Uniform-Cost Search (Dijkstra for large Graphs)

Uniform-Cost Search is a variant of Dijikstra’s algorithm. Here, instead of inserting all vertices into a priority queue, we insert only source, then one by one insert when needed. In every step, we check if the item is already in priority queue (using visited array). If yes, we perform decrease key, else we insert it.
This variant of Dijsktra is useful for infinite graphs and those graph which are too large to represent in the memory. Uniform-Cost Search is mainly used in Artificial Intelligence.


Input :

Output :
Minimum cost from S to G is =3

Uniform-Cost Search is similar to Dijikstra’s algorithm . In this algorithm from the starting state we will visit the adjacent states and will choose the least costly state then we will choose the next least costly state from the all un-visited and adjacent states of the visited states, in this way we will try to reach the goal state (note we wont continue the path through a goal state ), even if we reach the goal state we will continue searching for other possible paths( if there are multiple goals) . We will keep a priority queue which will give the least costliest next state from all the adjacent states of visited states .





// C++ implemenatation of above approach
#include <bits/stdc++.h>
using namespace std;
// graph
vector<vector<int> > graph;
// map to store cost of edges
map<pair<int, int>, int> cost;
// returns the minimum cost in a vector( if 
// there are multiple goal states)
vector<int> uniform_cost_search(vector<int> goal, int start)
    // minimum cost upto
    // goal state from starting
    // state
    vector<int> answer;
    // create a priority queue
    priority_queue<pair<int, int> > queue;
    // set the answer vector to max value
    for (int i = 0; i < goal.size(); i++)
    // insert the starting index
    queue.push(make_pair(0, start));
    // map to store visited node
    map<int, int> visited;
    // count
    int count = 0;
    // while the queue is not empty
    while (queue.size() > 0) {
        // get the top element of the 
        // priority queue
        pair<int, int> p = queue.top();
        // pop the element
        // get the original value
        p.first *= -1;
        // check if the element is part of
        // the goal list
        if (find(goal.begin(), goal.end(), p.second) != goal.end()) {
            // get the position
            int index = find(goal.begin(), goal.end(), 
                             p.second) - goal.begin();
            // if a new goal is reached
            if (answer[index] == INT_MAX)
            // if the cost is less
            if (answer[index] > p.first)
                answer[index] = p.first;
            // pop the element
            // if all goals are reached
            if (count == goal.size())
                return answer;
        // check for the non visited nodes
        // which are adjacent to present node
        if (visited[p.second] == 0)
            for (int i = 0; i < graph[p.second].size(); i++) {
                // value is multiplied by -1 so that 
                // least priority is at the top
                queue.push(make_pair((p.first + 
                  cost[make_pair(p.second, graph[p.second][i])]) * -1, 
        // mark as visited
        visited[p.second] = 1;
    return answer;
// main function
int main()
    // create the graph
    // add edge
    // add the cost
    cost[make_pair(0, 1)] = 2;
    cost[make_pair(0, 3)] = 5;
    cost[make_pair(1, 6)] = 1;
    cost[make_pair(3, 1)] = 5;
    cost[make_pair(3, 6)] = 6;
    cost[make_pair(3, 4)] = 2;
    cost[make_pair(2, 1)] = 4;
    cost[make_pair(4, 2)] = 4;
    cost[make_pair(4, 5)] = 3;
    cost[make_pair(5, 2)] = 6;
    cost[make_pair(5, 6)] = 3;
    cost[make_pair(6, 4)] = 7;
    // goal state
    vector<int> goal;
    // set the goal
    // there can be multiple goal states
    // get the answer
    vector<int> answer = uniform_cost_search(goal, 0);
    // print the answer
    cout << "Minimum cost from 0 to 6 is = " 
         << answer[0] << endl;
    return 0;



Minimum cost from 0 to 6 is = 3

Complexity: O( m ^ (1+floor(l/e)))
m is the maximum number of neighbor a node has
l is the length of the shortest path to the goal state
e is the least cost of an edge

My Personal Notes arrow_drop_up

Second year Department of Information Technology Jadavpur University

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