# Hey Sci.Math, Musatov here. I know I've posted a lot of weird stuff trying to figure things out, but I really think I am onto something here and need some bright minds to have a look at this, computer people, too. Thank you for your time.

A Serious Moment marty.musatov at gmail.com
Fri Apr 2 12:14:25 CEST 2010

```SPARSE COMPLETE SETS FOR NP:
SOLUTION OF A CONJECTURE
BY MARTIN MICHAEL MUSATOV *

for llP:

Sparse Comp1ete Sets
Solution of a Conjecture

In this paper we show if NP has a sparse complete
set under many-one reductions, then ? NP.

The result is extended to show NP is sparse reducible, then P =
ip.

The main technicues:technical cues and techniques of this paper
generalize the :;P 'recognizer' for the compliment of a sparse
complete set with census function to the case where the census
function is not
1'? own (c.f. [1? ii]) than a many-one reduction of tI: gives us the
language to the sparse set permits a polynomial time bounded
tree search as in [ti, [F], or [?:P].

Even without actual knowledge of the census, the algorithm utilizes
the properties of the true census to decide membership in SAT in
polynomial time.

Sparse Complete Sets for `LP:
Solution of a Conjecture
by Martin Michael Musatov

1. Computer Science

L. ? nan and J. 1? at?:i is [tH] under the assumption P? i?.

P all llP-complete sets are iso-morphic; i.e. there are polynomial
time, bijective reductions with polynomial time inverse reductions
between any two NP-cot-:complete sets.

A consequence of this conjecture is all NP-complete sets have
equivalent density; in particular, no sparse set could be i-complete
unless PP.

13e ? an EB] give a partial solution to the problem: by showing if a
subset of an SL i? language is NP-co:complete (a for ti or i, sparse),
then P = 1;P.

This result is strengthened by Fortune [F] showing if co-NP has a
sparse complete set, then P = 1p?

It is necessary to assume the satisfiable formulas reduce to a sparse
set since the proof uses the conjunctive self-reducibility of non-
satisfiable for:formulas and the sparse set to realize a polynomial
time algorithm.

N and P at e[?2] show similar results.

ll a rttq a n is and f[E14] extends the results of Fortune and NP at e
by showing if l IP has a sparse complete set with an easily computable
census function, then NP = co-I?P; P = tT.

P follows by Fortune 1s the or e?.

The question of how easy it is to compute census functions for l?-
complete sets is left open.

In light of Fortune's observation about co-I r 1 the original
conjecture by ten? a n and ll at i?a n is on reducing `?` to a sparse
set series temptingly close, however the tree search methods of [B],
, [i,:p] utilize the conjunctive self-reducibility of the co-l;P-
cot?NP-complete problem SAT0.

In this paper we settle the conjecture by showing if an L'P-complete
set is ? any-one reducible to a sparse set, then P = i?.

Thus determining the existence of a sparse co-??:complete set for l;p
is equivalent to solving the P = NP? problem.

We also show the census function of a sparse NP-co?i:complete set is
cota:computable in P.

Section 2 contains definitions an outline of the tree search time
method for showing sparse sets for co-NP i? implies P = ??1.

Section 3 contains the ? a in results; it assumes n familiarity with
the tree search methods.

?. Preliminaries

We will consider languages over an alphabet ? with two or or e sy?-
symbols.

We assume fa?:familiarity with NP--Hcor?:complete sets (cf. Ec], [K]
or
EAHU)).

All the reductions in this paper will be polynomial time many-one
reductions.

Definition: A subset 5 of is sparse if there is a polynomial so the
number of strings in 5 of size at most n is at most p.

We restate the following theorem (cf. [r] or [iP]) and s'etch: sketch
the proof.

Theorem 1.1. If SATc is reducible to a sparse set, then p = NP.

Proof.

Let f:SAT --> 5 be a reduction to & sparse set, 5, and let F be a
formula whose satisfiability is to be decided.

We search a binary tree formed from self-reductions of F as follows: F
is at the root; a for??:formula C with variables X1, ... , X occurring
in the tree will n have sons G0 a n? G1 where `: is replaced by false
and true, respectively, and trivial si?:simplifications are performed
( e.g. true or = true).

If the for i:?formula F has n variables, then the tree will have 2n?1
nodes. ? i e perform a depth-first search of the tree utilizing the
sparse set to prune the search.

