Communication Complexity. The dialogues of Alice and Bob...
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1 Communication Complexity The dialogues of Alice and Bob...
2 Alice and Bob make a date
3 Alice and Bob make a date Are you free on Friday?
4 Alice and Bob make a date Are you free on Friday? No, have to work at the Krusty Krab.
5 Alice and Bob make a date Are you free on Friday? No, have to work at the Krusty Krab. How about Saturday?
6 Alice and Bob make a date Are you free on Friday? No, have to work at the Krusty Krab. How about Saturday? No, have to work at the Krusty Krab.
7 Alice and Bob make a date Are you free on Friday? No, have to work at the Krusty Krab. How about Saturday? No, have to work at the Krusty Krab....
8 Alice and Bob make a date Are you free on Friday? No, have to work at the Krusty Krab. How about Saturday? No, have to work at the Krusty Krab.... How many s can it take to set a date?
9 Measures of Communication We want to quantify the amount of communication sent back and forth.
10 Measures of Communication We want to quantify the amount of communication sent back and forth. Several ways to do this: number of s, total number of characters, volume of breath...
11 Measures of Communication We want to quantify the amount of communication sent back and forth. Several ways to do this: number of s, total number of characters, volume of breath... Being computer scientists, we will use bits.
12 Alice s schedule Alice s schedule for the week: Mon Tue... Morn Aftn Eve Morn Aftn Eve... Free Busy Busy Busy Free Free...
13 Alice s schedule Alice s schedule for the week: Mon Tue... Morn Aftn Eve Morn Aftn Eve... Free Busy Busy Busy Free Free... We can convert this into a string of bits where 0=Busy, 1=Free
14 Alice s schedule Alice s schedule for the week: Mon Tue... Morn Aftn Eve Morn Aftn Eve... Free Busy Busy Busy Free Free... We can convert this into a string of bits where 0=Busy, 1=Free Any 21 bit string is a valid input.
15 Set Intersection Alice s schedule for the week: Mon Tue... Morn Aftn Eve Morn Aftn Eve Bob s schedule for the week: Mon Tue... Morn Aftn Eve Morn Aftn Eve
16 Set Intersection Alice s schedule for the week: Bob s schedule for the week:
17 Set Intersection Alice s schedule for the week: Bob s schedule for the week: Q: Is there a position where both strings have a 1?
18 Set Intersection Alice s schedule for the week: Bob s schedule for the week: Q: Is there a position where both strings have a 1? Known as the set intersection problem.
19 Set Intersection In general, Alice and Bob will each hold a n bit binary string.
20 Set Intersection In general, Alice and Bob will each hold a n bit binary string. Again, the task is to decide if these strings have a common position with a 1 or not.
21 Set Intersection In general, Alice and Bob will each hold a n bit binary string. Again, the task is to decide if these strings have a common position with a 1 or not. Notice the problem can always be solved with n+1 bits of communication---alice can send her entire input to Bob, he can produce the correct output.
22 Set Intersection In general, Alice and Bob will each hold a n bit binary string. Again, the task is to decide if these strings have a common position with a 1 or not. Notice the problem can always be solved with n+1 bits of communication---alice can send her entire input to Bob, he can produce the correct output. This is known as the trivial protocol.
23 Try It!?
24 Communication Matrix
25 010 Communication Matrix
26 Communication Protocol We also assume the communication is in bits
27 Communication Protocol We also assume the communication is in bits
28 Communication Protocol We also assume the communication is in bits
29 Communication Protocol We also assume the communication is in bits
30 Communication Protocol We also assume the communication is in bits
31 Communication Protocol We also assume the communication is in bits
32 Communication Protocol We also assume the communication is in bits
33 Communication Protocol We also assume the communication is in bits. 1 The last bit of the protocol is the answer!
34 Communication Protocol The first bit Alice sends depends only on her input. She knows nothing about Bob s schedule
35 Communication Protocol The first bit Alice sends depends only on her input. She knows nothing about Bob s schedule
36 Communication Protocol The first bit Alice sends depends only on her input. She knows nothing about Bob s schedule
37 Communication Protocol The first bit Alice sends depends only on her input. She knows nothing about Bob s schedule
38 Communication Protocol The first bit Alice sends depends only on her input. She knows nothing about Bob s schedule
39 Communication Protocol The first bit Alice sends depends only on her input. She knows nothing about Bob s schedule We can divide Alice s inputs into two groups: -those where she first says 1 -those where she first says 0
40 Communication Protocol The first bit Bob sends depends only on his input and the bit sent by Alice
41 Communication Protocol The first bit Bob sends depends only on his input and the bit sent by Alice
42 Communication Protocol The first bit Bob sends depends only on his input and the bit sent by Alice Conclusion: Bob groups his inputs into two sets, conditioned on Alice s message.
