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Monday, February 13, 2012

10 Challenging number pattern programs in C

10 Challenging number pattern programs in C



 
 
 
 
 
 
26 Votes
This is a continuation to the series of challenging pattern C programs in Interview Mantra. This set of 10 puzzling programs are of type number patterns.

Also read Print Pattern Programs

  1. Write a C program to print the following pattern: 
    1
0 1
1 0 1
0 1 0 1
1 0 1 0 1
  2. Write a C program to print the following pattern: 
    0
1 1
2 3 5
8 13 21
  3. Write a C program to print the following pattern: 
    1
121
12321
1234321
12321
121
1
  4. Write a C program to print the following pattern: 
           2
     4 5 6
   6 7 8 9 10
     4 5 6
       2
  5. Write a C program to print the following pattern: 
     1                         1
 3 3 3                 3 3 3
 5 5 5 5 5         5 5 5 5 5
 7 7 7 7 7 7 7 7 7 7 7 7 7 7
 5 5 5 5 5         5 5 5 5 5
 3 3 3                 3 3 3
 1                         1
  6. Write a C program to print the following pattern: 
        0
 -2-3 0
-4-3-2-1 0
 -2-3 0
    0
  7. Write a C program to print the following pattern: 
    77777777777
         7
        7
       7
      7
     7
    7
   7
  7
 7
7
  8. Write a C program to print the following pattern: 
     1                       1
 1 0                   1 0
 1 0 1               1 0 1
 1 0 1 0           1 0 1 0
 1 0 1 0 1       1 0 1 0 1
 1 0 1 0 1 0   1 0 1 0 1 0
 1 0 1 0 1 0 1 0 1 0 1 0 1
 1 0 1 0 1 0   1 0 1 0 1 0
 1 0 1 0 1       1 0 1 0 1
 1 0 1 0           1 0 1 0
 1 0 1               1 0 1
 1 0                   1 0
 1                       1
  9. Write a C program to print the following pattern: 
    1
2 4
3 6 9
2 4
1
  10. Write a C program to print the following pattern: 
     1
 1 0
 1 0 0
 1 0 0 0
 1 0 0 0 0
 1 0 0 0 0 0
 1 0 0 0 0 0 0
 1 0 0 0 0 0
 1 0 0 0 0
 1 0 0 0
 1 0 0
 1 0
 1
  1. Write a C program to print the following pattern:
    2. 1
0 1
1 0 1
0 1 0 1
1 0 1 0 1
    Program:
    #include <stdio.h>

int main(void) {
 int i, j;
 for (i = 0; i < 4; i++) {
  for (j = 0; j <= i; j++) {
   if (((i + j) % 2) == 0) {  // Decides on as to which digit to print.
    printf("0");
   } else {
    printf("1");
   }
   printf("\t");
  }
  printf("\n");
 }
 return 0;
}
    Download Code
    Explanation: This is a right angle triangle composed of 0′s and 1′s.
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    End of Question1 Start of Question2
  2. Write C program to print the following pattern: 
    0
1 1
2 3 5
8 13 21
    3. Program:
    #include <stdio.h>

int main(void) {
 int i, j, a = 0, b = 1, temp = 1;
 for (i = 1; i <= 4; i++) {
  for (j = 1; j <= i; j++) {
   if (i == 1 && j == 1) { // Prints the '0' individually first
    printf("0");
    continue;
   }
   printf("%d ", temp);  // Prints the next digit in the series
   //Computes the series
   temp = a + b;
   a = b;
   b = temp;
   if (i == 4 && j == 3) { // Skips the 4th character of the base
    break;
   }
  }
  printf("\n");
 }
 return 0;
}
    Download Code
    Explanation: This prints the Fibonacci series in a right angle triangle formation where the base has only three characters.
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    End of Question2 Start of Question3
  3. Write C program to print the following pattern: 
    1
121
12321
1234321
12321
121
1
    4. Program:
    #include <stdio.h>

void sequence(int x);
int main() {
 /* c taken for columns */
 int i, x = 0, num = 7;
 for (i = 1; i <= num; i++) {
  if (i <= (num / 2) + 1) {
   x = i;
  } else {
   x = 8 - i;
  }
  sequence(x);
  puts("\n");
 }
 return 0;
}

void sequence(int x) {
 int j;

 for (j = 1; j < x; j++) {
  printf("%d", j);
 }
 for (j = x; j > 0; j--) {
  printf("%d", j);
 }
}
    Download Code
    Back to top
    End of Question3 Start of Question4
  4. Write a C program to print the following pattern: 
           2
     4 5 6
   6 7 8 9 10
     4 5 6
       2
    5. Program:
    #include <stdio.h>

