{"id":8222,"date":"2018-06-04T18:41:14","date_gmt":"2018-06-04T18:41:14","guid":{"rendered":"https:\/\/assignment.essayshark.com\/blog\/?p=8222"},"modified":"2023-01-06T11:48:08","modified_gmt":"2023-01-06T11:48:08","slug":"relationship-graph-construction-example","status":"publish","type":"post","link":"https:\/\/assignmentshark.com\/blog\/relationship-graph-construction-example\/","title":{"rendered":"Relationship Graph Construction Example"},"content":{"rendered":"
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In the example you can read below, you can get information about the\u00a0<\/span>relationship graph\u00a0<\/span>solution that was written by one expert from AssignmentShark. If you have problems with understanding a particular math assignment, you can ask for help from one of our math gurus! “I look for someone to do my homework for free<\/a>. ” – Our help won’t be completely free, but you can learn from the examples yourself, or get advice for the future from our experts.<\/span><\/em><\/p>\n

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Don\u2019t give up even if you struggle with your\u00a0<\/span>relationship graph\u00a0<\/span>assignment and are ready for failure. You can rely on assignment help from online experts. For the AssignmentShark team we have gathered professionals from many fields: physics, math, programming, and many more. Here you can receive unique, high-quality homework assistance that will be delivered strictly on time. Make an order and stay calm: your homework will be done on time.<\/span><\/em><\/p>\n<\/blockquote>\n

Task:<\/span><\/strong><\/p>\n

Construct a graph of the following relationship:<\/span><\/p>\n

\"\"<\/a><\/span>\u00a0<\/span><\/p>\n

Define<\/span>\u00a0<\/span>its properties.<\/span><\/p>\n

Solution:<\/b><\/p>\n

Construct graph <\/span>G<\/span><\/i>(<\/span> X<\/span><\/i> )<\/span> with the set of vertices\u00a0\"\"<\/a>\u00a0<\/span>,<\/span> and two vertices<\/span> X <\/span><\/i>i<\/span><\/i> and <\/span>X<\/span><\/i> j<\/span><\/i> are connected by an edge if and only if\u00a0\u00a0\"\"<\/a><\/span>. Since the relation\u00a0\"\"<\/a>\u00a0<\/span> \u00a0<\/span>is symmetric, graph<\/span> G<\/span><\/i>( <\/span>X<\/span><\/i> ) <\/span>is undirected.<\/span><\/p>\n

\"\"<\/a><\/p>\n

Let\u2019s construct the adjacency matrix (vertices) A:<\/span><\/p>\n\n\n\n\n\n\n\n\n\n
<\/td>\nX<\/span><\/i>1<\/span><\/td>\nX<\/span><\/i> 2<\/span><\/td>\nX<\/span><\/i> 3<\/span><\/td>\nX<\/span><\/i> 4<\/span><\/td>\nX<\/span><\/i> 5<\/span><\/td>\nX<\/span><\/i> 6<\/span><\/td>\n<\/tr>\n
X<\/span><\/i>1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n<\/tr>\n
X <\/span><\/i>2<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n<\/tr>\n
X<\/span><\/i> 3<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n<\/tr>\n
X <\/span><\/i>4<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n<\/tr>\n
X <\/span><\/i>5<\/span><\/td>\n1<\/td>\n0<\/td>\n0<\/td>\n0<\/td>\n0<\/td>\n0<\/td>\n<\/tr>\n
X <\/span><\/i>6<\/span><\/td>\n1<\/td>\n0<\/td>\n0<\/td>\n0<\/td>\n0<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Here Aij element is the number of edges going from vertex\u00a0\u00a0X <\/span><\/i>i<\/span><\/i> \u00a0\u00a0<\/span><\/i>to vertex<\/span> X<\/span><\/i> j<\/span><\/i> . As our\u00a0<\/span>graph is undirected, the adjacency matrix is symmetric.<\/span><\/p>\n

