1 communication complexity מגישים: 317735678 מיכאל זמור: 01762926/2 אבי מינץ:...

1

Communication Complexity

:מגישים

:מיכאל זמור 317735678

:אבי מינץ 01762926/2

02801532/9ערן מנצור: ת.ז.

2


The Model

• 2 Computers : A,B

• All calculations for A are free

• All calculations for B are free

• Algorithm costs are measured by cost of communications.

• Cost is measured per bits

3


The Model


14, 29,53,28,284,348 39, 67,98,22,35,253

(48) 01110110

(98) 00100110

A B

4

Example

• Input

A has array of n numbers

B has array of n numbers

• Output

median of all 2n numbers

• Heuristic

numbers are O(log n) bits long

5

Naïve Algorithm

1. A sends all of his numbers to B

2. B calculates median of all 2n numbers

• Cost– Each number is O(log n) bits– N numbers are sent– Total cost is O(n*log n) bits

6


Naive algorithm


14, 29,53,28,284,348 39, 67,98,22,35,253

(284) 00011111

(348) 001110101(53) 101011

(29) 10111

(14) 001110

A B

Total cost is O(n*log n) bits

7


Naive algorithm


14, 29,53,28,284,348 39, 67,98,22,35,25314, 29,53,28,284,348

B calculate the median

A B

8

Better algorithm1. A sorts his array and sends his median ( )

to B

2. B sorts his array and sends his median ( ) to A.

3. Exercise :

define : r = real median

b = MAX{ }

s = MIN { }

prove :

AM

BM

BA M,M

BA M,Mb r s

9


Better algorithm


14,

28,29,53,284,348,50022, 35,39,67,98,253,300

A B

B sort his array and find his median

A sort his array and find his median

10


Better algorithm


14,

28,29,53,284,348,50022, 35,39,67,98,253,300

A B

67

53

B send his median to A

A send his median to B

11

4. If = then return ( = )

5. If > then A throws top (n/2) elements

B throws low (n/2) elements

6. And vice versa

• We reduces the size of the problem by half

7. Back to step 1, until size of arrays = 1

AM

AMAM

BMBMBM

12


Better algorithm


,53,284,348,500

A B

67>53

Then B throws the big half of his array

53<67

Then A throws the small half of his array

14, 2

8,29

22, 35,39,67

,98,253,300,98,253,300

13


Better algorithm


,53,284,348,500

A B

14, 2

8,29

22, 35,39,67

,98,253,300,98,253,300

Since equal amount of members were thrown from two sides of the median

So, new median not changed.

14


Better algorithm


,53,284,348,500

A B

14, 2

8,29

22, 35,39,67

,98,253,300,98,253,300

We will repeat this algorithm until the size of the array will be 1, while every loop the array is cut in half, and log n bits transferred

Total cost is O (Log^2(n)) bits

15

Calculation of cost

• Each number is O(log n) bits

• Each step 2 numbers are sent

• log(n) steps

• Total cost is O (Log^2(n)) bits

16


A B

17 22 27 31 4513 18 20 25 3213 18 20 25 32

- A calculates his median

- B calculates his median

another improvement

17

Two assumptions

1) The requested median is between the A’s median and B’s median.

13 18 20 25 32 17 22 27 31 45

13 17 18 20 22 25 27 31 32 45

A’s median B’s median

The requested median between A’s and

B’s median

* (we have an even amount of numbers, so we choose the the 6 th place of the 10 numbers)

18

Two assumptions2) In binary representation we divide the medians to

common segment and different segment.

Note: Sometimes there is no common segment, but if there is common for A’s and B’s medians then its common to the request median.

Decimal binary representation

27 - 1 1 0 1 1

25 - 1 1 0 0 1

20 - 1 0 1 0 0Common

segment

Different

segment

19



A B

After those assumptions we continue….

Every side send each bit of his median (start from MSB bit)

27 = 1 1 0 1 1 b 20 = 10100 b

17 22 27 31 4513 18 20 25 3213 18 20 25 321 b

1 b

20



A B

Every side compares the bit he gets with the bit of his median.

If equal, those bits are in the “common segment” (see slide 17) and this common to the request median then every side continue send the next bit of his median.

20 = 10100 b

1 b

27 = 1 1 0 1 1 b

1 b

21



A B

If not equal then A sees that his median is bigger, so he throws the bigger half of his array,

B sees that his median is smaller then A’s median, so he throws the smaller half of his array

27 = 1 1 0 1 1 b

0 b

20 = 10100 b

1 b

22



20 25 32

A B

13 1

8

17 22 27

31 45,

•After throwing half of his array, every side calculates his new median.

•Every side send bits of the new median, starting from the bit that there was different in the previous iteration ( and not from the MSB bit).

•Every side continues the algorithm.

23



20 25 32

A B

17 22 27

•The algorithm stops when one of two events occur.

1) the arrays of A and B contain one element (each).

2) all bits of the medians were sent.

24



20 25 32

A B

17 22 27

•If arrays contain one element so this element is the median.

•If all bits were sent so the medians are equal and this is the request median.

25

Complexity of algorithm

Every iteration one of this events occur

Bit is send to the other sides.(log n bits can be sent from every side)

The array become shorter by half.

(the array can reduce by half , log n times)

So, sum of bits can be sent limited by O(log n).

