Can you solve the Big Bang riddle? - James Tanton
Вставка
- Опубліковано 23 лют 2022
- Practice more problem-solving at brilliant.org/TedEd
--
It’s moments after the Big Bang and you’re still reeling. You’re a particle of matter, amidst a chaotic stew of forces, fusion, and annihilation. If you’re lucky and avoid being destroyed by antimatter, you’ll be the seed of a future galaxy. Can you ensure that you’re the last particle standing? James Tanton shows how.
Lesson by James Tanton, directed by Igor Coric, Artrake Studio.
Support Our Non-Profit Mission
----------------------------------------------
Support us on Patreon: bit.ly/TEDEdPatreon
Check out our merch: bit.ly/TEDEDShop
----------------------------------------------
Connect With Us
----------------------------------------------
Sign up for our newsletter: bit.ly/TEDEdNewsletter
Follow us on Facebook: bit.ly/TEDEdFacebook
Find us on Twitter: bit.ly/TEDEdTwitter
Peep us on Instagram: bit.ly/TEDEdInstagram
----------------------------------------------
Keep Learning
----------------------------------------------
View full lesson: ed.ted.com/lessons/can-you-su...
Dig deeper with additional resources: ed.ted.com/lessons/can-you-su...
Animator's website: www.artrake.com
----------------------------------------------
Thank you so much to our patrons for your support! Without you this video would not be possible! SookKwan Loong, Bev Millar, Lex Azevedo, Noa Shore, Michael Aquilina, Jason A Saslow, Dawn Jordan, Prasanth Mathialagan, Samuel Doerle, David Rosario, Dominik Kugelmann - they-them, Siamak H, Ryohky Araya, Mayank Kaul, Christophe Dessalles, Heather Slater, Sandra Tersluisen, Zhexi Shan, Bárbara Nazaré, Andrea Feliz, Victor E Karhel, Sydney Evans, Latora Slydell, Noel Situ, emily lam, Sid, Kent Logan, Alexandra Panzer, John Hellmann, Poompak Meephian, Chuck Wofford, Daniel Erickson, frank goto, Jayson Hauschild, J D Wallace, Marq Short, Chen Jun Xiang, Adam Pagan, Paul Schultz and Behzad Farhanieh.
Visit brilliant.org/TedEd to check out Brilliant’s 60+ courses in math, logic, science, and computer science. They feature storytelling, code-writing, interactive challenges, and plenty of puzzles for you to solve. And as an added bonus, the first 833 of you to use that link will receive 20% off the annual premium subscription fee.
It is very informative
"Stand to the left of the biggest patch of antimatter" was my strategy
reallly?
Very nice. Danks.
@@leo_the_cow u very nice
"Pick a spot to stand where you won't be annihilated." Every middle school dance and workplace birthday "party" I've ever been to.
I never comment on comments but for some reason this one made me laugh.
🤣
my mind immediately went to sh..tings
Hi I’m a 17 years old mom from Afghanistan. I take care of 7 homeless kids that became homeless and alone because Taliban kill they whole family in front of them.I’m making videos about children’s right and women rights in my channel.i need world know more about us,i hope one they it can help all of us…please help me and watch my shorts,my kids ask me to translate all the comments for them and I’m teaching them English
@@j-r-m7775 you're too busy getting gains to comment
I didn't expect to start my day with an existential crisis over a riddle. Thanks Ted Ed :D
Tell us you're antimatter without telling us you're antimatter.
Ah, it's fine. None of the matter or anti-matter was at all sentient so no one got hurt in the super hot mess at the start of the universe.
@@DemPilafian i am extremely complex & maybe doesn't exist
@@DemPilafian i'm already gone
@@DemPilafian You're antimatter without telliing us you're antimatter.