43 Communication Matrix
44 Communication Matrix 0 1
45 Communication Matrix
46 Communication Matrix
47 Communication Matrix
48 Observations Protocol partitions matrix into rectangles. With each additional bit of communication, a rectangle can be split into two. After c bits of communication, at most 2 c rectangles. For protocol to be correct, all inputs lying in the same rectangle must have the same output value. We say that such a rectangle is monochromatic.
49 Diagonal property of rectangles (x,y) (x,y ) If (x,y) and (x,y ) lie in the same rectangle...
50 Diagonal property of rectangles (x,y) (x,y ) (x,y) (x,y ) So must (x,y ) and (x,y).
51 010 Going back to SI
52 010 Going back to SI
53 010 Going back to SI
54 010 Going back to SI
55 010 Going back to SI
56 010 Going back to SI
57 010 Going back to SI
58 010 Going back to SI
59 Lower bound for SI Notice that all the entries on the anti-diagonal are 0. All the entries below the anti-diagonal are 1. No two anti-diagonal entries can be in the same monochromatic rectangle!
60 Lower bound for SI Notice that all the entries on the anti-diagonal are 0. All the entries below the anti-diagonal are 1. No two anti-diagonal entries can be in the same monochromatic rectangle! This means at least 2 n many monochromatic rectangles are needed---n bits of communication.
61 Recap We just showed a lower bound---any protocol with less than n bits of communication will incorrectly answer the set intersection problem on some input. For set intersection, the trivial protocol is optimal!
62 Bob goes to the moon
63 Bob goes to the moon Alice wishes to send a huge file M to Bob
64 Bob goes to the moon Alice wishes to send a huge file M to Bob
65 Bob goes to the moon Is the file corrupted along the way? Alice wishes to send a huge file M to Bob
66 Bob goes to the moon Is the file corrupted along the way? Alice wishes to send a huge file M to Bob Bob receives some file M. Does M=M?
67 Bob goes to the moon M M Bob could just send M back. This is very costly.
68 Checking Equality More abstractly, the problem is the following: Alice has an n bit string M, Bob has an n bit string M. They want to answer 1 if M=M and 0 otherwise.
69 010 Communication Matrix
70 010 Communication Matrix This is known as the identity matrix
71 Communication Matrix
72 Communication Matrix By the diagonal property of rectangles, no two pairs (x,x) and (y,y) can be in the same monochromatic rectangle.
73 Communication Matrix By the diagonal property of rectangles, no two pairs (x,x) and (y,y) can be in the same monochromatic rectangle. Again we need 2 n many monochromatic rectangles, and so n bits of communication.
74 Communication Matrix By the diagonal property of rectangles, no two pairs (x,x) and (y,y) can be in the same monochromatic rectangle. Again we need 2 n many monochromatic rectangles, and so n bits of communication. Is communication complexity always so boring?
75 Randomized Model Not if we allow randomness! Alice and Bob are each given the same book of random numbers. This is known as shared randomness. For every input, they must output the correct answer with probability 90%.
76 Randomized Protocol M M
77 Randomized Protocol Alice looks at string R=first n bits of book. M M
78 Randomized Protocol Alice looks at string R=first n bits of book. Then she computes a = M R = i M i R i mod 2. M M
79 Randomized Protocol Alice looks at string R=first n bits of book. Bob checks if a=a, where a = M R = i M ir i mod 2. M M
80 Randomized Protocol Alice looks at string R=first n bits of book. Bob checks if a=a, where a = M R = i M ir i mod 2. M M If they are equal he outputs 1, otherwise outputs 0.
81 Correctness of Protocol If M=M then we will always have a=a. No mistakes.
82 Correctness of Protocol If M=M then we will always have a=a. No mistakes. If M = M what is the probability that a=a?
83 Correctness of Protocol If M=M then we will always have a=a. No mistakes. If M = M what is the probability that a=a? As M = M, they must differ in some position. Say it is the first one.
