtained with a single change of a light in a 4 lights array. EX for Corrupt Disk is 1: 0110 (6) -> 0100 (4); 1110 (14) -> 1111 (15); 0101 (5) -> 0100 (4).
About to comment down in this same comment to react to the TEd-ed answer.
I am keeping my strat, but the Ted-ed answer is not only more easy to understand to follow and elegant, but easy to scale up. I still cannot explain how i come up with *that* tag assigment to each value. I just started from the extremes and, when I realized that it was a reflection, I just went to fill the gaps with the most obvious pattern. And worked lol.
@@sirnikkel6746 I did the same method, but I used a Karnaugh map to obtain the assignments. It's a table that is characterised by a switch of 1 digit when you move from a box to an adjacent one.
@@yogue3108 How you managed to even use a Karnaugh map to do it?
I was about to do a lengthy reply, and then I saw this comment. I did the EXACT same thing you did. I agree the Ted-ed answer is much more elegant. I'm curious why this solution works though.
Haha! Me too!
I managed to solve this one, but I came up with a much more complicated and less intuitive and generalizable answer.
Basically, if you go into the mainframe and there are 0 or 4 active lights, flip the corrupted disk.
If there are 2 on and 2 off, find the good disk that is the same setting as the corrupted one and flip it.
In these cases, your team will discover the lights split 1-3, and can destroy the odd man out.
If there are 1 or 3 lights on at the start, things get more complicated.
If the corrupted disk is the first one, flip the odd man out.
If it isn’t the first disk, then if it has a different setting from the first disk, flip whichever of the first and corrupted disks isn’t the odd man out.
If the corrupted disk isn’t the first one but has the same setting, then flip whichever of the other two disks has the same setting as the first disk.
Then, when your team goes in, they’ll either discover 2 on and 2 off (in which case they’ll check the setting of the first disk, find the other disk with the same setting, and destroy it), or all 4 the same setting (in which case they’ll destroy the first disk). For example, if the lights are initially set 0010, you could indicate disks 1, 2, 3, and 4 with 0000, 0011, 1010, and 0110 respectively.
so, if i understand correctly: use disc 11 as a NOT operation, discs 01 and 10 to point to a different disc, and disc 00 to not change anything if its already correct
I would have preferred this explanation as well. The "break the rules regard 2 as 0" thing seemed like an unnecessary complication, but I am speaking from a programming background.
“Don’t use “beef stew” as a computer password. Its not stroganoff.”
That right there, made my day ️
"It's a rational transaction. One life for billions." -Dr. Hans Zarkov, Flash Gordon
I came up with a different solution. I will use 0 for off and 1 for on.
2^4 = 16 possible starting points which are as follows:
--- 2 options of all the same number (0000/1111) -> then flip the faulty one
--- 6 options of two on and two off (eg. 1010) -> flip in a way that so that the faulty one is always the odd one (eg cont. 1st faulty: 1000, 2nd: 1011, 3rd: 0010, 4th: 1110)
--- 8 options of only one odd either on/off (eg.0100, 1110) -> there are 4 non-symmetric patterns that can be created by flipping only one light. The patterns are discussed and mapped with the team before the sacrifice.
For eg. : 1100/0011 -> 1st faulty
1010/0101 -> 2nd
1001/0110 -> 3rd
0000/1111 -> 4th
Hence, when the team comes in together to attack and they see that:
- only 1 light is on and all the rest are off (or vice versa), then the odd one is the faulty one.
- two lights are on and the other two are off, then they follow the mapping done above
The main thing I learned today was that "parity" is not spelled "parody".
Two crucially different concepts
the solution to this video is just a parody of the concept of parity
I can't believe I turned off the notifications, I missed the riddle but thank god UA-cam recommended it to me.
Remember children: 1+1 = 0 !
I mean… he’s out of line but he’s right!
So I broke this down into 5 cases (which is really 3, but whatever), based on how many lights are on when you go in. If all the lights are on, or all off, you can swap just the corrupted light and your team will know to hit the different one. If 2 are on and 2 are off, ensure that the corrupted disk is showing a different light from the other 3.
So then you have just the case where 1 or 3 lights are on at the start. In this case we will end up in one of 8 different end cases: all lights on, all off, or 6 patterns of 2 on 2 off. By assigning opposite cases to the same disk, you assign each pattern to one of the disks. So 1111 and 0000 would mean, perhaps, the right most disk, while 1010 and 0101 would mean, perhaps, the 2nd from right. In any case where 1 or 3 are on you can get to one of the settings for whichever disk is corrupted.
