Answer / c.p.senthil

Generic solution:

n = 5
loop 1: i = 0 to 4
loop 2: i = 5 to 0

Loop variables => i j k j exp k exp
* 0 5 spaces, 1 stars (5-0) (2*0)+1
*** 1 4 spaces, 3 stars (5-1) (2*1)+1
***** 2 3 spaces, 5 stars (5-2) (2*2)+1
******* 3 2 spaces, 7 stars (5-3) (2*3)+1
********* 4 1 spaces, 9 stars (5-4) (2*4)+1
*********** 5 0 spaces, 11 stars(5-5) (2*5)+1
********* 4 1 spaces, 9 stars (5-4) (2*4)+1
******* 3 2 spaces, 7 stars (5-3) (2*3)+1
***** 2 3 spaces, 5 stars (5-2) (2*2)+1
*** 1 1 spaces, 3 stars (5-1) (2*1)+1
* 0 5 spaces, 1 stars (5-0) (2*0)+1

generalising expressions => (n-i) (2*i)+1

void printPattern(int n)
{
int i, j, k;

for(i=0; i<n; i++)
{
for(j=0; j<=(n-i); j++)
printf(" ");

for(k=0; k<(2*i+1); k++)
printf("*");

printf("\n");
}

for(i=n; i>=0; i--)
{
for(j=0; j<=(n-i); j++)
printf(" ");

for(k=0; k<(2*i+1); k++)
printf("*");

printf("\n");
}
}

Answer / arka bandyopadhyay

#include<stdio.h>
#include<conio.h>
void main()
{ int i,j,k,a=4;

k=1;
clrscr();
printf("\n");
for(i=8;i>=0;i--)
{
for(j=0;j<=8 ;j++)


{
if(i>=4)
{
if(j<a ||j>(9-a-1) )
printf(" ");
else
printf(" *");



}
else
{
if(j>(8-k)||j< k)
printf(" ");
else
printf(" *");
}
}
--a;

if(i<4)k++;
printf("\n");
}


getch();
}

Answer / c.p.senthil

description for previous answer posted again

n = 5
loop 1: i = 0 to 4
loop 2: i = 5 to 0

Loop variables => i j k j exp k exp
.....*............0 5 spaces, 1 stars (5-0) (2*0)+1
....***...........1 4 spaces, 3 stars (5-1) (2*1)+1
...*****..........2 3 spaces, 5 stars (5-2) (2*2)+1
..*******.........3 2 spaces, 7 stars (5-3) (2*3)+1
.*********........4 1 spaces, 9 stars (5-4) (2*4)+1
***********.......5 0 spaces, 11 stars(5-5) (2*5)+1
.*********........4 1 spaces, 9 stars (5-4) (2*4)+1
..*******.........3 2 spaces, 7 stars (5-3) (2*3)+1
...*****..........2 3 spaces, 5 stars (5-2) (2*2)+1
....***...........1 1 spaces, 3 stars (5-1) (2*1)+1
.....*............0 5 spaces, 1 stars (5-0) (2*0)+1

..........generalising expressions => (n-i) (2*i)+1

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