5:36 That face of pure joy watching your own brethren VIOLENTLY EXPLODE AGAINST ONE ANOTHER
LOL
This is a pretty messed up idea for a game if you think about it, but I guess intergalactic ethics standards were different back then
cosmic suicide game
Russian roulette with all but one chamber filled. 😱
i mean like especially since they all just started existing like damn give it a few days first
Stop spoiling Squid Game season 2.
stuff like this would never hold up in front of intergalactic ethics boards today
I don't know how friendly the other particles can be if their idea of fun is Mutually Assured Destruction.
Personally I would accelerate so much into the other direction that I become the first photon.
I wonder why the phrase 'Mutually Assured Destruction' has been popping up so much lately... weird :/
@@speedyboi1471 haha Ukraine go boom funni
Thereby skipping the annihilation parts, remember photons are pure energy.
@@arushjain1103 what
@@moo8866 I was joking but I actually support Ukraine.
"you are a particle floating through the space..."
"If you're lucky, you will become a seed for a future galaxy..."
I wasn't expecting an existential crisis over a riddle
I'm with dignity and honour announcing to Ted Ed and groups that this is the only riddle I'm able to solve after so many years to facing hardships to even understand the solution of riddles
(〒﹏〒)
How? I guess you haven't watched all the series or you've become way better, because there were extremely easy riddles before, like the virus and the robot ants. This was fairly complicated.
@@MiguelAngel-fw4sk it’s harder than it looks for some people😊
@@annihilate2479 i can say from experience that it is harder than it looks for me
Did you see the ant riddle? That one was easy.
I took 1 minute and 37 seconds.
My strategy was similar to the one at the end of the video. It doesn't really matter how many anti-matter particles are to your left, and thus, you should maximize the number of anti-matter particles in a sequence to your left, and maximize the number of matter particles to your right. That removes a ton of the starting positions already, just compare the ones remaining and make sure that the number of matter particles minus anti-matter particles equals 0. Thankfully, this is one puzzle that I am actually able to solve.
Yeah, I think is is one of the easiest, if not the easiest riddle so far.
Im not even smart enough to understand the rules
I took 21 sec to find the same strategy
Yeah this one was quite easy i too got the strategy tho only at the last(meaning when pausing for the second one i got it)
I also came up with this same strategy pretty quickly
I recognized the strategy within 30 seconds! I realized that the condition was standing to the left of another antiparticle, and not to the right led to anihilation, so I reasoned that finding the longest string of gluons and standing to the leftmost position will leave only you standing. Typically though, I can't figure these questions out, so that made me very happy!
That doesn't work for the 24-particle circle though, right? Or the 10-particle circle, actually.
this doesn't work. imagine 2, -1, 5, -4, 1, -3 setup. with your logic, you'd stand left to 5, but you would still get destroyed, as the correct answer is standing left to 2
I see, looks like I still didn't figure it out. Looks like Ted-Ed is still too smart for me after all lol
I thought something similar I think, have the longest string of antimatter particles on your left - and it makes sense why this would work sometimes, but I tried a couple more situations & figured out it wouldn't always work
Someone beat me to it 1 year earlier... but there is no guarantee that he is still alive today...perhaps i am the one that still survive?😂
Never thought I’d be a particle going through such a roller coaster
My strategy for the end was to pick a random spot, add and subtract around the circle as shown, and keep track of the lowest number reached and where it was. I'll stand exactly where the lowest number was reached. Tied lowest numbers are all valid.
Don't need to test all spaces, just one test around the circle is enough.
I did the same ❤️ ..
yes!! this was how I thought about it too. important clarification when you're talking about programming a device :) don't want to have to repeat your program 50,000+ times
Well, your solution is a magnitude better than the O((0.5n)²) ones
Loved this one.
For the second riddle, you create a pair of coordinates.
First represent position, second a "total".
Where you start is 0, 0.
From then, you add 1 for every matter encountered.
Add -1 for every antimatter encountered.
You continue until your sum go below zero or you complete the circle.
If you complete the circle without going negative any moment, that's a safe spot.
If you get negative, it means that the position where you went negative and all previous positions (included the spot you choose) are not safe.
