Chapter 4 - Tail inequalities

Transcription

1 Chapter 4 - Tail inequalities Igor Gonopolskiy & Nimrod Milo BGU November 10, 2009 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

2 Outline 1 Definitions 2 Chernoff Bound Upper deviation Lower deviation 3 The wiring problem Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

3 Outline 1 Definitions 2 Chernoff Bound Upper deviation Lower deviation 3 The wiring problem Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

4 Experiment types Bernoulli trails Let X 1,, X n be { A set of n independent Bernoulli trails such that, for 1 with probability p 1 i n X i =. 0 otherwise Let X = n i=1 X i then X is said to have the binomial distribution. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

5 Experiment types Bernoulli trails Let X 1,, X n be { A set of n independent Bernoulli trails such that, for 1 with probability p 1 i n X i =. 0 otherwise Let X = n i=1 X i then X is said to have the binomial distribution. Poisson trails The general case of the above distribution is when each trail X i has a success probability p i. Let X 1,, X n be { A set of n independent Bernoulli trails such that, for 1 with probability pi 1 i n X i =. 0 otherwise Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

6 Type of questions we would like to answer We consider two questions regarding the deviation of X from its expectation µ = n i=1 p i. for a real number δ > 0: Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

7 Type of questions we would like to answer We consider two questions regarding the deviation of X from its expectation µ = n i=1 p i. for a real number δ > 0: what is the probability that X exceeds (1 + δ)µ? - useful in analyzing an algorithm, showing that the chance it fails to achieve a solution is small. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

8 Type of questions we would like to answer We consider two questions regarding the deviation of X from its expectation µ = n i=1 p i. for a real number δ > 0: what is the probability that X exceeds (1 + δ)µ? - useful in analyzing an algorithm, showing that the chance it fails to achieve a solution is small. how large must δ be in order that the tail probability is less then a prescribed value ɛ? - useful when designing an algorithm. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

9 Outline 1 Definitions 2 Chernoff Bound Upper deviation Lower deviation 3 The wiring problem Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

10 Chernoff Bound Theorem (4.1) Let X 1,, X n be independent Poisson trails such that, for 1 i n, Pr[X i = 1] = p i, Then for X = n i=1 X i, µ = E[X ] = n i=1 p i and any δ > 0, [ e δ ] µ Pr[X > (1 + δ)µ] < (1 + δ) (1+δ) Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

11 Chernoff Bound - Proof Pr[X > (1 + δ)µ] = Pr[exp(tX ) > exp(t(1 + δ)µ]) < E[exp(tx)] exp(t(1 + δ)µ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

12 Chernoff Bound - Proof Applying X exp(tx ) to both sides Pr[X > (1 + δ)µ] = Pr[exp(tX ) > exp(t(1 + δ)µ]) < E[exp(tx)] exp(t(1 + δ)µ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

13 Chernoff Bound - Proof Applying X exp(tx ) to both sides Pr[X > (1 + δ)µ] = Pr[exp(tX ) > exp(t(1 + δ)µ]) < E[exp(tx)] exp(t(1 + δ)µ Applying the Markov inequality Pr[X a] E[X ] a to the right-hand side. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

14 Chernoff Bound - Proof Pr[X > (1 + δ)µ] < E[exp(tx)] exp(t(1 + δ)µ = Πn i=1 E[exp(tX i)] exp(t(1 + δ)µ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

15 Chernoff Bound - Proof What we had before Pr[X > (1 + δ)µ] < E[exp(tx)] exp(t(1 + δ)µ = Πn i=1 E[exp(tX i)] exp(t(1 + δ)µ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

16 Chernoff Bound - Proof What we had before Pr[X > (1 + δ)µ] < E[exp(tx)] exp(t(1 + δ)µ = Πn i=1 E[exp(tX i)] exp(t(1 + δ)µ n E[exp(tX )] = E[exp(t X i )] = E[Πi=1exp(tX n i )] and because X i i=1 are independent exp(tx i ) are also independent. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

17 Chernoff Bound - Proof Pr[X > (1 + δ)µ] < Πn i=1 E[exp(tX i)] exp(t(1 + δ)µ = Πn i=1 [p ie t + 1 p i ] exp(t(1 + δ)µ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

