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Archive for the ‘Algorithm’ Category

Combination

with 12 comments

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

Pollard Rho Brent Integer Factorization

with 10 comments

Pollard Rho is an integer factorization algorithm, which is quite fast for large numbers. It is based on Floyd’s cycle-finding algorithm and on the observation that two numbers x and y are congruent modulo p with probability 0.5 after 1.177\sqrt{p} numbers have been randomly chosen.

Algorithm
Input : A number N to be factorized
Output : A divisor of N
If x mod 2 is 0
	return 2

Choose random x and c
y = x
g = 1
while g=1
	x = f(x)
	y = f(f(y))
	g = gcd(x-y,N)
return g

Note that this algorithm may not find the factors and will return failure for composite n. In that case, use a different f(x) and try again. Note, as well, that this algorithm does not work when n is a prime number, since, in this case, d will be always 1. We choose f(x) = x*x + c. Here’s a python implementation :

def pollardRho(N):
        if N%2==0:
                return 2
        x = random.randint(1, N-1)
        y = x
        c = random.randint(1, N-1)
        g = 1
        while g==1:             
                x = ((x*x)%N+c)%N
                y = ((y*y)%N+c)%N
                y = ((y*y)%N+c)%N
                g = gcd(abs(x-y),N)
        return g

In 1980, Richard Brent published a faster variant of the rho algorithm. He used the same core ideas as Pollard but a different method of cycle detection, replacing Floyd’s cycle-finding algorithm with the related Brent’s cycle finding method. It is quite faster than pollard rho. Here’s a python implementation :

def brent(N):
        if N%2==0:
                return 2
        y,c,m = random.randint(1, N-1),random.randint(1, N-1),random.randint(1, N-1)
        g,r,q = 1,1,1
        while g==1:             
                x = y
                for i in range(r):
                        y = ((y*y)%N+c)%N
                k = 0
                while (k<r and g==1):
                        ys = y
                        for i in range(min(m,r-k)):
                                y = ((y*y)%N+c)%N
                                q = q*(abs(x-y))%N
                        g = gcd(q,N)
                        k = k + m
                r = r*2
        if g==N:
                while True:
                        ys = ((ys*ys)%N+c)%N
                        g = gcd(abs(x-ys),N)
                        if g>1:
                                break
        
        return g    

-fR0DDY

Written by fR0DDY

September 18, 2010 at 11:51 PM

Miller Rabin Primality Test

with 15 comments

Miller Rabin Primality Test is a probabilistic test to check whether a number is a prime or not. It relies on an equality or set of equalities that hold true for prime values, then checks whether or not they hold for a number that we want to test for primality.

Theory

1> Fermat’s little theorem states that if p is a prime and 1 ≤ a < p then a^{p-1}\equiv1\pmod{p}.
2> If p is a prime and x^2\equiv1\pmod{p} or \left(x-1\right)\left(x+1\right)\equiv0\pmod{p} then x\equiv1\pmod{p} or x\equiv-1\pmod{p}.
3> If n is an odd prime then n-1 is an even number and can be written as 2^{s}.d. By Fermat’s Little Theorem either a^{d}\equiv1\pmod{n} or a^{2^r\cdot d}\equiv -1\pmod{n} for some 0 ≤ r ≤  s-1.
4> The Miller–Rabin primality test is based on the contrapositive of the above claim. That is, if we can find an a such that a^{d}\not\equiv1\pmod{n} and a^{2^r\cdot d}\not\equiv -1\pmod{n} for all 0 ≤ r ≤  s-1 then a is witness of compositeness of n and we can say n is not a prime. Otherwise, n may be a prime.
5> We test our number N for some random a and either declare that N is definitely a composite or probably a prime. The probably that a composite number is returned as prime after k itereations is 4^{-k}.

Algorithm

Input :A number N to be tested and a variable iteration-the number 
of 'a' for which algorithm will test N.
Output :0 if N is definitely a composite and 1 if N is probably a prime.

