Context Directed Sorted - Boise State University · Context Directed Sorted: Robustness, Complexity, and Games The cds Rescue Squad Mansi Bezbaruah, Henry Fessler, Leigh Foster, Marion

INTRODUCTION STRATEGIC MAP PARITY CUTS GAMES SUMMARY

Context Directed Sorted:Robustness, Complexity, and Games

The cds Rescue SquadMansi Bezbaruah, Henry Fessler, Leigh Foster, Marion Scheepers, George Spahn

July 27, 2018


INTRODUCTIONINTRODUCTION

Ciliates and PermutationsPointers, cdr, cds

STRATEGIC MAP

Terminal Points and the Strategic MapEffect of cds and cdrSortability CriteriaConstruction of Strategic Heaps

PARITY CUTS

Breakpoint and Overlap GraphsLaplacianTransformation

GAMES

Playing the CDR/CDS GameTypes of GamesGrundy Numbering

SUMMARY


CILIATES

Somatic Germline

Source: [4] J.Burns


WHAT IS A PERMUTATION?

DefinitionA permutation is an arrangement of all members of set in asequence.

For our purposes a permutation will be an arrangement of theintegers 1 to n.A signed permutation is an arrangement of the integers 1 to nwhere any entry can take on either a positive or negative sign.


EXAMPLE PERMUTATION

(1 2 3 4 5 6 7 83 −8 −2 −1 −7 4 −5 −6

)

[3, -8, -2, -1, -7, 4, -5, -6]


EXAMPLE PERMUTATION

(1 2 3 4 5 6 7 83 −8 −2 −1 −7 4 −5 −6

)

[3, -8, -2, -1, -7, 4, -5, -6]


SORTED, REVERSE SORTED, AND UNSORTED

PERMUTATIONS

DefinitionA permutation is sorted if it is written in the form[1, 2, . . . , (n − 1),n].

DefinitionA permutation is reverse sorted if it is written in the form[−n,−(n − 1), . . . ,−2,−1].

DefinitionA permutation is unsorted if it is not sorted and not reversesorted.


EXAMPLES

I Sorted Permutation: [1, 2, 3, 4, 5, 6]I Reverse Sorted Permutation: [-6, -5, -4, -3, -2, -1]I Unsorted Permutation: [-1, 2, -5, -3, 6, 4]


WHAT IS A POINTER?

Pointers help us understand what should be next to an entry ina permutation.

(3,4)

4

(4,5)

-(4,5)

-4

-(3,4)


WHAT IS A POINTER?


(3,4) 4

(4,5)

-(4,5)

-4

-(3,4)


WHAT IS A POINTER?


(3,4) 4 (4,5)

-(4,5)

-4

-(3,4)


WHAT IS A POINTER?


(3,4) 4 (4,5)

-(4,5) -4

-(3,4)


WHAT IS A POINTER?


(3,4) 4 (4,5)

-(4,5) -4 -(3,4)


CDR

DefinitionGiven a permutation with the pointer structure [. . . p . . .− p . . .]or [. . .− p . . . p . . .], the cdr operation cdrp is the reversal of thesection between p and −p. When this section is reversed itsentries are written in reverse order with the opposite sign oneach entry.


CDR EXAMPLE

[ 1 -5 3 -2 4 ](1,2) (1,2)

[ 1 2 -3 5 4 ](1,2) (1,2)


CDS

DefinitionGiven a permutation with the pointer structure[. . . p . . . q . . . p . . . q . . .], the cds operation cdspq is the swappingof the two sections surrounded by the p and q pointers.


CDS EXAMPLE

[ 5(1,2)

2(3,4)

-3 1(1,2) (3,4)

-4 ]

[ 5(3,4)

-4(3,4)

-3 1(1,2) (1,2)

2 ]


TERMINAL POINT

DefinitionA Terminal Point is a signed permutation on which no cdr andno cds operations can be performed.

[4, 1, 2, 3] is a Terminal Point.[-5, -2, 6, 1, -3, 4] is not a Terminal Point.


TERMINAL POINT


[4, 1, 2, 3] is a Terminal Point.

[-5, -2, 6, 1, -3, 4] is not a Terminal Point.


TERMINAL POINT


[4, 1, 2, 3] is a Terminal Point.[-5, -2, 6, 1, -3, 4] is not a Terminal Point.


