GOOGLE Interview Question Puzzle: 12 Men On An Island Hard Logic Puzzle
video description
[1] a see saw measuring is a comparison event and we are trying to get the maximum information per measurement. we want to be able to get the maximum three possible states (HL, LH or Equal. Call it 0, 1, 2. but we have to impose the rule that can must always measure the same number of men on each side (all test groups should be equal size) or else the state '2' will be impossible to achieve and the information per measurement is greatly reduced. this is an obvious conclusion most people will get but more on this later.
[2] since the state of the odd man is unknown ('0' or '1') at some point he will need to be measured against a known normal man. one move is reserved for this.
[3] we are left with two moves and within that we need to guarantee we find the normal men. time for groupings. no matter how i group them, at least one group needs to have participated twice (each versus a different group as per [1]. if it is the odd group that was tested twice, we would know the value of the oddness (is odd heavy or light) and all the other groups would be normal. if the odd group was not tested twice, we would identify ALL of the normal men but would not know the value of the oddness. so i realised that EVERYBODY needs to be tested twice in order to guarantee that the odd man is tested twice.
[4] we are left with two moves to test everybody twice. so i put 12 men evenly distributed in a circle like peperoni on a pizza and realised with two cuts creating quarters (two moves, this can be achieved. A&B will be tested against B&C and subsequently A&D will be tested against B&C. northern hemisphere vs southern hemisphere first then western versus eastern second, if you will. no matter where any man was, he would be tested twice in two moves. the group which received the same '0' or '1' state twice is the odd one and it would either be '0' both times or '1' both times which gives me its oddness.
[5] the only question remained was: will i be able to finish [2] in one step? and the answer is yes. each group would have 3 men. from the odd group which i found from [4], take any two men and perform the third and final test. if they are equal, then the untested man is the odd one. if they are not equal then take the information from [4] to determine which one is the odd man.
while i did manage to solve it, it was by luck instead of algorithm. clearly if the problem was extended to 100 men it would not work. but based on my original solution it can be done although i cannot prove it. tell me if i am wrong:
100 men can be narrowed down to 25 containing the odd man with two weighings per above. additionally since i also know the oddness, on the third move i can add 2 known normal men and out of the 27 split them into three groups of 9 and weigh two of those groups, narrowing down to 9. on the fourth move i narrow down to 3 men. on the fifth and last move i weigh any 2 of the 3 and done. 5 total moves.
for 1000 men again we quarter it to 250 in two moves and finding out the oddness at the same time. move #3 cuts this into a third at 84 remaining men containing the odd man. this is less than 100 men which we already know takes 5 moves. so total 8 moves.
Date: 2023-11-15
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Comments and reviews: 29
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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 = (X-, X-, X-, where X- can be R, L, 0, and X- = R (L) means this man appears on the right (left) side of the seesaw for the k-th turn, and X- = 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 X- must be the same as the number of L assigned to X-, 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 = (Y-, Y-, Y-, where Y- can be 'R', 'L', or '='. Y- = 'R'('L') means the seesaw resulting in Right (Left) side heavier, while Y- = '=' 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.
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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 = (X-, X-, X-, where X- can be R, L, 0, and X- = R (L) means this man appears on the right (left) side of the seesaw for the k-th turn, and X- = 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 X- must be the same as the number of L assigned to X-, 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 = (Y-, Y-, Y-, where Y- can be 'R', 'L', or '='. Y- = 'R'('L') means the seesaw resulting in Right (Left) side heavier, while Y- = '=' 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.
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TrixMTL
i can find the odd man in 0 tries.
given that this is a riddle, and the stipulations are all stated. there is no stipulation to what you are not allowed to do. but there are stipulations that suggest you -have- to use all the tools given.
in riddles often the answer is in the begining. like the riddle of -you are driving a bus. 12 people get on, on the first stop, 10 people get off on the second stop. what colour are the bus driver's eyes? -
so given this context and the stipulations given.
