2022-06-02 07:11:29 +03:00
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/*
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Exploring Dijkstra,
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The data example is from
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https://www.geeksforgeeks.org/dijkstras-shortest-path-algorithm-greedy-algo-7/
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by CCS
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Dijkstra's single source shortest path algorithm.
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The program uses an adjacency matrix representation of a graph
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This Dijkstra algorithm uses a priority queue to save
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the shortest paths. The queue structure has a data
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which is the number of the node,
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and the priority field which is the shortest distance.
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PS: all the pre-requisites of Dijkstra are considered
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$ v run file_name.v
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// Creating a executable
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$ v run file_name.v -o an_executable.EXE
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$ ./an_executable.EXE
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Code based from : Data Structures and Algorithms Made Easy: Data Structures and Algorithmic Puzzles, Fifth Edition (English Edition)
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pseudo code written in C
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This idea is quite different: it uses a priority queue to store the current
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shortest path evaluted
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The priority queue structure built using a list to simulate
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the queue. A heap is not used in this case.
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*/
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// a structure
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struct NODE {
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mut:
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data int // NUMBER OF NODE
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priority int // Lower values priority indicate ==> higher priority
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}
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// Function to push according to priority ... the lower priority is goes ahead
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// The "push" always sorted in pq
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fn push_pq<T>(mut prior_queue []T, data int, priority int) {
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mut temp := []T{}
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lenght_pq := prior_queue.len
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mut i := 0
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for (i < lenght_pq) && (priority > prior_queue[i].priority) {
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temp << prior_queue[i]
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i++
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}
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// INSERTING SORTED in the queue
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temp << NODE{data, priority} // do the copy in the right place
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// copy the another part (tail) of original prior_queue
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for i < lenght_pq {
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temp << prior_queue[i]
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i++
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}
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prior_queue = temp.clone() // I am not sure if it the right way
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// IS IT THE RIGHT WAY?
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}
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// Change the priority of a value/node ... exist a value, change its priority
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fn updating_priority<T>(mut prior_queue []T, search_data int, new_priority int) {
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mut i := 0
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mut lenght_pq := prior_queue.len
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for i < lenght_pq {
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if search_data == prior_queue[i].data {
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prior_queue[i] = NODE{search_data, new_priority} // do the copy in the right place
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break
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}
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i++
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// all the list was examined
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if i >= lenght_pq {
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2022-11-15 16:53:13 +03:00
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print('\n This data ${search_data} does exist ... PRIORITY QUEUE problem\n')
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2022-06-02 07:11:29 +03:00
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exit(1) // panic(s string)
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}
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} // end for
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}
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// a single departure or remove from queue
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fn departure_priority<T>(mut prior_queue []T) int {
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mut x := prior_queue[0].data
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prior_queue.delete(0) // or .delete_many(0, 1 )
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return x
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}
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// give a NODE v, return a list with all adjacents
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// Take care, only positive EDGES
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fn all_adjacents<T>(g [][]T, v int) []int {
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2022-07-08 15:35:45 +03:00
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mut temp := []int{}
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2022-06-02 07:11:29 +03:00
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for i in 0 .. (g.len) {
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if g[v][i] > 0 {
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temp << i
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}
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}
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return temp
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}
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// print the costs from origin up to all nodes
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fn print_solution<T>(dist []T) {
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print('Vertex \tDistance from Source')
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for node in 0 .. (dist.len) {
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2022-11-15 16:53:13 +03:00
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print('\n ${node} ==> \t ${dist[node]}')
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2022-06-02 07:11:29 +03:00
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}
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}
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// print all paths and their cost or weight
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fn print_paths_dist<T>(path []T, dist []T) {
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print('\n Read the nodes from right to left (a path): \n')
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for node in 1 .. (path.len) {
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2022-11-15 16:53:13 +03:00
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print('\n ${node} ')
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2022-06-02 07:11:29 +03:00
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mut i := node
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for path[i] != -1 {
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print(' <= ${path[i]} ')
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i = path[i]
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}
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print('\t PATH COST: ${dist[node]}')
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}
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}
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// check structure from: https://www.geeksforgeeks.org/dijkstras-shortest-path-algorithm-greedy-algo-7/
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// s: source for all nodes
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// Two results are obtained ... cost and paths
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fn dijkstra(g [][]int, s int) {
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mut pq_queue := []NODE{} // creating a priority queue
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push_pq(mut pq_queue, s, 0) // goes s with priority 0
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mut n := g.len
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mut dist := []int{len: n, init: -1} // dist with -1 instead of INIFINITY
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mut path := []int{len: n, init: -1} // previous node of each shortest paht
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// Distance of source vertex from itself is always 0
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dist[s] = 0
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for pq_queue.len != 0 {
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mut v := departure_priority(mut pq_queue)
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// for all W adjcents vertices of v
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mut adjs_of_v := all_adjacents(g, v) // all_ADJ of v ....