FORMS OF TERMINAL POINTS

Terminal Points have a limited number of forms.

[1, . . . , n]This is the sorted permutation.

[-n, . . . , -1]This is the reverse sorted permutation.

[(k+1), . . . , n, 1, . . . , k], where k is an integer greater than one.This is denoted tpk

[-(k-1), . . . , -1, -n, . . . , -k], where k is an integer greater than one.This is denoted tp−k






























STRATEGIC MAP

DefinitionGiven a permutation π = [a1, a2, . . . , an] define threepermutations of the set {−(n + 1),−n, . . . ,−1, 0, 1, . . . , n,n + 1},written in cycle notation as follows:

X = (−(n + 1)− n . . .− 1 0 1 . . . n(n + 1))Yπ = ((n + 1) an . . . a2 a1 0 − a1 − a2 . . . − an − (n + 1))

Then Cπ = Yπ ◦ X. We refer to Cπ as the Strategic Map of π.


STRATEGIC HEAP

DefinitionThe Strategic Heap of a permutation is the set of numbersbetween the entries n and 0 in the strategic map. If the entries 0and n are in different cycles, the strategic heap is said to beempty.


LET’S FIND A STRATEGIC HEAP

π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=(

0 6 4 2 -4 1 -6

)(

3 -3 -5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=(

0 6 4 2 -4 1 -6

)(

3 -3 -5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=(

0 6 4 2 -4 1 -6

)(

3 -3 -5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=(

0 6 4 2 -4 1 -6

)(

3 -3 -5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0

6 4 2 -4 1 -6

)(

3 -3 -5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6

4 2 -4 1 -6

)(

3 -3 -5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4

2 -4 1 -6

)(

3 -3 -5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4 2

-4 1 -6

)(

3 -3 -5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4 2 -4

1 -6

)(

3 -3 -5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4 2 -4 1

-6

)(

3 -3 -5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4 2 -4 1 -6 )(

3 -3 -5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4 2 -4 1 -6 )( 3

-3 -5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4 2 -4 1 -6 )( 3 -3

-5 -7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4 2 -4 1 -6 )( 3 -3 -5

-7 -1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4 2 -4 1 -6 )( 3 -3 -5 -7

-1 5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4 2 -4 1 -6 )( 3 -3 -5 -7 -1

5 -2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4 2 -4 1 -6 )( 3 -3 -5 -7 -1 5

-2

)(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4 2 -4 1 -6 )( 3 -3 -5 -7 -1 5 -2 )(

7

)



π =[-5, -2, 6, 1, -3, 4]

X=(-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7)

Yπ=(7 4 -3 1 6 -2 -5 0 5 2 -6 -1 3 -4 -7)

Cπ=( 0 6 4 2 -4 1 -6 )( 3 -3 -5 -7 -1 5 -2 )( 7 )


BUDDY MAP

DefinitionThe buddy map is a permutation on the set S = {-n-1, -n, . . ., 0,. . ., n-1, n} defined by B(x) = −x − 1. Note B(B(x)) = x, whichjustifies calling x and B(x) a buddy pair.


BUDDY CYCLES

DefinitionA cycle is a Buddy Cycle to a cycle A if each entry of the cycleis the buddy number of an entry in cycle A.

The buddy cycle of a given cycle A can be generated by writingthe cycle in reverse and replacing each entry with its buddy.


CDR

TheoremSuppose that π is a signed permutation. Let τ=cdr(x,x+1)(π). Thenthe only cycles in the cycle decomposition of Cπ that do not alsoappear in the cycle decomposition of Cτ are those that contain x or−(x + 1). Moreover, x and −(x + 1) are placed in single cycles by thepermutation Cτ , and the rest of the cycle decomposition for Cτ isobtained from Cπ by deleting x and −(x + 1) from the cycles in whichthey appear.


