166 lines
5.1 KiB
JavaScript
166 lines
5.1 KiB
JavaScript
'use strict';
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/******************************************************************************
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* Created 2008-08-19.
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*
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* Dijkstra path-finding functions. Adapted from the Dijkstar Python project.
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*
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* Copyright (C) 2008
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* Wyatt Baldwin <self@wyattbaldwin.com>
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* All rights reserved
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*
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* Licensed under the MIT license.
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*
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* http://www.opensource.org/licenses/mit-license.php
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*
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* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
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* IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
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* FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
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* AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
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* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
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* OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
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* THE SOFTWARE.
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*****************************************************************************/
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var dijkstra = {
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single_source_shortest_paths: function(graph, s, d) {
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// Predecessor map for each node that has been encountered.
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// node ID => predecessor node ID
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var predecessors = {};
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// Costs of shortest paths from s to all nodes encountered.
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// node ID => cost
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var costs = {};
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costs[s] = 0;
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// Costs of shortest paths from s to all nodes encountered; differs from
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// `costs` in that it provides easy access to the node that currently has
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// the known shortest path from s.
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// XXX: Do we actually need both `costs` and `open`?
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var open = dijkstra.PriorityQueue.make();
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open.push(s, 0);
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var closest,
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u, v,
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cost_of_s_to_u,
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adjacent_nodes,
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cost_of_e,
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cost_of_s_to_u_plus_cost_of_e,
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cost_of_s_to_v,
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first_visit;
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while (!open.empty()) {
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// In the nodes remaining in graph that have a known cost from s,
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// find the node, u, that currently has the shortest path from s.
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closest = open.pop();
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u = closest.value;
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cost_of_s_to_u = closest.cost;
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// Get nodes adjacent to u...
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adjacent_nodes = graph[u] || {};
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// ...and explore the edges that connect u to those nodes, updating
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// the cost of the shortest paths to any or all of those nodes as
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// necessary. v is the node across the current edge from u.
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for (v in adjacent_nodes) {
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if (adjacent_nodes.hasOwnProperty(v)) {
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// Get the cost of the edge running from u to v.
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cost_of_e = adjacent_nodes[v];
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// Cost of s to u plus the cost of u to v across e--this is *a*
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// cost from s to v that may or may not be less than the current
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// known cost to v.
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cost_of_s_to_u_plus_cost_of_e = cost_of_s_to_u + cost_of_e;
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// If we haven't visited v yet OR if the current known cost from s to
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// v is greater than the new cost we just found (cost of s to u plus
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// cost of u to v across e), update v's cost in the cost list and
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// update v's predecessor in the predecessor list (it's now u).
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cost_of_s_to_v = costs[v];
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first_visit = (typeof costs[v] === 'undefined');
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if (first_visit || cost_of_s_to_v > cost_of_s_to_u_plus_cost_of_e) {
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costs[v] = cost_of_s_to_u_plus_cost_of_e;
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open.push(v, cost_of_s_to_u_plus_cost_of_e);
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predecessors[v] = u;
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}
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}
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}
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}
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if (typeof d !== 'undefined' && typeof costs[d] === 'undefined') {
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var msg = ['Could not find a path from ', s, ' to ', d, '.'].join('');
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throw new Error(msg);
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}
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return predecessors;
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},
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extract_shortest_path_from_predecessor_list: function(predecessors, d) {
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var nodes = [];
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var u = d;
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var predecessor;
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while (u) {
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nodes.push(u);
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predecessor = predecessors[u];
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u = predecessors[u];
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}
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nodes.reverse();
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return nodes;
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},
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find_path: function(graph, s, d) {
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var predecessors = dijkstra.single_source_shortest_paths(graph, s, d);
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return dijkstra.extract_shortest_path_from_predecessor_list(
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predecessors, d);
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},
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/**
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* A very naive priority queue implementation.
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*/
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PriorityQueue: {
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make: function (opts) {
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var T = dijkstra.PriorityQueue,
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t = {},
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key;
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opts = opts || {};
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for (key in T) {
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if (T.hasOwnProperty(key)) {
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t[key] = T[key];
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}
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}
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t.queue = [];
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t.sorter = opts.sorter || T.default_sorter;
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return t;
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},
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default_sorter: function (a, b) {
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return a.cost - b.cost;
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},
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/**
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* Add a new item to the queue and ensure the highest priority element
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* is at the front of the queue.
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*/
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push: function (value, cost) {
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var item = {value: value, cost: cost};
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this.queue.push(item);
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this.queue.sort(this.sorter);
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},
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/**
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* Return the highest priority element in the queue.
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*/
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pop: function () {
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return this.queue.shift();
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},
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empty: function () {
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return this.queue.length === 0;
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}
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}
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};
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// node.js module exports
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if (typeof module !== 'undefined') {
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module.exports = dijkstra;
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}
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