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william wu
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Measuring Weights
« on: May 27th, 2003, 6:33pm » |
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A merchant's 40 pound measuring weight falls from a table and breaks into 4 pieces. When the pieces are weighed, it was found that each had an integral weight, and the 4 pieces could be used to weight every integral weight between 1 and 40 pounds. What were the weights?
Author: Claude Gaspard Bachet de Meziriac (1581-1638)
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| « Last Edit: May 30th, 2003, 1:48am by william wu » |
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Leo Broukhis
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Re: Measuring Weights
« Reply #1 on: May 27th, 2003, 7:46pm » |
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Duh: 1, 3, 9, and 27 Why is it medium?
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BNC
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Re: Measuring Weights
« Reply #2 on: May 28th, 2003, 3:16am » |
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And why twice?
PS: It's on the 3rd post here
And another note: look at the "inverse" problem: use 4 integer weights to measure any integer weight up to N. What is N, what are the weights (1,3,9,27 => N=40 is sub-optimal).
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| « Last Edit: Oct 18th, 2003, 3:49pm by BNC » |
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Leo Broukhis
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Re: Measuring Weights
« Reply #3 on: May 28th, 2003, 8:06am » |
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What's optimality in this case? They do, or they don't.
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wowbagger
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Re: Measuring Weights
« Reply #4 on: May 28th, 2003, 8:44am » |
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I guess BNC meant to ask for the maximal N possible with 4 weights.
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towr
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Re: Measuring Weights
« Reply #5 on: May 28th, 2003, 9:28am » |
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well, you can have the weight as a negative, positive or not, so 3^4 -1 = 80 is the theoretical maximum.
This limits the space enough to find the an answer brute-force.
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| « Last Edit: May 28th, 2003, 9:31am by towr » |
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towr
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Re: Measuring Weights
« Reply #6 on: May 28th, 2003, 10:09am » |
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hmm.. well, without a novel way of weighing I'm hardpressed to find any N over 40..
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| « Last Edit: May 28th, 2003, 10:23am by towr » |
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tohuvabohu
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There is one way to figure out more than 40 integral weights:
infer what you can't match
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Leo Broukhis
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Re: Measuring Weights
« Reply #9 on: May 28th, 2003, 12:54pm » |
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on May 28th, 2003, 9:28am, towr wrote:well, you can have the weight as a negative, positive or not, so 3^4 -1 = 80 is the theoretical maximum.
This limits the space enough to find the an answer brute-force. |
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You need to divide by two, because it does not matter on which side of the scale you put the object being weighted.
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towr
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Re: Measuring Weights
« Reply #10 on: May 28th, 2003, 1:35pm » |
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heh, yeah.. that's true of course..
So that's an elegant proof that 40 is the best, this way..
You can weigh integral weights from -40 to 40 (including lighter than air objects like helium-filled balloons). And you can determine between which integral weights any nonintegral weight on that range is.
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| « Last Edit: May 28th, 2003, 1:36pm by towr » |
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BNC
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Re: Measuring Weights
« Reply #11 on: May 28th, 2003, 4:36pm » |
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on May 28th, 2003, 12:47pm, towr wrote:| but then how do you know it's an integral weight |
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I did "say" any integer weight, but I gues I should have stressed that you have an a-priori knowledge of the "integerability" of the weights.
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