De Bruijn sequences#

A De Bruijn sequence is defined as the shortest cyclic sequence that incorporates all substrings of a certain length of an alphabet.

For instance, the \(2^3=8\) binary strings of length 3 are all included in the following string:

sage: DeBruijnSequences(2,3).an_element()
[0, 0, 0, 1, 0, 1, 1, 1]

>>> from sage.all import *
>>> DeBruijnSequences(Integer(2),Integer(3)).an_element()
[0, 0, 0, 1, 0, 1, 1, 1]

They can be obtained as a subsequence of the cyclic De Bruijn sequence of parameters \(k=2\) and \(n=3\):

sage: seq = DeBruijnSequences(2,3).an_element()
sage: print(Word(seq).string_rep())
00010111
sage: shift = lambda i: [(i+j)%2**3 for j in range(3)]
sage: for i in range(2**3):
....:    w = Word([b if j in shift(i) else '*' for j, b in enumerate(seq)])
....:    print(w.string_rep())
000*****
*001****
**010***
***101**
****011*
*****111
0*****11
00*****1

>>> from sage.all import *
>>> seq = DeBruijnSequences(Integer(2),Integer(3)).an_element()
>>> print(Word(seq).string_rep())
00010111
>>> shift = lambda i: [(i+j)%Integer(2)**Integer(3) for j in range(Integer(3))]
>>> for i in range(Integer(2)**Integer(3)):
...    w = Word([b if j in shift(i) else '*' for j, b in enumerate(seq)])
...    print(w.string_rep())
000*****
*001****
**010***
***101**
****011*
*****111
0*****11
00*****1

This sequence is of length \(k^n\), which is best possible as it is the number of \(k\)-ary strings of length \(n\). One can equivalently define a De Bruijn sequence of parameters \(k\) and \(n\) as a cyclic sequence of length \(k^n\) in which all substring of length \(n\) are different.

See also Wikipedia article De_Bruijn_sequence.

AUTHOR:

  • Eviatar Bach (2011): initial version

  • Nathann Cohen (2011): Some work on the documentation and defined the __contain__ method

class sage.combinat.debruijn_sequence.DeBruijnSequences(k, n)[source]#

Bases: UniqueRepresentation, Parent

Represents the De Bruijn sequences of given parameters \(k\) and \(n\).

A De Bruijn sequence of parameters \(k\) and \(n\) is defined as the shortest cyclic sequence that incorporates all substrings of length \(n\) a \(k\)-ary alphabet.

This class can be used to generate the lexicographically smallest De Bruijn sequence, to count the number of existing De Bruijn sequences or to test whether a given sequence is De Bruijn.

INPUT:

  • k – A natural number to define arity. The letters used are the integers \(0..k-1\).

  • n – A natural number that defines the length of the substring.

EXAMPLES:

Obtaining a De Bruijn sequence:

sage: seq = DeBruijnSequences(2, 3).an_element()
sage: seq
[0, 0, 0, 1, 0, 1, 1, 1]

>>> from sage.all import *
>>> seq = DeBruijnSequences(Integer(2), Integer(3)).an_element()
>>> seq
[0, 0, 0, 1, 0, 1, 1, 1]

Testing whether it is indeed one:

sage: seq in DeBruijnSequences(2, 3)
True

>>> from sage.all import *
>>> seq in DeBruijnSequences(Integer(2), Integer(3))
True

The total number for these parameters:

sage: DeBruijnSequences(2, 3).cardinality()
2

>>> from sage.all import *
>>> DeBruijnSequences(Integer(2), Integer(3)).cardinality()
2

Note

This function only generates one De Bruijn sequence (the smallest lexicographically). Support for generating all possible ones may be added in the future.

an_element()[source]#

Returns the lexicographically smallest De Bruijn sequence with the given parameters.

ALGORITHM:

The algorithm is described in the book “Combinatorial Generation” by Frank Ruskey. This program is based on a Ruby implementation by Jonas Elfström, which is based on the C program by Joe Sadawa.

EXAMPLES:

sage: DeBruijnSequences(2, 3).an_element()
[0, 0, 0, 1, 0, 1, 1, 1]

>>> from sage.all import *
>>> DeBruijnSequences(Integer(2), Integer(3)).an_element()
[0, 0, 0, 1, 0, 1, 1, 1]
cardinality()[source]#

Returns the number of distinct De Bruijn sequences for the object’s parameters.

EXAMPLES:

sage: DeBruijnSequences(2, 5).cardinality()
2048

>>> from sage.all import *
>>> DeBruijnSequences(Integer(2), Integer(5)).cardinality()
2048

ALGORITHM:

The formula for cardinality is \(k!^{k^{n-1}}/k^n\) [Ros2002].

sage.combinat.debruijn_sequence.debruijn_sequence(k, n)[source]#

The generating function for De Bruijn sequences. This avoids the object creation, so is significantly faster than accessing from DeBruijnSequence. For more information, see the documentation there. The algorithm used is from Frank Ruskey’s “Combinatorial Generation”.

INPUT:

  • k – Arity. Must be an integer.

  • n – Substring length. Must be an integer.

EXAMPLES:

sage: from sage.combinat.debruijn_sequence import debruijn_sequence
sage: debruijn_sequence(3, 1)
[0, 1, 2]

>>> from sage.all import *
>>> from sage.combinat.debruijn_sequence import debruijn_sequence
>>> debruijn_sequence(Integer(3), Integer(1))
[0, 1, 2]
sage.combinat.debruijn_sequence.is_debruijn_sequence(seq, k, n)[source]#

Given a sequence of integer elements in \(0..k-1\), tests whether it corresponds to a De Bruijn sequence of parameters \(k\) and \(n\).

INPUT:

  • seq – Sequence of elements in \(0..k-1\).

  • n, k – Integers.

EXAMPLES:

sage: from sage.combinat.debruijn_sequence import is_debruijn_sequence
sage: s = DeBruijnSequences(2, 3).an_element()
sage: is_debruijn_sequence(s, 2, 3)
True
sage: is_debruijn_sequence(s + [0], 2, 3)
False
sage: is_debruijn_sequence([1] + s[1:], 2, 3)
False

>>> from sage.all import *
>>> from sage.combinat.debruijn_sequence import is_debruijn_sequence
>>> s = DeBruijnSequences(Integer(2), Integer(3)).an_element()
>>> is_debruijn_sequence(s, Integer(2), Integer(3))
True
>>> is_debruijn_sequence(s + [Integer(0)], Integer(2), Integer(3))
False
>>> is_debruijn_sequence([Integer(1)] + s[Integer(1):], Integer(2), Integer(3))
False