32
votes
Accepted
Optimization of function taking a permutation
How about a Monte-Carlo-Metropolis search? I'll implement a simplistic version here. See complete universal code further down. Update: Cleaned-up code now available in the Wolfram Function Repository, ...
- 36.9k
28
votes
Accepted
Faster derangements?
Chunks of derangements
Since I've already written library link code generating permutations, generating derangements requires just few tweaks:
...
- 14.8k
25
votes
Permutations[Range[12]] produces an error instead of a list
Since Mathematica 8 it is possible generate the elements of any group one by one with GroupElements. Here's for example a randomly chosen element of the permutation ...
- 5,921
25
votes
21
votes
Fastest way to generate {{a,a},{a,b},{a,c},{b,b},{b,c},{c,c}} from {a,b,c}
Select[Tuples[{a,b,c},2],OrderedQ]
- 6,734
18
votes
Faster derangements?
Here is one way to generate them directly: it is based on a way to generate all permutations but discards invalid ones early:
...
- 8,574
18
votes
Accepted
Delete duplicates from list of lists as if on a necklace
A high-performance solution
Since you are planning to work with thousands of necklaces, it may be much faster to introduce a canonical form.
The main point is to write a necklace canonization ...
- 7,676
17
votes
Permutations of nested parentheses (Dyck words)
StringReplaceList
I just realized that there is a comparatively clean though not highly efficient way to write this using ...
- 265k
17
votes
Terse Method to Swap Lowest for Highest?
How about:
Module[{tmp = test},
With[{ord=Ordering[tmp]},
tmp[[ord]] = Reverse @ tmp[[ord]]];
tmp
]
{56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, ...
- 124k
17
votes
Accepted
StringContainsQ, but anywhere in order
StringMatchQ["aabbc", "*a*c*"]
True
StringMatchQ["aabbc", "*b*a*"]
False
You can also use ...
- 350k
16
votes
Accepted
Generating Tuples with restrictions
Select[
IntegerPartitions[24, {8}, Range[5]],
#.# == 86 &
]
...
Kuba♦
- 133k
16
votes
Accepted
ls Ordering[Ordering[list]] optimal?
No, Ordering[Ordering[list]] not optimal. And yes, there is a faster method:
...
- 96.3k
15
votes
Accepted
Check if two lists are equal in any order
Is it any permutation ? Then I think then this should do
areListsEqual[List1_,List2_]:=(Sort[List1]===Sort[List2]);
- 3,254
15
votes
Accepted
List all possible license plate numbers
I would use
a = Alphabet[]; (* letter *)
d = Range[0, 9]; (* digit *)
result = Tuples[{a, a, d, d, d, d}];
- 227k
14
votes
Permutations of lists of fixed even numbers
The permutation you described is called "derangement". There is a function Derangement in Combinatoricapackage.
...
- 7,681
14
votes
Accepted
Permutations of lists of fixed even numbers
Permutations where no element remains in its original place are called derangements. Counting them is easy enough: the number of derangements of a set of size $n$ is $!n$, or the subfactorial of $n$. ...
- 8,574
14
votes
StringContainsQ, but anywhere in order
Is this what you need?
StringContainsQ["aabbc","a" ~~ ___ ~~ "c"]
True
The following documentation pages should help you get going with string patterns in ...
- 18.8k
14
votes
Accepted
Fastest way to generate {{a,a},{a,b},{a,c},{b,b},{b,c},{c,c}} from {a,b,c}
GroupTheory`Tools`Multisets[{a, b, c}, 2]
{{a,a},{a,b},{a,c},{b,b},{b,c},{c,c}}
- 24.5k
13
votes
Accepted
Number by permutation
Just a note - any method generating the permutations and the searching will get very slow very quickly, and blow RAM soon after.
Something like this s/b much more efficient (e.g., on my goof-top, for ...
- 25.1k
13
votes
Delete duplicates from list of lists as if on a necklace
Preface
If one could create a function f that calculates a canonical form of a necklace that turns all equivalent necklaces ...
- 111k
13
votes
Accepted
12
votes
Check if two lists are equal in any order
If you have many duplicates (and the number of duplicates matter) then Tally can improve performance sufficiently
...
- 42.9k
12
votes
Accepted
What is the fastest way to get the nth distinct permutation of a list?
A simple recursive function I've used many times:
...
- 7,644
12
votes
Accepted
Cyclic and Non-cyclic Permutations
Per the request, I post my comment as an answer:
First question
cy := Permute[#, CyclicGroup[Length@#]] &
cy[Range@5]
{{1, 2, 3, 4, 5}, {2, 3, 4, 5, 1}, {3, 4,...
- 24.5k
12
votes
Accepted
Can we turn this for loop into a more elegant Mathematica code?
Apparently, you try to apply a permutation given by list indirection to a vector duplicate.
Here are several ways to do it, ...
- 96.3k
12
votes
Fastest way to generate {{a,a},{a,b},{a,c},{b,b},{b,c},{c,c}} from {a,b,c}
Another possibility is to use Pick:
...
- 124k
11
votes
Permutations on elements a>b && b>c
Subsets[Reverse@Range[5], {3}]
{{5, 4, 3}, {5, 4, 2}, {5, 4, 1}, {5, 3, 2}, {5, 3, 1}, {5, 2, 1},
{4, 3, 2}, {4, 3, 1}, {4, 2, 1}, {3, 2, 1}}
...
- 350k
11
votes
Accepted
Connectivity in a molecule and permutations
Getting the graph from the XYZ file
You have only a set of coordinates and atom types in an XYZ file. When you import it in Mathematica you can import 3 elements: the 3D plot, the coordinates, and ...
- 63.5k
11
votes
Faster derangements?
Here is a straightforward compiled implementation of Knuth's "Algorithm X" for lexicographically generating restricted permutations, specialized to the derangement case:
...
11
votes
Only top scored, non community-wiki answers of a minimum length are eligible