EXAMPLE CDR

π=[-5, -2, 6, 1, -3, 4]

Cπ=(0 6 4 2 -4 1 -6)(3 -3 -5 -7 -1 5 -2)(7)cdr(1,2)(π)=[-5, -2, -1, -6, -3, 4]

τ=[-5, -2, -1, -6, -3, 4]Cτ=(0 6 4 2 -4 -6)(3 -3 -5 -7 -1 5 )(7)(1)(-2)


EXAMPLE CDR

π=[-5, -2, 6, 1, -3, 4]Cπ=(0 6 4 2 -4 1 -6)(3 -3 -5 -7 -1 5 -2)(7)

cdr(1,2)(π)=[-5, -2, -1, -6, -3, 4]τ=[-5, -2, -1, -6, -3, 4]

Cτ=(0 6 4 2 -4 -6)(3 -3 -5 -7 -1 5 )(7)(1)(-2)


EXAMPLE CDR

π=[-5, -2, 6, 1, -3, 4]Cπ=(0 6 4 2 -4 1 -6)(3 -3 -5 -7 -1 5 -2)(7)

cdr(1,2)(π)=[-5, -2, -1, -6, -3, 4]

τ=[-5, -2, -1, -6, -3, 4]Cτ=(0 6 4 2 -4 -6)(3 -3 -5 -7 -1 5 )(7)(1)(-2)


EXAMPLE CDR

π=[-5, -2, 6, 1, -3, 4]Cπ=(0 6 4 2 -4 1 -6)(3 -3 -5 -7 -1 5 -2)(7)

cdr(1,2)(π)=[-5, -2, -1, -6, -3, 4]τ=[-5, -2, -1, -6, -3, 4]

Cτ=(0 6 4 2 -4 -6)(3 -3 -5 -7 -1 5 )(7)(1)(-2)


EXAMPLE CDR

π=[-5, -2, 6, 1, -3, 4]Cπ=(0 6 4 2 -4 1 -6)(3 -3 -5 -7 -1 5 -2)(7)

cdr(1,2)(π)=[-5, -2, -1, -6, -3, 4]τ=[-5, -2, -1, -6, -3, 4]

Cτ=(0 6 4 2 -4 -6)(3 -3 -5 -7 -1 5 )(7)(1)(-2)


EXAMPLE CDR

π=[-5, -2, 6, 1, -3, 4]Cπ=(0 6 4 2 -4 1 -6)(3 -3 -5 -7 -1 5 -2)(7)

cdr(1,2)(π)=[-5, -2, -1, -6, -3, 4]τ=[-5, -2, -1, -6, -3, 4]

Cτ=(0 6 4 2 -4 -6)(3 -3 -5 -7 -1 5 )(7)(1)(-2)


CDS

TheoremSuppose that π is a signed permutation. Let τ=cds(x,x+1)(y,y+1)(π).Then the only cycles in the cycle decomposition of Cπ that do not alsoappear in the cycle decomposition of Cτ are those that contain x,−(x + 1), y, or −(y + 1). Moreover, x, −(x + 1), y, and −(y + 1) areplaced in single cycles by the permutation Cτ , and the rest of the cycledecomposition for Cτ is obtained from Cπ by deleting x, −(x + 1), y,and −(y + 1) from the cycles in which they appear.


SORTABILITY CRITERIA

I A permutation is unsortable if in Cπ there is a cyclecontaining both n and 0.

I A permutation is reverse sortable if in Cπ there is a cyclecontaining both n and −1.

I A permutation is sortable if in Cπ there is a cyclecontaining n but not −1 and not 0.


TERMINAL POINT THEOREM

Given a signed permutation that is unsortable, n and 0 mustappear in the same cycle of the strategic map. We call this cyclethe strategic cycle.

TheoremThe set containing the last digits of all possible terminal points thatcan be reached is given by the numbers that appear between n and 0in the strategic cycle.

I π = [−2, 3, 1,−4,−5]I Strategic Cycle = (5,−5, 1,−3, 0)I [−4,−3,−2,−1,−5], [2, 3, 4, 5, 1], [−2,−1,−5,−4,−3]


MAXIMUM NUMBER OF TERMINAL POINTS

Fix n, there are 2 · (n − 1) terminal points.

Theorem (Buddy Cycle)

k and −(k + 1) can never appear in the same cycle of the strategicmap.

I Two terminal points with last numbers that are buddiescan never appear together.

I “Buddy terminal points”I Can never have more than (n − 1) reachable terminal

points.


