asked on

How to calculate this math problem?

/*--------------------------------
| | 7 | | | | 4 |
| | | | | | |
|-------------------------------|
| | | 1 | | | |
| | | | | | |
|-------------------------------|
| | | | | | 8 |
| | | | | | |
|---------------+---------------|
| 5 | | | 6 | | |
| | | | | | |
|-------------------------------|
| | | | | | |
| | | | | | |
|-------------------------------|
| 9 | | 2 | | 3 | |
| | | | | | |
|-------------------------------|

All empty boxes can be filled with digits from 1-9 and the sum of
each row or column equals to 30. In addition, each number
from 1-9 can only be used 4 times including those that
have already been put in place to start you off */

If using a two-demensional array x[7][7] to express this 6x6 box,
then one of the known facts is as below,
x[1][2]=7;
x[1][6]=4;
x[2][3]=1;
x[3][6]=8;
x[4][1]=5;
x[4][4]=6;
x[6][1]=9;
x[6][3]=2;
x[6][5]=3

But I found it very difficult to caculate. The number of possibilities is
about 10**36 !!! Is there any better way to solve this problem?

ozo

I count more like 10**21 possibilities. which a simple backtracking search should prune easily enough.
Why not post your code?
(you might even add more constraints, like getting a 15 sum in each quadrant, and solve it by hand)

Arvindtn

Is the total 30, hardcode ?

shiqi

ASKER

The total possiblities is more like 36 nines time together, that is 9**36 , which is nearly equivelent to 10**36.
I doubt it may take a very long time to calculate this problem. Is there better
solution.

As for 30, the sum of each row elements. Sorry to cause confusion.
Let me explain other given facts.
The sum of all numbers in each column or
row equals to 30. This applys to every column and row. That means,
for (i=1;i<=6;i++){
sum=0;
for (j=1;j<=6;j++) sum=sum+x[i][j];
if(sum!=30){printf("This does not match!\n"); exit}
else
printf("Matched 30 for row %d\n",i);
}
for(j=1;j<=6;j++){
sum=0;
for(i=1;i<=6;i++)sum=sum+x[i][j];
if(sum!=30){printf("This does match!\n"); exit}
else
printf("Matched 30 for column %d\n",j);
}

deighton

2 7 6 2 9 4
9 5 1 7 5 3
4 3 8 6 1 8
5 3 7 6 7 2
1 8 6 1 5 9
9 4 2 8 3 4

deighton

Here's code in .c rather than VB which I solved it in

#include<stdio.h>

main()
{

int digits[10],c[10],babort,row1,row2,row3,col1,col2,col3;
int i,a1,a2,a3,a4,a5,a6,a7,a8,a9;
int sq[37],q,q1,q2;

for(a1=1;a1<10;a1++)
{
for(a2=7;a2<=7;a2++)
{
for(a3=1;a3<10;a3++)
{
for(a4=1;a4<10;a4++)
{
for(a5=1;a5<10;a5++)
{
for(a6=1;a6<=1;a6++)
{
for(a7=1;a7<10;a7++)
{
for(a8=3;a8<=3;a8++)
{
for(a9=1;a9<10;a9++)
{

      digits[1] = a1;
      digits[2] = a2;
      digits[3] = a3;
      digits[4] = a4;
      digits[5] = a5;
      digits[6] = a6;
      digits[7] = a7;
      digits[8] = a8;
      digits[9] = a9;

      for(i=1;i<10;i++)
       c[i] = 0;

      babort = 0;

      for(i=1;i<10;i++)
      {
       ++c[digits[i]];
       if (c[digits[i]] == 2)
       {
            babort = 1;
            break;
       }
      }

      if (babort == 0)
      {
       row1 = digits[1] + digits[2] + digits[3];
       row2 = digits[4] + digits[5] + digits[6];
       row3 = digits[7] + digits[8] + digits[9];
       col1 = digits[1] + digits[4] + digits[7];
       col2 = digits[2] + digits[5] + digits[8];
       col3 = digits[3] + digits[6] + digits[9];

       if (row1 == 15 && row2 == 15 && row3 == 15
       && col1 == 15 && col2 == 15 && col3 == 15)
       {

             for(q1=0;q1<3;q1++){
             for(q2=0;q2<3;q2++){
              sq[q2*6 + q1 + 1] = digits[q2*3+q1+1];
             }}