1 person can ask another person how much they weigh, if he weighs the same, then ask, who doesnt weigh this much.
they would know who is lighter or heavier immediately in 0 uses of the seesaw.
12 men are part of the riddle. men can speak. men can also do math and weigh stuff. and ride a seesaw. the riddle doesnt ask for math, the riddle asks for who weighs different.
you can also use Variables, and argue the stipulations of -use- in the -3 uses max-, stipulation. to do it in 2 tries.
variables. X is lighter. Y is heavier, Z is 11 men that weigh the same, A is one side of the seesaw. B is the other side. U is unidentified.
so we start with 12 U and need to find X or Y. and have an A B axis that tilts to help us filter out.
on the first use. you fill the seesaw. 6U+6U = A down B up. with this first use you dont record when the seesaw stays the same but when the seesaw equalizes. you ask the men to come off in pairs.
so - (1U+1U) until A=B. you have then identified the pair of U and Z. weigh U and Z against 2 other Z and you have your answer if he was heavier or lighter, and on which side he started, identifying the person too.
or you over complicate rather than simplify, and probably starve to death on that island before calculating the outcome there. for the -nobility- of using the 3 uses with arbitrary stipulations.
which of the 3 ways is more clever or efficient. which way makes it look extra fancy with self imposed non existing rules. costing 3 times the money and time to produce. to get the same result.
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i can find the odd man in 0 tries.
given that this is a riddle, and the stipulations are all stated. there is no stipulation to what you are not allowed to do. but there are stipulations that suggest you -have- to use all the tools given.
in riddles often the answer is in the begining. like the riddle of -you are driving a bus. 12 people get on, on the first stop, 10 people get off on the second stop. what colour are the bus driver's eyes? -
so given this context and the stipulations given.
1 person can ask another person how much they weigh, if he weighs the same, then ask, who doesnt weigh this much.
they would know who is lighter or heavier immediately in 0 uses of the seesaw.
12 men are part of the riddle. men can speak. men can also do math and weigh stuff. and ride a seesaw. the riddle doesnt ask for math, the riddle asks for who weighs different.
you can also use Variables, and argue the stipulations of -use- in the -3 uses max-, stipulation. to do it in 2 tries.
variables. X is lighter. Y is heavier, Z is 11 men that weigh the same, A is one side of the seesaw. B is the other side. U is unidentified.
so we start with 12 U and need to find X or Y. and have an A B axis that tilts to help us filter out.
on the first use. you fill the seesaw. 6U+6U = A down B up. with this first use you dont record when the seesaw stays the same but when the seesaw equalizes. you ask the men to come off in pairs.
so - (1U+1U) until A=B. you have then identified the pair of U and Z. weigh U and Z against 2 other Z and you have your answer if he was heavier or lighter, and on which side he started, identifying the person too.
or you over complicate rather than simplify, and probably starve to death on that island before calculating the outcome there. for the -nobility- of using the 3 uses with arbitrary stipulations.
which of the 3 ways is more clever or efficient. which way makes it look extra fancy with self imposed non existing rules. costing 3 times the money and time to produce. to get the same result.
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Romain
Here is my idea:
1- you made three groups of 4 A, B and C
2- you compare B and C
If B=C, so A contains the odd person:
You separate 1 man from the group:
- you compare the 3 remaining with 3 regular men from B or C
- if it is balanced, then the man put aside is the odd one and you compare him alone with another one to know if he is heavier or lighter
- if not, you know that the odd man is in the three remaining persons and you know also if he is heavier or lighter (because you compared them with three known regular men. You just repeat the first step with 3 group of one to identify him.
If B! = C
- we take one of each group and put them aside (group D. Then we take two men from B and add them to C and one from C to add to B creating B- and C-
- if B-=C-, the odd man belongs to D. We weight one of D with a regular one to identify the odd man. The first comparison between B and C tells us if he is heavier or lighter
- If the balance has not changed, the odd man belong to those who hasn-t moved. Either the remaining man of B still in B- or one of the 2 of C still in C-. We compare these two. If balanced, the odd man is the man from B and the previous comparison tells about him being heavier or lighter. If not, the one with the same balanced as C is the odd man and we also know if he is heavier or not.