CONSTRUCTING (n − 1) REACHABLE TERMINAL

POINTS

I [6, 5, 4, 3, 2,−1,−7] has strategic cycle(7,−7,−5,−3, 1, 3, 5, 0)

I [(n − 1), (n − 2), . . . 2,−1,−n] has strategic heap of sizen − 1, containing all buddies with the same parity


WHAT IF WE SPECIFY EXACTLY WHAT THE (n − 1)TERMINAL POINTS ARE?

I Given a specific strategic cycle, easy to decide whether it ispossible

I Many strategic cycles correspond to the same set ofterminal points

I e.g. Rearrange the order of the strategic cycle

I Testing all possible strategic cycles is not efficient.


USUALLY IT’S POSSIBLE

TheoremIf your set of terminal points contains two elements of different sign,then there exists a permutation which can reach exactly that set.


A LEMMA

LemmaLet π = [a1, . . . k, k + 1, . . . , an]. Then the permutations[−k, . . .− a1, k + 1, . . . , an] and [a1, . . . k,−an, . . . ,−(k + 1)]obtained by reversals on π have −(k + 1) added to the strategic heap.


EXAMPLE

Say n = 5 and we want the terminal points 1,−3, 3 and 4.I [2, 3, 4, 5, 1]

I [2,−1,−5,−4,−3]I [2,−1,−5,−4, 3]I [2,−1,−5,−3, 4]

The first step requires a change in sign.


EXAMPLE

Say n = 5 and we want the terminal points 1,−3, 3 and 4.I [2, 3, 4, 5, 1]I [2,−1,−5,−4,−3]

I [2,−1,−5,−4, 3]I [2,−1,−5,−3, 4]



EXAMPLE

Say n = 5 and we want the terminal points 1,−3, 3 and 4.I [2, 3, 4, 5, 1]I [2,−1,−5,−4,−3]I [2,−1,−5,−4, 3]

I [2,−1,−5,−3, 4]



EXAMPLE

Say n = 5 and we want the terminal points 1,−3, 3 and 4.I [2, 3, 4, 5, 1]I [2,−1,−5,−4,−3]I [2,−1,−5,−4, 3]I [2,−1,−5,−3, 4]



ALL THE SAME SIGN?

If n is even, it works.

I [2, 4, 6, 8, 1, 3, 5, 7] has strategic cycle (8, 7, 6, 5, 4, 3, 2, 1, 0)

If n is odd, it doesn’t work.


WHAT IF WE WANT TO SPECIFY LESS THAN (n − 1)TERMINAL POINTS?

It is always possible, unless we have a set of the form

{k, k + 1, k + 2, . . . , k + i}

with an even number of elements


WHAT IS THE BREAKPOINT GRAPH?

Consider a Signed Permutation α. If we draw an arc betweenthe two occurrences of the pointer (i, i + 1) for all i appearing inthe framed permutation for α, we will obtain the BreakpointGraph of α.


WHAT IS THE OVERLAP GRAPH?

DefinitionAn Overlap Graph G = (V,E,(x,y)) of a permutation α is a finitetwo-rooted graph such that the vertices represent the pointersof the framed permutation of α. The two pointers framing thepermutation are assigned the roots.

Two vertices v and w in the Overlap Graph are connected if inthe Breakpoint graph of α the arcs connecting thecorresponding pointers cross.


WHAT ARE PARITY CUTS?

Definition (Parity Cut for a two-rooted graph (REU 2015))

A parity cut of a two-rooted graph G = (V,E, (x, y)) is apartition of the set of vertices V into V1 and V2 such that:

1. for all non-root v ∈ V1, v is adjacent to an even number ofvertices in V2

2. for all non-root w ∈ V2, w is adjacent to an even number ofvertices in V1.

Definition (Trivial Parity Cut)

A parity cut of a finite two-rooted graph that assigns everyvertex to the same parity set.


PROPERTY A

A Graph G with root x and root y has parity Property A ifI x ∈ V1

I y ∈ V2

I x is connected to an even number of vertices in V2

I y is connected to an even number of vertices in V1.


PROPERTY A


I y ∈ V2




PROPERTY A


I y ∈ V2




PROPERTY A


I y ∈ V2




PROPERTY B

A Graph G with root x and root y has parity Property B ifI x ∈ V1

I y ∈ V2

I x is connected to an odd number of vertices in V2

I y is connected to an odd number of vertices in V1.