At each node Ft encounters we compute
c
a label f(F').

We infer certain labels correspond to SAT by the following:

When a node with formula false is found, its label is assigned 1.

t'false."

ii. t

Then two sons of a node have ?:labels assigned V?: false, ?1 then the
label of the node is also assigned t'false."

(This is tl-ie:time conjunctive self-reducibility of non-?:satisfiable
for?:formulas.)

We prune the search by stopping if a leaf has for n?:l a true in which
case F is satisfiable by the as:s i2 n?:assistant on the p at?: part
to the leaf; and by not searching below a node whose label has already
been assigned ?false."

?1e follow? j in g:following le r2a establishes poly-not:polynomial
running time of the algorithn.

Lemna 1.2.

Let F be a for?:formula with n variables.

Let p(.) be bound

Let the density of 5 and let q(.) be a poly-no?:polynomial bounding
the increases in size under the reduction f.

Then the algorithm= above visits at u most interior nodes.

o + n + n * p (q(if?))

Proof.

(Er, hT).

Observe if a label is expanded ?or e U?:a n

Once, then t'j e expansions are all on the sa?:safe path from? the
root
since path len?:lengths are at L?:least n+l (with leaf), at ?:most n *
p(q(iF1)) interior nodes with label ?false? visited.

A satisfying assign-ent:assignment visits at ti os t another n nodes.
QED

?ote:note the algoritbim:algorithm does not require a sparse set of
labels for satisfiable fon:iulas:formulas.

The sparse set of labels reduces the number of unsatisfiable foi
ulas:formulas to be searcheci:searched.

?. Solution of tiie:time Con??c?ur?.:concur

Initially, we establish the result for a sparse LP-cornplete:complete
set.

The proof will be modified for the hypothesis that 1?P is sparse
reducible.

The outline of the proof below is as follo?.s:follows:

We first give an flP recognizer for a set si?'ilar:similar to the co?
pler:?ent:compiler component of the sparse set 5.

1.?ny-one:many-one reductions of this set to the sparse set are used
to prove the existence of a sparse set of labels for SATC; however,
tile conputation: time computation of this set of labels requires kno?
iing:knowing tie:time census of 5.

Finally, the depth--H first search is tiodified:modified to detei?
nine:determine satisfiability of a fon?ula:formula.

(without actually !:now in g:knowing the census value will generate
the sparse set of'?abels:labels for sATC).

*

For the following discussion let 5 c (0,1) be a sparse
coriplete:complete set for I?P under iiiany-one:many-one reductions.

Let l,I? be a non-detenninistic:non-deterministic

polynot:'ial:polynomial-tiiae:time recognizer of 5 and let
c(n) IS fl (A+?)ni ? p(n)
where c(.) is the true census function of 5, and p(.) is a polyno?
ial:polynomial
bound the size of the census.

We begin by coljstructing:constructing a non-deteit?
inistic:deterministic polynotiial:polynomial tiL?:time Turi%:Turing

?achine:machine to recognize a flp5e?do?co?ple?entfl:flip 5e do
compliment f1 of 5.

The inputs include a padding, #n, and a potential census, k.

Define the non-dete??inistic:non-deterministic
recognizer `A by the following procedure:

?(??,8,k)

Check			I 5 I			<- n;			o the n is e:otherwise reject.
Check			k <			p(n);			o the n? is e:otherwise reject.
Guess ?i' 5k ??
i.			for all i,			I 5 I <--H n.
ii. for all i and j,
iii. for all i, checit:check it 5. is accepted by II?
iv. check for all i s?s.

Leriraa:Lemma 3.1.

Let I si < n and k --H< p(n).

Then on input (#n,s1k)
the inachine:in a machine n' will:
a. accept if k <
b. reject if k > c(n); and
c. if k = c(n), then II accepts if and only if I4? rejects 5.

Proof of Lei??.:Lemma.

We show part c.

If 14 accepts, it will have
enun?erated:enumerated the elerients:elements of 5 up to size n,
verified they belong to 5, and shown 5 is distinct.

Since k is the true census, 11 accepts
if and only if & is not in 5.
