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 05232010, 10:49 AM #21
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You don't need recursion for generating combinations, e.g. given the 'word' ABCD (all letters alphabetized), you want all the words ABCD, ABC, ABD, AB, ACD, AC, AD, A, BCD, BC, BD, B, CD, C, D and possibly the empty words. Let met put this is a table:
Java Code:ABCD present?   ABCD 1111 ABC 1110 AB D 1101 AB 1100 A CD 1011 A C 1010 A D 1001 A 1000 BCD 0111 BC 0110 B D 0101 B 0100 CD 0011 C 0010 D 0001 0000
Counting backwards in binary is easy: the following method finds the next combination of booleans in an array; it either returns the array or null when there is no next combination:
Java Code:boolean[] next(boolean[] present) { for (int i= present.lenght; i > 0; ) if (present[i]) { for (present[i]= false; ++i < present.lenght; present[i]= true); return present; } return null; }
kind regards,
Jos
 05232010, 10:53 AM #22Senior Member
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You should also look into memoizing your results. That will make a big difference in efficiency. Keep track of the Strings you've seen already (hint: HashMap<String, String> or HashSet<String>) and either return the same result you got the first time, or simply return null if you get a repeat.
Gary
 05232010, 10:55 AM #23
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 05232010, 11:26 AM #24Member
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Wow JosAH the boolean arrays threw me off for a second there, I've mulled everything over, and well, from this standpoint, this is what I figure I have to do:
intialise a boolean array with true all over it,
write a toString method that takes a String and a boolean array and converts to a str,
and then this follows:
Java Code:boolean[]=b initialize(b); while ( next(b)!=null){ String str= toString(String query, boolean[] b); if(isInDictionary(str)){ return wordFoundInDictionary; } else{ return null; } }
 05232010, 11:57 AM #25Senior Member
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The longest results still don't come first with this method. AD comes before BCD, for instance. But it is a good, efficient way to generate the combinations. There will still be repeated combinations if there are repeated letters in the initial set, but definitely fewer repeats than the recursive method would generate. Recursion generates N! combinations, while the binary countdown method generates 2^N. Even with N=4 that's 24 vs. 16. When N=5, it's 120 vs. 32, when N=6 it's 720 vs. 64, and after that it starts to make a really big difference. :)
Gary
 05232010, 12:29 PM #26Member
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Alright, I just finished a 10minute joyjumping session. In the past 24 hours I've been literally running on caffeine and trying to beat this deadline. I am beside myself right now.
To Jos, Gary, I really can't thank you guys enough, all of this would have been completely impossible had it not been for you. If someday you find yourselves in the south of  nay, in France, just let me know. I owe you guys quite a few drinks. You guys are awesome. Just awesome.
 05232010, 01:21 PM #27
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Well, the absolute longest combination does come first because that's where the algorithm starts. If you want all the combinations in descending order w.r.t. their length a more complicated (iterative) method is needed. Here's an example:
Java Code:import java.util.Arrays; public class Combination { public static boolean combine(int[] c, int n) { for (int i= c.length; i >= 0;) { if (++c[i] <= n(c.lengthi)) { for (; ++i < c.length; ) c[i]= c[i1]+1; return true; } } return false; } public static void main(String[] args) { int N= 5; for (int m= N; m > 0; m) { int c[]= new int[m]; for (int i= 0; i < c.length; c[i]= i++); do { System.out.println(Arrays.toString(c)); } while(combine(c, N)); } } }
JosLast edited by JosAH; 05232010 at 01:39 PM.
 05232010, 02:13 PM #28
I've lost track of what you're trying to find. Is it the combination of n items taken n1, n2 ... to 1 times?
for "ABCD" we have "ABC", "ABD", "ACD", "AD", "AC", "AB" "BC", "BD", "CD", "A", "B", "C", "D"
Running: "C:\Program Files\Java\j2re1.4.2_08\bin\java.exe" cp D:\JavaDevelopment;. GetCombinations
possibilities=4 for 3 out of 4
A B C
A B D
A C D
B C D
created 4
possibilities=6 for 2 out of 4
A B
A C
A D
B C
B D
C D
created 6
possibilities=4 for 1 out of 4
A
B
C
D
created 4
0 error(s)
 05232010, 02:44 PM #29Member
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Hehe well, since I mapped my dictionary, the number of combinations of letters that i need to use has been drastically cut down, so I basically need all subsets of the original set, not taking into consideration the order in which elements of the subset/set are placed.
It's been solved, and rather elegantly I might add, thanks to you, Jos and Gary's help!
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