PROPERTY B


I y ∈ V2




PROPERTY B


I y ∈ V2




PROPERTY B


I y ∈ V2




PROPERTY C

A Graph G with root x and root y has parity Property C ifI x ∈ V1

I y ∈ V1




PROPERTY C


I y ∈ V1




PROPERTY C


I y ∈ V1




PROPERTY C


I y ∈ V1




LAPLACIAN MATRIX

DefinitionThe Laplacian Matrix for a finite graph G with n vertices is an × n matrix that has elements Li,j defined as follows:

I if i = j, Li,i = (degree of the ith vertex) mod 2I if i 6= j and the ith and the jth vertex have an edge, Li,j = 1I otherwise, Li,j = 0


KERNELS OF LAPLACIANS

DefinitionA kernel of a matrix is a vector subspace such that:A vector v belongs to the kernel of a matrix A if and only ifAv = 0.


DETERMINING PROPERTY A PARITY CUT

Given a graph G:I Compute the Laplacian MatrixI Compute the Kernels of the Laplacian Matrix mod 2.I If any element of the kernel has different entries

corresponding to the root nodes, it represents the paritycut of G.

Conclusion: Graph has Property A


DETERMINING PROPERTY B PARITY CUT

If a graph G does not have a parity cut of Property A.I Complement the edge between (0,1) and (n, n+1) in the

Overlap GraphI Compute the Laplacian MatrixI Compute the Kernels of the Laplacian Matrix mod 2.I If any element of the kernel has different entries


Conclusion: Graph has Property B


DETERMINING PROPERTY C PARITY CUT

If a graph G does not have a parity cut of Property A orProperty B.

I Introduce an artificial vertex to the Overlap GraphI Connect this artificial vertex to both rootsI Compliment the edge connecting the rootsI Compute the Laplacian MatrixI Compute the Kernels of the Laplacian Matrix mod 2.I If any element of the kernel has different entries


Conclusion: Graph has Property C


CDR AND CDS OPERATIONS PRESERVE PARITY

PROPERTIES

Theoremcdr operations preserve the parity property of the overlap graph of thepermutations (REU 2015).

Theoremcds operations preserve the parity property of the overlap graph of thepermutations (REU 2017).


CDS AND CDR SORTABILITY CONDITIONS

The parity property of an overlap graph determines thesortability of its generating permutation.

I If an overlap graph has property A, then its generatingpermutation is sortable.

I If an overlap graph has property B, then its generatingpermutation is reverse sortable.

I If an overlap graph has property C, then its generatingpermutation is unsortable.




















THE CDR-CDS GAME

Lets Play!

https://darthcalculus.github.io/cdr_cds_game/cdr_cds_game.html


WHAT IS THE CDR-CDS GAME?

Given a signed permutation, player ONE and player TWO takealternating legal cdr or cds moves. Play continues until nomore valid moves can be made.


HOW IS THIS VISUALIZED?

We can also play these games on graphs where every vertexwill represent a permutation and each edge will represent a cdror cds move leading to the next permutation.


NIM ON PERMUTATIONS

DefinitionThe cdr-cds NIM game on a permutation is played as follows:NIM(π, {cdr, cds}): Starting with player ONE, the playersalternately make cdr or cds operations on the current signedpermutation until a terminal is reached. The last player tomake a legal operation wins.


NIM GAME GRAPH

[3, 5, 2, 1, -4][3, -1, -2, -5, -4]

[3, 4, -1, -2, -5]

[1, 2, 3, 5, -4]

[3, 4, 5, 2, 1]

[1, -3, -2, -5, -4]

[1, -4, -3, -2, -5]

[1, 2, 3, -5, -4]

[1, 2, 3, 4, -5][1, 2, 3, 4, 5]


TERMINAL POINT GAME

The Terminal Point Game is a two player game played on apermutation.

Definition

TERM(π, {cdr, cds},A): A nonempty set A of terminal points inS±

n is given. In the game TERM(π, cdr, cds,A) players ONEand TWO operation alternately, making a legal cdr or a legalcds operation. The game ends when a terminal point isreached. ONE wins the play if this terminal point is in A, andotherwise TWO wins.


