6
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Inspired by this challenge, I have a similar, with stricter rules.

Find a word that can be decomposed into several other words (more than one). Those words are not allowed to be proper substrings of the starting word. Entries are scored by the length of the shortest word.

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6
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Despite the stricter rules, my entry there would be just as valid here.

Here's my word (again) with a score of 5.

TrIeNnIaLlY.

Made up of the words TINILY and renal with alternating letters. And neither word is a proper substring of the original.

EDIT:

Inspired from pacoverflow's answer to the original challenge, here's a word of score 11 with a small tweak of mine.

REinstitutionALIZATIONs

Made up of the words REALIZATION and institutions, with simply the pluralizing 's' at the end for the word institutions, thereby making this fit the new tougher rules.

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  • $\begingroup$ Reinstitutionalization is not countable, so does not have a plural. Your second suggestion is not a word. $\endgroup$ – Ian MacDonald Jul 13 '15 at 11:57
  • 2
    $\begingroup$ @IanMacDonald: Reinstitutionalization is a noun and can have a logical plural. Consider this usage: Despite repeated reinstitutionalizations, Jack's drug habit is far from over. $\endgroup$ – CodeNewbie Jul 13 '15 at 12:35
  • $\begingroup$ Damn, that word did you right! $\endgroup$ – JLee Jul 13 '15 at 15:47
  • $\begingroup$ @JLee: Hahaha, all thanks to you for teaching me that word. :-D $\endgroup$ – CodeNewbie Jul 13 '15 at 16:37
  • $\begingroup$ And thanks to SQL database queries for teaching it to me! $\endgroup$ – JLee Jul 13 '15 at 16:42
2
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Score 8

The highest-scoring word in the Wolfram dictionary is:

interrelatedness

Which consists of:

INTERRElateDness

and

interreLATEdNESS

Found using:

n = 8;
wl = ToLowerCase@DictionaryLookup[Repeated[_, {n}]];
wl2 = ToLowerCase@DictionaryLookup[Repeated[_, {2 n}]];
subwords = 
  Cases[Table[
    w -> Select[wl, 
      LongestCommonSequence[w, #] == # && 
        LongestCommonSubsequence[w, #] != # &], {w, wl2}], 
   Pattern[p, _ -> {_, __}]];
Cases[subwords, (w_ -> {___, a_, ___, b_, ___}) /; (Sort[
      Characters[w]] == Sort[Characters[a]~Join~Characters[b]]) :> {w,
    a, b}]
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