R

u

t

c

o

r

Research

R

e

p

o

r

t

RUTCOR

ffl

Rutgers

Center

for

Operations

Research

ffl

Rutgers

University

ffl

P.O.

Box

ffl

New

Brunswick

New

Jersey

ffl

Telephone

Telefax

Email

rrr@rutcor.rutgers.edu

Complexity

of

Identification

and

Dualization

of

Positive

Boolean

Functions

Jan

C.

Biocha

Toshihide

Ibarakib

RRR

September

aDepartment

of

Computer

Science

AI-section

Erasmus

University

Rotterdam

P.O.

Box

DR

Rotterdam

The

Netherlands

bioch@cs.few.eur.nl

bDepartment

of

Applied

Mathematics

and

Physics

Faculty

of

Engineering

Kyoto

University

Kyoto

Japan

ibaraki@kuamp.kyoto-u.ac.jp

Rutcor

Research

Report

RRR

September

Complexity

of

Identification

and

Dualization

of

Positive

Boolean

Functions

Jan

C.

Bioch

Toshihide

Ibaraki

Abstract.We

consider

in

this

paper

the

problem

of

identifying

minT

-LRB-

f

-RRB-

and

maxF

-LRB-

f

-RRB-

of

a

positive

Boolean

function

f

by

using

membership

queries

only

where

minT

-LRB-

f

-RRB-

-LRB-

maxF

-LRB-

f

-RRB-

-RRB-

denotes

the

set

of

minimal

true

vectors

-LRB-

maximal

false

vectors

-RRB-

of

f

It

is

shown

that

the

existence

of

an

incrementally

polynomial

algorithm

for

this

problem

is

equivalent

to

the

existence

of

the

following

algorithms

where

f

and

g

are

positive

Boolean

functions

-LRB-

i

-RRB-

An

incrementally

polynomial

algorithm

to

dualize

f

-LRB-

ii

-RRB-

An

incrementally

polynomial

algorithm

to

self-dualize

f

-LRB-

iii

-RRB-

A

polynomial

algorithm

to

decide

if

f

and

g

are

mutually

dual

-LRB-

iv

-RRB-

A

polynomial

algorithm

to

decide

if

f

is

self-dual

-LRB-

v

-RRB-

A

polynomial

algorithm

to

decide

if

f

is

saturated

The

equivalence

of

these

problems

with

other

problems

encountered

in

various

fields

such

as

hypergraph

theory

and

theory

of

coteries

-LRB-

used

in

distributed

systems

-RRB-

is

also

discussed

The

complexity

of

all

these

problems

is

still

unknown

Some

other

related

topics

are

also

included

Acknowledgements

This

paper

was

completed

while

the

first

author

visited

Kyoto

University

in

April

the

support

for

which

is

greatly

appreciated

The

authors

were

benefited

from

the

discussion

with

their

colleagues

E.

Boros

Y.

Crama

P.

L.

Hammer

T.

Kameda

K.

Makino

H.

Nagamochi

and

others

whose

cooperation

is

greatly

appreciated

This

research

was

partially

supported

by

the

Scientific

Grant-in-Aid

from

Ministry

of

Education

Science

and

Culture

of

Japan

The

second

author

gratefully

acknowledges

the

partial

support

given

to

Rutcor

by

the

Office

of

Naval

Research

-LRB-

grants

and

-RRB-

and

by

the

Air

Force

Office

of

Scientific

Research

-LRB-

grant

-RRB-

RRR

Page

Introduction

In

this

paper

we

are

interested

in

the

problem

of

identifying

a

given

Boolean

function

-LRB-

or

a

function

in

short

-RRB-

f

by

asking

membership

queries

to

an

oracle

whether

f

-LRB-

u

-RRB-

or

holds

for

some

selected

vectors

u

In

the

terminology

of

computational

learning

theory

-LSB-

-RSB-

this

is

nothing

but

the

exact

learning

of

a

Boolean

theory

f

by

membership

queries

only

It

is

also

a

process

of

forming

a

theory

that

explains

a

certain

phenomenon

by

collecting

positive

and

negative

data

-LRB-

in

the

sense

of

causing

and

not

causing

that

phenomenon

-RRB-

-LSB-

-RSB-

In

this

case

each

membership

query

is

interpreted

as

conducting

an

experiment

with

a

given

specification

to

observe

its

outcome

We

emphasize

here

that

by

identification

we

mean

to

acquire

explicit

knowledge

of

both

T

-LRB-

f

-RRB-

-LRB-

set

of

true

vectors

of

f

-RRB-

and

F

-LRB-

f

-RRB-

-LRB-

set

of

false

vectors

of

f

-RRB-

Although

this

may

appear

to

be

redundant

since

F

-LRB-

f

-RRB-

can

be

computed

from

T

-LRB-

f

-RRB-

in

principle

we

consider

that

obtaining

both

T

-LRB-

f

-RRB-

and

F

-LRB-

f

-RRB-

is

essential

During

the

identification

process

which

we

shall

consider

in

this

paper

only

partial

knowledge

of

T

-LRB-

f

-RRB-

and

F

-LRB-

f

-RRB-

is

available

In

order

to

ensure

that

this

partial

knowledge

actually

has

become

the

full

knowledge

it

is

necessary

to

know

both

T

-LRB-

f

-RRB-

and

F

-LRB-

f

-RRB-

In

this

sense

maintaining

both

T

-LRB-

f

-RRB-

and

F

-LRB-

f

-RRB-

is

natural

We

also

note

in

this

regard

that

the

complexity

of

computing

F

-LRB-

f

-RRB-

from

T

-LRB-

f

-RRB-

is

in

general

nontrivial

As

an

example

consider

that

a

function

fG

is

defined

for

a

graph

G

-LRB-

V

E

-RRB-

by

fG

-LRB-

x

-RRB-

if

x

x

-LRB-

e

-RRB-

for

some

e

E

and

otherwise

where

x

-LRB-

e

-RRB-

for

e

-LRB-

vi

vj

-RRB-

E

is

the

vector

with

xi

xj

and

xk

for

k

i

j

In

other

words

fG

is

a

representation

of

graph

G

Now

it

is

not

difficult

to

see

that

x

F

-LRB-

fG

-RRB-

if

and

only

if

the

set

of

vertices

fvi

V

j

xi

is

an

independent

set

of

G.

Computing

all

independent

sets

of

a

graph

G

is

not

a

trivial

task

though

it

is

known

that

it

can

be

done

in

incrementally

polynomial

time

-LRB-

e.g.

-LSB-

-RSB-

-RRB-

If

we

do

not

assume

any

property

for

the

function

f

to

be

identified

identification

is

not

possible

unless

all

values

of

f

-LRB-

v

-RRB-

for

v

are

explicitly

tested

To

avoid

this

situation

we

assume

in

this

paper

that

f

is

known

to

be

positive

-LRB-

i.e.

monotone

-RRB-

since

this

is

the

case

in

many

applications

-LRB-

The

definition

of

a

positive

functions

and

related

concepts

will

be

given

in

subsequent

sections

-RRB-

For

example

if

f

-LRB-

x

-RRB-

describes

the

result

of

diagnosing

a

disease

from

the

given

data

-LRB-

i.e.

symptoms

-RRB-

x

e.g.

denotes

whether

temperature

is

high

-LRB-

-RRB-

or

not

-LRB-

-RRB-

and

denotes

whether

blood

presure

is

high

-LRB-

-RRB-

or

not

-LRB-

-RRB-

etc.

it

would

be

natural

to

assume

that

we

somehow

know

the

direction

of

each

variable

that

tends

to

cause

the

disease

to

appear

In

this

case

by

adjusting

the

polarities

of

variables

if

necessary

the

function

f

-LRB-

x

-RRB-

can

be

assumed

to

be

positive

without

loss

of

generality

If

f

is

a

positive

function

T

-LRB-

f

-RRB-

and

F

-LRB-

f

-RRB-

can

be

compactly

represented

by

minT

-LRB-

f

-RRB-

-LRB-

set

of

minimal

true

vectors

-RRB-

and

maxF

-LRB-

f

-RRB-

-LRB-

set

of

maximal

false

vectors

-RRB-

Therefore

we

want

to

compute

minT

-LRB-

f

-RRB-

and

maxF

-LRB-

f

-RRB-

instead

of

T

-LRB-

f

-RRB-

and

F

-LRB-

f

-RRB-

themselves

in

our

identification

process

The

complexity

of

this

type

of

enumeration

algorithm

is

usually

measured

in

its

lengths

of

input

and

output

A

particularly

interesting

class

from

this

view

point

is

that

of

Page

RRR

incrementally

polynomial

algorithms

-LSB-

-RSB-

An

algorithm

to

enumerate

items

ap

is

called

incrementally

polynomial

-LRB-

i

-RRB-

if

it

iterates

the

following

procedure

for

i

p

output

the

i-th

item

ai

from

the

knowledge

of

its

input

and

items

ai

generated

by

then

and

-LRB-

ii

-RRB-

if

the

time

required

for

the

i-th

iteration

is

polynomial

in

the

input

length

and

the

sizes

of

ai

If

an

algorithm

is

incrementally

polynomial

it

also

satisfies

the

criterion

of

P-enumerability

-LSB-

-RSB-

-LRB-

i.e.

polynomial

time

in

the

lengths

of

input

and

output

-RRB-

Our

main

question

is

to

know

whether

the

identification

of

a

positive

f

has

an

incrementally

polynomial

algorithm

or

not

Unfortunately

we

could

not

resolve

this

question

in

this

paper

However

we

could

relate

the

existence

of

such

an

algorithm

with

many

other

interesting

problems

discussed

not

only

in

Boolean

theory

but

also

in

different

fields

such

as

hypergraph

theory

-LSB-

-RSB-

theory

of

coteries

-LRB-

used

in

distributed

systems

-RRB-

-LSB-

-RSB-

database

theory

-LSB-

-RSB-

and

artificial

intelligence

-LSB-

-RSB-

In

particular

we

show

that

an

incrementally

polynomial

identification

algorithm

exists

if

and

only

if

dualization

of

a

positive

function

has

an

incrementally

polynomial

algorithm

Also

such

algorithms

exist

if

and

only

if

problems

such

as

deciding

if

two

positive

functions

are

mutually

dual

deciding

if

a

positive

function

is

self-dual

deciding

if

a

simple

hypergraph

is

saturated

and

deciding

if

a

coterie

is

nondominated

can

be

solved

in

polynomial

time

All

these

problems

are

polynomially

equivalent

but

their

complexity

is

still

left

open

It

is

again

noted

here

that

our

problem

is

different

from

that

of

identifying

only

minT

-LRB-

f

-RRB-

or

only

maxF

-LRB-

f

-RRB-

since

it

is

known

-LSB-

-RSB-

that

no

incrementally

polynomial

algorithm

exists

for

the

latter

problem

For

several

special

classes

of

positive

functions

it

is

known

that

some

problems

in

the

above

list

of

equivalent

problems

can

be

solved

in

polynomial

time

Such

classes

include

-LRB-

i

-RRB-

the

class

of

positive

functions

-LSB-

-RSB-

which

include

as

a

special

case

the

class

of

positive

threshold

functions

-LSB-

-RSB-

-LRB-

ii

-RRB-

the

class

of

positive

k-DNF

-LSB-

-RSB-

-LRB-

though

the

algorithm

is

explained

in

hypergraph

terminology

-RRB-

where

a

positive

function

f

is

k-DNF

if

each

minimal

true

vector

has

at

most

k

elements

of

and

-LRB-

iii

-RRB-

other

classes

-LRB-

e.g.

