Diamond made of numbers in C++ using only recursion? - c++

Question

I currently have code that displays a number triangle:
#include <iostream>
using namespace std;
void RowNumbers(int n, int max) {
if (n < max) {
cout << n << ' ';
RowNumbers(n + 1, max);
}
cout << n << ' ';
}
void PrintRhombus(int n, int space = 0) {
if (n > 1) {
PrintRhombus(n - 1, space + 2);
}
cout << string(space, ' ');
RowNumbers(1, n);
cout << "\n";
}
int main() {
int a;
cout << "Enter a number [1-9]: " << endl;
cin >> a;
if (a > 0 && a < 10) {
PrintRhombus(a, 0);
}
else
cout << "Wrong input." << endl;
}
The output is displayed here if the input is two.
1
1 2 1
This is only half of what I want. I want to make a full diamond, so I would want a "1" as the third line of this output. The RowNumbers function generates each line, and the PrintRhombus function is responsible for creating the number triangle. I only want to use recursions to create the bottom half of the triangle. So should I create another recursion function that creates the bottom half of the diamond? Or should I build upon the PrintRhombus function, and start something else when the value reaches one.
So once again my goal is to create a number diamond, so if you input 3, then the output is:
1
1 2 1
1 2 3 2 1
1 2 1
1

Answer 1

Here's what I would do:
First, write PrintRhombus in the form of a single loop:
void PrintRhombus(int n) {
for(int i = 1; i < 2 * n; i++){
int numbers_in_line = i < n ? i : 2 * n - i;
int space = (n - numbers_in_line) * 2;
cout << string( space, ' ' );
RowNumbers( 1, numbers_in_line );
cout << endl;
}
}
Test your loop and make sure you got it right. It's usually much easier to reason about a loop than a recursive function.
Now transform the loop into a tail-recursive function. This transformation is trivial to perform:
void PrintRhombus(int n, int i = 1){
// ^ loop variable --> extra parameter
if(i == 2 * n) return; // <- terminating condition --> return
// loop body remains unchanged
int numbers_in_line = i < n ? i : 2 * n - i;
int space = (n - numbers_in_line) * 2;
cout << string( space, ' ' );
RowNumbers( 1, numbers_in_line );
cout << endl;
// Now do recursive call with incremented loop variable
PrintRhombus(n, i+1);
}
Demo.

Answer 2

It is because you only iterate to the top of the rhombus, then print out the top half. I could not find a solution that used only one function, but here are two functions, one for the top and one for the bottom, that work:
void PrintRhombusTop(int n, int space = 0)
{
if (n > 1)
{
PrintRhombusTop(n - 1, space + 2);
}
cout << string(space, ' ');
RowNumbers(1, n);
cout << "\n";
}
void PrintRhombusBottom(int n, int space)
{
if (n >= 1)
{
cout << string(space, ' ');
RowNumbers(1, n);
cout << "\n";
PrintRhombusBottom(n - 1, space + 2);
}
}
Call them in main like this:
if (a > 0 && a < 10)
{
PrintRhombusTop(a, 0);
PrintRhombusBottom(a - 1, 2);
}
You need to pass (a - 1, 2) to PrintRhombusBottom so you don't print the middle row twice and so the row after the middle row is indented correctly.

Answer 3

to follow the same logic as RowNumber, you probably want something like that:
void PrintRhombus(int min, int max)
{
if (min < max)
{
RowNumber(1, min);
PrintRhombus(min+1, max);
RowNumber(1, min);
}
else
RowNumber(1, max);
}
Just add the necessary spaces

Answer 4

Another one for fun:
#include <iostream>
#include <string>
void row(int n, int m) {
std::cout << n << " ";
if (n < m) {
row(n + 1, m);
std::cout << n << " "; } }
void line(int n, int r, int m) {
std::cout << std::string(2 * (m - n), ' ');
row(1, r);
std::cout << "\n"; }
void diamond(int n, int r, int m) {
line(n, r, m);
if (r < m) {
diamond(n + 1, r + 1, m);
line(n, r, m); } }
int main(int, char* []) {
std::cout << "Number? ";
int n = 0;
std::cin >> n;
diamond(1, 1, n);
return 0; }