Help with counting Permutations

[jusfry01]

I am having problems using the permutation formula in my code. I need to use P(n,k) to count k permutations in a string of n characters.
Example:
Input: P(3,2)
Output: "A string of 3 characters has 6 permutations of length 2"

I know the formula is 'P(n,k) = n! / (n-k)!' But I get compiler errors when trying to execute.

Here is the section of my code related to my problem.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()

int n, k, p;

printf("Enter one of the following options: \n\
             Q - Quit the program\n\
             Eval(<exp>) - Evaluate a simple expression\n\
             ! - Evaluate an integer factorial\n\
             P(n,k) - Count k-permutations of a string of n characters\n\
           \n\ Enter your option: ");

if(strstr(userInput,"P") != NULL){
        sscanf(userInput,"%*c%*c%d%*c%d%*c", &n, &k);
        p = n! / (n - k)!;
        printf("A string of %d characters has %d permutations of length %d\n", n, p, k);
         }

How can I get the code to compile correctly?

Thank you,
Justin

[Adak]

You have omitted:

1) Curly Braces for main()
2) Any values for your variables
3) a return int for main()
4) how you were going to figure the factorial of n, to prepare for your equation
5) any code logic, whatsoever.

You need to put some "skin' into this assignment. You're a smart guy, I'm sure you'll find something to get the ball rolling.

You don't have a C problem just yet. You have a "done really nothing", problem. Come back when you have a C problem.

[jusfry01]

Sorry, I was just posting the portion of my code for which I was having issues, here is my full code. I have the first 3 printf options working properly, but I am having trouble with the final 'else if' section. I am supposed to take input in the form of "P(3,2)", for example, and the program output would calculate the permutations of three characters using just two at a time.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define bufSize 2048

int main()
{
  char userInput[bufSize];
  int opers[5];
  char operA;
  int exprA;
  int numA, idx;
  double result;
  char doneFlag = 'f';
 
  do {     
      printf("Enter one of the following options: \n\
             Q - Quit the program\n\
             Eval(<exp>) - Evaluate a simple expression\n\
             ! - Evaluate an integer factorial\n\
             P(n,k) - Count k-permutations of a string of n characters\n\
           \n\ Enter your option: ");
      
      fgets(userInput,bufSize,stdin);
      
    if(strstr(userInput,"Q") != NULL) {
        doneFlag ='t';
    } else if(strstr(userInput,"q") != NULL) {
        doneFlag ='t';
    } else if(strstr(userInput,"Eval") != NULL){
      sscanf(userInput,"%*c%*c%*c%*c%*c%d %c %d%*c", &opers[0], &operA, &opers[1]);
      switch(operA) {
                   case '+':
                   exprA = opers[0] + opers[1];
                   printf("%d %c %d = %d\n", opers[0], operA, opers[1], exprA);
                   break;
                   case '-':
                   exprA = opers[0] - opers[1];
                   printf("%d %c %d = %d\n", opers[0], operA, opers[1], exprA);
                   break;
                   case '/':
                   exprA = opers[0] / opers[1];
                   printf("%d %c %d = %d\n", opers[0], operA, opers[1], exprA);
                   break;
                   case '*':
                   exprA = opers[0] * opers[1];
                   printf("%d %c %d = %d\n", opers[0], operA, opers[1], exprA);
                   break;
                   case '%':
                   exprA = opers[0] % opers[1];
                   printf("%d %c %d = %d\n", opers[0], operA, opers[1], exprA);
                   break;
                   case '^':
                   exprA = pow(opers[0],  opers[1]);
                   printf("%d %c %d = %d\n", opers[0], operA, opers[1], exprA);
                   break;
                   default:
                           exprA = 0;
                           printf("?'%c' is not a valid operator", operA);
                   }
    } else if(strstr(userInput,"!") != NULL){               // Calculate the Factorial of !
        sscanf(userInput,"%*c%d", &opers[2]);
          result = 1.0;
          for(idx = 1; idx <= numA; idx++) {
            result *= (double)idx;
          }
          printf("%d! = %.0lf\n", opers[2], result);
    } else if(strstr(userInput,"P") != NULL){            // Calculate P(n,k)
        sscanf(userInput,"%*c%*c%d%*c%d%*c", &opers[3], &opers[4]);
        opers[5] = opers[3] / (opers[3] - opers[4]);
        printf("A string of %d characters has %d permutations of length %d\n", opers[3], opers[5], opers[4]);
          }
    
   } while(doneFlag == 'f');


 return 0;
}

[Adak]

I know the problem is well known, but it missed me, and I don't have the time to jump into it, atm.

<< Bump for a smart problem >>

[jusfry01]

I got it figured out, it was all too simple, and I was just over-thinking it.
Thanks

[jusfry01]

Nevermind, It does not work. I thought I had the P(n,k) encoded correctly, but I get a compiler error. I am sure that my 'for() && () {' line is not formatted correctly, can someone help me correct it?

