{"id":26495,"date":"2017-12-20T21:25:11","date_gmt":"2017-12-20T15:55:11","guid":{"rendered":"https:\/\/www.wikitechy.com\/technology\/?p=26495"},"modified":"2017-12-20T21:25:11","modified_gmt":"2017-12-20T15:55:11","slug":"c-programming-backtracking-hamiltonian-cycle","status":"publish","type":"post","link":"https:\/\/www.wikitechy.com\/technology\/c-programming-backtracking-hamiltonian-cycle\/","title":{"rendered":"C Programming – Backtracking | Set 6 (Hamiltonian Cycle)"},"content":{"rendered":"

Hamiltonian Path<\/a> in an undirected graph is a path that visits each vertex exactly once. A Hamiltonian cycle (or Hamiltonian circuit) is a Hamiltonian Path such that there is an edge (in graph) from the last vertex to the first vertex of the Hamiltonian Path.<\/span> Determine whether a given graph contains Hamiltonian Cycle or not. If it contains, then print the path. Following are the input and output of the required function.<\/p>\n

Input:<\/em>
\nA 2D array graph[V][V] where V is the number of vertices in graph and graph[V][V] is adjacency matrix representation of the graph. A value graph[i][j] is 1 if there is a direct edge from i to j, otherwise graph[i][j] is 0.<\/p>\n

Output:<\/em>
\nAn array path[V] that should contain the Hamiltonian Path. path[i] should represent the ith vertex in the Hamiltonian Path. The code should also return false if there is no Hamiltonian Cycle in the graph.<\/p>\n[ad type=”banner”]\n

For example, a Hamiltonian Cycle in the following graph is {0, 1, 2, 4, 3, 0}. There are more Hamiltonian Cycles in the graph like {0, 3, 4, 2, 1, 0}<\/p>\n

