# COME ON CODE ON

A blog about programming and more programming.

## Combination

In mathematics a combination is a way of selecting several things out of a larger group, where (unlike permutations) order does not matter. More formally a k-combination of a set S is a subset of k distinct elements of S. If the set has n elements the number of k-combinations is equal to the binomial coefficient. In this post we will see different methods to calculate the binomial.

1. Using Factorials
We can calculate nCr directly using the factorials.
nCr = n! / (r! * (n-r)!)

#include<iostream>
using namespace std;

long long C(int n, int r)
{
long long f[n + 1];
f[0]=1;
for (int i=1;i<=n;i++)
f[i]=i*f[i-1];
return f[n]/f[r]/f[n-r];
}

int main()
{
int n,r,m;
while (~scanf("%d%d",&n,&r))
{
printf("%lld\n",C(n, min(r,n-r)));
}
}

But this will work for only factorial below 20 in C++. For larger factorials you can either write big factorial library or use a language like Python. The time complexity is O(n).
If we have to calcuate nCr mod p(where p is a prime), we can calculate factorial mod p and then use modular inverse to find nCr mod p. If we have to find nCr mod m(where m is not prime), we can factorize m into primes and then use Chinese Remainder Theorem(CRT) to find nCr mod m.

#include<iostream>
using namespace std;
#include<vector>

/* This function calculates (a^b)%MOD */
long long pow(int a, int b, int MOD)
{
long long x=1,y=a;
while(b > 0)
{
if(b%2 == 1)
{
x=(x*y);
if(x>MOD) x%=MOD;
}
y = (y*y);
if(y>MOD) y%=MOD;
b /= 2;
}
return x;
}

/* 	Modular Multiplicative Inverse
Using Euler's Theorem
a^(phi(m)) = 1 (mod m)
a^(-1) = a^(m-2) (mod m) */
long long InverseEuler(int n, int MOD)
{
return pow(n,MOD-2,MOD);
}

long long C(int n, int r, int MOD)
{
vector<long long> f(n + 1,1);
for (int i=2; i<=n;i++)
f[i]= (f[i-1]*i) % MOD;
return (f[n]*((InverseEuler(f[r], MOD) * InverseEuler(f[n-r], MOD)) % MOD)) % MOD;
}

int main()
{
int n,r,p;
while (~scanf("%d%d%d",&n,&r,&p))
{
printf("%lld\n",C(n,r,p));
}
}

2. Using Recurrence Relation for nCr
The recurrence relation for nCr is C(i,k) = C(i-1,k-1) + C(i-1,k). Thus we can calculate nCr in time complexity O(n*r) and space complexity O(n*r).

#include<iostream>
using namespace std;
#include<vector>

/*
C(n,r) mod m
Using recurrence:
C(i,k) = C(i-1,k-1) + C(i-1,k)
Time Complexity: O(n*r)
Space Complexity: O(n*r)
*/

long long C(int n, int r, int MOD)
{
vector< vector<long long> > C(n+1,vector<long long> (r+1,0));

for (int i=0; i<=n; i++)
{
for (int k=0; k<=r && k<=i; k++)
if (k==0 || k==i)
C[i][k] = 1;
else
C[i][k] = (C[i-1][k-1] + C[i-1][k])%MOD;
}
return C[n][r];
}
int main()
{
int n,r,m;
while (~scanf("%d%d%d",&n,&r,&m))
{
printf("%lld\n",C(n, r, m));
}
}

We can easily reduce the space complexity of the above solution by just keeping track of the previous row as we don’t need the rest rows.

