[pymnConyelell]

what is your word you have in mind?

[Zhgpavye]

Does any language count?

[iDzcs7TU]

Originally posted by Ari Rahikkala
Well, hell. I must say that I managed to impress myself with Haskell. Here's mine. A combination of C and bash:

First I took a wordlist (only 5 letter words) with one word on each line. Then I used gedit to add in a series of bash commands (using find/replace on the newlines) so I got a second file like this.

5letscr.sh

Code:

Code

#!/bin/bash  bash jmb2.sh abaca . . . bash jmb2.sh zymes

jmb2.sh

Code:

Code

#!/bin/bash gcc -o jmb2.out jmb2.c echo $1 > jmbtmpfile ./jmb2.out < jmbtmpfile > words2.sh rm jmbtmpfile bash ./words2.sh

jmb2.c

Code: [code]#include #include int rvd5(int a, int b, int c, int d, int e); int rvd6(int a, int b, int c, int d, int e, int f); main() { char word[80]; int count, tag, i, j, k, l, m, n; for (count=0; (word[count] = getchar()) != '\n'; ++count) ; tag = count; printf ("#!/bin/bash\n"); printf ("rm -f jmbtmpfile\n"); if (tag == 5) { for (i = 0; i

[tinetttstation]

I'm a shitty programmer

[ufUUZCnc]

Mine was originally written to cheat on the jumble in the paper. That's why it's roundabout. I already had something written. And that's why it's called jmb2.c and jmb2.sh...

[natahololll]

You're probably better than I am a bash scripting. I just use linux for Matlab and C.

[Coededgeme]

Originally posted by Kuciwalker
You're probably better than I am a bash scripting. I just use linux for Matlab and C. I wrote my first bash script two days ago

[viagra_generic]

I particularly liked how instead of writing something that counted +1 every time it found a match instead I wrote to a temp file and grep -c for 1s in that temp file.

[Obsententicab]

I'm lazy and copy-paste is way too easy.

[Lxbsvksl]

It took me ~45 seconds to write that function. Thinking would have cost extra time.

[Siuchingach]

You could cut down the number of permutation that need to be checked by atleast an order of magnitude by forcing two of the letters to be vowels.

[Voitramma]

Never mind that n is completely unused, that it always returns 0, and this:

i+1

I'm not sure why n went there.

Anyway, it's easier to detect bugs in mine than his, since his could be missing a line and you wouldn't notice.

[boanuatiguali]

- it can deal with types for which equality is defined but order isn't, that is, you can pass it stuff like complex numbers

You can order complex numbers

[SmuffNuSMaxqh]

I didn't say it was an ordered field

[kristloken]

Good idea . It took the execution time of my Haskell program for going through /usr/share/dict/words (234937 lines, 2486824 bytes) from ~1.7 seconds to ~1.5 seconds. Perhaps you can go even further, how do these ideas sound to you.

- drop all matching vowels
- drop very rare consinents like x and z
- restrict the 3 other letters to consinents only
- drop all matching consinents

The base posibilities are 26^5th power = 11,881,376

With agressive culling its 5 * 4 * 19 * 18 * 17 = 116,280

A full two orders of magnitude reduction

[Zoxeeoy]

Radix sort. Order them first by real part, then imaginary

[WFSdZuP3]

Originally posted by Kuciwalker
Radix sort. Order them first by real part, then imaginary Does not obey rules of total order.

[Ettiominiw]

Nerds...

[egexgfczc]

So we've seen Linux geek code (Ari), physicist/mathematician code (KH), and now for some software engineer code (in Java for variety):

Code:

Code

import java.io.IOException;  public class Anagram  {   int size, count;   char[] arr;    public static void main(String[] args) throws IOException    {     String input = args[0];     size = input.length();     count = 0;     arr = new char[size];     for (int i = 0; i < size; i++)     	arr[i] = input.charAt(i);     anagram(size);   }          public static void rotate(int anagramSize)    {     int i;     int position = size - anagramSize;     char temp = arr[position]; // 1st letter      for (i = position + 1; i < size; i++) // push the rest left       arr[i - 1] = arr[i];      arr[i - 1] = temp; // now put the 1st on the end   }     public static void anagram(int anagramSize) {     int limit;     if (anagramSize != 1) // if 1, this is useless this is useless so don't do anything     { 	for (int i = 0; i < anagramSize; i++) { 	      anagram(anagramSize - 1); // anagram the last bits 	      if (anagramSize == 2) // the innermost anagram 	      { 	      	for (int i = 0; i < size; i++) // display it 		     System.out.print(arr[i]); 		System.out.println(); 	      } // end if 	     else 	      	rotate(anagramSize); // rotate word 	} // end for     }  // end if   } // end anagram() }

[bertanu]

^^Asher

The right way to do it