TERM GAME GRAPH

[4, 2, -1, -5, -3]

[4, 5, 1, -2, -3]

[5, 1, -2, -4, -3]

[4, 2, 3, 5, 1]

[4, -2, -1, -5, -3]

[2, -1, -5, -4, -3]

[4, 5, 1, 2, -3]

[5, 1, 2, -4, -3]

[2, 3, 4, 5, 1]

[-2, -1, -5, -4, -3]

[4, 5, 1, 2, 3]

[5, 1, 2, 3, 4]


POSITIVE/NEGATIVE GAME

DefinitionThe Positive/Negative Game is a special case of the TerminalPoint Game. Each player is assigned either all of the positive ornegative Terminal Point. Play then proceeds as in the FixedPoint Game.


POSITIVE/NEGATIVE GAME GRAPH

[4, 2, -1, -5, -3]

[4, 5, 1, -2, -3]

[5, 1, -2, -4, -3]

[4, 2, 3, 5, 1]

[4, -2, -1, -5, -3]

[2, -1, -5, -4, -3]

[4, 5, 1, 2, -3]

[5, 1, 2, -4, -3]

[2, 3, 4, 5, 1]

[-2, -1, -5, -4, -3]

[4, 5, 1, 2, 3]

[5, 1, 2, 3, 4]


WHAT IS GRUNDY NUMBERING?

DefinitionThe Grundy Numbering of a graph can be assigned as follows:

1. If a vertex is a terminal position, then the Grundy numberof that vertex is 0.

2. If a vertex leads only to vertices that are not zero, thatvertex is assigned a Grundy number of 0.

3. If a vertex leads to at least one vertex with a Grundynumber of 0, then this vertex gets assigned the lowestpositive value it is not connected to.


WHAT DOES THIS TELL US?

A Grundy Number tells us information about which player hasa winning strategy.

If a player can move to a position labeled with a 0 then thatplayer will win if appropriate moves are made.


WHAT DOES THIS TELL US?

A Grundy Number tells us information about which player hasa winning strategy.If a player can move to a position labeled with a 0 then thatplayer will win if appropriate moves are made.


GRUNDY NUMBERING GAME GRAPH

[4, 2, -1, -5, -3]

2

[4, 5, 1, -2, -3]

0

[5, 1, -2, -4, -3]

0

[4, 2, 3, 5, 1]

1

[4, -2, -1, -5, -3]

0

[2, -1, -5, -4, -3]

1

[4, 5, 1, 2, -3]

1

[5, 1, 2, -4, -3]

1

[2, 3, 4, 5, 1]

0

[-2, -1, -5, -4, -3]

0

[4, 5, 1, 2, 3]

0

[5, 1, 2, 3, 4]

0



[4, 2, -1, -5, -3]

2

[4, 5, 1, -2, -3]

0

[5, 1, -2, -4, -3]

0

[4, 2, 3, 5, 1]

1

[4, -2, -1, -5, -3]

0

[2, -1, -5, -4, -3]

1

[4, 5, 1, 2, -3]

1

[5, 1, 2, -4, -3]

1

[2, 3, 4, 5, 1]

0

[-2, -1, -5, -4, -3]

0

[4, 5, 1, 2, 3]

0

[5, 1, 2, 3, 4]0



[4, 2, -1, -5, -3]

2

[4, 5, 1, -2, -3]

0

[5, 1, -2, -4, -3]

0

[4, 2, 3, 5, 1]

1

[4, -2, -1, -5, -3]

0

[2, -1, -5, -4, -3]

1

[4, 5, 1, 2, -3]

1

[5, 1, 2, -4, -3]

1

[2, 3, 4, 5, 1]

0

[-2, -1, -5, -4, -3]

0

[4, 5, 1, 2, 3]0

[5, 1, 2, 3, 4]0



[4, 2, -1, -5, -3]

2

[4, 5, 1, -2, -3]

0

[5, 1, -2, -4, -3]

0

[4, 2, 3, 5, 1]

1

[4, -2, -1, -5, -3]

0

[2, -1, -5, -4, -3]

1

[4, 5, 1, 2, -3]

1

[5, 1, 2, -4, -3]

1

[2, 3, 4, 5, 1]

0

[-2, -1, -5, -4, -3]0

[4, 5, 1, 2, 3]0

[5, 1, 2, 3, 4]0



[4, 2, -1, -5, -3]