- If the balance has changed, it is the same as previously with only those who have moved.
I hope it remains understandable (quite hard without a scheme, but it should be right. I hope the video will present a more elegant solution-
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Here is my idea:
1- you made three groups of 4 A, B and C
2- you compare B and C
If B=C, so A contains the odd person:
You separate 1 man from the group:
- you compare the 3 remaining with 3 regular men from B or C
- if it is balanced, then the man put aside is the odd one and you compare him alone with another one to know if he is heavier or lighter
- if not, you know that the odd man is in the three remaining persons and you know also if he is heavier or lighter (because you compared them with three known regular men. You just repeat the first step with 3 group of one to identify him.
If B! = C
- we take one of each group and put them aside (group D. Then we take two men from B and add them to C and one from C to add to B creating B- and C-
- if B-=C-, the odd man belongs to D. We weight one of D with a regular one to identify the odd man. The first comparison between B and C tells us if he is heavier or lighter
- If the balance has not changed, the odd man belong to those who hasn-t moved. Either the remaining man of B still in B- or one of the 2 of C still in C-. We compare these two. If balanced, the odd man is the man from B and the previous comparison tells about him being heavier or lighter. If not, the one with the same balanced as C is the odd man and we also know if he is heavier or not.
- If the balance has changed, it is the same as previously with only those who have moved.
I hope it remains understandable (quite hard without a scheme, but it should be right. I hope the video will present a more elegant solution-
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Fish
My solution:
1) Take only 6 people and weight them 3 v 3
If all are even then you know the odd man is in the other group (call that neutral group, group A)
If the scale tips in any direction, then you know the odd man is in this group of 6 (call it Group B)
2) Take 3 from group B and 3 from group A (the known neutral weight group)
If the scales is even then the odd guy in in the other part of Group B (call it guy 4-6)
If the scale tip, then you know if they are HEAVY or LIGHt, and you know they are in this group (call it guy 1-3.
3) Weigh 2 of the guys from the tipped group (guys 1-3. So guy 1 vs guy 2.
If the scale remains the same, the guy is guy 3 (and you already know if he is heavy or light.
If the scale tips, then you know it's guy 1 or 2, and since you already know Heavy/light, you just pick that guy.
done.
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My solution:
1) Take only 6 people and weight them 3 v 3
If all are even then you know the odd man is in the other group (call that neutral group, group A)
If the scale tips in any direction, then you know the odd man is in this group of 6 (call it Group B)
2) Take 3 from group B and 3 from group A (the known neutral weight group)
If the scales is even then the odd guy in in the other part of Group B (call it guy 4-6)
If the scale tip, then you know if they are HEAVY or LIGHt, and you know they are in this group (call it guy 1-3.
3) Weigh 2 of the guys from the tipped group (guys 1-3. So guy 1 vs guy 2.
If the scale remains the same, the guy is guy 3 (and you already know if he is heavy or light.
If the scale tips, then you know it's guy 1 or 2, and since you already know Heavy/light, you just pick that guy.
done.
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ChannelMath
Puzzle extension (Information theory):
The puzzle asks you for 4. 58496 bits of information (1 out of 12 people is log2(12) bits, plus one bit for whether the person is heavier or lighter.
How many bits does weighing get you? Since it can show if two sides weigh the same, it seems to tell you 1 out of 3 possibilities, or log_2(3) = 1. 58496 bits, and so 3 weighings give 3 times the info, or 4. 75489 bits.
since 4. 75489 is just greater than the 4. 58496 bits you are asked for, a solution possible, but just barely.
Is this a correct Information theory analysis? and if so, does it prove it's possible to solve the riddle? And if so, what does it say (if anything) about how likely a random strategy is to yield enough information for a solution?
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Puzzle extension (Information theory):
The puzzle asks you for 4. 58496 bits of information (1 out of 12 people is log2(12) bits, plus one bit for whether the person is heavier or lighter.