Write N as 2^{s}.d
For each iteration
	Pick a random a in [1,N-1]
	x = a^{d} mod n
	if x =1 or x = n-1 
		Next iteration
	for r = 1 to s-1
		x  = x^{2} mod n
		if x = 1
			return false
		if x = N-1
			Next iteration
	return false
return true

Here’s a python implementation :

import random
def modulo(a,b,c):
        x = 1
        y = a
        while b>0:
                if b%2==1:
                        x = (x*y)%c
                y = (y*y)%c
                b = b/2
        return x%c
        
def millerRabin(N,iteration):
        if N<2:
                return False
        if N!=2 and N%2==0:
                return False
        
        d=N-1
        while d%2==0:
                d = d/2
        
        for i in range(iteration):
                a = random.randint(1, N-1)
                temp = d
                x = modulo(a,temp,N)
                while (temp!=N-1 and x!=1 and x!=N-1):
                        x = (x*x)%N
                        temp = temp*2
                
                if (x!=N-1 and temp%2==0):
                        return False
        
        return True

-fR0DDY

Written by fR0DDY

September 18, 2010 at 12:23 PM

Knuth–Morris–Pratt Algorithm (KMP)

with one comment

Knuth–Morris–Pratt algorithm is the most popular linear time algorithm for string matching. It is little difficult to understand and debug in real time contests. So most programmer’s have a precoded KMP in their kitty.

To understand the algorithm, you can either read it from Introduction to Algorithms (CLRS) or from the wikipedia page. Here’s a sample C++ code.

void preKMP(string pattern, int f[])
{
        int m = pattern.length(),k;
        f[0] = -1;
        for (int i = 1; i<m; i++)
        {
                k = f[i-1];
                while (k>=0)
                {
                        if (pattern[k]==pattern[i-1])
                                break;
                        else
                                k = f[k];
                }
                f[i] = k + 1;
        }
}
 
bool KMP(string pattern, string target)
{ 
        int m = pattern.length();
        int n = target.length();
        int f[m];
        
        preKMP(pattern, f);
        
        int i = 0;
        int k = 0;
        
        while (i<n)
        {
                if (k==-1)
                {
                        i++;
                        k = 0;
                }
                else if (target[i]==pattern[k])
                {
                        i++;
                        k++;
                        if (k==m)
                                return 1;
                }
                else
                        k=f[k];
        }
        return 0;
}

NJOY!
-fR0DDY

Written by fR0DDY

August 29, 2010 at 12:20 PM

The Z Algorithm

with one comment

In this post we will discuss an algorithm for linear time string matching. It is easy to understand and code and is usefull in contests where you cannot copy paste code.

Let our string be denoted by S.
z[i] denotes the length of the longest substring of S that starts at i and is a prefix of S.
α denotes the substring.
r denotes the index of the last character of α and l denotes the left end of α.

To find whether a pattern(P) of length n is present in a target string(T) of length m, we will create a new string S = P$T where $ is a character present neither in P nor in T. The space taken is n+m+1 or O(m). We will compute z[i] for all i such that 0 < i < n+m+1. If z[i] is equal to n then we have found a occurrence of P at position i – n – 1. So we can all the occurrence of P in T in O(m) time. To calculate z[i] we will use the z algorithm.

The Z Algorithm can be read from the section 1.3-1.5 of book Algorithms on strings, trees, and sequences by Gusfield. Here is a sample C++ code.

bool zAlgorithm(string pattern, string target)
{
	string s = pattern + '$' + target ;
	int n = s.length();
	vector<int> z(n,0);
	
	int goal = pattern.length();
	int r = 0, l = 0, i;
	for (int k = 1; k<n; k++)
	{
		if (k>r)
		{
			for (i = k; i<n && s[i]==s[i-k]; i++);
			if (i>k)
			{
				z[k] = i - k;
				l = k;
				r = i - 1;
			}
		}
		else
		{
			int kt = k - l, b = r - k + 1;
			if (z[kt]>b)
			{
				for (i = r + 1; i<n && s[i]==s[i-k]; i++);
				z[k] = i - k;
				l = k;
				r = i - 1;
			}
		}
		if (z[k]==goal)
			return true;
	}
	return false;
}

NJOY!
-fR0D

Written by fR0DDY

August 29, 2010 at 12:03 AM

All Pair Shortest Path (APSP)

with one comment

Question : Find shortest paths between all pairs of vertices in a graph.

