2020-07-19 16:36:52 -0500 received badge ● Good Answer (source) 2020-07-15 22:16:59 -0500 received badge ● Nice Question (source) 2019-11-11 20:30:14 -0500 received badge ● Good Answer (source) 2019-08-27 14:54:11 -0500 received badge ● Nice Answer (source) 2019-06-10 09:25:32 -0500 received badge ● Nice Answer (source) 2019-03-12 15:11:08 -0500 received badge ● Taxonomist 2018-06-30 20:42:10 -0500 commented question How can I change the sage console prompt? If relevant to anyone with a dark background in their terminal wanting to change the color in the prompt, an option is to go to colors under src/sage/repl/prompts.py. Possible values to return are any of [u'Neutral', u'NoColor', u'LightBG', u'Linux']. 2018-06-30 19:05:01 -0500 commented answer Drawing Auslander-Reiten quivers with sage possible? You are welcome :-) 2018-04-03 02:35:49 -0500 received badge ● Nice Answer (source) 2017-10-10 21:09:17 -0500 edited question What are the following commands telling us? R=Integers() [R.ideal([a,b]) == R.ideal([gcd(a,b)]) for a in range(1,20) for b in range(1,20)] What are the following commands telling us? R=Integers() [R.ideal([a,b]) == R.ideal([gcd(a,b)]) for a in range(1,20) for b in range(1,20)]  2017-10-08 15:58:45 -0500 commented answer Finding all simply laced Dynkin graphs with a given number of vertices up to isomorphism @maresage Would you please be so kind to try and donate via paypal to the address listed here? This info is also available here(They're collecting money for helping repair damage from the quake in Oaxaca). If this is not possible, then donating to sagemath would be nice. You can donate through the top right link at sagemath.org or via this direct link. You may contact me via the fidelbarrera+asksage gmail account for the receipt or to let me know if none of this worked. Thanks! 2017-10-06 12:12:15 -0500 answered a question Finding all simply laced Dynkin graphs with a given number of vertices up to isomorphism Here is some code that might do the job. Please feel free to test it and let me know if this works. def directg(g6): import subprocess sp = subprocess.Popen("directg -o", shell=True, stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.PIPE, close_fds=True) stdout = sp.communicate(input='{0}\n'.format(g6)) d6_strings = stdout.split() for d6 in d6_strings: yield DiGraph(d6[1:]) def A_base(n): return graphs.PathGraph(n).graph6_string() def D_base(n): g = graphs.PathGraph(n-1) g.add_edge(n-3,n-1) return g.graph6_string() def E_base(n): g = graphs.PathGraph(n-1) g.add_edge(2,n-1) return g.graph6_string() def generate_DynkinQuivers(n,kind='A'): if kind not in 'AED': raise ValueError("Invalid type of Dynkin quiver, should be one of 'A', 'D' or 'E'.") if kind == 'A': g6_base = A_base(n) elif kind=='D': g6_base = D_base(n) elif kind=='E': g6_base = E_base(n) output_template = 'DynkinQuiver("{kind}",{size},[{orientation}])' for digraph in directg(g6_base): orientation = [] for i in range(n-2): if digraph.has_edge(i,i+1): orientation.append('"r"') elif digraph.has_edge(i+1,i): orientation.append('"l"') else: raise ValueError("Format conversion error.") if kind in 'AD': if kind=='A': other = n-2 else: other = n-3 if digraph.has_edge(other,n-1): orientation.append('"r"') elif digraph.has_edge(n-1,other): orientation.append('"l"') else: raise ValueError("Format conversion error. Last edge of kind {0} quiver.".format(kind)) if kind =='E': if digraph.has_edge(n-1,2): orientation.append('"d"') elif digraph.has_edge(2,n-1): orientation.append('"u"') else: raise ValueError("Format conversion error. Last edge of kind E quiver.") yield output_template.format(kind=kind,size=n,orientation=','.join(orientation)) def DynkinQuivers(n,kind='A'): return '[{0}]'.format(','.join(generate_DynkinQuivers(n,kind)))  Sample usage: DynkinQuivers(3,'A')  produces '[DynkinQuiver("A",3,["r","r"]),DynkinQuiver("A",3,["r","l"]),DynkinQuiver("A",3,["l","r"])]'  Which can be saved to a file. It takes about half a second to go up to n=13 %time output=DynkinQuivers(13,'E') CPU time: 0.41 s, Wall time: 0.53 s  2017-10-06 11:29:04 -0500 commented question Finding all simply laced Dynkin graphs with a given number of vertices up to isomorphism Guess the purpose of $E_n$ is having the $n$-th vertex anchored at the third vertex of the path from left to right, correct? 