r,for-loop,plot,dataframes,graph-coloring

Your last line of the loop will overwrite all previous operations. V(g)$color <- ifelse(V(g)$name %in% c(aa,nnn), rainbow(i), "white") It will check for correspondence with the last i, and make everything else white. Use the following one instead, it does nothing when if is not answered. V(g)$color <- if(V(g)$name %in% c(aa,nnn))...

c++,boost,graph,boost-graph,graph-coloring

Hehe. This is gonna be the third boost bug to report today. The dimacs parsing code looks truly evil to me. The interface is very inconvenient and the implementation... Well, as you found it's very wrong: In read_edge_line there is this loop: if ('e' == linebuf[0]) { while (*tmp !=...

r,pie-chart,igraph,graph-coloring

vertex.pie requires a list of values/proportions that each color will have in the pie. You need to convert your color_* variables to that. To do so, you can first bind them together: v<-cbind(V(test_graph)$color_1,V(test_graph)$color_2,V(test_graph)$color_3,V(test_graph)$color_4) Then for each vertex, make a vector of 6 values (one for each color) containing 0 and...

Replace: colorGraph(graph, bindings) :- colorItem(graph,bindings). With: colorGraph(Graph, Bindings) :- colorItem(Graph, Bindings). Variables start with either an underscore or an upper case letter in Prolog. Same issue apparently with some of the other predicates....

r,logistic-regression,graph-coloring

Does this do what you want ? plot(predict(reg),residuals(reg),col=ifelse(residuals(reg)<0,"blue","red")) I just test whether the residual is larger than 0 or not. The main idea is to create a vector of colour of the same length of your data....

graph,discrete-mathematics,graph-coloring

in here it is so clear,by the way the sentence is In an optimal coloring there must be at least one of the graphâ€™s m edges between every pair of color classes I'll break it down and make every part more clear. In an optimal coloring : this means that...