GOOGLE Interview Question || Puzzle : 12 Men On An Island || Hard Logic Puzzle
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- Опубліковано 18 чер 2020
- GOOGLE technical interview puzzle : This puzzle is asked in the software or technical porgramming Google interviews. This video explains the questions or requirements and two different answers.
It's a hard logic riddle.
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You can find the solution with 4 men at this google drive link below:
drive.google.com/file/d/1KgRf...
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This interview puzzle has a mind blowing GOOGLE twist inside, so watch the complete video as you will see a smarter approach to solve this puzzle.
PUZZLE :
There are 12 men on an island. Eleven weigh exactly the same amount but one of them is either heavier or lighter than others. There is a see-saw on the island which can be used to find out the odd man, but the catch here is that you can use the see-saw only 3 times.
So using the see-saw only 3 times... Your task is to find out
- the odd man (i.e. the man with different weight).
- and Also find out if the odd man is heavier or lighter.
There exists several variations of this interview puzzle such as:
- 12 (twelve) balls and 3 (three) weighing/scale
- 12 marbles and 3 scale
- Twelve coins and three weighing (12 coins riddle)
In all these variations the intention is to identify the defective object by using the scale only three times. Also, you would need to figure out if the defective object is lighter or heavier compared to others.
This google interview coding question or puzzle was also featured in a TV series called Brooklyn 99 (Brooklyn Nine-Nine ) in the same format of 12 men as in this video.
Google is known for asking tricky puzzles and hard riddles in the interviews. They check the optimization skills of a candidate with his approach of solving a puzzle. If a candidate has good optimization skills then it will benefit the organization in writing optimized programming code that eventually improves the response time of software applications and websites.
In the video I have explained a natural approach followed by a fresh approach which is expected by google interviewer. I also have explained why the first approach is not upto the mark for google and how we can arrive at the second approach in most logical way possible.
So if you are preparing for google interview questions for software engineer then you can watch all google puzzles on my channel.
You are most welcome to share puzzle, math problems or any topics for upcoming videos.
Gmail : logicreloaded@gmail.com
Facebook(message) : / mohammmedammar
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It's a war between space and time complexity. Earlier days memory was too costly but we had time, hence we had less space consuming but more time consuming apps. But now space is cheap but time is costly, that's why we are having more space consuming and less time consuming apps. Dynamic Programming literally means using more space to reduce the time required for computation.
So it's simple, the first approach takes more time but less space, and the second approach takes more space but less time. The solution depends relative to the problem always which is why we don't have 1 definite solution for all.
Both the approaches are correct and you would get a job if you are able to explain it.
Murali... that was a valuable suggestion/thought to understand that the present era has changed a lot, and now everything is about time (responsive design).
I wonder if you could get away with O(n) space with this algorithm. The original algorithm still needs O(n) space, because that's the only way to compare weights in this scenario. Here, the naive approach is to load in all weight mappings in advance for O(cubedroot(n)*n) space. But if you could somehow determine the final #/h/l mapping without having to back-reference the list and count L/R/= signs, then it's pretty trivial to switch to O(n) space by loading in one weighing at a time.
For the 1st riddle cant u split the group in half (6 on each side) then let the men get off in pairs (1from each side) until its is balanced then u will know which pair was uneven and have 2 suspects. Now balance 1 of the suspects against any of the other 10 we know r normal weight and if he is same weight then with ur last attempt at balance him against any of the remaining eleven to tell if he is lighter or heavier if suspect 1 was uneven u would have observed if he is lighter or heavier and still had 1 attempt remaining
@@danielbinns2802 No, because each time you remove a pair, you are weighing a different combination which means it counts towards the 3 times you can weigh a combination.
@@alidhan1299 the riddle said u can only use the seesaw 3 times it did not say that u can only weigh different combinations 3 times and if they get on the seesaw they have to come off it. I'm just giving them a specific order to come off instead of having them all come off at once. Same can be done for the order they go on the seesaw because they can all simultaneously go on so I would observe the weights while they go on or get off. This is my way of thinking outside the box
If we weren’t all so awkward we could simply ask the people their weights, thus eliminating the need for a seesaw
@Panso Pe ink gir gai Having extra or less weight doesn't make you an imposter.
@Panso Pe ink gir gai Don't forget ,they are man, and man don't get shy telling their weight
That is only fair if the imposter cannot lie or he turns into a chicken.
Imagine the question was about coins. You can't ask a coin for its weight.
@@OskaIvanovichSmirnov but the question isn't about coins, why would we need to imagine a completely different scenario
Like always, while credit is given to those who solved the question, I always admired those who came up with these questions in the first place....
Well, those who came up with the questions in the first place are the first to solve them I guess🤔
The question looks like it was written by a 12 year old that just finished their first semester of programming electives.
If you want a search/sort algorithm ask me for a search sort algorithm--don't make up some retarded scenario that would never ever happen and pretend that there aren't hundreds of ways to solve it without a search sort algorithm.
To put it clearly--no one in their right mind would solve this problem in the real world the way google is expecting you to solve it. It's just stupid.