-LSB-

-RSB-

-RRB-

By

the

above

results

therefore

all

the

problems

in

the

list

can

be

solved

in

polynomial

time

for

these

classes

in

particular

positive

functions

in

such

classes

can

be

identified

by

incrementally

polynomial

algorithms

Another

special

class

called

read-once

functions

also

deserve

mentioning

here

since

it

is

shown

in

-LSB-

-RSB-

that

this

class

has

an

algorithm

to

identify

its

read-once

formula

-LRB-

i.e.

an

expression

of

T

-LRB-

f

-RRB-

-RRB-

in

polynomial

time

in

its

formula

length

-LRB-

note

however

that

the

framework

for

identification

is

different

from

that

of

this

paper

-RRB-

Finally

although

not

discussed

in

this

paper

other

classes

different

from

that

of

positive

functions

such

as

general

threshold

functions

Horn

functions

and

general

k-DNF

functions

may

be

fruitful

to

study

Some

other

results

related

to

our

main

question

are

also

included

in

this

paper

It

turns

out

that

some

relevant

problems

are

solvable

in

polynomial

time

while

others

are

NP-complete

This

may

indicate

that

our

problem

is

located

somewhere

close

to

the

border

between

polynomiality

and

NP-hardness

RRR

Page

Identification

of

Positive

Boolean

Functions

Definitions

and

an

Algorithm

A

Boolean

function

or

a

function

in

short

is

a

mapping

f

where

v

is

called

a

Boolean

vector

-LRB-

a

vector

in

short

-RRB-

If

f

-LRB-

v

-RRB-

-LRB-

resp

-RRB-

then

v

is

called

a

true

-LRB-

resp

false

-RRB-

vector

of

f

The

set

of

all

true

vectors

-LRB-

false

vectors

-RRB-

is

denoted

by

T

-LRB-

f

-RRB-

-LRB-

F

-LRB-

f

-RRB-

-RRB-

Two

special

functions

with

T

-LRB-

f

-RRB-

and

F

-LRB-

f

-RRB-

are

respectively

denoted

by

f

-LRB-

x

-RRB-

and

f

-LRB-

x

-RRB-

A

function

f

is

positive

if

v

w

always

implies

f

-LRB-

v

-RRB-

f

-LRB-

w

-RRB-

A

positive

function

is

also

called

monotone

-LRB-

However

we

do

not

use

the

latter

name

since

some

people

use

it

to

imply

a

function

that

is

either

monotone

nonincreasing

or

monotone

nondecreasing

-RRB-

A

true

vector

v

of

f

is

minimal

if

there

is

no

other

true

vector

w

such

that

w

v

and

let

minT

-LRB-

f

-RRB-

denote

the

set

of

all

minimal

true

vectors

of

f

A

maximal

false

vector

is

symmetrically

defined

and

maxF

-LRB-

f

-RRB-

denotes

the

set

of

all

maximal

false

vectors

of

f

Two

vectors

u

and

v

are

incomparable

if

neither

u

v

nor

v

u

holds

A

set

of

vectors

U

is

called

incomparable

if

every

pair

of

vectors

u

v

U

is

incomparable

It

is

clear

that

the

sets

minT

-LRB-

f

-RRB-

and

maxF

-LRB-

f

-RRB-

are

both

incomparable

If

f

is

positive

it

is

known

-LSB-

-RSB-

that

f

has

the

unique

disjunctive

normal

form

-LRB-

DNF

-RRB-

consisting

of

all

prime

implicants

There

is

one-to-one

correspondence

between

prime

implicants

and

minimal

true

vectors

For

example

a

positive

function

f

where

stands

for

and

so

on

has

prime

implicants

which

correspond

to

minimal

true

vectors

-LRB-

-RRB-

-LRB-

-RRB-

-LRB-

-RRB-

respectively

In

other

words

the

input

length

to

describe

a

positive

function

f

is

O

-LRB-

njminT

-LRB-

f

-RRB-

j

-RRB-

if

it

is

represented

in

this

manner

Furthermore

the

sets

minT

-LRB-

f

-RRB-

and

maxF

-LRB-

f

-RRB-

conversely

define

T

-LRB-

f

-RRB-

and

F

-LRB-

f

-RRB-

by

T

-LRB-

f

-RRB-

fv

j

v

w

for

some

w

minT

-LRB-

f

-RRB-

g

F

-LRB-

f

-RRB-

fv

j

v

w

for

some

w

maxF

-LRB-

f

-RRB-

g

Since

jminT

-LRB-

f

-RRB-

j

jT

-LRB-

f

-RRB-

j

and

jmaxF

-LRB-

f

-RRB-

j

jF

-LRB-

f

-RRB-

j

we

ask

in

our

identification

algorithm

to

compute

minT

-LRB-

f

-RRB-

and

maxF

-LRB-

f

-RRB-

instead

of

T

-LRB-

f

-RRB-

and

F

-LRB-

f

-RRB-

Thus

our

problem

is

stated

as

follows

Problem

IDENTIFICATION

Input

an

oracle

for

a

positive

function

f

Output

minT

-LRB-

f

-RRB-

and

maxF

-LRB-

f

-RRB-

Now

let

MT

and

MF

respectively

denote

the

partial

knowledge

of

minT

-LRB-

f

-RRB-

and

maxF

-LRB-

f

-RRB-

currently

at

hand

i.e.

MT

minT

-LRB-

f

-RRB-

and

MF

maxF

-LRB-

f

-RRB-

-LRB-

-RRB-

Define

T

-LRB-

MT

-RRB-

fv

j

v

w

for

some

w

MTg

F

-LRB-

MF

-RRB-

fv

j

v

w

for

some

w

MFg

-LRB-

-RRB-

Page

RRR

By

assumption

-LRB-

-RRB-

T

-LRB-

MT

-RRB-

T

-LRB-

f

-RRB-

and

F

-LRB-

MF

-RRB-

F

-LRB-

f

-RRB-

and

hence

T

-LRB-

MT

-RRB-

F

-LRB-

MF

-RRB-

-LRB-

-RRB-

holds

A

vector

u

is

called

an

unknown

vector

if

u

n

-LRB-

T

-LRB-

MT

-RRB-

-LSB-

F

-LRB-

MF

-RRB-

-RRB-

-LRB-

-RRB-

since

it

is

not

known

at

the

current

stage

whether

u

is

a

true

vector

or

a

false

vector

of

f

There

is

no

unknown

vector

if

and

only

if

T

-LRB-

MT

-RRB-

-LSB-

F

-LRB-

MF

-RRB-

holds

i.e.

MT

minT

-LRB-

f

-RRB-

and

MF

maxF

-LRB-

f

-RRB-

hold

for

some

positive

function

f

The

general

procedure

of

identifying

a

positive

function

f

can

now

be

described

as

follows

Algorithm

IDENTIFY

Input

an

oracle

for

a

positive

function

f

Output

minT

-LRB-

f

-RRB-

and

maxF

-LRB-

f

-RRB-

Start

with

appropriate

sets

MT

-LRB-

minT

-LRB-

f

-RRB-

-RRB-

and

MF

-LRB-

maxF

-LRB-

f

-RRB-

-RRB-

-LRB-

e.g.

MT

MF

;-RRB-

Test

if

T

-LRB-

MT

-RRB-

-LSB-

F

-LRB-

MF

-RRB-

holds

If

so

output

MT

and

MF

and

halt

Otherwise

go

to

Find

an

unknown

vector

u

and

ask

an

oracle

if

f

-LRB-

u

-RRB-

or

f

-LRB-

u

-RRB-

If

f

-LRB-

u

-RRB-

then

compute

a

new

minimal

true

vector

y

from

u

and

let

MT

MT

-LSB-

fyg

On

the

other

hand

if

f

-LRB-

u

-RRB-

compute

a

new

maximal

false

vector

y

from

u

and

let

MF

MF

-LSB-

fyg

Return

to

Analysis

of

Algorithm

IDENTIFY

We

are

particularly

interested

in

whether

algorithm

IDENTIFY

can

be

incrementally

polynomial

or

not

For

this

Step

is

crucial

and

we

define

Problem

EQ

Input

Incomparable

sets

MT

MF

-LRB-

-RRB-

such

that

T

-LRB-

MT

-RRB-

F

-LRB-

MF

-RRB-

Question

Does

T

-LRB-

MT

-RRB-

-LSB-

F

-LRB-

MF

-RRB-

-LRB-

i.e.

no

unknown

vector

-RRB-

hold

Unfortunately

the

complexity

of

problem

EQ

is

not

known

yet

but

will

be

related

to

other

well

known

problems

in

Section

We

show

below

however

that

if

problem

EQ

-LRB-

which

is

a

decision

problem

-RRB-

is

solved

in

time

polynomial

in

its

input

length

n

-LRB-

jMT

j

jMF

j

-RRB-

then

finding

an

unknown

vector

in

Step

can

also

be

done

in

polynomial

time

For

this

introduce

the

following

notations

Let

a

-LRB-

ak

-RRB-

for

some

k

n

and

define

MT

-LSB-

a

-RSB-

MinSetf

-LRB-

vk

vn

-RRB-

j

v

MT

and

vi

ai

i

kg

minT

-LRB-

f

-RRB-

-LSB-

a

-RSB-

MinSetf

-LRB-

vk

vn

-RRB-

j

v

minT

-LRB-

f

-RRB-

and

vi

ai

i

kg

MF

-LSB-

a

-RSB-

MaxSetf

-LRB-

vk

vn

-RRB-

j

v

MF

and

vi

ai

i

kg

maxF

-LRB-

f

-RRB-

-LSB-

a

-RSB-

MaxSetf

-LRB-

vk

vn

-RRB-

j

v

maxF

-LRB-

f

-RRB-

and

vi

ai

i

kg

-LRB-

-RRB-

RRR

Page

where

MinSet

-LRB-

A

-RRB-

-LRB-

MaxSet

-LRB-

A

-RRB-

-RRB-

denotes

the

set

of

minimal

-LRB-

maximal

-RRB-

vectors

in

A.

