arunayyar's profile - activity

overview network karma followed questions activity

2022-06-02 20:22:03 +0200	received badge	● Nice Question (source)
2021-01-31 15:05:23 +0200	received badge	● Notable Question (source)
2019-10-26 13:26:22 +0200	received badge	● Popular Question (source)
2017-06-20 01:08:55 +0200	received badge	● Famous Question (source)
2016-05-09 20:18:52 +0200	received badge	● Notable Question (source)
2016-05-09 20:18:52 +0200	received badge	● Popular Question (source)
2014-08-13 20:54:13 +0200	commented question	enumerate integer points in a polytope You are right. It took one full day for the code to finish running. It gave about 75500 vectors. I found that the problem has another constraint. I am only looking for integer vectors that belong to $S$ and have 4 or less non-zero elements. But $\sum_{i=1}^n \|x_i\| \leq 4$ is a convex constraint. Polyhedron seems to take only linear equality and inequality constraints.. Is there a way to add the L_1 norm constraint?
2014-08-06 18:42:23 +0200	received badge	● Nice Question (source)
2014-08-03 05:48:46 +0200	asked a question	enumerate integer points in a polytope This is a followup question to http://ask.sagemath.org/question/2366... $A$ is a $5 \times 19$ matrix. $S=${-1,0,1}. As suggessted in the solution I compute integral points within a polytope. I have the following code: eq3=[ (0,6,10,10,6,6,6,3,3,0,3,21,21,0,0,3,3,3,3,3), (0,12,16,15,13,13,14,7,7,2,8,27,27,1,2,7,7,5,4,6), (0,6,13,10,9,9,12,6,6,4,10,14,14,0,1,7,7,6,3,3), (0,0,5,5,0,0,0,0,0,0,0,7,7,0,0,0,0,0,0,0), (0,0,3,2,1,1,2,1,1,1,2,2,2,0,0,1,1,1,0,0)]; ineq3=[ (-lb,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0),(ub,-1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0), (-lb,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0),(ub,0,-1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0), (-lb,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0),(ub,0,0,-1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0), (-lb,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0),(ub,0,0,0,-1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0), (-lb,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0),(ub,0,0,0,0,-1,0,0,0,0,0,0,0,0,0,0,0,0,0,0), (-lb,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0),(ub,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,0,0,0,0,0), (-lb,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0),(ub,0,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,0,0,0,0), (-lb,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0),(ub,0,0,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,0,0,0), (-lb,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0),(ub,0,0,0,0,0,0,0,0,-1,0,0,0,0,0,0,0,0 ... (more)
2014-08-02 22:09:47 +0200	answered a question	Solving system of linear Diophantine equations Dear Dima, Thank you for the solution. It works well. I am putting the example code here. lb=-1; ub=1; A=matrix(ZZ,[(8,2,10,0,12,2,0),(4,6,9,1,14,5,2),(2,0,1,1,0,1,0),(2,1,3,0,4,0,1)]); eq1=[(0,8,2,10,0,12,2,0),(0,4,6,9,1,14,5,2),(0,2,0,1,1,0,1,0),(0,2,1,3,0,4,0,1)]; ineq1=[ (-lb,1,0,0,0,0,0,0),(ub,-1,0,0,0,0,0,0), (-lb,0,1,0,0,0,0,0),(ub,0,-1,0,0,0,0,0), (-lb,0,0,1,0,0,0,0),(ub,0,0,-1,0,0,0,0), (-lb,0,0,0,1,0,0,0),(ub,0,0,0,-1,0,0,0), (-lb,0,0,0,0,1,0,0),(ub,0,0,0,0,-1,0,0), (-lb,0,0,0,0,0,1,0),(ub,0,0,0,0,0,-1,0), (-lb,0,0,0,0,0,0,1),(ub,0,0,0,0,0,0,-1)]; p=Polyhedron(eqns=eq1,ieqs=ineq1,base_ring=QQ) p.integral_points() returns the answer ((-1, -1, 1, 1, 0, 0, 0), (0, -1, -1, 1, 1, 0, 0), (0, -1, 0, -1, 0, 1,1), (0, 0, 0, 0, 0, 0, 0), (0, 1, 0, 1, 0, -1, -1), (0, 1, 1, -1, -1, 0,0), (1, 1, -1, -1, 0, 0, 0)) which is exactly what i needed. Thanks again.
2014-08-02 22:05:14 +0200	received badge	● Scholar (source)
2014-08-02 12:31:26 +0200	received badge	● Student (source)
2014-08-02 12:23:16 +0200	received badge	● Editor (source)
2014-08-02 12:17:47 +0200	asked a question	Solving system of linear Diophantine equations I have an $m \times n$ integer matrix $A$ with $m>n$, and a set $S \subset \mathbb{Z}$, for example $S=\{-1,0,1\}$. I want to enumerate all possible $X \in S^n$ such that $AX=0$. I tried using smith_form(), but then i could not figure out how to force the solution to belong to elements of $S$. EDIT. Thank you Dima for the solution. It works well. I am adding an example here to illustrate your solution. lb = -1 ub = 1 A = matrix(ZZ, [(8,2,10,0,12,2,0), (4,6,9,1,14,5,2), (2,0,1,1,0,1,0), (2,1,3,0,4,0,1)]) eq1 = [(0,8,2,10,0,12,2,0), (0,4,6,9,1,14,5,2), (0,2,0,1,1,0,1,0), (0,2,1,3,0,4,0,1)] ieq1 = [(-lb,1,0,0,0,0,0,0), (ub,-1,0,0,0,0,0,0), (-lb,0,1,0,0,0,0,0), (ub,0,-1,0,0,0,0,0), (-lb,0,0,1,0,0,0,0), (ub,0,0,-1,0,0,0,0), (-lb,0,0,0,1,0,0,0), (ub,0,0,0,-1,0,0,0), (-lb,0,0,0,0,1,0,0), (ub,0,0,0,0,-1,0,0), (-lb,0,0,0,0,0,1,0), (ub,0,0,0,0,0,-1,0), (-lb,0,0,0,0,0,0,1), (ub,0,0,0,0,0,0,-1)] p = Polyhedron(eqns=eq1, ieqs=ieq1, base_ring=QQ) p.integral_points() returns the answer `((-1, -1, 1, 1, 0, 0, 0), (0, -1, -1, 1, 1, 0, 0), (0, -1, 0, -1, 0, 1,1), (0, 0, 0, 0, 0, 0, 0), (0, 1, 0, 1, 0, -1, -1), (0, 1, 1, -1, -1, 0,0), (1, 1, -1, -1, 0, 0, 0))` which is exactly what i needed. Thanks again.