int main(void) {
 int prnt;
 int i, j, k, r, s, sp, nos = 3, nosp = 2; //nos n nosp controls the spacing factor
 // Prints the upper triangle
 for (i = 1; i <= 5; i++) {
  if ((i % 2) != 0) {
   for (s = nos; s >= 1; s--) {
    printf("  ");
   }
   for (j = 1; j <= i; j++) {
    if (i == 5 && j == 5) { //Provides the extra space reqd betn 9 n 10
     printf(" ");        // as 10 is a 2 digit no.
    }
    prnt = i + j;
    printf("%2d", prnt);
   }
  }
  if ((i % 2) != 0) {
   printf("\n");
   nos--;
  }
 }
 // Prints the lower triangle skipin its base..
 for (k = 3; k >= 1; k--) {
  if ((k % 2) != 0) {
   for (sp = nosp; sp >= 1; sp--) {
    printf("  ");
   }
   for (r = 1; r <= k; r++) {
    prnt = k + r;
    printf("%2d", prnt);
   }
  }
  if ((k % 2) != 0) {
   printf("\n");
   nosp++;
  }
 }
 return 0;
}
    Download Code
    Explanation: This is a diamond formation composed of numbers. The numbers are in the following order next_no=i+j where
    next_no = The next no to be printed
    i = index of the outer for loop
    j = index of the inner for loop
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    End of Question4 Start of Question5
  5. Write a C program to print the following pattern: 
     1                           1
 3 3 3                   3 3 3
 5 5 5 5 5           5 5 5 5 5
 7 7 7 7 7 7 7   7 7 7 7 7 7 7
 5 5 5 5 5           5 5 5 5 5
 3 3 3                   3 3 3
 1                           1
    6. Program:
    #include <stdio.h>

int main(void) {
 int i, j, k, s, p, q, sp, r, c = 1, nos = 13;
 for (i = 1; c <= 4; i++) {
  if ((i % 2) != 0) {   // Filters out the even line nos.
   for (j = 1; j <= i; j++) { // The upper left triangle
    printf("%2d", i);
   }
   for (s = nos; s >= 1; s--) {  // The spacing factor
    printf("  ");
   }
   for (k = 1; k <= i; k++) { // The upper right triangle
    printf("%2d", i);
   }
   printf("\n");
   nos = nos - 4;  // Space control
   ++c;
  }
 }
 nos = 10;  // Space control re intialized
 c = 1;
 for (p = 5; (c < 4 && p != 0); p--) {
  if ((p % 2) != 0) {  // Filters out the even row nos
   for (q = 1; q <= p; q++) {  // Lower left triangle
    printf("%2d", p);
   }
   for (sp = nos; sp >= 1; sp--) { // Spacing factor
    printf(" ");
   }
   for (r = 1; r <= p; r++) {  // Lower right triangle
    printf("%2d", p);
   }

   printf("\n");
   --c;
   nos = nos + 8;  // Spacing control.
  }
 }

 return 0;
}
    Download Code
    Explanation: Here we are printing only the odd row nos along with thier respective line number. This structure can divided into four identical right angle triangles which are kind of twisted and turned placed in a particular format .
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    End of Question5 Start of Question6
  6. Write a C program to print the following pattern: 
        0
 -2-3 0
-4-3-2-1 0
 -2-3 0
    0
    7. Program:
    #include <stdio.h>

int main(void) {
 int i, j, k, r, s, sp, nos = 2, nosp = 1;
 for (i = 1; i <= 5; i++) {
  if ((i % 2) != 0) {
   for (s = nos; s >= 1; s--) {  //for the spacing factor.
    printf("  ");
   }
   for (j = 1; j <= i; j++) {
    printf("%2d", j-i);
   }
  }
  if ((i % 2) != 0) {
   printf("\n");
   nos--;
  }
 }
 for (k = 3; k >= 1; k--) {
  if ((k % 2) != 0) {
   for (sp = nosp; sp >= 1; sp--) {  // for the spacing factor.
    printf("  ");
   }
   for (r = 1; r <= k; r++) {
    printf("%2d", r-k);
   }
  }
  if ((k % 2) != 0) {
   printf("\n");
   nosp++;
  }
 }
 return 0;
}
    Download Code
    Explanation:This can be seen as a diamond composed of numbers. If we use the conventional nested for loop for its construction the numbers can be seen to flowing the following function f(x) -> j-i
    where
    j= inner loop index
    i= outer loop index
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    End of Question6 Start of Question7
  7. Write a C program to print the following pattern: 
    77777777777
         7
        7
       7
      7
     7
    7
   7
  7
 7
7
    8. Program:
    #include <stdio.h>

