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# How to calculate this math problem?

Posted on 2000-02-18
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/*--------------------------------
|     |  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?

0
Question by:shiqi
[X]
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LVL 84

Expert Comment

ID: 2537516
I count more like 10**21 possibilities. which a simple backtracking search should prune easily enough.
(you might even add more constraints, like getting a 15 sum in each quadrant, and solve it by hand)
0

Expert Comment

ID: 2537873
Is the total 30, hardcode ?
0

Author Comment

ID: 2541318
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);
}
0

LVL 18

Expert Comment

ID: 2542051
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
0

LVL 18

Expert Comment

ID: 2542439
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();
}
0

LVL 18

Expert Comment

ID: 2542569
deighton changed the proposed answer to a comment
0

LVL 18

Accepted Solution

deighton earned 80 total points
ID: 2542573
#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();
}

0

Author Comment

ID: 2553598
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?

0

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