Haim Kermany

Learning Decision Treesusing the Fourier Spectrum

By

Eyal Kushilevitz

Yishay Mansour

( )h x

What is Learning

( )f x( ,?)a

( )f a

What are we Learning?

Boolean Function: :{0,1 1} { ,1}nf

1 2 3 4 5 1 2 3 4 1: ( , , , , ) ( ) ( )ex f x x x x x x x x x x

•Term (and of literals(

1 2 3 4 5 1 2 4 1 2 4: ( , , , , )ex f x x x x x x x x x x x

(01011) (0 1) (0 1 0) 1f

•DNF (or of terms)

1 2 3 4 5 1 2 4 2 3 4 5: ( , , , , )ex f x x x x x x x x x x x x

•DT , : See latter

Randomized Algorithm

( ) ( ) ?h x f x Does always All inputs

( )Prob ( )h x f x We Want:

( ) 1h x

( ) 1f x

( ) ( )h x f xDo we always succeed ?

Prob =

Pr ( )ob Pr ( )ob

fail

h x f x 0 Deterministic Algorithem

0 ( ) ( )h x f x

( ) 1,1sign g x

2E ( ) ( )f x g x

:g approcximates f

approximation

P rob ( ) ( )f x sign g x

( ) and thenif f x boolean g approcximates f

What we want

Fourier Transform

i

i1

1 if mod 2 = 0

1 if mod 2 = 1( ... ) i i

i i

z nx z

x zx x

1 0 ( )1

01110

011010110 =+1

(1+ 1)mod 2=0

0,1

ˆ ( )n

z z

z

f f x f x

{0,1}nz x z

t-phase functiont-phase function – a function that has at most t Fourier coefficient

2 g approximates f

2

then th O approcximates f

1 g t phase functionif that:g

ˆ0,1

ˆ ˆ( ) ( )i

nz ti

z z z zfz

f x f x h x f x

Only big coefficients

Very small

( ) has all coefficient that start with f x

:{0,1}n kf

0,1

ˆ( )n k

f x f x

{0,1}k

( ) ( ) , is the empty stringf x f x

all coefficient in ( ) = +f x1all coefficient in ( ) f x

0all coefficient in ( ) f x

( )f x

Tree f x

1f x

11f x

0f x

10f x 00f x 01f x

1f x

1f x

n

f x

2

n

What we need

Finding only big coefficients

ˆif f if output n

else 0 ; 1 ;Coef Coef

Coef

Analyzing coef()

for any , 0,1 :k

k 2 2ˆif then f E f

{0,1}

1[ ] ( ) ( )

2 nnx

E fg f x g x

Changing coef()

2if B if output n

else 0 ; 1 ;Coef Coef

Coef 2B E f

Approximating 2E f x

{0,1}

( )kyf x E f xy y

( )f x

2, , ( ) ( )x y zE f x E f xy f xz y z

2

{0,1}

2

{0,1}

, ,

, ,

( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

n k

k

x

x y

x y y

x y z

x y z

x y z

E f x

E E f xy y

E E f xy y E f xy y

E E f xy y E f xz z

E f xy y f xz z

E f xy f xz y z

Proof :

Approx do times:m

choose 0,1 ;k

iy choose 0,1 ;

n k

ix

1

Let ;i i i i i ii m

B AVG f y x f y z y z

Approximating 2E f x

2

approximating

E f x

MQ

choose 0,1 ;k

iy

Changing coef()

if output n

else 0 ; 1 ;Coef Coef

Coef

2f 2i B

B Approx Save Side

Finding

m 22 1

4 222Pr 4m

E f B e

41 1logm O

with probability 1 : 22 2 2E f B

2 22 4 2E f B

22 2

4E f

22 2 2E f B

Coef() output

22 2ˆ 2ˆ will be output

z z z

z

z f E f B

f

ˆevery will be outputzf

Coaf() time

2 22

1 f E f

for any :k

2 22 4 2There wont be a recursive call

E f B

222

4 4nE f

22

2 2

2

1 f E f

for any :k

2

22

4 4f E f

for any :k

24There will be at most recursive call

(every call work )

n

m

There will be

coefficients

1( )poly

Running time is

1 1( , , )poly n

Finding

h x

ˆoutput z zz Z

h x f

Z Coef

ˆ approximate zz Z f

find h x

ˆ [ ]z zf E f

Conclusion

2 g approximates f 1 g t phase function

if that:g

1 1

Then there is exist a

that output a function , such that with

probability 1 .

The algorithm run in ( , , , )

randomized algorithm

h

O approximates f

poly n t log

ˆ0,1 ( )1

ˆ ˆ( ) ( )i

nzi

z z z zfz L f

f x f x h x f x

f h approximates f

1ˆ( ) zz

L f f

1( )L fHow to find h(x)

use Coaf() again

Best algorithm ever

1 11

There is a

that for boolean function

output a function , such that with

probability 1 .

The algorithm run in

any

( , , , )

randomized algorithm

f

h

h approximates f

poly n L f log

1L f

Can be exp(n) 1L f

Decision Tree (DT)

+1-1

-1

-1-1-1 +1

+1

+1

+1

+1

01011

11100

10101

11001

10010

00010

0001010010 11100

11100

1 ( )f DT L f m

the of nodes in the treem number

Decision Tree (DT)

Proof:

1 1

There is a

that for any output a function ,

such that with probability 1

.

The algorithm run in ( , , , )

randomized algorithm

f DT h

h approximates f

poly n m log

Exacting

depth of the tree f fd T T

ˆ ˆ

2 fz z d T

kf k f

ˆ ˆz z zf g

:g approcximates f

choose:

21 12 2d

1 1

2 2 fd T ˆ zround g

Final Result

1

There is a

that for any

output a function , such that

with probability 1

.

The algor

ithm run in ( , , log )

randomized algorithm

f DT

h

poly n m

h f