All possible combinations of a string 1. Finds all possible cominations of a given string. The total number of combinations with recurrences is the factorial of the string length. 2. Have given the word 'post' as an example since if I included a cin >> word; then a user may unconscientiously enter a long word which will take toooo long to end. To change to another word change the string str. 3. A 9 letter word like 'aeroplane' can take around 40 seconds on a 1GHz machine. A 10 letter word will take 9 x 40 s - thats 6 mins! So watch out for the string length. 4. This may time according to the length of the string str Code: // Combination #include<iostream.h> #include<string.h> #include<conio.h> unsigned long fact(unsigned long); // finds factorial of a number void swap(int[],int,int); void display(int[],char[]); void Add(int[],int); void main() { // clrscr(); // Remove the first comment tag if running in TC++ or BC++ DOS mode int j,k,n; unsigned long i; char str[]="post"; // small word n=strlen(str); int*key=new int[n+1]; key[0]=n; // key[0] contains the total no: of letters for (i=1;i<=key[0];i++) key[i]=i; display(key,str); swap(key,key[0],key[0]-1); display(key,str); for (i=3;i<=fact(key[0]);i+=2) { for (j=key[0]-1;j>=0;j--) { if ((i-1)%fact(j)==0) { Add(key,key[0]-j); for (k=1;k<=key[0]-j-1;k++) if (key[k]==key[key[0]-j]) {Add(key,key[0]-j);k=0;} } } display(key,str); swap(key,key[0],key[0]-1); display(key,str); } cout << "\nTotal no: of combinations = " << key[0] << "! = " << i-1; getch(); } unsigned long fact(unsigned long f) { return f==0?1:f*fact(f-1); } void swap(int k[],int a,int b) { int t=k[a];k[a]=k[b];k[b]=t; } void display(int k[],char s[]) { for(int i=1;i<=k[0];i++) cout << s[k[i]-1]; cout << '\t'; } void Add(int k[],int i) { k[i]++; if (k[i]==k[0]+1) k[i]=1; }
First of all welcome to the go4expert forums and yes a nice code also. Can you mention which compiler it will compile as its not meant for Visual Studio as it has clrscr method. Thanks Shabbir Bhimani
I wrote this using TC++ 3.1. It compiled and worked in TC++ 3.1 and Borland C++ 3.11. Remove clrscr() and it compiles and executes in vc++6 - checked.
Thank you... with a little modification, I was able to make it work for my C# app that I am working on. Great code... too bad google ranks it so low in its engine
Google is slowly loosing the share and the quality in Search domain. In sitepoint forum in some thread I read recently that within couple of years Google has lost from 90% of the share traffic to 48% and the rest have moved to Yahoo and MSN. Its probably because they are more interested in satisfying there Clients of adwords.
hello this is realy a very good code ... can any one tell me that what is the basic process of function void add (int[] , int); i need its processing flow plz if anybody can help me i'll be thankful
I should've documented this when I wrote this in college 4 yrs back. I'll try to remember and repost this code with documentation.
hello one more question is that plz tell me the logic of "if ((i-1)%fact(j)==0)" this condition in this code.. and plz send that documentation as soon as possible
I spent a lot of time try to get this algo in college days. But never bothered to write down the steps. I'll defintely post the documentation here in a few weeks time after I recollect the steps.
I have documented the code completely Code: // Combination #include<iostream.h> #include<string.h> #include<conio.h> unsigned long fact(unsigned long); // finds factorial of a number void swap(int[],int,int); void display(int[],char[]); void Add(int[],int); /***********************************************************************/ /// <b>Function: main</b> /// /// \param NONE /// /// \return void /// /// \remarks main function /// /***********************************************************************/ void main() { int j,k,n; unsigned long i; char str[]="G4EF"; // sample small word n=strlen(str); // Get the length of the string int*key=new int[n+1]; // Allocate buffer of size n+1 key[0]=n; // key[0] contains the total no: of letters // Assign all the keys an index from the string for (i=1;i<=key[0];i++) key[i]=i; display(key,str); // Displays the string swap(key,key[0],key[0]-1); // swaps the key values n and n-1 display(key,str); // Displays the string // Added by Shabbir // store the factorial once and so it does not calulate each time unsigned long num = fact(key[0]); // Loop till factorial n so we get all the combination of strings // It starts with i=3 as we already have 2 combination // Its increments by 2 as we get 2 combination with the help of swap for (i=3;i<=num;i+=2) { // Loop for all the elements of the string for (j=key[0]-1;j>=0;j--) { // Edited by Shabbir // Can be avoided //if ((i-1)%fact(j)==0) { Add(key,key[0]-j); // Increment the key in a cyclic form for (k=1;k<=key[0]-j-1;k++) { // If kth index in key is euqal to n-jth index if (key[k]==key[key[0]-j]) { Add(key,key[0]-j); // Increment the key in a cyclic form k=0; } } } } display(key,str); // Displays the string swap(key,key[0],key[0]-1); // swaps the key values display(key,str); // Displays the string } cout << "\nTotal no: of combinations = " << key[0] << "! = " << i-1; cout<<endl; getch(); } /***********************************************************************/ /// <b>Function: fact</b> /// /// \param long f (in) /// /// \return unsigned long /// /// \remarks returns the factorial of parameter f using recursion /// /***********************************************************************/ unsigned long fact(unsigned long f) { return f==0?1:f*fact(f-1); } /***********************************************************************/ /// <b>Function: swap</b> /// /// \param k[] (out) /// /// \param a (in) /// /// \param b (in) /// /// \return void /// /// \remarks Swaps the index of the key. /// /***********************************************************************/ void swap(int k[],int a,int b) { int t=k[a];k[a]=k[b];k[b]=t; } /***********************************************************************/ /// <b>Function: display</b> /// /// \param k[] (in) /// /// \param s[] (in) /// /// \return void /// /// \remarks Displays the string using the key /// /***********************************************************************/ void display(int k[],char s[]) { for(int i=1;i<=k[0];i++) cout << s[k[i]-1]; cout << '\t'; } /***********************************************************************/ /// <b>Function: Add</b> /// /// \param k[] (out) /// /// \param i (in) /// /// \return void /// /// \remarks Increments the key values of the index i. If it crosses /// the max limit resets back to 1. i.e. in cyclic form. /// /***********************************************************************/ void Add(int k[],int i) { k[i]++; if (k[i]==k[0]+1) k[i]=1; } Also I have removed some unnecessary calculations which now speeds up the process. I have added "Added/Edited by shabbir" so that its easily trackable.
Hats off to you shabbir ! I cant belive you figured this out so soon. You are a true algo expert ! Thanks