It doesn't scale as well as your answer though, I didn't get to the idea of assigning the disks themselves values and adding them.
Sorry dude it's totally wrong 3 people have made the same MISTAKE IN THE CHATS AND I HAVE HAD TO REPPY 3 TIMES !! LETS ASSUME THE TWO LIGHT ON AND 2 OFF SCENARIO ,LETS SAY LIGHT NO 1 AND 2 ARE ONE BUT NOT CORRUPT , NO 3 IS JUST LIKE 1 AND 2,meaning not corrupt , BUT ITS SWITHCED OFF AND SO THAT LEAVES LIGHT NO 4 AS CORRUPT AND ALSO SWITCHED OFF , THEN ACCORDING TO YOU THE LIGHT WHICH IS CORRUPTED MUST BE MADE ODD ONE OUT ISNT IT ?? WHICH IN THIS SCENARIO IS LIGHT NO 4 . AND THE ONLY POSSIBLE WAY TO MAKE LIGHT NO4 ODD ONE IS TO CHANGE THE STATUS OF LIGHT 1 AND 2 WHICH ARE BOTH ONE INTO OFF AND THEN LEAVE LIGHT 3 AS IT IS AND THEN TURN ON LIGHT NO 4 AS ON bcoz that's the corrupt one to be made an odd one out so that's 3 flips dude
No, I can't solve the riddle. I never can. I don't know why I keep clicking on this type of videos
Okay, I have another solution. I haven't wrote every possiblity down to check if that would work, but I am curious to have your opinion on this :
What we need is to communicate a number : 1, 2, 3 or 4. If we think in binary, we only need 2 bits (lights) to write those numbers : they usually allow us to write numbers between 0 and 3, but let's say that we want our values between 1 and 4 so we'll say that 00 is 1, 01 is 2, 10 is 3 and 11 is 4. We will therefore use the first 2 lights to communicate the number of the corrupted disk.
We have several possibilities now : let's say that when I come in I am told which disk is corrupted and realize that I need to switch exactly one of the first 2 lights to get the correct number : that's what I do, great.
But now imagine that I need to switch both lights (for example, the corrupted disk is the 4th one and both lights are off). Our solution is to invert our code : if we decide that lights off don't mean 0 anymore but 1 and lights on mean 0, then it's the same as switching both lights. Now how do I tell my team what means 0 and what means 1 ? Third light is here to help. The situation of this light will indicate what configuration means 1. So in our example, if I want to indicate the number 4 with both my first 2 lights off, I need my 3rd light off too. If it is on when I come in, I switch if off and problem solved. But what if it is already off ? Everything is how I want it, I don't need to change anything. Well, let's switch 4th light. This is the useless light, doesn't provide any information. I could just leave everything the way it is and my team would be able to determine the solution. But the rules say I have to switch a light, the 4th one will serve in that case.
So to synthetize :
if, when I come in, and considering the version of the code the 3rd light is currently telling me :
- I need to turn exactly one of the first 2 lights for them to give me the correct number : I switch it.
- I need to switch both of them : I switch the 3rd one, this is equivalent.
- I don't need to switch anything, I switch the 4th one.
What do you think ?
can i can i really?!
You can do it in a less abstract way by making a code:
If 0, 2 or 4 lights are on you can make the one different from the others;
If it's 3 and 1 (or 1 and 3, same thing) you can make 4 patterns: alternative, grouped, middle/edges and all the same, which would correspond to 4 predetermined positions positions;
It's nice how it doesn't matter whether lights are on or off, just their relation to the other lights.
The given solution cooler and much more practical, but at this scale you can find unintended solutions still, which makes the riddle more fun as you have less to go on.
You can just number the lights 0,1,2,3 then work out the sum of them mod 4 and it works in exactly the same way. More than that it will work for any number of lights, not just powers of 2.
