Communication Complexity 317735678 017629262 028015329 1 Communication Complexity

Communication Complexity מגישים : מיכאל זמור 317735678 : 01762926/2 אבי מינץ : ערן מנצור : ת. ז 02801532/9. 1

Communication Complexity 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 2

Communication Complexity The Model 14, 29, 53, 284, 348 (98) 00100110 39, 67, 98, 22, 35, 253 (48) 01110110 B A Communication Complexity 3

Example • Input A has array of n numbers B has array of n numbers • Output median of all 2 n numbers • Heuristic numbers are O(log n) bits long 4

Naïve Algorithm 1. A sends all of his numbers to B 2. B calculates median of all 2 n numbers • Cost – Each number is O(log n) bits – N numbers are sent – Total cost is O(n*log n) bits 5

Communication Complexity Naive algorithm 14, 29, 53, 284, 348 (14) 001110 (53) 101011 (29) 10111 (348) 001110101 39, 67, 98, 22, 35, 253 (284) 00011111 B A Communication Complexity Total cost is O(n*log n) bits 6

Communication Complexity Naive algorithm 14, 29, 53, 28, 284, 348 39, 67, 98, 22, 35, 253 B calculate the median B A Communication Complexity 7

Better algorithm 1. 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 : 8 ) )

Communication Complexity Better algorithm A sort his array and find his median B sort his array and find his median 14, 28, 29, 53, 284, 348, 500 22, 35, 39, 67, 98, 253, 300 B A Communication Complexity 9

Communication Complexity Better algorithm B send his median to A A send his median to B 67 14, 28, 29, 53, 284, 348, 500 22, 35, 39, 67, 98, 253, 300 53 B A Communication Complexity 10

4. If 5. If = > then return (= ) 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 11

Communication Complexity 53<67 Better algorithm Then A throws the small half of his array 67>53 Then B throws the big half of his array , 53, 284, 348, 500 22, 35, 39, 67 14, 28, 29 3, 300 , 98, 25 B A Communication Complexity 12

Communication Complexity Better algorithm , 53, 284, 348, 500 Since equal amount of members were thrown from two sides of the median 22, 35, 39, 67 14, 28, 29 3, 300 , 98, 25 So, new median not changed. B A Communication Complexity 13

Communication Complexity Better algorithm , 53, 284, 348, 500 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 22, 35, 39, 67 14, 28, 29 3, 300 , 98, 25 Total cost is O (Log^2(n)) bits B A Communication Complexity 14

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 15

another improvement - A calculates his median - B calculates his median 13 18 20 25 32 17 22 27 31 45 B A Communication Complexity 16

Two assumptions 1) The requested median is between the A’s median and B’s median. 13 18 20 25 32 A’s median The requested median between A’s and B’s median 13 17 18 20 22 17 22 27 31 45 B’s median 25 27 31 32 45 17 * (we have an even amount of numbers, so we choose the 6 th place of the 10 numbers)

Two assumptions 2) 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 - 11011 25 - 11001 20 - 10100 Common Different segment 18

Communication Complexity After those assumptions we continue…. Every side send each bit of his median (start from MSB bit) 20 = 10100 b 27 = 1 1 0 1 1 b 1 b 13 18 20 25 32 1 b 17 22 27 31 45 B A Communication Complexity 19

Communication Complexity 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 27 = 1 1 0 1 1 b 1 b 1 b B A Communication Complexity 20

Communication Complexity 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 20 = 10100 b 27 = 1 1 0 1 1 b 1 b 0 b B A Communication Complexity 21

Communication Complexity • 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. 20 25 32 17 22 27 13 18 , 31 45 B A Communication Complexity 22

Communication Complexity • 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. 20 25 32 17 22 27 B A Communication Complexity 23

Communication Complexity • 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. 20 25 32 17 22 27 B A Communication Complexity 24

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). 25

Communication complexity The 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. X ? Y X=Y B A Communication Complexity 26

Communication complexity Naïve algorithm A send X to B. B compares X to Y and return yes/no X is log. X bits long so cost is log. X X X Y B A Communication Complexity 27

Communication complexity New random algorithm - GLOBAL CC In this algorithm we have a random number R, that can be changed and known for both sides. X ? Y X=Y B A Communication Complexity 28

