DePaul Discoveries
Volume 2 | Issue 1
Article 2
2013
Enumerating Alternating Permutations with One
Alternating Descent
Stacey Wagner
[email protected]
Follow this and additional works at: http://via.library.depaul.edu/depaul-disc
Part of the Physical Sciences and Mathematics Commons
Acknowledgements
Faculty Advisor: Bridget Tenner, Department of Mathematical Sciences
Recommended Citation
Wagner, Stacey (2013) "Enumerating Alternating Permutations with One Alternating Descent," DePaul Discoveries: Vol. 2: Iss. 1,
Article 2.
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Wagner: Enumerating Alternating Permutations with One Alternating Descent
D E PA U L D I S C O V E R I E S ( 2 O 1 3 )
Enumerating Alternating Permutations
with One Alternating Descent
Stacey Wagner*
Department of Mathematical Sciences
This paper introduces a new statistic for alternating permutations, called an alternating descent.
ABSTRACT
Specifically this paper focuses on alternating permutations with one alternating descent. We then enumerate these
permutations by decomposing them into four sets.
A permutation w may be expressed as a word where
INTRODUCTION
Permutations as combinatorial objects will be the basis
w  w(1) w(2) ... w(n).
for this paper. Two of their most basic attributes are
ascents and descents. We will examine in this paper
DEFINITION 1.2.
a statistic for alternating permutations. Alternating
An ascent in a permutation is a position i such that w(i)
permutations have been studied by Stanley[2] and
 w(i1). A descent in a permutation is a position i
Dulucq and Simion[1]. This statistic is based on the
such that w(i)  w(i1).
concept of a descent, called alternating descents. We will
focus our attention, however, solely on the alternating
DEFINITION 1.3.
permutations containing one alternating descent.
An alternating permutation is a permutation with
ascents in all odd positions and descents in all even
In the first section we define this statistic and its
positions.
notation. In Section 2, we introduce the decomposition
of these permutations into four subsets that will then be
enumerated in Sections 3 through 6.
E X A M P L E 1 .4 .
The alternating permutation of S1 is {1}.
The alternating permutation of S2 is {12}.
The alternating permutations of S3 are {132, 231}.
1 . A LT E R N AT I N G D E S C E N TS
D E F I N I T I O N 1 .1 .
A permutation is a specific ordering of a set of elements.
The alternating permutations of S4 are {1324, 1423,
2314, 2413, 3412}.
Since we consider finite sets this definition coincides
with the definition of a permutation as a one-to-one
Permutations can also be displayed visually. In these
and onto map.
graphs, one can see the relative differences between
consecutive values. For alternating permutations, what
* Faculty Advisor: Bridget Tenner, Department of Mathematical Sciences
Academic year of completion: 2012-2013
Author contact: [email protected]
is immediately apparent is that they zig-zag. These
graphs can be formed by plotting for each i (i, w(i)), then
connecting each plotted point.
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1
DePaul Discoveries, Vol. 2 [2013], Iss. 1, Art. 2
E N U M E R AT I N G A LT E R N AT I N G P E R M U TAT I O N S W I T H O N E A LT E R N AT I N G D E S C E N T
EXAMPLE 1.5.
EXAMPLE 1.8.
Consider w362415; w can be expressed visually with
For
each point representing a term in w:
w24153867
from
Example
1.6,
we
have
w63124587. There are three descents in w, so
Example 1.8. For w = 24153867 from Example 1.6, we have w = 63124587. The
Example 1.8. For
w = 24153867 from Example 1.6, we have w = 63124587. The
altdes(w)2.
three descents in w , so altdes(w) = 3.
three descents in w , so altdes(w) = 3.
EXAMPLE 1.9.
There are no alternating permutations with 1 alternating
Example 1.9. There are no alternating permutations with 1 alternating descent in eit
Example 1.9. There
are in
noeither
alternating
because therewith
are 1noalternating
descents descent in eit
descent
S1 or Spermutations
2
or S2 because there
are no(1)
descents
in
either (1) = 1 or (12) = 12.
in
either
1
or
(12)12.
or S2 because there are no descents in either (1) = 1 or (12) = 12.