Denote

by

fa

the

function

obtained

from

f

by

fixing

variables

xi

to

ai

for

i

k

Then

the

above

minT

-LRB-

f

-RRB-

-LSB-

a

-RSB-

and

maxF

-LRB-

f

-RRB-

-LSB-

a

-RSB-

represent

minT

-LRB-

fa

-RRB-

and

maxF

-LRB-

fa

-RRB-

respectively

Now

as

noted

previously

MT

minT

-LRB-

f

-RRB-

and

MF

maxF

-LRB-

f

-RRB-

hold

if

and

only

if

T

-LRB-

MT

-RRB-

-LSB-

F

-LRB-

MF

-RRB-

Furthermore

if

MT

minT

-LRB-

f

-RRB-

and

MF

maxF

-LRB-

f

-RRB-

then

MT

-LSB-

a

-RSB-

minT

-LRB-

f

-RRB-

-LSB-

a

-RSB-

and

MF

-LSB-

a

-RSB-

maxF

-LRB-

f

-RRB-

-LSB-

a

-RSB-

for

any

a.

Also

MT

-LSB-

a

-RSB-

minT

-LRB-

f

-RRB-

-LSB-

a

-RSB-

and

MF

-LSB-

a

-RSB-

maxF

-LRB-

f

-RRB-

-LSB-

a

-RSB-

hold

if

and

only

if

T

-LRB-

MT

-LSB-

a

-RSB-

-RRB-

-LSB-

F

-LRB-

MF

-LSB-

a

-RSB-

-RRB-

k

-LRB-

-RRB-

If

the

current

MT

-LSB-

a

-RSB-

and

MF

-LSB-

a

-RSB-

do

not

satisfy

condition

-LRB-

-RRB-

then

it

follows

that

at

least

one

of

-LRB-

a

-RRB-

and

-LRB-

a

-RRB-

does

not

satisfy

condition

-LRB-

-RRB-

for

a

ai

-LRB-

i

-RRB-

since

T

-LRB-

MT

-LSB-

a

-RSB-

-RRB-

T

-LRB-

f

-LRB-

w

-RRB-

j

w

MT

-LSB-

-RSB-

g

-RRB-

-LSB-

T

-LRB-

f

-LRB-

w

-RRB-

j

w

MT

-LSB-

-RSB-

g

-RRB-

F

-LRB-

MF

-LSB-

a

-RSB-

-RRB-

F

-LRB-

f

-LRB-

w

-RRB-

j

w

MF

-LSB-

-RSB-

g

-RRB-

-LSB-

F

-LRB-

f

-LRB-

w

-RRB-

j

w

MF

-LSB-

-RSB-

g

-RRB-

by

definition

Based

on

this

property

the

following

algorithm

finds

an

unknown

vector

u

by

systematically

searching

vectors

a

that

do

not

satisfy

the

above

condition

-LRB-

-RRB-

by

increasing

the

dimension

of

a

at

each

iteration

Algorithm

UNKNOWN

Input

Incomparable

sets

MT

MF

-LRB-

-RRB-

such

that

T

-LRB-

MT

-RRB-

F

-LRB-

MF

-RRB-

and

T

-LRB-

MT

-RRB-

-LSB-

F

-LRB-

MF

-RRB-

where

n

Output

An

unknown

vector

u.

Let

-LRB-

-RRB-

-LRB-

-RRB-

and

k

Go

to

If

k

n

then

at

least

one

of

n

-LRB-

MT

-LSB-

-RSB-

-LSB-

MF

-LSB-

-RSB-

-RRB-

and

n

-LRB-

MT

-LSB-

-RSB-

-LSB-

MF

-LSB-

-RSB-

-RRB-

is

nonempty

If

and

b

let

u

-LRB-

b

-RRB-

and

halt

Otherwise

let

u

-LRB-

b

-RRB-

where

b

and

halt

If

k

n

go

to

Test

if

T

-LRB-

MT

-LSB-

-RSB-

-RRB-

-LSB-

F

-LRB-

MF

-LSB-

-RSB-

-RRB-

k

holds

-LRB-

i.e.

solve

problem

EQ

-RRB-

If

no

let

-LRB-

-RRB-

-LRB-

-RRB-

and

k

k

Return

to

Otherwise

-LRB-

i.e.

yes

-RRB-

let

-LRB-

-RRB-

-LRB-

-RRB-

and

k

k

Return

to

The

running

time

of

UNKNOWN

is

O

-LRB-

nTEQ

-LRB-

n

-LRB-

jMT

j

jMF

j

-RRB-

-RRB-

-RRB-

where

n

-LRB-

jMT

j

jMF

j

-RRB-

is

the

input

length

to

problem

EQ

and

TEQ

is

the

time

to

solve

problem

EQ

if

we

ignore

the

time

for

MinSet

and

MaxSet

operations

of

-LRB-

-RRB-

for

MT

-LSB-

ai

-RSB-

and

MF

-LSB-

ai

-RSB-

in

Steps

and

for

the

time

being

At

this

point

recall

the

two

kinds

of

queries

restricted

equivalence

query

and

equivalence

query

well

studied

in

computational

leaning

theory

-LRB-

e.g.

-LSB-

-RSB-

-RRB-

A

restricted

equivalent

query

asks

whether

the

current

knowledge

of

f

completely

characterizes

f

or

not

while

an

equivalence

query

asks

in

addition

to

that

to

provide

a

counterexample

if

it

does

not

characterize

f

In

our

framework

a

restricted

equivalence

query

is

essentially

equal

to

solving

problem

EQ

since

MT

and

MF

completely

characterize

f

if

and

only

if

T

-LRB-

MT

-RRB-

-LSB-

F

-LRB-

MF

-RRB-

while

an

equivalence

query

is

nothing

but

to

solve

problem

EQ

and

then

to

execute

algorithm

UNKNOWN

since

the

latter

provides

a

counterexample

when

problem

EQ

has

answer

Page

RRR

no

The

above

time

complexity

of

UNKNOWN

says

that

a

restricted

equivalence

query

and

an

equivalence

query

do

not

have

an

essential

difference

-LRB-

modulo

polynomial

time

-RRB-

in

our

problem

setting

Now

we

consider

the

second

half

of

Step

of

IDENTIFY

Asking

an

oracle

if

f

-LRB-

u

-RRB-

or

f

-LRB-

u

-RRB-

is

called

a

membership

query

Computing

a

minimal

true

vector

or

a

maximal

false

vector

y

from

an

unknown

vector

u

can

be

done

as

described

in

such

papers

as

-LSB-

-RSB-

For

example

if

f

-LRB-

u

-RRB-

a

maximal

false

vector

y

-LRB-

u

-RRB-

not

in

F

-LRB-

MF

-RRB-

can

be

computed

as

follows

where

y

-LSB-

yi

-RSB-

denotes

the

vector

y

with

its

i-th

element

yi

fixed

to

Algorithm

MAXIMAL

Input

An

incomparable

set

MF

maxF

-LRB-

f

-RRB-

a

vector

u

F

-LRB-

f

-RRB-

nF

-LRB-

MF

-RRB-

and

an

oracle

for

a

positive

function

f

Output

a

maximal

vector

y

such

that

y

F

-LRB-

f

-RRB-

n

F

-LRB-

MF

-RRB-

y

u.

For

i

n

let

y

y

-LSB-

yi

-RSB-

if

f

-LRB-

y

-LSB-

yi

-RSB-

-RRB-

-LRB-

membership

queries

are

used

here

-RRB-

Output

y.

A

minimal

true

vector

can

be

analogously

computed

and

its

algorithm

is

called

MINIMAL

Both

algorithms

issue

O

-LRB-

n

-RRB-

membership

queries

before

outputting

the

final

y

If

each

membership

query

in

step

is

answered

in

constant

time

by

an

oracle

the

running

time

of

MAXIMAL

or

MINIMAL

is

O

-LRB-

n

-RRB-

The

sum

of

the

above

running

times

in

IDENTIFY

becomes

O

-LRB-

-LRB-

jminT

-LRB-

f

-RRB-

j

jmaxF

-LRB-

f

-RRB-

j

-RRB-

n

TEQ

-LRB-

n

-LRB-

jMT

j

jMF

j

-RRB-

-RRB-

-RRB-

O

-LRB-

mnTEQ

-LRB-

mn

-RRB-

-RRB-

-LRB-

-RRB-

where

m

is

defined

by

m

jminT

-LRB-

f

-RRB-

j

jmaxF

-LRB-

f

-RRB-

j

-LRB-

-RRB-

Before

proceeding

further

recall

that

we

have

postponed

the

discussion

on

the

time

complexity

of

MinSet

and

MaxSet

operations

of

-LRB-

-RRB-

required

in

Steps

and

of

UNKNOWN

For

simplicity

we

explain

how

to

facilitate

this

computation

for

MT

since

MF

can

be

analogously

treated

Prepare

data

PAIR

-LRB-

MT

k

-RRB-

k

n

defined

by

-LRB-

u

v

-RRB-

PAIR

-LRB-

MT

k

-RRB-

if

-LRB-

a

-RRB-

u

v

MT

-LRB-

b

-RRB-

-LRB-

uk

un

-RRB-

-LRB-

vk

vn

-RRB-

and

-LRB-

c

-RRB-

-LRB-

ui

un

-RRB-

and

-LRB-

vi

vn

-RRB-

are

incomparable

for

all

i

k.

PAIR

-LRB-

MT

k

-RRB-

for

all

k

n

can

be

obtained

in

O

-LRB-

njMT

-RRB-

time

Using

this

data

MinSet

operations

to

obtain

MT

-LSB-

-RSB-

and

MT

-LSB-

-RSB-

in

Step

can

be

carried

out

in

O

-LRB-

jPAIR

-LRB-

MT

k

-RRB-

j

-RRB-

time

for

each

k

and

Pk

O

-LRB-

jPAIR

-LRB-

MT

k

-RRB-

j

-RRB-

O

-LRB-

njMT

-RRB-

time

for

all

k.