2

[4, 5, 1, -2, -3]

0

[5, 1, -2, -4, -3]

0

[4, 2, 3, 5, 1]

1

[4, -2, -1, -5, -3]

0

[2, -1, -5, -4, -3]

1

[4, 5, 1, 2, -3]

1

[5, 1, 2, -4, -3]

1

[2, 3, 4, 5, 1]0

[-2, -1, -5, -4, -3]0

[4, 5, 1, 2, 3]0

[5, 1, 2, 3, 4]0



[4, 2, -1, -5, -3]

2

[4, 5, 1, -2, -3]

0

[5, 1, -2, -4, -3]

0

[4, 2, 3, 5, 1]

1

[4, -2, -1, -5, -3]

0

[2, -1, -5, -4, -3]

1

[4, 5, 1, 2, -3]

1

[5, 1, 2, -4, -3]1

[2, 3, 4, 5, 1]0

[-2, -1, -5, -4, -3]0

[4, 5, 1, 2, 3]0

[5, 1, 2, 3, 4]0



[4, 2, -1, -5, -3]

2

[4, 5, 1, -2, -3]

0

[5, 1, -2, -4, -3]

0

[4, 2, 3, 5, 1]

1

[4, -2, -1, -5, -3]

0

[2, -1, -5, -4, -3]

1

[4, 5, 1, 2, -3]1

[5, 1, 2, -4, -3]1

[2, 3, 4, 5, 1]0

[-2, -1, -5, -4, -3]0

[4, 5, 1, 2, 3]0

[5, 1, 2, 3, 4]0



[4, 2, -1, -5, -3]

2

[4, 5, 1, -2, -3]

0

[5, 1, -2, -4, -3]

0

[4, 2, 3, 5, 1]

1

[4, -2, -1, -5, -3]

0

[2, -1, -5, -4, -3]1

[4, 5, 1, 2, -3]1

[5, 1, 2, -4, -3]1

[2, 3, 4, 5, 1]0

[-2, -1, -5, -4, -3]0

[4, 5, 1, 2, 3]0

[5, 1, 2, 3, 4]0



[4, 2, -1, -5, -3]

2

[4, 5, 1, -2, -3]

0

[5, 1, -2, -4, -3]

0

[4, 2, 3, 5, 1]

1

[4, -2, -1, -5, -3]0

[2, -1, -5, -4, -3]1

[4, 5, 1, 2, -3]1

[5, 1, 2, -4, -3]1

[2, 3, 4, 5, 1]0

[-2, -1, -5, -4, -3]0

[4, 5, 1, 2, 3]0

[5, 1, 2, 3, 4]0



[4, 2, -1, -5, -3]

2

[4, 5, 1, -2, -3]

0

[5, 1, -2, -4, -3]

0

[4, 2, 3, 5, 1]1

[4, -2, -1, -5, -3]0

[2, -1, -5, -4, -3]1

[4, 5, 1, 2, -3]1

[5, 1, 2, -4, -3]1

[2, 3, 4, 5, 1]0

[-2, -1, -5, -4, -3]0

[4, 5, 1, 2, 3]0

[5, 1, 2, 3, 4]0



[4, 2, -1, -5, -3]

2

[4, 5, 1, -2, -3]

0

[5, 1, -2, -4, -3]0

[4, 2, 3, 5, 1]1

[4, -2, -1, -5, -3]0

[2, -1, -5, -4, -3]1

[4, 5, 1, 2, -3]1

[5, 1, 2, -4, -3]1

[2, 3, 4, 5, 1]0

[-2, -1, -5, -4, -3]0

[4, 5, 1, 2, 3]0

[5, 1, 2, 3, 4]0



[4, 2, -1, -5, -3]

2

[4, 5, 1, -2, -3]0

[5, 1, -2, -4, -3]0

[4, 2, 3, 5, 1]1

[4, -2, -1, -5, -3]0

[2, -1, -5, -4, -3]1

[4, 5, 1, 2, -3]1

[5, 1, 2, -4, -3]1

[2, 3, 4, 5, 1]0

[-2, -1, -5, -4, -3]0

[4, 5, 1, 2, 3]0

[5, 1, 2, 3, 4]0



[4, 2, -1, -5, -3]2

[4, 5, 1, -2, -3]0

[5, 1, -2, -4, -3]0

[4, 2, 3, 5, 1]1

[4, -2, -1, -5, -3]0

[2, -1, -5, -4, -3]1

[4, 5, 1, 2, -3]1

[5, 1, 2, -4, -3]1

[2, 3, 4, 5, 1]0

[-2, -1, -5, -4, -3]0

[4, 5, 1, 2, 3]0

[5, 1, 2, 3, 4]0


REDUCTION

Reductions help us to shorten permutations and analyzesimilar structures of longer permutations.

