Unshuffling a square is NP-hard

Unshu!ing a Square is NP-hard

Sam Buss and Michael Soltys

February 6, 2013

Unshuffling - Buss & Soltys Slides - Feb 6, 2013 Title - 1/27

Shu!e

w is the shu!e of u, v : w = u ! v

w = 0110110011101000

u = 01101110

v = 10101000

w = 0110110011101000

Unshuffling - Buss & Soltys Slides - Feb 6, 2013 Definition - 2/27

Shu!e

w is the shu!e of u, v : w = u ! v

w = 0110110011101000

u = 01101110

v = 10101000

w = 0110110011101000

w is a shu!e of u and v provided:

u = x1x2 · · · xk

v = y1y2 · · · yk

and w obtained by “interleaving” w = x1y1x2y2 · · · xkyk .


Square Shu!e

w is a square provided it is equal to a shu!e of a u with itself, i.e.,"u s.t. w = u ! u

The string w = 0110110011101000 is a square:

w = 0110110011101000

andu = 01101100 = 01101100


Result

given an alphabet ", |"| # 7,

Square = {w : "u(w = u ! u)}

is NP-complete.

What we leave open:

! What about |"| = 2 (for |"| = 1, Square is just the set ofeven length strings)

! What about if |"| = $ but each symbol cannot occur moreoften than, say, 6 times (if each symbol occurs at most 4times, Square can be reduced to 2-Sat – see P. AustrinStack Exchange post bit.ly/WATco3)


Bipartite graph

w = (c1x3c2)2(c1x2c2)2(c1xc2)2 is a square:

w = u ! u where u = c1xxxc2xc2c1xc1xc2.


Bipartite graph



Bipartite graph G associated with particular solution w = u ! u:

c1 x x x c2 c1 x x x c2 c1 x x c2 c1 x x c2 c1 x c2 c1 x c2


Bipartite graph



Bipartite graph G associated with particular solution w = u ! u:

c1 x x x c2 c1 x x x c2 c1 x x c2 c1 x x c2 c1 x c2 c1 x c2

We also have w = v ! v with v = c1x3c2c1x2c2c1xc2

which would have its associated bipartite graph


Nesting

a b a b a b b a

G has the “non-nesting” property: if G contains an edge from wk

to w! and an edge from wp to wq, then it is not the case thatk < p < q < !

Claim: There is a complete bipartite graph G of degree one (i.e., aperfect matching) on the symbols of w which is non-nesting, i# wcan be expressed as a square shu!e.


! Formal languages! Ginsburg & Spanier in ’65

! Modelling sequential execution of concurrent processes! Riddle (’73 & ’79), Shaw (’78), Hoare

! Shu!e on its own! Mansfield (’82 & ’83; polytime dynamic prog. alg. for

w = u1 ! u2 ! · · ·! uk for constant k ; when k varies,NP-complete)

! Warmuth & Haussler (’84; show w = u ! u ! · · ·! uNP-complete by reduction from 3-Partition)

! More complexity! Buss & Yianilos (’98; “Monge condition”)! Erickson (2010; how hard is "u,w = u ! u?); as mentioned,

Austrin gives polytime algorithm when each symbol occurs atmost 4 times. Problem of the year on Stack Exchange.

! Soltys (2012; shu!e in AC1 but not AC0.)

Unshuffling - Buss & Soltys Slides - Feb 6, 2013 History - 7/27

Monge

The Monge condition (or “quasi-convexity” condition) allowsnested edges but prohibits crossing edges; widely studied formatching and transportation problems.

Many problems that satisfy the Monge condition are known tohave e$cient polynomial time algorithms: Buss & Yianilos ’98

Thus, the non-nesting property on our G can be viewed as an“anti-Monge.” Fewer algorithms for this case:

This is another reason why we find the NP-completeness of thesquare problem to be interesting: it provides a hardness result foranti-Monge matching in a very simple and abstract situation.

Unshuffling - Buss & Soltys Slides - Feb 6, 2013 History - 8/27

PDA

A queue automaton is a PDA but with a queue instead of a stack.

Reads the input w fromleft to right.

Queue is initially emptyand supports theoperations push-bottomand pop-top.

Automaton accepts if itsqueue is empty after thelast symbol of w hasbeen read.

w

Pop

Push

Unshuffling - Buss & Soltys Slides - Feb 6, 2013 PDAs - 9/27

PDA algorithm for Square

Repeatedly do one of the following:

a Read the next input symbol " and push it onto thequeue, or

b If the next input symbol " is the same as the symbolat the top of the queue, read past the inputsymbol " and pop " from the queue.

We can view an accepting computation as imposing a non-nestingmatching.

! When either step a. or b. is performed, we say that the inputsymbol " has been consumed.

! In case b., we say that the symbol " on the queue has beenmatched by the input symbol.


w = 0110110011101000

andu = 01101100 = 01101100


# 0110110011101000


0 110110011101000


01 10110011101000


011 0110011101000


11 110011101000


1 10011101000


# 0011101000


0 011101000


# 11101000


1 1101000


# 101000


1 01000


10 1000


0 000


# 00


0 0


# # accepted


Square is NP-complete

Many-one logspace reduction of 3-Partition to Square.

