2021-10-22 17:06:38 +0200 | commented answer | Non-Symmetric Macdonald expansion Essentially it defines the ideal $J$ generated by the Macdonald polynomials, and computes a representation of the given |

2021-10-22 17:05:09 +0200 | commented answer | Non-Symmetric Macdonald expansion Essentially it defines the ideal $J$ generated by the Macdonald polynomials, and computes a representation of the given |

2021-10-22 17:04:28 +0200 | commented answer | Non-Symmetric Macdonald expansion Essentially it defines the ideal $J$ generated by the Macdonald polynomials, and computes a representation of the given |

2021-10-22 17:03:56 +0200 | commented answer | Non-Symmetric Macdonald expansion |

2021-10-22 17:01:20 +0200 | commented answer | Non-Symmetric Macdonald expansion Essentially it defines the ideal $J$ generated by the Macdonald polynomials, and computes representation of the given po |

2021-10-22 17:01:11 +0200 | commented answer | Non-Symmetric Macdonald expansion Essentially it defines the ideal $J$ generated by the Macdonald polynomials, and computes representation of the given po |

2021-10-22 04:39:14 +0200 | answered a question | Non-Symmetric Macdonald expansion Here is a sample code that find such a representation for a given polynomial pol: from sage.combinat.sf.ns_macdonald im |

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2021-10-11 01:27:38 +0200 | edited answer | Subposets of the Boolean lattice via Sage First notice that for $x,y\in P$, if $x\subseteq y$ then $x\cap \overline{y}=\emptyset$ and thus $y\setminus x = \overl |

2021-10-10 21:55:49 +0200 | edited answer | Subposets of the Boolean lattice via Sage First notice that for $x,y\in P$, if $x\subseteq y$ then $x\cap \overline{y}=\emptyset$ and thus $y\setminus x = \overl |

2021-10-10 21:52:12 +0200 | answered a question | Subposets of the Boolean lattice via Sage First notice that for $x,y\in P$, if $x\subseteq y$ then $x\cap \overline{y}=\emptyset$ and thus $y\setminus x = \overl |

2021-10-09 17:56:16 +0200 | edited answer | [SOLVED] Integer to the 1/3 power representation To prevent 24^(1/3) from evaluation one can use SR(24).power(1/3, hold=True) instead. |

2021-10-09 17:55:28 +0200 | answered a question | [SOLVED] Integer to the 1/3 power representation To prevent '24^(1/3)' from evaluation one can use 'SR(24).power(1/3, hold=True)' instead. |

2021-10-08 23:22:18 +0200 | commented answer | The meaning of arguments of is_primitive I assume the meaning of $n$ is to provide a multiple of the order of $x$ modulo $f$ - in some cases (when it's known and |

2021-10-08 23:21:10 +0200 | commented answer | The meaning of arguments of is_primitive I assume the meaning of 'n' is to provide a multiple of the order of 'x' modulo 'f' - in some cases (when it's known and |

2021-10-06 22:14:09 +0200 | commented answer | Consider the class of simple, connected unicyclic graphs on $n$ vertices (a graph on $n$ vertices is unicyclic, if it has $n$ edges). To remove restriction on the number of edges use for g in graphs.nauty_geng(options=f'-c {n}'): |

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2021-10-06 00:36:51 +0200 | commented question | [SOLVED] Integer to the 1/3 power representation Notice that $24 = 2^3\cdot 3$ and thus $24^{1/3} = 2\cdot 3^{1/3}$ - what is your concern here? |

2021-10-05 15:44:50 +0200 | edited answer | Running a search algorithm You can speed up things by factoring the denominator of $f(x,y)$ and see if for each factor you can find its multiple of |

2021-10-05 15:10:53 +0200 | commented answer | Consider the class of simple, connected unicyclic graphs on $n$ vertices (a graph on $n$ vertices is unicyclic, if it has $n$ edges). Yes - restricting to singular adjacency matrices and nonzero eigenvalues can be achieved by changing the if f == h: cond |

2021-10-05 15:08:25 +0200 | commented answer | Consider the class of simple, connected unicyclic graphs on $n$ vertices (a graph on $n$ vertices is unicyclic, if it has $n$ edges). Yes - restricting to singular adjacency matrices and nonzero eigenvalues can be achieved by changing the if condition to |

2021-10-04 23:54:40 +0200 | edited answer | Consider the class of simple, connected unicyclic graphs on $n$ vertices (a graph on $n$ vertices is unicyclic, if it has $n$ edges). Something like this will do the job: def find_graphs(n): for g in graphs.nauty_geng(options=f'-c {n} {n}:{n}'): f |

2021-10-04 23:27:31 +0200 | answered a question | Consider the class of simple, connected unicyclic graphs on $n$ vertices (a graph on $n$ vertices is unicyclic, if it has $n$ edges). Something like this will do the job: def find(n): for g in graphs(n, size=n): if not g.is_connected(): cont |