E X A M P L E 1 .1 0 .
with
one alternating
is 132
Example 1.10. The
The alternating
alternating permutation
permutation of
of S
with one
alternating descent
one alternating
descent is 132
Example 1.10. The
alternating
permutation
of
SS333 with
If we care only about ascents and descents,
not
descent is 132 where (132)213.
= 213.
(132)and
(132) = 213.
the relative differences between consecutive values, we
can also capture this zig-zag property without using a
E X A M P L E 1 .1 1 .
The alternating
one
alternating
graph. We will use this form in later analysis.
with
one
alternating descent are
Example 1.11. The
alternatingpermutations
permutationsofofS Swith
Example 1.11. The alternating permutations of 4S44 with one alternating descent are
descent
are
1324, 2413, and 3412 where (1324)2134,
2413, and 3412 where (1324) = 2134, (2413) = 1243, and (3412) = 1342.
2413, and 3412 where
(1324) = 2134, (2413) = 1243, and (3412) = 1342.
EXAMPLE 1.6.
(2413)1243, and (1342)1342.
If w is the alternating permutation w24153867 then w
may be written asw
w
=
2
4
1
5
3
8
6
7
.
2 .2.D EC
O M P O S I N G E N,1
Decomposing
E
2. Decomposing En,1
n,1
Let n,1 be the set of alternating permutations
of size
Let En,1 be thenset
of alternating
permutations
ofwill
size show
n with
1 alternating
descen
with
1
alternating
descent.
We
that
the
Counting the number of descents in an alternating
Let En,1 be the set of alternating permutations of size n with 1 alternating
descen
set
of
alternating
permutations
of
n
letters
with
one
permutation is uninteresting because wewill
willshow
always
that the set of alternating permutations of n letters with one alternating d
will show that the set of alternating permutations of n letters with one alternating d
alternating descent, n,1, can be decomposed as a union
know exactly how many there are: if the alternating
n,1 , can be decomposed as a union of the following pairwise disjoint sets:
 n − 1 E
,
can
be
decomposed
as a union
of thedisjoint
following
pairwise disjoint sets:
E
n,1
ascents
and
of the following
pairwise
sets:
permutation w  S , then there are 

n
 2 
 n − 1 descents. So we will examine a different statistic.
 2 


D E F I N I T I O N 1 .7.
En,1 = An Bn Cn Dn .
En,1 = An Bn Cn Dn
Given
that
|S| is the the
number
of elements
of S,
we
have: of S, we
Given the alternating permutation w, readGiven
the that
odd |S| Given
thatnumber
| S | is the
the number
of we
elements
is the the
of elements
of S,
have:
positions of w from right to left and then the even
have:
|An | = |En−1,1 |
positions from left to right. This mapping of w of size
|An | = |En−1,1 |
n will be called w, where its size is also n. Equivalently,
|Bn | = |Bn−1 | + |Cn−1 | + |Dn−1 |
|Bn | = |Bn−1 | + |Cn−1 | + |Dn−1 |
if the alternating permutation is in the zig-zag form,
|Cn | = |Bn−2 | + |Cn−2 | + |Dn−2 |
read the letters starting from the bottom right corner
|Cn | = |Bn−2 | + |Cn−2 | + |Dn−2 |
and continuing clockwise all the way around to the top
|D | = 1 for n > 3.
|Dnn | = 1 for n > 3.
right. The number of alternating descents of w, denoted
altdes(w), is defined as altdes(w)des(w), the number
of descents of w.
We can also find w and altdes(w) for non-alternating
permutations w, but here will focus only on the
alternating permutations.
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2
Wagner: Enumerating Alternating Permutations with One Alternating Descent
D E PA U L D I S C O V E R I E S ( 2 O 1 3 )
and onto, therefore the cardinality of An is equal to the
3. SET An
The set An is the set of all alternating permutations
cardinality of n-1,1. If we assume that w does not lie in
with one alternating descent ending in n if n is even
An, then w has the alternating descent in the top row if n
and ending in 1 if n is odd. The set An will be acquired
is even and in the bottom row if n is odd.
directly from the permutations w in n-1,1 via the map
 (w), defined as follows:
4. SET Bn
• If n is even, then obtain  (w) by inserting n as the
The set B will be acquired from the permutations in
n-1,1 An-1. This means that the permutations in B will
last letter of each permutation in n-1,1 .
