Improved Testing Algorithms For Monotonicity

By Range Reduction

Presented By

Daniel Sigalov

Introduction

• The main idea of the article is to prove that there exist a tester of monotonicity with query and time complexity

log lognO

The theorem of range reduction• Consider the task of checking monotonicity of functions

defined over partially ordered set S.

Suppose that for some distribution on pairs

with and for every function

where C defends on S only. Then for every and every function for pairs selected according to the same distribution

,x y S S

Sx y : 0,1f S

( )Pr ( ) ( ) M f

f x f yC

:f S

( )Pr ( ) ( )

logM f

f x f yC

Basic definitions• For each 2 functions

- the fraction of instances

On which

• - the minimum distance between function and any other monotone function

• - the probability that a pair selected according to witnesses that is not monotone.

nx

( ) ( )f x g x

f

( )M f

, : nf g

( , )Dist f g

: ng

( , )~( , ) Pr ( ) ( )x y DDetect f D f x f y ( , )x y D

f

Monotonicity How we do it?

• Incrementally transform into a monotone function, while insuring that for each repaired violated edge, the value of the function changed only in a few points.

f

1 2 3f f f f

Operators (1) MON(f)

- arbitrary monotone function

at distance from MON f

( )M f f

: 0, 1 , 0, 1f S r a b r a b

Operators(2) SQUASH

: 0, 1 , 0, 1f S r a b r a b

( )

, , ( ) ( )

( )

a if f x a

SQUASH f a b x b if f x b

f x otherwise

, ,VIOL SQUASH f a b VIOL f

Operators (3) CLEAR

: 0, 1 , 0, 1f S r a b r a b

, , ( )

, , ( ) , , ( ) , , ( )

( )

MON SQUASH f a b x

CLEAR f a b x if MON SQUASH f a b x SQUASH f a b x

f x otherwise

, , , , ,MDist f CLEAR f a b SQUASH f a bClaim:

Proof: by the definition of CLEAR

by the definition of MON

, , ,

, , , , ,

, ,M

Dist f CLEAR f a b

Dist SQUASH f a b MON SQUASH f a b

SQUASH f a b

Interval of a violated edge with respect to function -

More definitions..

( , )x yf ( ), ( )f y f x

two intervals cross if they intersect in more than one point.

example: [2,3], [4,6]

0 1 2 3 4

6

5

4

3

2

1

0 1 2 3 4 5 6

[1,6]

Lemma 1 - Clear • Lemma: The function has the

following properties:

1.

2. has no violated edges whose intervals cross .

3. The interval of a violated edge with respect to is contained in the

interval of this edge with respect to .

, ,CLEAR f a b

, ,VIOL CLEAR f a b VIOL f

, ,CLEAR f a b

,a b

, ,CLEAR f a b

f

Proof of the LemmaDefine

Note: 1. is monotone and takes values from

2.

3.

We will check the 4 possibilities for :

1. - not possible. Why?

2. - agree on is violated by and . Proves (1) & (3).

If cross

Contradiction to the monotonicity of

, , , , ,

( , ) . . ( ) ( )

g MON SQUASH f a b h CLEAR f a b

Let x y be an edge violated by h i e h x h y

g ,a b

( ) , ( ) ( )

( ) , ( ) ( )

if h x a b h x f x

if h x a b h x g x

( ), ( ) ,h x h y a b

( ), ( )h x h y

( ), ( ) ,h x h y a b ,h f , ,x y x yf ( ), ( ) ( ), ( )h y h x f y f x

( ) ( ), ( ) ( ), ( , )h x g x h y g y g monotone h cannot violate x y

CLEAR definition

( ), ( )h y h x , ( ) , ( ) ( ) , ( )a b h x b h y a g x b g y a

g

Proof of the Lemma (cont.)3. - is violated

Therefore intersects

in one point only - . This proves (2)

In case (1) & (3) follows.

If not then

(1) & (3) follows.

4. - symmetric to case 3.

( ) , , ( ) ,h x a b h y a b

( ) , , ( ) ,h x a b h y a b

( , )x y

( )h x b ( ) ( )f x h x b ( )g x b

g monotone ( ) ( )g y g x b ( ) ( )h y g y b

( ), ( )h x h y ,a bb

( ) ( )f y h y b

( ) , , ( )b g y SQUASH f a b y , , ( ) ( ) ( ), ( ) ( )SQUASH f a b y b f y b h y f x h x b

• Lemma: given define:

Those functions have the following properties:

1.

2.

3.

Lemma 2 - Range reductionDefining the functions

1

' , 1,2 2

, 1,2 2

r rf SQUASH f

r rf CLEAR f

: 0, 1f S r

1 1

2 1

' , 0, 12

,0, 12

rf SQUASH f

rf CLEAR f

2 2

3 2

' , , 12

, , 12

rf SQUASH f r

rf CLEAR f r

( , ) ( ', )Detect f D Detect f D

1 2( , ) ( ', ) ( ', )Detect f D Detect f D Detect f D

1 2( ) ( ') ( ') ( ')M M M Mf f f f

Proof of the Range reduction lemma (1)

1. The SQUASH operator never adds new violated edges

'VIOL f VIOL f

( , ) ( ', )Detect f D Detect f D

Proof of the Range reduction lemma (2)

2. Note:

1 2( , ) ( ', ) ( ', )Detect f D Detect f D Detect f D

1 2' , 'VIOL f VIOL f VIOL f

1 2' 'VIOL f VIOL f

Proof of the Range reduction lemma (3)

3. Note: Why?

the distance from to the set of monotone functions is at most the distance to a particular monotone function :

1 1 2 2 3

1 2

( ) , , ,

( ') ( ') ( ')M

M M M

f dist f f dist f f dist f f

f f f

3 !f is monotone

f

3f

, , , , ,MDist f CLEAR f a b SQUASH f a b

Proof of The theorem of range reduction

We will prove by induction on

that for every function

the following hypothesis:

• Base case :

In the theorem we assumed -

By the definition of detect we get the hypothesis.

: , 2sf S

( ) ( , )M f C Detect f D s 1s

s

( )Pr ( ) ( ) M f

f x f yC

• Lets assume the hypothesis holds for and prove it for :

Proof of The theorem of range reduction (cont.)

1s s

1 2

1 2

( ) ( ') ( ') ( ')

( ', ) ( ', ) 1 ( ', ) 1

( , ) ( , ) 1

( , )

M M M Mf f f f

C Detect f D C Detect f D s C Detect f D s

C Detect f D Detect f D s

C Detect f D s

Testing monotonicity

Questions?