“… then that spot is safe”
*big flashing arrow points to most obviously unsafe antimatter arc in the entire ring*
5:31
Solving the first part took me definitely less than a minute. I did not time it unfortunately, but here's how I did it:
A viable position is always characterized by an antimatter particle followed by a matter particle (counter-clockwise/right direction).
Reason: if you position yourself left of an antimatter particle, you immediately annihilate; if you position yourself right of a matter particle, this particle will never annihilate before you, since the annihilation occurs to the right.
Then it's a matter of counting (if you're a machine) or pattern matching (if you're good at that):
If matter particles are +1 and antimatter particles -1, go through the ring summing them up counter-clockwise from your chosen position. If the sum never goes below 0 => you found a winning position.
For the code:
point the device at any position x
we will denote this position as 0 (x = 0)
assuming the "order of particles" is an array, where index 0 is the first particle to the right of the chosen position x
While x < number of particles (there are as many positions as particles): (we'll call this loop1)
if the first particle in the sequence (right of x) is antimatter:
increment x
continue with the next iteration of loop1
if the last particle in the sequence (left of x) is matter:
increment x
continue with the next iteration of loop1
set n = 0
set sum = 0
while n < number of particles: (loop2)
if particle at position n is matter:
add 1 to sum
else if the particle at position n is antimatter:
subtract 1 from sum
if add < 0:
increment x
continue with the next iteration of loop1
increment n
(if loop2 completes successfully:)
break out of loop1 and return position x as safe OR
if the goal is to find all safe positions: save/print value of x, increment x, continue with next iteration of loop1
I like your funny words magic man
I didn’t take much time to fully double check, but I think ur idea can still be simplified.
Consider what would happen if you moved the starting position one to the right. The entire circle’s “states” would all increase or decrease by 1 depending on if you just shifted over a matter or antimatter.
Considering the lack of true change in states, we can simply just start anywhere. Run the calculations and choose the spot with the lowest value.
The logic being that we can always shift the starting spot to the “lowest state” and now every state will be positive.
Might be wrong though didn’t double check
@@P0is0ndagger127 Hmm, do you mean something like this:
(beginning at any particle in the ring)
n = 0
sum = 0
min_sum = 0
safe_positions = variable-size array []
while n < number_of_particles:
if particle at position n is matter:
sum = sum + 1
else if the particle at position n is antimatter:
sum = sum - 1
if sum < min_sum:
min_sum = sum
overwrite safe_positions with [n]
else if sum == min_sum:
append n to safe_positions
increment n
=> positions right after/to the right of any particle in safe_positions is safe
@@TheDisasterMo something like that, at least if it works it would save some time
@@P0is0ndagger127 I thought of the same strategy and it works. As for implementation, you don't even need to store the min value, just change the sum back to 0:
sum = 0
safe_positions = [-1]
for pos, type in particles:
sum += type == matter ? 1 : -1
if sum == -1:
sum = 0
safe_positions = []
if sum == 0:
append pos to safe_positions
Part 1 took about 60 seconds for me. I found a simple algorithm to find safe spots: pick any particle in the circle and label it "1". Now traverse the circle from that particle by going right. If you are visiting a matter particle, label it one more than the previous particle. Otherwise, label it one less. Continue until every particle has a label. The safe spots are to the right of all particles that share the smallest value. I used mspaint on a screenshot of the puzzles to label the particles which sped things up a lot.
The safe-unsafe checker will likely do something similar: traverse to the right from your starting position, starting at the value 0. Add one if you see matter, subtract one if you see anti-matter. If you ever end up in the negatives, the spot isn't safe. If you get back to your starting position, it's safe.
