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I would like to use the Quinapalus Matcher to look for words that are made (with some prior jiggerypokery) from two letter sets, let's call them $word_a$ and $word_b$.

The letters in $word_a$ can be in any order (i.e. anagramed), and between $1$ and $n-1$ from $word_b$ in front and enter remaining $n-1$ to $1$ words behind.

E.g. if $word_a$ was set and $word_b$ was stamen, then statesmen would match, as it is sta(set)*men, where the anagramed/scrambled set is used as tes in the middle.

I've had a look at both the Compound patterns and wildcards, but I can't figure out how to search for my use case, if it is possible at all.

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  • $\begingroup$ Deleted my answer - I was wrong! I misunderstood your problem, thinking that you were looking for the two letter sets to themselves be variables -- if the sets are fixed, then athin and msh have perfectly workable answers. $\endgroup$
    – Deusovi
    Oct 25 at 3:49
  • $\begingroup$ @deusovi, that would be cool, but a step beyond what I need right now. $\endgroup$ Oct 25 at 8:12
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I'm no expert in Qat but, is this the answer you're looking for?

ABC; AC=stamen; B=(/set)

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  • $\begingroup$ To be fair, if both words are variables (as Deusovi said), I don't think there will be a workaround for it. The "stamen" part can be anything (e.g. it can be "......" for any 6-letters word), but the "set" part might be put in manually as the anagram in Qat cannot handle a variable. $\endgroup$
    – athin
    Oct 25 at 4:30
  • $\begingroup$ re your comment $\endgroup$
    – msh210
    Oct 25 at 6:21
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You can write an "… or … or …" list to get what you want. In your example:

s(/set)tamen|st(/set)amen|sta(/set)men|stam(/set)en|stame(/set)n

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  • $\begingroup$ Go upvote athin's answer: it's more useful than this one. I'll leave this in place, though, since my solution works also. $\endgroup$
    – msh210
    Oct 25 at 4:24

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