Word Puzzle ( yet another thread)
Wednesday, January 14, 2004 by Rocky Nimda | Discussion: WinCustomize Talk
The Problem: You have a set of Balance Scales like the one shown here:
You have 9 identical looking marbles. You can't tell the difference between any of the 9 marbles by looking at them.
One of the 9 marbles is ever so slightly denser than the others (thus it weighs more). The only way to find out which marble is the one that is denser is to use the balance scales.
One way to do this would be to put 1 marble on one side and 1 marble on the other. If one was heavier than the other, you would then know which of the nine was the heaviest. If they were the same, you could then take one of the marbles off the scales, and put another on. As you can see, this could take quite a few measurements. Worst case: too many to count.
Now, your task is to figure out how few measurements you can make to find the 1 dense marble.
Keep in mind; you can put more than 1 marble on each side of the scales at a time. Also, keep in mind the "worst case" for your solution. The solution I mentioned above could be solved in 1 measurement, but that's NOT worse case (it could take many more).
When you post your solution, don't just say I can find the dense marble with 6 measurements. You need to describe your solution. I have to be able to determine if the number of measurements you suggest is truly worst case or not.
By the way "6" is just used as an example here; it may or may not be the correct answer.
After this puzzle is solved, someone else could add another word puzzle to the thread.
You have 9 identical looking marbles. You can't tell the difference between any of the 9 marbles by looking at them.
One of the 9 marbles is ever so slightly denser than the others (thus it weighs more). The only way to find out which marble is the one that is denser is to use the balance scales.
One way to do this would be to put 1 marble on one side and 1 marble on the other. If one was heavier than the other, you would then know which of the nine was the heaviest. If they were the same, you could then take one of the marbles off the scales, and put another on. As you can see, this could take quite a few measurements. Worst case: too many to count.
Now, your task is to figure out how few measurements you can make to find the 1 dense marble.
Keep in mind; you can put more than 1 marble on each side of the scales at a time. Also, keep in mind the "worst case" for your solution. The solution I mentioned above could be solved in 1 measurement, but that's NOT worse case (it could take many more).
When you post your solution, don't just say I can find the dense marble with 6 measurements. You need to describe your solution. I have to be able to determine if the number of measurements you suggest is truly worst case or not.
By the way "6" is just used as an example here; it may or may not be the correct answer.
After this puzzle is solved, someone else could add another word puzzle to the thread.
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Rocky Nimda
Reply #2 Wednesday, January 14, 2004 6:18 PM
Reply #2 Wednesday, January 14, 2004 6:18 PM
Close, very close, but I can beat it. Can anyone else?
Corio
Reply #3 Wednesday, January 14, 2004 6:26 PM
Reply #3 Wednesday, January 14, 2004 6:26 PM
Hmmm....Thought I had that one.
Ok! 2 measurments. Divide the marbles into 3 X 3. Weigh 3 on each scale. If they're equal, they're in the left out group. If they're not, take out the light ones, put aside on of the 3 "heavies" and compare the other two!
Did anybody understand that explanation?
Ok! 2 measurments. Divide the marbles into 3 X 3. Weigh 3 on each scale. If they're equal, they're in the left out group. If they're not, take out the light ones, put aside on of the 3 "heavies" and compare the other two!
Did anybody understand that explanation?
Chris TH
Reply #4 Wednesday, January 14, 2004 6:37 PM
Reply #4 Wednesday, January 14, 2004 6:37 PM
I think I get it...
divide into 3 groups of 3 as Ordos said, test 2 of the groups
Determine which group the heavy marble is in: either in one of the two groups on the balance, or in the remaining untested group (if the balance is equal).
Take that group and weigh any 2 marbles, the heavy one is easily seen if it's on the balance, or it's the remaining one, again if the balance is equal.
Therefore only 2 tests required.
What do I and Ordos of course win?
[Message Edited]
divide into 3 groups of 3 as Ordos said, test 2 of the groups
Determine which group the heavy marble is in: either in one of the two groups on the balance, or in the remaining untested group (if the balance is equal).
Take that group and weigh any 2 marbles, the heavy one is easily seen if it's on the balance, or it's the remaining one, again if the balance is equal.
Therefore only 2 tests required.
What do I and Ordos of course win?
[Message Edited]
Ingui
Reply #6 Wednesday, January 14, 2004 7:24 PM
Reply #6 Wednesday, January 14, 2004 7:24 PM
Being Lazy and Dishonest, I take no measurements. I select one marble at random and conveniently dispose of the other eight. I cleverly fabricate some obscure, mathematical equation demonstrating how the denser marble was clearly visible by the smaller shadow it cast on the table.