I'd though of the same, but it only took me like 10-20sec because instead of counting I "estimated" the points that are safe
Exactly, find the min value. This is an O(n) algorithm compared to an O(n²) solution in the video
4:07 “Russian doll of fundamental particles “ Definitely not the sentence I expected to hear today 😂
Wow. I’m just going to say that this is the only riddle I’ve managed to actually understand and solve after 2 years of watching TED Ed. I was so proud of myself haha
My algorithm to always find the spot: start your counting and go leftwards(clockwise) around the circle (from anywhere with zero). For every anti matter you encounter subtract 1 and for every matter add 1. When you go arround you are bound to get to zero (the number you started with). Now the best place to stand is to the left of the particle with the highest positive score. Haven't seen the actual solution but I'm positive about this one.
Note that if you encounter a maximum twice (or more), then it shouldn't matter, stand to the left of any of the maximums and you should survive.
This is how I solved it, I think is the easiest way. I did prove it with the 3 examples and it worked perfectly :)
This is exactly how I solved it. Feels great reading my exact thoughts on a comment written by someone else. 😄👍
Funny that I found someone who also used this method.
A point worth mentioning is that this method is actually more efficient than the one mentioned in the video. This is because firstly it lets you know of the safe spots just by assigning numbers once. The equivalence of both the results is found by realizing that if I had started my counting from the maximum then since it is an extremum the curve made would lie completely below the 0 point and hence would never hit zero. Note: The maximum will only be one point after you add a +1 to a shared maximum spot by adding your matter particle over there.
If you have ever been trained for the Olympiad in Informatics, the riddle will seem much less difficult. In fact, the core concept of it is what we called "prefix sum" method. The riddle itself is interesting, and if we take a further step like asking ourselves "What'll happen if two matters of cube annihilate with only one antimatter of cube", we'll find ourselves observing the question at a higher level.
This was the best story behind a riddle….scientific story+mathematical riddle
Isn't this basically the Josephus problem but with particles?
4:02 minutes. My strategy was look for the amount of antimatter to the right of matter. Be the one in the last spot to get eliminated so long as the antimatter to the left of you has an equal number of matter to the left of it
My strategy before unpausing: choose any matter particle and assign it value 0. Moving clockwise, add one if matter or subtract one if antimatter. Keep track of which particle had the highest value seen thus far. The spot to the left of that particle is safe.
If you see a big group of matter particles, the rightmost one is a good starting point (making negative values less likely). You could actually assign any starting value; only the relative differences matter. If the max value appears multiple times, there are multiple safe spots. Finally, you can end early if you're sure the current max value can't be surpassed.
For my fellow nerds, this algorithm is O(n).
I had the exact same strategy, but reversed: moving anti-clockwise, keep track of the lowest value, stand in the spot right of that particle. 👍
Even I did the same
Looking for someone who did it like this in comments section :D
My solution was to choose a random spot between a matter and anti-matter cube (anti-matter to the left of matter), begin summing while counting clockwise (minus for anti-matter, plus for matter) until I hit a spot which fulfills the conditions:
- count is zero
- spot is between an antimatter and a matter cube
That means the device should be programmed to count until it hits zero THEN minus 1 and declare the spot previous to it safe
More efficient but complicated
01:15 love that Analog Synthesizer sound 😍
1:34 every TED-ED riddle goes "let's work backwards from the result"
As a developer, this is like tracking stacks of bracket pairs on an arbitrarily long line that loops back on itself and trying to find out what's the outermost scope :D Doing the +1, -1 and taking the spot with the minimum value does the trick.
Normally, I’m barely smart enough to understand the solution when it’s explained. This time I actually figured it out on my own.
Start from each matter particle and count the particles going clockwise. If at no point there is more antimatter than matter, you can go to the starting particle's immediate (counterclockwise) left.
I love these. I mean, I can’t solve them, but it’s cool to see the solution as if I did.
This riddle reminds me a lot about the STACK data structure. On the circle, when performing the *sum*, every time you add 1 its like pushing something to it, and every time you subtract 1 its like popping something, if the stack ever gets empty then its not a safe spot
I did both riddles, i always loved these riddles man its so nostalgic 💗💗 my first was the bridge riddle, it was like around 6 years ago.