My answer: Google search which man is heavier or lighter since all of their data is online
😂
Searching answer in Google to get job at Google😂😂
You don't get any info abt 12 men on a deserted island. That's sad
🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣
@@kaimadamantonyshejin2034 You need to know how your company works before joining it.
(Works has dual meaning here)
I puzzled over and gave up. The first solution is so amazing how it saves every bit of useful information from each step to combine later to reach a result quickly. The second solution washes all that away and looks at the general problem, showing how it all just falls together naturally and without anything special. The richness of the solutions to such a simple problem are what brings me back to mathematics every time. So much hidden wonder. Even though I didn't solve it myself I found this inspiring.
We had this question back in winter term 1989 on our very first weekly exercise sheet at the very beginning of our Computer Sciences studies, at Karlsruhe University, Germany. We did come up with solutions similar to this one, but the real catch came when we were shown and told that this can be modelled as an error-correcting Hamming code with a 12x3 matrix over Galois Field GF(3). One matrix multiplication and bang, you get the position and weight of the odd man out. Oh, that was an ideal precursor to the advanced stuff that was to follow in the next couple of years, and a sure way to whet our appetite. Happy memories.
I remember my uncle, a physics teacher, gave me this question when I was 12, it took me a few hours to solve it. I still remember I intuitively went for dividing them into 3 groups right away (anything else felt useless in terms of revealing the maximum amount of information). Fir the last/third weighing, I didn't know any better than brute forcing. I was so fascinated that later I proved a formula for the generic case of N men for the minimum number of weighings required. Back then with no internet, this kept me busy for a few weeks.
if every thing won't go as usual you will be selected.
my father came 1989 to Karlsruhe for his phd in computer science (FZI). he and you probably knew each other
You just whet my appetite. That's awesome
@@NaderHGhanbariwhat was the formula you proved with minimum number of weighings
I watch the first minute of this video and then I paused it. I realized that I was presented with this puzzle in 1989. However it was presented to us as 12 balls and a balance scale that could only be used three times.
The professor wanted us to come up with the best route that will result in the highest potential of coming up with a solution. I came up with a solution the next day. When I presented it to him he said that I must have made a mistake. The reason was; when he was presented with the puzzle, it was presented to them as something that does not have solution. I'm glad that he had forgotten to tell us that little detail.
Not at google but I was asked this question years ago for a server lab job. I did not know the answer and did not get the job lol. He said not one interviewee knew the answer. I still believe to this day if I knew this riddle I would have gotten that job.
I solved this puzzle in other way around
12 yrs back..
W-weigh, we give number to each on
Scenario 1, W1: four on each side 1234-5678 if it is balanced then it is simple to find the odd one from remaining 4, as follows W2: keep a person(12) aside n weigh like this 9,10-11, 1, if it is balanced then compare 12 with 1, ll know whether he is lighter n heavier. If it isn't balnced, and tilting towards 9,10 means either one of them is heavier r 11 is lighter, then W3:compare 9&10, which way it goes is the heavier one, if balanced 11 is lighter.
Scenarios 2: W1: 1,2,3,4 - 5,6, 7,8 unbalanced n shifted towards 1234, W2: 1,2,5 - 3,6,9 if still shift to 1,2,5 it means either one from 1&2 is heavier r 6 is lighter.. Then compare 1&2 which side it tilts is the heavier if balanced 6 is lighter. If 1,2,5- 3,6,9 is balanced means eeither 4 is heavier r one from 7&8 is lighter, then compare 7 with 8. if in Scenario 2 balance shifted to 3,6,9 side that means either 3 is heavier r 5 is lighter... Compare any one from 3&5 with normal one to get to know the odd one...
Ramesh Kiran's solution (almost same as mine) seems to be the best solution, so I've written it up below in clearer terms ("W#n" means the nth weighing):
W#1: Weigh 1,2,3,4 vs. 5,6,7,8.
If W#1 is balanced (odd one is 9,10,11,12), then:
W#2: Weigh 9,10 vs. 11,1.
A. If balanced, then W#3: weigh 12 vs. 1 to see whether 12 is light or heavy.
B. If unbalanced, then W#3: weigh 9 vs. 10; if W#2 had 9,10 heavier, then whichever side is heavier in W#3 is the heavy one (or if W#3 balanced, 11 is light); or if W#2 had 9,10 lighter, then whichever side is lighter in W#3 is the light one (or if W#3 balanced, 11 is heavy).
--------------
If W#1 has 1,2,3,4 heavier:
W#2: Weigh 1,2,5 vs. 3,6,9:
A. If 1,2,5 heavier (1 heavy or 2 heavy or 6 light), then W#3: weigh 1 vs. 2; whichever side is heavier is the heavy one (or if balanced, 6 is light).
B. If 3,6,9 heavier (3 heavy or 5 light), then W#3: weigh 3 vs. 12 to see whether 3 heavy (if balanced, then 5 is light).
C. If W#2 balanced (4 heavy or 7 light or 8 light), then W#3: weigh 7 vs. 8; whichever side is lighter is the light one (or if balanced, 4 is heavy).
--------------
If W#1 has 1,2,3,4 lighter, then the process is same as above, switching corresponding numbers.
