Data Streams, Dyck Languages, and Detecting Dubious Data Structures

Transcription

1 Data Streams, Dyck Languages, and Detecting Dubious Data Structures Amit Chakrabarti! Graham Cormode! Ranganath Kondapally! Andrew McGregor! Dartmouth College AT&T Research Labs Dartmouth College University of Massachusetts, Amherst

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5 At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted

6 5 MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5)

7 5 3 MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3)

8 extract min! 5 3 MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3)

9 3 5 MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3) ext(3)

10 3 5 6 MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3) ext(3) ins(6)

11 MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3) ext(3) ins(6) ins(7)

12 extract min! MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3) ext(3) ins(6) ins(7)

13 MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3) ext(3) ins(6) ins(7) ext(5)

14 extract min! MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3) ext(3) ins(6) ins(7) ext(5)

15 653 7 MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3) ext(3) ins(6) ins(7) ext(5) ext(6)

16 extract min! MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3) ext(3) ins(6) ins(7) ext(5) ext(6)

17 5376 MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3) ext(3) ins(6) ins(7) ext(5) ext(6) ext(7)

18 5376 MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3) ext(3) ins(6) ins(7) ext(5) ext(6) ext(7)? Challenge: Without remembering all the interaction, can you verify the priority queue performed correctly?

19 5376 MEMORY At each step, user inserts a value into the memory or asks that the smallest currently stored value is extracted ins(5) ins(3) ext(3) ins(6) ins(7) ext(5) ext(6) ext(7)? Challenge: Without remembering all the interaction, can you verify the priority queue performed correctly? Motivation: Memory checking useful when using cheap commodity hardware. [Blum, Evans, Gemmell, Kannan, Naor 94]

20 PQ Language Problem

21 PQ Language Problem Let PQ be set of legitimate transcripts of a priority queue that starts and ends empty.

22 PQ Language Problem Let PQ be set of legitimate transcripts of a priority queue that starts and ends empty. ins(5), ins(3), ext(3), ins(7), ext(5), ext(7) PQ

23 PQ Language Problem Let PQ be set of legitimate transcripts of a priority queue that starts and ends empty. ins(5), ins(3), ext(3), ins(7), ext(5), ext(7) PQ ins(5), ext(3), ins(5), ins(7), ext(7), ext(5) PQ

24 PQ Language Problem Let PQ be set of legitimate transcripts of a priority queue that starts and ends empty. ins(5), ins(3), ext(3), ins(7), ext(5), ext(7) PQ ins(5), ext(3), ins(5), ins(7), ext(7), ext(5) PQ PQ Problem: Given streaming access to length N transcript, determine if it s in PQ using o(n) space.

25 PQ Language Problem Let PQ be set of legitimate transcripts of a priority queue that starts and ends empty. ins(5), ins(3), ext(3), ins(7), ext(5), ext(7) PQ ins(5), ext(3), ins(5), ins(7), ext(7), ext(5) PQ PQ Problem: Given streaming access to length N transcript, determine if it s in PQ using o(n) space. Previous results: If each extract is annotated with insert time, Õ( N) space suffices. [Chu, Kannan, McGregor 07] ins(5), ins(3), ext(3,2), ins(7), ext(5,1), ext(7,4) PQ+

26 PQ Language Problem Let PQ be set of legitimate transcripts of a priority queue that starts and ends empty. ins(5), ins(3), ext(3), ins(7), ext(5), ext(7) PQ ins(5), ext(3), ins(5), ins(7), ext(7), ext(5) PQ PQ Problem: Given streaming access to length N transcript, determine if it s in PQ using o(n) space. Previous results: If each extract is annotated with insert time, Õ( N) space suffices. [Chu, Kannan, McGregor 07] ins(5), ins(3), ext(3,2), ins(7), ext(5,1), ext(7,4) PQ+? Big Question: Is annotation necessary for sub-linear space?