I did the challenge in 6:30 nice and easy
Another way to look at this is starting from any position and moving towards the right step by step and keeping in track the number of antimatter and matter cubes separately, we can see that any instant the number of matter cubes must be greater than or equal to no of antimatter cubes(also counting the cube itself from where we start) for that spot to be a valid position. If at any instant no of antimatter cubes becomes more, then that spot is not a solution. +1, -1 approach was a bit nicer though.
Yeah, and from a programming standpoint, it’s neater, since you only have to keep track of one variable, and you have a fixed point to compare to.
@@samuelding7854 Yup
Imagine being one of the ANTIparticles who can’t do anything about being annihilated.
I had come up with a slightly different method. start with sum = 0. index clockwise from any point. If the current cube is matter, add one, if it is antimatter, subtract one. Keep a value for the maximum sum you've seen so far and the index at which it was seen. If the sum is greater than or equal to the previous max, save the new max value and the index where you reached it. After one cycle, your safe spot is one spot clockwise from the stored index. It doesn't matter which cube you choose to index as the first, it will always work.
My strategy was basically:
1) Pick a random spot was was to the right of antimatter and the the left of matter.
2) Check left pretty much doing the +-1 thing.
3a) If I got a +1 aka a matter to my left, I repeat step 1 to the next spot to that matter's left that meets the conditions.
3b) If i go through the whole circle without reaching +1 and end up with a 0, thats my spot.
Muito obrigado pela tradução Mafalda e Margarida!
Start at any particle and go left and count particles. Matter particles are counted up and antimatter counted down. Each time you count a particle, you assign them that number. After assigning every particle with a number, insert yourself left of the highest number.
For example if we start at top left for this circle:
--++--
+ +
+ +
--+--+
We count and assign numbers assign numbers as follows:
-1 -2 -1 0 -1 -2
0 -1
-1 0
-2 -1 0 -1 0 1
So the best place to insert yourself will be left of the matter particle on the bottom right.
5:10 No need to do it for every spot. Just pick a spot at random and start the running sum. See where that running sum hits minima. That minima will have antimatter at one side and matter on the other. Just sit between those two particles.
I came to the comments to say this is the missing step to the solution (finding the minimum).
You can make the program more effiecient to tell you the correct spot without check each one individually
If you start counting from 0 from any given spot subtracting 1 at anti matter and adding 1 at matter simply find the spot which has the lowest value and you know it will be safe. As you know that if you counted from that value the number would never go down
You can start anywhere from 0, and going clockwise add 1 for every particle, subtract 1 for antiparticle. Each maximum is a safe spot.
Except... each *_minimum_* is a safe spot. ;-)
The second part was amazing 👏🏻 😍
Holy moly Brilliant shells out sponsorship money. I wish I could get some of that!
The first full riddle video I got right! :))
Genius
Usually most of us don't get it but still watch the video till the end cause we love TED ed
Ted Ed, can we have easy riddles or puzzles like your very first one? The zombie crossing bridge one?
4 min to solve all! Thanks TED-Ed!
3:45 run through the sequence to your right. +1 for glouns, -1 for antimatter. As you go down the sequence if you ever get to negative the spot is unsafe
I actually did the final solution just like they did. That’s a first I wasn’t expecting
“Can you solve the Big Bang riddle?”
Me: “no. But I like watching the answer!”
You can think of it as sets of brackets cancelling out, where a (+) is an opening bracket and a (-) is a closing bracket. So the sum hitting zero means that the next bracket that will be matched is you, because you’re on the zeroth level.
the way I described the riddle in my head to solve it was to find a line of anti matter with a line of matter next to them, then stand behind them and push them to annihilate each other like a long conga line chain. I don't know why I chose the word conga line, but if it works, it works.
Thank you Josephus!
Haven't seen it yet, but it looks fascinating as always!
No need to repeat the calculation for each possible starting spot. Just start anywhere and keep adding/removing 1 until you are back where you started. As there are the same number of matter and antimatter cubes, the running count will be back to zero. The safe spot is next to the maximum value ever reached by the running count (there may be ties).