I have a way simpler solution. You can tell your crew the plan... If the lights are 1,1,1,1 then you can switch one off to leave an odd one out. Likewise, if the lights are 0,0,0,0 you can switch one on to leave an odd one out. If the lights have 2 on and 2 off, then you can ALWAYS switch one light in a way to leave the special light as the odd one out (try a mental experiment to prove this example if it's 1,0,0,1 and the special light is in position 2, you can simply switch position three to make position 2 the odd one out). The tricky part is when there is 3 of one kind and 1 of another when you first get there. If this happens, you can create one of four patterns that you've agreed with your team will correspond to a specific light. For example, if you switch the odd light, you will make uniform lights, and that could correspond to spot one. The other three combinations are alternating (ex 1,0,1,0), same at either end (ex 1,1,0,0) or same at The opposite end (ex 1,0,0,1). You can create those four combinations for any sequence in which you have three of the lights the same and one of the lights different (run a mental experiment to prove this). I hope that was helpful!
You can say that this riddle is LIT
Alternative answer:
If there are 0, 2 or 4 lights on, you can flip one so the odd one out (the only one which is on/off) is corrupted.
If there are 1 or 3 on, you can ignore the first light, and make the odd one out corrupted (if the corrupted is already odd one out, swap the first light)
If there are 1 or 3 on, and the first one is corrupted, you can make all of the lights the same (all on or all off)
For the decoder:
If there is 1 or 3 on, the odd one out is corrupted.
If there are 2 on, the one matching the first light is corrupted.
If there are 0 or 4 on, the first light is corrupted
I feel like my CS degree is worth it finally.
I arrived at the same answer through a different thought process, feels nice to see divergent thinking converge to the same answer
4 bits:
Useless bit: flip if what u want is already done
The "negative" bit: take the reverse of the following bits
Last 2 bits: 2 bit number
Taking the reverse here means that u treat 0 as 1 and 1 as 0 if the second bit is a 1, else treat them as normal. This is just a rudimentary thinking of parity flipping ig
I used patterns to solve it. If you come in and 0, 2 or 4 lights are on, just flip one so that the corrupted disc is the odd one out. Now, if you come in and 1 or 3 lights are on it's a bit trickier, but the math still works out. You have to disregard the exact values and instead use patterns. All these can be got from any 1/3 lights on:
0000 and 1111 -> disc 1
0011 and 1100 -> disc 2
0110 and 1001 -> disc 3
0101 and 1010 -> disc 4
You just have to agree which disc corresponds to which pattern beforehand.
Check if you have green eyes
Ask the virus to leave
😂😂
Your solution was much neater and more extendable than mine, but I think mine still works:
If you see all four lights on or all four lights off, flip the corrupted switch (so that it becomes the odd one out)
If you see two lights on and two lights off, flip whichever is the same as the corrupted switch (so that the corrupted switch again becomes the odd one out)
If you see three lights on and one light off, or vice versa, there is always one switch you can pull to get the lights into each of the following four patterns:
- All lights the same (all on or all off)
- Lights alternate between on and off
- Left two lights on and right two lights off, or vice versa
- Middle two lights on and edge two lights off, or vice versa
So you can assign each of those four patterns to one of the disks. If your team comes in and sees one light is the odd one out, they know that's the right one. Otherwise they can identify the correct pattern and act accordingly.
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I would not let myself watch the solution until I figured it out. This is way more elegant than what I came up with.
I treated the 4 mainframes as a bit in a number from 0 to 15 represented as binary. When you walk in the 4 disks can represent any one of those 16 numbers.
If the team comes in and sees the number 8 (1000) it has to mean the same thing regardless of how you found it or which switch you flipped to make it that way. So 16 possible numbers need to be grouped into 4 groups of 4 numbers. And each group must contain a possible outcome for any scenario you walk into.
The possibilities look like this
START | Switch 1 , Switch 2, Switch 3, Switch 4
0 | 8,4,2,1
1 | 9,5,3,0
2 | 10,6,0,3
3 | 11,7,1,2
4 | 12,0,6,5
5 | 13,1,7,4
6 | 14,2,4,7
7 | 15,3,5,6
8 | 0,12,10,9
9 | 1,13,11,8
10 | 2,14,8,11
11 | 3,15,9,10
12 | 4,8,14,13
13 | 5,9,15,12
14 | 6,10,12,15
15 | 7,11,13,14
The logic works in reverse as well. For example, the only starting numbers that would let you get to 15 by flipping a switch are 7,11,13,14
A pattern emerges where the 16 numbers fall into groups that are exclusive, meaning only one number from the group is possible in any starting position.
Group A: 1,2,15,16
Group B: 2,3,12,13
Group C: 4,5,10,11
Group D: 6,7,8,9
You tell the team that if the corrupted server is the first you will leave a number from group A, 2nd Group B and so on.