Thanks buddy@qc1okay
Isn’t this a ted Ed puzzle
Noelle Rhee yup
@@qc1okay
Y these many scenarios ,
It's simple
11 are having equal weight 1 guy is over weight
3 times we have to use that sea saw
1) first divide the 12 members into 2 grps 6-6 weight them one side shall be heavier.
2)take the heavier grp ie 6 ....now divide them again into 3-3 as grps and again weigh them .you will get the heavier grpWho are only 3
3) Now of those 3 members tell two members to be on seasaw ..... If they weight equal ....then the 3 Rd guy is the different one .
And if any one of those 2 are weighing heavier ......he is the guy as other all 11 members are of equal weight.
When we are going to attend for aptitude classes.....we can easily test the trainer by giving this question....🙋♂️
Evarikaina ichavaa aa question?
Are you going to attend English classes?
teachers are also watching this video
for the first few minutes, "Ahh I got this, simply do it as TedEd has explained".
the second explanation, "Ohh wait, what was that? how could you... but... nevermind."
But i believe, even the first approach is not easy.. if compared with TedEd... because in their version of puzzle we don't have to figure out the DEFECT (if the object is heavier or lighter).
@@LOGICALLYYOURS when 2 set of four men left , try with balancing 1-1 in
@@LOGICALLYYOURS In this problem there is a trick that is "lever" thats why I would just say it cannot be solved. Cuz as you can imagine at seesaw the nearest man close to the middle, will be the lightest. But its true just for this question.
@@grayfurt95 Even if you factor in the lever, that implies that the force of a person's weight is relative to their position on the scale, as long as people or objects being weighed are evenly distributed along the scale on the scale and both sides have the same number of people, the problem is still solvable. If all people on the scale weigh the same, it will balance. If a single lighter or heavier person is on the scale, it will be uneven regardless of the outlier's position as there will be a difference in the total amount of downward force between the two sides of the scale.
@@LOGICALLYYOURS bro I actually found a realistic way to solve
Use the sand on the island
Measure the amount of compression one guy makes then see all compressions and compare the deeper or shallower one
Man u think out of the box 💯💯respect
Where would you find an apparatus for measuring the compression in the sand? Think realistically.
@@krishbedi6410 a stick. It doesnt need to be a ruler lol
Epik using a stick would not yield accurate results and also the compression of sand particles could result in the elasticity of the the sand as a whole
Hampering of the elasticity*
At first I thought this was one of those bs "Only 1% with 6000 IQ can solve" puzzles but it was a genuine programming problem. Thank you algorithm
This is based on a problem from a brazilian math professor Júlio César de Mello e Souza. during the 1930 decade. My father was one of his students and told that he was a wornderfull mathematician. I'd lije to suggest that the Google give credit to him. He wrote a book called "O homem que calculava" Something like "The man that used to do calculus"
The key idea is the "balancing with known normal men" part. That's what I didn't see initially when I considered how many to place on the see-saw.
Well after strong focusing on the procedure I still do not understand nothing 😂
Ya bro,same here with me🥴I made the first method myself and then I didn't get the second one explained by him😑
Don't apply for the job at google.... Hope you understand this..
@@manassinha9145 you don’t need to Ans to actually get a job
that means you understand everything if you don't understand nothing
First
I might be slow
But when there's an upload
I click so fast I glow!
When u answered 1 but friend told you answer is Egypt
Just found this: my father gave me this one when I was a young teenager. His only clue was "you must get the maximum amount of information from each weighing". When I couldn't do it, he amended this to "what does each weighing tell you?". I did it - in the end. Later on I did programming and got to know about the "even more logical" approach mapped out here.
I want to get job in Google
After watching this video,this has totally inspired me
Please complete the sentence...
Inspired you to ……?
I think in general you should balance the following three quantities as good as possible in every step (while making sure Left and Right contain the same number of people by filling up with Confirmed Persons):
A) 2 * Unknowns Outside + Suspects Outside
B) Unknowns Inside + Heavy Suspects Left + Light Suspects Right
C) Unknowns Inside + Light Suspects Left + Heavy Suspects Right
If not perfectly possible you should prefer to have the Inside a bit larger.
Reasoning: This optimizes the worst possible outcome of the scaling step.
Google taking indiana jones' puzzles as interview questions XD
I did group of 4 vs 4. Then shift out 3 pleople from the left and shift in three to the right. Weigh again. By tracking which way scale tilts you can tell which group of three has lighter/hevier person. If last guy is left alone just weigh him against anyone else. .
Not going to lie, I had a migraine trying to solve this puzzle, but it was the good kind of migraine. Thank you for the challenge!
Thanks Freddy:) I appreciate your comments on many of the videos :)
My solution to the problem was to divide all the 12 men into two groups and put them on either sides of the see saw, obviously it will be imbalanced. Then we ask one person from each side to step down together and when the see saw becomes balanced this way, we select the pair which stepped down at last. The suspect is one of them. Then we can weigh the two of them one by one with a safe person to know who it is and if he is lighter or heavier.
too many uses of the see saw
12 men into 2 groups
6-6.
One side will be unbalanced
Take everyone off the see saw. Take the 6 from the unbalanced side and split em 3-3. One side will be unbalanced.
Take everyone off. Take the 3 left, put 2 on the seasaw. If it's equal then the one not on the sea saw is the odd man. If it's not equal then you've found your odd man.
