PermutationList

Cycles back to a one-line word: the image list $\{\sigma(1), \sigma(2), \dots\}$. PermutationList(perm, n) pads that list with fixed points out to length $n$.

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PermutationList(perm)the one-line word, as long as perm's own support

PermutationList(perm, n)the one-line word, padded with fixed points to length $n$

Typed signatures

Type	Package	Applies
(expression<Cycles>, integer?) -> list<integer>	groupalgebra	always

Domain: Permutations

Details

• Only takes Cycles — a one-line word is already this head's own answer, so there is nothing to convert
• n shorter than the cycles' own largest point leaves the call unevaluated — there is no room for that point in the result
• The inverse of PermutationCycles's no-argument form, which reads a one-line word back into cycle notation

Examples

Basic #

the cycle $(1\,3\,2)$'s image list runs out to its own largest point, 3

live ↗

In$\mathrm{PermutationList}(\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 3, 2\bigr\rbrack\bigr\rbrack))$Out$\bigl\lbrack3, 1, 2\bigr\rbrack$

points 4 and 5 are fixed, so padding to $n = 5$ just appends them

live ↗

In$\mathrm{PermutationList}(\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack2, 3\bigr\rbrack\bigr\rbrack), 5)$Out$\bigl\lbrack1, 3, 2, 4, 5\bigr\rbrack$

Expands cycles to a one-line list, sized to the largest point

live ↗

In$\mathrm{PermutationList}(\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack3, 2\bigr\rbrack, \bigl\lbrack1, 6, 7\bigr\rbrack\bigr\rbrack))$Out$\bigl\lbrack6, 3, 2, 4, 5, 7, 1\bigr\rbrack$

A length argument pads the list with fixed points

live ↗

In$\mathrm{PermutationList}(\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack3, 2\bigr\rbrack, \bigl\lbrack1, 6, 7\bigr\rbrack\bigr\rbrack), 9)$Out$\bigl\lbrack6, 3, 2, 4, 5, 7, 1, 8, 9\bigr\rbrack$

Scope #

Length follows the largest point moved, here 9

live ↗

In$\mathrm{PermutationList}(\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 5\bigr\rbrack, \bigl\lbrack2, 9, 3\bigr\rbrack\bigr\rbrack))$Out$\bigl\lbrack5, 9, 2, 4, 1, 6, 7, 8, 3\bigr\rbrack$

The identity with no length gives an empty list

live ↗

In$\mathrm{PermutationList}(\mathrm{Cycles}(\bigl\lbrack\bigr\rbrack))$Out$\bigl\lbrack\bigr\rbrack$

The identity at a given length is $1, 2, \dots, 10$

live ↗

In$\mathrm{PermutationList}(\mathrm{Cycles}(\bigl\lbrack\bigr\rbrack), 10)$Out$\bigl\lbrack1, 2, 3, 4, 5, 6, 7, 8, 9, 10\bigr\rbrack$

Properties #

PermutationList and PermutationCycles undo each other

live ↗

In$\mathrm{PermutationCycles}(\mathrm{PermutationList}(\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 5\bigr\rbrack, \bigl\lbrack4, 6, 8, 9\bigr\rbrack\bigr\rbrack)))$Out$\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 5\bigr\rbrack, \bigl\lbrack4, 6, 8, 9\bigr\rbrack\bigr\rbrack)$

Inverse of PermutationCycles: the round trip recovers the list

live ↗

In$\mathrm{PermutationList}(\mathrm{PermutationCycles}(\bigl\lbrack6, 3, 5, 4, 2, 7, 8, 1\bigr\rbrack))$Out$\bigl\lbrack6, 3, 5, 4, 2, 7, 8, 1\bigr\rbrack$

Each position holds the image of that point

live ↗

In$\mathrm{PermutationList}(\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 2, 3\bigr\rbrack, \bigl\lbrack4, 5\bigr\rbrack\bigr\rbrack))$Out$\bigl\lbrack2, 3, 1, 5, 4\bigr\rbrack$

Untouched point 3 stays in place

live ↗

In$\mathrm{PermutationList}(\mathrm{Cycles}(\bigl\lbrack\bigl\lbrack1, 5, 6\bigr\rbrack, \bigl\lbrack2, 4\bigr\rbrack\bigr\rbrack))$Out$\bigl\lbrack5, 4, 3, 2, 6, 1\bigr\rbrack$

About

Examples	11
Checked against	Wolfram Language 11 agree
Crosswalk	Wolfram Language

See also: Cycles, PermutationCycles, Permute, PermutationReplace