2017-10-06 11:27:58 -0500 commented question Finding all simply laced Dynkin graphs with a given number of vertices up to isomorphism What exactly do we mean by "fast code"? Any memory usage constraints? Is a 1 minute wait reasonable for $n=13$? 2017-10-06 11:02:42 -0500 commented question Finding all simply laced Dynkin graphs with a given number of vertices up to isomorphism Is $E_n$ defined for all $n\geq 6$? Or is it just for $n\in{6,7,8}$? 2017-09-27 16:33:28 -0500 commented question How to run maple in cocal? Provided you manage to install it, it might be possible. 2017-09-24 16:34:55 -0500 commented question Finding posets of a certain form 2 Depending on what you are planning to do with these posets, it might be worth filtering them up to a certain size and store the output. So that you can just go back and read from the stored list. However, if you are interested in say, counting them, then maybe generating them will be the way to go. 2017-09-24 16:31:52 -0500 commented question Finding posets of a certain form 2 This looks like a very particular set of posets. Maybe you'll need to generate them manually if you do not want to filter from the list of all posets. 2017-09-24 16:09:02 -0500 commented question Finding posets of a certain form 2 By Hasse quiver, do you mean Hasse diagram? If so, then you may use the DiGraph methods on the diagram to check the out-degree and the in-degree. See for example out_degree and in_degree. 2017-09-24 16:05:14 -0500 edited question Finding posets of a certain form 2 I try to obtain the connected posets with n elements having a top and bottom such that in the Hasse quiver, in every point there start at most two arrows and at every point there end at most two arrows (thus for example all distributive lattices are of this form). Is there a quick command to obtain this as a list? What condition does one has to add in posets = [ p for p in Posets(n) if p.is_connected() and p.has_top() and p.has_bottom()]  Actually a quick method would be cool which works up to n=10 or 11. (filtering things out of the set of all posets, seems to be very slow in SAGE) 2017-09-24 16:01:36 -0500 commented answer Output too long in SAGE @maresage L[0:40] will output all elements in L with indices i such that 0<=i<40. Thus L will not be contained in L[0:40]. However L[40:80] will contain L. 2017-09-23 19:57:22 -0500 commented question Output too long in SAGE [Bug report (?)] Should P.is_lattice be P.is_lattice() in the second to last line in the code above? 2017-09-23 19:54:27 -0500 answered a question Output too long in SAGE To save to a file, you may try the following. with open('filename.txt','w') as outfile: outfile.write("{0}\n".format(what_you_want(8)))  Searching for "write to file in python" might give some insight into this aspect of Python. If you wish to process a list by chunks, then looking up "slicing in Python" might be useful. Given a list L, you may access the second chunk of length 40 by using L[40:80]. So using slicing within a for loop could produce the desired result. 2017-09-23 00:20:07 -0500 answered a question Call error for integers (when I haven't declared any.) Your code contains syntax errors. One of the problems seems to be: sin^2(j*2*pi/N)  This is being interpreted as sin^(2(j*2*pi/N)), which is problably not what you meant. Did you mean sin(j*2*pi/N)^2? Seems like this is the only error. 2017-09-22 05:29:43 -0500 answered a question Stochastic block model Assuming the input consists of n, r, and the probabilities for edges within a community; and assuming that the probability of edges between vertices in different communities is 0, then here is a possible way to generate such a graph. community_sizes = Partitions(n,length=r).random_element() H = Graph() for comm_size,Pr in zip(community_sizes,probs): H=H.disjoint_union(graphs.RandomGNP(comm_size,Pr),labels='integers')  This is assuming that n, r and probs (assumed to be a list of length r containing $P_1,P_2,\ldots,P_r$) are given as part of the input. To see the resulting graph you can use H.show(). 