27 Dyck Language Problem [Magniez, Mathieu, Nayak 10]

28 Dyck Language Problem [Magniez, Mathieu, Nayak 10] DYCK2 is the set of strings of balanced brackets when there are two different types of brackets: ((([])()[])) DYCK 2 ([([]])[])) DYCK2

29 Dyck Language Problem [Magniez, Mathieu, Nayak 10] DYCK2 is the set of strings of balanced brackets when there are two different types of brackets: ((([])()[])) DYCK 2 ([([]])[])) DYCK2 DYCK Problem: Given streaming access to length N string, determine if it s in DYCK2 using o(n) space.

30 Dyck Language Problem [Magniez, Mathieu, Nayak 10] DYCK2 is the set of strings of balanced brackets when there are two different types of brackets: ((([])()[])) DYCK 2 ([([]])[])) DYCK2 DYCK Problem: Given streaming access to length N string, determine if it s in DYCK2 using o(n) space. Previous results: Given one pass, Õ( N) space suffices. If you re allowed one forward pass followed by a reverse pass, Õ(log N) space suffices.

31 Dyck Language Problem [Magniez, Mathieu, Nayak 10] DYCK2 is the set of strings of balanced brackets when there are two different types of brackets: ((([])()[])) DYCK 2 ([([]])[])) DYCK2 DYCK Problem: Given streaming access to length N string, determine if it s in DYCK2 using o(n) space. Previous results: Given one pass, Õ( N) space suffices. If you re allowed one forward pass followed by a reverse pass, Õ(log N) space suffices.? Big Question: Does Õ(log N) space suffice if we re only allowed multiple forward passes?

32 Outline

33 Outline I. Priority Queues and Memory Checking Algorithms Õ( N) space algorithm for PQ with no annotations! Extensions to stacks, double-ended queues, etc.

34 Outline I. Priority Queues and Memory Checking Algorithms Õ( N) space algorithm for PQ with no annotations! Extensions to stacks, double-ended queues, etc. II. Multipass Stream Lower Bounds Any constant pass algorithm for DYCK2 or PQ that only uses forward passes requires Ω( N) space

35 Outline I. Priority Queues and Memory Checking Algorithms Õ( N) space algorithm for PQ with no annotations! Extensions to stacks, double-ended queues, etc. II. Multipass Stream Lower Bounds Any constant pass algorithm for DYCK2 or PQ that only uses forward passes requires Ω( N) space III. Information Complexity Trade-offs for Augmented Index Even multi-round protocol leak information.

36 I. Memory Checking II. Lower Bounds III. Augmented Indexing

37 I. Memory Checking

38 PQ Algorithm

39 PQ Algorithm Thm: There exists a O( N log N) space algorithm with O(log N) amortized update time for recognizing PQ.

40 PQ Algorithm Thm: There exists a O( N log N) space algorithm with O(log N) amortized update time for recognizing PQ. Prelim: Easy to check that set of values inserted equals set of values extracted using fingerprinting: f S (x) = a S(x a) mod p

41 PQ Algorithm Thm: There exists a O( N log N) space algorithm with O(log N) amortized update time for recognizing PQ. Prelim: Easy to check that set of values inserted equals set of values extracted using fingerprinting: f S (x) = a S(x a) mod p! For this talk: Assume inserted elements are distinct and that inserts come before their corresponding extract. I.e., we re trying to identify the following bad pattern: ins(u)... ext(v)... ext(u) for some u < v

42 Epochs and Local Bad Patterns...

43 Epochs and Local Bad Patterns... Increasing Value Increasing Time

44 Epochs and Local Bad Patterns... Increasing Value Increasing Time Split length N sequence into N epochs of length N

45 Epochs and Local Bad Patterns... Epoch-1 Epoch-2 Epoch-3 Epoch-4 Epoch-5 Epoch-6 Increasing Value Increasing Time Split length N sequence into N epochs of length N

46 Epochs and Local Bad Patterns... Epoch-1 Epoch-2 Epoch-3 Epoch-4 Epoch-5 Epoch-6 Increasing Value Increasing Time Split length N sequence into N epochs of length N Defn: Bad pattern ins(u)... ext(v)... ext(u) is local if ins(u) and ext(v) occur in same epoch and long-range otherwise.

47 Epochs and Local Bad Patterns... Epoch-1 Epoch-2 Epoch-3 Epoch-4 Epoch-5 Epoch-6 Increasing Value Increasing Time Split length N sequence into N epochs of length N Defn: Bad pattern ins(u)... ext(v)... ext(u) is local if ins(u) and ext(v) occur in same epoch and long-range otherwise. Using O( N) space, we can buffer each epoch and check for local bad patterns.