18 Chernoff Bound - Proof What we had before Pr[X > (1 + δ)µ] < Πn i=1 E[exp(tX i)] exp(t(1 + δ)µ = Πn i=1 [p ie t + 1 p i ] exp(t(1 + δ)µ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

19 Chernoff Bound - Proof Pr[X > (1 + δ)µ] < Πn i=1 E[exp(tX i)] exp(t(1 + δ)µ = Πn i=1 [p ie t + 1 p i ] exp(t(1 + δ)µ The random variable e tx i the value 1 with 1 p i. assumes the value e t with probability p i and Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

20 Chernoff Bound - Proof Π n i=1 E[exp(tX i)] exp(t(1 + δ)µ = Πn i=1 [p ie t + 1 p i ] exp(t(1 + δ)µ = Πn i=1 [1 + p i(e t 1)] exp(t(1 + δ)µ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

21 Chernoff Bound - Proof Pr[X > (1 + δ)µ] < Πn i=1 [1 + p i(e t 1)] exp(t(1 + δ)µ = Πn i=1 exp(p i(e t 1)) exp(t(1 + δ)µ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

22 Chernoff Bound - Proof What we had before Pr[X > (1 + δ)µ] < Πn i=1 [1 + p i(e t 1)] exp(t(1 + δ)µ = Πn i=1 exp(p i(e t 1)) exp(t(1 + δ)µ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

23 Chernoff Bound - Proof Pr[X > (1 + δ)µ] < Πn i=1 [1 + p i(e t 1)] exp(t(1 + δ)µ = Πn i=1 exp(p i(e t 1)) exp(t(1 + δ)µ we use the inequality 1 + x < e x with x = p i (e t 1). Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

24 Chernoff Bound - Proof Π n i=1 [1 + p i(e t 1)] exp(t(1 + δ)µ = Πn i=1 exp(p i(e t 1)) exp(t(1 + δ)µ = exp( n i=1 p i(e t 1)) exp(t(1 + δ)µ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

25 Chernoff Bound - Proof Π n i=1 [1 + p i(e t 1)] exp(t(1 + δ)µ = exp( n i=1 p i(e t 1)) exp(t(1 + δ)µ = exp((et 1)µ) exp(t(1 + δ)µ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

26 Chernoff Bound - Proof Until now we have proved that: Pr[X > (1 + δ)µ] < exp((et 1)µ) exp(t(1 + δ)µ for any possible t to get the best value of t we can differentiate the last expression with respect to t and set to zero. the value that we get is t = ln(1 + δ) which is positive for δ > 0. applying the value to the above formula we obtain our theorem: [ e δ Pr[X > (1 + δ)µ] < (1 + δ) (1+δ) ] µ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

27 Chernoff Bound - Proof Until now we have proved that: Pr[X > (1 + δ)µ] < exp((et 1)µ) exp(t(1 + δ)µ for any possible t to get the best value of t we can differentiate the last expression with respect to t and set to zero. the value that we get is t = ln(1 + δ) which is positive for δ > 0. applying the value to the above formula we obtain our theorem: [ e δ Pr[X > (1 + δ)µ] < (1 + δ) (1+δ) Definition 4.1 we define an upper tail bound function for the sum of Poisson trails. F + (µ, δ) = [ e δ (1 + δ) 1+δ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28 ] µ ] µ

28 Example example The Aardvarks win each game with probability of 1/3. assuming that all the games are independent we want to calculate the chance that they have a winning season (n games in a season) Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

29 Example example The Aardvarks win each game with probability of 1/3. assuming that all the games are independent we want to calculate the chance that they have a winning season (n games in a season) solution Let X i be 1 if the Aardvarks win the ith game and 0 otherwise. Let Y n = n i=1 X i. by applying 4.1 to Y n we get: Pr[Y n > n/2] < F + (n/3, 1/2) < (0.965) n Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

30 Chernoff bound - below expectation We now want to consider deviation of X below its expectation µ. Theorem (4.6) Let X 1,, X n be independent Poisson trails such that, for 1 i n, Pr[X i = 1] = p i, Then for X = n i=1 X i, µ = E[X ] = n i=1 p i and any 0 < δ 1, Pr[X < (1 δ)µ] < e µδ2 2 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