int main(void) {
 int i, j;
 for (i = 11; i >= 1; i--) {
  for (j = 1; j <= i; j++) {
   if (i == 11) {
    printf("7");  // Makes sure the base is printed completely
    continue;
   } else if (j == i) { // Hollows the rest
    printf("7");
   } else {
    printf(" ");
   }
  }
  printf("\n");
 }
 return 0;
}
    Download Code
    Explanation: This can be seen as a hollow right-angled triangle composed of 7′s
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    End of Question7 Start of Question8
  8. Write a C program to print the following pattern: 
     1                       1
 1 0                   1 0
 1 0 1               1 0 1
 1 0 1 0           1 0 1 0
 1 0 1 0 1       1 0 1 0 1
 1 0 1 0 1 0   1 0 1 0 1 0
 1 0 1 0 1 0 1 0 1 0 1 0 1
 1 0 1 0 1 0   1 0 1 0 1 0
 1 0 1 0 1       1 0 1 0 1
 1 0 1 0           1 0 1 0
 1 0 1               1 0 1
 1 0                   1 0
 1                       1
    9. Program:
    #include <stdio.h>

int main(void) {
    int i,j,k,s,nos=11;
    for (i=1; i<=7; i++) {
        for (j=1; j<=i; j++) {
            if ((j%2)!=0) {   // Applying the condition
                printf(" 1");
            } else {
                printf(" 0");
            }
        }
        for (s=nos; s>=1; s--) {  // Space factor
            printf("  ");
        }
        for (k=1; k<=i; k++) {
            if(i==7 && k==1)  // Skipping the extra 1
            {
                continue;
            }
            if ((k%2)!=0) {  // Applying the condition
                printf(" 1");
            } else {
                printf(" 0");
            }
        }
        printf("\n");
        nos=nos-2;  // Space Control
    }
     nos=1;
     for ( i=6; i>=1; i--) {  // It shares the same base
         for (j=1; j<=i; j++) {
             if (j%2!=0) {
                 printf(" 1");
             } else {
                 printf(" 0");
             }
         }
         for(s=nos; s>=1; s--)  // Spacing factor
         {
             printf("  ");
         }
         for (k=1; k<=i; k++) {
             if (k%2!=0) {
                 printf(" 1");
             } else {
                 printf(" 0");
             }
         }
         printf("\n");
         nos=nos+2;
     }
    return 0;
}
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    Back to top
    End of Question8 Start of Question9
  9. Write a C program to print the following pattern: 
    1
2 4
3 6 9
2 4
1
    10. Program:
    #include <stdio.h>
int main(void) {
    int i,j;
    for (i=1; i<=3 ; i++) {
        for (j=1; j<=i; j++)  {
            printf("%2d", (i*j));
        }
        printf("\n");
    }
    for (i=2; i>=1; i--) { // As they share the same base
        for (j=1; j<=i; j++)  {
            printf("%2d",i*j);
        }
        printf("\n");
    }
    return 0;
}
    Download Code
    Explanation: This can be seen as two right angle triangles sharing th same base
    The numbers are following the following function f(x) = i *j
    where
    i = Index of the Outer loop
    j = Index of the inner loop
    Back to top
    End of Question9 Start of Question10
  10. Write a C program to print the following pattern: 
     1
 1 0
 1 0 0
 1 0 0 0
 1 0 0 0 0
 1 0 0 0 0 0
 1 0 0 0 0 0 0
 1 0 0 0 0 0
 1 0 0 0 0
 1 0 0 0
 1 0 0
 1 0
 1
    11. Program:
    #include <stdio.h>

int main(void) {
    int i,j;
    for (i=1; i<=7; i++) {
        for (j=1; j<=i; j++) {
            if (j==1) {           // Applying the condition
                printf(" 1");
            } else {
                printf(" 0");
            }
        }
        printf("\n");
    }
    for (i=6; i>=1; i--) {  //As it shares the same base i=6
        for (j=1; j<=i; j++) {
            if (j==1) {   // Applying the condition
                printf(" 1");
            } else {
                printf(" 0");
            }
        }
        printf("\n");
    }
    return 0;
}
    Download Code
    Explanation: This can be seen as two right angle triangles sharing the same base which is composed of 0′s n 1′s. The first column is filled with 1′s and rest with 0′s
    Back to top End of Question10
Posted by at 2/13/2012 08:00:00 PM

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