#include<iostream>
using namespace std;
#include<vector>

/*
Time Complexity: O(n*r)
Space Complexity: O(r)
*/
long long C(int n, int r, int MOD)
{
vector< vector<long long> > C(2,vector<long long> (r+1,0));

for (int i=0; i<=n; i++)
{
for (int k=0; k<=r && k<=i; k++)
if (k==0 || k==i)
C[i&1][k] = 1;
else
C[i&1][k] = (C[(i-1)&1][k-1] + C[(i-1)&1][k])%MOD;
}
return C[n&1][r];
}

int main()
{
int n,r,m,i,k;
while (~scanf("%d%d%d",&n,&r,&m))
{
printf("%lld\n",C(n, r, m));
}
}

3. Using expansion of nCr
Since

C(n,k) = n!/((n-k)!k!)
[n(n-1)...(n-k+1)][(n-k)...(1)]
= -------------------------------
[(n-k)...(1)][k(k-1)...(1)]


We can cancel the terms: [(n-k)…(1)] as they appear both on top and bottom, leaving:

    n (n-1)     (n-k+1)
- ----- ... -------
k (k-1)       (1)


which we might write as:

C(n,k) = 1,                 if k = 0
= (n/k)*C(n-1, k-1), otherwise


#include<iostream>
using namespace std;

long long C(int n, int r)
{
if (r==0) return 1;
else return C(n-1,r-1) * n / r;
}

int main()
{
int n,r,m;
while (~scanf("%d%d",&n,&r))
{
printf("%lld\n",C(n, min(r,n-r)));
}
}

4. Using Matrix Multiplication
In the last post we learned how to use Fast Matrix Multiplication to calculate functions having linear equations in logarithmic time. Here we have the recurrence relation C(i,k) = C(i-1,k-1) + C(i-1,k).
If we take k=3 we can write,
C(i-1,1) + C(i-1,0) = C(i,1)
C(i-1,2) + C(i-1,1) = C(i,2)
C(i-1,3) + C(i-1,2) = C(i,3)
Now on the left side we have four variables C(i-1,0), C(i-1,1), C(i-1,2) and C(i-1,3).
On the right side we have three variables C(i,1), C(i,2) and C(i,3).
We need those two sets to be the same, except that the right side index numbers should be one higher than the left side index numbers. So we add C(i,0) on the right side. NOw let’s get our all important Matrix.

(.   .   .   .)  ( C(i-1,0) )   ( C(i,0) )
(.   .   .   .)  ( C(i-1,1) ) = ( C(i,1) )
(.   .   .   .)  ( C(i-1,2) )   ( C(i,2) )
(.   .   .   .)  ( C(i-1,3) )   ( C(i,3) )

The last three rows are trivial and can be filled from the recurrence equations above.

(.   .   .   .)  ( C(i-1,0) )   ( C(i,0) )
(1   1   0   0)  ( C(i-1,1) ) = ( C(i,1) )
(0   1   1   0)  ( C(i-1,2) )   ( C(i,2) )
(0   0   1   1)  ( C(i-1,3) )   ( C(i,3) )

The first row, for C(i,0), depends on what is supposed to happen when k = 0. We know that C(i,0) = 1 for all i when k=0. So the matrix reduces to

(.   .   .   .)  ( C(i-1,0) )   ( C(i,0) )
(1   1   0   0)  ( C(i-1,1) ) = ( C(i,1) )
(0   1   1   0)  ( C(i-1,2) )   ( C(i,2) )
(0   0   1   1)  ( C(i-1,3) )   ( C(i,3) )

And this then leads to the general form:

               i
(.   .   .   .)  ( C(0,0) )   ( C(i,0) )
(1   1   0   0)  ( C(0,1) ) = ( C(i,1) )
(0   1   1   0)  ( C(0,2) )   ( C(i,2) )
(0   0   1   1)  ( C(0,3) )   ( C(i,3) )

For example if we wan’t C(4,3) we just raise the above matrix to the 4th power.

              4
(1   0   0   0)  ( 1 )   ( 1 )
(1   1   0   0)  ( 0 ) = ( 4 )
(0   1   1   0)  ( 0 )   ( 6 )
(0   0   1   1)  ( 0 )   ( 4 )

Here’s a C++ code.