After watching the second approach I'm like toote udh gye bhaishab 😂😂
Toote ,kabutar
Sab udgaye bro
🥴
Thnks a lot, it's really beneficial to all CSE students...This is a very good concept for dynamic programming
Thanks Sachin :)
I don't want your job, Google.
The hell of a question and hell of an explanation.. we need to get to the 2nd approach to make computers work... Hoping to get these type of puzzles 🤛
This puzzle actually gives a glimpse into how an AI model is trained. It compares known data to the test data. ;-)
Actually, the 2nd "solution" is a great example of poor optimization - that is, optimizing the WRONG THING. The 2nd solution only works better on an ideal machine - one which has no performance limits based on data set size, and no memory performance limits. Such machines, of course, don't exist in the Real World, and thus, the optimization done in the 2nd solution will perform significantly worse than a solution that uses the 1st solution as it's base algorithm, but which is tuned to understand how limits of hardware affect actual computation.
@@eriktrimble8784 for optimization, both explanations in the video are wrong. Elimination by consistent half is always mathematically the fastest procedure for narrowing. and also finds the Odd man in three steps in this situation. this approach would not always tell the weight of the odd man. but it would be more proficient to weight the odd man at the end instead of considering the weight of everyone during the process.
My first solution had a slight difference: in the second weighting, on both sides I put two suspects to be heavy and one suspect to be light. This removes case 2b and adds an additional case 2c.
I've seen this question many times before Google even existed...This is a famous puzzle.
The big problem that threw me off track when I first was confronted with this problem in the late 90s is the amount of if/then/else branching you have to do. And if you know the 9 ball / 1 heavier problem, it actually makes it harder for you to solve this since you start off trying to solve it the same way, which does not work.
This is totally out of topic but isn't this the same riddle given by captain holt in the show B99 😕
Read the description
It's definitely older than that. The first time I heard this puzzle was from my uncle when I was a kid. It was years before the release of the show. Now I wonder about the origins of this puzzle.
Yes, but it wasn’t invented by Brooklyn writers.
Yeah obviously it must be much older of a puzzle ..... I was just pointing it out that this was mentioned in that show as well 😅
I mean I'm here cause I saw the title and thought of B99 😂
"Hey Guys, I brought donuts, have one, they taste great!"
"You're odd..."
Great presentation and a very good explanation of the google solution! As a computer programmer I often use these kinds of solutions in programs.
The solution i came up with :
Just devide them into 2 equal groups and place them on the seesaw
Then let them jump down one by one ,
1 man from each side at the same time
When the pair with oddman jumps down the seasaw will be balanced
Then we know one of them is heavier or lighter
For the 2nd time
Just weigh one of them with a normal man
If he is the oddman we will know
And if the seesaw balances then the other man is odd
And we just have to weigh him with a normal to know if he is heavier or lighter
You can't perform such jumping
every time you let people jump off the seesaw you are essentially weighing the remaining people, counting against the three weightings that are allowed.
haha;;{{)) love it,... yeah, you're right,...
Someone needs to make a game with all these difficult logic games. Ooo and in VR
Great video, I'd like to see a more generalized solution with the max number of men for each number of weighings, but this seems much more difficult
It is a variation of the "fake coin" problem that we were solving in a high school. Our nice solution puts the people in a grid 3x3 with the extra 3 diagonally. Then you weigh two columns against each other and depending on a result you pick two rows to weigh against each other. When you look at it you'll see the solution right away without a need for any spreadsheets. Still nice though.
Me: *Doesn't understand something*
Random indian man: I shall reach you.
Next time you're stuck on an island with 10 other unknown men who weigh the same, remember this 😌
Then the first thing that will come to my mind is "how to get the hell out of here"..
This problem was the climax of _With a Tangled Skein_ by Piers Anthony. The solution given was the same if the first weighing balanced, but it had an interesting concept for when the scale was unbalanced.
Suppose at first weighing, you have:
W1: H1 H2 H3 H4 vs L1 L2 L3 L4
Meaning the side with Hs is heavier and the side with Ls is lighter.
Then proceed as follows:
W2: H1 L1 L2 L3 vs N1 N2 N3 L4
If this balances, the odd one is one of H2, H3, or H4.
Weigh:
W3: H2 vs H3
If this balances, the odd one is H4.
If this does not balance, then the odd one is the heavier one of H2 and H3.
But if W2: H1 L1 L2 L3 vs N1 N2 N3 L4 does not balance, there are two scenarios:
If the H1 L1 L2 L3 side is still heavier, then the odd one is either H1 or L4.
Weigh:
W3: H1 vs N1
If this balances, then the odd one is L4.
If this does not balance, then the odd one is H1
If the H1 L1 L2 L3 side is now lighter, then the odd one is one of L1, L2, or L3
Weigh:
W3: L1 vs L2
If this balances, then the odd one is L3.
If this does not balance, then the odd one is the lighter of L1 and L2.
This puzzle features prominently in the climax of Piers Anthony's book "With a Tangled Skein", which I read about thirty years ago. The protagonist had made a bet with Satan and was weighing demons.
Obviously the puzzle was a memorable one, since I still remember it!