48 Catching Long-Range Bad Patterns... 1/2

49 Catching Long-Range Bad Patterns... 1/2 f(1) f(2) Defn: Let ft(i) be maximum value extracted between end of i-th epoch and present time t.

50 Catching Long-Range Bad Patterns... 1/2 f(1) f(2) Defn: Let ft(i) be maximum value extracted between end of i-th epoch and present time t.

51 Catching Long-Range Bad Patterns... 1/2 f(1) f(2) f(3) Defn: Let ft(i) be maximum value extracted between end of i-th epoch and present time t.

52 Catching Long-Range Bad Patterns... 1/2 f(1) f(2) f(3) Defn: Let ft(i) be maximum value extracted between end of i-th epoch and present time t.

53 Catching Long-Range Bad Patterns... 1/2 f(1) f(2) f(3) Defn: Let ft(i) be maximum value extracted between end of i-th epoch and present time t.

54 Catching Long-Range Bad Patterns... 1/2 f(1) f(2) f(3) Defn: Let ft(i) be maximum value extracted between end of i-th epoch and present time t. Defn: Each insert or extract is adopted by k-th epoch where k = min{j : f(j) u} where we assume f(current epoch)=0.

55 Catching Long-Range Bad Patterns... 2/2

56 Catching Long-Range Bad Patterns... 2/2 Lemma: If ins(u)... ext(v)... ext(u) is a long-range bad pattern then ins(u) and ext(u) are adopted by different epochs.

57 Catching Long-Range Bad Patterns... 2/2 Lemma: If ins(u)... ext(v)... ext(u) is a long-range bad pattern then ins(u) and ext(u) are adopted by different epochs. Proof: i. Let ins(u) be adopted by k-th epoch.

58 Catching Long-Range Bad Patterns... 2/2 Lemma: If ins(u)... ext(v)... ext(u) is a long-range bad pattern then ins(u) and ext(u) are adopted by different epochs. Proof: i. Let ins(u) be adopted by k-th epoch. ii. After v is extracted f(k) v>u and hence ext(u) will be adopted by k -th epoch for some k >k.

59 Catching Long-Range Bad Patterns... 2/2 Lemma: If ins(u)... ext(v)... ext(u) is a long-range bad pattern then ins(u) and ext(u) are adopted by different epochs. Proof: i. Let ins(u) be adopted by k-th epoch. ii. After v is extracted f(k) v>u and hence ext(u) will be adopted by k -th epoch for some k >k. Lemma: If there are no bad patterns, every ins(u) and ext(u) pair get adopted by the same epoch.

60 Catching Long-Range Bad Patterns... 2/2 Lemma: If ins(u)... ext(v)... ext(u) is a long-range bad pattern then ins(u) and ext(u) are adopted by different epochs. Proof: i. Let ins(u) be adopted by k-th epoch. ii. After v is extracted f(k) v>u and hence ext(u) will be adopted by k -th epoch for some k >k. Lemma: If there are no bad patterns, every ins(u) and ext(u) pair get adopted by the same epoch. Algorithm: Using fingerprinting to check: {(u,k) : ins(u) adopted by k} = {(u,k) : ext(u) adopted by k}.

61 Finishing Up

62 Finishing Up Some steps to removing assumptions: i. Within buffered epoch, rearrange terms such that all inserts follow a series of increasing extracts: ensures all extracts are adopted by previous epochs. ii. Keep track of number of times epoch k adopts ins(u) while f(k)<u and while f(k)=u separately and whether epoch k has ever adopted more ext(u) s than ins(u) s.

63 Finishing Up Some steps to removing assumptions: i. Within buffered epoch, rearrange terms such that all inserts follow a series of increasing extracts: ensures all extracts are adopted by previous epochs. ii. Keep track of number of times epoch k adopts ins(u) while f(k)<u and while f(k)=u separately and whether epoch k has ever adopted more ext(u) s than ins(u) s. Thm: There exists a O( N log N) space algorithm with O(log N) amortized update time for recognizing PQ.