#include<iostream>
using namespace std;

/*
C(n,r) mod m
Using Matrix Exponentiation
Time Complexity: O((r^3)*log(n))
Space Complexity: O(r*r)
*/

long long MOD;

template< class T >
class Matrix
{
public:
int m,n;
T *data;

Matrix( int m, int n );
Matrix( const Matrix< T > &matrix );

const Matrix< T > &operator=( const Matrix< T > &A );
const Matrix< T > operator*( const Matrix< T > &A );
const Matrix< T > operator^( int P );

~Matrix();
};

template< class T >
Matrix< T >::Matrix( int m, int n )
{
this->m = m;
this->n = n;
data = new T[m*n];
}

template< class T >
Matrix< T >::Matrix( const Matrix< T > &A )
{
this->m = A.m;
this->n = A.n;
data = new T[m*n];
for( int i = 0; i < m * n; i++ )
data[i] = A.data[i];
}

template< class T >
Matrix< T >::~Matrix()
{
delete [] data;
}

template< class T >
const Matrix< T > &Matrix< T >::operator=( const Matrix< T > &A )
{
if( &A != this )
{
delete [] data;
m = A.m;
n = A.n;
data = new T[m*n];
for( int i = 0; i < m * n; i++ )
data[i] = A.data[i];
}
return *this;
}

template< class T >
const Matrix< T > Matrix< T >::operator*( const Matrix< T > &A )
{
Matrix C( m, A.n );
for( int i = 0; i < m; ++i )
for( int j = 0; j < A.n; ++j )
{
C.data[i*C.n+j]=0;
for( int k = 0; k < n; ++k )
C.data[i*C.n+j] = (C.data[i*C.n+j] + (data[i*n+k]*A.data[k*A.n+j])%MOD)%MOD;
}
return C;
}

template< class T >
const Matrix< T > Matrix< T >::operator^( int P )
{
if( P == 1 ) return (*this);
if( P & 1 ) return (*this) * ((*this) ^ (P-1));
Matrix B = (*this) ^ (P/2);
return B*B;
}

long long C(int n, int r)
{
Matrix<long long> M(r+1,r+1);
for (int i=0;i<(r+1)*(r+1);i++)
M.data[i]=0;
M.data[0]=1;
for (int i=1;i<r+1;i++)
{
M.data[i*(r+1)+i-1]=1;
M.data[i*(r+1)+i]=1;
}
return (M^n).data[r*(r+1)];
}

int main()
{
int n,r;
while (~scanf("%d%d%lld",&n,&r,&MOD))
{
printf("%lld\n",C(n, r));
}
}

5. Using the power of prime p in n factorial
The power of prime p in n factorial is given by
εp = ⌊n/p⌋ + ⌊n/p2⌋ + ⌊n/p3⌋…
If we call the power of p in n factorial, the power of p in nCr is given by
e = countFact(n,i) – countFact(r,i) – countFact(n-r,i)
To get the result we multiply p^e for all p less than n.

#include<iostream>
using namespace std;
#include<vector>

/* This function calculates power of p in n! */
int countFact(int n, int p)
{
int k=0;
while (n>0)
{
k+=n/p;
n/=p;
}
return k;
}

/* This function calculates (a^b)%MOD */
long long pow(int a, int b, int MOD)
{
long long x=1,y=a;
while(b > 0)
{
if(b%2 == 1)
{
x=(x*y);
if(x>MOD) x%=MOD;
}
y = (y*y);
if(y>MOD) y%=MOD;
b /= 2;
}
return x;
}

long long C(int n, int r, int MOD)
{
long long res = 1;
vector<bool> isPrime(n+1,1);
for (int i=2; i<=n; i++)
if (isPrime[i])
{
for (int j=2*i; j<=n; j+=i)
isPrime[j]=0;
int k = countFact(n,i) - countFact(r,i) - countFact(n-r,i);
res = (res * pow(i, k, MOD)) % MOD;
}
return res;
}

int main()
{
int n,r,m;
while (scanf("%d%d%d",&n,&r,&m))
{
printf("%lld\n",C(n,r,m));
}
}

6. Using Lucas Theorem
For non-negative integers m and n and a prime p, the following congruence relation holds:
$\binom{m}{n}\equiv\prod_{i=0}^k\binom{m_i}{n_i}\pmod p,$
where
$m=m_kp^k+m_{k-1}p^{k-1}+\cdots +m_1p+m_0,$
and
$n=n_kp^k+n_{k-1}p^{k-1}+\cdots +n_1p+n_0$
are the base p expansions of m and n respectively.
We only need to calculate nCr only for small numbers (less than equal to p) using any of the above methods.

