2021-10-24 20:38:31 +0100 received badge ● Student (source) 2021-10-24 20:31:29 +0100 received badge ● Popular Question (source) 2019-07-04 16:29:35 +0100 commented question Need help converting python code to sage compatible anyone please ???? 2019-07-02 20:42:37 +0100 commented question Need help converting python code to sage compatible is it possible to convert above code and use it in sage? does sage do iterations fast? 2019-07-02 19:50:34 +0100 commented question Need help converting python code to sage compatible i am not solving anything, but generating a result from elements. elements are in shape of list - tuple (000000000, 000000001, ........ 333333332, 333333333) i need to multiply each element with all elements to get the result for single element with respect to all elements. you can say, i am making a grid of element as row and columns and their results in do_calculations, i am solving an algebric like equation. I hope you understand what i am doing 2019-07-02 14:14:18 +0100 asked a question Need help converting python code to sage compatible I have a list of 262144 elements in a list created through "itertools.product". Now I have to loop over these elements and multiply it with all other elements, which is taking too much time. (I don't have any issue of memory / cpu) elements = [] for e in itertools.product(range(4), repeat=9): elements.append(e) for row in elements: for col in elements: do_calculations(row, col) def do_calculations(ro, co): t = {} t[0] = [multiply(c=ro[0], r=co[0])] for i in range(1, len(ro)): _t = [] for j in range(i+1): _t.append(multiply(c=ro[j], r=co[i-j])) t[i] = _t for vals in t.values(): nx = len(vals) _co = ro[nx:] _ro = co[nx:] for k in range(len(_ro)): vals.append(multiply(c=_co[k], r=_ro[k])) _t = [] for k in t.values(): s = k[0] for j in range(1, len(k)): s = addition(c=s, r=k[j]) _t.append(s) return _t def addition(c, r) -> int: __a = [[0, 3, 1, 2], [3, 2, 0, 1], [0, 3, 2, 1], [1, 0, 2, 3]] return __a[c][r] def multiply(c, r) -> int: __m = [[0, 0, 0, 0], [0, 1, 2, 3], [0, 3, 1, 2], [0, 2, 3, 1]] return __m[c][r]  it is taking too much time to process single col with rows.... can any one help me in this? regards