How many bits does weighing get you? Since it can show if two sides weigh the same, it seems to tell you 1 out of 3 possibilities, or log_2(3) = 1. 58496 bits, and so 3 weighings give 3 times the info, or 4. 75489 bits.
since 4. 75489 is just greater than the 4. 58496 bits you are asked for, a solution possible, but just barely.
Is this a correct Information theory analysis? and if so, does it prove it's possible to solve the riddle? And if so, what does it say (if anything) about how likely a random strategy is to yield enough information for a solution?
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Murali
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.
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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.
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Simon
My solution to the 4 men puzzle:
cmp(a, b(--
-left down-: cmp(a, c(--
-left down-: -a is heavier-, -
-flat-: -b is lighter-, -
-left up-: -paradox-, -
-, -
-flat-: cmp(c, d(--
-left down-: cmp(a, c(--
-left down-: -paradox-, -
-flat-: -d is lighter-, -
-left up-: -c is heavier-, -
-, -
-flat-: -paradox-, -
-left up-: cmp(a, c(--
-left down-: -c is lighter-, -
-flat-: -d is heavier-, -
-left up-: -paradox-, -
-, -
-, -
-left up-: cmp(b, c(--
-left down-: -b is heavier-, -
-flat-: -a is ligher-, -
-left up-: -paradox-, -
-, -
-)
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My solution to the 4 men puzzle:
cmp(a, b(--
-left down-: cmp(a, c(--
-left down-: -a is heavier-, -
-flat-: -b is lighter-, -
-left up-: -paradox-, -
-, -
-flat-: cmp(c, d(--
-left down-: cmp(a, c(--
-left down-: -paradox-, -
-flat-: -d is lighter-, -
-left up-: -c is heavier-, -
-, -
-flat-: -paradox-, -
-left up-: cmp(a, c(--
-left down-: -c is lighter-, -
-flat-: -d is heavier-, -
-left up-: -paradox-, -
-, -
-, -
-left up-: cmp(b, c(--
-left down-: -b is heavier-, -
-flat-: -a is ligher-, -
-left up-: -paradox-, -
-, -
-)
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ashwani
I have a better solution: make 2 groups of 6-6each and wiegh them on the sea saw, the group that's heavier or lighter will be the odd group, now deselect 2 random people from the odd group and wiegh the 4 people left 2-2 each side, if the sea aaw balances, then the 2 deselected people shall be weighed and one will be normal while the other heavier or lighter ( we'll know heavy or light as when we weighed the first time with 6-6. We'll know if it was heavy or light), if it doesn't balance out, then wiegh the group thats odd this time of two and you shall have the heavy or light one
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I have a better solution: make 2 groups of 6-6each and wiegh them on the sea saw, the group that's heavier or lighter will be the odd group, now deselect 2 random people from the odd group and wiegh the 4 people left 2-2 each side, if the sea aaw balances, then the 2 deselected people shall be weighed and one will be normal while the other heavier or lighter ( we'll know heavy or light as when we weighed the first time with 6-6. We'll know if it was heavy or light), if it doesn't balance out, then wiegh the group thats odd this time of two and you shall have the heavy or light one
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Aditya
I'd have 6 go on each side first, reducing it to just 6 after observation. Then I would put 3 on each side leaving me with 3 after observation. Now I would pick 2 of 3 and put them on each side. If they balance then the remaining 1 man is the odd one out. If they dont balance then either of them could be the odd one ---either lighter or heavier, I'll pick either one of them with the remaining 1 man, which would help me single out the odd one out.
In the worst case this solution would take 1+1+1+1=4 uses of the see saw
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I'd have 6 go on each side first, reducing it to just 6 after observation. Then I would put 3 on each side leaving me with 3 after observation. Now I would pick 2 of 3 and put them on each side. If they balance then the remaining 1 man is the odd one out. If they dont balance then either of them could be the odd one ---either lighter or heavier, I'll pick either one of them with the remaining 1 man, which would help me single out the odd one out.
