Kuntal Majumder Kuntal Majumder - 1 year ago 81
C++ Question

Matrix Exponentiation for calculating number of routes possible

Recently I came across this problem in the Iarcs website. Here is the problem statement:

It is well known that the routing algorithm used on the Internet is
highly non-optimal. A "hop", in Internet jargon, is a pair of nodes
that are directly connected - by a cable or a microwave link or
whatever. The number of hops that a packet may take in going from one
node to another may be far more than the minimum required.

But the routing algorithm used by the Siruseri Network company is
worse. Here, a packet sent from one node to another could even go
through the same node twice or even traverse the same hop twice before
it eventually finds its way to its destination. Sometimes a packet
even goes through the destination more than once before it is
considered "delivered". Suppose the network in Siruseri consisted of
the following nodes and cables: Figure

There are 5 nodes and 8 cable links. Note that a pair of nodes may be
connected by more than one link. These are considered to be different
hops. All links are bidirectional. A packet from node 1 to node 5 may,
for example, travel as follows: 1 to 2, 2 to 1, 1 to 3, 3 to 2, 2 to
1, 1 to 4, 4 to 5, 5 to 4, 4 to 5. This routing is of length 9 (the
number of hops is the length of a given routing). We are interested in
counting the number of different routings from a given source to a
target that are of a given length.

For example, the number of routings from 1 to 2 of length 3 are 7.
They are as follows (separated by ;): 1 to 2, 2 to 1 and 1 to 2; 1 to
3, 3 to 1 and 1 to 2; 1 to 4, 4 to 1 and 1 to 2; 1 to 5, 5 to 1 and 1
to 2; 1 to 4, 4 to 3 (via the left cable) and 3 to 2; 1 to 4, 4 to 3
(via the right cable) and 3 to 2; 1 to 2, 2 to 3 and 3 to 2.

You will be given a description of the network at Siruseri as well as
a source, a target and the number of hops, and your task is to
determine the number of routings from the source to the target which
have the given number of hops. The answer is to be reported modulo

So as discussed on this thread , the solution is to calculate the given matrix to the power k where k is the number of routes given.

Here I did the same:

#include <iostream>
#include <vector>

std::vector<std::vector<int> >MatrixMultiplication(std::vector<std::vector<int> >matrix1,std::vector<std::vector<int> >matrix2,int n){
std::vector<std::vector<int> >retMatrix(n,std::vector<int>(n));

for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
for(int k=0;k<n;k++){
retMatrix[i][j] = retMatrix[i][j] + matrix1[i][k] * matrix2[k][j];

return retMatrix;

std::vector<std::vector<int> >MatrixExponentiation(std::vector<std::vector<int> >matrix,int n,int power){
if(power == 0 || power == 1){
return matrix;

if(power%2 == 0){
return MatrixExponentiation(MatrixMultiplication(matrix,matrix,n),n,power/2);
return MatrixMultiplication(matrix,MatrixExponentiation(MatrixMultiplication(matrix,matrix,n),n,(power-1)/2),n);

int main (int argc, char const* argv[])
int n;
std::cin >> n;
std::vector<std::vector<int> >matrix(n,std::vector<int>(n));
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
std::cin >> matrix[i][j];
int i ,j ,power;
std::cin >> i >> j >> power;
std::vector<std::vector<int> >retMax(n,std::vector<int>(n));
retMax = MatrixExponentiation(matrix,n,power);
std::cout << matrix[i-1][j-1] << std::endl;
return 0;

But the output doesnt matches even for the example case , am I missing something here , or I have to try another approach for this problem?

Edit : As @grigor suggested I changed the code for power == 0 to return the identity matrix but the code still produces wrong outputs ,

if(power == 0){
std::vector<std::vector<int> >retMatrix(n,std::vector<int>(n));
for(int i=0;i<n;i++){
retMatrix[i][i] = 1;
return retMatrix;

Note : I havent wrote the code for modulo thing , do you think it will effect for the example testcase?

Answer Source

I think you are just printing out the wrong value, change:

std::cout << matrix[i-1][j-1] << std::endl;


std::cout << retMax [i-1][j-1] << std::endl;
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