64 Finishing Up Some steps to removing assumptions: i. Within buffered epoch, rearrange terms such that all inserts follow a series of increasing extracts: ensures all extracts are adopted by previous epochs. ii. Keep track of number of times epoch k adopts ins(u) while f(k)<u and while f(k)=u separately and whether epoch k has ever adopted more ext(u) s than ins(u) s. Thm: There exists a O( N log N) space algorithm with O(log N) amortized update time for recognizing PQ. Extensions: Sub-linear space streaming recognition of other data structures like stacks, double-ended queues...

65 I. Memory Checking II. Lower Bounds III. Augmented Indexing

66 II. Lower Bounds

67 Augmented Index

68 Augmented Index Many space lower bounds in data stream model are based on reductions from communication complexity.

69 Augmented Index x {0,1} n y {0,1} k-1, k [n], c {0,1}, Many space lower bounds in data stream model are based on reductions from communication complexity. Augmented Index: Alice has x {0,1} n and Bob has a prefix y {0,1} k-1 of x and c {0,1}. Bob wants to check if c=xk.

70 Augmented Index x {0,1} n y {0,1} k-1, k [n], c {0,1}, Many space lower bounds in data stream model are based on reductions from communication complexity. Augmented Index: Alice has x {0,1} n and Bob has a prefix y {0,1} k-1 of x and c {0,1}. Bob wants to check if c=xk. Thm: Any 1/3-error, one-way protocol from Alice to Bob for AIn requires Ω(n) bits sent. [Miltersen et al. JCSS 98]

71 Augmented Index x {0,1} n y {0,1} k-1, k [n], c {0,1}, Many space lower bounds in data stream model are based on reductions from communication complexity. Augmented Index: Alice has x {0,1} n and Bob has a prefix y {0,1} k-1 of x and c {0,1}. Bob wants to check if c=xk. Thm: Any 1/3-error, one-way protocol from Alice to Bob for AIn requires Ω(n) bits sent. [Miltersen et al. JCSS 98] Our main result concerns multi-way protocols but we ll cover the relevance to DYCK2 and PQ first...

72 Multi-player Augmented Index

73 Multi-player Augmented Index x 1 y 1, k 1, c 1 x 2 y 2, k 2, c 2 x 3 y 3, k 3, c 3 We now have 2m players A1,..., Am, B1,..., Bm where each Ai and Bi have an instance (x i,k i,c i ) of AIn

74 Multi-player Augmented Index x 1 y 1, k 1, c 1 x 2 y 2, k 2, c 2 x 3 y 3, k 3, c 3 x 3 x 2 x 1 We now have 2m players A1,..., Am, B1,..., Bm where each Ai and Bi have an instance (x i,k i,c i ) of AIn Want to determine if any of the AI instances are false using private messages communicated in the order A1 B1 A2 B2... Am Bm Am Am-1... A1

75 Multi-player Augmented Index x 1 y 1, k 1, c 1 x 2 y 2, k 2, c 2 x 3 y 3, k 3, c 3 x 3 x 2 x 1 We now have 2m players A1,..., Am, B1,..., Bm where each Ai and Bi have an instance (x i,k i,c i ) of AIn Want to determine if any of the AI instances are false using private messages communicated in the order A1 B1 A2 B2... Am Bm Am Am-1... A1 Thm: Any 1/3-error, p-round protocol for MULTI-AIm,n needs ps=ω(min m,n) where s is max message length.

