# Printing concentric squares of numbers

I wrote a code which prints the following pattern :

5 5 5 5 5 5 5 5 5
5 4 4 4 4 4 4 4 5
5 4 3 3 3 3 3 4 5
5 4 3 2 2 2 3 4 5
5 4 3 2 1 2 3 4 5
5 4 3 2 2 2 3 4 5
5 4 3 3 3 3 3 4 5
5 4 4 4 4 4 4 4 5
5 5 5 5 5 5 5 5 5


The code is as follows :

#include <stdio.h>

void main(){
int n,i,k,b,a;
printf("ENTER OUTER NUMBER:");
scanf("%d",&n);
for(i=0;i<2*n-1;i++){
printf("%d",n);
}
printf("\n");
int p=n;
for(i=0;i<n-1;i++){
int a=0;
for(k=0;k<=i;k++){
printf("%d",n-a);
a++;
}
for(b=0;b>i;b++){
printf(" ");
}
for(k=1;k<=2*p-3;k++){
printf("%d",n-i-1);
}
p--;
for(k=i;k>=0;k--){
printf("%d",n-k);
}
printf("\n");
}
int q=1;
for(i=0;i<n-2;i++){
for(k=n;k>i+1;k--){
printf("%d",k);
}
for(k=1;k<=2*q-1;k++){
printf("%d",q+1);
}
q++;
for(k=2+i;k<=n;k++){
printf("%d",k);
}
printf("\n");
}
for(i=0;i<2*n-1;i++){
printf("%d",n);
}
getchar();
}


I printed the first and last line independently.Then,I divided the pattern into two halves and considered some common triangular patterns as follows:

5                 5    2 3 4 5
5 4             4 5      3 4 5
5 4 3         3 4 5        4 5
5 4 3 2     2 3 4 5          5


The code ran well in the CODE BLOCKS IDE.But, I think it's too long and I complicated the coding.My query is:

1. Can I shorten the code with the similar logic of breaking the pattern into parts or mine is ok ?

2. Is there any alternative to this ?

• Use a bit more spaces. It will make your code more readable.

• Try to be coherent with your indents

• You should sanitize input. What happen if user enter a number greater than 9 ? Or 0, 1, or 2? Or negative number? Or not a number? Never trust user, They all try to broke your program.

• When you just print on char, prefer putchar() instead of printf().

Here's my solution, maybe not the most optimal, but clean and short. I compute the distance from the current coordinate to nearest side and then remove it to the base number.

#include <stdio.h>
int main(void) {
int n = 5;
int a;
int b;
const int m = 2*n-1;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < m; ++j) {
a = (i >= n) ? m - i - 1: i;
b = (j >= n) ? m - j - 1 : j;
putchar('0' + n - ((a < b) ? a : b));
}
putchar ('\n');
}
return 0;
}


To improve this, you could split the inner loop to avoid the two conditional assignments. I let you trying to do it yourself.

The same pattern can be produced by a single instance of nested for loops, where outer loop variable is line and inner one - horizontal position.

I do not want to spoil your fun in finding the function at the heart of the loop (at least yet), which given two coordinates produces the value at those coordinates.

• Did you speak about the method as mentioned in the abovw answer? Nov 14, 2018 at 2:25
• Yes. The method above is that. Nov 14, 2018 at 4:07