Green goblin - 11 months ago 50

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 Source

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.