TESTS REQUIRED : 0
CORONAS CONSUMED : 3
TESTS REQUIRED : 0
CORONAS CONSUMED : 3
kongit
Reply #7 Wednesday, January 14, 2004 7:32 PM
Reply #7 Wednesday, January 14, 2004 7:32 PM
travis being the bright entrepreneur takes the marbles and sells them as magic stones.
Rocky Nimda
Reply #9 Wednesday, January 14, 2004 8:11 PM
Ordos and Chris TH
Yes, two measurments is the correct answer. You win the satisfaction of figuring it out.
Now does someone else have another puzzle?
Sorry Ingui, while your answer is less than theirs, I can't consider it a winner. Quite ingenious though.
[Message Edited]
Reply #9 Wednesday, January 14, 2004 8:11 PM
Ordos and Chris TH
Yes, two measurments is the correct answer. You win the satisfaction of figuring it out.
Now does someone else have another puzzle?
Sorry Ingui, while your answer is less than theirs, I can't consider it a winner. Quite ingenious though.
[Message Edited]
kongit
Reply #10 Wednesday, January 14, 2004 8:18 PM
Reply #10 Wednesday, January 14, 2004 8:18 PM
The same riddle as above with 12 total balls, 11 the same weight and 1 not (it is lighter or heavier). But you can use the scale only three times. How can you find out which ball is the different one and if it is lighter or heavier.
Ingui
Reply #11 Wednesday, January 14, 2004 8:19 PM
Reply #11 Wednesday, January 14, 2004 8:19 PM
Divide $177 (in whole $ increments) into a number of bags so that I can ask for any amount between $1 and $177, and you can give me the proper amount by giving me a certain number of these bags without opening them. What is the minimum number of bags you will require?
Rocky Nimda
Reply #12 Wednesday, January 14, 2004 8:22 PM
Reply #12 Wednesday, January 14, 2004 8:22 PM
Travis,
Three groups of 4. One group on each side of the scales and one group on the table. You have now narrowed it down to one of four marbles. Put two of the four marbles on one side and two on the other. Now you have narrowed it down to two marbles. Put one on each side for the third measurement and you have your answer.
[Message Edited]
Three groups of 4. One group on each side of the scales and one group on the table. You have now narrowed it down to one of four marbles. Put two of the four marbles on one side and two on the other. Now you have narrowed it down to two marbles. Put one on each side for the third measurement and you have your answer.
[Message Edited]
kongit
Reply #13 Wednesday, January 14, 2004 8:26 PM
Reply #13 Wednesday, January 14, 2004 8:26 PM
yes but you don't know if the marble is heavier or lighter so how would you know which group of four was wrong.
Rocky Nimda
Reply #14 Wednesday, January 14, 2004 8:27 PM
Reply #14 Wednesday, January 14, 2004 8:27 PM
OK Ingui, now that one seems quite the challenge. I will have to give some thought to that one. 
kongit
Reply #15 Wednesday, January 14, 2004 8:28 PM
Reply #15 Wednesday, January 14, 2004 8:28 PM
if you got lucky and they were equal on the scale then you would know which group of four (but this involves luck so it isn't applicable). Also once you got it down to two marbles the same effect would apply
Rocky Nimda
Reply #16 Wednesday, January 14, 2004 8:29 PM
Well, either the first two groups of four are equal or their not. If they are, it's the third group.
Reply #16 Wednesday, January 14, 2004 8:29 PM
| yes but you don't know if the marble is heavier or lighter so how would you know which group of four was wrong. |
Well, either the first two groups of four are equal or their not. If they are, it's the third group.
Rocky Nimda
Reply #17 Wednesday, January 14, 2004 8:31 PM
OK, I get your point.
Reply #17 Wednesday, January 14, 2004 8:31 PM
| if you got lucky and they were equal on the scale then you would know which group of four (but this involves luck so it isn't applicable). |
OK, I get your point.
Koasati
Reply #18 Wednesday, January 14, 2004 8:32 PM
Reply #18 Wednesday, January 14, 2004 8:32 PM
| #11 by Ingui - 1/14/2004 8:19:44 PM Divide $177 (in whole $ increments) into a number of bags so that I can ask for any amount between $1 and $177, and you can give me the proper amount by giving me a certain number of these bags without opening them. What is the minimum number of bags you will require? |
177 bags.
Rocky Nimda
Reply #20 Wednesday, January 14, 2004 8:46 PM
Travis, your good. Did you already know that one? You sure came up with an answer pretty fast. I guess we will wait for Ingui to see if it is the right one or not.
Reply #20 Wednesday, January 14, 2004 8:46 PM
| 8 bags: 88,44,22,11,5,4,2,1 |
Travis, your good. Did you already know that one? You sure came up with an answer pretty fast. I guess we will wait for Ingui to see if it is the right one or not.
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Reply #1 Wednesday, January 14, 2004 6:15 PM