permuta

Permuta is a Python library for working with perms (short for permutations), patterns, and mesh patterns.

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Installing

To install Permuta on your system, run:

pip install permuta

It is also possible to install Permuta in development mode to work on the source code, in which case you run the following after cloning the repository:

./setup.py develop

To run the unit tests:

pip install -r test_requirements.txt
./setup.py test

Using Permuta

Once you've installed Permuta, it can be imported by a Python script or an interactive Python session, just like any other Python library:

>>> from permuta import *

Importing * supplies you with the 'Perm' and 'PermSet' classes along with the 'AvoidanceClass' class (with alias 'Av') for generating perms avoiding a set of patterns. It also gives you the 'MeshPatt' class and some other submodules which we will not discuss in this readme.

Creating a single perm

Permutations are zero-based in Permuta and can be created using any iterable.

>>> Perm()  # Empty perm
Perm(())
>>> Perm([])  # Another empty perm
Perm(())
>>> Perm((0, 1, 2, 3)) # The zero-based version of 1234
Perm((0, 1, 2, 3))
>>> Perm((2, 1, 3)) # Warning: it will initialise with any iterable
Perm((2, 1, 3))

Permutations can also be created using some specific class methods.

>>> Perm.from_string("201")  # strings
Perm((2, 0, 1))
>>> Perm.one_based((1, 3, 2, 4)) # one-based iterable of integers
Perm((0, 2, 1, 3))
>>> Perm.to_standard("a2gsv3") # standardising any iterable using '<'
Perm((2, 0, 3, 4, 5, 1))
>>> Perm.from_integer(210) # an integer between 0 and 9876543210
Perm((2, 1, 0))
>>> Perm.from_integer(321) # any integer given is standardised
Perm((2, 1, 0))
>>> Perm.from_integer(201)
Perm((2, 0, 1))

Printing perms gives zero-based strings.

>>> print(Perm(()))
ε
>>> print(Perm((2, 1, 0)))
210
>>> print(Perm((6, 2, 10, 9, 3, 8, 0, 1, 5, 11, 4, 7)))
(6)(2)(10)(9)(3)(8)(0)(1)(5)(11)(4)(7)

To get an iterator of all permutations of a certain length you can use

>>> Perms4 = Perm.of_length(4)

You can run a for-loop over this iterator if you need to do something with all the permutations of this size. If you just want a specific permutation of this size you might be better off using the unrank function.

>>> Perm.unrank(23,4)
Perm((3, 2, 1, 0))

The avoids, contains, and occurrence methods enable working with patterns:

>>> p = Perm((0,2,1,3))
>>> p.contains(Perm((2, 1, 0)))
False
>>> p.avoids(Perm((0, 1)))
False
>>> list(p.occurrences_of(Perm((1, 0))))
[(1, 2)]
>>> list(Perm((0, 1)).occurrences_in(p))
[(0, 1), (0, 2), (0, 3), (1, 3), (2, 3)]

The basic symmetries are implemented:

>>> [p.reverse(), p.complement(), p.inverse()]
[Perm((3, 1, 2, 0)), Perm((3, 1, 2, 0)), Perm((0, 2, 1, 3))]

To take direct sums and skew sums we use + and -:

>>> q = Perm((0, 1, 2, 3, 4))
>>> p + q
Perm((0, 2, 1, 3, 4, 5, 6, 7, 8))
>>> p - q
Perm((5, 7, 6, 8, 0, 1, 2, 3, 4))

There are numerous practical methods available:

>>> list(p.fixed_points())
[0, 3]
>>> list(p.ascents())
[0, 2]
>>> list(p.descents())
[1]
>>> list(p.inversions())
[(1, 2)]
>>> p.major_index()
2

Creating a perm class

Perm classes are created by first specifying a basis and then calling the 'Av' class, to create the set of permutations avoiding the basis:

>>> basis = Basis(Perm((1, 0, 2)), Perm((1, 2, 0)))
>>> basis
Basis((Perm((1, 0, 2)), Perm((1, 2, 0))))
>>> perm_class = Av(basis)
>>> perm_class
Av(Basis((Perm((1, 0, 2)), Perm((1, 2, 0)))))

You can ask whether a perm belongs to the perm class:

>>> Perm((3, 2, 1, 0)) in perm_class
True
>>> Perm((0, 2, 1, 3)) in perm_class
False

You can get its enumeration up to a fixed length.

>>> perm_class.enumeration(10)
[1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512]
>>> perm_class.count(11)
1024

You can also look to see if some well know enumeration strategies apply to a given class.