Floyd-Warshall Algorithm
It is one of the easiest algorithms, and just involves simple dynamic programming. The algorithm can be read from this wikipedia page.

#define SIZE 31
#define INF 1e8
double dis[SIZE][SIZE];
void init(int N)
{
	for (k=0;k<N;k++)
		for (i=0;i<N;i++)
			dis[i][j]=INF;
}
void floyd_warshall(int N)
{
        int i,j,k;
        for (k=0;k<N;k++)
                for (i=0;i<N;i++)
                        for (j=0;j<N;j++)
                                dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
}

int main()
{
	//input size N
	init(N);
	//set values for dis[i][j]
	floyd_warshall(N);
}

We can also use the algorithm to

  1. find the shortest path
    • we can use another matrix called predecessor matrix to construct the shortest path.
  2. find negative cycles in a graph.
    • If the value of any of the diagonal elements is less than zero after calling the floyd-warshall algorithm then there is a negative cycle in the graph.
  3. find transitive closure
    • to find if there is a path between two vertices we can use a boolean matrix and use and-& and or-| operators in the floyd_warshall algorithm.
    • to find the number of paths between any two vertices we can use a similar algorithm.

NJOY!!
-fR0DDY

Written by fR0DDY

August 7, 2010 at 1:53 PM

Number of Cycles in a Graph

with 2 comments

Question : Find the number of simple cycles in a simple graph.

Simple Graph – An undirected graph that has no loops and no more than one edge between any two different vertices.
Simple Cycle – A closed (simple) path, with no other repeated vertices or edges other than the starting and ending vertices.

Given a graph of N vertices and M edges, we will look at an algorithm with time complexity O(2NN2). We will use dynamic programming to do so. Let there be a matrix map, such that map[i][j] is equal to 1 if there is a edge between i and j and 0 otherwise. Let there be another array f[1<<N][N] which denotes the number of simple paths.

Let,
i denote a subset S of our vertices
k be the smallest set bit of i
then f[i][j] is the number of simple paths from j to k that 
contains vertices only from the set S. 

In our algorithm first we will find f[i][j] and then check if there is a edge between k and j, if yes, we can complete every simple path from j to k into a simple cycle and hence we add f[i][j] to our result of total number of simple cycles.Now how to find f[i][j].

For very subset i we iterate through all edges j. Once we have set k, we look for all vertices 'l' that can be neighbors of j in our subset S. So if l is a vertex in subset S and there is edge from j to l then f[i][j] = f[i][j] + the number of simple paths from l to i in the subset {S – j}. Since a simple graph is undirected or bidirectional, we have counted every cycle twice and so we divide our result by 2. Here's a sample C++ code which takes N, M and the edges as input.

#include<iostream>
using namespace std;

#define SIZE 20

bool map[SIZE][SIZE],F;
long long f[1<<SIZE][SIZE],res=0;

int main()
{
    int n,m,i,j,k,l,x,y;
    scanf("%d%d",&n,&m);
    for (i=0;i<m;i++)
    {
        scanf("%d%d",&x,&y);
        x--;y--;
        if (x>y)
           swap(x,y);
        map[x][y]=map[y][x]=1;
        f[(1<<x)+(1<<y)][y]=1;
    }
    
    for (i=7;i<(1<<n);i++)
    {
        F=1;
        for (j=0;j<n;j++)
            if (i&(1<<j) && f[i][j]==0)
            {
               if (F)
               {
                  F=0;
                  k=j;
                  continue;
               }
               for (l=k+1;l<n;l++)
               {
                   if (i&(1<<l) && map[j][l])
                      f[i][j]+=f[i-(1<<j)][l];
               }
               if (map[k][j])
                  res+=f[i][j];
            }
    }
    printf("%lld\n",res/2);
}

NJOY!
-fR0DDY

Written by fR0DDY

June 7, 2010 at 1:14 PM

Longest Common Increasing Subsequence (LCIS)

with 4 comments

Given 2 sequences of integers, we need to find a longest sub-sequence which is common to both the sequences, and the numbers of such a sub-sequence are in strictly increasing order.
For example,
2 3 1 6 5 4 6
1 3 5 6
the LCIS is 3 5 6.