Although

constructing

PAIR

becomes

necessary

whenever

IDEN

TIFY

calls

UNKNOWN

if

we

note

that

only

one

of

MT

and

MF

is

augmented

by

one

vector

at

each

such

call

the

new

PAIR

can

be

constructed

by

appending

the

pairs

involving

this

new

vector

to

the

old

PAIR

in

O

-LRB-

njMT

j

-RRB-

time

By

this

gimmick

the

overall

time

required

in

IDENTIFY

to

construct

PAIR

tables

for

both

MT

and

MF

is

O

-LRB-

njminT

-LRB-

f

-RRB-

njmaxF

-LRB-

f

-RRB-

-RRB-

O

-LRB-

-RRB-

and

so

is

the

overall

time

needed

for

MinSet

and

MaxSet

operations

This

time

bound

O

-LRB-

-RRB-

is

dominated

by

the

RRR

Page

above

time

bound

-LRB-

-RRB-

if

we

assume

TEQ

-LRB-

mn

-RRB-

O

-LRB-

mn

-RRB-

For

this

reason

we

shall

not

consider

explicitly

the

time

for

MinSet

and

MaxSet

operations

in

the

subsequent

discussion

As

a

summary

of

the

above

discussion

we

have

the

next

theorem

Theorem

Problem

IDENTIFICATION

has

an

incrementally

polynomial

algorithm

if

and

only

if

problem

EQ

can

be

solved

in

polynomial

time

Proof

The

if-part

is

immediate

from

-LRB-

-RRB-

To

prove

the

converse

execute

the

incrementally

polynomial

algorithm

A

for

problem

IDENTIFICATION

-LRB-

note

that

A

may

be

quite

different

from

IDENTIFY

-RRB-

in

which

a

positive

function

f

is

given

by

minT

-LRB-

f

-RRB-

MT

until

either

-LRB-

i

-RRB-

it

halts

or

-LRB-

ii

-RRB-

it

iterates

jMT

j

jMF

j

times

In

case

of

-LRB-

i

-RRB-

there

is

no

unknown

vector

if

and

only

if

the

obtained

maxF

-LRB-

f

-RRB-

satisfies

maxF

-LRB-

f

-RRB-

MF

In

case

of

-LRB-

ii

-RRB-

there

is

an

unknown

vector

since

jmaxF

-LRB-

f

-RRB-

j

jMF

j

holds

In

any

case

problem

EQ

is

solved

Since

a

membership

query

for

u

can

be

answered

in

time

O

-LRB-

njMT

j

-RRB-

by

directly

checking

the

property

f

-LRB-

u

-RRB-

-LRB-

-RRB-

u

v

for

some

v

MT

without

resorting

to

an

oracle

the

time

to

reach

-LRB-

i

-RRB-

or

-LRB-

ii

-RRB-

is

O

-LRB-

-LRB-

jMT

j

jMF

j

-RRB-

p

-LRB-

n

-LRB-

jMT

j

jMF

j

-RRB-

-RRB-

-LRB-

njMT

j

-RRB-

-RRB-

where

p

-LRB-

-RRB-

is

the

time

for

one

iteration

of

algorithm

A

The

third

factor

njMT

j

takes

into

account

the

time

to

answer

membership

queries

-LRB-

which

is

constant

if

an

oracle

is

used

-RRB-

The

time

bound

is

polynomial

in

the

input

length

n

-LRB-

jMT

j

jMF

j

-RRB-

of

problem

EQ

As

mentioned

in

Section

an

incrementally

polynomial

algorithm

for

IDENTIFICA

TION

is

a

polynomial

time

algorithm

in

the

length

of

input

and

output

i.e.

in

O

-LRB-

n

m

-RRB-

However

it

is

not

known

yet

whether

the

existence

of

the

latter

algorithm

means

the

existence

of

the

former

algorithm

or

not

We

note

here

that

the

latter

algorithm

exists

if

and

only

if

there

is

an

algorithm

to

compute

only

minT

-LRB-

f

-RRB-

in

time

polynomial

in

O

-LRB-

n

m

-RRB-

The

only-if-part

is

obvious

The

if-part

can

be

shown

as

follows

The

oracle

for

membership

queries

of

f

can

also

be

used

as

an

oracle

for

f

and

hence

the

algorithm

for

minT

-LRB-

f

-RRB-

can

be

regarded

as

an

algorithm

for

maxF

-LRB-

f

-RRB-

if

applied

to

g

f

-LRB-

with

appropriate

modifications

since

g

is

the

complement

of

a

positive

function

-RRB-

Therefore

if

minT

-LRB-

f

-RRB-

can

be

computed

in

time

polynomial

in

O

-LRB-

n

m

-RRB-

then

so

are

both

minT

-LRB-

f

-RRB-

and

maxF

-LRB-

f

-RRB-

Dual

Functions

and

Related

Problems

Definitions

The

dual

of

a

Boolean

function

f

denoted

fd

is

defined

by

fd

-LRB-

x

-RRB-

f

-LRB-

x

-RRB-

where

f

and

x

denote

the

complement

of

f

and

x

respectively

As

is

well

known

the

DNF

expression

of

fd

is

obtained

from

that

of

f

by

exchanging

-LRB-

or

-RRB-

and

-LRB-

and

-RRB-

as

well

as

constants

and

and

then

expanding

the

resulting

formula

-LRB-

The

symbol

of

and

is

usually

omitted

unless

confusion

occurs

-RRB-

For

example

the

dual

of

f

is

fd

-LRB-

-RRB-

-LRB-

-RRB-

-LRB-

-RRB-

As

a

special

case

functions

f

and

f

are

mutually

dual

Denote

f

g

if

these

functions

satisfy

f

-LRB-

v

-RRB-

g

-LRB-

v

-RRB-

for

all

v

It

is

immediate

from

definition

that

f

gd

-LRB-

-RRB-

fd

g

and

f

g

-LRB-

-RRB-

fd

gd

If

f

is

positive

another

characterization

of

fd

is

given

as

follows

y

minT

-LRB-

fd

-RRB-

if

and

only

if

y

is

a

minimal

vector

with

property

Page

RRR

y

v

-LRB-

-RRB-

for

all

v

minT

-LRB-

f

-RRB-

where

y

v

-LRB-

yn

vn

-RRB-

and

is

and

operator

For

this

reason

the

vectors

in

T

-LRB-

fd

-RRB-

-LRB-

minT

-LRB-

fd

-RRB-

-RRB-

are

called

the

-LRB-

minimal

-RRB-

transversals

of

f

A

function

f

is

called

dual-minor

if

f

fd

dual-major

if

f

fd

and

self-dual

if

f

fd

For

example

the

above

function

f

is

self-dual

These

definitions

have

the

following

characterizations

-LSB-

-RSB-

-LRB-

i

-RRB-

f

is

dual-minor

if

and

only

if

at

most

one

of

v

and

v

is

true

for

all

v

-LRB-

ii

-RRB-

f

is

dual-major

if

and

only

if

at

least

one

of

v

and

v

is

true

for

all

v

-LRB-

iii

-RRB-

f

is

self-dual

if

and

only

if

exactly

one

of

v

and

v

is

true

for

all

v

In

this

paper

we

mostly

deal

with

positive

functions

As

mentioned

in

Section

the

length

to

input

a

positive

function

is

O

-LRB-

njminT

-LRB-

f

-RRB-

j

-RRB-

The

lengths

of

f

and

its

dual

fd

may

be

substantially

different

however

as

exemplified

by

the

following

functions

-LSB-

-RSB-

f

xpyp

-LRB-

length

O

-LRB-

p

-RRB-

-RRB-

fd

zp

j

zi

fxi

yig

i

pg

-LRB-

length

O

-LRB-

-RRB-

-RRB-

-LRB-

-RRB-

The

definition

of

the

dual

fd

-LRB-

x

-RRB-

f

-LRB-

x

-RRB-

implies

that

T

-LRB-

fd

-RRB-

f?v

j

v

F

-LRB-

f

-RRB-

g

i

e

F

-LRB-

f

-RRB-

f?v

j

v

T

-LRB-

fd

-RRB-

g

-LRB-

-RRB-

Therefore

this

tells

that

for

a

positive

function

f

the

lengths

of

jminT

-LRB-

f

-RRB-

j

and

jmaxF

-LRB-

f

-RRB-

j

can

be

very

different

It

also

supports

our

claim

in

Section

that

both

minT

-LRB-

f

-RRB-

and

maxF

-LRB-

f

-RRB-

should

be

explicitly

output

in

the

identification

process

-LRB-

Since

the

function

f

in

-LRB-

-RRB-

is

DNF

the

results

in

-LSB-

-RSB-

and

this

paper

show

that

it

can

be

identified

in

our

sense

by

an

incrementally

polynomial

algorithm

-RRB-

The

complexity

of

deciding

if

a

given

positive

function

f

is

dual-major

and/or

self-dual

will

be

discussed

in

the

next

subsection

Here

we

note

that

it

can

be

checked

in

polynomial

time

whether

f

is

dual-minor

or

not

Lemma

-LSB-

-RSB-

-LRB-

i

-RRB-

Let

f

be

a

positive

function

f

is

dual-minor

if

and

only

if

every

pair

of

prime

implicants

mi

and

mj

of

f

has

at

least

one

literal

in

common

-LRB-

ii

-RRB-

The

property

in

-LRB-

i

-RRB-

can

be

checked

in

O

-LRB-

n

jminT

-LRB-

f

-RRB-

-RRB-

time

Complexity

of

Positive

Functions

We

call

problems

A

and

B

polynomially

equivalent

if

A

and

B

are

mutually

reducible

-LSB-

-RSB-

This

tells

in

particular

that

if

one

of

them

is

solvable

in

polynomial

time

then

so

is

the

other

In

relation

with

the

notions

introduced

in

the

previous

subsection

we

consider

the

following

problems

Problem

MD

Input

Positive

functions

f

and

g

of

n

variables

Question

Are

f

and

g

mutually

dual

i.e.

f

gd

-LRB-

hence

g

fd

-RRB-

RRR

Page

Problem

SD

Input

A

positive

function

f

of

n

variables

Question

Is

f

self-dual

Theorem

Problems

EQ

MD

SD

are

polynomially

equivalent

Proof

-LRB-

a

-RRB-

MD

and

SD

SD

is

a

special

case

of

MD

i.e.

when

g

f

and

this

shows

that

SD

is

reducible

to

MD.

To

show

the

converse

given

an

instance

of

MD

consider

function

h

fx

gy

xy

-LRB-

-RRB-

where

of

f

and

g

contain

x

or

y

as

a

variable

Then

hd

-LRB-

fd

x

-RRB-

-LRB-

gd

y

-RRB-

-LRB-

x

y

-RRB-

gdx

fdy

xy

-LRB-

-RRB-

and

h

hd

holds

-LRB-

i.e.

h

is

self-dual

-RRB-

if

and

only

if

f

and

g

are

mutually

dual

i.e.

f

gd

-LRB-

hence

fd

g

-RRB-

-LRB-

b

-RRB-

EQ

and

MD.

First

property

-LRB-

-RRB-

implies

minT

-LRB-

fd

-RRB-

f?v

j

v

maxF

-LRB-

f

-RRB-

g

Now

given

MT

and

MF

of

problem

EQ

define

two

functions

f

and

g

by

minT

-LRB-

f

-RRB-

MT

-LRB-

i

e

T

-LRB-

f

-RRB-

T

-LRB-

MT

-RRB-

-RRB-

minT

-LRB-

g

-RRB-

f?v

j

v

MFg

Assumption

T

-LRB-

MT

-RRB-

F

-LRB-

MF

-RRB-

in

problem

EQ

implies

F

-LRB-

f

-RRB-

F

-LRB-

MF

-RRB-

i.e.

fd

g

-LRB-

hence

f

gd

-RRB-

Therefore

T

-LRB-

MT

-RRB-

-LSB-

F

-LRB-

MF

-RRB-

holds

if

and

only

if

F

-LRB-

f

-RRB-

F

-LRB-

MF

-RRB-

i.e.

f

and

g

are

mutually

dual

This

shows

that

EQ

is

reducible

to

MD.

To

show

the

converse

reduction

first

check

if

gd

f

holds

for

given

positive

functions

f

and

g

of

MD.

