I was wondering if there was a short way to generate lists of random non-zero elements of GF(q) ? What I have so far is:
L=[];
for i in range(10):
a=GF(25).random_element();
while a==0:
a=GF(25).random_element();
L.append(a);
If you really want something short, yo can do something along the lines:
sage: from itertools import ifilter, islice
sage: list(islice(ifilter(lambda x: x !=0 , (GF(25).random_element() for _ in ZZ)), 10))
Explanataion:
(GF(25).random_element() for _ in ZZ) creates an infinite iterator of random elements of GF(25)ifilter(lambda x: x !=0 , (GF(25).random_element() for _ in ZZ)) creates an iterator that filters the nonzero elementsislice(ifilter(lambda x: x !=0 , (GF(25).random_element() for _ in ZZ)), 10) creates an iterator that produces only the first 10 elements of the previous iteratorlist(islice(ifilter(lambda x: x !=0 , (GF(25).random_element() for _ in ZZ)), 10)) transforms the previous iterator into a listAs for the first line, itertools is a standard module with very nice tools for creating iterators.
That said, if you do:
sage: import this
You will see that Readability counts, so your code is probably the best since you will be able to understand and modify it later.
EDIIT, if you want to use it when Sage will use Python3, you just have to replace the imported ifilter with the builtin filter:
sage: from itertools import islice
sage: list(islice(filter(lambda x: x !=0 , (GF(25).random_element() for _ in ZZ)), 10))
@FrédéricC - thanks for watching! This got me started on an alternative solution, see my answer.
Nice answer, however, I would not recommend the use of underscore as a counter in the loop. You will not be able to use it as the output of the last evaluated expression. Also, instead of ZZ it may be better to use iter(int,1); see https://stackoverflow.com/questions/5...
Let $q = p^n$ for some prime $p$ and some integer $n \ge 1$, and let $F_q$ be the field with $q$ elements, and $F_p$ the field with $p$ elements.
Recall that $F_q$ is a vector space over the prime field $F_p$, and that for any generator $z$ of $F_q$ as a field extension of $F_p$, the family $(1, z, ..., z^{n-1})$ is a basis of $F_q$ as a vector space over $F_p$.
One way to build a list of random nonzero elements in $F_q$ is to pick coefficients $a_0$, ..., $a_{n-1}$ at random, all between $0$ and $p-1$, but not all zero, and to take the element $\sum a_k z^k$ in $F_q$.
One way to pick such a collection of coefficients is to pick an integer at random between $1$ and $q-1$ and to let $a_k$ be its $k$-th digit in base $p$.
Choose $p$ and $n$, define $q = p^n$, and let $F$ be the finite field with $q$ elements.
sage: p = 5
sage: n = 3
sage: q = p^n
sage: F = GF(q)
Choose $m$, the number of random elements to pick.
sage: m = 10
Produce a list of length $m$ of random nonzero elements in $F$.
sage: L = [F(ZZ.random_element(1, q).digits(base=p)) for _ in range(m)]
r = ZZ.random_element(1, q) picks a random integer between $1$ and $q - 1$d = r.digits(base=p) gives the list of its digits in base $p$u = F(d) turns this list into a field element, seen a polynomial in $z$
(where $z$ is the generator of $F_q$ as a field extension of $F_p$)
with coefficients given by the list.Defining the finite field takes on the order of 0.1 ms.
It is probably worth defining $F$ once and for all, and then picking random
elements in $F$, rather than calling GF(25) inside the loop, which spends
time initializing the finite field at each iteration of the loop.
Of course if you're looping only ten times it doesn't matter much.
Asked: 2018-09-13 11:14:36 +0100
Seen: 2,029 times
Last updated: Sep 14 '18
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You don't actually need the semicolons at the end of the lines, by the way.