#include<iostream>
using namespace std;
#include<vector>

long long SmallC(int n, int r, int MOD)
{
vector< vector<long long> > C(2,vector<long long> (r+1,0));

for (int i=0; i<=n; i++)
{
for (int k=0; k<=r && k<=i; k++)
if (k==0 || k==i)
C[i&1][k] = 1;
else
C[i&1][k] = (C[(i-1)&1][k-1] + C[(i-1)&1][k])%MOD;
}
return C[n&1][r];
}

long long Lucas(int n, int m, int p)
{
if (n==0 && m==0) return 1;
int ni = n % p;
int mi = m % p;
if (mi>ni) return 0;
return Lucas(n/p, m/p, p) * SmallC(ni, mi, p);
}

long long C(int n, int r, int MOD)
{
return Lucas(n, r, MOD);
}

int main()
{

int n,r,p;
while (~scanf("%d%d%d",&n,&r,&p))
{
printf("%lld\n",C(n,r,p));
}
}

7. Using special n! mod p
We will calculate n factorial mod p and similarly inverse of r! mod p and (n-r)! mod p and multiply to find the result. But while calculating factorial mod p we remove all the multiples of p and write
n!* mod p = 1 * 2 * … * (p-1) * 1 * 2 * … * (p-1) * 2 * 1 * 2 * … * n.
We took the usual factorial, but excluded all factors of p (1 instead of p, 2 instead of 2p, and so on). Lets call this strange factorial.
So strange factorial is really several blocks of construction:
1 * 2 * 3 * … * (p-1) * i
where i is a 1-indexed index of block taken again without factors p.
The last block could be not full. More precisely, there will be floor(n/p) full blocks and some tail (its result we can compute easily, in O(P)).
The result in each block is multiplication 1 * 2 * … * (p-1), which is common to all blocks, and multiplication of all strange indices i from 1 to floor(n/p).
But multiplication of all strange indices is really a strange factorial again, so we can compute it recursively. Note, that in recursive calls n reduces exponentially, so this is rather fast algorithm.
Here’s the algorithm to calculate strange factorial.

int factMOD(int n, int MOD)
{
long long res = 1;
while (n > 1)
{
long long cur = 1;
for (int i=2; i<MOD; ++i)
cur = (cur * i) % MOD;
res = (res * powmod (cur, n/MOD, MOD)) % MOD;
for (int i=2; i<=n%MOD; ++i)
res = (res * i) % MOD;
n /= MOD;
}
return int (res % MOD);
}

But we can still reduce our complexity.
By Wilson’s Theorem, we know $(n-1)!\ \equiv\ -1 \pmod n$ for all primes n. SO our method reduces to:

long long factMOD(int n, int MOD)
{
long long res = 1;
while (n > 1)
{
res = (res * pow(MOD - 1, n/MOD, MOD)) % MOD;
for (int i=2, j=n%MOD; i<=j; i++)
res = (res * i) % MOD;
n/=MOD;
}
return res;
}

Now in the above code we are calculating (-1)^(n/p). If (n/p) is even what we are multiplying by 1, so we can skip that. We only need to consider the case when (n/p) is odd, in which case we are multiplying result by (-1)%MOD, which ultimately is equal to MOD-res. SO our method again reduces to:

long long factMOD(int n, int MOD)
{
long long res = 1;
while (n > 0)
{
for (int i=2, m=n%MOD; i<=m; i++)
res = (res * i) % MOD;
if ((n/=MOD)%2 > 0)
res = MOD - res;
}
return res;
}

Finally the complete code here:

#include<iostream>
using namespace std;
#include<vector>

/* This function calculates power of p in n! */
int countFact(int n, int p)
{
int k=0;
while (n>=p)
{
k+=n/p;
n/=p;
}
return k;
}

/* This function calculates (a^b)%MOD */
long long pow(int a, int b, int MOD)
{
long long x=1,y=a;
while(b > 0)
{
if(b%2 == 1)
{
x=(x*y);
if(x>MOD) x%=MOD;
}
y = (y*y);
if(y>MOD) y%=MOD;
b /= 2;
}
return x;
}

/* 	Modular Multiplicative Inverse
Using Euler's Theorem
a^(phi(m)) = 1 (mod m)
a^(-1) = a^(m-2) (mod m) */
long long InverseEuler(int n, int MOD)
{
return pow(n,MOD-2,MOD);
}

long long factMOD(int n, int MOD)
{
long long res = 1;
while (n > 0)
{
for (int i=2, m=n%MOD; i<=m; i++)
res = (res * i) % MOD;
if ((n/=MOD)%2 > 0)
res = MOD - res;
}
return res;
}

long long C(int n, int r, int MOD)
{
if (countFact(n, MOD) > countFact(r, MOD) + countFact(n-r, MOD))
return 0;

return (factMOD(n, MOD) *
((InverseEuler(factMOD(r, MOD), MOD) *
InverseEuler(factMOD(n-r, MOD), MOD)) % MOD)) % MOD;
}

int main()
{
int n,r,p;
while (~scanf("%d%d%d",&n,&r,&p))
{
printf("%lld\n",C(n,r,p));
}
}

-fR0D

Written by fR0DDY

July 31, 2011 at 5:30 PM

Posted in Algorithm

## Recurrence Relation and Matrix Exponentiation

Recurrence relations appear many times in computer science. Using recurrence relation and dynamic programming we can calculate the nth term in O(n) time. But many times we need to calculate the nth in O(log n) time. This is where Matrix Exponentiation comes to rescue.

We will specifically look at linear recurrences. A linear recurrence is a sequence of vectors defined by the equation Xi+1 = M Xi for some constant matrix M. So our aim is to find this constant matrix M, for a given recurrence relation.

Let’s first start by looking at the common structure of our three matrices Xi+1, Xi and M. For a recurrence relation where the next term is dependent on last k terms, Xi+1 and Xi+1 are matrices of size 1 x k and M is a matrix of size k x k.

| f(n+1)   |       | f(n)   |
|  f(n)    |       | f(n-1) |
| f(n-1)   | = M x | f(n-2) |
| ......   |       | ...... |
| f(n-k+1) |       | f(n-k) |

Let’s look at different type of recurrence relations and how to find M.

1. Let’s start with the most common recurrence relation in computer science, The Fibonacci Sequence. Fi+1 = Fi + Fi-1.

| f(n+1) | = M x | f(n)   |
| f(n)   |       | f(n-1) |

Now we know that M is a 2 x 2 matrix. Let it be

| f(n+1) | = | a b | x | f(n)   |
| f(n)   |   | c d |   | f(n-1) |

Now a*f(n) + b*f(n-1) = f(n+1) and c*f(n) + d*f(n-1) = f(n). Solving these two equations we get a=1,b=1,c=1 and d=0. So,

| f(n+1) | = | 1 1 | x | f(n)   |
| f(n)   |   | 1 0 |   | f(n-1) |

2. For recurrence relation f(n) = a*f(n-1) + b*f(n-2) + c*f(n-3), we get

| f(n+1) | = | a b c | x | f(n)   |
| f(n)   |   | 1 0 0 |   | f(n-1) |
| f(n-1) |   | 0 1 0 |   | f(n-2) |

3. What if the recurrence relation is f(n) = a*f(n-1) + b*f(n-2) + c, where c is a constant. We can also add it in the matrices as a state.

| f(n+1) | = | a b 1 | x | f(n)   |
| f(n)   |   | 1 0 0 |   | f(n-1) |
| c      |   | 0 0 1 |   | c      |

4. If a recurrence relation is given like this f(n) = f(n-1) if n is odd, f(n-2) otherwise, we can convert it to f(n) = (n&1) * f(n-1) + (!(n&1)) * f(n-2) and substitute the value accordingly in the matrix.

