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Two porters have to carry six pieces of luggage of unknown weight. Each piece weighs a different amount, and they are labeled in order of weight from A to F, with A being the lightest and F the heaviest. Each piece weighs up to 10 lbs., and the total weight is 40 lbs. or less.

Each porter can carry up to 20 lbs. at once, and between them they want to carry it all in one trip. How should they divide up the luggage between them?

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  • $\begingroup$ Have you used any hint coins? $\endgroup$ – Adam Jul 2 at 11:41
  • $\begingroup$ @Adam I have used and the last hint basically contains the solution. However, I want to understand ii / see how I can deduce it without using any hints. $\endgroup$ – Layton Player Jul 2 at 11:43
  • $\begingroup$ integer weights? $\endgroup$ – Oray Jul 2 at 11:45
  • $\begingroup$ @Oray Professor Layton puzzles always use whole numbers $\endgroup$ – Adam Jul 2 at 11:47
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    $\begingroup$ Also, if non-integer weights were allowed, the puzzle would have no solution: there's no way to divide up 1.0, 7.6, 7.7, 7.8, 7.9, 8.0 into two sums not exceeding 20. $\endgroup$ – aschepler Jul 2 at 23:33
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That particular puzzle is meant to be solved by trial and error. The Super Hint for that puzzle says:

enter image description here

"Thankfully, it doesn't matter if you get it wrong and make the porters buckle under the strain."

If you input an incorrect configuration, it shows which of the two porters is carrying too much and lets you try again. The solution is not deducible from only the text given, because it's not meant to be; you're supposed to figure it out by trying various configurations.

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  • $\begingroup$ Btw, should The Super Hint be spoilered, as it is a spoiler for the game itself? >< $\endgroup$ – athin Jul 2 at 13:25
  • $\begingroup$ It looks like Deusobi used the spoiler tag (you can see the pinkish background). But I think pressing "Enter" ruins the spoiler (while <br/> does not), which is a weird bug. $\endgroup$ – shoopi Jul 3 at 11:31
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    $\begingroup$ @shoopi I just used a quote block, not a spoiler tag. I don't think it's necessary to spoiler it, because the question is explicitly about the puzzle in the game. $\endgroup$ – Deusovi Jul 3 at 13:25
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The way it is written currently:

There isn't a clear, universal answer:

$\begin{array}{cccccc|cc}a&b&c&d&e&f&P1&P2\\4&5&6&7&8&10&ACF&BDE\\3&4&6&8&9&10&ADE&BCF\\3&5&6&7&9&10&ADF&BCE\\2&5&6&8&9&10&ADF&BCE\\2&4&7&8&9&10&ADF&BCE\\2&5&6&8&9&10&ADF&BCE\\1&5&7&8&9&10&AEF&BCD\end{array}$


Just go for ADF / BCE and pray, I guess.
Given that OP said intended solution is ACF BDE we may be missing information.
However, going for that solution would (according to my possibilities) give a 6/7 chance one of them is going to break his arm!


If the question means up to but not including 10 lbs then:

Highest weight solution would be 4 5 6 7 8 9 (A-F); this totals 39 lbs.
This would leave porter 1 with 19 lbs to carry (ACF) and porter 2 with 20 lbs (BDE).
Given OP comment I would assume therefore this is what the puzzle meant.

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  • $\begingroup$ Thanks a lot Rook $\endgroup$ – BMS21 Jul 4 at 8:24
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I think this logic gets you there:

If a solution works for a case where all values are maxed out, it will work for any lower value cases. Since each box has a different weight from 1 to 10, and the total sum can not exceed 40, we know 10+9+8+7+6+5= 45 is out. However, 10+8+7+6+5+4=40 is allowed. Since they are labeled in ascending order of weight, this gives A=4, B=5, etc. Since each porter is able to carry 20 lbs, we want to use the largest possible values to make 20 from this set, which is 10+6+4 or A+C+F. Since these are the largest integers from the set of largest allowed values, this is the upper limit and this solution will always work for the constraints given. If one porter carries A+C+F, the other must necessarily carry B+D+E for all the cases to make it up in one go.

Edit: As @BMS21 points out, this doesn't quite work when the weight is more distributed to the middle or bottom as in @Oray's examples.

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    $\begingroup$ Problem is that's not the only way to distribute the mass negative. I can list 7 variants of the distribution multiple of which have more than one solution when applied $\endgroup$ – BMS21 Jul 2 at 12:18
  • $\begingroup$ Oh, no I see your point, if you weight the bottom of the distribution disproportionately it doesn't work. You're right $\endgroup$ – JProblems Jul 2 at 12:22
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The most logical way to do this would be, ABCDEF being the weights (for simplification I will use 2 decimals but this is valid for any number of decimals):

1st person gets the lightest and the heaviest (A,F)
Second person gets the next lightest and the next heaviest (B,E)
Then each picks up one of the remaining ones (C,D)

Since the max per weight is 10 and the total is 40 the maximum value of extremes E and F you can have is 19.99 (10 + 9.99). That means there is an amount of maximum 20.01 to be split into 4 more weights. Considering the maximum possible scenario where A B are minimum values, subtracting them from 20.01 still amounts less than 10 per unit. Let’s say A and B are 0.01 and 0.02, summing up to 0.03. Now in the case of person 1, we have 10.01 and in the case of person 2 we have 10.01 also. That will leave 19.98 to be distributed. Since a weight cannot have 9.99 because we used that up already (for E), it means it must have at most 9.98.  9.98 + 10.1 (A+F or B+E) will not possibly exceed 20.

Note: with integers only, the solution would be easier to understand.

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  • $\begingroup$ What about the counter example that aschepler noted in his comment on the question: {1.0, 7.6, 7.7, 7.8, 7.9, 8.0 } cannot be split into two sets that sum to 20. Basically, the question does not work at all with non-integer weights. I think that all you've shown is that the porter with the heaviest box can choose two more boxes to carry, but that does not mean that the other porter can lift the remaining boxes. $\endgroup$ – Jaap Scherphuis Jul 3 at 13:11
  • $\begingroup$ For such example there is no solution so we either must assume that all weights my comply so that there is a solution. The whole point is to 1st get the lightest and the heaviest. For integers, clearly my selection solution cannot fail no matter the values. $\endgroup$ – Overmind Jul 4 at 5:09
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The deduction method I used when encountering the puzzle to work out at least part of the answer sans trial and error:

Counter-weights. We know A to F are in ascending weight order (A<B<C<D<E<F). So we know at a minimum one porter must carry the lightest A to offset the heaviest of F, and another must carry lighter B to offset heavier E. So one porter carries A+F and another B+E. The unanswerable part is which porter carries C and which carries D, which is boiled down to a 50/50 guess. ACF and BDE would be a logical first attempt, the other being ADF and BCE.

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  • $\begingroup$ What you “know” isn’t necessarily true.  Yesterday Oray gave the example 1, 5, 7, 8, 9, 10, for which the only solution is ‘‘B+C+D’’ and ‘‘A+E+F’’.  Beyond that, you’re basically just rehashing what others have said. $\endgroup$ – Peregrine Rook Jul 3 at 18:03

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