>>> from permuta.enumeration_strategies import find_strategies
>>> basis = [Perm((3, 2, 0, 1)), Perm((1, 0, 2, 3))]
>>> for strat in find_strategies(basis):
...     print(strat.reference())
The insertion encoding of permutations: Corollary 10
>>> basis = [Perm((1, 2, 0, 3)), Perm((2, 0, 1, 3)), Perm((0, 1, 2, 3))]
>>> for strat in find_strategies(basis):
...     print(strat.reference())
Enumeration of Permutation Classes and Weighted Labelled Independent Sets: Corollary 4.3
>>> basis = [Perm((1, 3, 0, 2)), Perm((2, 0, 3, 1))]
>>> for strat in find_strategies(basis):
...     print(strat.reference())
Enumeration of Permutation Classes and Weighted Labelled Independent Sets: Corollary 4.6
The class contains only finitely many simple permutations

The output is the name of a paper, followed by the statement in the paper where the enumeration strategy is discussed or stated.

Permutation statistics

With the PermutationStatistic class we can look for distributions of statistics for classes and look for statistics preservations (or transformation) either for two classes or given a bijection. First we need to import it.

>>> from permuta.permutils.statistics import PermutationStatistic

To see a distribution for a given statistic we grab its instance and provide a length and a class (no class will use the set of all permutations).

>>> PermutationStatistic.show_predefined_statistics() # Show all statistics with id
[0] Number of inversions
[1] Number of non-inversions
[2] Major index
[3] Number of descents
[4] Number of ascents
[5] Number of peaks
[6] Number of valleys
[7] Number of cycles
[8] Number of left-to-right minimas
[9] Number of left-to-right maximas
[10] Number of right-to-left minimas
[11] Number of right-to-left maximas
[12] Number of fixed points
[13] Order
[14] Longest increasing subsequence
[15] Longest decreasing subsequence
[16] Depth
[17] Number of bounces
[18] Maximum drop size
[19] Number of primes in the column sums
[20] Holeyness of a permutation
[21] Number of stack-sorts needed
[22] Number of pop-stack-sorts needed
[23] Number of pinnacles
[24] Number of cyclic peaks
[25] Number of cyclic valleys
[26] Number of double excedance
[27] Number of double drops
[28] Number of foremaxima
[29] Number of afterminima
[30] Number of aftermaxima
[31] Number of foreminima

>>> depth = PermutationStatistic.get_by_index(16)
>>> depth.distribution_for_length(5)
[1, 4, 12, 24, 35, 24, 20]
>>> depth.distribution_up_to(4, Av.from_string("123"))
[[1], [1], [1, 1], [0, 2, 3], [0, 0, 3, 7, 4]]

Given a bijection as a dictionary, we can check which statistics are preserved with check_all_preservations and which are transformed with check_all_transformed

>>> bijection = {p: p.reverse() for p in Perm.up_to_length(5)}
>>> for stat in PermutationStatistic.check_all_preservations(bijection):
...     print(stat)
Number of peaks
Number of valleys
Holeyness of a permutation
Number of pinnacles

We can find all (predefined) statistics equally distributed over two permutation classes with equally_distributed. We also support checks for joint distribution of more than one statistics with jointly_equally_distributed and transformation of jointly distributed stats with jointly_transformed_equally_distributed.

>>> cls1 = Av.from_string("2143,415263")
>>> cls2 = Av.from_string("3142")
>>> for stat in PermutationStatistic.equally_distributed(cls1, cls2, 6):
...     print(stat)
Major index
Number of descents
Number of ascents
Number of peaks
Number of valleys
Number of left-to-right minimas
Number of right-to-left maximas
Longest increasing subsequence
Longest decreasing subsequence
Number of pinnacles

The BiSC algorithm

The BiSC algorithm can tell you what mesh patterns are avoided by a set of permutations. Although the output of the algorithm is only guaranteed to describe the finite inputted set of permutations, the user usually hopes that the patterns found by the algorithm describe an infinite set of permutatations. To use the algorithm we first need to import it.

>>> from permuta.bisc import *

A classic example of a set of permutations described by pattern avoidance are the permutations sortable in one pass through a stack. We use the function stack_sortable which returns True for permutations that satisfy this property. The user now has two choices: Run auto_bisc(Perm.stack_sortable) and let the algorithm run without any more user input. It will try to use sensible values, starting by learning small patterns from small permutations, and only considering longer patterns when that fails. If the user wants to have more control over what happens that is also possible and we now walk through that: We input the property into bisc and ask it to search for patterns of length 3.

>>> bisc(Perm.stack_sortable, 3)
I will use permutations up to length 7
{3: {Perm((1, 2, 0)): [set()]}}

When this command is run without specifying what length of permutations you want to consider, bisc will create permutations up to length 7 that satisfy the property of being stack-sortable. The output means: There is a single length 3 pattern found, and its underlying classical pattern is the permutation Perm((1, 2, 0)). Ignore the [set()] in the output for now. We can use show_me to get a better visualization of the patterns found. In this call to the algorithm we also specify that only permutations up to length 5 should be considered.

