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  #1  
Old 01-14-2008, 11:38 PM
Kristi Kristi is offline
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Programming Exercise #2

Problem: Make an anagram finder.

Say what?!?
An anagram is a word that uses the same letters as another word. For example, santa and satan are anagrams.

The exercise is to propose and/or code the best and most efficient way to find all anagrams for a word. For our purposes, assume you're using an english dictionary.

Go go go
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  #2  
Old 01-15-2008, 02:08 AM
Novo Novo is offline
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PHP Code:
function anagramsOfword )
{
  if ( 
word == "jew" )
    return {
"jew""wej""wje""jwe""ewj""ejw" };
  else echo(
"This feature isn't implemented yet.")

=D Is O(1)! Wait? Do anagrams have to be dictionary words?

PHP Code:
function anagramsOfword )
{
  if ( 
word == "jew" )
    return {
"jew"};
  else echo(
"This feature isn't implemented yet.")

Still O(1)! I'm good.
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  #3  
Old 01-15-2008, 02:09 AM
Kristi Kristi is offline
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Quote:
Originally Posted by Novo View Post
PHP Code:
function anagramsOfword )
{
  if ( 
word == "jew" )
    return {
"jew""wej""wje""jwe""ewj""ejw" };
  else echo(
"This feature isn't implemented yet.")

=D Is O(1)!
those aren't even real words. Fail
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Old 01-15-2008, 02:23 AM
Novo Novo is offline
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Quote:
Originally Posted by Kristi View Post
those aren't even real words. Fail
Umm... You crazy? Or you just neglected my improvement, punk?
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  #5  
Old 01-15-2008, 04:02 AM
Tolnaftate2004 Tolnaftate2004 is offline
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Quote:
Originally Posted by Novo View Post
Still O(1)! I'm good.
With any luck, we should be able to get O(N!) [for single-word anagrams], or so, I'm thinking.

edit: Recursion can be used to get the arrangements, working on that...
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Last edited by Tolnaftate2004; 01-15-2008 at 04:43 AM..
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  #6  
Old 01-15-2008, 04:23 AM
Kristi Kristi is offline
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Quote:
Originally Posted by Tolnaftate2004 View Post
With any luck, we should be able to get O(N!) [for single-word anagrams], or so, I'm thinking.

The hard part is getting all of the arrangements in N!...
what does n represent
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  #7  
Old 01-15-2008, 04:36 AM
Tolnaftate2004 Tolnaftate2004 is offline
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Quote:
Originally Posted by Kristi View Post
what does n represent
The length of the word passed.
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  #8  
Old 01-15-2008, 05:31 AM
Kristi Kristi is offline
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Quote:
Originally Posted by Tolnaftate2004 View Post
The length of the word passed.
yes but you also have to find which arrangements are actually words in the dictionary.... fun fun fun
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  #9  
Old 01-15-2008, 05:57 AM
DustyPorViva DustyPorViva is offline
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I'm not gonna write it out, but I say just have a script that links to a website that solves anagrams :o
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  #10  
Old 01-15-2008, 06:10 AM
Tolnaftate2004 Tolnaftate2004 is offline
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Quote:
Originally Posted by Kristi View Post
yes but you also have to find which arrangements are actually words in the dictionary.... fun fun fun
Are we given said dictionary?

edit:
NPC Code:
function arrange(letters,first) {
temp.len = letters.length();
if (first == 0) first = "";
if (temp.len > 1) {
temp.returns.clear();
temp.htable = {0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,1};
for (temp.i=0; temp.i<temp.len; temp.i++) {
temp.letter = letters.charat(temp.i).tolower();
temp.hindex = (max(96,min(123,getascii(temp.letter)))-97)%27; // maps characters outside of a-z to htable[26]
if (temp.htable[temp.hindex] == 0) {
temp.htable[temp.hindex] = 1;
temp.pref= first @ temp.letter;
temp.endings = arrange(letters.substring(0,temp.i) @ letters.substring(temp.i+1),temp.pref);
temp.returns.addarray(temp.endings);
}
}
return temp.returns;
}
else {
temp.word = first @ letters.tolower();
if (inDictionary(temp.word))
return {temp.word};
else return char(0);
}
}



Now I just need some inDictionary(word).
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Last edited by Tolnaftate2004; 01-15-2008 at 09:03 AM..
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  #11  
Old 01-15-2008, 08:07 AM
Kristi Kristi is offline
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Quote:
Originally Posted by Tolnaftate2004 View Post
Are we given said dictionary?
assumption says theres a dictionary. you cant just return all possible combinations of the letters
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Old 01-15-2008, 06:01 PM
Tolnaftate2004 Tolnaftate2004 is offline
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Quote:
Originally Posted by Kristi View Post
assumption says theres a dictionary. you cant just return all possible combinations of the letters
Well is this dictionary stored by my choice of method?
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  #13  
Old 01-15-2008, 09:14 AM
napo_p2p napo_p2p is offline
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Quote:
Originally Posted by Tolnaftate2004 View Post
Now I just need some inDictionary(word).
Just say that you've made this.wordlist a sorted array of all known words in the english dictionary. Then for inDictionary() do a binary search or something on that list.

Is that cheating? Although, it probably isn't very efficient in terms of the memory needed to hold such a list...
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Old 01-15-2008, 08:44 AM
Dan Dan is offline
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I guess we should just load those word combinations generated, with some dictionairy script (it's around here somewhere).

*swings magic wand to gather post*

Quote:
Originally Posted by fowlplay4 View Post
Dictionary
Heres my functions , while playing with this script you'll find many funny definitions. LOL

PHP Code:
function Define(word) {
 
temp.link "http://www.onelook.com/?w=" word "&ls=a";
 echo(
"Requesting Definition of" SPC word ":");
 
temp.req requesturl(temp.link);
 
this.catchevent(temp.req,"onReceiveData","onDefine");
}
function 
onDefine(obj) {
 
temp.toks obj.fulldata.tokenize("\n");
 for(
btemp.toks) {
  if (
== "<LI>") continue;
  if (
b.starts("<LI>") > 0) {
   
temp.ntoks b.tokenize(";");
   
temp.npos temp.ntoks[2].pos("(");
   
temp.defin temp.ntoks[2].substring(0,temp.npos);
   if (
temp.defin.length() > 1) echo(" - " temp.defin);
   
temp.msg++;
  }
 }
 if (
temp.msg == 0) echo("No Definitions Found!");

Perhaps change that into returning 'temp.msg > 0'. Could need some better dictionary maybe.
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  #15  
Old 01-15-2008, 08:47 AM
Kristi Kristi is offline
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Quote:
Originally Posted by Dan View Post
I guess we should just load those word combinations generated, with some dictionairy script (it's around here somewhere).

*swings magic wand to gather thread*
thats definitely not the most efficient way at all. just getting it done and doing it efficiently are two different things. this is often your faults
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