Communication complexity GLOBAL CC Lets define inner product of A and B: A [0…. . n-1], B [0…. . n-1] Y X B A Communication Complexity 29

Communication complexity GLOBAL CC A: calculates B: calculates (X*R)mod 2 (Y*R)mod 2 Y X B A Communication Complexity 30

Communication complexity GLOBAL CC b. X=(X*R)mod 2 If X=Y then always b. X= b. Y If x ≠ Y then Prob(b. X= b. Y)=1/2 b. Y=(Y*R)mod 2 Y X B A Communication Complexity 31

Communication complexity GLOBAL CC Every side sends his result (b) and compares the results. If the results are not equal so the numbers are not equal. b. Y b. X=(X*R)mod 2 X b. Y=(Y*R)mod 2 Y B A Communication Complexity 32

Communication complexity GLOBAL CC b. Y=(Y*Rnew)mod 2 b. X=(X*Rnew)mod 2 We choose a new R and repeat the algorithm. X We do so C times. Y B A Communication Complexity 33

Communication complexity GLOBAL CC If X=Y then all C times b. X = b. Y. If X≠Y the probability that all C times b. X = b. Y is 2 -c. Y X B A Communication Complexity 34

Communication complexity Complexity of global CC Each time every side transfers b. X/Y by length of 1 bit. There are C times so every side transfers C bits. Complexity = O ( C ) Y X B A Communication Complexity 35

Communication complexity Another random algorithm 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, …. . , a 1, a 0] Y X B A Communication Complexity 36

Communication complexity Step 1: Polynom of side A will be: Poly. A(T)= an-1*Tn-1+ an-2*Tn-2+…. . + a 0*T 0 X [an-1, an-2, …. . , a 1, a 0] Y [bn-1, bn-2, …. . , b 1, b 0] Y X B A Communication Complexity 37

Communication complexity Step 1: Like side A polynom B will be: Polyb(T)= bn-1*Tn-1+ bn-2*Tn-2+…. . + b 0*T 0 X [an-1, an-2, …. . , a 1, a 0] Y [bn-1, bn-2, …. . , b 1, b 0] Y X B A Communication Complexity 38

Communication complexity Step 2: A chooses random prime p : n 2 <p < n 3 A sends p to B X [an-1, an-2, …. . , a 1, a 0] Y [bn-1, bn-2, …. . , b 1, b 0] Y X B A Communication Complexity 39

Communication complexity 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+…. . + b 0*R 0 B sends (R, (polyb(R) mod p) ) to A. Y X B A Communication Complexity 40

Communication complexity Step 4: A calculates polya(R) mod p A compares (polya(R) mod p) with (polyb(R) mod p) Y X B A Communication Complexity 41

Communication complexity Analysis of algorithm If polya(R) polyb(R) Then X Y Y X B A Communication Complexity 42

Communication complexity Analysis of algorithm 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)) Y X B A Communication Complexity 43

Communication complexity Analysis of algorithm 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. Y X B A Communication Complexity 44

Communication complexity Analysis of algorithm 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. Y X B A Communication Complexity 45

Communication complexity Analysis of algorithm We saw that n 2 <p < n 3 So: n-1/p>(n-1)/ n 2 1/n F=prob(X Y and polya(R) = polyb(R))<1/n The probability of mistake is: 1/n Y X B A Communication Complexity 46

Communication complexity Analysis of algorithm Communication complexity: The sides transferred p and R only: n 2 <p < n 3 |p|=O(log n) And R<p So the total complexity is : O(log n) X Y B A Communication Complexity 47

Communication complexity CC and Global CC What is the complexity relation between CC and GCC Note: CC is the last alogorithm we had seen. Y X B A Communication Complexity 48

Communication complexity solution A and B have a deterministic Turing machine which generates (deterministically) K random r’s. (while K= ) Y X B A Communication Complexity 49

Communication complexity solution Define R to be the chain of K’s random r’s X R Y B A Communication Complexity 50

Communication complexity 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) Y X B A Communication Complexity 51

Communication complexity Claim: For each subset of R there is wrong for each (x, y). in the subset that is never Y X B A Communication Complexity 52

Communication complexity Proof: Y X B A Communication Complexity 53

Communication complexity Proof: Y X B A Communication Complexity 54

Communication complexity So: Y X B A Communication Complexity 55

Communication complexity Conclusion: Y X B A Communication Complexity 56