76 Reduction to Dyck

77 Reduction to Dyck ([( ( [ (

78 Reduction to Dyck ([( )][( ( [ ( ) ( ][

79 Reduction to Dyck ([( )][( [[( ( [ ( ) ( ][ ( [ [

80 Reduction to Dyck ([( )][( [[( )])([( ( [ [ ) ( ] [ )( ( [ ( ) ( ][

81 Reduction to Dyck ([( )][( [[( )])([( (([ [ ( ( ( [ [ ) ( ] [ )( ( [ ( ) ( ][

82 Reduction to Dyck ([( )][( [[( )])([( (([ ][ [ ( ( ][ ( [ [ ) ( ] [ )( ( [ ( ) ( ][

83 Reduction to Dyck ([( )][( [[( )])([( (([ ][ ])) [ ( ( ][ ] ) ) ( [ [ ) ( ] [ )( ( [ ( ) ( ][

84 Reduction to Dyck ([( )][( [[( )])([( (([ ][ ])) )]] [ ( ( ][ ] ) ) ( [ [ ) ( ] [ )( ) ] ] ( [ ( ) ( ][

85 Reduction to Dyck ([( )][( [[( )])([( (([ ][ ])) )]] )]) [ ( ( ][ ] ) ) ( [ [ ) ( ] [ )( ) ] ] ( [ ( ) ( ][ ) ] )

86 Reduction to Dyck ([( )][( [[( )])([( (([ ][ ])) )]] )]) [ ( ( ][ ] ) ) ( [ [ ) ( ] [ )( ) ] ] ( [ ( ) ( ][ Ascension Problem [Magniez, Mathieu, Nayak 10] ) ] )

87 Reduction to Dyck

88 Reduction to Dyck Thm: A constant-pass, algorithm for DYCK2 that fail with probability at most 1/3 requires Ω( N) space.

89 Reduction to Dyck Thm: A constant-pass, algorithm for DYCK2 that fail with probability at most 1/3 requires Ω( N) space. Proof: i. Let A be a p-pass stream algorithm using s space.

90 Reduction to Dyck Thm: A constant-pass, algorithm for DYCK2 that fail with probability at most 1/3 requires Ω( N) space. Proof: i. Let A be a p-pass stream algorithm using s space. ii. Use A to construct a p-round protocol for MULTI-AI N, N where max message is s-bits: Each player simulates A on it s part of input using Magniez et al. reduction and forwards memory state to next player.

91 Reduction to Dyck Thm: A constant-pass, algorithm for DYCK2 that fail with probability at most 1/3 requires Ω( N) space. Proof: i. Let A be a p-pass stream algorithm using s space. ii. Use A to construct a p-round protocol for MULTI-AI N, N where max message is s-bits: Each player simulates A on it s part of input using Magniez et al. reduction and forwards memory state to next player. iii. Therefore s is Ω( N) as required.

92 Lower Bounds Summary

93 Lower Bounds Summary Thm: Any constant pass algorithm for recognizing PQ or DYCK2 requires a Ω( N) space.

94 Lower Bounds Summary Thm: Any constant pass algorithm for recognizing PQ or DYCK2 requires a Ω( N) space. Consequences: i. Multiple forward passes have no significant advantage for recognizing the languages considered. ii. One forward pass + one reverse pass is exponentially more powerful than two forward passes.

95 I. Memory Checking II. Lower Bounds III. Augmented Indexing

96 III. Augmented Indexing

97 Information Complexity [Chakrabarti, Shi, Wirth, Yao 01]

98 Information Complexity [Chakrabarti, Shi, Wirth, Yao 01] Entropy and Mutual Information: H(X ) = Σ Pr[X = x]lgpr[x = x] H(X Y ) = Σ Pr[X = x, Y = y]lgpr[x = x Y = y] I (X ; Y ) = H(X ) H(X Y ) = H(Y ) H(Y X ) I (X ; Y Z) = H(X Z) H(X Y, Z)

99 Information Complexity [Chakrabarti, Shi, Wirth, Yao 01] Entropy and Mutual Information: H(X ) = Σ Pr[X = x]lgpr[x = x] H(X Y ) = Σ Pr[X = x, Y = y]lgpr[x = x Y = y] I (X ; Y ) = H(X ) H(X Y ) = H(Y ) H(Y X ) I (X ; Y Z) = H(X Z) H(X Y, Z)

100 Information Complexity [Chakrabarti, Shi, Wirth, Yao 01] Entropy and Mutual Information: H(X ) = Σ Pr[X = x]lgpr[x = x] H(X Y ) = Σ Pr[X = x, Y = y]lgpr[x = x Y = y] I (X ; Y ) = H(X ) H(X Y ) = H(Y ) H(Y X ) I (X ; Y Z) = H(X Z) H(X Y, Z) Information cost method: Consider mutual information between random input for a communication problem and the communication transcript: I (transcript; input) length of transcript