I had forgotten that until you mentioned it. Piers Anthony is one of my favorite authors. Thanks for the memory.
Wow ! Soo Difficult !! I am grateful Google doesn't ask these questions nowadays otherwise every applicant will get the job !!! 🤣🤣🤣
Damn..... It was totally awesome.. tho i was lost
The moment when you thought you were smart and are proven wrong...
Then it means that you've learned something new, so it's a win.
Thanks man. It was an amazing puzzle
Lesson learned :
Always take a weight machine with you, when you go to an island. 🙏
😂
True af
I’m Quite Sure That This Is. a Teded Riddle. I remember doing a similar riddle like this
You need to identify whether the defective object is lighter or heavier.... that makes it difficult.... whereas in ted-ed version you don't have to find out the defect.
Yes ur right
Yup the coin riddle
kabir md in ted-Ed version, you don't know whether the fake coin is lighter or heavier too.
@@Thaplayer1209 that video gives the path to determining both the odd coin and whether or not it is heavier or lighter
My version with coloured hats (all hats have equal weight)
Strategy behind this is, roughly, that you have 1/3 of non-greens off-scale, 1/3 of them staying on their side and 1/3 of them switching side. If the scale balances the odd one is off the scale, if the scale switches the odd one is in the group that switched and if the scale stays the odd one is in the group that stayed.
1. Everyone gets a white hat (white like blank paper)
2. Whenever the scale is balanced, everyone on it trades their hat for a green one (green like all good, not the odd one out)
3. Whenever the scale is imbalanced, everyone NOT on the scale gets a green hat (because the odd one out must be on the scale)
4. On the "lighter" side, if any, white hats are traded for blue hats (blue like air) and black hats (black like ground) are traded for green hats.
5. On the "heavier side", if any, white hats are traded for black hats and blue hats are traded for green hats
6. Start by weighing 4 vs. 4. You get 8green/4white or 4blue/4black/4green.
7. Continue weighing 3 white vs. 3 green or 3blue+1black vs. 3green+1blue
8. You now have 1 white or 3 blue or 3 black or 1 of each blue and black, all the others are green (this is a bit tedious to check, but straightforward).
9. weigh 1 blue vs. 1 blue or 1 black vs. 1 black, if possible. If not you weigh a non-green vs. a green.
10. You now have 11 greens and a blue or a black (even more tedious to verify, sorry).
I get the optimized approach and I even get how you get to it but... I would never just think that up on the fly. Pretty smart though.
amazing!!
Appreciable effort
I've finally managed to do it. Man, that was really hard, but what a satisfaction, I can hardly believe I came up with the solution xd
I hear you David! I knew there was a perspective to approach from, I kinda created my own math language to write out the hypothetical situations. Once you really understand the rules, it seems super simple. At first I wanted to be creative with unequal numbers of men/not even using the seesaw. But even after trial and error, I think anyone can slowly learn how this puzzle works. Glad you figured it out! I usually feel like I can’t solve some problems, but yes very satisfying!
Thanks bro for questions actually its improving my logical thinking I actually solved it using COM(physics) concept. 👍👍
I ask this puzzles I'm my tution and no one is able to answer
You're your tution?
It's basically a mathematical induction of finding out the Imposter in Among Us.
This would have taken me about 25 million factorial years to solve the way you did! The brute force approach however is very simple and easy to implement, and takes less than a second even for large sets.
Honest man , same goes with me.
I created the same puzzle to challenge myself when I was a high school student back to 50 years ago. 12 coins with one fake without knowing it is heavier or lighter, the measurement tool is a balance scale. The solution I found was divided them into 3 groups, 4 each. 1,2,3,4 5,6,7,8 and, 9,10 ,11,12. The first measurement is to place 1,2,3,4 on left side and 5,6,7,8 on right side. If it is balanced, 9,10,11,12 has one fake coin, it's easy to find it with 2 measurements. If it is not balanced, then the tricky part is to do a rotation. Left side to be 1,6,7,8 right side to be 5,10,11,12. Then you'll find 2,3,4 or 6,7,8 or 1,5 has a fake coin in it. You can easily find it with 3rd measurement. Then I ask myself what is the maximum coin I can inspect with 4 measurements? How about 5 and more? Then I found the answer is 12*3^（n-3). I read most of the comments, I like the one solution to use 3x3 grid.
I remember reading a solution in verse to the 12 coin problem in the early 1970s, possibly by H E Dudeney:
"F" set the coins out in a row
And chalked on each a letter, so
To form the words: F AM NOT LICKED
(An idea in his brain had clicked)
And now his mother he'll enjoin:
MA, DO / LIKE
ME TO / FIND
FAKE / COIN !
Thank you very much sir for sharing such a important and valuable problem with solution and all your videos helped me a lot in improving my knowledge and I am now able to ace many of the aptitude test and I am really thankful to you a lot sir and plz keep posting videos. I am your big fan sir
I'm really happy to see your comment. Much appreciated:)
Thank you sir
Turns out I'm not interested in any job that throws puzzles at you when hiring.
Yah fuck logical reasoning
Somebody send this video to Capt. Holt. He’ll finally get that lunch.