For

this

consider

again

h

and

hd

of

-LRB-

-RRB-

and

-LRB-

-RRB-

It

is

not

difficult

to

show

that

gd

f

-LRB-

hence

fd

g

-RRB-

holds

if

and

only

if

hd

h

i.e.

h

is

dual-minor

As

noted

in

Lemma

the

last

condition

can

be

checked

in

polynomial

time

If

h

is

not

dual-minor

-LRB-

i.e.

gd

f

does

not

hold

-RRB-

then

f

and

g

are

not

mutually

dual

Therefore

assume

that

h

is

dual-minor

In

this

case

define

MT

minT

-LRB-

f

-RRB-

MF

f?v

j

v

minT

-LRB-

g

-RRB-

g

where

T

-LRB-

MT

-RRB-

F

-LRB-

MF

-RRB-

holds

by

property

gd

f

By

an

argument

similar

to

the

above

it

is

now

straightforward

to

see

that

T

-LRB-

MT

-RRB-

-LSB-

F

-LRB-

MF

-RRB-

holds

if

and

only

if

f

and

g

are

mutually

dual

Thus

MD

is

reducible

to

EQ

The

equivalence

of

problems

MD

and

SD

was

already

proved

in

-LSB-

-RSB-

in

hypergraph

setting

The

complexity

of

problem

SD

has

been

one

of

the

important

topics

in

distributed

system

theory

To

explain

this

note

that

a

positive

function

fC

can

be

used

as

a

convenient

tool

to

Page

RRR

represent

an

incomparable

family

C

of

subsets

S

ng

by

letting

w

minT

-LRB-

fC

-RRB-

if

and

only

if

w

cv

-LRB-

S

-RRB-

for

some

S

C

where

y

cv

-LRB-

S

-RRB-

is

the

characteristic

vector

of

S

defined

by

yi

if

i

S

and

otherwise

A

family

C

is

said

to

be

incomparable

if

no

two

S

C

satisfy

S

or

S

An

incomparable

family

is

also

called

a

family

or

an

antichain

Let

C

be

an

incomparable

family

It

is

a

coterie

if

it

also

satisfies

the

intersecting

property

S

for

all

S

C

Coteries

have

been

studied

in

conjunction

with

mutual

exclusion

in

a

distributed

system

-LSB-

-RSB-

A

coterie

C

is

nondominated

if

there

is

no

coterie

C

C

such

that

for

each

S

C

there

is

a

subset

C

with

S.

Nondominated

coteries

are

important

in

practical

applications

and

are

closely

related

to

problem

SD

as

shown

in

the

next

lemma

Lemma

-LSB-

-RSB-

Let

fC

be

the

positive

function

representing

an

incomparable

family

C

of

subsets

in

ng

Then

-LRB-

i

-RRB-

C

is

a

coterie

if

and

only

if

fC

is

dual-minor

and

-LRB-

ii

-RRB-

C

is

a

nondominated

coterie

if

and

only

if

fC

is

self-dual

Further

properties

of

positive

self-dual

functions

are

found

in

-LSB-

-RSB-

In

concluding

this

subsection

we

point

out

that

the

following

problem

closely

related

to

problem

SD

is

NP-complete

Problem

NDMAJOR

Input

A

positive

function

f

of

n

variables

Question

Is

f

not

dual-major

Theorem

Problem

NDMAJOR

is

NP-complete

Proof

We

reduce

the

following

problem

which

is

known

to

be

NP-complete

-LSB-

-RSB-

to

NDMAJOR

Problem

SET

SPLITTING

Input

A

family

C

of

subsets

of

V

ng

Question

Is

there

a

partition

of

V

into

two

subsets

and

such

that

no

subset

in

C

is

entirely

contained

in

either

or

First

note

that

we

can

assume

without

loss

of

generality

that

C

is

incomparable

This

is

because

if

S

C

satisfy

S

ae

then

can

be

deleted

without

changing

the

result

since

Vi

-LRB-

i

or

-RRB-

implies

S

Vi

Then

consider

the

positive

function

fC

defined

after

Theorem

In

view

of

characterization

-LRB-

ii

-RRB-

in

Section

fC

is

not

dual-major

if

and

only

if

there

is

a

pair

of

vectors

w

and

w

both

belonging

to

F

-LRB-

fC

-RRB-

i.e.

there

is

a

partition

of

V

as

stated

in

problem

SET

SPLITTING

-LRB-

define

and

by

cv

-LRB-

-RRB-

w

and

cv

-LRB-

-RRB-

w

-RRB-

Since

NDMAJOR

is

obviously

in

class

NP

this

proves

the

lemma

Essentially

the

same

proofs

are

also

found

in

-LSB-

-RSB-

though

they

are

stated

in

different

terminologies

-LRB-

i.e.

coteries

and

hypergraphs

-RRB-

RRR

Page

It

may

be

worth

noting

at

this

point

that

the

problem

SD

is

asking

whether

a

dual-minor

positive

function

f

is

at

the

same

time

dual-major

The

restriction

that

f

is

dual-minor

may

make

it

easier

to

decide

However

we

do

not

know

yet

whether

this

is

in

fact

the

case

or

not

Complexity

of

General

Functions

In

this

subsection

we

consider

the

problems

associated

with

dual

functions

of

general

Boolean

functions

In

order

to

deal

with

general

functions

we

first

discuss

how

to

represent

them

since

the

DNF

expression

with

prime

implicants

is

not

unique

in

this

case

Here

we

assume

that

a

general

function

is

represented

either

by

a

DNF

-LRB-

disjunctive

normal

form

-RRB-

of

implicants

-LRB-

monomials

i.e.

conjunctions

of

literals

-RRB-

or

by

a

CNF

-LRB-

conjunctive

normal

form

-RRB-

of

clauses

-LRB-

i.e.

disjunctions

of

literals

-RRB-

In

the

former

case

we

use

letters

f

g

h

and

so

on

while

in

the

latter

case

we

use

OE

For

example

f

OE

-LRB-

-RRB-

-LRB-

-RRB-

-LRB-

-RRB-

-LRB-

-RRB-

where

i

and

i

stands

for

xi

and

xi

respectively

For

the

subsequent

discussion

we

introduce

some

more

definitions

The

contra-dual

f

of

f

is

defined

by

f

-LRB-

x

-RRB-

f

-LRB-

x

-RRB-

-LSB-

-RSB-

For

example

the

contra-dual

of

f

is

f

A

function

f

is

called

anti-dual

if

f

f

For

example

f

is

anti-dual

Functions

such

as

ff

fd

f

and

f

f

are

all

anti-dual

Lemma

-LRB-

i

-RRB-

f

is

dual-minor

if

and

only

if

ff

-LRB-

i.e.

T

-LRB-

f

-RRB-

T

-LRB-

f

-RRB-

;-RRB-

-LRB-

ii

-RRB-

f

is

anti-dual

if

and

only

if

there

is

a

dual-minor

function

g

such

that

f

g

g

where

denotes

disjoint-or

-LRB-

i.e.

T

-LRB-

g

-RRB-

T

-LRB-

g

-RRB-

;-RRB-

Proof

-LRB-

i

-RRB-

Note

that

T

-LRB-

f

-RRB-

F

-LRB-

fd

-RRB-

by

-LRB-

-RRB-

and

the

definition

of

f

Therefore

f

fd

implies

T

-LRB-

f

-RRB-

F

-LRB-

fd

-RRB-

i.e.

T

-LRB-

f

-RRB-

T

-LRB-

f

-RRB-

The

converse

is

also

similar

-LRB-

ii

-RRB-

The

if-part

follows

from

f

fd

gd?g

-LRB-

by

-LRB-

g

-RRB-

d

g

-RRB-

g

g

-LRB-

by

gd

g

and

g

gd

-RRB-

f

To

show

the

converse

represent

f

by

f

fx

f

x

for

a

variable

x

Then

g

fx

is

dual-minor

since

gd

fd

x

x

fx

g

and

g

f?x

f

x

-LRB-

by

the

anti-duality

of

f

-RRB-

Now

Lemma

for

positive

functions

can

be

extended

to

general

functions

Page

RRR

Theorem

-LRB-

i

-RRB-

Let

a

function

f

be

given

by

f

mp

where

mi

are

implicants

of

f

Then

f

is

dual-minor

if

and

only

if

every

pair

of

implicants

mi

and

mj

has

at

least

one

literal

in

common

-LRB-

ii

-RRB-

The

property

in

-LRB-

i

-RRB-

can

be

checked

in

O

-LRB-

-RRB-

time

Proof

-LRB-

i

-RRB-

ff

Wi

mi

Wj

m?j

Wi

j

mim?j

where

if

mi

then

m?i

and

so

on

Therefore

ff

-LRB-

i.e.

f

is

dual-minor

by

Lemma

-LRB-

i

-RRB-

-RRB-

if

and

only

if

mim?j

for

all

i

and

j

which

holds

if

and

only

if

every

pair

of

implicants

mi

and

mj

has

a

common

literal

-LRB-

ii

-RRB-

Immediate

from

-LRB-

i

-RRB-

Now

we

cite

the

following

well

known

NP-complete

problems

-LSB-

-RSB-

Problem

SAT

-LRB-

satisfiability

-RRB-

Input

A

general

function

OE

of

n

variables

in

CNF

Question

Is

OE

Problem

NAE-SAT

-LRB-

not-all-equal

SAT

-RRB-

Input

A

general

function

OE

of

n

variables

in

CNF

Question

Is

there

a

vector

v

such

that

OE

-LRB-

v

-RRB-

OE

-LRB-

v

-RRB-

-LRB-

i.e.

OEOE

-RRB-

The

problem

NDMAJOR

for

a

general

function

f

is

NP-complete

since

it

is

already

NP-complete

for

positive

functions

-LRB-

Theorem

-RRB-

However

we

repeat

this

result

below

since

its

proof

reveals

an

interesting

equivalence

with

NAE-SAT

Theorem

The

problem

of

deciding

if

a

given

function

f

is

not

dual-major

is

NP

complete

Proof

Given

an

instance

OE

of

NAE-SAT

consider

the

corresponding

f

OEd

in

DNF

f

is

obtained

from

OE

by

simply

interchanging

-LRB-

or

-RRB-

and

-LRB-

and

-RRB-

Obviously

OE

fd

and

OE

f

This

OE

is

NAE-satisfiable

if

and

only

if

OEOE

Then

OEOE

-LRB-

-RRB-

fd

f

-LRB-

-RRB-

fd

f

-LRB-

i

e

f

is

not

dual-major

-RRB-

Since

our

problem

is

obviously

in

class

NP

this

proves

the

theorem

It

may

be

interesting

to

note

that

if

we

admit

the

NP-completeness

of

the

above

problem

as

a

corollary

to

Theorem

the

proof

can

be

easily

modified

to

show

that

NAE-SAT

is

NP-complete

The

resulting

proof

is

much

simpler

than

the

original

one

due

to

Schaefer

-LSB-

-RSB-

Up

to

this

point

the

complexity

for

positive

functions

and

general

functions

do

not

differ

However

the

next

result

may

indicate

a

difference

between

them

Theorem

The

problem

of

deciding

if

a

dual-minor

function

f

is

not

dual-major

-LRB-

i.e.

f

is

not

self-dual

-RRB-

is

NP-complete

Proof

Let

OE

be

an

instance

of

problem

SAT

Let

f

OEd

-LRB-

hence

f

OE

-RRB-

where

f

and

f

are

in

DNF

Introduce

h

ff

which

is

anti-dual

as

easily

shown

from

definition

An

RRR

Page

expression

of

h

in

DNF

can

be

computed

in

O

-LRB-

-RRB-

time

if

f

-LRB-

and

hence

f

-RRB-

has

p

implicants

By

Lemma

-LRB-

ii

-RRB-

h

can

be

represented

as

h

g

g

for

a

dual-minor

function

g

Then

OE

-LRB-

-RRB-

h

-LRB-

since

f

and

f

if

and

only

if

OE

-RRB-

-LRB-

-RRB-

g

g

-LRB-

-RRB-

gd

g

i.e.