31 Chernoff Bound Proof We will not prove this version of the Chernoff bound because the proof is similar to the upper tail with some small differences. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

32 Chernoff Bound Proof We will not prove this version of the Chernoff bound because the proof is similar to the upper tail with some small differences. Definition 4.2 we define an lower tail bound function for the sum of Poisson trails. F (µ, δ) = µδ2 e 2 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

33 Example example The Aardvarks hire a new coach, and now the chance to win each game is what is the probability that they suffer a losing season? Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

34 Example example The Aardvarks hire a new coach, and now the chance to win each game is what is the probability that they suffer a losing season? solution Let X i be 1 if the Aardvarks win the ith game and 0 otherwise. Let Y n = n i=1 X i. by applying 4.2 to Y n we get: Pr[Y n < n/2] < F (0.75n, 1/3) < (0.9592) n Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

35 answering the second type of questions Until now we have answered the first kind of questions. to answer the second type: how large must δ be in order that the tail probability is less then a prescribed value ɛ? We make the following definitions Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

36 answering the second type of questions Until now we have answered the first kind of questions. to answer the second type: how large must δ be in order that the tail probability is less then a prescribed value ɛ? We make the following definitions Definition 4.7 for any positive µ and ɛ, Let + (µ, ɛ) be the value of δ that satisfies: F + (µ, + (µ, ɛ)) = ɛ In other words we want a δ = + (µ, ɛ) deviation from µ suffices to keep Pr[X > (1 + δ)µ] below ɛ Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

37 answering the second type of questions Until now we have answered the first kind of questions. to answer the second type: how large must δ be in order that the tail probability is less then a prescribed value ɛ? We make the following definitions Definition 4.7 for any positive µ and ɛ, Let + (µ, ɛ) be the value of δ that satisfies: F + (µ, + (µ, ɛ)) = ɛ In other words we want a δ = + (µ, ɛ) deviation from µ suffices to keep Pr[X > (1 + δ)µ] below ɛ Definition 4.8 (µ, ɛ) can be defined in the same way. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

38 Outline 1 Definitions 2 Chernoff Bound Upper deviation Lower deviation 3 The wiring problem Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

39 The wiring problem - definition The input of the problem is: Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

40 The wiring problem - definition The input of the problem is: a n n matrix of gates (total of n gates) arranged at regularly spaced points in the plane. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

41 The wiring problem - definition The input of the problem is: a n n matrix of gates (total of n gates) arranged at regularly spaced points in the plane Figure: a 9 gate example 3 3 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

42 The wiring problem - definition The input of the problem is: a n n matrix of gates (total of n gates) arranged at regularly spaced points in the plane. a set of nets - two gates connected by a wire. all wires run above gates in Manhattan form (parallel to the axis) Figure: a 9 gate example 3 3 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

43 The wiring problem - definition The input of the problem is: a n n matrix of gates (total of n gates) arranged at regularly spaced points in the plane. a set of nets - two gates connected by a wire. all wires run above gates in Manhattan form (parallel to the axis) < 2, 7 >, < 8, 3 >, < 4, 9 >, < 7, 3 > Figure: a 9 gate example 3 3 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

44 the wiring problem general solution A solution S is an assignment of global routes to all the given nets such that each wire has at most one turn. also we would like to limit the maximum number of wires that pass in a boundary b between two gates in the array. Let denote the limit as w. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

45 the wiring problem general solution A solution S is an assignment of global routes to all the given nets such that each wire has at most one turn. also we would like to limit the maximum number of wires that pass in a boundary b between two gates in the array. Let denote the limit as w. definition for a boundary b between two gates in the array let w S (b) be the number of wires that pass through b in a solution S. Let W S = max b W s (b) be the maximum number of wires through any boundary in the solution S. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

46 the wiring problem general solution A solution S is an assignment of global routes to all the given nets such that each wire has at most one turn. also we would like to limit the maximum number of wires that pass in a boundary b between two gates in the array. Let denote the limit as w. definition for a boundary b between two gates in the array let w S (b) be the number of wires that pass through b in a solution S. Let W S = max b W s (b) be the maximum number of wires through any boundary in the solution S. observation If we can minimize the value of w S then we can decide on the feasibility of the problem. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

