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So I had an idea for a number-based door puzzle for a TTRPG campaign that could readjust itself every time a wrong guess is made. Here's the basic premise:

Given two numbers, find two more numbers in such a way that:

  • no number differs from any other number in the same way,
  • and none of these differences are part of the series themselves,
  • and no number is bigger than it has to be, but not zero.

It seems fully deterministic to me, so let me give you an example:

Given are two numbers (1, 4). The difference is 3.

2 is out because its difference from 1 is part of the series, and in itself would be the difference from 4. 3 is out because of its difference from 2 and 4 being the same, and that difference would be 1, which is additionally part of the series. 5 is not valid either, but 6 is. 7 is not valid due to its difference to 6 being part of the series, and 8 is invalid because of the difference between (6, 8) and (4, 6). 9-6=4-1, and 10-6=4. 11-6=5, 11-1=10, 11-4=7, but 11-6=6-1. 12-6=6, but 13 works: the correct sequence is (1, 4, 6, 13).

It seems simple enough for a door puzzle, and given the above set of rules, it seems absolutely deterministic. I was considering to give the following (more vague) clue to its solution:

No siblings differ alike, nor match they their differences, and none grow taller than they must.

I need it to sound like a mystical/fantasy riddle, but still be concise and accurate and its description of the puzzle. If anyone has any suggestions, I would greatly appreciate the help.

I suppose I have four questions for this community:

  • Is there a unique solution for any two starting numbers below 10?
  • Can you think of a more interesting variant, like perhaps by minimising the sum of the numbers (but not having any predetermined numbers given - (1, 4, 6, 13) being the best solution I found so far)?
  • It seems to me there should be some way to determine the two missing numbers in a way that minimizes the sum of all four, which is not the same as choosing the lowest possible numbers one after the other. Is there such a way?
  • Is there a better way to phrase the problem in a single sentence to my players, without being too obvious or sounding too scientific?

And some more example solutions for your convenience:

  • (3, 4) → (9, 11)
  • (2, 5) → (6, 13)
  • (1, 9) → (3, 13)
  • (1, 3) → (7, 12)
  • (1, 5) → (7, 15)
  • (4, 6) → (1, 13)
  • (8, 1) → (3, 12)
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    $\begingroup$ Your second and fourth questions are pretty opinion-based ("more interesting" could have any number of answers/interpretations, a "better way" to phrase is entirely opinion-based). The other two look fine, though. Welcome to the site! $\endgroup$
    – bobble
    Mar 6 at 19:21
  • $\begingroup$ Fair enough, I figured maybe someone has a good idea for how to describe it. But yes, they are heavily based on personal preference and opinion. However, I think the need to be concise and clear is important (especially since I will be posing the puzzle to non-mathematicians), so I'd happily welcome any feedback on whether the current phrasing is too ambiguous :) $\endgroup$
    – Chivalry
    Mar 6 at 19:24
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    $\begingroup$ Why is $(1, 4, 6, 9)$ valid? One has $9 - 6 = 4 - 1$. Does this break the first rule? Also for $(1, 5, 7, 13)$ we have $13 - 7 = 7 - 1$. $\endgroup$
    – WhatsUp
    Mar 6 at 19:52
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    $\begingroup$ And for $(8, 1)$ I get $(8, 1, 3, 12)$ which seems to be smaller. Am I missing something? $\endgroup$
    – WhatsUp
    Mar 6 at 19:58
  • $\begingroup$ @WhatsUp you are of course correct, sorry I made these mistakes. Corrected the sequences now. $\endgroup$
    – Chivalry
    Mar 6 at 20:01
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Here are some experimental results.


The "heuristic algorithm", namely choosing smallest numbers one after the other, does not always produce the smallest possible sum.

Examples:

$[1, 9]$: heuristic: $[1, 9, 3, 13]$; optimal: $[1, 9, 4, 11]$;

$[3, 5]$: heuristic: $[3, 5, 9, 16]$; optimal: $[3, 5, 11, 12]$;

and many more.


The example of $[3, 5, 11, 12]$ probably has the largest sum of two added numbers, being $11 + 12 = 23$.

The second largest is $[1, 5, 7, 15]$, being $7 + 15 = 22$.

I checked these for all starting values $\leq 50$. One can probably prove that above $50$ there must exist small numbers that fulfill the task (in many cases, the smallest pair $[1, 3]$ works).

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  • $\begingroup$ That is good to know - but besides approaching it via coding, there doesn't seem to be a concise rule for how to choose the optimal sequences based on the starting numbers, is there? I appreciate you testing it, and I'm ready to accept your answer for the time being :) $\endgroup$
    – Chivalry
    Mar 6 at 21:30

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