Triangle made with printf, consists of inner frames - c

Question

I posted a question before about a similar idea but with a rectangle,
this time, I got an assignment in which I'm supposed to draw a isosceles right triangle in this way:
I get the size of the triangle's edge as an input from the user;
the triangle's angle will be at the top left corner of the screen, the triangle's frame will be made of the letter a, the inner triangle's frame will be made of b's, the inner frame of the inner traingle of the inner triangle will be made of c and so on...
for example:
aaaaaaaaaaaa
abbbbbbbbba
abccccccba
abcdddcba
abcddcba
abcdcba
abccba
abcba
abba
aba
aa
a
This is what I got so far:
#include <stdio.h>
int main()
{
int length, originalLength;
int counterRows, counterLength;
scanf("%d", &length);
originalLength=length;
for (counterRows=0 ; length > 0; --length, ++counterRows)
{
for(counterLength=0; counterLength<length ;++counterLength)
{
if (counterRows == 0 || counterLength >= originalLength - 2)
printf("%c",'a');
else if (counterRows == 1 || counterRows == originalLength -3)
if (counterLength == 0 || counterLength == length -1)
printf("%c",'a');
else
printf("%c",'a'+1);
else
if (counterLength == 0 || counterLength == length - 1)
printf("a");
else if (counterLength == 1 || counterLength == length - 2)
printf("%c",'a'+1);
else
printf("%c",'a'+2);
}
printf("\n");
}
return 0;
}
and then i noticed that i keep repeating myself and i can't really do this until i reach the letter z. what other better way is there for such algorithm? thanks!

Answer 1

One way to work out a general construction is to observe that there's a triangular pattern within the rows of the triangle. Consider a triangle of size 10, meaning that the top row is 10 characters long. Look at the top four rows:
aaaaaaaaaa
abbbbbbba
abccccba
abcdcba
abccba
abcba
abba
aba
aa
a
There are 10 'a' characters in the top row, 7 'b' characters in the second row, 4 'c' characters in the third row, and 1 'd' character in the fourth row. We can decompose these rows into a middle segment and outer segments:
aaaaaaaaaa aaaaaaaaaa ..........
abbbbbbba .bbbbbbb. a.......a
abccccba ..cccc.. ab....ba
abcdcba ...d... abc.cba
The middle segment has length size-3*i, where size is the length of the top row and i is the row index, starting from 0 at the top. The outer segments consist of characters from 'a' through 'a'+i-1 on the left side, and the same sequence in reverse on the right side.
What about the bottom portion of the triangle? It starts when size-3*i is negative.
abccba
abcba
abba
aba
aa
a
Each row has a rising and falling sequence of characters. In the row of six characters, for example, we can compute each character by sequentially adding 0, 1, 2, 2, 1, 0 to the base character 'a'.
a b c c b a
0 1 2 2 1 0
In the row of five characters, the sequence is 0, 1, 2, 1, 0:
a b c b a
0 1 2 1 0
In general, for a row of length characters, the character at index j has the value 'a'+j or 'a'+length-1-j, whichever is smaller.
Let's illustrate this with a row of six characters.
n = 6:
j = 0 1 2 3 4 5
n-1-j = 5 4 3 2 1 0
min(j, n-1-j) = 0 1 2 2 1 0
Putting it all together, we can generate a triangle with the following code.
for (i = 0; i < size; ++i) {
middle = size-3*i;
if (middle >= 0) { /* Upper portion of the triangle. */
for (j = 0; j < i; ++j) {
printf("%c", 'a'+j); /* Left segment: abc... */
}
for (j = 0; j < middle; ++j) {
printf("%c", 'a'+i); /* Middle: size-3*i characters */
}
for (j = i-1; j >= 0; --j) {
printf("%c", 'a'+j); /* Right segment: ...cba */
}
} else { /* Now we're in the lower portion. */
length = size-i;
for (j = 0; j < length; ++j) {
printf("%c", 'a'+(j < length-1-j ? j : length-1-j));
} /* Rising and falling sequence. */
}
printf("\n");
}
Below is a complete program that gets the size of the triangle from the command line. You can easily modify it to read the desired size with scanf, as in the code you showed in the question.
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char **args) {
int size, middle, length, i, j;
if (argc != 2) {
printf("must enter the triangle size\n");
return 0;
}
size = atoi(args[1]);
if (size < 1 || size > 26) {
printf("the size must be in the range [1, 26]\n");
return 0;
}
printf("size = %d\n\n", size);
for (i = 0; i < size; ++i) {
middle = size-3*i;
if (middle >= 0) { /* Upper portion of the triangle. */
for (j = 0; j < i; ++j) {
printf("%c", 'a'+j); /* Left segment: abc... */
}
for (j = 0; j < middle; ++j) {
printf("%c", 'a'+i); /* Middle: size-3*i characters */
}
for (j = i-1; j >= 0; --j) {
printf("%c", 'a'+j); /* Right segment: ...cba */
}
} else { /* Now we're in the lower portion. */
length = size-i;
for (j = 0; j < length; ++j) {
printf("%c", 'a'+(j < length-1-j ? j : length-1-j));
} /* Rising and falling sequence. */
}
printf("\n");
}
return 0;
}

Answer 2

#include <stdio.h>
#define min(a,b) a < b ? a : b
void print_line(int len, int row, int level){
int deep = min(row, level);
if(len > 1){
putchar('a' + deep);
print_line(len-2, row, level + 1);
putchar('a' + deep);
} else if(len == 1){
putchar('a' + deep);
}
}
int main(){
int length, originalLength;
int i;
scanf("%d", &length);
originalLength=length;
for(i=0; i< originalLength; ++i){
print_line(length--, i, 0);
putchar('\n');
}
return 0;
}