OE

is

satisfiable

if

and

only

if

a

dual-minor

function

g

is

not

dual-major

Since

our

problem

is

obviously

in

class

NP

this

completes

the

proof

Incrementally

Polynomial

Algorithms

Problems

We

again

return

to

positive

functions

In

conjunction

with

duality

and

self-duality

consider

the

following

problems

Problem

DUALIZATION

Input

A

positive

function

f

of

n

variables

Output

Its

dual

g

fd

i.e.

minT

-LRB-

fd

-RRB-

Problem

SELF-DUALIZATION

Input

A

positive

dual-minor

function

f

of

n

variables

Output

A

positive

self-dual

function

g

-LRB-

i.e.

minT

-LRB-

g

-RRB-

-RRB-

such

that

f

g

fd

Recall

that

the

inputs

to

these

problems

are

described

by

set

minT

-LRB-

f

-RRB-

-LRB-

i.e.

prime

implicants

of

f

-RRB-

and

outputs

are

also

given

by

minT

-LRB-

g

-RRB-

Therefore

algorithms

to

solve

these

problems

are

enumerative

in

nature

i.e.

have

to

enumerate

all

vectors

in

minT

-LRB-

g

-RRB-

This

means

that

since

the

length

of

fd

can

be

exponential

in

the

length

of

f

as

noted

in

Section

such

enumeration

algorithms

can

not

be

polynomial

in

the

length

of

the

input

In

other

words

what

is

interesting

is

the

incremental

polynomiality

as

explained

in

Section

We

show

in

this

section

that

incrementally

polynomial

algorithms

for

DUALIZATION

and

SELF-DUALIZATION

exist

if

and

only

if

the

corresponding

problems

MD

and

SD

can

be

solved

in

polynomial

time

respectively

The

relation

of

these

results

with

the

original

problem

of

IDENTIFICATION

will

also

become

clear

Algorithm

for

DUALIZATION

We

first

consider

the

problem

DUALIZATION

Assume

that

a

set

TD

-LRB-

minT

-LRB-

fd

-RRB-

-RRB-

has

been

enumerated

so

far

and

define

a

positive

function

g

by

minT

-LRB-

g

-RRB-

TD

-LRB-

-RRB-

Then

g

fd

by

TD

minT

-LRB-

fd

-RRB-

and

g

fd

holds

if

and

only

if

f

and

g

are

mutually

dual

This

mutual

duality

can

be

checked

by

solving

problem

MD

for

f

and

g.

Furthermore

if

g

fd

-LRB-

i.e.

g

fd

-RRB-

we

can

resort

to

algorithm

UNKNOWN

of

Section

with

MT

minT

-LRB-

f

-RRB-

and

MF

f?v

j

v

TDg

Page

RRR

where

T

-LRB-

MT

-RRB-

F

-LRB-

MF

-RRB-

by

g

fd

-LRB-

recall

relation

-LRB-

-RRB-

-RRB-

to

obtain

an

unknown

vector

u

n

-LRB-

T

-LRB-

MT

-RRB-

-LSB-

F

-LRB-

MF

-RRB-

-RRB-

This

u

satisfies

u

T

-LRB-

MT

-RRB-

-LRB-

i.e.

u

F

-LRB-

f

-RRB-

-RRB-

and

u

F

-LRB-

MF

-RRB-

From

u

compute

a

maximal

false

vector

y

maxF

-LRB-

f

-RRB-

by

algorithm

MAXIMAL

and

let

w

y

Then

y

maxF

-LRB-

f

-RRB-

implies

w

minT

-LRB-

fd

-RRB-

and

u

F

-LRB-

MF

-RRB-

implies

y

maxF

-LRB-

MF

-RRB-

i.e.

w

TD

From

these

observations

we

see

that

the

following

algorithm

can

be

used

to

solve

the

problem

DUALIZATION

Algorithm

DUALIZE

Input

A

positive

function

f

i.e.

minT

-LRB-

f

-RRB-

Output

minT

-LRB-

fd

-RRB-

Start

with

MT

minT

-LRB-

f

-RRB-

and

TD

Solve

problem

MD

with

f

and

g

where

g

is

defined

by

-LRB-

-RRB-

If

f

and

g

are

mutually

dual

then

output

TD

and

halt

Otherwise

go

to

Let

MF

f?v

j

v

TDg

Execute

algorithm

UNKNOWN

with

MT

and

MF

to

obtain

an

unknown

vector

u

F

-LRB-

f

-RRB-

Then

compute

a

maximal

vector

y

satisfying

y

u

and

y

F

-LRB-

f

-RRB-

by

algorithm

MAXIMAL

and

let

w

y.

Let

TD

TD

-LSB-

fwg

and

return

to

The

time

to

solve

problem

MD

in

Step

is

TEQ

-LRB-

n

-LRB-

jMT

j

jMF

j

-RRB-

-RRB-

as

obvious

from

the

reduction

of

MD

into

EQ

in

the

proof

of

Theorem

-LRB-

b

-RRB-

In

Step

algorithm

UNKNOWN

requires

O

-LRB-

nTEQ

-LRB-

n

-LRB-

jMT

j

jMF

j

-RRB-

-RRB-

-RRB-

time

as

discussed

in

Section

Algorithm

MAXIMAL

then

requires

O

-LRB-

j

-RRB-

time

since

instead

of

using

membership

queries

we

directly

compute

the

value

of

f

-LRB-

y

-LSB-

yi

-RSB-

-RRB-

by

the

property

f

-LRB-

y

-LSB-

yi

-RSB-

-RRB-

if

there

is

v

MT

such

that

v

y

-LSB-

yi

-RSB-

and

otherwise

The

time

for

checking

the

last

condition

is

O

-LRB-

njMT

j

-RRB-

Consequently

one

iteration

of

Steps

and

is

done

in

O

-LRB-

nTEQ

-LRB-

n

-LRB-

jMT

j

jMF

j

-RRB-

-RRB-

-RRB-

time

-LRB-

assuming

TEQ

-LRB-

n

-LRB-

jMT

j

jMF

j

-RRB-

-RRB-

O

-LRB-

n

-LRB-

jMT

j

jMF

j

-RRB-

-RRB-

-RRB-

and

the

total

time

of

DUALIZE

is

O

-LRB-

n

jminT

-LRB-

fd

-RRB-

j

TEQ

-LRB-

n

-LRB-

jminT

-LRB-

f

-RRB-

j

jminT

-LRB-

fd

-RRB-

j

-RRB-

-RRB-

-RRB-

Therefore

DUALIZE

is

incrementally

polynomial

if

problem

MD

-LRB-

i.e.

problem

EQ

-RRB-

can

be

solved

in

polynomial

time

Theorem

Problem

DUALIZATION

has

an

incrementally

polynomial

algorithm

if

and

only

if

problem

MD

can

be

solved

in

polynomial

time

Proof

The

if-part

was

already

proved

above

To

prove

the

converse

apply

the

incrementally

polynomial

algorithm

A

for

DUALIZATION

to

f

until

either

-LRB-

i

-RRB-

it

halts

or

-LRB-

ii

-RRB-

it

outputs

jminT

-LRB-

g

-RRB-

j

vectors

in

order

to

solve

problem

MD

with

input

data

f

and

g

In

case

of

-LRB-

i

-RRB-

we

obtain

minT

-LRB-

fd

-RRB-

and

hence

we

can

directly

check

if

g

fd

holds

or

not

In

case

of

-LRB-

ii

-RRB-

g

fd

is

immediately

concluded

In

any

case

problem

MD

is

solved

The

time

required

for

this

is

O

-LRB-

jminT

-LRB-

g

-RRB-

j

p

-LRB-

n

-LRB-

jminT

-LRB-

f

-RRB-

j

jminT

-LRB-

g

-RRB-

j

-RRB-

-RRB-

where

p

-LRB-

-RRB-

is

the

time

for

one

iteration

of

algorithm

A

This

is

polynomial

in

the

input

length

n

-LRB-

jminT

-LRB-

f

-RRB-

j

jminT

-LRB-

g

-RRB-

j

-RRB-

of

problem

MD.

Since

problem

DUALIZATION

is

a

fundamental

topic

in

Boolean

theory

its

relation

with

problem

MD

has

been

recognized

in

various

papers

including

-LSB-

-RSB-

The

above

RRR

Page

argument

may

be

interesting

in

the

sense

that

it

reveals

a

similarity

between

algorithms

DUALIZE

and

IDENTIFY

Although

DUALIZE

starts

with

MT

minT

-LRB-

f

-RRB-

and

MF

-LRB-

f?v

j

v

TDg

-RRB-

IDENTIFY

starts

with

MT

MF

DUALIZE

then

computes

the

final

MF

maxF

-LRB-

f

-RRB-

from

the

knowledge

of

MT

minT

-LRB-

f

-RRB-

but

IDENTIFY

computes

both

MT

minT

-LRB-

f

-RRB-

and

MF

maxF

-LRB-

f

-RRB-

by

issuing

membership

queries

to

the

oracle

of

f

The

iteration

of

both

algorithms

to

find

a

new

vector

is

based

on

essentially

the

same

idea

as

evident

from

the

above

discussion

Although

the

complexity

of

DUALIZATION

is

not

known

at

present

there

are

some

special

classes

of

positive

Boolean

functions

for

which

polynomial

time

algorithms

are

known

such

as

positive

threshold

functions

and

regular

functions

-LSB-

-RSB-

and

positive

k-DNF

functions

-LSB-

-RSB-

Algorithm

for

SELF-DUALIZATION

For

a

dual-minor

function

f

the

whole

set

is

partitioned

into

three

disjoint

sets

T

-LRB-

f

-RRB-

T

-LRB-

fd

f

-RRB-

-LRB-

T

-LRB-

fd

-RRB-

n

T

-LRB-

f

-RRB-

-RRB-

and

F

-LRB-

fd

-RRB-

Furthermore

T

-LRB-

f

-RRB-

and

F

-LRB-

fd

-RRB-

are

related

by

-LRB-

-RRB-

Therefore

-LRB-

a

-RRB-

v

T

-LRB-

f

-RRB-

if

and

only

if

v

F

-LRB-

fd

-RRB-

and

-LRB-

b

-RRB-

v

T

-LRB-

fd

f

-RRB-

if

and

only

if

v

T

-LRB-

fd

f

-RRB-

Also

condition

-LRB-

-RRB-

implies

maxF

-LRB-

fd

-RRB-

f?v

j

v

minT

-LRB-

f

-RRB-

g

-LRB-

-RRB-

Without

loss

of

generality

we

assume

f

in

this

section

since

otherwise

any

self-dual

function

can

be

output

For

a

dual-minor

function

f

a

vector

v

is

called

a

counterexample

to

f

if

f

-LRB-

v

-RRB-

f

-LRB-

v

-RRB-

since

this

indicates

that

f

is

not

self-dual

-LRB-

see

the

characterizations

-LRB-

i

-RRB-

and

-LRB-

iii

-RRB-

in

Section

-RRB-

The

self-duality

can

be

checked

by

solving

problem

SD

If

f

is

not

self-dual

i.e.

