6
$\begingroup$

I have got a puzzle as a Christmas present from a friend of mine and I have managed to solve it while skipping this part, but I would still be interested in how this represents the information that I have figured out by myself as well.

The piece of paper I have.

The only instruction for it in the accompanying letter was this: "You will require some algebra."

What I've also got was a crossword and clues for the missing words. After getting all the words I arranged them into the crossword through trial and error, and after solving their order I came back to this piece of paper.

What I know right now is that the correct order was 3, 23, 10, 13, 20, 24, 16, 8, 12, 9, 14, 11, 21, 15, 2, 5, 22, 6, 19, 4, 7, 18, 17 which, as I have noticed, corresponds to the sequence that can be read by removing the 15 from every pair and then going left-to-right, top-to-bottom in a row major consecutive manner starting from the (3, 15) entry, inserting 15 at the end and then wrapping around.

I have also noticed that the pairs are ordered such that the lower number is always on the left and as I have already mentioned, the number 15 is present in every pair. (Also, there are only 22 pairs for 23 words.)

The only thing I have found to use this parenthesis pair notation is for denoting the greatest common divisor of two numbers, but that doesn't really seem to be of much help here.

$\endgroup$
3
  • 1
    $\begingroup$ Can you describe a bit more how the you define the order of the crossword clues? Also, did you deduce the correct order of the numbers using the crossword? $\endgroup$
    – hexomino
    May 10, 2019 at 12:12
  • $\begingroup$ If you solved the puzzle you were given without reference to this part, can it be reverse-engineered from the solution? Or was it a red herring? $\endgroup$ May 10, 2019 at 18:54
  • $\begingroup$ The reverse-engineering process led me to noticing the things I've mentioned in the end of the post, but those still don't clarify to me the mathematical operator I should've used. $\endgroup$
    – Isti115
    May 10, 2019 at 20:58

1 Answer 1

3
$\begingroup$

I've posted this same thing on reddit's /r/puzzles as well and a helpful commenter (who goes by the username /u/Ibot02) managed to crack the meaning behind the numbers there. The answer is as follows:

Discussion: This looks to me like a permutation, written as the composition of transpositions.

In this way, (2 15) represents the permutation that swaps 2 and 15 (and does not change anything else). (2 15) (5 15) is then the composition of two such transpositions, so you'd swap 2 and 15, then 5 and 15, resulting in the permutation that maps 2 to 5, 5 to 15 and 15 to 2, also written (2 5 15). (Composition is sometimes written the other way around, but this is consistent with the result you found).

In the end, putting everything together, you'd get (2 5 22 6 19 4 7 18 17 3 23 10 13 20 24 16 8 12 9 14 11 21 15), or (3 23 10 13 20 24 16 8 12 9 14 11 21 15 2 5 22 6 19 4 7 18 17) (these are the same permutation since the notation is cyclical - you'd map every element to the next in the cycle, wrapping around).

This represents the permutation that maps 2 to 5, 3 to 23, 4 to 7…

I hope this isn't too confusing, if you still have questions or some of this is unclear, please ask.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.