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Idea from Longest word with adjacent letters on a keyboard and Longest word with adjacent letters on a keyboard (without using any letters twice)

What is the longest word found in the Merriam-Webster online dictionary with NO adjacent or repeated letters, when typed on GOTO 0's keyboard?

Adjacency counts throughout the word, so RAT is invalid for example.

 ┌───┬───┬───┬───┬───┬───┬───┬───┬───┬───┐
 | Q | W | E | R | T | Y | U | I | O | P |
 └─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┘
   | A | S | D | F | G | H | J | K | L |
   └─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴───┘
     | Z | X | C | V | B | N | M |
     └───┴───┴───┴───┴───┴───┴───┘
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  • $\begingroup$ This should either have the [no-computers] tag, in which case I think we have to accept that a demonstrably best solution will require (for the demonstration) something that solutions aren't allowed; or else not have the [no-computers] tag and say explicitly that computer search is permitted. $\endgroup$ – Gareth McCaughan Apr 21 '20 at 15:03
  • $\begingroup$ Actually, demonstrably-best might be impossible unless someone has what they know is an exact copy of the MW wordlist, or at least something known to contain all its words. $\endgroup$ – Gareth McCaughan Apr 21 '20 at 15:04
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    $\begingroup$ ... Oh, wait, "adjacency counts throughout the word" which means we can't actually have all that many letters, so if we get lucky there might be an optimality proof that doesn't require a complete wordlist. $\endgroup$ – Gareth McCaughan Apr 21 '20 at 15:05
  • $\begingroup$ How do the no-computer rules work? I'm interested in this as a programming problem. If I solve it, should I just abstain from posting the answer? $\endgroup$ – Alexander Apr 21 '20 at 22:47
  • $\begingroup$ So for the record, what's the website that tells you the answer to questions like this? $\endgroup$ – Mazura Apr 21 '20 at 23:18
10
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I can get

eight with EMPATHIC.

This might

be optimal just because I'm not sure you can get more than eight nonadjacent letters. [EDITED to add:] No, you can just about get nine: e.g., QZECTBUMP. But maybe all the nines are obviously impossible?

Let's see. Suppose

there's a 9-letter word. We can cover the 26 keys with the following groups of mutually-adjacent letters: QAW ZSX EDR CFV TGY BHN UJI MK OLP. A 9-letter word will have to have exactly one from each group. (Note: this isn't enough on its own; e.g., ROMANTICS, ALCHEMIST, SHIFTWORK; thank you, Qat.)

Now suppose first of all that

we pick K rather than M; then from OLP we have to take P, and from UJI we have to take U. So we have KPU plus one from each of QAW ZSX EDR CFV TGY BHN. Qat's largest wordlist doesn't find anything matching this other than SPUNKGAVE which I assume is actually a pair of words whose space was somehow missed somewhere in some corpus or other, since it certainly isn't a word. Neither is anything fitting these criteria in a ~400k wordlist I have on my computer.

Otherwise

we have M rather than K; eliminating adjacent letters gives us QAW ZSX EDR CFV TGY BH UI M OLP. Having one of BH rules out G, so we have QAW ZSX EDR CFV TY BH UI M OLP. This still leaves quite a number of possibilities according to Qat, and some of them are even words I know, so let's subdivide according to which we take from TY and on whether we have an A or not. That leaves these cases: [T, no A] QW ZSX ED CV T BH UI M OLP. [T and A] A X ED V T BH UI M OLP. [Y, no A] QW ZSX EDR CFV Y B I M OLP. [Y and A] A X EDR FV Y B I M OLP.

Of these:

the first (T, no A) has only CHOWTIMES in Qat's union list, and that uses both O and I which are adjacent; the same goes for my list. The other three match no words in either Qat's list or mine.

So

8 letters does appear to be the best one can do. But this isn't a watertight proof because I haven't checked whether e.g. QRCZYBMIP is in M-W :-).

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  • $\begingroup$ You can get at most 9 letters, and there are 42 ways. $\endgroup$ – RobPratt Apr 21 '20 at 15:56
  • $\begingroup$ Sounds plausible. 42 times 9! is too many to look up by hand in the M-W dictionary, though... $\endgroup$ – Gareth McCaughan Apr 21 '20 at 15:57
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    $\begingroup$ Nice! There is one anagram of your word too. $\endgroup$ – Weather Vane Apr 21 '20 at 16:44
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    $\begingroup$ There most definitely is. $\endgroup$ – Gareth McCaughan Apr 21 '20 at 18:37
4
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For starters:

climate, 7 letters
mixable, 7 letters (technically, it redirects, not sure if it counts) or the three anagrams
timbale, 7 letters
bimetal, 7 letters
limbate, 7 letters

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Probably many 8 letter ones like

Megabuck

Combated

If I understand it correctly

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  • $\begingroup$ I don't think you do understand correctly. You can't have both D and E, for instance. Nor both B and G. $\endgroup$ – Gareth McCaughan Apr 21 '20 at 15:38
  • $\begingroup$ I guess i do not then $\endgroup$ – DrD Apr 21 '20 at 15:49
  • $\begingroup$ Diagonal counts as adjacent. $\endgroup$ – GC_ Apr 22 '20 at 13:02

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