In the worst case this solution would take 1+1+1+1=4 uses of the see saw
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Marco
The diagrams with items at different distance from the scale-center will not give you the correct outcomes (a weight closer to the scale-center will have a smaller effect and may give you false information. Make sure the men are standing side-by-side at the end of the see-saw to make this work as described. That's why actual balancing scales hang a bowl from a single point on the balancing arm to make it much easier to weigh a number of items without having to position them too precisely on the scale.
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The diagrams with items at different distance from the scale-center will not give you the correct outcomes (a weight closer to the scale-center will have a smaller effect and may give you false information. Make sure the men are standing side-by-side at the end of the see-saw to make this work as described. That's why actual balancing scales hang a bowl from a single point on the balancing arm to make it much easier to weigh a number of items without having to position them too precisely on the scale.
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Noel
I have not read all comments, so this may be covered already. The puzzle clearly makes the simplification of not taking account of the distance of individual men from the fulcrum. Further away from fulcrum greater lever arm, and vice versa. If this simplification were removed, this would make things way harder.
Maybe it should be noted in the intro that somehow all men stand at equal distance from fulcrum - thus eliminating the effect of differing lever arms (force x distance)
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I have not read all comments, so this may be covered already. The puzzle clearly makes the simplification of not taking account of the distance of individual men from the fulcrum. Further away from fulcrum greater lever arm, and vice versa. If this simplification were removed, this would make things way harder.
Maybe it should be noted in the intro that somehow all men stand at equal distance from fulcrum - thus eliminating the effect of differing lever arms (force x distance)
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Kurt
This is stoopid I wasted my time, to analyze this on an actual manner with 99. 99efficiency putt all man on the seesaw 6 man on each side and then alternately remove 1 from each side and put them back until the seesaw balances and now you have your suspect but there are two of them what should you do? Change a man from 1 side with the 1 one your suspect until the seesaw imbalances finally you have your suspect with using only 1 attempt of using the seesaw tadaaa its magic
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This is stoopid I wasted my time, to analyze this on an actual manner with 99. 99efficiency putt all man on the seesaw 6 man on each side and then alternately remove 1 from each side and put them back until the seesaw balances and now you have your suspect but there are two of them what should you do? Change a man from 1 side with the 1 one your suspect until the seesaw imbalances finally you have your suspect with using only 1 attempt of using the seesaw tadaaa its magic
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yackawaytube
And how does solving this puzzle demonstrate that you are a good software developer who has the ability to write good quality code quickly that can adapt to the rapid changing business environment. Answer: Not at all.
My company stops using puzzle solving in interviews because it doesn't help us to find good software developers at all. Maybe puzzles like this are good for picking new college grads because they are all equally bad in software development.
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And how does solving this puzzle demonstrate that you are a good software developer who has the ability to write good quality code quickly that can adapt to the rapid changing business environment. Answer: Not at all.
My company stops using puzzle solving in interviews because it doesn't help us to find good software developers at all. Maybe puzzles like this are good for picking new college grads because they are all equally bad in software development.
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Obakeng
My approach was like this
1st split the 12 men into 2 groups and weigh them
Obviously the seesaw won't be balanced
Remove one from each side and observe a change
If the seesaw is still unbalanced repeat
If the seesaw gets balanced that means one of those I just removed is the odd man
I then remove the rest from the seesaw and weigh the last 2 I just removed against each other
Therefore we will know if the odd man is heavier or lighter
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My approach was like this
1st split the 12 men into 2 groups and weigh them
Obviously the seesaw won't be balanced
Remove one from each side and observe a change
If the seesaw is still unbalanced repeat
If the seesaw gets balanced that means one of those I just removed is the odd man
I then remove the rest from the seesaw and weigh the last 2 I just removed against each other
Therefore we will know if the odd man is heavier or lighter
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KK
1st weight - divide 12 into 6men(a) and 6men (b) - find out which group is heavier.