QED

Intuitively1: Intuitively 1 for k = c(n), :?` is a recognizer of 5
cor::?lement.:correct or corresponding elements

1?oreover1:1 moreover ii accepts its 1 an?uage: 1n usage in non-
detenainistic polynot?ial:polynmial-ti L ie:time

(the input #n is ? padding to ensure this).

We require labellin'?:labelling functions for pruning tree searches.

The following discussion sho?,s:shows how to construct such functions
from the sparse set 5 and many-one reductions of L(I):

Since N is an UP ??chine:UP Machine and 5 is NP-co?plete:NP-complete,
there is a P-time r:iany-one:many-one reduction:

g:L(II) --H> 5

so for some monotonic polynorial:polynomial q(.), inputs two !! of
size n are
reduced to strings of size at ?ost:most q(n) (cf. [c] and [:3). Si?
ilarly:Yes, similarly,

for the i'?-coniplete:i-complete problem SAT, there is a P-ti?e:P-time
r?any-one:many-one reduction

f:SAT -? 5 (f:SAT minus question five)

and a monotonic polynomial r(.) bounding ti ie:time increase in size.

Let F of size m be a f on n ul a:formula to be decided.

T?ien:Time ?ny:Many for?nula:formulas F'

occurring in the tree of all self reductions will have size <

Regarding I t:it as a possible value for c(n), we define:

Ln,k (F1) =

which will be the labelling function.

Lemma 3.2.

Let F be a f o i?ul a:formula of size r a.

Let n rC?):near C?; i.e. n

is a polynotial:polynomial upper bound on the size of f(F') where IF 1
I ? iii.(If 1 I question 3i)

Finally, Let k = c(n), the true census. Then the function:

for forulas:formulas F' of size at most n satisfies:

i. F' is not satisfiable if and only if L(F') is in 5;
ii. The unsatisfiable for?uljs:formulas of size at i most:i-most in/
are mapped by L to at most:

p(q(2n+c'log(n))) ? p(q(3n))

distinct strings of 5 where c' is a constant dependin0':depending 0
o?'ily:o question il y on p(.)

Proof: Part i. is imiuediate:immediate froi?:from question

LeLTh?:Lemma 2.1.

For p?rt ?i..:For part question i

observe 2 n + CI i o g(n) ? 3 n: 2n + CI og(n) question 3n is a bound
on the size of (#nf(P'), k).

Applying p 0 q gives the census of strings in 5 the triple could
nap:map onto.
QED

We now know a suitable labelling function e'?iSt5:exists for k but we
do not know c(n)! Tiie:Time algorithir:algorithm in the following
theorem shows how we can try (Ln,k:Link) for all (k ? pCn:k question
pCn).

Theoren:Theorem 3.3.

If UP has a sparse complete set, then P UP.

Proof: We give a deterrainistic:deterministic procedure to recognize
SAT.

Let F be a for ra ul a:formula of size m.

Apply the follouin0-:following zero minus algorithrn:algorithm:

begin
For k 0 to p(r(n)) do
?ecute:question execute the depth first search algoriti??:allgorithm
question with

at each node F' encountered in the pruned search tree.

If a satisfying assignment is found,
then halt; F is satisfiable.

If a tree search visits rr?re: are two questions regarding than:then

n + n * pCq(3 rCr))) internal nodes,
then halt the search for this k.
end;

F is not satisfiable;
end

The algorith?:algorithm clearly runs in polyno?ial:polynomial-tir?
e:time

since the loop is e':ecuted:executed

at most p(r(n)) tines:times and each iteration of the loop visits at
most a polynonial (in i?):(in i question) (nur?er):number of nodes.

The correctness of the algorithn:algorithm is established in the
following:

lei?ia:let i question i a

Lemr?:Lemma 3.4.

If F is satisfiable, then for k c(r(n)) the search

will find a satisfying assigni':?nt:assignment.

Proof: By Lenina:Lemma 2.2.

This k gives a lab?lling:labelling function maps

the unsatisfiable fofl??1as:f of one two question 1 as f of size at no
s t:most ta: (at or to) a polynowially:polynomially bounded set.

Fortune shows that thL:the depth first search (?:ill):(question:I
will)

find a satisfying
assignment visiting at most
internal nodes.