f

fd

then

a

vector

u

satisfying

u

n

-LRB-

T

-LRB-

f

-RRB-

-LSB-

F

-LRB-

fd

-RRB-

-RRB-

-LRB-

-RRB-

can

be

found

by

algorithm

UNKNOWN

executed

with

MT

minT

-LRB-

f

-RRB-

and

MF

f?v

j

v

minT

-LRB-

f

-RRB-

g

where

T

-LRB-

MT

-RRB-

F

-LRB-

MF

-RRB-

T

-LRB-

f

-RRB-

F

-LRB-

fd

-RRB-

by

assumption

f

fd

This

u

is

a

counterexample

i.e.

f

-LRB-

u

-RRB-

f

-LRB-

u

-RRB-

because

u

T

-LRB-

fd

f

-RRB-

by

-LRB-

-RRB-

and

hence

u

T

-LRB-

fd

f

-RRB-

by

property

-LRB-

b

-RRB-

above

Given

a

counterexample

u

we

then

obtain

a

minimal

vector

y

u

such

that

y

T

-LRB-

fd

-RRB-

by

applying

MINIMAL

with

MT

minT

-LRB-

f

-RRB-

in

which

condition

-LRB-

y

-LRB-

yi

-RRB-

-RRB-

T

-LRB-

fd

-RRB-

in

Step

of

MINIMAL

is

checked

by

condition

F

-LRB-

fd

-RRB-

-LRB-

i.e.

there

is

no

vector

v

maxF

-LRB-

fd

-RRB-

such

that

v

-RRB-

in

O

-LRB-

njMT

j

-RRB-

time

since

maxF

-LRB-

fd

-RRB-

is

provided

by

-LRB-

-RRB-

instead

of

the

explicit

data

of

minT

-LRB-

fd

-RRB-

We

then

define

a

new

function

g

satisfying

f

g

fd

by

minT

-LRB-

g

-RRB-

MinSet

-LRB-

minT

-LRB-

f

-RRB-

-LSB-

fyg

-RRB-

Page

RRR

We

claim

that

this

g

is

positive

and

dual-minor

The

positivity

is

an

immediate

consequence

from

definition

To

show

that

g

is

dual-minor

assume

otherwise

i.e.

there

is

a

pair

of

vectors

w

and

w

such

that

g

-LRB-

w

-RRB-

g

-LRB-

w

-RRB-

These

vectors

belong

to

T

-LRB-

fd

f

-RRB-

by

the

above

properties

-LRB-

a

-RRB-

and

-LRB-

b

-RRB-

and

hence

satisfy

w

y

and

w

y

by

the

above

definition

of

g

But

this

implies

y

-LRB-

-RRB-

i.e.

g

which

is

a

contradiction

to

assumption

f

Therefore

g

is

positive

and

dual-minor

The

following

algorithm

repeats

the

above

procedure

until

a

self-dual

function

is

obtained

Algorithm

SELF-DUALIZE

Input

A

positive

dual-minor

function

f

i.e.

minT

-LRB-

f

-RRB-

Output

A

positive

self-dual

function

g

i.e.

minT

-LRB-

g

-RRB-

such

that

f

g

fd

Start

with

MT

minT

-LRB-

f

-RRB-

and

g

f

Solve

problem

SD

with

g

If

g

is

self-dual

then

output

MT

and

halt

Otherwise

go

to

Let

MF

f?v

j

v

MTg

Execute

algorithm

UNKNOWN

with

MT

and

MF

to

obtain

an

unknown

vector

u

-LRB-

T

-LRB-

g

-RRB-

-LSB-

F

-LRB-

gd

-RRB-

-RRB-

Then

compute

a

minimal

vector

y

such

that

y

u

and

y

T

-LRB-

gd

-RRB-

nT

-LRB-

g

-RRB-

by

algorithm

MINIMAL

and

let

MT

MinSet

-LRB-

MT

-LSB-

fyg

-RRB-

Define

the

function

g

by

minT

-LRB-

g

-RRB-

MT

Return

to

Similarly

to

algorithm

DUALIZE

it

can

be

shown

that

SELF-DUALIZE

requires

O

-LRB-

nTEQ

-LRB-

njMT

j

-RRB-

-RRB-

-LRB-

O

-LRB-

nTEQ

-LRB-

n

-LRB-

jminT

-LRB-

f

-RRB-

j

jminT

-LRB-

g

-RRB-

j

-RRB-

-RRB-

-RRB-

-RRB-

time

in

each

iteration

Therefore

this

is

an

incrementally

polynomial

algorithm

if

problem

SD

-LRB-

i.e.

problem

EQ

-RRB-

can

be

solved

in

polynomial

time

The

total

time

is

O

-LRB-

n

jminT

-LRB-

g

-RRB-

j

TEQ

-LRB-

n

-LRB-

jminT

-LRB-

f

-RRB-

j

jminT

-LRB-

g

-RRB-

j

-RRB-

-RRB-

-RRB-

where

g

in

this

formula

is

the

last

g

computed

by

SELF-DUALIZE

Theorem

Problem

SELF-DUALIZATION

has

an

incrementally

polynomial

algorithm

if

and

only

if

problem

SD

can

be

solved

in

polynomial

time

Proof

The

if-part

follows

from

the

above

argument

To

prove

the

only-if-part

first

check

if

f

is

dual-minor

-LRB-

which

can

be

done

in

polynomial

time

by

Lemma

-RRB-

If

f

is

not

dual-minor

it

is

not

self-dual

If

f

is

dual-minor

apply

the

incrementally

polynomial

algorithm

A

for

SELF-DUALIZATION

until

either

-LRB-

i

-RRB-

it

halts

or

-LRB-

ii

-RRB-

it

iterates

jminT

-LRB-

f

-RRB-

j

times

In

case

of

-LRB-

i

-RRB-

f

is

self-dual

if

and

only

if

the

obtained

function

g

satisfies

g

f

In

case

of

-LRB-

ii

-RRB-

f

is

not

self-dual

since

each

vector

y

generated

by

A

belongs

to

minT

-LRB-

g

-RRB-

In

any

case

problem

SD

is

solved

The

time

required

for

this

is

O

-LRB-

jminT

-LRB-

f

-RRB-

j

p

-LRB-

njminT

-LRB-

f

-RRB-

j

-RRB-

-RRB-

where

p

-LRB-

-RRB-

is

the

time

for

one

iteration

of

algorithm

A

This

is

polynomial

in

the

input

length

n

-LRB-

jminT

-LRB-

f

-RRB-

j

-RRB-

of

problem

SD

Combining

Theorems

and

with

Theorem

we

obtain

the

next

result

Corollary

Three

problems

IDENTIFICATION

DUALIZATION

and

SELF-DUALIZATION

are

equivalent

in

the

sense

that

if

one

of

them

has

an

incrementally

polynomial

algorithm

then

so

does

the

rest

of

them

RRR

Page

For

a

given

dual-minor

function

f

there

are

in

general

many

self-dual

functions

g

satisfying

f

g

fd

The

above

algorithm

SELF-DUALIZE

obtains

only

one

of

them

If

other

unknown

vectors

are

used

in

Step

-LRB-

e.g.

u

instead

of

u

-RRB-

the

resulting

function

g

may

become

different

Saturated

Positive

Functions

In

the

course

of

investigating

the

complexity

of

problems

MD

and

SD

some

other

problems

have

been

found

to

be

polynomially

equivalent

We

list

one

problem

extensively

studied

in

-LSB-

-RSB-

in

the

hypergraph

setting

A

family

C

of

subsets

of

V

ng

is

sometimes

referred

to

as

a

hypergraph

H

-LRB-

V

C

-RRB-

where

each

S

C

is

called

a

hyperedge

A

hypergraph

H

is

simple

if

C

is

incomparable

Problem

SIMPLE-H-SAT

-LRB-

simple

hypergraph

saturation

-RRB-

Input

A

simple

hypergraph

H

-LRB-

V

C

-RRB-

Question

Is

there

no

subset

S

of

V

such

that

H

-LRB-

V

C

-LSB-

fSg

-RRB-

is

also

a

simple

hypergraph

In

-LSB-

-RSB-

it

is

shown

that

this

problem

is

polynomially

equivalent

to

problem

SD

It

also

contains

interesting

applications

found

in

such

fields

as

artificial

intellgence

and

database

theory

In

this

section

we

define

the

same

problem

in

Boolean

terminology

and

again

show

its

polynomial

equivalence

with

SD

As

the

methods

of

Boolean

algebra

become

available

in

this

way

the

proof

is

much

simplified

and

easy

to

understand

Some

new

characterizations

such

as

Lemma

-LRB-

iii

-RRB-

below

are

also

derived

Furthermore

some

concepts

introduced

in

its

proof

turn

out

to

be

very

useful

to

understand

properties

of

self-dual

functions

as

will

be

discussed

in

a

companion

paper

-LSB-

-RSB-

For

a

positive

function

f

call

a

vector

w

incomparable

with

f

if

w

v

and

w

v

hold

for

all

v

minT

-LRB-

f

-RRB-

A

positive

function

f

is

saturated

if

there

is

no

vector

w

incomparable

with

f

Problem

ST

Input

A

positive

function

f

of

n

variables

Question

Is

f

saturated

Before

proving

the

polynomial

equivalence

of

ST

and

SD

we

introduce

two

positive

functions

f

s

and

fc

associated

with

f

First

fc

is

defined

by

minT

-LRB-

fc

-RRB-

f?v

j

v

minT

-LRB-

f

-RRB-

g

-LRB-

-RRB-

The

DNF

expression

of

fc

by

prime

implicants

is

fc

fm?v

j

v

minT

-LRB-

f

-RRB-

g

-LRB-

-RRB-

Page

RRR

where

mw

for

w

denotes

the

conjunction

of

literals

xi

satisfying

wi

e.g.

if

w

-LRB-

-RRB-

then

mw

-LRB-

abbreviation

of

-RRB-

For

example

f

i

e

minT

-LRB-

f

-RRB-

f

-LRB-

-RRB-

-LRB-

-RRB-

g

fc

i

e

minT

-LRB-

fc

-RRB-

f

-LRB-

-RRB-

-LRB-

-RRB-

g

-LRB-

-RRB-

Next

let

L

-LRB-

w

-RRB-

denote

the

set

of

vectors

obtained

from

w

by

changing

one

element

wi

to

wi

For

example

for

w

-LRB-

-RRB-

L

-LRB-

w

-RRB-

f

-LRB-

-RRB-

-LRB-

-RRB-

-LRB-

-RRB-

g.