2021-10-04 16:14:08 +0200 | edited answer | Running a search algorithm You can speed up things by factoring the denominator of $f(x,y)$ and see if for each factor you can find its multiple of |

2021-10-04 16:11:45 +0200 | edited answer | Running a search algorithm You can speed up things by factoring the denominator of $f(x,y)$ and see if for each factor you can find its multiple of |

2021-10-04 16:10:23 +0200 | answered a question | Running a search algorithm |

2021-09-29 16:12:23 +0200 | received badge | ● Nice Answer (source) |

2021-09-25 20:48:09 +0200 | edited answer | Polynomial Substitution for only higher terms If your polynomial is assumed to be zero, then it is also possible to entirely eliminate $y$ by computing resultant of y |

2021-09-24 19:26:25 +0200 | edited answer | Polynomial Substitution for only higher terms It is also possible to entirely eliminate $y$ by computing resultant of your polynomial and $y^2-(x^3+x)$ with respect t |

2021-09-24 19:25:31 +0200 | edited answer | Polynomial Substitution for only higher terms If under substitution you mean to entirely eliminate $y$, you can do so by computing resultant of your polynomial and $y |

2021-09-24 19:24:48 +0200 | answered a question | Polynomial Substitution for only higher terms If under substitution you mean to entirely eliminate $y$, you can do so by computing resultant of your polynomial and $y |

2021-09-16 15:51:58 +0200 | commented question | Weird rounding error in n and N It may be possible that digits=4 is applied to every term in the expression before it's evaluated, resulting in the loss |

2021-09-16 15:47:48 +0200 | commented question | ValueError: Cannot pickle code objects from closures Python version 3.9.5 here and there. So, it does not explain the different behavior. |

2021-09-16 15:47:42 +0200 | commented answer | ValueError: Cannot pickle code objects from closures Why it does not like capturing S? And why SageCell does not have this issue? |

2021-09-16 05:03:58 +0200 | commented question | Weird rounding error in n and N As a temporary workaround you can use n(n(number),digits=4) |

2021-09-16 04:34:39 +0200 | commented question | Weird rounding error in n and N Something fishy is going on here. If we take the numerical value directly. it's rounded correctly: sage: a = 12.0947570 |

2021-09-16 04:30:19 +0200 | commented answer | Weird rounding error in n and N I doubt that 4 "is not a determinant factor". Typically rounding happens to a nearest number (eg., per IEEE754) from a g |

2021-09-16 04:29:31 +0200 | commented answer | Weird rounding error in n and N I doubt that 4 "is not a determinant factor". Typically rounding happens to a nearest number (eg., per IEEE754) from a g |

2021-09-16 04:22:21 +0200 | commented answer | Weird rounding error in n and N I doubt that 4 "is not a determinant factor". Typically rounding happens to a nearest number from a given set (such inte |

2021-09-16 04:21:02 +0200 | commented answer | Weird rounding error in n and N I doubt that 4 "is not a determinant factor". Typically rounding happens to a nearest number from a given set (such inte |

2021-09-16 02:48:15 +0200 | asked a question | ValueError: Cannot pickle code objects from closures ValueError: Cannot pickle code objects from closures The following code produces a strange error ValueError: Cannot pick |

2021-09-15 14:56:16 +0200 | edited answer | Working with unlabelled graphs It's unclear what do you mean under "distinguish". If it's about creation/generation of unlabeled graphs, then many grap |

2021-09-14 23:24:03 +0200 | commented answer | Working with unlabelled graphs Canonical representatives of the equivalence classes are produced by .canonical_label(). |

2021-09-14 23:23:45 +0200 | commented answer | Working with unlabelled graphs Representatives of the equivalence classes are produced by .canonical_label(). |

2021-09-14 22:41:32 +0200 | edited answer | Working with unlabelled graphs It's unclear what do you mean under "distinguish". If it's about creation/generation of unlabeled graphs, then both Grap |

2021-09-14 22:41:07 +0200 | answered a question | Working with unlabelled graphs It's unclear what do you mean under "distinguish". If it's about creation/generation of unlabeled graphs, then both Grap |

2021-09-12 09:51:40 +0200 | marked best answer | unexpected TypeError while creating a polyhedron over ZZ with Normaliz The following code gives me The error disappears if |

2021-09-12 09:51:40 +0200 | received badge | ● Nice Answer (source) |

2021-09-09 15:56:09 +0200 | commented question | Sturm polynomials with polynomial coefficients. How? The coefficient (2*k+1)/(k-1) is not a polynomial but a rational function in k. So, "polynomial coefficients" are irrele |

2021-09-09 15:51:35 +0200 | commented question | how to install the package LattE on sage in ubuntu What version of Sage do you use? Try to install Latte package by running sage -pip install latte_int |

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