Hello, which operator should I use instead of ":=" (assignment statement) so that the graph H remains unchanged in the code below?

with(GraphTheory);
G := CompleteGraph(4);
DrawGraph(G);
H := G;
DeleteEdge(G, {1, 2}, inplace = true);
DrawGraph(G);
DrawGraph(H);

Is it possible to change the caption of RiemannSum command in Student[Calculus1] package with animation output option, keeping approximate value and number of partitions outputs?  I tried to use typeset command, but I haven’t succeeded.

Best regards,

Emrah

 

Hi friends, In my maple procedure I m calculating large amount of point then I m plotting them. Although I don’t know how many points will be calculated, I have a logical upper bound for number of points. I was using the method 1: for i from 1 to N do if something is okey then x := ... y := ... data := [op(data), [x,y]]: end if; od: but I have relalized that this is really slow. Then I started to use Method 2: data := array(1..N): j := 0; for i from 1 to N do if something is okey then x := ... y := ... j := j + 1; data[j] := [x,y]:

In a loop I m calculating x and y coordinates of a point. Then, I m plotting this points with plot command. But my loop takes too much time so I haven’t seen anything until the loop end. Is there any way to see these points immediately? My code is like: data := []; For i from 1 to LargeInteger do … x := … y := … … data := [ op(data), [x,y] ]; … od; plot(data, style=point); Thanks...

Hi I have a problem. As you know the number of strings of length n, composed of k given elements is k^n. I want to list all this k^n strings. For example if the length is 3 (n=3) and elements are choosing from {0,1} we obtain: 000 001 010 011 100 101 110 111 Actually, for n=3 the solution is simple: B:=[0,1]: for i from 1 to nops(B) do for j from 1 to nops(B) do for k from 1 to nops(B) do print(B[i],B[j],B[k]); od: od: od; I need a code working for all n any idea? Emrah AKYAR Anadolu University Faculty of Science Department of Mathematics