Pollard Rho Brent Integer Factorization

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 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

All Pair Shortest Path (APSP)

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

Number of Cycles in a Graph

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(2^NN²). 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