So now, regardless of what the lights are when you walk in, you can flip a switch that shows the "Group" translating to the corrupted server.
I realize after watching the solution that this is a very convoluted way to get the same binary math, but I wanted to do it on my own...
If you need help solving this riddle, a good idea is to simplify the problem down to its basics - two switches.
A switch can either be on or off. It's a binary choice. We can represent off with a zero and on with a one.
Now we can represent each configuration of lights as binary. 00, 01, 10 and 11. 00 is when both are off, 11 is when both are on.
And with these binary numbers, we can also convert them into decimal to make it easier for us to understand. 0, 1, 2 and 3.
As for the disks, we'll label them A and B, to keep them separate from all the numbers.
Let's say we enter the mainframe and see 00 (0 in decimal). If we flip one switch, we get 01 (1) and if we flip the other switch, we get 10 (2).
Now, let's say if disk A is broken, we turn 00 into 01. Therefore, if disk B is broken, we MUST turn 00 into 10 instead. Otherwise, we have NO way to communicate if disk B is broken.
If we enter the mainframe and see 11 (3), we can flip a switch to get 01 or flip a switch to get 10, just like before.
However, if we see 01, we can't make 01 or 10. Instead, we can make 00 (0) or 11 (3) so we must assign one of these to disk A and the other one MUST be with disk B.
It doesn't matter if you assign 00 to A or B, as long as you set its partner, 11, to the opposite disk.
There are two unique plans.
You can assign 00 (0) and 01 (1) to A, while 10 (2) and 11 (3) to B. Here you will notice that A is 2 or =2.
Or you can assign 00 (0) and 10 (2) to A, while 01 (1) and 11 (3) to B. Here you will notice that A is even while B is odd.
Going back to binary, the right-most digit is the one that determines if the number is even or odd.
Likewise, the left-most digit is the one that determines if the number is less than 2.
In decimal, we have the ones, tens, hundreds, thousands digits. In binary, this is ones, twos, fours, eights. Instead of getting 10 times bigger, it gets 2 times bigger, as each digit can only be one of two things instead of the whole 0-9.
You lost me after the dad joke in the beginning, I'll have you know it was worthy of a hearty chuckle
which joke are you even talking about
oh u mean the quote nevermind
knew this one for once cause of error correcting codes. surprised they didnt mention that, this same concept is what makes the internet and your computers work today, and the guy who invented them won a turing award for this.
There are actually 567 (=24*24) strategies that work. Here is a simple strategy I came up with:
The corrupted disk is the odd one out of the first 3; if the first 3 lights are all the same, the last disk is corrupted.
Example: If the virus shows you 0101 (second and fourth lights on) and tells you the first disk is corrupted, you can make it the odd one out of the first 3 by flipping the third light: 0111.
Example 2: Same lights, but now the fourth disk is corrupted. The first 3 lights must be the same, so now we flip the second light: 0001.
Each possible strategy comes down to distributing the 16 possible light combinations into 4 groups: (for example)
light 1 corrupted = 1000 / 0111 / 0110 / 1001
light 2 corrupted = 0100 / 1011 / 0101 / 1010
light 3 corrupted = 0010 / 1101 / 0011 / 1100
light 4 corrupted = 0001 / 1110 / 0000 / 1111
All groups must be reachable from any starting combination, therefore if two combinations are 2 flips apart, they must be in different groups!
A consequence of this rule is that a combination and its complement (i.e. all lights flipped) have to be in the same group.
Another consequence is that each group must have 2 even combinations (i.e. even number of 1's and 0's) and two uneven combinations.
Consider these uneven combinations: 1000, 0100, 0100, 0001. They are all separated by two flips, so they must all be in different groups. You can assign them in 24 ways (4 possible choices for the first one, 3 for the second, and 2 for the third; 4*3*2 = 24). Assign the complements of each combination to the correct group, which is forced. This takes care of all the uneven combinations.
Now for the even combinations: 0000, 0011, 0101, 1001. For the same reason as above, these can only be assigned in 24 ways. The complements are again forced.
Multiply the 24 choices from the uneven combinations with the 24 choices from the even combinations to get the total number of possible groupings: 567.
Finally, you can try to come up with a more intuitive method based on the grouping (like I did above).
