Ask Your Question

Max Alekseyev's profile - activity

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

Essentially it defines the ideal $J$ generated by the Macdonald polynomials, and computes a representation of the given

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

2021-10-21 15:22:08 +0200 received badge  Popular Question (source)
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}'):

2021-10-06 15:40:22 +0200 received badge  Notable Question (source)
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

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

2021-09-29 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

Polyhedron( ieqs=[(-1, 1, -1, 0, -4), (2, -1, 1, 0, 4), (0, 1, 1, 2, -2), (2, -1, -1, -2, 2), (0, 0, -1, 1, 1), (1, 0, 1, -1, -1), (0, 1, 0, 0, 1), (1, -1, 0, 0, -1)], base_ring=ZZ, backend='normaliz' )

gives me TypeError: no conversion of this rational to integerin Sage 9.4. It can also be seen in SagMathCell - https://sagecell.sagemath.org/?q=soiohx

The error disappears if base_ring is changed to QQ. What's wrong?

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