       }

      }

}}}}}}}}}

for(a1=1;a1<10;a1++)
{
for(a2=1;a2<10;a2++)
{
for(a3=4;a3<=4;a3++)
{
for(a4=1;a4<10;a4++)
{
for(a5=1;a5<10;a5++)
{
for(a6=1;a6<10;a6++)
{
for(a7=1;a7<10;a7++)
{
for(a8=1;a8<10;a8++)
{
for(a9=8;a9<=8;a9++)
{

      digits[1] = a1;
      digits[2] = a2;
      digits[3] = a3;
      digits[4] = a4;
      digits[5] = a5;
      digits[6] = a6;
      digits[7] = a7;
      digits[8] = a8;
      digits[9] = a9;

      for(i=1;i<10;i++)
       c[i] = 0;

      babort = 0;

      for(i=1;i<10;i++)
      {
       ++c[digits[i]];
       if (c[digits[i]] == 2)
       {
            babort = 1;
            break;
       }
      }

      if (babort == 0)
      {
       row1 = digits[1] + digits[2] + digits[3];
       row2 = digits[4] + digits[5] + digits[6];
       row3 = digits[7] + digits[8] + digits[9];
       col1 = digits[1] + digits[4] + digits[7];
       col2 = digits[2] + digits[5] + digits[8];
       col3 = digits[3] + digits[6] + digits[9];

       if (row1 == 15 && row2 == 15 && row3 == 15
       && col1 == 15 && col2 == 15 && col3 == 15)
       {
             for (q = 1; q<10;q++)
              sq[((q-1) / 3) * 6 + (q-1) % 3 + 4] = digits[q];
       }

      }

}}}}}}}}}

for(a1=5;a1<=5;a1++)
{
for(a2=1;a2<10;a2++)
{
for(a3=1;a3<10;a3++)
{
for(a4=1;a4<10;a4++)
{
for(a5=1;a5<10;a5++)
{
for(a6=1;a6<10;a6++)
{
for(a7=9;a7<=9;a7++)
{
for(a8=1;a8<10;a8++)
{
for(a9=2;a9<=2;a9++)
{

      digits[1] = a1;
      digits[2] = a2;
      digits[3] = a3;
      digits[4] = a4;
      digits[5] = a5;
      digits[6] = a6;
      digits[7] = a7;
      digits[8] = a8;
      digits[9] = a9;

      for(i=1;i<10;i++)
       c[i] = 0;

      babort = 0;

      for(i=1;i<10;i++)
      {
       ++c[digits[i]];
       if (c[digits[i]] == 2)
       {
            babort = 1;
            break;
       }
      }

      if (babort == 0)
      {
       row1 = digits[1] + digits[2] + digits[3];
       row2 = digits[4] + digits[5] + digits[6];
       row3 = digits[7] + digits[8] + digits[9];
       col1 = digits[1] + digits[4] + digits[7];
       col2 = digits[2] + digits[5] + digits[8];
       col3 = digits[3] + digits[6] + digits[9];

       if (row1 == 15 && row2 == 15 && row3 == 15
       && col1 == 15 && col2 == 15 && col3 == 15)
       {
             for (q = 1; q<10;q++)
              sq[((q-1) / 3 +3) * 6 + (q-1) % 3 + 1] = digits[q];
       }

      }

}}}}}}}}}

for(a1=6;a1<=6;a1++)
{
for(a2=1;a2<10;a2++)
{
for(a3=1;a3<10;a3++)
{
for(a4=1;a4<10;a4++)
{
for(a5=1;a5<10;a5++)
{
for(a6=1;a6<10;a6++)
{
for(a7=1;a7<10;a7++)
{
for(a8=3;a8<=3;a8++)
{
for(a9=1;a9<10;a9++)
{

      digits[1] = a1;
      digits[2] = a2;
      digits[3] = a3;
      digits[4] = a4;
      digits[5] = a5;
      digits[6] = a6;
      digits[7] = a7;
      digits[8] = a8;
      digits[9] = a9;

      for(i=1;i<10;i++)
       c[i] = 0;

      babort = 0;

      for(i=1;i<10;i++)
      {
       ++c[digits[i]];
       if (c[digits[i]] == 2)
       {
            babort = 1;
            break;
       }
      }

      if (babort == 0)
      {
       row1 = digits[1] + digits[2] + digits[3];
       row2 = digits[4] + digits[5] + digits[6];
       row3 = digits[7] + digits[8] + digits[9];
       col1 = digits[1] + digits[4] + digits[7];
       col2 = digits[2] + digits[5] + digits[8];
       col3 = digits[3] + digits[6] + digits[9];

       if (row1 == 15 && row2 == 15 && row3 == 15
       && col1 == 15 && col2 == 15 && col3 == 15)
       {
              for(q1=0;q1<3;q1++){
              for(q2=0;q2<3;q2++){
                   sq[q2*6 + q1 + 18+4] = digits[q2*3+q1+1];
                         }}

       }

      }

}}}}}}}}}

for (i=0;i<6;i++)
{
printf("\n");
for (q=0;q<6;q++){
printf("%i ",sq[i*6+q+1]);
}}

getch();
}

deighton

deighton changed the proposed answer to a comment

ASKER CERTIFIED SOLUTION

deighton

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See answer

shiqi

ASKER

Are you adopting an algorithm to break 6x6 matrix into four separated
3x3 matrix and making an assumption that the sum of each row element and the sum of each column element must equal 15 within each smaller matrix? Will this lose some other possibilities? I ran you program and
only got one choice. Is there only one answer for this question? Are there any other alternatives to solve this problem?
Can you explain your program in detail?