(0)--(1)--(2)\r\n |   \/ \\   |\r\n |  \/   \\  | \r\n | \/     \\ |\r\n(3)-------(4)\r\n<\/pre>\n
And the following graph doesn\u2019t contain any Hamiltonian Cycle.<\/p>\n
(0)--(1)--(2)\r\n |   \/ \\   |\r\n |  \/   \\  | \r\n | \/     \\ |\r\n(3)      (4) \r\n<\/pre>\n
\n
Recommended: Please solve it on “PRACTICE<\/u><\/i><\/b>” first, before moving on to the solution.<\/a><\/h2>\n<\/div>\n
Naive Algorithm<\/strong>
\nGenerate all possible configurations of vertices and print a configuration that satisfies the given constraints. There will be n! (n factorial) configurations.<\/p>\n
while there are untried conflagrations\r\n{\r\n   generate the next configuration\r\n   if ( there are edges between two consecutive vertices of this\r\n      configuration and there is an edge from the last vertex to \r\n      the first ).\r\n   {\r\n      print this configuration;\r\n      break;\r\n   }\r\n}\r\n<\/pre>\n
Backtracking Algorithm<\/strong>
\nCreate an empty path array and add vertex 0 to it. Add other vertices, starting from the vertex 1. Before adding a vertex, check for whether it is adjacent to the previously added vertex and not already added. If we find such a vertex, we add the vertex as part of the solution. If we do not find a vertex then we return false.<\/p>\n[ad type=”banner”]\n
Implementation of Backtracking solution<\/strong>
\nFollowing are implementations of the Backtracking solution.<\/p>\n
C++ program:<\/strong><\/p>\n\n\n\n
\n
\n
\/* C\/C++ program for solution of Hamiltonian Cycle problem<\/code><\/div>\n
\u00a0\u00a0\u00a0<\/code>using backtracking *\/<\/code><\/div>\n
#include<stdio.h><\/code><\/div>\n
<\/div>\n
\/\/ Number of vertices in the graph<\/code><\/div>\n
#define V 5<\/code><\/div>\n
<\/div>\n
void<\/code> printSolution(<\/code>int<\/code> path[]);<\/code><\/div>\n
<\/div>\n
\/* A utility function to check if the vertex v can be added at<\/code><\/div>\n
\u00a0\u00a0\u00a0<\/code>index 'pos' in the Hamiltonian Cycle constructed so far (stored<\/code><\/div>\n
\u00a0\u00a0\u00a0<\/code>in 'path[]') *\/<\/code><\/div>\n
bool<\/code> isSafe(<\/code>int<\/code> v, <\/code>bool<\/code> graph[V][V], <\/code>int<\/code> path[], <\/code>int<\/code> pos)<\/code><\/div>\n
{<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>\/* Check if this vertex is an adjacent vertex of the previously<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>added vertex. *\/<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>if<\/code> (graph [ path[pos-1] ][ v ] == 0)<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>return<\/code> false<\/code>;<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>\/* Check if the vertex has already been included.<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>This step can be optimized by creating an array of size V *\/<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>for<\/code> (<\/code>int<\/code> i = 0; i < pos; i++)<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>if<\/code> (path[i] == v)<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>return<\/code> false<\/code>;<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>return<\/code> true<\/code>;<\/code><\/div>\n
}<\/code><\/div>\n
<\/div>\n
\/* A recursive utility function to solve hamiltonian cycle problem *\/<\/code><\/div>\n
bool<\/code> hamCycleUtil(<\/code>bool<\/code> graph[V][V], <\/code>int<\/code> path[], <\/code>int<\/code> pos)<\/code><\/div>\n
{<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>\/* base case: If all vertices are included in Hamiltonian Cycle *\/<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>if<\/code> (pos == V)<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>{<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>\/\/ And if there is an edge from the last included vertex to the<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>\/\/ first vertex<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>if<\/code> ( graph[ path[pos-1] ][ path[0] ] == 1 )<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>return<\/code> true<\/code>;<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>else<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>return<\/code> false<\/code>;<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>}<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>\/\/ Try different vertices as a next candidate in Hamiltonian Cycle.<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>\/\/ We don't try for 0 as we included 0 as starting point in in hamCycle()<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>for<\/code> (<\/code>int<\/code> v = 1; v < V; v++)<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>{<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>\/* Check if this vertex can be added to Hamiltonian Cycle *\/<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>if<\/code> (isSafe(v, graph, path, pos))<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>{<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>path[pos] = v;<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>\/* recur to construct rest of the path *\/<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>if<\/code> (hamCycleUtil (graph, path, pos+1) == <\/code>true<\/code>)<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>return<\/code> true<\/code>;<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>\/* If adding vertex v doesn't lead to a solution,<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>then remove it *\/<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>path[pos] = -1;<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>}<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>}<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>\/* If no vertex can be added to Hamiltonian Cycle constructed so far,<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>then return false *\/<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>return<\/code> false<\/code>;<\/code><\/div>\n
}<\/code><\/div>\n
<\/div>\n
\/* This function solves the Hamiltonian Cycle problem using Backtracking.<\/code><\/div>\n
\u00a0\u00a0<\/code>It mainly uses hamCycleUtil() to solve the problem. It returns false<\/code><\/div>\n
\u00a0\u00a0<\/code>if there is no Hamiltonian Cycle possible, otherwise return true and<\/code><\/div>\n
\u00a0\u00a0<\/code>prints the path. Please note that there may be more than one solutions,<\/code><\/div>\n