47 example < 2, 7 >, < 8, 3 >, < 4, 9 >, < 7, 3 > Figure: a 9 gate example 3 3 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

48 example < 2, 7 >, < 8, 3 >, < 4, 9 >, < 7, 3 > Figure: a 9 gate example 3 3 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

49 example < 2, 7 >, < 8, 3 >, < 4, 9 >, < 7, 3 > Figure: a 9 gate example 3 3 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

50 example < 2, 7 >, < 8, 3 >, < 4, 9 >, < 7, 3 > Figure: a 9 gate example 3 3 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

51 example b < 2, 7 >, < 8, 3 >, < 4, 9 >, < 7, 3 > Figure: a 9 gate example 3 3 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

52 example b < 2, 7 >, < 8, 3 >, < 4, 9 >, < 7, 3 > w S (b ) = 1 Figure: a 9 gate example 3 3 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

53 example b b < 2, 7 >, < 8, 3 >, < 4, 9 >, < 7, 3 > w S (b ) = 1 Figure: a 9 gate example 3 3 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

54 example b b < 2, 7 >, < 8, 3 >, < 4, 9 >, < 7, 3 > w S (b ) = 1 w S (b ) = 2 Figure: a 9 gate example 3 3 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

55 zero-one Integer programming In general, a zero-one integer linear programming problem can be defined as: Optimizing a function w = c 1 x 1 + c 2 x 2 s.t. some constraints like: a 1 1x 1 + a 1 2x 2 b 1 a 2 1x 1 + a 2 2x 2 b 2 and so on... also we demand that all the variables x i are integers from a range {a..b} Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

56 Casting the Wiring problem to integer programming definitions routes for a net i we use two variables x i0 and x i1 to indicate which one of the routes will be used (horizontal first vs. vertical first). thus x i0 will be 1 if the wire goes horizontally first starting from the left end-point of net i, and 0 otherwise. boundary For each boundary b we define: T by = {i net ipasses through b if x iy = 1} where y 0, 1 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

57 Applying the IP programming to our problem With the last definitions our IP can be expressed as: minimize function w where x i0 + x i1 w i T b0 i T b1 s.t. where x i0 + x i1 = 1( net i) x i0, x i1 {0, 1}( net i) Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

58 Applying the IP programming to our problem With the last definitions our IP can be expressed as: minimize function w where x i0 + x i1 w i T b0 i T b1 s.t. where x i0 + x i1 = 1( net i) x i0, x i1 {0, 1}( net i) Question so what is the problem? Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

59 Applying the IP programming to our problem With the last definitions our IP can be expressed as: minimize function w where x i0 + x i1 w i T b0 i T b1 s.t. where x i0 + x i1 = 1( net i) x i0, x i1 {0, 1}( net i) Question so what is the problem? Answer The IP problem in NP hard Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

60 The solution - relaxation plus randomization relaxation Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

61 The solution - relaxation plus randomization relaxation we want to allow a less restrictive set of routes Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

62 The solution - relaxation plus randomization relaxation we want to allow a less restrictive set of routes allow x i0 and x i1 to get real numbers between 0 and 1. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

63 The solution - relaxation plus randomization relaxation we want to allow a less restrictive set of routes allow x i0 and x i1 to get real numbers between 0 and 1. let xˆ i0 and xˆ i1 for 1 i n be the solutions given by the linear program. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

64 The solution - relaxation plus randomization relaxation we want to allow a less restrictive set of routes allow x i0 and x i1 to get real numbers between 0 and 1. let xˆ i0 and xˆ i1 for 1 i n be the solutions given by the linear program. we denote ŵ the solution of the relaxation and w 0 the result of the IP. it is clear that w 0 ŵ. notice that each value can be fractional values so the solution may not be feasible. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

65 The solution - relaxation plus randomization relaxation we want to allow a less restrictive set of routes allow x i0 and x i1 to get real numbers between 0 and 1. let xˆ i0 and xˆ i1 for 1 i n be the solutions given by the linear program. we denote ŵ the solution of the relaxation and w 0 the result of the IP. it is clear that w 0 ŵ. notice that each value can be fractional values so the solution may not be feasible. rounding Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

