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I found this weird-looking puzzle in the 2007 Tournament of the Towns.

For each letter A to Z, I allot a specific English word containing (but not necessarily starting with) that letter. Then I write pages as follows:

  • The first page consists only of the word corresponding to the letter A.
  • On each subsequent page, I replace each letter of the preceding page by its corresponding word.

The opening of the 40th page begins with:

Till whatsoever star that guides my moving.

Prove that this sentence is written again later on the same page.


Bonus question: what properties of the above-quoted sentence are essential to the proof? Or in other words, what other sentences could we replace this by so that the result still holds?

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  • $\begingroup$ Do you have the link? $\endgroup$ Feb 19, 2018 at 0:16
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    $\begingroup$ I don't get it. If the 40th page opens with those words, the 39th page must open with the first letters of those words, but "Twstgmm" isn't a word (at least not in English), isn't the start of a word, and doesn't start with a word. Can you please explain further how this is supposed to work? $\endgroup$
    – Lawrence
    Feb 19, 2018 at 7:39
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    $\begingroup$ @Lawrence: No, not the first letters. A word that corresponds to a letter is only required to contain that letter at any position. (Otherwise, the first word on each page would start with A.) $\endgroup$
    – M Oehm
    Feb 19, 2018 at 7:40
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    $\begingroup$ @ibrahimmahrir 1) The 41st page will have well over 36 words, because the quoted sentence is just the start of the 40th page. 2) Yes, the number of words grows very fast. 3) On the same page means part of the words written on the same page - in this case, the 40th page. $\endgroup$ Feb 21, 2018 at 22:59
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    $\begingroup$ @Lawrence I edited to clarify that more explicitly. $\endgroup$ Feb 21, 2018 at 22:59

3 Answers 3

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First, note that

every time a page begins with a certain letter, the next page will always begin with the same word. The first letter of that word will determine the first word on the next page.

There are 26 letters in the alphabet. Therefore,

If we look at the first words of pages, they must form a cycle of length 26 at most. This is the 40th page; therefore, we are well into at least the second such cycle.

The first word of the present page is "Till".

Because we are in the second (or subsequent) cycle, there must be a prior page that began with "Till". We know that the first word of the preceding page was not "Till" or the third and fourth word on this page would be the same. Therefore, we know that "till" is not the word corresponding to T. It must then be the word corresponding to I or L.

Consider the first word of the 39th page. Call it 39th.

Since we are so far into the cycle, we know that 39th must also be appearing for at least the second time. 39th must begin with either I or L. None of the seven letters on this page begins with I or L, therefore none of them is 39th. Because each word must contain the letter it corresponds to, that word must appear in its own expansion. Therefore, 39th must be more than seven letters long. Since 39th appears in its own expansion and it has occurred as the first word at least once before, we know that it occurs in some position after the seventh on every subsequent page including the 39th page. Therefore, the expansion of its first seven letters occurs later on this page.

For the bonus,

I got nothing, other than the above. Definitely, it is sufficient that none of the words in the sentence could be the first word on the preceding page but but I feel certain that there is something more intelligent to be said.

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Random observations:

Each page contains a copy of the previous page. The word assigned to A cannot start with A - otherwise the words on all the pages would start with A, and the 40th page starts with "Till". This means that there exist two pages at most 26 pages apart that start with the same letter. Let's call those pages a and b. Then pages a+1 and b+1 would start with the same word, etc, and pages a+n and b+n would start with the same n words. This may have nothing to do with what we are trying to prove, who knows.

EDIT: Just ate some ice cream, my mind is clear now.

So the part I was missing was that the copy of page i that page (i+1) has does not start with the content of page i. Therefore page 40 contains some page (14-40) that starts with "Till whatsoever star that guides my moving" not from the beginning, so the sentence must appear somewhere else.

As for the bonus:

Since the pages are at most 26 pages apart, the maximum length of the sentence would be 13 words, so any sentence from 1-13 words that does not start with the letter A(so each page can't be the same) satisfies this property.