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// print('\n ADJ ${v} is ${adjs_of_v}')
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mut new_dist := 0
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for w in adjs_of_v {
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new_dist = dist[v] + g[v][w]
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if dist[w] == -1 {
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dist[w] = new_dist
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push_pq(mut pq_queue, w, dist[w])
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path[w] = v // collecting the previous node -- lowest weight
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}
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if dist[w] > new_dist {
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dist[w] = new_dist
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updating_priority(mut pq_queue, w, dist[w])
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2022-07-08 15:35:45 +03:00
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path[w] = v
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2022-06-02 07:11:29 +03:00
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}
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}
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}
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// print the constructed distance array
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print_solution(dist)
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// print('\n \n Previous node of shortest path: ${path}')
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print_paths_dist(path, dist)
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}
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/*
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Solution Expected
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Vertex Distance from Source
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0 0
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1 4
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2 12
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3 19
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4 21
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5 11
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6 9
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7 8
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8 14
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*/
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fn main() {
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// adjacency matrix = cost or weight
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graph_01 := [
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[0, 4, 0, 0, 0, 0, 0, 8, 0],
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[4, 0, 8, 0, 0, 0, 0, 11, 0],
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[0, 8, 0, 7, 0, 4, 0, 0, 2],
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[0, 0, 7, 0, 9, 14, 0, 0, 0],
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[0, 0, 0, 9, 0, 10, 0, 0, 0],
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[0, 0, 4, 14, 10, 0, 2, 0, 0],
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[0, 0, 0, 0, 0, 2, 0, 1, 6],
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[8, 11, 0, 0, 0, 0, 1, 0, 7],
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[0, 0, 2, 0, 0, 0, 6, 7, 0],
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]
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graph_02 := [
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[0, 2, 0, 6, 0],
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[2, 0, 3, 8, 5],
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[0, 3, 0, 0, 7],
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[6, 8, 0, 0, 9],
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[0, 5, 7, 9, 0],
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]
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// data from https://www.geeksforgeeks.org/prims-minimum-spanning-tree-mst-greedy-algo-5/
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/*
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The graph:
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2 3
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(0)--(1)--(2)
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| / \ |
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6| 8/ \5 |7
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| / \ |
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(3)-------(4)
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9
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*/
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/*
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Let us create following weighted graph
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From https://www.geeksforgeeks.org/kruskals-minimum-spanning-tree-algorithm-greedy-algo-2/?ref=lbp
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10
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0--------1
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| \ |
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6| 5\ |15
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| \ |
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2--------3
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4
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*/
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graph_03 := [
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[0, 10, 6, 5],
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[10, 0, 0, 15],
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[6, 0, 0, 4],
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[5, 15, 4, 0],
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]
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// To find number of coluns
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// mut cols := an_array[0].len
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mut graph := [][]int{} // the graph: adjacency matrix
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// for index, g_value in [graph_01, graph_02, graph_03] {
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for index, g_value in [graph_01, graph_02, graph_03] {
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graph = g_value.clone() // graphs_sample[g].clone() // choice your SAMPLE
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// allways starting by node 0
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start_node := 0
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2022-11-15 16:53:13 +03:00
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println('\n\n Graph ${index + 1} using Dijkstra algorithm (source node: ${start_node})')
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2022-06-02 07:11:29 +03:00
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dijkstra(graph, start_node)
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}
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println('\n BYE -- OK')
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}
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