There is an alternative way describing the 2nd method:
Again, we use the seesaw independently. Instead of encoding on the results, we encode the men.
We have total 3 times to use the seesaw. So, a man is encoded as X = (X₁, X₂, X₃), where Xₖ can be R, L, 0, and Xₖ = R (L) means this man appears on the right (left) side of the seesaw for the k-th turn, and Xₖ = 0 means the men does not appear on the seesaw for the k-th turn, k = 1, 2, 3.
Now, the rule of encoding:
0. Of course, each man must have a distinct encoding.
1. Each turn, the number of men on the right and left sides must be equal, i.e. the number of R assigned to Xₖ must be the same as the number of L assigned to Xₖ, for all k = 1, 2, 3
2. Each man must appear at least once on the seesaw, so you can not encode (0, 0, 0) to a man. (because even though we find everyone else having equal weights, we cannot make sure this odd man to be heavier or lighter: he never went on the seesaw!)
3. If an encoding is assigned to a men, the inverse encoding (R L, e.g. X=(R, 0, R) vs X=(L, 0, L)) will be dropped (cannot be assigned to another man), otherwise since the two men always take the opposite positions, if one of the two men is the odd man, we cannot distinguish them (the results would be either A is lighter of B is heavier, but we cannot make sure which man of the two is odd).
Similar to the arguments in the video, it's obvious that we need to drop (0, 0, 0), (L, L, L), and (R, R, R) to make number of R's and L's balanced. Similarly, we have 27-3 = 24 combinations, and by Rule #3, we only use half of them, and drop all the inverse encodes: So we use only 12 combinations, and assigned to 12 men.
We further define the 3 results Y = (Y₁, Y₂, Y₃), where Yₖ can be 'R', 'L', or '='. Yₖ = 'R'('L') means the seesaw resulting in Right (Left) side heavier, while Yₖ = '=' means the seesaw balances, at the k-th turn.
After we assigned the men's encoding, it's obvious, for example, if the results of the three times are Y = (L, R, =), it must be some men encoded either X=(L, R, 0) or X=(R, L, 0) to be the odd man. In this case, If we found a men is encoded as (L, R, 0) , we can also confirm this man is heavier (because the man stayed each time on the heavier side), while if we found men is encoded as (R, L, 0), we can confirm that man is lighter.
Omg! Decades ago our middle school math teacher gave us this puzzle, except there were coins and balance scales.
Here is a simpler algorithm for the general case: For each step, balance the Positions in such a way that 2 * Unknowns + 1 * Suspects is the same no matter what the scale does. (If not perfectly possible, place the remainder ON the scale.)
This does not account for cases where there are not enough Confirmed Persons to make sure both sides of the scale have the same number of persons.
Working solution. I've solved this puzzle the same way.
The biggest problem of this interview question is that it does not tell you anything about the quality of the candidate, no matter if his answer is right or not. That is why Google failed in more projects then he successfully finished 🙂
Actually this type of question does help know more bout a person. It helps Google know whether the person is a creative, quick or logical thinker. By answering the question in a creative way, maybe google would want to hire them, for creative solutions to problems. If they can answer the question quickly, then that person is great for quick responses to problems. And if they are logical in there answer then, there’s a great chance that the person makes smart decisions. Also if the person maybe takes longer to answer, they may be the type of person to make an informed choice to a decision. All these types of answering a simple question could really show how the person thinks when asked a spontaneous question. Although I do agree that only 1 question does not determine how the person actually makes decisions or choices, it gives a good gist on how the person thinks in an interview setting, with the stress. 🙂
Literally any question can at least lead to inference of something qualitative. You have no clue what position the person is applying for or what qualities that are looking for, just made an empty negative claim.
@@TheJacklikesvideos During my carrer, I have interviewed dozens of people and yet I am far from to be an expert but I have at least SOME experience. May I ask you what real experience you have?
Appreciate your effort.
Actually, I didn't know the first solution; but I learned the spreadsheet-solution in the late 90s.
This is indeed a nice puzzle which tests your logic and patience.
Me(when he said there is a twist at the end):are they blind???..... Can't they figure out just by seeing!!! Lol i was wrong.. 😂😂
Just to add on this, coding this is actually very trivial, because if you assign L=0, (=) = 1, R = 2 than the numbering of the people will match the weighting scheme if you look at the numbers in base 3
I've been asked this before in interviews, usually with quarters. It's just a simple recursion algorithm. 😉
Nice Explanation. Just a bit of correction though. Last part (3^x-3) that you wrote by the justifying -3 as 1 for === and 2 for equalling L and R. That should be (3^x - 2*no. of men). Because you see for 12 men it eventually turned out to be 27-24=3 (Making extra 3 possibilities so -3) but for 100 men it will be 243-200=43 extra possibilities. Love your videos. Thank you.
he said to remove at least 3 rows which means greater or equal (43>3) so no need for a correction . if you try this with just 13 people you'll need to discard 3^4-26 = 55 possibilities since 3^3-3 < 13*2
that equation was to find out number of weighing required.
we can eliminate extra lines later.
now you add this algorithm for pooling covid-19 tests, and we're done
Eh, wouldn't it only work if there was only one sample that actually had Covid?
