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Build as many words as possible from the given letters

twofx.github.io

3 points by re-framer 2 years ago · 6 comments

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genezeta 2 years ago

Given that you so generously keep the dictionary there, one can just open the console and paste:

    words.filter((word) => word.split('').every((letter) => "egtdrnq"].indexOf(letter) !== -1) && word.indexOf('d') !== -1 && word.length>3)
But, of course, that spoils the fun :)
  • genezeta 2 years ago

    More generally (speed may need to be adjusted):

      var letters = $$('.hex').map((x)=>x.textContent.toLowerCase()).join('');
      var obLetter = $('.hex.yellow').textContent.toLowerCase();
      var sol = words.filter((word) => word.split('').every((letter) => letters.indexOf(letter) !== -1) && word.indexOf(obLetter) !== -1 && word.length>3);
    
      let tick = 0;
      let nextWord = ()=>{
        $('#textInput').value=sol[tick].toUpperCase();
        $('.enter-button').click();
        tick++;
        if(tick<sol.length) setTimeout(nextWord, 50);
      };
      setTimeout(nextWord, 30);
smcin 2 years ago

Which dictionary are you using? Words not in it: DREG, TREND(ED), DEDENT(ED), DENNED...

beardyw 2 years ago

Q without a U is more or less a dead letter.

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