Let\u2019s construct the incidence matrix (ribs) R:<\/span><\/p>\n

 <\/p>\n\n\n\n\n\n\n\n\n\n
<\/td>\ng<\/span><\/i>1<\/span><\/td>\ng<\/span><\/i>2<\/span><\/td>\ng<\/span><\/i>3<\/span><\/td>\ng<\/span><\/i>4<\/span><\/td>\ng<\/span><\/i>5<\/span><\/td>\ng<\/span><\/i>6<\/span><\/td>\ng<\/span><\/i>7<\/span><\/td>\ng<\/span><\/i>8<\/span><\/td>\ng<\/span><\/i>9<\/span><\/td>\ng<\/span><\/i>10<\/span><\/td>\ng<\/span><\/i>11<\/span><\/td>\ng<\/span><\/i>12<\/span><\/td>\n<\/tr>\n
x1<\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n<\/tr>\n
x2<\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n<\/tr>\n
x3<\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n<\/tr>\n
x4<\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n<\/tr>\n
x5<\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n<\/tr>\n
x6<\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Here, the element Rij is 1 if the top vertex of the X i incident to edge g j and 0 otherwise.<\/p>\n

Let\u2019s construct the distance matrix D:<\/p>\n\n\n\n\n\n\n\n\n\n
<\/td>\nx<\/span><\/i>1<\/span><\/td>\nx<\/span><\/i>2<\/span><\/td>\nx<\/span><\/i>3<\/span><\/td>\nx<\/span><\/i>4<\/span><\/td>\nx<\/span><\/i>5<\/span><\/td>\nx<\/span><\/i>6<\/span><\/td>\n<\/tr>\n
x1<\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n<\/tr>\n
x2<\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n2<\/span><\/td>\n<\/tr>\n
x3<\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n2<\/span><\/td>\n2<\/span><\/td>\n<\/tr>\n
x4<\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n2<\/span><\/td>\n2<\/span><\/td>\n<\/tr>\n
x5<\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n<\/tr>\n
x6<\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n0<\/span><\/td>\n1<\/span><\/td>\n0<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Here, the element Dij is the length the shortest path from vertex X i to vertex X j .<\/p>\n

Since our graph is undirected, the distance matrix is symmetric.<\/p>\n

We find the distance vector d, each of which is defined as the component
\n\"\"<\/a> (the maximum distance from the vertex X i to every other vertex).<\/p>\n

Distance vector d = (1, 2, 2, 2, 2, 2). Center is the peak vertex X1 , since it corresponds to the smallest distance (1 = d1 < d j , j = 2,…,6) .<\/p>\n

Peripheral tops: X 2 , X 3 , X 4 , X 5 , X 6 , since it corresponds to the maximum remoteness (d j = 2, j = 2,…,6) .<\/p>\n

Graph radius G( X ) is the center distance, r(G) = 1 .<\/p>\n

Graph diameter G( X ) is the removal of the peripheral vertex, Diam(G) = 2 .<\/p>\n

Let\u2019s find the number for the internal and external stability graph. The largest set of internal stability for our graph has the form S = {X 4 , X 5 , X 6 } (the addition of any other peaks will receive adjacent vertices). Accordingly, graph G( X ) number is equal to the internal stability card(T ) = 1.<\/p>\n

The smallest set for external stability for our graph is T X1 (as any other vertex (not owned to T) is connected to the apex of X1 from T). The number for the external sustainability graph G( X ) is equal to card(T ) 1.<\/p>\n","protected":false},"excerpt":{"rendered":"

In the example you can read below, you can get information about the\u00a0relationship graph\u00a0solution that was written by one expert from AssignmentShark. If you have problems with understanding a particular math assignment, you can ask for help from one of our math gurus! “I look for someone to do my homework for free. ” – […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[55,35],"tags":[],"class_list":["post-8222","post","type-post","status-publish","format-standard","hentry","category-math","category-samples"],"_links":{"self":[{"href":"https:\/\/assignmentshark.com\/blog\/wp-json\/wp\/v2\/posts\/8222","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/assignmentshark.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/assignmentshark.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/assignmentshark.com\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/assignmentshark.com\/blog\/wp-json\/wp\/v2\/comments?post=8222"}],"version-history":[{"count":34,"href":"https:\/\/assignmentshark.com\/blog\/wp-json\/wp\/v2\/posts\/8222\/revisions"}],"predecessor-version":[{"id":13407,"href":"https:\/\/assignmentshark.com\/blog\/wp-json\/wp\/v2\/posts\/8222\/revisions\/13407"}],"wp:attachment":[{"href":"https:\/\/assignmentshark.com\/blog\/wp-json\/wp\/v2\/media?parent=8222"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/assignmentshark.com\/blog\/wp-json\/wp\/v2\/categories?post=8222"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/assignmentshark.com\/blog\/wp-json\/wp\/v2\/tags?post=8222"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}