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  • $\begingroup$ Your bonus answer is definitely not quite correct as it stands. If the word corresponding to A were A and the sentence were also just A, then the condition in your last spoiler block would hold, but there would be only one copy of that sentence on page 40. I'm not sure I'm convinced by the rest of your argument, but I may very well just be being slow. $\endgroup$
    – Gareth McCaughan
    Feb 19, 2018 at 0:50
  • $\begingroup$ @GarethMcCaughan hmm, nice catch. would adding the condition that the sentence doesn't start with the letter A suffice? $\endgroup$
    – Quintec
    Feb 19, 2018 at 0:54
  • $\begingroup$ I think this argument (minus the bonus, as Gareth points out) is essentially correct, but I'm having trouble intuiting the details of it. I had similar trouble with the solution I read at the source, which is probably a good sign for your answer, but is there any chance you could explain it more clearly? $\endgroup$ Feb 19, 2018 at 20:22
  • $\begingroup$ I think it may turn out that making this clear and rigorous reveals another property of the given sentence that's required. (I have a specific nother property in mind.) I could, however, very easily be wrong. $\endgroup$
    – Gareth McCaughan
    Feb 19, 2018 at 22:57
  • $\begingroup$ @Randal'Thor I can try... Let's say the sentence "hello my name is joe" corresponds to the letters E, Y, A, S, O and is on page 3. Then page 2 starts with the "word" eyaso. Let's also assume that the cycle is length 20 - i.e. the letter e corresponds to a word with a first letter of h, etc etc 20 times which takes us back to e being the first letter. Then on page 22, the first letter will be e. On page 23, the first word will be "hello", the same first word as page 3. ... $\endgroup$
    – Quintec
    Feb 20, 2018 at 13:22
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Hugh Meyers' answer seems right to me. But I think the premise of the puzzle is actually flawed!

We are told that there's a function f mapping letters to English words, such that for each of the 26 English letters @, the word f(@) contains the letter @. We're also told that

  • f(@1) = TILL
  • f(@2) = WHATSOEVER
  • f(@3) = STAR
  • f(@4) = THAT
  • f(@5) = GUIDES
  • f(@6) = MY
  • f(@7) = MOVING

and finally we're told that 39th[:7] = @1@2@3@4@5@6@7 itself forms a prefix of a string of English words which themselves are values in the codomain of f. This is something like a cryptarithm.

Hugh Meyers correctly deduces the following lemmas:

@1 ≠ T. Therefore @1 is I or L.
Since all seven words are distinct, all the letters in 39th[:7] must be distinct too.
39th must contain some letter @ such that f(@) = 39th. But 39th is definitely not TILL, or WHATSOEVER, or STAR, or THAT, or GUIDES, or MY, or MOVING! Therefore 39th must be longer than 7 letters.

For example: Suppose for the sake of contradiction that the 39th page began "Let him not speak." Then f(L) = TILL, f(E) = WHATSOEVER, f(T) = STAR, and so on. But then what would be the letter @ such that f(@) = LET? It can't be L, E, or T — so we have a contradiction! By this kind of reasoning, in fact 39th must be longer than 7 letters.

So: The 39th page starts with a word of at least 8 letters, the first 7 of which are all distinct; the first letter is I or L; the sixth letter is M or Y; and so on.

A computer search of /usr/share/dict/words turns up zero English words matching these criteria.

A computer search of my crossword dictionary, including proper names and pop-culture references, turns up only one "word" matching these criteria: maybe appropriately, it's "IHATEYOU" — which is not really a word! (Per above, we know we can't have f(I) = TILL, because then there would be no @ for which f(@) = I.)

So I think the puzzle is actually ill-formed, and we should replace the above sentence with another sentence — such that the result will still hold, and so will the premise!


Concretely, the 40th page might start with "These lovely caves, these round enchanting pits," as demonstrated by this Python program:

f = {
  'A': 'CAVES', 'C': 'GRACE', 'D': 'AID', 'E': 'LOVELY', 'G': 'SEASONING', 'H': 'HAD',   
  'I': 'PITS',  'L': 'LIE', 'N': 'ENCHANTING', 'O': 'ROUND', 'P': 'APE', 'R': 'DRY', 
  'S': 'THESE', 'T': 'ROT', 'U': 'URN', 'V': 'EVE', 'Y': 'IVY'
}
assert all(k in v for k,v in f.items())
text='A'
for i in range(1, 41):
  print(i, text)
  text = ' '.join(f[c] for c in text if c.isupper())[:200]
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