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U PROP BY THERE, USE PARS PAR THERE PROP
BY THERE IYBSM PAR YUCR THER, USE YH HS.
THINE BP AUQQRS PAUP PAR HGBMBSUN PROP
BY GRUTARE UP YHCR YPUMR?

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1 Answer 1

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Answer still in progress

I have decoded the cryptogram as it stands (not manually, but with a website) and it results in this:

A TEXT IS CODED, AND THEN THE CODED TEXT IS CODED USING THE SAME CODE, AND SO ON. COULD IT HAPPEN THAT THE ORIGINAL TEXT IS REACHED AT SOME STAGE?

I haven't found the answer to this yet, though. Just wanted to get this up so others could use it. I have a very vague hypothesis that it will happen after 26 iterations, but this is completely unsupported.

EDIT: Thanks to @Ross Millikan in the comments, the answer is actually 1260 - the highest LCM of a set of numbers with a sum of 26. The reason this works is because there can be "loops" where A is substituted with B, B with C, and C with A. This loop has a size of 3. The loops in a substitution cipher can sum to 26, and the 1260 figure comes from loops of 4, 5, 7, and 9 and an extra letter that is substituted with itself. If this isn't legal, then the actual amount is something else - I'll find that eventually.

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  • $\begingroup$ That would be the upper bound, right? (11*17*19*21*23*24*45*26, if anyone's interested) $\endgroup$ Commented Mar 3, 2015 at 18:14
  • $\begingroup$ @EngineerToast Yes - it can happen earlier, but it will always happen after that many iterations. $\endgroup$
    – mdc32
    Commented Mar 3, 2015 at 18:32
  • $\begingroup$ The maximum cycle length is the greatest number that is the LCM of a set of numbers summing to 26. This is given in oeis.org/A000793 as $1260=4\cdot 9 \cdot 5\cdot 7$ If you break the permutation into cycles, the overall cycle length is the LCM of the lengths of the various cycles. $\endgroup$ Commented Mar 3, 2015 at 20:23
  • $\begingroup$ If you don't need an exact answer, then you could say that it will happen by the 26th iteration $\endgroup$
    – KSmarts
    Commented Mar 3, 2015 at 21:24
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    $\begingroup$ @xnor As far as I know, yes. I don't really see the point in the Caesar clue or the cryptogram itself. $\endgroup$
    – mdc32
    Commented Mar 4, 2015 at 2:48

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