Actually the algo you are talking about wouldn't require the complexity cause you wouldn't need the high/low variable in the end.
you would simply eliminate by half until you narrowed the odd man out. The CDC does this for you. They post all the data on their website.
you can see when and where covid is happening. That where you have to go for data and info. Liberal TV radio only tell people how to feel.
Took me around 30 seconds. Put 4 on each side and leave 4 out. If the scale tips remove the 4 left out and the high side. If it balances remove those 8. Then take the 4 that were on the low side or the 4 that were not on the scale. Place those 4 men on the scale, 2 on each end. Remove the two from the high side. That’s the second scale move. Then take the remaining two men and put them on the scale. The heavy guy drops the scale to his side. That’s the third scale move. I then watched the video and the explanation. His explanation had me shaking my head
This is a good puzzle. You can solve it for 13 if you don’t need to know whether the odd man out is heavier or lighter. One can solve for 14 men if you have a 15th man that you know ahead of time isn’t the odd one out. Classic riddle.
Wonderful video!
It's clear that with independent weighing, the numbers of steps required is bound log(3)(2N), but is it optimal?
In another word, is it possible that some non-independent weighing scheme (like in solution #1), can beat that estimate?
Can you either prove it's impossible? or come up with a counter example?
Given that the balance only has 3 output states, using it n times only gives enough information to differentiate 3^n situations, no matter how you use it. So there is an upper bound there. If you have to determine the bit of information about being heavier or lighter, the single state where the scale is always balanced doesn't help, so the upper limit is 3^n-1 states, of (3^n-1)/2 people.
The second technique described here only works with a number of people that is a multiple if 3, because otherwise the groupings aren't the same size and the weightings don't balance. This means that you can differentiate 3*floor((3^n-1)/2/3) people this way. (Due to the nature of powers of 3, this appears to always be (3^n-3)/2). 0 when n=1, 3 when n=2, and 12 when n=3.
I can't seem to find a way around that divisibility constraint, but I also can't find a way to demonstrate it must apply to all strategies either. I think it might be fruitful to attempt to prove that by showing that conditional weighing strategies have to map directly back to an unconditional one, but where you have added people ruled "normal" back on both sides. I can't seem to come up with a satisfactory map though. You can definitely do 4 for n=2 and 13 for n=3 if you add in a rock that weighs the same as the "normal" people to balance out the scales, even though that problem has the same information upper bound.
Good one...
Took the one who gave me the puzzle 20 years ago 2 weeks to solve it. I did it in 2 days. Didn't make me very popular with him. I guess there are people who, without being trained in such logical structures, just take a few seconds.
Great interview question... if your are applying to a job where you'll later on will solving logic puzzles. Yet I bet, 99.9% of all Google employees never are going to have to do anything like that, not even if they work 40 years for the company. Wouldn't it make more sense to ask them questions that really relate to what they will be doing on their new job?
Good point. It's better to ask shorter questions and those closer to the type of work they are actually going to be doing, otherwise you're burning interview time. The original poster is taking 20 minutes to explain this elegant generalized solution, he probably thought about this for hours, and that's not something you can expect a candidate to come up with in 10-20 minutes. Also if this is to help people with interviews, it should give tips on how to answer it to show they are committed to solving it for themselves as opposed to Goggling the solution.
I guess i'm not working for Google in this lifetime then
But when given in an interview, how much time do we have to answer and do we get pen and paper?
We get pen 🖊️ and paper 📓 but only three minutes
@@haio7710
Three?!!!
No thx
Jesus did anyone ever solved it without knowing the solution beforehand?
@@guitaek4100 Ofc not. Thats why youll never have a riddle on a job interview. It doesnt provide any info about the candidate. Its pointless. Talking as a Psychologist.
@@svilenacarapica4491 Interesting. In my job interview I had a riddle (not google) I didn't found the solution my own however I found an own (not as good as) solution. I guess it should test you wether you're creative. So why do you think it's pointless?
If this truly were an interview question Google would have no employees 😂
New Google hire brilliantly solves the puzzle!
First task: "Could you help us maintain this Excel spreadsheet that tracks business metrics?"
I haved watched almost all of your videos...and this video could be hardest one😂
The solutions presented are quite impressive. My personal solution involved some lateral thinking:
1. Take half of the 12 men randomly, and weigh them against each other (3 vs 3). If the seesaw is balanced, we know the other half of the men contains the suspect, otherwise he is in the half that was weighed. So now we have identified the suspect group, and the even group, each with half of all the men.
2. Take the even group and put them all on one side of the seesaw, and the suspect group on the other side. The seesaw will be unbalanced slightly, and now we will know whether the suspect is lighter or heavier .
3. With all the men still on the seesaw, instead of having everyone get off at once, have just 1 man from each side get off at the same time, and continue the disembarkation in this manner. When the seesaw becomes balanced, pause to acknowledge that the last man who got off the seesaw from the suspect side is the one with the odd weight, unless it is the last man from the suspect group in which case it will remain unbalanced when there is one man left from each side.
This technically only uses the seesaw twice, and can be scaled up to any number of men assuming the seesaw is big enough to support them all. The method of each side taking it turns to let a man off makes sense too because the first man to get off will make the other side heavier and sink closer to the ground, making it easier for them to leave.
