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.
Examples:
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 .
CPP
// C++ implemenatation of above approach#include <bits/stdc++.h>using namespace std;// graphvector<vector<int> > graph;// map to store cost of edgesmap<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++) answer.push_back(INT_MAX); // 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 queue.pop(); // 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) count++; // if the cost is less if (answer[index] > p.first) answer[index] = p.first; // pop the element queue.pop(); // 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, graph[p.second][i])); } // mark as visited visited[p.second] = 1; } return answer;}// main functionint main(){ // create the graph graph.resize(7); // add edge graph[0].push_back(1); graph[0].push_back(3); graph[3].push_back(1); graph[3].push_back(6); graph[3].push_back(4); graph[1].push_back(6); graph[4].push_back(2); graph[4].push_back(5); graph[2].push_back(1); graph[5].push_back(2); graph[5].push_back(6); graph[6].push_back(4); // 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 goal.push_back(6); // 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;} |
Python3
# Python3 implemenatation of above approach# returns the minimum cost in a vector( if# there are multiple goal states)def uniform_cost_search(goal, start): # minimum cost upto # goal state from starting global graph,cost answer = [] # create a priority queue queue = [] # set the answer vector to max value for i in range(len(goal)): answer.append(10**8) # insert the starting index queue.append([0, start]) # map to store visited node visited = {} # count count = 0 # while the queue is not empty while (len(queue) > 0): # get the top element of the queue = sorted(queue) p = queue[-1] # pop the element del queue[-1] # get the original value p[0] *= -1 # check if the element is part of # the goal list if (p[1] in goal): # get the position index = goal.index(p[1]) # if a new goal is reached if (answer[index] == 10**8): count += 1 # if the cost is less if (answer[index] > p[0]): answer[index] = p[0] # pop the element del queue[-1] queue = sorted(queue) if (count == len(goal)): return answer # check for the non visited nodes # which are adjacent to present node if (p[1] not in visited): for i in range(len(graph[p[1]])): # value is multiplied by -1 so that # least priority is at the top queue.append( [(p[0] + cost[(p[1], graph[p[1]][i])])* -1, graph[p[1]][i]]) # mark as visited visited[p[1]] = 1 return answer# main functionif __name__ == '__main__': # create the graph graph,cost = [[] for i in range(8)],{} # add edge graph[0].append(1) graph[0].append(3) graph[3].append(1) graph[3].append(6) graph[3].append(4) graph[1].append(6) graph[4].append(2) graph[4].append(5) graph[2].append(1) graph[5].append(2) graph[5].append(6) graph[6].append(4) # add the cost cost[(0, 1)] = 2 cost[(0, 3)] = 5 cost[(1, 6)] = 1 cost[(3, 1)] = 5 cost[(3, 6)] = 6 cost[(3, 4)] = 2 cost[(2, 1)] = 4 cost[(4, 2)] = 4 cost[(4, 5)] = 3 cost[(5, 2)] = 6 cost[(5, 6)] = 3 cost[(6, 4)] = 7 # goal state goal = [] # set the goal # there can be multiple goal states goal.append(6) # get the answer answer = uniform_cost_search(goal, 0) # prthe answer print("Minimum cost from 0 to 6 is = ",answer[0])# This code is contributed by mohit kumar 29 |
Minimum cost from 0 to 6 is = 3
Complexity: O( m ^ (1+floor(l/e)))
where,
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
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