84 Correctness of Protocol M=0 M =1
85 Correctness of Protocol M=0 M =1 Consider the random strings R in pairs (0r, 1r).
86 Correctness of Protocol M=0 M =1 Consider the random strings R in pairs (0r, 1r). Now notice that M 0r = M 1r M 0r = M 1r
87 Correctness of Protocol M=0 M =1 Consider the random strings R in pairs (0r, 1r). Now notice that M 0r = M 1r M 0r = M 1r So either
88 Correctness of Protocol M=0 M =1 Consider the random strings R in pairs (0r, 1r). Now notice that M 0r = M 1r M 0r = M 1r So either M 0r = M 0r
89 Correctness of Protocol M=0 M =1 Consider the random strings R in pairs (0r, 1r). Now notice that M 0r = M 1r M 0r = M 1r So either M 0r = M 0r or
90 Correctness of Protocol M=0 M =1 Consider the random strings R in pairs (0r, 1r). Now notice that M 0r = M 1r M 0r = M 1r So either M 0r = M 0r or M 1r = M 1r
91 Correctness of Protocol M=0 M =1 Consider the random strings R in pairs (0r, 1r). Now notice that M 0r = M 1r M 0r = M 1r So either M 0r = M 0r or M 1r = M 1r For half the random strings, we get different answers.
92 Randomized Protocol Protocol Recap:
93 Randomized Protocol Protocol Recap: Alice computes a = M R = i M i R i mod 2.
94 Randomized Protocol Protocol Recap: Alice computes a = M R = i M i R i mod 2. Bob computes a = M R = i M ir i mod 2.
95 Randomized Protocol Protocol Recap: Alice computes a = M R = i M i R i mod 2. Bob computes a = M R = i M ir i mod 2. If M=M then a=a for any R.
96 Randomized Protocol Protocol Recap: Alice computes a = M R = i M i R i mod 2. Bob computes a = M R = i M ir i mod 2. If M=M then a=a for any R. If M = M then a=a for at most half of the R s.
97 Deterministic vs. Random Here we have a case where with randomness we can provably do better than without. Deterministically, we must send the entire input, n bits! But if Alice and Bob share a book of randomness, constant communication suffices.
98 On the model You may object that sharing a book of randomness is a lame assumption.
99 On the model You may object that sharing a book of randomness is a lame assumption. Now we will see how to do away with this assumption and just use private randomness.
100 On the model You may object that sharing a book of randomness is a lame assumption. Now we will see how to do away with this assumption and just use private randomness. As a result, the amount of communication needed will go from constant to order log n.
101 Equality with random primes Idea: With her own (private) randomness Alice chooses a random prime number between 1 and n 2.
102 Equality with random primes Idea: With her own (private) randomness Alice chooses a random prime number between and. 1 n 2 She sends Bob p and a=m mod p.
103 Equality with random primes Idea: With her own (private) randomness Alice chooses a random prime number between and. 1 n 2 She sends Bob p and a=m mod p. This is known as a fingerprint of M.
104 Equality with random primes Idea: With her own (private) randomness Alice chooses a random prime number between and. 1 n 2 She sends Bob p and a=m mod p. This is known as a fingerprint of M. How many bits does this take?
105 Equality with random primes Idea: With her own (private) randomness Alice chooses a random prime number between and. 1 n 2 She sends Bob p and a=m mod p. This is known as a fingerprint of M. How many bits does this take? Again Bob checks if a=m mod p.
106 Correctness of the protocol If M=M then again clearly a=a.
107 Correctness of the protocol If M=M then again clearly a=a. Suppose that M = M yet it happens that a=a : M mod p = M mod p
108 Correctness of the protocol If M=M then again clearly a=a. Suppose that M = M yet it happens that a=a : M mod p = M mod p This means that M M mod p =0.
109 Correctness of the protocol If M=M then again clearly a=a. Suppose that M = M yet it happens that a=a : M mod p = M mod p This means that M M mod p =0. How many prime divisors can M-M have?
110 There are lots of primes! Prime number theorem (1896): Asymptotically, the number of primes less than N is N ln(n).
111 There are lots of primes! Prime number theorem (1896): Asymptotically, the number of primes less than N is N ln(n). In other words, if we choose a random n bit number, the probability it is prime is at least 1 n.
112 There are lots of primes! Prime number theorem (1896): Asymptotically, the number of primes less than N is N ln(n). In other words, if we choose a random n bit number, the probability it is prime is at least 1 n. There are about n 2 primes less than 2 ln(n) n 2 and only n of them are bad.
113 Questions? How do you choose a random prime efficiently? Can you find an n-bit prime efficiently deterministically? Can every protocol with shared randomness be modified to use private randomness in this way?
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