• If n is odd, then obtain  (w) by inserting 1 as
the last letter of each permutation in n-1,1 , shifting
come from those in n-1,1 that do not end in n if n1 is
even or 1 if n1 is odd.
every other letter’s value up by one.
E X A M P L E 4 .1
E X A M P L E 3 .1 .
3,1 contains only the permutation 132. 4,1 contains the
For w  24351, we have n  6, and so  (w)243516.
permutations 1324, 3412, and 2413. Because  (132)
1324, the permutations of 4,1 A4 are 2413 and 3412.
EXAMPLE 3.2.
For w  143526, we have n  7, and so  (w)2546371.
Before we can show how set B is generated, we must
first understand the process of complementing a
permutation.
PROPOSITION 3.3.
For any w  n-1,1, we have  (w)  n,1.
DEFINITION 4.2.
Proof. When n is even, there is a descent in the final
Consider the permutation ww(1)w(1)w(2) …. w(n). To
position of w, that is w(n2)  w(n1) . Adding n
after the final position of w, we must surely have
n1 w(i). This operation will be expressed as .
complement a permutation, w, replace each w(i) with
w (n1)  n. This also guarantees that w(n1) 
n, which means there is no new alternating descent.
EXAMPLE 4.3.
Since we begin with an alternating permutation with
For w  1365247, we have
 7523641.
one alternating descent and only insert n in the final
position, we preserve its alternating structure and its
alternating descent. Therefore  (w) is in n,1.
When n is odd, there is an ascent in the final position
E X A M P L E 4 .4 .
For the alternating permutationw
w=
2
6
7
5
3
= 4 2 1 .
have w 
4
1
6
3
7
5
, we
of w, that is w(n2)  w(n1). So when 1 is inserted
Note in the second example how complementing an
after the final position of w and every other letter in w
alternating permutation preserves its zig-zag quality.
is shifted up by one, it is certain that w(n1)  1. Since
we begin with an alternating permutation with one
The set B will be generated from the permutations w 
alternating descent and only add 1 to the final position
n–1,1 An–1, acquired by the map, (w), defined as follows:
and shift every other letter up by one, we preserve
• Consider the permutation, w, in its zig-zag form.
its alternating structure and its alternating descent.
If n is even, then to obtain (w), first find . Then
Therefore  (w) is in n,1.
shift each letter in the top row of
that is to the left
of the alternating descent, including the position
It is easy to check that the map  (w) is one-to-one
of alternating descent, to the left by one position.
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DePaul Discoveries, Vol. 2 [2013], Iss. 1, Art. 2
E N U M E R AT I N G A LT E R N AT I N G P E R M U TAT I O N S W I T H O N E A LT E R N AT I N G D E S C E N T
This includes moving the top leftmost letter of
value of the descent. The alternating structure of the
to the bottom leftmost position. Then insert in the
permutation is preserved because we know the value of
remaining slot of , which will be immediately to
n must be greater than every other letter. Thus we have
the right of the descent, the letter n.
 (w)  n,1.
• Consider the permutation, w, in its zig-zag from.
If n is odd, then to obtain (w), first find . Then
Consider w in its zig-zag form. When n is odd, w has an
shift each letter in the top row of
to the left by
alternating descent in its top row. By complementing the
one entry. This includes moving the top leftmost
permutation, this alternating descent now comes in the
to the bottom leftmost position. Finally,
bottom row. Therefore when we shift each letter of the
insert in the remaining slot of , which will be the
top row over by one position and shift the top leftmost
top rightmost position, the letter n.
letter to the bottom row, no new alternating descent is
letter of
added. The alternating structure is preserved because
EXAMPLE 4.5
3
6
7
For w 2 1 5
4,
6
7
3
4
, which is 5
.
2
1
w=
the alternating descent occurs in the bottom row of
we have n8. First find
, the top row must be in ascending order, which means
Then move the top leftmost
that there will be an ascent in the first position of (w).
letter to the bottom leftmost position and shift each letter
Additionally, when we insert n into the top rightmost
that is left of the alternating descent
in the top row of
to the left by one position, so
becomes 6
7
5
2
3
4
1
position, n 
(n3) and n 
(n1) must be true.
Thus we have  (w)  n,1.
. Finally, insert n in the empty remaining top position,
so we have 6
7
5
8
2
3
1
4
. So (w)  67582314.