>>> SG = bisc(Perm.stack_sortable, 3, 5)
>>> show_me(SG)
There are 1 underlying classical patterns of length 3
There are 1 different shadings on 120
The number of sets to monitor at the start of the clean-up phase is 1
<BLANKLINE>
Now displaying the patterns
<BLANKLINE>
 | | |
-+-●-+-
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-●-+-+-
 | | |
-+-+-●-
 | | |
<BLANKLINE>

We should ignore the The number of sets to monitor at the start of the clean-up phase is 1 message for now.

We do not really need this algorithm for sets of permutations described by the avoidance of classical patterns. Its main purpose is to describe sets with mesh patterns, such as the West-2-stack-sortable permutations

>>> SG = bisc(Perm.west_2_stack_sortable, 5, 7)
>>> show_me(SG)
There are 2 underlying classical patterns of length 4
There are 1 different shadings on 1230
There are 1 different shadings on 2130
The number of sets to monitor at the start of the clean-up phase is 1
There are 1 underlying classical patterns of length 5
There are 1 different shadings on 42130
<BLANKLINE>
Now displaying the patterns
<BLANKLINE>
 | | | |
-+-+-●-+-
 | | | |
-+-●-+-+-
 | | | |
-●-+-+-+-
 | | | |
-+-+-+-●-
 | | | |
<BLANKLINE>
 |▒| | |
-+-+-●-+-
 | | | |
-●-+-+-+-
 | | | |
-+-●-+-+-
 | | | |
-+-+-+-●-
 | | | |
<BLANKLINE>
 |▒| | | |
-●-+-+-+-+-
 | |▒| | |
-+-+-+-●-+-
 | | | | |
-+-●-+-+-+-
 | | | | |
-+-+-●-+-+-
 | | | | |
-+-+-+-+-●-
 | | | | |
<BLANKLINE>

This is good news and bad news. Good because we quickly got a description of the set we were looking at, that would have taken a long time to find by hand. The bad news is that there is actually some redundancy in the output. To understand better what is going on we will start by putting the permutations under investigation in a dictionary, which keeps them separated by length.

>>> A, B = create_bisc_input(7, Perm.west_2_stack_sortable)

This creates two dictionaries with keys 1, 2, ..., 7 such that A[i] points to the list of permutations of length i that are West-2-stack-sortable, and B[i] points to the complement. We can pass the A dictionary directly into BiSC since only the permutations satisfying the property are used to find the patterns. We can use the second dictionary to check whether every permutation in the complement contains at least one of the patterns we found.

>>> SG = bisc(A, 5, 7)
>>> patterns_suffice_for_bad(SG, 7, B)
Starting sanity check with bad perms
Now checking permutations of length 0
Now checking permutations of length 1
Now checking permutations of length 2
Now checking permutations of length 3
Now checking permutations of length 4
Now checking permutations of length 5
Now checking permutations of length 6
Now checking permutations of length 7
Sanity check passes for the bad perms
(True, [])

In this case it is true that every permutation in B, up to length 7, contains at least one of the patterns found. Had that not been the case a list of permutations would have been outputted (instead of just the empty list).

Now, we claim that there is actually redundancy in the patterns we found, and the length 4 mesh patterns should be enough to describe the set. This can occur and it can be tricky to theoretically prove that one mesh pattern is implied by another pattern (or a set of others, as is the case here). We use the dictionary B again and run

>>> bases, dict_numbs_to_patts = run_clean_up(SG, B)
<BLANKLINE>
The bases found have lengths
[2]

There is one basis of mesh patterns found, with 2 patterns

>>> show_me_basis(bases[0], dict_numbs_to_patts)
<BLANKLINE>
Displaying the patterns in the basis
<BLANKLINE>
 | | | |
-+-+-●-+-
 | | | |
-+-●-+-+-
 | | | |
-●-+-+-+-
 | | | |
-+-+-+-●-
 | | | |
<BLANKLINE>
 |▒| | |
-+-+-●-+-
 | | | |
-●-+-+-+-
 | | | |
-+-●-+-+-
 | | | |
-+-+-+-●-
 | | | |
<BLANKLINE>

This is the output we were expecting. There are several other properties of permutations that can be imported from permuta.bisc.perm_properties, such as smooth, forest-like, baxter, simsun, quick_sortable, etc.

Both bisc and auto_bisc can accept input in the form of a property, or a list of permutations (satisfying some property).

License

BSD-3: see the LICENSE file.

Citing

If you found this library helpful with your research and would like to cite us, you can use the following BibTeX or go to Zenodo for alternative formats.