Actually, each observation is considered a trial, so this doesn’t work in the allotted trials. Good logical thinking, though. 👍🏻👨🏼🏫
@@TomCee53that depends on definition of 'use' in the problem. Which opens the possibility for many interpretations
pick one to stand on seesaw...and line up everyone else..........done!
That is why a computer programming definition is more precise. You can call a function measure(men_left_side, men_right_side) which gives you the output L, R, =, and you can only call it 3 times. No ambiguity.
And I had literally paused the video to find the answer in my head....🤣🤣😁😁
My first answer would be to ask each of the men their weight, then if that doesnt lend results, I would start analysing it mathematically
Ok so In an interview, face to face in person your telling me they ask this riddle and that what determines your apptitude, whether or not you can think out a puzzle in your head without any prior knowledge of it? Yeah now i see why Google's algorithm has been extra fucked up lately lol
I think this is a perfect question for a programmers job. Dont know if you ever programmed, but its this type of issues you will have to solve. And he who solves it best makes the most efficient software. Consider the amount of weighing's you need to be the amount of work the processor needs to do to solve the problem. (and thats espescially important when its a problem that needs to be solved very, very often)
3 friends A B and C,C knows how much time A and B individually watched Titanic movie.C said, between a and b, one of them watched the movie 1time more...
A said B= video watched one time more than me?
B replied- No,Did You?
A said- No,but i got it, how much time you watched Titanic movie.
B said,- O really? Then I also know how much time you watched Titanic movie...
Question is how much times A and B watched Titanic movie?
A: once
B: twice
But I think this one is also a possibility
A: twice
B: three times
isn't it?
If b didn't know that at start how many times a watched , how could he ans the first question ?
@@anksssssssss I think the writer meant to write "A to B: Did you know whether you watch video more than me?"
I came up with a totally diferent approach. If the defenition of using the scale is to put objects on it to see the result I would:
1, Place everyone on the midle and ask one person on each side to walk to the end. The scale will be balanced until we find the odd one. then we have two candidates.
2, Compare the first candidate to the control group in the midle.
3, Compare the second one to the control group if needed.
Man, everyone thinks they've stumbled into greatness pretending constant adjustment and observation is a single instance of weighing.
My brain with the first case: oh well, duh! Just what I was thinking!
My brain with the second case: 😵💫
I think this is your longest video ever...
Yes Rownak, you're right!
Where or in which real-life system this algorithm can be used.
I'm just curious
It is not, they just want to find a person with right problem solving thinking that souts for the position they are hiring.
@@stefanpodgorsek7687 What if my problem solving skills are average but I binged watched all the Ted-Ed puzzle/riddle videos, and just happened to see the one about finding the counterfeit coin in a pile of 12 coins using a scale?
It is more of a test to see if you can recognize a situation that can be solved with a hash map
What if you position men on top of each other or shoulder to shoulder facing the fulcrum on a wide seesaw, then you could measure 1 person on the end vs 2 at the half way point. That how I initially solved it in my head, I split the men into 4 groups of 3, measured 6 in the middle vs 3 on the end, then did a similar heavy and light suspect thing that you did. Still required 3 measures.
This is useful in saving a measure in the case of 4 men, since 3^2 - 3 < 2*4 < 3^3 - 3, your spreadsheet method would require 3 measurements, but mine requires 2.
My solution: Let the 4 men be called A,B,C,D and weigh B on the end of his side against C and D in the middle of their side. Case 1: If they balance, A is the odd man, weigh A against any one of them to find if he's heavy or light. Case 2: The CD side goes down, weigh them against each other. 2a: if balanced then B is the heavy guy. 2b: if unbalanced, the lighter side has the light guy. Case 3: The CD side goes up, weigh them against each other. 3a: if balanced then B is the light guy. 3b: if unbalanced, the heavier side has the heavy guy. Only required 2 weighings.
I haven't done the working out to prove the minimum number of weighting for N people when this rule is allowed, but since the men are split into 4 groups instead of 3 each time, maybe the bound is ceiling[log_4(3N+4)] instead of ceiling [log_3(2N+3)]. I'll work on finding a bound and edit this if I do.
Old problem. Great one. Took me 6 months. Took my sister 10 minutes without paper and pencil.
Amazing
Okay, Now I got why I am not in Google. 🙄😟😅😂
The classic version of this puzzle, which is at least 50 but more probably closer to 100 (?) years old, uses ball bearing balls, i.e. they all look & feel the same but one of them is slightly wrong, either heavier or lighter. What is the minimum number of weighings needed?
Since each weighing can be used to split the men/balls into three groups (heavy/light/ok), we get 3 to the 3rd power or 27 as the maximum amount of information that it is possible to get in 3 rounds, and since this is larger than the 24 (12*2) possible outcomes, it is theoretically solvable. I do remember spending quite a bit of time actually finding a working solution back around 1970. :-)
A seesaw does not compare weights. It compares torque. It will balance = only if torque is same on both sides and that's highly unlikely as men will not be at exactly the same distances from the center. The puzzle makes much more sense (cents?) with coins on a balance scale where the weight is concentrated at the pan's suspension point.