combinatorial function c++ using Recursion - c++

Question

Now this is the combinatorial function if you don't know it:
C(n,k)= { 1 if k=0 or k = n
C(n−1,k−1)+C(n−1,k) otherwise
Now, What I really need is to use recursion to print a Pascal's triangle.
Ok,so what I've done so far is this simple recursion function:
#include <iostream>
using namespace std;
int Pas(int r, int c) {
if (c == 0 || c == r) {
return 1;
} else {
return Pas(r - 1, c - 1) + Pas(r - 1, c);
}
}
int main(){
cout << Pas(4,2) << endl;
return 0;
}
Now this function computes perfectly for example:
Pas(4,2) = 6
But I'm having problem using it to print the whole Pascal's triangle, because I'm new into C++ and especially recursion ..
I'd appreciate any feedback, and I hope that someone would help figure out this problem. But I'd appreciate it more if you guys don't just give me the whole answer (code) just like that; I want to learn.
Thanks!

Answer 1

Something similar to this might to the job
void printTriangle(int printRows, int a = 1, int b = 0)
{
if (a > printRows) return;
int val = Pas(a, b);
cout << val << " ";
if (a == b) {
cout << endl;
printTriangle(printRows, a + 1, 0);
} else {
printTriangle(printRows, a, b + 1);
}
}
Running printTriangle(7) should print the first 7 rows.

Answer 2

Tail recursion is the recursive equivalent to iterative loops. The following function when called with sum(0, 5)
int sum(int start, int end, int resultSoFar = 0) {
if (start == end) return resultSoFar;
return sum(start + 1, end, resultSoFar + start);
}
is equivalent to the iterative function called with sum(0, 5).
int sum(int start, int end) {
int resultSoFar = 0;
for (int i = start; i < end; i++) {
resultSoFar += i;
}
return resultSoFar;
}

Answer 3

If you are allowed to use your recursive function in a loop, the easiest way would be something like:
int n=4;
for (int i=0; i<=n; i++) {
for (int j=0; j<=i; j++) {
cout << Pas(i,j)<<" ";
}
cout <<endl;
}
If you want to reinforce the layout, you coud also #include <iomanip> and <limits> to use a fixed size number output, using the number of digits required to display an integer: just replace the output statement with:
cout << setw(numeric_limits<int>::digits10+1)<<Pas(i,j);
Edit:
You could easily build a recursive function to print lines of the triangle:
void PrintPas(int r) {
if (r==1)
cout << Pas(1,0);
else {
PrintPas(r-1);
for (int c=0; c<=r; c++)
cout << Pas(r,c)<< " ";
}
cout <<endl;
}
Edit 2
If you want a fully recursive version:
void PrintPas(int r, int c) {
if (r==1)
cout << Pas(1,0)<<" ";
else if (c==-1) {
PrintPas(r-1,r-1);
}
else {
PrintPas(r,c-1);
cout << Pas(r,c)<< " ";
}
if (r==c)
cout <<endl;
}
Online demo