By the way, based on the video, it’s likely all attendees are now dead. Annihilation is much more efficient than nuclear fusion or fission, and standing near a blast of hard gamma radiation in such large amounts, depending on the distance, will vaporize the attendees, crush them in the ensuing pressure wave, or give them radiation sickness.
Minimum value, not maximum value
I first had a bad strategy but it kind of worked for the three initial examples, I pick a spot, draw an imaginary line that goes trhough the center of the circle and divides it in two, then check that the portion at my left has more antimater than the portion at the right. Then i realized that it doesn´t have to work but it was a fast way to visually pick a candidate spot.
First riddle: Oh I solved, I'm so smart
Second riddle: let's not talk about the second riddle
This is actually the first one ive been able to get :)
I miss these riddles :D
Took me about 3 seconds to pick my spots based on my strategy, which was a fairly simple “always be to the left of the longest line of mater because I figured all shorter lines would fully deplete before your longer line. I realize after the fact that there could be a case where two shorter line are divided by just 1 antimatter and could combine after the first round to be longer. Taking this into consideration and spending another 2-3 seconds looking at the rings I still like my original spots. Time to see how I did.
Well, looks like that only worked on the second circle. My quick double check was off by one pair.
At the end, I like how he’s smiling when he’s ‘enjoying the fireworks’.
finally one i sort of understood
(but it's because another channel, this one i think, did a video about the josephus problem)
Took me about 15 seconds to figure it out, but I only got the second one right. My strategy was to just be exactly to the left of fthe biggest number of matter particles
This strategy doesn't work. You can see the first circle at 1:27 would put you in a losing spot.
So did you actually figure it out? I don't get it.
Yes! Finally a riddle that I could solve again!
this is similar to a problem numberphile covered some time ago about a circle of warriors killing the odd numbered ones to their right
So, we used to be regular people, then wildebeest, then pilgrims, then vampire hunters, then sausage spies, then cannonballs, and now we’re just particles. I’m not Camilo, Ted-Ed, i can’t shapeshift.
I got the first one in, like, a minute and I'm so happy because this is the first riddle I've actually gotten lol
Cosmic Whistle? How clever! 😍
Thanks for the Challanged Version. It took mee 1:26 minute to get all 3.
This reminds me of the Josephus problem which numberphile did a great video on
I didn't start a timer but it took me around 30 seconds. My strategy was to start by writing the circle as a long binary number. Like the first one was 1110110000. Then I crossed off any 0 followed by a 1. Repeat till last pair remains then you stand to the right of the last 1.
Thank you.
I actually used the second approach from the start and figured it out in like a minute, so it only took like 5 to 10 seconds for each ring. That's probably, because a similar concept is used in many combinatorics problems in mathematics competitions. So for me it's basically as if I've had already known the solution. But many people don't have a mathematics background like I do, so you shouldn't feel discouraged if it took you longer or you couldn't figure it out on your own. If anything I should probably do something more productive with my time instead of watching youtube. um... yeah. Thanks for listening (I mean reading, I guess) to my TED-talk.
Scanning every possible insertion point, where at each point makes +1/-1 through the whole ring would take O(n^2) time. However, we can pick any insertion point as a starting point (don't insert yourself into that point yet). Mark that starting point to have sum = 0. Run through the whole ring once, calculate accumulate sum of +1/-1 of each step. Every time the sum get lower than the previous seen sum, update that spot to be a candidate. After finish running through the whole ring, the candidate spot is the point of insertion. This algorithm run in O(n).
Wow, what a great riddle, i can't believe you can devise it.
I introduce a counter starting at zero and choose a random spot, then I move to the spot to its immediate left
if I pass an antimatter particle I lower the counter by 1
if I pass a matter particle I raise it by one
then I continue to move to the next spot to the left until I reach my original spot
once I circled the entire circle I choose my spot as the one where the counter was the highest
4:17 including the time I took to actually figure out a strategy during the round
Took me a minute to figure out for all of them, not too hard once you get the idea.