101 Information Complexity [Chakrabarti, Shi, Wirth, Yao 01] Entropy and Mutual Information: H(X ) = Σ Pr[X = x]lgpr[x = x] H(X Y ) = Σ Pr[X = x, Y = y]lgpr[x = x Y = y] I (X ; Y ) = H(X ) H(X Y ) = H(Y ) H(Y X ) I (X ; Y Z) = H(X Z) H(X Y, Z) Information cost method: Consider mutual information between random input for a communication problem and the communication transcript: I (transcript; input) length of transcript

102 Information Complexity [Chakrabarti, Shi, Wirth, Yao 01] Entropy and Mutual Information: H(X ) = Σ Pr[X = x]lgpr[x = x] H(X Y ) = Σ Pr[X = x, Y = y]lgpr[x = x Y = y] I (X ; Y ) = H(X ) H(X Y ) = H(Y ) H(Y X ) I (X ; Y Z) = H(X Z) H(X Y, Z) Information cost method: Consider mutual information between random input for a communication problem and the communication transcript: I (transcript; input) length of transcript Can restrict to partial transcript and subsets of input: useful for proving direct-sum arguments.

103 Information Complexity of AIn

104 Information Complexity of AIn Defn: Let P be a protocol for AIn using public random string R. Let T be the transcript and (X, K, C)~ξ. Define icost A ξ (P) = I (T : X K, C, R) icost B ξ (P) = I (T : K, C X, R)

105 Information Complexity of AIn Defn: Let P be a protocol for AIn using public random string R. Let T be the transcript and (X, K, C)~ξ. Define icost A ξ (P) = I (T : X K, C, R) icost B ξ (P) = I (T : K, C X, R) Thm: Let P be a randomized protocol for AIn with error 1/3 under the uniform distribution μ. Then, icost A µ 0 (P) = Ω(n) or icost B µ 0 (P) = Ω(1) where μ0 is μ conditioned on XK=C.

106 MULTI-AIm,n versus AIn

107 MULTI-AIm,n versus AIn Defn: Let Q be a protocol for MULTI-AIm,n using public random string R. Let T be transcript and (X i,k i,c i )i [m]~ξ. icost ξ (Q) = I(T m :K 1,C 1,...,K m,c m X 1,...,X m,r) where Tm is the set of messages sent by Bm.

108 MULTI-AIm,n versus AIn Defn: Let Q be a protocol for MULTI-AIm,n using public random string R. Let T be transcript and (X i,k i,c i )i [m]~ξ. icost ξ (Q) = I(T m :K 1,C 1,...,K m,c m X 1,...,X m,r) where Tm is the set of messages sent by Bm. Thm (Direct Sum): If there exists a p-round, s-bit, ε-error protocol Q for MULTI-AIm,n then there exists a p-round, ε-error randomized protocol P for AIn where i. Alice sends at most ps bits ii. m icost B µ 0 (P) icost µ m 0 (Q)

109 Putting it all together...

110 Putting it all together... Thm: Any p-round, s-bit, 1/3-error protocol Q for MULTI- AIm,n requires ps=ω(min m,n).

111 Putting it all together... Thm: Any p-round, s-bit, 1/3-error protocol Q for MULTI- AIm,n requires ps=ω(min m,n). Proof: i. By direct sum theorem, there exists ε-error, p-pass protocol P for AIn such that: p s icost µ m(q) m icost B µ 0 0 (P) p s icost A µ 0 (P)

112 Putting it all together... Thm: Any p-round, s-bit, 1/3-error protocol Q for MULTI- AIm,n requires ps=ω(min m,n). Proof: i. By direct sum theorem, there exists ε-error, p-pass protocol P for AIn such that: p s icost µ m(q) m icost B µ 0 0 (P) p s icost A µ 0 (P) ii. By information complexity of AIn max(m icost B µ 0 (P), icost A µ 0 (P)) = Ω(min(m, n))

113 Summary Memory Checking: Sub-linear space recognition of various datastructure transcript languages is possible without annotation! Theory of Stream Computation: Forward + reverse pass can be much more useful than many forward passes! Further Work: Annotations, stream language recognition,... Thanks!

114