This is getting a little frustrating, as I am new to C, and have been looking everywhere for answers. I can calculate the first n!, but I am lost in regards to calculating (n-k)!, how do I modify my code to enable it to calculate correctly?

Please Help.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define bufSize 2048

int main()
{
  char userInput[bufSize];
  int opers[4];
  char operA;
  int exprA;
  int idxA;
  int idxB, idxC, resultB, resultC, resultD;
  double resultA;
  char doneFlag = 'f';
 
  do {     
      printf("Enter one of the following options: \n\
             Q - Quit the program\n\
             Eval(<exp>) - Evaluate a simple expression\n\
             ! - Evaluate an integer factorial\n\
             P(n,k) - Count k-permutations of a string of n characters\n\
           \n\ Enter your option: ");

fgets(userInput,bufSize,stdin);

if(strstr(userInput,"!") != NULL){               // Calculate the Factorial of !
        sscanf(userInput,"%*c%d", &opers[2]);
          resultA = 1.0;
          for(idxA = 1; idxA <= opers[2]; idxA++) {
            resultA *= (double)idxA;
          }
          printf("%d! = %.0lf\n", opers[2], resultA);
    } else if(strstr(userInput,"P") != NULL){              // CALCULATE  P(n,k)
        sscanf(userInput,"%*c%*c%d%*c%d%*c", &opers[3], &opers[4]);
        resultB = 1;
        for(idxB = 1; idxB <= opers[3]; idxB++) && (idxC = opers[3] - opers[4]; <= opers[4]; idxC++) {
             resultB = idxB / idxC;
        }  
        printf("A string of %d characters has %d permutations of length %d\n", opers[3], resultD, opers[4]);
          
          }
   } while(doneFlag == 'f');


 return 0;
}

[tabstop]

Why would you compute (n-k)! differently than n! ? It's the same function, to be computed in the same way.

[dwks]

There's also no such syntax as this.

for(idxB = 1; idxB <= opers[3]; idxB++) && (idxC = opers[3] - opers[4]; <= opers[4]; idxC++) {

Maybe you're trying to create a nested for loop?

for(idxB = 1; idxB <= opers[3]; idxB++) {
    for(idxC = opers[3] - opers[4]; <= opers[4]; idxC++) {
        // ...
    }
}

By the way, there's a builtin macro in stdio.h called BUFSIZ that you could make use of . . . .

[jusfry01]

I understand that n! and (n-k)! are the same function, but I am having trouble doing the factorial of the result of (n-k). To this point, I have only learned to do a factorial on a single value, and am not sure how to generate the factorial of the (n-k).

My structure is in a nested for statement now, and the program reads all of my input, but tells me that I have only 1 permutation no matter what numbers I put in.

Here is the section of my code that I am having trouble with... Please help me sort this out

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define bufSize 2048

int main()
{
  char userInput[bufSize];
  double opers[1];
  int num[3];
  char operA;
  double exprA;
  int idxA;
  int idxB, idxC, resultB;
  double resultA;
  char doneFlag = 'f';
 
  do {     
      printf("Enter one of the following options: \n\
             Q - Quit the program\n\
             Eval(<exp>) - Evaluate a simple expression\n\
             ! - Evaluate an integer factorial\n\
             P(n,k) - Count k-permutations of a string of n characters\n\
           \n\ Enter your option: ");
      
      fgets(userInput,bufSize,stdin);

 if(strstr(userInput,"P") != NULL){              // Calculate P(n,k)
        sscanf(userInput,"%*c%*c%d%*c%d%*c", &num[1], &num[2]);
        resultB = 1;
        for(idxB = 1; idxB <= num[1]; idxB++) {
                 for(idxC = num[1] - num[2]; idxC <= num[2]; idxC++) {
             resultB = idxB / idxC;
        }
        }  
        printf("A string of %d characters has %d permutations of length %d\n", num[1], resultB, num[2]);
          
          }
   } while(doneFlag == 'f');


 return 0;
}

[thomasekugm]

here is the mathmatical formula for calculating permutations if it helps. By the way, I am in the same class and this assignment is a B!%$&.

P(n,k) = n! / (n - k)! =
120

[claudiu]

Well IMO you guys should have a separate function that computes factorial of some generic parameter m.

Then if you want to calculate the factorial of n you just call that function(assume it's called factorial) with the argument n or n-k.

i.e.

long result = factorial(n);
long result2 = factorial(n-k);

[Adak]

If you have a function that will calculate the factorial, then just have the calling function make the subtraction first, and then send that number, to the factorial function.

I'd be glad to help you sort out your problem with this program, but I don't see what problem you are having. Once you have the factorial logic worked out, you've got the heavy lifting done, (don't you?).