Question

I have to write a C program (for my Discrete Mathematics assignment) that finds the number of onto functions from set A (|A| = m) to set B (|B|=n) and to display all those functions. Number of onto functions I calculated using the code:

for(k=0; k<n; k++)
    x = x + pow( -1, k) * combination( n, n - k ) * pow( ( n - k ), m);

Where combination is a function that finds the Number of possible combinations.

For example if A = {1,2,3}, B={a,b,c} then the number of onto functions evaluated from the formula is
3^3 – 3(2^3) + 3 = 6.
One possible solution is f = {(1,a),(2,b),(3,c)} [I know this is a solution].

But my problem is: How to display each and every solution!?
This is just a trivial example. But if m and n values are increased (provided m>=n) then the number of possible onto functions increases exponentially!!
For example if m=7 and n=4 there are 8400 functions!

I can’t think of any method to display each and every function that exists between A and B.

Answer 1

I answered a similar problem sometime ago but m and n were equal m = n.(You must think recursively to solve this), by your comment I think the possible answers are: {(1,a)(2,b)(3,c)}, {(2,a)(3,b)(1,c)}, {(3,a)(1,b)(2,c)}, {(3,a)(2,b)(1,c)}, {(2,a)(1,b)(3,c)} and {(1,a)(3,b)(2,c)} then this is my recipe:

1. Set 2 arrays with their initial value let’s call them letters and numbers.

   *---*---*---*                          *---*---*---*
   | a | b | c | <---letters.             | 1 | 2 | 3 | <---numbers.
   *---*---*---*                          *---*---*---*
   

2. Choose one of the arrays to be your pivot, I chose the letters, it will be statical.

   *---*---*---*                          *---*---*---*
   | a | b | c | <---STATIC.              | 1 | 2 | 3 | <---DYNAMIC.
   *---*---*---*                          *---*---*---*
   

3. Rotate the dynamic array counter-clockwise or clockwise as you wish, you must print the i element of numbers with the i element of letters.

   *---*---*---*               *---*---*---*                 *---*---*---* 
   | 1 | 2 | 3 |   -(Print)->  | 2 | 3 | 1 |    -(Print)->   | 3 | 1 | 2 |
   *---*---*---*               *---*---*---*                 *---*---*---*
   

So you get at this point: {(1,a)(2,b)(3,c)}, {(2,a)(3,b)(1,c)}, {(3,a)(1,b)(2,c)}, 3 are missing.

1. Swap the i element with the n element of the dynamic array.

   *---*---*---*                                     *---*---*---*
   | 1 | 2 | 3 |   ---------( Swap (0<->2) )-------> | 3 | 2 | 1 | 
   *---*---*---*                                     *---*---*---*
   

2. Repeat the step 3.

   *---*---*---*               *---*---*---*                 *---*---*---* 
   | 3 | 2 | 1 |   -(Print)->  | 2 | 1 | 3 |    -(Print)->   | 1 | 3 | 2 |
   *---*---*---*               *---*---*---*                 *---*---*---*
   

So you get the missed subsets:{(3,a)(2,b)(1,c)}, {(2,a)(1,b)(3,c)} and {(1,a)(3,b)(2,c)}.

If you have more than 3 example 4. Easy: 1234 (rotate N times where N is the number of variables and print with each movement), swap 1 and 4 -> 4231 (Rotate and Print), swap 2 and 3 -> 4321 (Rotate and Print), swap 4 and 1 –> 1324 (Rotate and Print).

I hope this helped.