5.If there are more than one recurrence relation, g(n) = a*g(n-1) + b*g(n-2) + c*f(n), and, f(n) = d*f(n-1) + e*f(n-2). We can still define the matrix X in following way

| g(n+1) |   | a b c 0 |   | g(n)   |
| g(n)   | = | 1 0 0 0 | x | g(n-1) |
| f(n+2) |   | 0 0 d e |   | f(n+1) |
| f(n+1) |   | 0 0 1 0 |   | f(n)   |

Now that we have got our matrix M, how are we going to find the nth term.
Xi+1 = M Xi
(Multiplying M both sides)
M * Xi+1 = M * M Xi
Xi+2 = M^2 Xi
..
Xi+k = M^k Xi

So all we need now is to find the matrix M^k to find the k-th term. For example in the case of Fibonacci Sequence,

M^2 = | 2 1 |
| 1 1 |

Hence F(2) = 2.

Now we need to learn to find M^k in O(n3 log b) time. The brute force approach to calculate a^b takes O(b) time, but using a recursive divide-and-conquer algorithm takes only O(log b) time:

• If b = 0, then the answer is 1.
• If b = 2k is even, then ab = (ak)2.
• If b is odd, then ab = a * ab-1.

We take a similar approach for Matrix Exponentiation. The multiplication part takes the O(n3) time and hence the overall complexity is O(n3 log b). Here’s a sample code in C++ using template class:

#include<iostream>
using namespace std;

template< class T >
class Matrix
{
public:
int m,n;
T *data;

Matrix( int m, int n );
Matrix( const Matrix< T > &matrix );

const Matrix< T > &operator=( const Matrix< T > &A );
const Matrix< T > operator*( const Matrix< T > &A );
const Matrix< T > operator^( int P );

~Matrix();
};

template< class T >
Matrix< T >::Matrix( int m, int n )
{
this->m = m;
this->n = n;
data = new T[m*n];
}

template< class T >
Matrix< T >::Matrix( const Matrix< T > &A )
{
this->m = A.m;
this->n = A.n;
data = new T[m*n];
for( int i = 0; i < m * n; i++ )
data[i] = A.data[i];
}

template< class T >
Matrix< T >::~Matrix()
{
delete [] data;
}

template< class T >
const Matrix< T > &Matrix< T >::operator=( const Matrix< T > &A )
{
if( &A != this )
{
delete [] data;
m = A.m;
n = A.n;
data = new T[m*n];
for( int i = 0; i < m * n; i++ )
data[i] = A.data[i];
}
return *this;
}

template< class T >
const Matrix< T > Matrix< T >::operator*( const Matrix< T > &A )
{
Matrix C( m, A.n );
for( int i = 0; i < m; ++i )
for( int j = 0; j < A.n; ++j )
{
C.data[i*C.n+j]=0;
for( int k = 0; k < n; ++k )
C.data[i*C.n+j] = C.data[i*C.n+j] + data[i*n+k]*A.data[k*A.n+j];
}
return C;
}

template< class T >
const Matrix< T > Matrix< T >::operator^( int P )
{
if( P == 1 ) return (*this);
if( P & 1 ) return (*this) * ((*this) ^ (P-1));
Matrix B = (*this) ^ (P/2);
return B*B;
}

int main()
{
Matrix<int> M(2,2);
M.data[0] = 1;M.data[1] = 1;
M.data[2] = 1;M.data[3] = 0;

int F[2]={0,1};
int N;
while (~scanf("%d",&N))
if (N>1)
printf("%lld\n",(M^N).data[0]);
else
printf("%d\n",F[N]);
}

Written by fR0DDY

May 8, 2011 at 5:26 PM

Posted in Programming