The sequence a1, a2, …, an is called increasing, if ai<ai+1 for i<n.The sequence s1, s2, …, sk is called the subsequence of the sequence a1, a2, …, an, if there exist such a set of indexes 1 ≤ i1 < i2 < … < ik ≤ n that aij = sj. In other words, the sequence s can be derived from the sequence a by crossing out some elements.

A nice tutorial on the algorithm is given on CodeChef blog here. If you read the blog you can see that instead of looking for LCIS in a candidate matrix, we can keep an array that stores the length of LCIS ending at that particular element. Also we keep a lookup previous array that gives the index of the previous element of the LCIS, which is used to reconstruct the LCIS.

For every element in the two arrays
1> If they are equal we check whether the LCIS formed will be bigger than any previous such LCIS ending at that element. If yes we change the data.
2> If the jth element of array B is smaller than ith element of array A, we check whether it has a LCIS greater than current LCIS length, if yes we store it as previous value and it’s LCIS length as current LCIS length.

Here’s a C++ code.

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

void LCIS(vector<int> A, vector<int> B)
{
    int N=A.size(),M=B.size(),i,j;
    vector<int> C(M,0);
    vector<int> prev(M,0);
    vector<int> res;
    
    for (i=0;i<N;i++)
    {
        int cur=0,last=-1;
        for (j=0;j<M;j++)
        {
            if (A[i]==B[j] && cur+1>C[j])
            {
               C[j]=cur+1;
               prev[j]=last;
            }
            if (B[j]<A[i] && cur<C[j])
            {
               cur=C[j];
               last=j;
            }
        }
    }
    
    int length=0,index=-1;
    for (i=0;i<M;i++)
        if (C[i]>length)
        {
           length=C[i];
           index=i;
        }
    printf("The length of LCIS is %d\n",length);
    if (length>0)
    {
       printf("The LCIS is \n");
       while (index!=-1)
       {
             res.push_back(B[index]);
             index=prev[index];
       }
       reverse(res.begin(),res.end());
       for (i=0;i<length;i++)
           printf("%d%s",res[i],i==length-1?"\n":" ");
    }
}


int main()
{
    int n,m,i;
    scanf ("%d", &n);
    vector<int> A(n,0);
    for (i = 0; i < n; i++)
        scanf ("%d", &A[i]);
    scanf ("%d", &m);
    vector<int> B(m,0);
    for (i = 0; i < m; i++)
        scanf ("%d", &B[i]);
    LCIS(A,B);
}

NJOY!
-fR0DDY

Written by fR0DDY

June 1, 2010 at 12:47 PM

Number of Binary Trees

with 3 comments

Question : What is the number of rooted plane binary trees with n nodes and height h?

Solution :
Let tn,h denote the number of binary trees with n nodes and height h.

Lets take a binary search tree of n nodes with height equal to h. Let the number written at its root be the mthnumber in sorted order, 1 ≤ m ≤ n. Therefore, the left subtree is a binary search tree on m-1 nodes, and the right subtree is a binary search tree on n-m nodes. The maximum height of the left and the right subtrees must be equal to h-1. Now there are two cases :

  1. The height of left subtree is h-1 and the height of right subtree is less than equal to h-1 i.e from 0 to h-1.
  2. The height of right subtree is h-1 and the height of left subtree is less than equal to h-2 i.e from 0 to h-2, since we have considered the case when the left and right subtrees have the same height h-1 in case 1.

Therefore tn,h is equal to the sum of number of trees in case 1 and case 2. Let’s find the number of trees in case 1 and case 2.

