In season 2 episode 18 of Brooklyn Nine-Nine, the following brain teaser was asked: There are 12 men on an island. The riddle, as you'll recall, is this: There are twelve men on an island. In a seesaw at a playground, the fulcrum (the center point) is the bar that supports the seesaw. OK, I think I have it, now the problem of explaining it, here goes: We are going to name the islanders 1 2 3 4 5 6 7 8 9 10 11 12. seesaw: R1 L2 L3 L4 L1 S2 S3 S4 (balanced) You signed in with another tab or window. The second measurement was unequal, but the balance of weight shifted. The riddle 12 men live on an island. (First Use). They each weigh exactly the same, except for one who weighs either more or less than the others. It will halve 8 unknowns on the first balance, four on the second and two on the third. SPOILER ALERT! If he is lighter then it is the one being measured and he is lighter. if balanced it is L4 - else it is the heavier of L2 and L3. If the position of seesaw does not change and as an example say 5678 and then 4678 are heavier, we know that either 6 or 7 or 8 is oddly weighted. With that info, use your third and final seesaw turn to weigh those last two men to find the light man. test all the cases: go test . sideline: L1 L2 L3 S4 L4 R4 This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. :D, Brooklyn 99 riddle: Weighing Islanders [duplicate]. Thus in principle we can get log227 bits from the three comparisons. You can figure out both in three steps. How do you do it? Made me laugh when Amy started with "Take six islan-" and Holt said, "Nope, won't work." You have a seesaw to determine who is the odd one out but you That's not true. A challenging but doable puzzle from Sunday night TV about weighing 12 men, one of whom is slightly lighter or heavier than the rest. seesaw: S1 S2 S3 (heavy) R1 R2 R3 You must figure out which. Weigh A against B. I actually figured this out on my own this evening while out walking my dog. So, what is the 12 people on an island riddle and what is the answer? we know it is either R1 or L1. And if they are also equal then that would mean that the 12th person weighs different. Beach House Semis in Port Union Village. Would the reflected sun's radiation melt ice in LEO? Questions: There are 2 possible outcomes, either they are equal, or unequal. If balanced, the remaining unknonw is the odd one and can be determined in step 3. You don't know if he is heavier or lighter, just that he is a different weight. now compare R1 against a neutral islander (S1 for example). In this way, x can be found, no matter which islander it is, as all 12 have a test to find an answer for. seesaw: R1 L2 L3 L4 L1 S2 S3 S4 (heavy) Use the third and final seesaw use to determine if the 12th person is the heavier or lighter islander (i.e. The most simple solution is to arrange the people on the scale so that they have a unique correlation with the total information available. {{#media.media_details}} {{#media.focal_point}}, First weigh 1, 2, 3, 4 vs 5, 6, 7, 8 on the seesaw. The island has no scales but there is a see-saw. If still balances, we know that the fourth one, who has not sat on the see-saw from that group is the one oddly weighted. Death by roo roo. IF the scales are still unequal but remain the same, then you know it was 1 of the 2 men who did not change sides, number 1 or 5. The seesaw will balance perfectly. A standard see-saw must be used and only three times. If they are the same, 8 is the one we are looking for (and lighter). (Detailed . You must use a see-saw to figure out whose weight is different, and you may only use the see-saw 3 times. All of the islanders individually weigh exactly the same amount, except for one, who either weighs more or less than the other 11. There are no scales or other weighing device on the island. (Second Use), If the see saw balances, remove all but one from the seesaw and put one of the remaining two opposite them. Are you sure you want to create this branch? (Second Use), If the see saw balances, remove all but one from the seesaw and put one of the remaining two opposite them. Ackermann Function without Recursion or Stack. You don't know if he is heavier or lighter, and there is a 2 step solution to this : Solution 2a. Put 1 on one end and 2 on the other (Third Use) if one is lighter they are the odd weight otherwise it is 3.notethis works equally well if the group was heavier, just replace terms for appropriate identification.'. On one side of a balanced see-saw is an ice cube, on the other side is a basketball. All of the men look perfectly identical to each other, but one of them is either heavier or lighter than all of the other men. Proceed to step 2. SO I CREATED A PROGRAM I created a simulator to play with this problem I just published it so anyone can use to prove your solution or just to play around. Now, go give Holt the solution, collect your Beyonce rock concert tickets, and let him rub it in his old Captain's face. There are 12 men on an island. I like this answer other than the fact that if "Condition 2.1.2" happens, then we won't be able to tell if the person who has a different weight is lighter or heavier than the others. Jordan's line about intimate parties in The Great Gatsby? Otherwise the one who stayed is the oddly weighted one. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. Weigh 5 against 6. *** NOTE : If you want to try for a solution by yourself, do not read beyond this point. Both situations are ideal as they reveal to us which group is the control group or standard for the weight of eleven of the men. Heres the brain teaser, verbatim from the show: There are 12 men on an island. Learn more. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. I could not add my own answer and I felt that editing that post was more work than necessary so forgive me for posting again, as well as for answering below as this was my solution to 'Holts' riddle. (seesaw measurement number 3), 2.B.2 if balanced a) If 1,2,5,6,7 is heavier, then either 1 or 2 heavier, or 8 is lighter. Applications of super-mathematics to non-super mathematics. How can you find out which islander is the one that has a different weight? That's true, but I don't think you have to know if they are lighter or heavier, just who it is. This way of wording will help make it easier to figure out the puzzle. Problem solved. The pattern for all islanders is below: person: A B C . Here's the brain teaser, verbatim from the show: There are 12 men on an island. Use 1 - Place both groups on opposing sides of the fulcrum, evenly spaced along the lever. Another coin weighing puzzle. It is a good write up though. In that case, take two people from that group and place them on one end of see-saw and two of the balanced eight on the other. Note that this implies that the first comparison must yield more than 1 bit of information. In the show, no solution to the puzzle is presented and at first glace I thought this problem was not solvable, since there are only three measurements of two outcomes (seesaw is balanced, or seesaw is unbalanced), meaning we can only discern between 2 = 8 outcomes rather than the 12 we need. You don't know heavier or lighter, but you still have 1 remaining move. Comparing (1,2,3,4) to (5,6,7,8) does exactly this. Drift correction for sensor readings using a high-pass filter. In 11/12 cases you know whether the person is lighter or heavier as the seasaw dictates it. Take note of whether it is heavier or lighter. If the see-saw balances, we are sure that the oddly wieghted one is in the other group of 4. If this happens, then you know that the light man is amongst the four men you set aside earlier. Note : I am not looking for a new way to solve this. 2) They are different. b) If 1,2,5,6,7 is lighter, then one of 5,6,7 is different and lighter. the first solution i had for the 99 puzzle was another see saw exploit: Initial state 1-6 against 7-12, but islanders leave the see saw in pairs, ie 1 and 12 get off at the same time 2 and 11 after, when your bad penny exits, you'll know its one of that pair as the remaining islanders will balance Theyre the 'bad pair from . {{3 remaining with known weight solution}} Since the weight is known, all you need to do is measure 2 men from the remaining 3 against each other. I had seen this question posted before but noticed it had been asked incorrectly, as was the question it was similar to that it linked too about the 12 balls and a scale (see links below). Now put 7 on one end and 8 on the other. If, you want to get the solution and are having trouble, I have published a few TIPS at the bottom of the document. Now put 7 on one end and 8 on the other. We know {9,10,11,12} all weigh the same. All our articles and reviews are written independently by the Netmums editorial team. Show Answer PREV NEXT by Mikhael Panju v1. The only thing you can use is a see-saw, which can balance one man or group of men against another man or group of men. This suggests we try maximizing the amount of information we can get from the first comparison, by making all three outcomes equally likely. Thus in principle we can get log 2 27 bits from the three comparisons. Solution 2c. If balanced it is the heavy L1 else if unbalanced it is light R1. sideline: S1 R2 R3 R4, we know that the light islander is amongst R2 R3 R4 A classic logic puzzle has resurfaced online, one which stumped the characters of Brooklyn Nine-Nine back in season 2. (We could always relabel the men so that this is true). 11 of them weigh the same and one of them is either slightly heavier or slightly lighter. So in principle, we should be able to solve the problem. The essence of the riddle is: "There are 12 men on an island. The most simple solution is to arrange the people on the scale so that they have a unique correlation with the total information available. You must figure out which. (seesaw measurement number 3), 2.B.3 if right heavy All of the men look perfectly identical to each other, but one of them is either heavier or lighter than all of the other men. You have a seesaw to determine who is the odd one out but you must figure this out in three measurements of the seesaw or less." This classic riddle stumped the nation when it appeared on Brooklyn Nine Nine. If one is heavier they are the odd one otherwise it is 6. 11 weigh exactly the same amount, but one of them is slightly lighter or heavier. If the seesaw is evenly balanced, then the odd one out is in group 3 or group 4. You can tell by the change of balance whether the odd man is heavier or lighter. 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