A permutation can be reduced if two consecutive integers n,n + 1 appear in that order. This block is rewritten as thenumber closer to zero and each entry with a larger absolutevalue is relabeled as necessary.


REDUCTION EXAMPLE

[5, 1, -2, -4, -3]

[5, 1, -2, -3][4, 1, -2, -3]


REDUCTION EXAMPLE

[5, 1, -2, -4, -3]

[5, 1, -2, -3][4, 1, -2, -3]


REDUCTION EXAMPLE

[5, 1, -2, -4, -3][5, 1, -2, -3]

[4, 1, -2, -3]


REDUCTION EXAMPLE

[5, 1, -2, -4, -3][5, 1, -2, -3]

[4, 1, -2, -3]


REDUCTION EXAMPLE

[5, 1, -2, -4, -3][5, 1, -2, -3][4, 1, -2, -3]


RECALL: GRUNDY NUMBERING GAME GRAPH

[4, 2, -1, -5, -3]2

[4, 5, 1, -2, -3]0

[5, 1, -2, -4, -3]0

[4, 2, 3, 5, 1]0

[4, -2, -1, -5, -3]0

[2, -1, -5, -4, -3]1

[4, 5, 1, 2, -3]1

[5, 1, 2, -4, -3]1

[2, 3, 4, 5, 1]0

[-2, -1, -5, -4, -3]0

[4, 5, 1, 2, 3]0

[5, 1, 2, 3, 4]0


REDUCED GRUNDY NUMBERED GAME GRAPH

[4, 2, -1, -5, -3]2

[4, 1, -2, -3]0

[3, 2, 4, 1]1

[3, -1, -4, -2]0

[2, -1, -3]1

[3, 1, -2]1

[2, 1]0

[-1, -2]0


RESULTS

TheoremIf two nodes can be combined through reduction, then they haveisomorphic lower sub-graphs.

Corollary

The Grundy number of a permutation is maintained throughreduction.


SUMMARY

Strategic Map Parity CutsUnsortable n and 0 appear in the same cycle property c

Reverse Sortable n and -1 appear in the same cycle property bSortable n appears in a cycle without 0 or -1 property a

I The strategic map method takes linear time.I The parity cut method takes polynomial time.

Sortable and reverse sortable: Applications of cdr and cdsproduce only sorted or reverse sorted permutations. No otherterminal points are reachable.Unsortable: The entries appearing between n and 0 in a singlecycle in the strategic map correspond with the possibleterminal points that are reachable.


REFERENCES

1. D.M. Prescott, A. Ehrenfeucht and G. Rozenberg,Template-guided recombination for IES elimination andunscrambling of genes in stichotrichous ciliates, Journal ofTheoretical Biology 222 (2003) 323330

2. K.L.M. Adamyk, E.T. Holmes, G. Mayfield, D.J. Moritz, M.Scheepers, B.E. Tenner and H.C. Wauck, Sortingpermutations: Games, Genomes and Cycles, DiscreteMathematics, Algorithms and Applications 9:5 (2017),1750063

3. J. Burns. Ciliate Biology . Princeton University, TheTrustees of Princeton University,oxytricha.princeton.edu/mdsiesdb/ciliate.html.

4. J. Burns. Ciliate Biology . Princeton University, TheTrustees of Princeton University,oxytricha.princeton.edu/mdsiesdb/ciliate.html.


ACKNOWLEDGEMENTS

I 2012-2017 teams for their workI Boise State University Mathematics REUI NSF - REU funding via grant DMS-1659872

Context Directed Sorted - Boise State University · Context Directed Sorted: Robustness, Complexity, and Games The cds Rescue Squad Mansi Bezbaruah, Henry Fessler, Leigh Foster, Marion

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