  1. The height of the left subtree is equal to h-1. There are tm-1,h-1 such trees. The right subtree can have any height from 0 to h-1, so there are \displaystyle\sum_{i=0}^{h-1}t_{n-m,i} such trees. Therefore the total number of such tress are t_{m-1,h-1}\displaystyle\sum_{i=0}^{h-1}t_{n-m,i}.
  2. The height of the right subtree is equal to h-1. There are tn-m,h-1 such trees. The left subtree can have any height from 0 to h-2, so there are \displaystyle\sum_{i=0}^{h-2}t_{m-1,i} such trees. Therefore the total number of such tress are t_{n-m,h-1}\displaystyle\sum_{i=0}^{h-2}t_{m-1,i}.

Hence we get the recurrent relation
t_{n,h}= \displaystyle\sum_{m=1}^{n}{(t_{m-1,h-1}\displaystyle\sum_{i=0}^{h-1}t_{n-m,i})} + \displaystyle\sum_{m=1}^{n}{(t_{n-m,h-1}\displaystyle\sum_{i=0}^{h-2}t_{m-1,i})}
or
t_{n,h}= \displaystyle\sum_{m=1}^{n}{(t_{m-1,h-1}\displaystyle\sum_{i=0}^{h-1}t_{n-m,i} + t_{n-m,h-1}\displaystyle\sum_{i=0}^{h-2}t_{m-1,i})}
where t0,0=1 and t0,i=ti,0=0 for i>0.
Here’s a sample C++ code.

#include<iostream>
using namespace std;

#define MAXN 35

int main()
{
    long long t[MAXN+1][MAXN+1]={0},n,h,m,i;
    
    t[0][0]=1;
    for (n=1;n<=MAXN;n++)
        for (h=1;h<=n;h++)
            for (m=1;m<=n;m++)
            {
                for (i=0;i<h;i++)
                    t[n][h]+=t[m-1][h-1]*t[n-m][i];
                
                for (i=0;i<h-1;i++)
                    t[n][h]+=t[n-m][h-1]*t[m-1][i];
            }
            
    while (scanf("%lld%lld",&n,&h))
          printf("%lld\n",t[n][h]);
}

Note :

  1. The total number of binary trees with n nodes is \frac{2n!}{n!(n+1)!}, also known as Catalan numbers or Segner numbers.
  2. tn,n = 2n-1.
  3. The number of trees with height not lower than h is \displaystyle\sum_{i=h}^{n}{t_{n,i}}.
  4. There are other recurrence relation as well such as
    t_{n,h}= \displaystyle\sum_{i=0}^{h-1}{(t_{n-1,h-1-i}(2*\displaystyle\sum_{j=0}^{n-2}t_{j,i} + t_{n-1,i}))}.

NJOY!
-fR0DDY

Written by fR0DDY

April 13, 2010 at 11:44 AM

Posted in Algorithm, Maths, Programming

Tagged with , , , , ,

Convert a number from decimal base to any Base

with 6 comments

Convert a given decmal number to any other base (either positive or negative).

For example, 100 in Base 2 is 1100100 and in Base -2 is 110100100.

Here’s a simple algorithm to convert numbers from Decimal to Negative Bases :

def tonegativeBase(N,B):
   digits = []
   while i != 0:
	i, remainder = divmod (i, B)
	if (remainder < 0):
	    i, remainder = i + 1, remainder + B*-1
	digits.insert (0, str (remainder))
   return ''.join (digits)

We can just tweak the above algorithm a bit to convert a decimal to any Base. Here’s a sample code :

#include<iostream>
using namespace std;

void convert10tob(int N,int b)
{
     if (N==0)
        return;
     
     int x = N%b;
     N/=b;
     if (x<0)
        N+=1;
        
     convert10tob(N,b);
     printf("%d",x<0?x+(b*-1):x);
     return;
}


int main()
{
    int N,b;
    while (scanf("%d%d",&N,&b)==2)
    {
          if (N!=0)
          {
              convert10tob(N,b);
              printf("\n");
          }
          else
              printf("0\n");
    }
}

NJOY!
-fR0DDY

Written by fR0DDY

February 17, 2010 at 6:06 PM

Posted in Algorithm, Maths, Programming

Tagged with , , , , ,

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