5. SET Cn
Similar to set B, the permutations of set C will be
EXAMPLE 4.6
5
7
8
6
For w 4
3
2
1 , we
6
7
8 .
, which is 5
4
2
1
3
w=
in the top leftmost position of
generated from the permutations in n-2,1 An-2 .
have n9. First we find
The set C will generated from the permutations
Then we move the letter
w  n-2,1 An-2, acquired by the map, (w), defined as
to the bottom leftmost
follows:
to the
• Consider the permutation, w, in its zig-zag form.
.
If n is odd, then to obtain (w), examine the
Finally, we insert n in the empty remaining top position,
bottom row of w. Insert ni1 in the position
position and shift each letter of the top row of
left by one position, which gives
yielding 5
6
4
7
2
8
1
9
5
6
4
7
2
8
1
3
immediately right of the descent, where i is
3 . So (w)  564728193.
the number of letters that are to the right of the
P R O P O S I T I O N 4 .7.
descent on the bottom row. Preserving the value of
For any w  n-1,1 An-1, we have (w)  n,1.
the letters in the bottom row, increase the values of
the letters of the top row by at least one, making
Proof. Consider w in its zig-zag form. When n is
sure not to acquire a new letter in the top row that
even, w has an alternating descent in its bottom row.
has the same value as one in the bottom row, and
By complementing the permutation, this alternating
also inserting ascending letters in the remaining
descent now comes in the top row. Therefore when we
top position.
shift every letter to the left of the alternating descent
• Consider the permutation, w, in its zig-zag form.
to the left by one, including moving the top leftmost
If n is even, then to obtain (w), first insert the
letter to the bottom leftmost position, we add no new
letter 1 to the right of the bottom rightmost letter
alternating descent. Inserting the letter n into this
of w. Then insert n1 to the right of the top
permutation does not add a descent, but increases the
rightmost letter. Next, increase each letter in the
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4
Wagner: EnumeratingExample
Alternating 1.8.
Permutations
with24153867
One Alternating
Descent 1.6, we have w = 63124587. The
For w =
from Example
D E PA U L D I S C O V E R I E S ( 2 O 1 3 )
three descents in w , so altdes(w) = 3.
permutation by one except the letter n2 which
increase the letter n1 by one and decrease the
we increase by two, not including the
two letters
letter
by one. permutations with 1 alternating descent in eit
Example
1.9. There are
nonalternating
we have just inserted.
EXA
M Pno
L E descents
6.3.
or S2 because there
are
in either (1) = 1 or (12) = 12.
For n6, we havedd6 6
= 3 4 2 5 1 6 , so (d6) 
E X A M P L E 5 .1 .
4
5
For w 1 3
2
s
dthat
(d 6 ) = 3 4 2 6 1 5 .
,
we
have
n7.
First
we
find
w=
Example 1.10. The alternating permutation of S3 with one alternating descent is 132
i2, and ni14 because there are two letters right
P R O P O S I T I O N 6 .4 .
= 213.
(132)yielding
of the descent. Next we insert this letter,
For any dn with n  3, we have (dn)  n,1 .
4
5
1
4
3 2 . We then increase the letters in the top
This proposition can be easily verified, so the proof here
row and insert the letter n, or in this case 7, to find (w)
5
6
7
is omitted.
Example
1.11.
The
alternating permutations of S4 with one alternating descent are
(w) = 1 4 3 2 .
2413, and 3412 where (1324) = 2134, (2413) = 1243, and (3412) = 1342.
7. C O N C LU S I O N
EXAMPLE 5.2.
For
ww
=
3
4
2
1
, we have n6. First we insert the letter
Given that the sets An, Bn, Cn, and Dn are defined as
in Sections 2 through
5, we arrive Eat
the following
2. Decomposing
n,1
7. Conclusion
conclusion.
. We then insert the letter n1, or in
yields
3
1
1
Let En,1 be the set of alternating permutations of size n with 1 alternating descen
this case 5, to the right of the top rightmost
letter,
which
Given
that
the sets An , Bn , Cn , and Dn are defined as in Sections 2 through 5, we arriv
T H EO R E M 7.1 .
4
2
5
will
show
. Next we increase all the letters that
by the set of alternating permutations of n letters with one alternating d
yields
3
1
1
at the following conclusion.
