Comments on: C++ Algorithms: next_permutation() http://marknelson.us/2002/03/01/next-permutation/ Programming, mostly. Tue, 07 Feb 2012 16:05:51 +0000 hourly 1 http://wordpress.org/?v=3.2.1 By: Prashant http://marknelson.us/2002/03/01/next-permutation/comment-page-1/#comment-435046 Prashant Mon, 26 Dec 2011 17:20:48 +0000 /2002/03/01/next-permutation/#comment-435046 Superb...I just loved it...:) Was looking for a similar stuff for a few days. I have implemented a recursive version but for distinct values, was trying to get it adapted for repeated values. But this algorithm is really awesome...thanks...:) Superb…I just loved it…:)
Was looking for a similar stuff for a few days. I have implemented a recursive version but for distinct values, was trying to get it adapted for repeated values. But this algorithm is really awesome…thanks…:)

]]> By: Refrences « Belbesy M. Adel http://marknelson.us/2002/03/01/next-permutation/comment-page-1/#comment-401259 Refrences « Belbesy M. Adel Wed, 16 Nov 2011 13:43:36 +0000 /2002/03/01/next-permutation/#comment-401259 [...] Generator | next_permutation() | next_permutation() 2 | Algorithms Questions GA_googleAddAttr("AdOpt", "1"); GA_googleAddAttr("Origin", "other"); [...] [...] Generator | next_permutation() | next_permutation() 2 | Algorithms Questions GA_googleAddAttr("AdOpt", "1"); GA_googleAddAttr("Origin", "other"); [...]

]]> By: SPEEDY http://marknelson.us/2002/03/01/next-permutation/comment-page-1/#comment-399917 SPEEDY Mon, 14 Nov 2011 15:41:59 +0000 /2002/03/01/next-permutation/#comment-399917 1. The lazybone method: Just store all permutations generated in a map that is a hash table, and check for the first time generated or not by a hash or so called a dictionary ! 2. If there's no handy hash, just generate an ordered list of numbers somewhat like an odometer with input chars ordered in ASCII is fine! 1. The lazybone method: Just store all permutations generated in a map that is a hash table, and check for the first time generated or not by a hash or so called a dictionary !

2. If there’s no handy hash, just generate an ordered list of numbers somewhat like an odometer with input chars ordered in ASCII is fine!

]]> By: SPEEDY http://marknelson.us/2002/03/01/next-permutation/comment-page-1/#comment-375292 SPEEDY Sat, 17 Sep 2011 14:29:06 +0000 /2002/03/01/next-permutation/#comment-375292 I love to play with variants of this kind of problems. I am thinking if the program is required not to produce duplicate output strings when there might be some repeated characters in the input string. For example, "good_dog" is fed into the program! I love to play with variants of this kind of problems.
I am thinking if the program is required not to produce duplicate output strings when there might be some repeated characters in the input string. For example, “good_dog” is fed into the program!

]]> By: Mark Nelson http://marknelson.us/2002/03/01/next-permutation/comment-page-1/#comment-371577 Mark Nelson Sun, 14 Aug 2011 18:06:49 +0000 /2002/03/01/next-permutation/#comment-371577 @Thomas: Nice approach. On problems like this I have a tendency to go for the brute force first. A more elegant solution like yours is always more satisfying though. - Mark @Thomas:

Nice approach. On problems like this I have a tendency to go for the brute force first. A more elegant solution like yours is always more satisfying though.

- Mark

]]> By: Thomas Nygreen http://marknelson.us/2002/03/01/next-permutation/comment-page-1/#comment-371471 Thomas Nygreen Sat, 13 Aug 2011 16:19:55 +0000 /2002/03/01/next-permutation/#comment-371471 Nice article! The original problem can be solved by first listing the seven possible combinations of numbers in the range 1–9 such that no numbers are repeated and the sum is 17: [code] i: {1, 7, 9} ii: {2, 6, 9} iii: {2, 7, 8} iv: {3, 5, 9} v: {3, 6, 8} vi: {4, 5, 8} vii: {4, 6, 7} [/code] Exactly five of these must be included in the solution. Let's label the positions a–i, as Hugo did, and start with the diagonal c-e-g. Note that both c and g must occur in three different combinations, the candidates being 6, 7, 8 and 9. Using the fact that the figure is symmetric, let c < g. Then we have c = 6 or c = 7. As there are no combinations containing both 5 and 6 or both 5 and 7, the only positions left for 5 are d and h. Again we use symmetry, and let d = 5. This in turn means a must be 3 or 4, and g must be 8 or 9. There is only one combination containing 1, so 1 cannot be in any corner, and there are no combinations with both 8 and 9, so h and i cannot be 9. That leaves only e and h as possible positions for 1. Now if we were to let e = 1, we would get c = 7, g = 9, a = 3, and b would need to be 7, which is already taken by c. Therefore h = 1. Then follows g = 9, i = 7, a = 3, c = 6, b = 8, f = 4 and e = 2. Nice article!

The original problem can be solved by first listing the seven possible combinations of numbers in the range 1–9 such that no numbers are repeated and the sum is 17:

PLAIN TEXT
CODE:
  1. i:   {1, 7, 9}
  2. ii:  {2, 6, 9}
  3. iii: {2, 7, 8}
  4. iv:  {3, 5, 9}
  5. v:   {3, 6, 8}
  6. vi:  {4, 5, 8}
  7. vii: {4, 6, 7}

Exactly five of these must be included in the solution. Let's label the positions a–i, as Hugo did, and start with the diagonal c-e-g. Note that both c and g must occur in three different combinations, the candidates being 6, 7, 8 and 9. Using the fact that the figure is symmetric, let c < g. Then we have c = 6 or c = 7.