Function

f

s

which

is

called

the

subdual

of

f

is

then

defined

by

minT

-LRB-

f

s

-RRB-

MinSet

-LRB-

-LSB-

fL

-LRB-

v

-RRB-

j

v

minT

-LRB-

f

-RRB-

g

-RRB-

-LRB-

-RRB-

Its

DNF

expression

is

given

by

f

s

fm?vmdv

j

v

minT

-LRB-

f

-RRB-

g

-LRB-

-RRB-

where

mdv

denotes

the

dual

of

mv

e.g.

if

mv

then

mdv

-LRB-

-RRB-

For

the

example

of

-LRB-

-RRB-

we

have

f

s

-LRB-

-RRB-

-LRB-

-RRB-

minT

-LRB-

f

s

-RRB-

MinSetf

-LRB-

-RRB-

-LRB-

-RRB-

-LRB-

-RRB-

-LRB-

-RRB-

-LRB-

-RRB-

g

f

-LRB-

-RRB-

-LRB-

-RRB-

-LRB-

-RRB-

-LRB-

-RRB-

g

-LRB-

-RRB-

Now

recall

property

-LRB-

-RRB-

that

holds

between

maxF

-LRB-

fd

-RRB-

and

minT

-LRB-

f

-RRB-

Then

the

definition

of

fc

implies

minT

-LRB-

fc

-RRB-

maxF

-LRB-

fd

-RRB-

Since

any

w

minT

-LRB-

f

s

-RRB-

is

larger

than

some

y

maxF

-LRB-

fd

-RRB-

by

definitions

-LRB-

-RRB-

and

-LRB-

-RRB-

it

satisfies

w

z

for

any

z

maxF

-LRB-

fd

-RRB-

implying

w

T

-LRB-

fd

-RRB-

This

proves

f

s

fd

i

e

-LRB-

f

s

-RRB-

d

f

-LRB-

-RRB-

It

is

also

clear

from

definitions

-LRB-

-RRB-

and

-LRB-

-RRB-

that

f

s

fc

-LRB-

-RRB-

Therefore

f

s

fdfc

The

converse

is

also

true

since

fdfc

-LRB-

f

-RRB-

m?v

-LRB-

-LRB-

f

-RRB-

mw

-RRB-

d

v

m?v

-LRB-

w

mdw

-RRB-

v

m?vmdv

f

s

which

proves

f

s

fdfc

-LRB-

-RRB-

Lemma

For

a

positive

function

f

the

following

conditions

are

equivalent

RRR

Page

-LRB-

i

-RRB-

f

is

saturated

-LRB-

ii

-RRB-

fd

fc

-LRB-

iii

-RRB-

f

s

fd

Proof

A

vector

w

is

incomparable

with

f

if

and

only

if

-LRB-

a

-RRB-

w

T

-LRB-

f

-RRB-

and

-LRB-

b

-RRB-

w

v

-LRB-

-RRB-

-LRB-

i.e.

w

v

-RRB-

for

all

v

minT

-LRB-

f

-RRB-

However

condition

-LRB-

b

-RRB-

is

equal

to

saying

that

w

T

-LRB-

-LRB-

fc

-RRB-

d

-RRB-

-LRB-

see

definition

-LRB-

-RRB-

of

fc

and

the

transversality

of

a

dual

function

mentioned

in

Section

-RRB-

Therefore

such

an

incomparable

w

does

not

exists

if

and

only

if

T

-LRB-

f

-RRB-

T

-LRB-

-LRB-

fc

-RRB-

d

-RRB-

-LRB-

-RRB-

f

-LRB-

fc

-RRB-

d

-LRB-

-RRB-

fd

fc

This

proves

the

equivalence

of

-LRB-

i

-RRB-

and

-LRB-

ii

-RRB-

The

equivalence

of

-LRB-

ii

-RRB-

and

-LRB-

iii

-RRB-

is

then

obvious

from

property

-LRB-

-RRB-

For

example

function

f

of

-LRB-

-RRB-

is

not

saturated

since

fd

-LRB-

-RRB-

-LRB-

-RRB-

is

not

equal

to

f

s

of

-LRB-

-RRB-

In

fact

a

vector

-LRB-

-RRB-

is

incomparable

with

f

Theorem

Problems

ST

and

SD

are

polynomially

equivalent

Proof

-LRB-

a

-RRB-

Reduction

from

ST

to

SD

Let

f

be

a

positive

function

and

x

y

be

variables

not

contained

in

f

From

this

f

construct

the

following

h

h

fx

f

sy

xy

i.e.

by

-LRB-

-RRB-

hd

-LRB-

f

s

-RRB-

dx

fdy

xy

Then

-LRB-

-RRB-

shows

that

h

is

dual-minor

i.e.

h

hd

and

furthermore

h

hd

holds

if

and

only

if

f

s

fd

Thus

h

is

self-dual

if

and

only

if

f

is

saturated

-LRB-

b

-RRB-

Reduction

from

SD

to

ST.

Assume

that

f

is

positive

and

dual-minor

-LRB-

dual-minority

can

be

checked

in

polynomial

time

by

Lemma

-RRB-

From

f

define

g

f

fc

xy

-LRB-

-RRB-

where

x

and

y

are

variables

not

contained

in

f

Then

minT

-LRB-

g

-RRB-

f

-LRB-

v

-RRB-

-LRB-

v

-RRB-

j

v

minT

-LRB-

f

-RRB-

g

-LRB-

-RRB-

where

the

last

two

elements

of

vectors

respectively

correspond

to

x

and

y

because

f

fd

implies

v

w

-LRB-

-RRB-

for

any

v

w

minT

-LRB-

f

-RRB-

-LRB-

by

Lemma

-RRB-

and

hence

v

w

for

any

v

w

minT

-LRB-

f

-RRB-

Property

-LRB-

-RRB-

says

that

w

minT

-LRB-

g

-RRB-

-LRB-

-RRB-

w

minT

-LRB-

g

-RRB-

and

hence

g

gc

Therefore

g

is

saturated

-LRB-

-RRB-

gd

gc

-LRB-

by

Lemma

-LRB-

ii

-RRB-

-RRB-

-LRB-

-RRB-

gd

g

-LRB-

-RRB-

fd

-LRB-

-LRB-

fc

-RRB-

d

x

y

-RRB-

f

fc

xy

The

last

inequality

implies

fd

f

-LRB-

take

x

and

y

-RRB-

i.e.

fd

f

-LRB-

self-dual

-RRB-

by

assumption

f

fd

On

the

other

hand

since

f

g

-LRB-

i.e.

gd

fd

-RRB-

follows

from

-LRB-

-RRB-

fd

f

implies

gd

fd

f

g

gc

i.e.

g

is

saturated

by

Lemma

-LRB-

ii

-RRB-

Consequently

g

is

saturated

if

and

only

if

f

is

self-dual

Page

RRR

Conclusion

Starting

from

the

problem

of

identifying

minT

-LRB-

f

-RRB-

and

maxF

-LRB-

f

-RRB-

of

a

positive

function

f

-LRB-

i.e.

problem

IDENTIFICATION

-RRB-

it

is

shown

that

its

complexity

is

essentially

-LRB-

modulo

polynomial

time

-RRB-

equal

to

other

problems

such

as

DUALIZATION

SELF-DUALIZATION

EQ

MD

SD

and

ST.

Some

of

these

problems

have

been

well

studied

in

respective

fields

but

unfortunately

their

complexity

is

still

left

open

In

this

sense

the

most

urgent

goal

is

to

clarify

whether

all

of

these

problems

or

of

them

permit

polynomial

-LRB-

or

incrementally

polynomial

-RRB-

algorithms

As

noted

in

Section

and

other

places

there

are

several

special

classes

of

positive

functions

for

which

polynomial

-LRB-

or

incrementally

polynomial

-RRB-

algorithms

are

known

One

of

our

companion

papers

-LSB-

-RSB-

investigates

this

direction

by

using

a

new

tool

called

the

maximum

latency

which

is

a

complexity

measure

for

finding

an

unknown

vector

and

provides

some

polynomially

solvable

classes

Another

possible

avenue

would

be

to

introduce

approximate

concepts

of

dual

self-dual

saturation

and

so

forth

so

that

they

become

computationally

tractable

The

concept

of

subdual

introduced

in

Section

is

interesting

in

this

respect

since

it

is

a

subfunction

of

the

dual

and

is

polynomially

computable

for

all

positive

functions

The

concept

of

almost

self-duality

which

is

a

close

approximation

of

the

self-duality

defined

in

terms

of

subduality

-LRB-

hence

is

polynomially

testable

-RRB-

and

related

topics

are

extensively

studied

in

another

companion

paper

-LSB-

-RSB-

Acknowledgments

The

authors

were

benefited

from

the

discussion

with

their

colleagues

E.

Boros

Y.

Crama

P.

L.

Hammer

T.

Kameda

K.

Makino

H.

Nagamochi

and

others

whose

cooperation

is

greatly

appreciated

This

research

was

partially

supported

by

the

Scientific

Grant-in-Aid

from

Ministry

of

Education

Science

and

Culture

of

Japan

The

second

author

gratefully

acknowledges

the

partial

support

given

to

Rutcor

by

the

Office

of

Naval

Research

-LRB-

grants

and

-RRB-

and

by

the

Air

Force

Office

of

Scientific

Research

-LRB-

grant

-RRB-

References

-LSB-

-RSB-

D.

Angluin

Queries

and

concept

learning

Machine

Learning

-LRB-

-RRB-

-LSB-

-RSB-

D.

Angluin

L.

Hellerstein

and

M.

Karpinski

Learning

read-once

formulas

with

queries

J.

ACM

-LRB-

-RRB-

-LSB-

-RSB-

M.

Anthony

and

N.

Biggs

Computational

Learning

Theory

Cambridge

University

Press

-LSB-

-RSB-

P.

Bertolazzi

and

A.

Sassano

An

O

-LRB-

mn

-RRB-

time

algorithm

for

regular

set-covering

problems

Theoretical

Computer

Science

-LRB-

-RRB-

RRR

Page

-LSB-

-RSB-

J.

C.

Bioch

Decompositions

of

coteries

and

Boolean

functions

Tech

Rep.

Dept.

of

Computer

Science

No.

Erasmus

University

Rotterdam

The

Netherlands

January

-LSB-

-RSB-

J.

C.

Bioch

and

T.

Ibaraki

Decompositions

of

positive

self-dual

Boolean

Functions

May

submitted

to

Discrete

Mathematics

-LSB-

-RSB-

J.

C.

Bioch

and

T.

Ibaraki

Generating

and

approximating

positive

Boolean

functions

Tech

Rep.

Dept.

of

Computer

Science

No.

Erasmus

University

Rotterdam

The

Netherlands

-LSB-

-RSB-

E.

Boros

P.

L.

Hammer

T.

Ibaraki

and

K.

Kawakami

Identifying

positive

Boolean

functions

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