You can use it for many word games: to create or to solve crosswords, arrowords (crosswords with arrows), word puzzles, Grep -v "" just_heterograms.txt | grep "^.Lots of Words is a word search engine to search words that match constraints (containing or not containing certain letters, starting or ending letters, Grep -v "" just_heterograms.txt | grep "^.$" | tr '\n' ' ' echoīsdhyg bsfmgt crwths crypts flysch ftncmd glyphs kfrsch khlyst lymphs nymphs schftz schwyz strych wrycht ![]() Without vowels grep -v "" just_heterograms.txt | \ PTRN=`for z in $(seq 1 $V) do echo -n '' done` \īeauing beauish beauism codiaeum couaism coueism coueist cypraeoid douai euoiĪcquiet actious adeuism adieu adieus adieux aeolic aeolics aeolid aeolightĪbdiel abduction abductions abeigh abeu abie abied abiegh abient abies Strings of vowels seq 5 -1 2 | while read V do \ Grep "^$$" _tmp2 | head -n 5 | tr '\n' ' ' \ġ4 : ambidextrously benzhydroxamic dermatoglyphic hydromagnetics hydropneumaticġ3 : amphigenously brachydontism bridgehampton chimneyboards chromeplatingġ2 : absorptively adjunctively adrenolytics adsorptively ambidextrousġ1 : abolishment abridgments abruptiones absolutized achondritesġ0 : abductions abductores abjections abjunctive abortivelyĩ : abducting abduction abductors abjecting abjectionĨ : abditory abdomens abducens abducent abducingħ : abderus abdomen abduces abducts abelson Here's the first 5 from each see: seq 15 -1 1 | while read LEN do \ The resulting file have these many lines ( = words) wc -l only_lower_a_z.txt just_heterograms.txtĮasy now to see the counts and distribution: cat just_heterograms.txt | tr '' 'x' | sort | uniq -c | grep -n ^ Only_lower_a_z.txt > just_heterograms.txt ![]() Next, remove from that list repeating letters, leaving heterograms: grep -vE \ It might be especially useful to use SCHMALTZ or SCHMALZY/ SHMALTZY to use a Z.įirst, get a word list, simplified to only have letters a-z D=/usr/share/dict/british-english-insane Some possibilities are BRIGHTLY GRYPHONS HYDRANTS PSYCHING SHREWDLY SPLOTCHY. I'm currently working on the eight-letter question (and I want to do it without writing a search script), but as a hint to other solvers: there are only five vowels not including Y (which I just used as a word), so one of the eight-letter words will need to have only one "regular" vowel, and will probably need to use Y as a vowel. In this case, one can list 26 distinct words with no letters in common between any of them! ![]() That is, as a word that identifies the letter, as in " C is for Cookie" or "The L Word". The second answer relies on a more flexible meaning of "word": every letter can be regarded as a self-referential word. Strictly speaking, these are the only one-letter words in English ("O" is an alternate spelling of the interjection "Oh"). The first is that there are three different words: For the 1-letter bonus, there are two possible answers:
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