Row of a Pascal's Triangle using recursion

Given a positive integer 'm', I'm writing a code to display the m'th row of Pascal's Triangle. By definition, R m (the m'th row) has m elements, being the first and the last elements equal to 1. The remaining elements are computed by the recursive relationship: R m(i) =R m-1(i-1) + R m-1(i) for i = 2,...,m-1. What I've done so far is :

function [row] = PascalRow(m)

PascalRow(1) = 1;
if m == 1
    row = 1;
elseif m == 2;
    row = [1 1];
else
    row = [1,1];

      for i=2:m-1
          row(i) = row(i-1) + row(i);
          row(m)=1;
      end
  end
  end

but I'm getting 1221, 12221, 122221 for m>3. What am I doing wrong? I haven't caught the hang of this recursion thing yet =(.

[Image Analyst]

There's no recursion there. With recursion, PascalRow() would call itself inside itself. And since it's a function, you can't assign 1 to it's output

PascalRow(1) = 1; % Not legal

[Gijs de Wolf]

But what should he have done to fix this?

Answer 1

[John D'Errico]

Simplest is to use conv. In terms of your code, I have no idea what you think the line

PascalRow(1) = 1;

does, but it does not belong in a function called PascalRow.

Anyway, this will suffice, as only a single line change to the rest of your code.

function [row] = PascalRow(m)
if m == 1
    row = 1;
elseif m == 2;
    row = [1 1];
else
    row = [1,1];
      for i=2:m-1
        row = conv([1 1],row);
      end
  end
end

As a test,

PascalRow(12)
ans =
     1    11    55   165   330   462   462   330   165    55    11     1

Could you have done it without use of conv? Of course. For example, this would also have worked, replacing the line I wrote using conv.

        row = [row,0] + [0,row];

One final point is, I would probably describe that first "row" of the triangle as the zeroth row. Why? because we can think of the nth row of Pascal's triangle as the coefficients of the polynomial (x+1)^n. As such, the polynomial (x+1)^0 is 1.