Used a different solution: based on the limited configurations, you can use parity to create two different rule sets, one for even number of lights on and off (or just all on or off), and 3/1 lights on/off (and vice versa). So, say you show up and it’s a 2/2; play odd-man out, and flick the flight that makes the corrupt one not match the other three lights. If you show up and it’s a 4/0, then same rule. But for 3/1, we need a slight variation: it does not matter what the last light is. Do odd man out on the first three. If the last light is the corrupt, make all 3 others match.
Ex: you find 1001. team shows up to 0001. It’s odd parity, meaning it came from a 2/2 or a 0/4; finding the odd light out it’s the last light. Ex2: you find 0010. The first is corrupt. You flip to make 0110. Your team shows up. Seeing a 2/2, they ignore the last light, and see that the first 0 is the odd light out.
It has been a long time since TedEd ever given us a riddle like this.
Me a intellectual: alright imma just toggle the one that’s infected
Lots of possible solutions for this one. Here was mine: I need to deliver two binary pieces of information, which could give four possible outcomes in total, enough to communicate which disk. Now, consider two pairs of lights. I can't control the initial state, but if we consider the output of a pair not based on which lights are on, but on whether both lights in that pair are the same or different. But how to manipulate two pairs with one action? Overlap the pairs, with one unit in common. Now, with one flip, I can change the sameness of both pairs (flip the common unit), either (flip the one belonging to that pair) or neither (flip the last unit.)
So, instructions left with the team: if the two units on the left are both on, or both off, the corrupted unit is on the left side, and if they're different, it's on the right. If the two units in the middle are both on or off, the bad drive is toward the inside of the row, otherwise on an end. From any initial arrangement, I can see which of those pairs is already appropriately same/different and which need to change to indicate the right drive, and can change either pair, both, or neither with one flip.
Found a different solution before looking at the original one:
Condition 1: If all are off or all are on/two are on.
Then flip one bulb so only the corrupted one remains (1 off or 1 on). The team will recognize that 1 of the lights is unique so they know that is the one.
Condition two: 3 are on/1 is on.
This one is a bit more difficult, but still easily solvable. Solution: if it’s 4, turn all of them on or off (depending on wether 3 lights are on or off). If it’s 3, flip one bulb until the split is 1-2/3-4 (if 1-2 is on, 3-4 should be off and vice versa). If it’s 2, flip one bulb until the split is 1-3/2-4. If it’s 1, split should be 1-4/2-3. The team will recognize that 2 lights are on and will destroy the corrupted one based on the split, or if all are on/off.
Should work :)
[SPOILERY]
This can also be thought of as an exclusive or (XOR) operation, where an even number of 1s outputs 0 and an odd number of 1s outputs 1.
I had a solution that only works with the four disks that is different.
I'll represent the lights that are ON with an O and the lights that are OFF with an I. The corrupted disk will be in brackets.
Let's say this was the arrangement numbering them 1 to 4 from left to right:
1 2 3 4
I O I (O)
I would indicate to my team to look at the centre two disks (2 and 3) and see whether the corrupted disk is on the left(1, 2) or on the right (3, 4). If they are both ON/OFF then the corrupted disk is on the right(3, 4). If one is ON and one is OFF the corrupted disk is on the left(1, 2).
Once we figure out the side the corrupted disk is on (left/right), we can indicate between the two odd and even numbers which one it is. If both lights on the side the corrupted disk is on are ON/OFF the even numbered disk is corrupted (2 or 4). If one is ON and the other is OFF, it's the odd number (1 or 3).
Applying those rules to the light arrangement looking at the two in the middle, they indicate the light is on the left (1 or 2) currently. To change that we need to either turn OFF 2, or turn ON 3. The corrupted disk is an even number so 3 and 4 BOTH must be ON/OFF. For this scenario turning ON 3 indicates the corrupted disk is on the RIGHT and is EVEN. So then it has to be 4.
If this was the new arrangement:
1 2 3 4
I O O (O)
You would switch ON 1 because it wouldn't matter.
If this doesn't work for an arrangement of lights let me know. I tested it out for a couple and they all seemed to work.
This one is basically the XOR command, very cool!
Hi was hoping if you could make a video on how we could try solving these . I solved one of your riddles ( the catching the rare fish Einstein problem one ) it took me quite a while but once I figured out how to solve it I pieced the puzzle also your videos are extremely engaging and keep my brain on its toes , hunting for answers constantly . Thank you for the amazing content . Have a great day !
Loving the Tron references in the animation!
Spectacular video as always!