Green goblin - 7 months ago 39

C Question

Approach 1:

C(n,r) = n!/(n-r)!r!

Approach 2:

In the book Combinatorial Algorithms by wilf, i have found this:

C(n,r) can be written as

`C(n-1,r) + C(n-1,r-1)`

e.g.

`C(7,4) = C(6,4) + C(6,3)`

= C(5,4) + C(5,3) + C(5,3) + C(5,2)

. .

. .

. .

. .

After solving

= C(4,4) + C(4,1) + 3*C(3,3) + 3*C(3,1) + 6*C(2,1) + 6*C(2,2)

As you can see, the final solution doesn't need any multiplication. In every form C(n,r), either n==r or r==1.

Here is the sample code i have implemented:

`int foo(int n,int r)`

{

if(n==r) return 1;

if(r==1) return n;

return foo(n-1,r) + foo(n-1,r-1);

}

See output here.

In the approach 2, there are overlapping sub-problems where we are calling recursion to solve the same sub-problems again. We can avoid it by using Dynamic Programming.

I want to know which is the better way to calculate C(n,r)?.

Answer

Both approaches will save time, but the first one is very prone to integer overflow.

**Approach 1:**

This approach will generate result in shortest time (in at most `n/2`

iterations), and the possibility of overflow can be reduced by doing the multiplications carefully:

```
long long C(int n, int r) {
if(r > n / 2) r = n - r; // because C(n, r) == C(n, n - r)
long long ans = 1;
int i;
for(i = 1; i <= r; i++) {
ans *= n - r + i;
ans /= i;
}
return ans;
}
```

This code will start multiplication of the numerator from the smaller end, and as the product of any `k`

consecutive integers is divisible by `k!`

, there will be no divisibility problem. But the possibility of overflow is still there, another useful trick may be dividing `n - r + i`

and `i`

by their GCD before doing the multiplication and division (and *still* overflow may occur).

**Approach 2:**

In this approach, you'll be actually building up the Pascal's Triangle. The dynamic approach is much faster than the recursive one (the first one is `O(n^2)`

while the other is exponential). However, you'll need to use `O(n^2)`

memory too.

```
# define MAX 100 // assuming we need first 100 rows
long long triangle[MAX + 1][MAX + 1];
void makeTriangle() {
int i, j;
// initialize the first row
triangle[0][0] = 1; // C(0, 0) = 1
for(i = 1; i < MAX; i++) {
triangle[i][0] = 1; // C(i, 0) = 1
for(j = 1; j <= i; j++) {
triangle[i][j] = triangle[i - 1][j - 1] + triangle[i - 1][j];
}
}
}
long long C(int n, int r) {
return triangle[n][r];
}
```

Then you can look up any `C(n, r)`

in `O(1)`

time.

If you need a particular `C(n, r)`

(i.e. the full triangle is not needed), then the memory consumption can be made `O(n)`

by overwriting the same row of the triangle, top to bottom.

```
# define MAX 100
long long row[MAX + 1]; // initialized with 0's by default
int C(int n, int r) {
int i, j;
// initialize by the first row
row[0] = 1; // this is the value of C(0, 0)
for(i = 1; i <= n; i++) {
for(j = i; j > 0; j--) {
// from the recurrence C(n, r) = C(n - 1, r - 1) + C(n - 1, r)
row[j] += row[j - 1];
}
}
return row[r];
}
```

The inner loop is started from the end to simplify the calculations. If you start it from index 0, you'll need another variable to store the value being overwritten.

Source (Stackoverflow)