66 The solution - relaxation plus randomization relaxation we want to allow a less restrictive set of routes allow x i0 and x i1 to get real numbers between 0 and 1. let xˆ i0 and xˆ i1 for 1 i n be the solutions given by the linear program. we denote ŵ the solution of the relaxation and w 0 the result of the IP. it is clear that w 0 ŵ. notice that each value can be fractional values so the solution may not be feasible. rounding we will use randomized rounding Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

67 The solution - relaxation plus randomization relaxation we want to allow a less restrictive set of routes allow x i0 and x i1 to get real numbers between 0 and 1. let xˆ i0 and xˆ i1 for 1 i n be the solutions given by the linear program. we denote ŵ the solution of the relaxation and w 0 the result of the IP. it is clear that w 0 ŵ. notice that each value can be fractional values so the solution may not be feasible. rounding we will use randomized rounding independently for each i, set x i0 to 1 and x i1 to 0 with probability xˆ i0 ; otherwise set x i0 to 0 and x i1 to 1. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

68 Correctness of the algorithm Theorem (4.8) Let ɛ be a real number such that 0 < ɛ < 1. Then with probability 1 ɛ the randomized algorithm provide a solution S satisfies: w S ŵ(1 + + (ŵ, ɛ/2n)) w 0 (1 + + (w 0, ɛ/2n)) Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

69 Correctness of the algorithm Theorem (4.8) Let ɛ be a real number such that 0 < ɛ < 1. Then with probability 1 ɛ the randomized algorithm provide a solution S satisfies: w S ŵ(1 + + (ŵ, ɛ/2n)) w 0 (1 + + (w 0, ɛ/2n)) And now in words! with at least 1 ɛ, no boundary in the array has more than w 0 (1 + + (w 0, ɛ/2n)) (while w 0 is the optimal solution. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

70 proof We will prove for any boundary b that w(b) S > w 0 (1 + + (w 0, ɛ/2n)) is at most ɛ/2n. then since there are O(2n) boundaries we can sum the probability for w S. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

71 proof Consider a boundary b, since the solution of the LP satisfies the constraints, we have xˆ i0 + xˆ i1 ŵ (1) i T b0 i T b1 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

72 proof i T b0 xˆ i0 + i T b1 xˆ i1 ŵ (1) as before The number of wires passing through b in a solution S is: w(b) S = x i0 + x i1 (2) i T b0 i T b1 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

73 proof i T b0 xˆ i0 + w(b) S = i T b0 i T b1 xˆ i1 ŵ (1) x i0 + i T b1 x i1 (2) because x i0, x i1 are independent Poisson trails with probability of xˆ i0, xˆ i1 we apply the following on the second equation: E[w(b) S ] = E[ x i0 ] + E[ x i1 ] (3) i T b0 i T b1 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

74 proof i T b0 xˆ i0 + w(b) S = i T b0 i T b1 xˆ i1 ŵ (1) x i0 + i T b1 x i1 (2) E[w(b) S ] = E[ x i0 ] + E[ x i1 ] (3) i T b0 i T b1 notice that E[ x i0 ] = xˆ i0 (the same for x i1 ). E[w(b) S ] = xˆ i0 + xˆ i1 ŵ (4) i T b0 i T b1 Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

75 proof i T b0 xˆ i0 + w(b) S = i T b0 i T b1 xˆ i1 ŵ (1) x i0 + i T b1 x i1 (2) E[w(b) S ] = E[ x i0 ] + E[ x i1 ] (3) i T b0 i T b1 E[w(b) S ] = xˆ i0 + xˆ i1 ŵ (4) i T b0 By the definition of + (ŵ, ɛ/2n) we get i T b1 Pr[w(b) S > ŵ(1 + + (ŵ, ɛ/2n))] ɛ/2n Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

76 real time performance Question How good is the bound on + (ŵ, ɛ/2n)? Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28

77 real time performance Question How good is the bound on + (ŵ, ɛ/2n)? Answer It depends on the value of w 0. we can show that if w 0 = n c where c is a positive constant that we find a solution with an additive term that is vanishingly small as n grows. on the other end, if w 0 = 20 then w S = O(logn/loglogn). thus the algorithm is likely to perform well provided w 0 is not too small. Igor Gonopolskiy & Nimrod Milo (BGU) Chapter 4 November 10, / 28