@vs140580 I already answered your (modified) question about finding the length of the shortest walk that returns to the same vertex. Please look through the answer for the procedure and explanation.

@Carl Love 

Download FindCycles.mw

@Carl Love The last one in particular is nice. This operation is one I need to do fairly often and am always frustrated there isn't some simpler (to the average user) way. The help page for solve,details has an example using eval(r, op(indets(r)) = 3) but you have to know it's the third op. My answer has the same problem. Yours allow _Z11 etc, so again you have to know what possibililities might occur. This is compounded by the fact that extra solve commands lead to higher numbers, so the first one isn't always _Z1. To really to do it properly probably some slicing and dicing of the name is necessary.

@Samir Khan Thanks. Most other software I have, the corner handle resizes keeping the aspect ratio the same and handles on the vertical and horizontal edges change one dimension and therefore the aspect ratio.

@Carl Love But this replaces _Z4 as well, which is not what the OP wanted.

@rasmusgs At least for integrals the answer is simple; just add the option numeric, e.g. 

int(sin(x), x = 0 .. Pi, numeric)

gives 2.0000

(But sometimes, as here the exact value is nice, so maybe you want to try without numeric first.)

@vs140580 Here some minimal changes. I made it a row Vector rather than 1x15 Matrix. To see more entries of a Vector or Matrix use for example interface(rtablesize=20) at the beginning of your worksheet.

The last line of the procedure would normally be the return value. (I could have used A rather than return A.) Then you would print outside the procedure; normally print is not used within procedures to provide results.

To convert entries to floats use evalf(A). 

To work with all edges in a single line of code use map or map2. For an example see my other answer here.

Download Try_Degree_based.mw

@jud If I understand your question, then this isn't possible in general. GaloisGroup only gives the permutations between the roots, which is not enough to reconstruct the roots. Here is a simpler example that is my take on it.

Download Galois.mw

@Sphericalmoments A general method would be

1. Find all cycles in the corresponding undirected graph.
2. Find the 0, 1 or 2 directed cycles corresponding to each cycle.
3. Count them.
The DirectedCycles routine I gave earlier does step 2. Step 1 is the difficult step. It is possible to find all cycles from the cycle basis, but the algorithm is nontrivial. So here I give a very inefficient way of doing it, which is probably OK for small graphs.

[Edit: Updated routine works by finding all potential directed cycles on n vertices (rotationally distinct permutations on 1..n (with smallest integer first)) and then sees if the graph contains these edges. Still very inefficient]

Download Cycles4.mw

@Christian Wolinski I've just starting using the Logic package, and saw that in a long expression that I'd Imported and assumed it was an extra variable that was always true, added to put the expression in some sort of standard form. But here is a smaller example, showing it is a bug. I'll submit an SCR.

Download x_true.mw

@lcz I agree, my code actually tests for a cutset, which is a minimal cut, but not all cuts are minimal. Apologies; I sorted all this out before, but forgot about this distinction.

"A cutset S of a connected graph G is a minimal set of edges of G such that removal of S [dis]connects G into exactly two components", p.43, K. Thulasiraman and M.N.S. Swamy, "Graphs, Theory and Algorithms", Wiley 1992. doi:10.1002/9781118033104.

The stackexchange algorithm involving contraction detecting extra edges with loops is not easy to implement in Maple with its Contract or ContractSubgraph commands because it doesn't keep track of where loops came from. But one can just check for extra edges that are in one component before contracting. Here it is: 

Download IsCut.mw

@vv @Christian Wolinski Thanks! Interestingly, I was using a more complicated expression earlier, and I got an error message:

with(Logic):
q:=&or((&not x[2]) &and (&not x[8]),(&not x[2]) &and (&not x[true]),&and(x[3],x[9]
,&not x[true]),&and(&not x[3],&not x[9],&not x[true]),&and(x[2],x[3],x[8],x[9])
,&and(x[2],x[8],&not x[3],&not x[9]),&and(x[3],x[true],&not x[8],&not x[9]),
&and(x[9],x[true],&not x[3],&not x[8])):
SymmetryGroup(q);

gives: Error, (in Logic:-SymmetryGroup) not in conjunctive normal form

so it can test. So then I didn't think about the form, which was a mistake. I'll submit an SCR about this (and Normalize)

@mmcdara Not sure what the OP wants; let's see. I would think a cycle in the underlying (undirected) graph could correspond to 0,1, or 2 directed cycles (neither direction works, clockwise works or counterclockwise works but not both, both directions work). Perhaps like this:

Download Cycles.mw

@sursumCorda I guess that's what the help means by "The isolve command has some limited ability to deal with inequalities." In my experience isolve is limited. In this case, as you probably already tried, you can just inelegantly iterate through all the _Z1 etc values, (Iterator:-Multiseq might be a little faster.)

Download algeqns.mw

@Christian Wolinski Definitely unexpected to get a simpler RootOf here - documentation for convert,radical says "The conversion can fail if Maple cannot find radical expressions for the roots or if the correct radical expression cannot be selected. If the conversion fails, the RootOf remains unchanged." [my bold].