Bioinformatics

over thanksgiving, i figured it'd be a good hack to solve the jumble . Originally, i tried to do all permutation of letter orders in the word but got confused and just did letter frequency: import sys word = len(sys.argv) > 1 and (sys.argv[1]).strip("\n").lower() or "egaugnal" words = [w.strip("\n").lower() for w in open('/usr/share/dict/web2') if len(w) == len(word)+1] def lfreq(w): wfreq = {} for letter in w: wfreq[letter] = letter in wfreq and wfreq[letter]+1 or 1 return wfreq wfreq = lfreq(word) match = [w for w in words if lfreq(w) == wfreq] print match that will print all matches for the word send in on the command line, assuming your dictionary file is in /usr/share/dict/web2 :-( . I tried for a while to do a recursive permute function which would take a word or list of letters and return all possible permutations: permute('abc') -> ['abc','acb','bca',bac',cba',cab'], which

this is the AJAX implementation i use: function jfetch(url,t,o) { var req = jfetch.xhr(); req.open("GET",url,true); req.onreadystatechange = function() { if(req.readyState == 4){ var rsp = req.responseText; if(t.constructor == Function) return t.apply(o,[rsp]); t = document.getElementById(t); t[t.value ==undefined ? 'innerHTML': 'value'] = rsp; req = null; } }; req.send(null); } jfetch.xhr = (window.ActiveXObject) ? function(){ return new ActiveXObject("Microsoft.XMLHTTP"); } : function(){ return new XMLHttpRequest()}; it's short, it only checks for the transport (ActiveX or XHR) once, and it takes either an element id or a call back function. and, i can understand it.