The set  can be decomposed as the union of four
one, except the letter n2, which we increase
by two,
be decomposed asn,1a union of the following pairwise disjoint sets:
En,1 , can
pairwise disjoint sets:
and those that we have just inserted, to find (w)
1 to the right of the bottom rightmost letter, which
4
s (w) =
4
6
2
3
1
2
5
.
Theorem 7.1. The set En,1 can be decomposed as the union of four pairwise disjoint set
En,1 = An Bn Cn Dn .
En,1 = An Bn Cn Dn .
PROPOSITION 5.3.
Given that |S| The
is theproof
the of
number
of elements of S, we have:
this theorem is omitted here due its length
For any w  n-2,1 An-2, we have (w)  n,1.
and technicality.
The proof of this theorem is omitted
due |its length and technicality.
|An | here
= |En−1,1
The proof is omitted here as it is too technical to
of| alternating
permutations
one alternating desce
If we let |En,1 | = en,1If, the
 e , |the
number of with
alternating
wenumber
let ||B
n,1 n | = |Bn,1
include, but it uses the same methods as the proof
of
n−1 + |Cn−1 | + |Dn−1 |
permutations
with
one
alternating
descent
can
be
found
can be found by the recurrence relation defined as follows [3]:
Propositions 3.3 and 4.7.
= |Bn−2defined
| + |Cn−2
+ |Dn−2[3]:
|
|Cn | relation
by the recurrence
as| follows
6. SET Dn
D E F I N I T I O N 6 .1 .
e|D
en−1,1 + F (n) − 1
n,1 =
n | = 1 for n > 3.
where F(n) is the function giving the Fibonacci numbers.
where
The alternating permutation of size n with
no F (n) is the function giving the Fibonacci numbers.
alternating descents has w(i) i. We willFurther,
call thisif we let F (n) − 1 = f , we can express this recursively as [4]:
n
Further, if we let F(n)1n, we can express this
permutation dn, with n   equal to its size.
recursively as [4]:
fn = fn−1 + fn−2 + 1
nn1n21
EXAMPLE 6.2.
For n7, we have d7  4536271 where dFrom
  1234567.
7
Examples 1.9, 1.10 and 1.11, we know that |E1,1 | = |E2,1 | = 0, |E3,1 | = 1 and |E4,1 | =
From Examples 1.9, 1.10 and 1.11, we know that | 1,1 |
we let e3,1
1, e4,1 = 3, f1 =
therefore we will beginthis
sequence
enumerating
|  0, | by
| 1 and | 4,1|E| 3,1
 |.3,Iftherefore
we=will
| 2,1
3,1
The permutation in set Dn will be generated from dn.
begin
this sequence by enumerating| 3,1 |. If we let e3,1
0, we
the sequence:
f2 =
)
This permutation will be generated and
by the
map
(dacquire
n
1, e4,1  3, 1  0, and 2  0, we acquire the sequence:
defined as follows:
1 3 7 14 26 46 79 · · ·
• For all permutations of size n, to obtain (dn),
1 3 7 14 26 46 79 ….
References
[1] S. Dulucq and R.
— Simion.
198 — Combinatorial statistics on alternating permutations. J. Alg. Comb., 8:169-19
Published by Via Sapientiae, 2013
1998.
[2] R. P. Stanley. A survey of alternating permutations. Contemp. Math., 531:165-196, 2010.
5
DePaul Discoveries, Vol. 2 [2013], Iss. 1, Art. 2
E N U M E R AT I N G A LT E R N AT I N G P E R M U TAT I O N S W I T H O N E A LT E R N AT I N G D E S C E N T
REFERENCES
[1] S. Dulucq and R. Simion. Combinatorial statistics on alternating
[3] The On-Line Encyclopedia of Integer Sequences, published electronically
permutations. J. Alg. Comb., 8:169-191, 1998.
[2] R. P. Stanley. A survey of alternating permutations. Contemp. Math.,
at http://oeis.org, 2013, Sequence A001924 .
[4] The On-Line Encyclopedia of Integer Sequences, published
531:165-196, 2010.
electronically at http://oeis.org, 2013, Sequence A000071 .
Matthew Francis, Brittany Fitzpatrick, and Kate Dovenberg take a quick break from presenting their research at the 10th Annual
Natural Sciences, Mathematics, and Technology Showcase.
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