Can't Understand this line in Pascal's Triangle for C++

Question

I have to write a code that generates a pascal's triangle with 12 rows.

I've written everything on my own except one part, That's the formula we use to generate numbers. And the problem is I don't understand what's the connection between our counters and generated numbers (since we're using our counters.).

#include <iostream>
#include <string>

using namespace std;

int main() {
    const int rows=12;
    int padding, value, fxValue;

    for(int rowCounter=0; rowCounter<rows; rowCounter++)
    {
        fxValue=1;
        cout << string((rows-rowCounter)*6, ' ');

        for(int fxCounter=0; fxCounter<=rowCounter; fxCounter++)
        {
            value=fxValue;
            fxValue = fxValue*(rowCounter-fxCounter)/(fxCounter+1);

//          cout << "fxCounter: "<< fxCounter << endl
//               << "rowCounter: " << rowCounter << endl
//               << "fxCounter: " << fxCounter << endl
//               << "fxValue: " << fxValue << endl;

            padding=fxValue/10;

            if(padding==0) cout << value << string(11, ' ');
            else if(10>padding) cout << value << string(10, ' ');
            else if(padding>10) cout << value << string(9, ' ');
        }
        cout << endl;
    }
    return 0;
}

Here's the problem:

fxValue = fxValue*(rowCounter-fxCounter)/(fxCounter+1);

Can someone please explain how did the author came up with the idea of using these variables and how it works fine?

`I've written everything on my own except one part, That's the formula we use to generate numbers` Isn't that part the most important part in your assignment? — PaulMcKenzie, Jan 06 '15 at 10:43

score 0 · Accepted Answer · edited Jan 06 '15 at 11:14

0

This works because Pascal's triangle can be represented with binomial coefficients :

enter image description here

This formula in your code is based on the fact that, on the same n-index (in the pascal's triangle case, same line), in order to get the next element (k -> k+1), we need to multiply the current value by (n-k)/(k+1):

enter image description here

It's fairly easy to prove if you want to convince yourself about that. So you can get the next value from the previous one with this operation.

edited Jan 06 '15 at 11:14

Veritas

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answered Jan 06 '15 at 10:36

Dimitri Mockelyn

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Note that this is one case where left-associativity of integer multiplicative operations is important, and (stating the same thing differently) `fxValue *= (rowCounter-fxCounter)/(fxCounter+1);` would **not** give the correct result (the division then is usually not exact). So "multiply the current value by (n-k)/(k+1)" should be taken in the mathemaical, not computational, sense. – Marc van Leeuwen Jan 06 '15 at 10:45
@MarcvanLeeuwen You mean I should use double instead of integer or what? I didn't get you! (excuse my english) – Jan 06 '15 at 10:50
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I mean that (for integer variables `n,k,val`) the statement `val = val * (n-k) / (k-1)` does not have the same effect as `val *= (n-k) / (k-1)`. The latter is equivalent to `val = val * ((n-k) / (k-1))`, in which the now parenthesised division is performed _before_ the multiplication is done. But in the latter form the division is not exact, and the remainder gets dropped, which causes the result to be wrong. However in the former form `val = (val * (n-k)) / (k-1)` (where I have added the implicit parenthesis for clarity) the mutliplication is done first, whence the division is always exact. – Marc van Leeuwen Jan 06 '15 at 10:57

Can't Understand this line in Pascal's Triangle for C++

1 Answers1