\u00a0\u00a0<\/code>this function prints one of the feasible solutions. *\/<\/code><\/div>\n
bool<\/code> hamCycle(<\/code>bool<\/code> graph[V][V])<\/code><\/div>\n
{<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>int<\/code> *path = <\/code>new<\/code> int<\/code>[V];<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>for<\/code> (<\/code>int<\/code> i = 0; i < V; i++)<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>path[i] = -1;<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>\/* Let us put vertex 0 as the first vertex in the path. If there is<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>a Hamiltonian Cycle, then the path can be started from any point<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>of the cycle as the graph is undirected *\/<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>path[0] = 0;<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>if<\/code> ( hamCycleUtil(graph, path, 1) == <\/code>false<\/code> )<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>{<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>printf<\/code>(<\/code>\"\\nSolution does not exist\"<\/code>);<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>return<\/code> false<\/code>;<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>}<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>printSolution(path);<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>return<\/code> true<\/code>;<\/code><\/div>\n
}<\/code><\/div>\n
<\/div>\n
\/* A utility function to print solution *\/<\/code><\/div>\n
void<\/code> printSolution(<\/code>int<\/code> path[])<\/code><\/div>\n
{<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>printf<\/code> (<\/code>\"Solution Exists:\"<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>\" Following is one Hamiltonian Cycle \\n\"<\/code>);<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>for<\/code> (<\/code>int<\/code> i = 0; i < V; i++)<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>printf<\/code>(<\/code>\" %d \"<\/code>, path[i]);<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>\/\/ Let us print the first vertex again to show the complete cycle<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>printf<\/code>(<\/code>\" %d \"<\/code>, path[0]);<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>printf<\/code>(<\/code>\"\\n\"<\/code>);<\/code><\/div>\n
}<\/code><\/div>\n
<\/div>\n
\/\/ driver program to test above function<\/code><\/div>\n
int<\/code> main()<\/code><\/div>\n
{<\/code><\/div>\n
\u00a0\u00a0\u00a0<\/code>\/* Let us create the following graph<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>(0)--(1)--(2)<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>|\u00a0\u00a0 \/ \\\u00a0\u00a0 |<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>|\u00a0 \/\u00a0\u00a0 \\\u00a0 |<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>| \/\u00a0\u00a0\u00a0\u00a0 \\ |<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>(3)-------(4)\u00a0\u00a0\u00a0 *\/<\/code><\/div>\n
\u00a0\u00a0\u00a0<\/code>bool<\/code> graph1[V][V] = {{0, 1, 0, 1, 0},<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>{1, 0, 1, 1, 1},<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>{0, 1, 0, 0, 1},<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>{1, 1, 0, 0, 1},<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>{0, 1, 1, 1, 0},<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>};<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>\/\/ Print the solution<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>hamCycle(graph1);<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0<\/code>\/* Let us create the following graph<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>(0)--(1)--(2)<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>|\u00a0\u00a0 \/ \\\u00a0\u00a0 |<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>|\u00a0 \/\u00a0\u00a0 \\\u00a0 |<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>| \/\u00a0\u00a0\u00a0\u00a0 \\ |<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>(3)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (4)\u00a0\u00a0\u00a0 *\/<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>bool<\/code> graph2[V][V] = {{0, 1, 0, 1, 0},<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>{1, 0, 1, 1, 1},<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>{0, 1, 0, 0, 1},<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>{1, 1, 0, 0, 0},<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>{0, 1, 1, 0, 0},<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/code>};<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>\/\/ Print the solution<\/code><\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>hamCycle(graph2);<\/code><\/div>\n
<\/div>\n
\u00a0\u00a0\u00a0\u00a0<\/code>return<\/code> 0;<\/code><\/div>\n
}<\/code><\/div>\n
\n
Output:<\/p>\n
Solution Exists: Following is one Hamiltonian Cycle\r\n 0  1  2  4  3  0\r\n\r\nSolution does not exist\r\n<\/pre>\n<\/div>\n<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n[ad type=”banner”]\n","protected":false},"excerpt":{"rendered":"
C Programming – Backtracking – Hamiltonian Cycle – Create an empty path array and add vertex 0 to it. Add other vertices, starting from the vertex 1<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[73904,69866,1,73906,78521],"tags":[78964,75506,78968,78963,78966,78965,78967],"class_list":["post-26495","post","type-post","status-publish","format-standard","hentry","category-backtracking","category-c-programming","category-coding","category-graph-algorithms","category-hard-problems","tag-hamiltonian-circuit-problem-algorithm","tag-hamiltonian-cycle-algorithm-using-backtracking-in-c","tag-hamiltonian-cycle-algorithm-using-backtracking-pdf","tag-hamiltonian-cycle-algorithm-using-backtracking-ppt","tag-hamiltonian-cycle-algorithm-using-backtracking-time-complexity","tag-hamiltonian-cycle-example","tag-hamiltonian-path-algorithm-pseudocode"],"_links":{"self":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/26495","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/comments?post=26495"}],"version-history":[{"count":0,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/26495\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/media?parent=26495"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/categories?post=26495"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/tags?post=26495"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}