2nd weight - divide heavier 6men group into 3men (x) and 3men (y) - find out which group is heavier
3rd weight - take two people from the 3men heavier group and take two persons and weigh them - If they show any difference in weight then that person is heavier. If they both are showing any weight difference the 3rd person who is excluded is the heavier person.
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1st weight - divide 12 into 6men(a) and 6men (b) - find out which group is heavier.
2nd weight - divide heavier 6men group into 3men (x) and 3men (y) - find out which group is heavier
3rd weight - take two people from the 3men heavier group and take two persons and weigh them - If they show any difference in weight then that person is heavier. If they both are showing any weight difference the 3rd person who is excluded is the heavier person.
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Simone
you can do it in 2 if lucky or 3 if not if you split the groups 5 2 5 and weight the 5v5 first.
5v5 stable. weight the 2 men standing and find your 1
5v5 unstable. 2 3 2 weight the 2v2
stable. weight the remaining 1v1 unstable its one of them, stable is the one you dont weight.
unstable, weight 1v1.
you have a better chance to do it in LESS than the 4v4v4 solution given that ALWAYS needs 3 steps.
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you can do it in 2 if lucky or 3 if not if you split the groups 5 2 5 and weight the 5v5 first.
5v5 stable. weight the 2 men standing and find your 1
5v5 unstable. 2 3 2 weight the 2v2
stable. weight the remaining 1v1 unstable its one of them, stable is the one you dont weight.
unstable, weight 1v1.
you have a better chance to do it in LESS than the 4v4v4 solution given that ALWAYS needs 3 steps.
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Josh
6 4 2.
Took about 2 minutes-
How is this not a basic answer?
Weigh 6 out of 12, 3 on each side. That discerns a group of 6 that has the offweight person. Or a group of 3 which is easily solved in time.
Weigh 4 of the 6 from that new group, 2 on each side. That discerns a group of 2 that has the heavier person.
Weigh 1 on each side from the group of 2.
Solved for all scenarios.
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6 4 2.
Took about 2 minutes-
How is this not a basic answer?
Weigh 6 out of 12, 3 on each side. That discerns a group of 6 that has the offweight person. Or a group of 3 which is easily solved in time.
Weigh 4 of the 6 from that new group, 2 on each side. That discerns a group of 2 that has the heavier person.
Weigh 1 on each side from the group of 2.
Solved for all scenarios.
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BASIC
Why not real world questions?
How would you standardize all municipal Annual Treasury Reports such that taxpayers could see where their taxes are spent in their city and throughout the state?
Here is some semicolon delimited data where spending data=vendor (space) amount; vendor (space) amount; vendor (space) amount; vendor (space) amount; vendor (space) amount; , etc.
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Why not real world questions?
How would you standardize all municipal Annual Treasury Reports such that taxpayers could see where their taxes are spent in their city and throughout the state?
Here is some semicolon delimited data where spending data=vendor (space) amount; vendor (space) amount; vendor (space) amount; vendor (space) amount; vendor (space) amount; , etc.
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Chris
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.
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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.
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Paiboon
I first found this puzzle at the time of Gulf war ( about 1990. Very fun to solve. Later, I found a math text showing how to calculate general solutions for puzzles with N men ( how many times ( minimum ) we need to weigh, but I cannot remember the book name. Anyone ever come across where the general solution can be found, please tell me. Thanks
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I first found this puzzle at the time of Gulf war ( about 1990. Very fun to solve. Later, I found a math text showing how to calculate general solutions for puzzles with N men ( how many times ( minimum ) we need to weigh, but I cannot remember the book name. Anyone ever come across where the general solution can be found, please tell me. Thanks
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Quantum
1. weigh 3 vs 3 from the 12
if unstable,
2a. weigh 1 vs 1 from the heavier group
if stable, it's the third guy, otherwise, it's the heavier of the two
2b. weigh the 3 vs 3 that weren't weighed previously
3. weigh 1 vs 1 from the heavier group
if stable, it's the third guy, otherwise, it's the heavier of the two
NEXT!