2017-09-22 05:06:05 -0500 commented question Stochastic block model Are the sizes of the communities given as part of the input? Maybe this is what the comment above is trying to address. Is the probability of edges between vertices in different communities equal to 0? 2017-09-21 13:46:10 -0500 commented answer eigenvector corresponding to largest eigenvalue @A You can always upvote/accept if it helped ;-) 2017-09-21 10:32:40 -0500 answered a question eigenvector corresponding to largest eigenvalue Why not just use the following? sorted(A.eigenvectors_left(),reverse=True)  Here A.eigenvectors_left() returns a list of triples (ev, EVS,n), where ev is an eigenvalue, EVS is the list of associated eigenvectors, and n is the algebraic multiplicity of ev. Output seems to be sorted, but just in case we wrap the output inside a call to sorted (in reverse order). In your example this outputs: [ (1.000000000000000?, 1.000000000000000?, 0.732050807568877?, 0.267949192431123?, 0.732050807568877?, 0.267949192431123?) ]  The proposed code does not work since A -3.732050807568878*E  is invertible (determinant very close to zero). Note that 3.732050807568878 is just an approximation to the eigenvalue, not the actual value. 2017-09-20 18:20:05 -0500 commented answer Obtaining all posets in a certain form with SAGE @dan_fulea Seems like OP requires both. For a poset P, produce a list of 3 lists. First list being the list of points, second list the list of cover relations and the third list being the list of comparabilities. This was updated a couple of hours ago to reflect the change :) 2017-09-20 18:15:14 -0500 commented answer Obtaining all posets in a certain form with SAGE @maresage Did you mean posets.DivisorLattice(3)? Regarding the previous question, this output already contains Z. See for example the chain on three elements: [["'x1'", "'x2'", "'x3'"], [["'x1'", "'x2'"], ["'x2'", "'x3'"]], [["'x1'", "'x2'"], ["'x1'", "'x3'"], ["'x2'", "'x3'"]]]  Which is the middle poset in the output above. 2017-09-20 15:55:49 -0500 commented answer Obtaining all posets in a certain form with SAGE @maresage Seems like your comment is truncated. Answer has been updated with the format you provided. 2017-09-20 15:55:08 -0500 commented answer Obtaining all posets in a certain form with SAGE @dan_fulea Were these questions meant to be on the original post? 2017-09-20 15:54:11 -0500 commented answer Obtaining all posets in a certain form with SAGE @maresage Other answer has been updated, it reflects the update on format. 2017-09-20 09:37:36 -0500 answered a question Obtaining all posets in a certain form with SAGE Not entirely clear what the question was. Or was this an implementation request? In case you were looking for an implementation, here is a possible way of doing it. :-) le_relations = lambda P: [(a,b) if P.le(a,b) else (b,a) for a,b in P.comparability_graph().edges(labels=None)] cover_relations = lambda P: [(a,b) if P.le(a,b) else (b,a) for a,b in P.hasse_diagram().edges(labels=None)] format_pt = lambda k: "'x{0}'".format(k+1) format_relations = lambda relations: [[format_pt(a),format_pt(b)] for a,b in relations] format_pts = lambda P: [format_pt(a) for a in P] format_poset = lambda P: [format_pts(P), format_relations(cover_relations(P)), format_relations(le_relations(P))] what_you_want = lambda n: [format_poset(P) for P in posets(n) if P.is_connected()]  Now you should be able to just issue: what_you_want(3) [[["'x1'", "'x2'", "'x3'"], [["'x1'", "'x2'"], ["'x1'", "'x3'"]], [["'x1'", "'x2'"], ["'x1'", "'x3'"]]], [["'x1'", "'x2'", "'x3'"], [["'x1'", "'x2'"], ["'x2'", "'x3'"]], [["'x1'", "'x2'"], ["'x1'", "'x3'"], ["'x2'", "'x3'"]]], [["'x2'", "'x1'", "'x3'"], [["'x1'", "'x3'"], ["'x2'", "'x3'"]], [["'x1'", "'x3'"], ["'x2'", "'x3'"]]]]  Observe that if you just wish to format a single poset, you can do so via format_poset. Hope this is useful. Update: Adapted to new output format. 2017-09-14 02:55:26 -0500 received badge ● Nice Answer (source) 2017-09-13 10:15:33 -0500 answered a question Symbolic to numerical array Would [n(x) for x in ( sin(1), cos(1), pi/4 )]  count as a single command? If L is the list of numbers, say L=( sin(1), cos(1), pi/4), then map(n,L) should do the job.