As there are no combinations containing both 5 and 6 or both 5 and 7, the only positions left for 5 are d and h. Again we use symmetry, and let d = 5. This in turn means a must be 3 or 4, and g must be 8 or 9.

There is only one combination containing 1, so 1 cannot be in any corner, and there are no combinations with both 8 and 9, so h and i cannot be 9. That leaves only e and h as possible positions for 1. Now if we were to let e = 1, we would get c = 7, g = 9, a = 3, and b would need to be 7, which is already taken by c. Therefore h = 1.

Then follows g = 9, i = 7, a = 3, c = 6, b = 8, f = 4 and e = 2.

]]> By: Quick Facts http://marknelson.us/2002/03/01/next-permutation/comment-page-1/#comment-335444 Quick Facts Fri, 29 Oct 2010 22:33:51 +0000 /2002/03/01/next-permutation/#comment-335444 You you could change the webpage subject title C++ Algorithms: next_permutation() to more catching for your content you create. I enjoyed the the writing still. You you could change the webpage subject title C++ Algorithms: next_permutation() to more catching for your content you create. I enjoyed the the writing still.

]]> By: Sven Forstmann http://marknelson.us/2002/03/01/next-permutation/comment-page-1/#comment-324522 Sven Forstmann Thu, 08 Apr 2010 14:05:17 +0000 /2002/03/01/next-permutation/#comment-324522 Well, lets say you have 120 permutations and you want permutation number 60. Using STL you woule have to walk through all of them step by step - wont you ? So I've written an implementation that gives you any permutation immideately, without recursion: [c] #include <string> int main(int,char**) { std::string default_str = "12345"; int perm=1, digits=default_str.size(); for (int i=1;i<=digits;perm*=i++); for (int a=0;a<perm;a++) { std::string avail=default_str; for (int b=digits,div=perm;b>0; b--) { div/=b; int index = (a/div)%b; printf("%c", avail[index] ); avail.erase(index,1) ; } printf("\n"); } printf("permutations:%d\n",perm); while(1); } [/c] (c) Sven Forstmann Well, lets say you have 120 permutations and you want permutation number 60. Using STL you woule have to walk through all of them step by step - wont you ?

So I've written an implementation that gives you any permutation immideately, without recursion:

PLAIN TEXT
C:
  1. #include <string>
  2.  
  3. int main(int,char**)
  4. {
  5.     std::string default_str = "12345";
  6.  
  7.     int perm=1, digits=default_str.size();
  8.     for (int i=1;i<=digits;perm*=i++);
  9.     for (int a=0;a<perm;a++)
  10.     {
  11.         std::string avail=default_str;
  12.  
  13.         for (int b=digits,div=perm;b>0; b--)
  14.         {
  15.             div/=b;
  16.             int index = (a/div)%b;
  17.             printf("%c", avail[index] );
  18.             avail.erase(index,1) ;
  19.         }
  20.         printf("\n");
  21.     }
  22.     printf("permutations:%d\n",perm);
  23.     while(1);
  24. }

(c) Sven Forstmann

]]> By: gaston770 http://marknelson.us/2002/03/01/next-permutation/comment-page-1/#comment-314728 gaston770 Sun, 13 Sep 2009 05:13:26 +0000 /2002/03/01/next-permutation/#comment-314728 http://code.google.com/codejam/contest/dashboard?c=186264#s=p1 This Google Code Jam 2009 problem solves in less than 30 lines with next_permutation.. I love it :D, Great post dear Mark. ;) [c] int main (void) { int T; string N; cin >> T; for (int count = 1; count <= T; count++) { cin >> N; if (!next_permutation (N.begin(), N.end())) { int xx = 0; while (N[xx++] == '0'); swap (N[--xx],N[0]); N.insert (N.begin(),'0'); swap (N[0], N[1]); } cout << "Case #" << count << ": "<< N << endl; } return 0; [/c] http://code.google.com/codejam/contest/dashboard?c=186264#s=p1

This Google Code Jam 2009 problem solves in less than 30 lines with next_permutation.. I love it :D, Great post dear Mark. ;)

PLAIN TEXT
C:
  1. int main (void) {
  2.     int T;
  3.     string N;
  4.     cin>> T;
  5.     for (int count = 1; count <= T; count++) {
  6.         cin>> N;
  7.         if (!next_permutation (N.begin(), N.end())) {
  8.             int xx = 0;
  9.             while (N[xx++] == '0');
  10.             swap (N[--xx],N[0]);
  11.             N.insert (N.begin(),'0');
  12.             swap (N[0], N[1]);
  13.         }
  14.         cout <<"Case #" <<count << ": "<<N <<endl;
  15.     }
  16.     return 0;

]]> By: Mark Nelson http://marknelson.us/2002/03/01/next-permutation/comment-page-1/#comment-310890 Mark Nelson Sun, 26 Jul 2009 18:55:07 +0000 /2002/03/01/next-permutation/#comment-310890 @lzzy88: I don't know where the algorithm came from, but you can pretty much figure out next_permutation by examining the source code. A recursive version of a permutation function is trivial, the iterative version is not quite as obvious and requires a little bit of work. - Mark @lzzy88:

I don't know where the algorithm came from, but you can pretty much figure out next_permutation by examining the source code. A recursive version of a permutation function is trivial, the iterative version is not quite as obvious and requires a little bit of work.

- Mark

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