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2022-11-24 16:14:53 +0200 | answered a question | Solving a system of equations over the field of rational functions in Q I've figured it out now. Syntactically, we can treat r, g, b as if they were numbers. I had not anticipated that at firs |
2022-11-24 11:31:23 +0200 | commented question | Solving a system of equations over the field of rational functions in Q Yes, that's true. |
2022-11-24 10:30:01 +0200 | asked a question | Solving a system of equations over the field of rational functions in Q Solving a system of equations over the field of rational functions in Q I have a system of equations over the field of r |
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2016-01-16 03:21:55 +0200 | asked a question | I keep losing my node labels on graph export Hello! I often export Sage graphs by using networkx. This is a typical example: However, when I open the graphml-file in the YEd Graph Editor (which is my favorite software for displaying graphs) I find that the node labels (that are clearly existent in the Sage version of the graph) have disappeared forever. Everything else works fine. Assuming that this is Sage's fault rather than the graph editor's fault: What can I do to export the node labels as well? Thank you very much for your help. Malte |
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2015-12-28 22:08:34 +0200 | commented answer | Counting cycles of induced permutations Merci, Vincent, for that quick and profound answer. I am surprised to learn that answering my question by doing maths is easier than by doing Sage. Your answer completely satisfies my needs, as far as I understand things by now. The reason I was thinking of this as a Sage problem was that I couldn't find a way to make Sage tell me anything about the cycle decomposition of a permutation operating on anything other than integers (here, it's sets of integers instead), and I thought maybe I needed a hint in that direction. Do you happen to know of a general theorem cited in a mathbook that discusses your mathematical solution? Anyway: Thank you very much for your kind answer. Vive la France, Malte |
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2015-12-28 17:56:15 +0200 | asked a question | Counting cycles of induced permutations In order to do some sophisticated counting in graph theory, I need to count the cycles of some particular permutations. In my situation, $n$ is an integer greater than 1, and $K_n$ is the set of all two-element sets {a,b} with $a, b$ being integers not greater than $n$. Now any element $\pi$ of the symmetric group $S_n$ induces a permutation $\overline{\pi}$ of $K_n$ in a natural way, i.e. $\overline{\pi}$ maps any set {a,b} of $K_n$ onto {$\pi(a)$, $\pi(b)$}. What I want to figure out with the help of SAGE is the number of cycles that the permutation $\overline{\pi}$ has. If you can help me, please do not forget to mention those little extra things that need to be done and that might appear obvious to you (e.g. importing packages and so forth), since I am a relative novice to SAGE. Thank you very much. Malte |
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2015-11-25 15:50:41 +0200 | answered a question | Installing optional package Thank you very much, both of you. It works now. Usually I am not quite as stupid as I am probably appearing now, but yesterday I really could not find the command terminal within VirtualBox. Maybe there was some confusion in the configuration of the keyboard within the the virtual machine, because I could swear Thank you very much indeed. Malte |
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2015-11-24 18:18:19 +0200 | asked a question | Installing optional package I am running Sage 6.9 (notebook) on a Windows 10 machine via VirtualBox (Oracle VM VirtualBox Version 5.0.10 r104061, GUI-based). For some purpose I need the "database_gap" extension package which has to be installed separately. This, however, is a huge problem to me. I tried to implement every piece of advice I could find in this forum or elsewhere on the internet. First, I typed Surely a sophisticated piece of software such as Sage will offer a more obvious and user-friendly way to install optional packages. But what exactly must I do? I seek your advice and kindly ask for some sort of help that can be understood even by those who are not computer experts (like me). Thanks, Malte |