26

Communication complexityThe previous subject talked about problem of finding median of array that was divided to two parts.

Now we consider a new problem : Each side has a number and we want to know if the numbers are equal.


A B

X Y?

X=Y

27

Communication complexityNaïve algorithm

A send X to B.

B compares X to Y and return yes/no

X is logX bits long so cost is logX


A B

X YX

28

Communication complexityNew random algorithm - GLOBAL CC


A B

X Y?

X=Y

In this algorithm we have a random number R, that can be changed and known for both sides.

29

Communication complexityGLOBAL CC


A B

X Y

Lets define inner product of A and B:

A[0…..n-1], B[0…..n-1]

1

0

**n

iii BABA

30



A B

X Y

A: calculates

(X*R)mod 2

B: calculates

(Y*R)mod 2

31



A B

X Y

bX=(X*R)mod 2 bY=(Y*R)mod 2If X=Y then always bX= bY

If x ≠ Y then Prob(bX= bY)=1/2

32



A B

X Y

bX=(X*R)mod 2 bY=(Y*R)mod 2

Every side sends his result (b) and compares the results.

If the results are not equal so the numbers are not equal.

bX

bY

33



A B

X Y

bX=(X*Rnew)mod 2 bY=(Y*Rnew)mod 2

We choose a new R and repeat the algorithm.

We do so C times.

34



A B

X Y

If X=Y then all C times bX = bY.

If X≠Y the probability that all C times bX = bY is 2-c.

35

Communication complexityComplexity of global CC


A B

X Y

Each time every side transfers bX/Y by length of 1 bit.

There are C times so every side transfers C bits.

Complexity = O ( C )

36

Communication complexityAnother random algorithm


A B

X Y

In this algorithm every side constructs a polynom from his number according to these steps:

Every number consist of n bits.X[an-1, an-2,…..,a1,a0] Y[an-1, an-2,…..,a1,a0]

37

Communication complexity


A B

X Y

Polynom of side A will be:

PolyA(T)= an-1*Tn-1+ an-2*Tn-2+…..+ a0*T0

X[an-1, an-2,…..,a1,a0] Y[bn-1, bn-2,…..,b1,b0]

Step 1:

38



A B

X Y

Like side A polynom B will be:

Polyb(T)= bn-1*Tn-1+ bn-2*Tn-2+…..+ b0*T0

X[an-1, an-2,…..,a1,a0] Y[bn-1, bn-2,…..,b1,b0]

Step 1:

39



A B

X Y

X[an-1, an-2,…..,a1,a0] Y[bn-1, bn-2,…..,b1,b0]

Step 2:

A chooses random prime p : n2 <p < n3

A sends p to B

40



A B

X Y

Step 3:

B chooses 1≤R ≤ p-1 and calculates (polyb(R) mod p)

Remind: Polyb(R)= bn-1*Rn-1+ bn-2*Rn-2+…..+ b0*R0

B sends (R, (polyb(R) mod p) ) to A.

41



A B

X Y

Step 4:

A calculates polya(R) mod p

A compares (polya(R) mod p) with (polyb(R) mod p)

42



A B

X Y

Analysis of algorithm

If polya(R) polyb(R)

Then X Y

43



A B

X Y


But if polya(R) = polyb(R) there is some probability that X Y.

We will calculate F:

F=prob(X Y and polya(R) = polyb(R))

44



A B

X Y


polya(R) = polyb)R) polya(R) - polyb)R)=0

F=prob[X Y and polya(R) = polyb)R(]= prob[X Y and polya(R)-polyb)R(=0].

polya(R)-polyb)R(=0 is a polynom of degree n it has n-1 roots.

45



A B

X Y


Because there are n-1 R’s s.t. polya(R)-polyb)R(=0.

And 1<=R<=p-1.

So the probability to choose one of those R’s is: (n-1)/p.

46



A B

X Y


We saw that n2 <p < n3

So: n-1/p>(n-1)/ n21/n

F=prob(X Y and polya(R) = polyb(R))<1/n

The probability of mistake is:1/n

47



A B

X Y


Communication complexity:

The sides transferred p and R only:

n2 <p < n3 |p|=O(log n) And R<p

So the total complexity is : O(log n)

48



A B

X Y

CC and Global CC

What is the complexity relation between CC and GCC

Note: CC is the last alogorithm we had seen.

49



A B

X Y

solutionA and B have a deterministic Turing machine which generates (deterministically) K random r’s.

(while K= )2n

50



A B

X Y

solutionDefine R to be the chain of K’s random r’s

2*nrR i1r 2r ir 2n

r

R

51



A B

X Y

Now, instead of A sending r to B, A only has to send

index in R.

The index can be sent with O(log( ) bits

=O(log n)

sri '2n

52



A B

X Y

Claim:

For each subset of R there is in the subset that is never wrong for each (x,y).

ir

53



A B

X Y

Proof:

54



A B

X Y

Proof:

55



A B

X Y

So:

0-1 wrong)isr that no is r thereprob(-1

g)never wron is that (

rprob

56



A B

X Y

Conclusion:

)GCCn) O(log(CC

indexsending GCC

running

1 communication complexity מגישים: 317735678 מיכאל זמור: 01762926/2 אבי מינץ:...

Documents

b slide

model communication

array of n numbers b

median slide

real median b

poly b r slide

b total cost

n2 elements b