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1. weigh 3 vs 3 from the 12
if unstable,
2a. weigh 1 vs 1 from the heavier group
if stable, it's the third guy, otherwise, it's the heavier of the two
2b. weigh the 3 vs 3 that weren't weighed previously
3. weigh 1 vs 1 from the heavier group
if stable, it's the third guy, otherwise, it's the heavier of the two
NEXT!
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Ravi
My simple brain got simple answer.
Divide 12 people into 2groups 6 each and keep them on seasaw, obviously it wouldn't be balanced so on the next use pick the hevier side and divide into 2 groups with 3 each. Then after pick 2 from the hevier side, if one of them is hevier the he is the odd one else the remaining 1 is odd one of the group-
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My simple brain got simple answer.
Divide 12 people into 2groups 6 each and keep them on seasaw, obviously it wouldn't be balanced so on the next use pick the hevier side and divide into 2 groups with 3 each. Then after pick 2 from the hevier side, if one of them is hevier the he is the odd one else the remaining 1 is odd one of the group-
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John
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.
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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.
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Prasun
Through them all in the water. Check the water level, it will be the same. Then throw the interviewers in the water. The water level will change. Why? The interviewers ego is too big. Then beat him up with the see saw. That's my answer. They have to accept your answer now you have the plank of the see saw. Sorted and solved.
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Through them all in the water. Check the water level, it will be the same. Then throw the interviewers in the water. The water level will change. Why? The interviewers ego is too big. Then beat him up with the see saw. That's my answer. They have to accept your answer now you have the plank of the see saw. Sorted and solved.
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George
If there's no limit per side, put half on each side and discard the lighter side. Repeat this one more time.
Then have one of your remaining 3 stand aside and weigh the other 2 men. If they balance evenly, the standby is the heavy man. Else the scale will show you.
This isn't all that challenging.
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If there's no limit per side, put half on each side and discard the lighter side. Repeat this one more time.
Then have one of your remaining 3 stand aside and weigh the other 2 men. If they balance evenly, the standby is the heavy man. Else the scale will show you.
This isn't all that challenging.
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education
why is this so overcomplicated? just weigh 6 on each side to find out which 6 is lighter/heavier, then with second try place 3 on each side and you are left with 3, try 2 of them on each side - if they balance that means the third one is odd, if they dont balance then you see which one it is
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why is this so overcomplicated? just weigh 6 on each side to find out which 6 is lighter/heavier, then with second try place 3 on each side and you are left with 3, try 2 of them on each side - if they balance that means the third one is odd, if they dont balance then you see which one it is
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OriginalKKB
Just because they give you hatd logic questions in the interview doesn't mean you'll ever need the ability to solve these problems in the job you are interviewing for. From personal experiene it is infact the opposite. The more insane the interview the more boring and easy the job.
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Just because they give you hatd logic questions in the interview doesn't mean you'll ever need the ability to solve these problems in the job you are interviewing for. From personal experiene it is infact the opposite. The more insane the interview the more boring and easy the job.
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Romain
Just looked at the case-1. To me, you don't even have to use normal men (unless you want to have fun xD. As there is only 4 man left and 2 tries on the seesaw, you can simply put 2 men from the 4. if the seesaw is balanced, put the last 2 men on it and you will have your imposter; )
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Just looked at the case-1. To me, you don't even have to use normal men (unless you want to have fun xD. As there is only 4 man left and 2 tries on the seesaw, you can simply put 2 men from the 4. if the seesaw is balanced, put the last 2 men on it and you will have your imposter; )
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Jack
You over complicated this. Start with 6&6 heavier side moves on then 3&3 heavier side moves on then 1&1. If it-s balance heavier person is on the ground if it-s not balanced then heavier person is on the down side. Forget coloring and adding neutral people back. Smarter not harder
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You over complicated this. Start with 6&6 heavier side moves on then 3&3 heavier side moves on then 1&1. If it-s balance heavier person is on the ground if it-s not balanced then heavier person is on the down side. Forget coloring and adding neutral people back. Smarter not harder
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