S = %ni : 1 & i & 3m' such that the ni ’s are given in unarynotation and such that B = (

!3mi=1 ni )/m is an integer.

The question is: can S be partitioned into m disjoint subsequencesS1, . . . ,Sm such that each Sk has exactly three elements with thesum of the three members of Sk equal to B?

Assume ni given in non-increasing order.

Unshuffling - Buss & Soltys Slides - Feb 6, 2013 Proof - 13/27

Reduction

The many-one reduction to Square constructs a string wS overthe alphabet

" = {a1, a2, b, e0, e, c1, c2, x , y},

such that wS is a square i# S is a “yes” instance of 3-Partition.

The string wS consists of three parts:

wS := %loaderS'%distributorS'%verifierS '.


%loaderS' = e0

m"

i=1

(b2Be)

%distributorS' = e0

m"

i=1

((a1bBa2)

3e)

%verifierS' =3m"

k=1

[v4k!3Dkv4k!3v4k!2Dkv4k!2v4k!1Ekv4k!1v4kFkv4k ]

where

v! = c1x!y !c2

Dk = (a21bnk a22)

3m!k+1

Ek = (a21bBa22)

3m!k(a1bnka2)(a

21b

Ba22)3m!k

Fk = (a21bBa22)

2(3m!k)


Action of %loaderS'

e0b2Beb2Be...

b2Be


Action of %distributorS'

b2Be QQ a1bi1a2

a1bi2a2a1bi3a2

(a1bBa2)3e

where i1 + i2 + i3 = B .


After %loaderS'%distributorS ' has been scanned, we have thefollowing configuration:

3m"

k=1

(a1bika2)(%verifierS'

The role of the verifier is to check wheter %ik'3mk=1 is a re-orderingof %nk'3mk=1.


Correctness

Given any sequence of non-negative integers %ik'3mk=1 such that

)j * {1, 2, . . . ,m}, i3j!2 + i3j!1 + i3j = B , (1)

there exists a computation

#(wS +"3m"

k=1


Conversely, if#(wS +"W (%verifierS '

then W must be of the form#3m

k=1(a1bika2), so that condition (1)

holds.


Therefore it su$ces to show that

3m"

k=1


is accepted ,-%ik'3mk=1 is a permutation of %nk'3mk=1.


If %ik'3mk=1 is a permutation of %nk'3mk=1, then it is a matter ofkeeping track of lots of details to show that

3m"

k=1

(a1bika2)(%verifierS '+"#(#,

and hence wS * Square.

The other direction is more complicated because we have to showthat if a computation accepts, then the ik ’s must be a permutationof the nk ’s.


Recall that:

%verifierS' =3m"

k=1

[v4k!3Dkv4k!3v4k!2Dkv4k!2v4k!1Ekv4k!1v4kFkv4k ]

where

v! = c1x!y !c2

Dk = (a21bnk a22)

3m!k+1

Ek = (a21bBa22)

3m!k(a1bnka2)(a

21b

Ba22)3m!k

Fk = (a21bBa22)

2(3m!k)


The role of the v ’s is that of delimiters; they impose a scope onthe edges.

The role of the D’s is to make sure that at each of the k steps, theremaining ij ’s are all less than or equal to the next nk .

The role of the Ek and Fk is to cycle the queue until the righta1bixa2 is found for a1bnk a2; i.e., such that ix = nk .


The main idea is that if n1 corresponds to ix then:

a1bi1a2 a1bixa2a1bi2aa a1bix+1a2 a1bix+1a2 a1bB!i1a2 a1bi1a2

......

......

...a1bixa2 a1bi3ma2 a1bi3ma2 a1bB!ix!1a2 a1bix!1a2a1bix+1a2 a1bB!i1a2 a1bB!i1a2 a1bB!ix+1a2 a1bix+1a2

......

......

...a1bi3ma2 a1bB!ix!1a2 a1bB!ix!1a2 a1bB!ima2 a1bima2

The changes to the queue while applying E1,F1 with theassumption that ix = n1.


If a computation accepts, then the ik ’s must be a permutation ofthe nk ’s.

Elements of the proof:

! If y is consumed by u yielding the resultant z , then y = u! z .

u QQ z

y

so a subsequence of u has to be a subsequence of y .

! A string z has k alternations of the symbols a1, a2 provided(a1a2)k is a subsequence of z but (a1a2)k+1 is not.


a1bija2 Q

Q Zj = a1bB!m1#!

s=2(a22a

21b

B!ms )a2

Yj = (a21bBa22)

! * Ek

where m1 +m2 + · · ·+m! = ij . Note that Yj and Zj both have !alternations of a1, a2.

Zj = a1bB!m1#!

s=2(a22a

21b

B!ms )a2 QQ a1bija2

Yj = (a21bBa22)

! * Fk


! Buss: bit.ly/Sxsr5I

! Soltys: bit.ly/UyD7lh

! arXiv: arxiv.org/abs/1211.7161

! Stack Exchange Post: bit.ly/TzD7hH

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Unshuffling a square is NP-hard

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