n + n * p(q(3n))
QED

It is interesting to note here we have not cor?uted:corrupted or
computed the census: a satisfying assignrAent:assignment could be
found ?-ith:question minus it h any nu?ber:number of k's;

similarly* if no satisfying assignrent:assignment (CA'?5t5*):(See A
has question five t five times) many of the trees could be searched
but the tree with k = c(r(j?)) is not distinguished.

The method of conducting many tree searches is parallelled in the
un i for a:uniform algorith'n:algorithm technique by Karp and Lipton
[KL].

They show if NP could be accepted in (P ?jith):(P question j it h)
log(space) advice, then P = iP.

The census function `night:might be co?pared:compared to a (log( )-):
(Log (inclusive space) minus) advisor to the polynoraial (in for-
? at i on):(in form at i on) in the set 5.

It is not necessary to (assur:?):(assume you are without question), (a
n flP):(an n-flip) recognizer for the sparse set:

just at 5 is NP-hard.

Le?'na:Lemma 3.5.

If 5 is sparse and (1?P-hard*):(one question p minus hard times) then
there is a set s# sparse,

(I;P-co?plete*):(I semi p minus see or question p let e)

has a P-ti:e:P-time reduction: SAT --> S#
length increasing.

Proof.

Let f: SAT --> 5 be a P-time reduction and let # be a new
-:minus symbol.

Define f#: SAT --> S# by
=:equals

v-here p = max( 0, If(F)I - IFI ). Clearly S? is sparse.

The mapping f# reduces SAT to S#. `&ia:2?ership:membership of 5 in S#
is verified by guessing a satisfiable fon-?ula:formula raps:maps to 5
and verifyin?:verifying question satisfiability.
QED

Corollary 3.6.

If NP is sparse reducible, then P = NP

Lastly we renark:remark the census, c(n), of a sparse NP-co?plete:NP
minus complete set is coj-:?utable:computable in polynot?
ial:polynomial ti?'.e:time.

Indeed. assw?ing:assuming P = I?,

the census

of any sparse set in :?p:question p can be corputed:computed by stand?
rd:standard techniques.

If 5 is sparse and (t;P-cot?plete*):(t; P-Complete) then P = NI' by

Theorei?:Theorem 3.3 so the census of 5

is cot:putablc:computable (iI'):(two I have) polynoi.ial:polynomial
tiI-Ae:time. WL':We have proved:

Corollary 3.7.

If ip has a sparse complete set S1 then the census
of ?:question is computable in P.

?:question period. Discussion

Although the isonorphism:isomorphism results [t"tl]:[t quote t l] are
the direct ancestry of the work discussed here, the concept of
sparseness has another (Lotiva tion):motivation.

Can a "sparse a?ount:amount of (informatior?):(question information)
be used to solve IT problems in

polyno?ial:(polynomial) tir??:(time?)(time to question)

The approach here (assur:?es):(assures and assumes) the information is
given as a (in any-one):(in any or many-one) reduction to a sparse
set.

For Turing reductions, the infon?ation:information is given as a
sparse oracle.

A. ii'eyer: (two i have eye are) a has sho?rn:shown a sparse oracle
for rP:NP is equivalent to the existence of polynomial size circuits
to solve i;P:NP (cf. [13113).

The recent work by `?arp:(have question) Karp, Lipton and Si-?ser:
(Sipser) [KL] has shown if ip has polynomial size circuits, then the
polynomial time hierarchy collapses

p
to

Their result has a weaker hypothesis than Theorem 3.3.

It is an important open problem to determine if polynomial size
circuits for i?:i question implies P = NP.

Ackno?lec'?ement.:Acknowledgement Period

I ar?:am greatly indebted to Juris U. artmanis: j you are is you
period art man is

and

Vivian Sewelson: Viv an sew el son

for

numerous discussions lent insight into the methods developed in this
paper.

The u n i for?:(you and I for, or uniform) al0orithm :(algorithm)
techniques of [KL] suggested the methed:(me the method) in Theorei_:
(the or e i underscore or) Theorem 3.3.

I am grateful to Richard '1:arp:one' Karp and Richard Lipton

?,ho:(questionably) circulate an early version

(AHU] Aho, A., ilopcroft, J., and UlThian, J., The Design and Analysis
of
(?1):(one question) Berman, L. and hartmanis, J., 11On(one hundred and
ten n) Isomorphists and Density of NP
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(April 1980).
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