I'm a kid and I love Ted Ed riddles!
Start the sum from any point. Mark cumulative sum for each point, then all the points where sum is minimum (max negative), are the safe spots.
(0:37) Cosmic Whistle sounds like the name of a low-class alcohol brand.
My time was 2:23 but I only got the first two right. Messed up on the last one :(
I tried to find some kind of trick for it, but it turns out there isn't one... you literally have to count particles, starting from each point to the left (clockwise) of a group of + ones, to see if you'd make it all the way around without running out of buffer. When there's a small number of them, you can sort of do that by intuition or grouping, but it's effectively the same process.
Kimari, thank you. I tried this for over an hour, and I also found no strategy.
Thanks
I know this is slightly late, but I took a slightly different approach.
I selected any point on the circle at random (without myself in it), and begin to count anti-clockwise, with a + adding 1, and a - subtracting 1. The difference with my approach was that I calculated the value at which the most negative value was calculated. When I was done with the circle, I would simply move to the space that had the most negative value calculated.
To use the safe method, I would do something as follows:
Pick any gap between two particles at random.
Set lowest number seen to be -1
-- Loop starts here
While counting in a clockwise direction:
If sum is lower than the lowest number we've ever seen, print safe.
Then record that number as the lowest number ever seen.
Select the point that the minimum occurred and loop again.
Else if there is no sum that is lower than the number we've ever seen, then this point is the safest point. Stay here, and stop looping.
-- Loop ends here.
Thought it was a fun way to solve. Thanks for the challegne :D
If you store the block where the minimum value is, you can immediately output the safe spot
Awesome 👍
1:34. Took me a minute of thinking, but lots of fun!
Goodjob
had a couple minute think about a strategy only to conclude that i couldn't find a 100% foolproof trick and decided on winging it with one piece of strategy; to go the furthest left of which there are the most amount of normal matter to my right. That actually really helped as i solved all three in under a minute total using a slightly varied approach after actually seeing the circles, which was to go as far right as possible of which there are the most amount of anti matter to my left, pretty much the same thing but slightly different, to be exact it took me 56 seconds, pretty happy with it!
Riddles. My favourite
Numberphile already tackled this on their channel in their video on the Josephus problem, but it's nice to see it mentioned elsewhere :)
Maybe it's just me but in the past, all of these riddles stumped me. This time, it seemed mind numbingly obvious off the bat because it's just finding pairs to shield you. I'm not even sure how this is a riddle it's just a math concept/addition and subtraction. First time I've been underwhelmed by one of these...
43 seconds, strategie was combining 4 things that are needed
1. anti immigetly to left
2. the amount of anti to the left needs to be >= the amount of glutons to the left of the antimather.
3.Gluton to the immigetly right.
4.The amount of gluton to the right needs to be >= the amount of anti to the right of them.
whit this alot of spots dissappear immigetly, so by trying a spot in the head that followed these rules there is high chance it works. the first two circles I got whit the first spot guess while the last circle needed two guesses to find the correct spot.
*strategy *immediately *gluon *with *a lot. :-B
To make it more efficient, start at one spot and keep track of where the sum was the least. If the sum is always positive, that's a safe spot. Else, pick any spot where the sum was the least at.
This reminds me of the Dr. NIM problem!
1:30 sec about because I feel like it’s intuitive once you analyze a small scale circle
Good to know
about 1, 1, and 3. My strategy was the same as suggested :)
The one math Ted Riddle I'll ever actually figure out lol
And quite quickly too, I'm surprised
1 minute 16 seconds, yay.
For the second question, pick a point, and move to the right until you reach the start again. For each matter, increase the "buffer particles" by 1, for each antimatter, decrease it by 1.
If the buffer goes negative before it completes the loop, it's unsafe. Otherwise, it's safe.
edit: Close enough, my take starts at zero and can't go negative, instead of starting at +1 and can't be 0.