Polynomial-Time Problems Lecture 4: The polynomial method Part II: - - PowerPoint PPT Presentation

Complexity Theory of Polynomial-Time Problems

Lecture 4: The polynomial method Part II: All-pairs shortest paths

Sebastian Krinninger

Overview on APSP Algorithms

• Floyd-Warshall algorithm: 𝑃(𝑜3)

Inserts one node at a time 𝑜 iterations, each taking time 𝑃(𝑜2)

• Faster algorithms for sparse graphs
• Directed graphs:
• Single-source shortest paths: 𝑃(𝑛 + 𝑜 log 𝑜) (Dijkstra with Fibonacci heap/Hollow heap)
• ⇒ All-pairs shortest paths: 𝑃(𝑛𝑜 + 𝑜2 log 𝑜), improved to 𝑃(𝑛𝑜 + 𝑜2 log log 𝑜) [Pettie 02]
• Undirected graphs:
• Single-source shortest paths: 𝑃(𝑛) [Thorup 97]
• ⇒ All-pairs shortest paths: 𝑃(𝑛𝑜)
• Pseudopolynomial algorithms
• Today: Fastest “general-purpose” algorithm

History of slightly subcubic algorithms

Running Time Author(s) Year(s) 𝑜3 Floyd, Warshall 1962 𝑜3/ log1/3 𝑜 Fredman 1975 𝑜3/ log1/2 𝑜 Dobosiewicz, Takaoka 1990, 1991 𝑜3/ log5/7 𝑜 Han 2004 𝑜3/ log 𝑜 Takaoka, Zwick, Chan 2004, 2005 𝑜3/ log5/4 𝑜 Han 2006 𝑜3/ log2 𝑜 Chan, Han/Takaoka 2007, 2012 𝑜3/2Ω log 𝑜 1/2 Williams 2014 Grows faster than any polylogarithmic factor

Problem definition: desired output

• Can create instances such that for every pair of nodes 𝑣, 𝑤 shortest

path from 𝑣 to 𝑤 consists of Ω(𝑜) nodes

• ⇒ Cannot output all shortest paths explicitly in time 𝑝(𝑜3)
• Distance matrix: output size 𝑜2
• Shortest path matrix 𝐓𝐐: output size 𝑜2

For every pair of nodes 𝑣, 𝑤, SP 𝑣, 𝑤 = next node on shortest path from 𝑣 to 𝑤

Machine model: Real RAM

Floyd-Warshall: 𝑃(𝑜3) with only additions and comparisons Ω(𝑜3) lower bound if only additions and comparisons allowed [Kerr 70] Real RAM:

• Additions and comparisons of reals: unit cost
• Other operations: logarithmic cost

Tools

Tool 1: Razborov-Smolensky

Represent AND of 𝑒 variables 𝑦1 ∧ ⋯ ∧ 𝑦𝑒 by low-degree polynomial

• Parameter 𝑟
• For every 𝑗 = 1, … , 𝑟, 𝑘 = 1, … , 𝑒: Set 𝑠

𝑗,𝑘 = 0 or 1 with probability 1 2

𝐵 𝑦1, … , 𝑦𝑒 =

𝑗=1 𝑟

1 ⊕

𝑘=1 𝑒

𝑠

𝑗,𝑘 ⋅ 𝑦𝑘 ⊕ 1

By distributive law, 𝐵 can be written as XOR of 1 + 𝑒 𝑟 monomials Lemma: Pr

𝑠𝑗,𝑘 𝐵 𝑦1, … , 𝑦𝑒 = 𝑦1 ∧ ⋯ ∧ 𝑦𝑒 ≥ 1 − 1 2𝑟

𝐵 𝑦1, … , 𝑦𝑒 =

𝑗=1 𝑟

1 ⊕

𝑘=1 𝑒

𝑠

𝑗,𝑘 ⋅ 𝑦𝑘 ⊕ 1

Lemma: Pr

𝑠𝑗,𝑘 𝐵 𝑦1, … , 𝑦𝑒 = 𝑦1 ∧ ⋯ ∧ 𝑦𝑒 ≥ 1 − 1 2𝑟

Proof: 𝑦1 ∧ ⋯ ∧ 𝑦𝑒 = 1: Each 𝑦𝑘 must be 1. Clearly, 𝐵 𝑦1, … , 𝑦𝑒 = 1 𝑦1 ∧ ⋯ ∧ 𝑦𝑒 = 0: First, fix some 𝑗 𝑇: subsets of 𝑦𝑘’s that are 0 𝑇′: subsets of 𝑦𝑘’s that are 0 and additionally 𝑠

𝑗,𝑘 = 1

Bad event: 𝑗-th component of outer AND is 1 ⇔ |𝑇′| is even Each subset of 𝑇 has same probability of being picked as 𝑇′ Pr |𝑇′| is odd = Pr |𝑇′| is even = 1 2 𝐵 𝑦1, … , 𝑦𝑒 = 1 only if bad event happens for each component of outer AND ⇒ Error probability ≤ 1

2𝑟

Observation: For every set 𝑇, #odd subsets = #even subsets (by binary encoding of the 2|𝑇| subsets)

Tool 2: Fast rectangular matrix multiplication

Theorem: There is an algorithm for multiplying an 𝑜 × 𝑜0.17 matrix with an 𝑜0.17 × 𝑜 matrix in time 𝑃(𝑜2 log2 𝑜). 𝑜0.1 𝑜 Also works for finite fields such as 𝐺2! × =

Fast evaluation of polynomial

Given: Polynomial 𝑄(𝑦[1], … , 𝑦[𝑒], 𝑧[1], … , 𝑧[𝑒]) over 𝐺2

• With 𝑛 ≤ 𝑜0.1 monomials
• Two sets of inputs:

𝑌 = 𝑦1, … , 𝑦𝑜 ⊆ 0,1 𝑒 𝑍 = 𝑧1, … , 𝑧𝑜 ⊆ 0,1 𝑒

(𝑦𝑗 = 𝑦𝑗[1], … , 𝑦𝑗[𝑒] ) (𝑧𝑘 = 𝑧𝑘[1], … , 𝑧𝑘[𝑒] )

Lemma: There is an algorithm for evaluating 𝑄 on all pairs 𝑦𝑗, 𝑧𝑘 ∈ 𝑌 × 𝑍 (simultaneously) in time 𝑃 𝑜2poly log 𝑜 .

Restrictions of monomials

Shape of polynomial 𝑄:

• 𝑄 = 𝑞1 + ⋯ + 𝑞𝑛
• each 𝑞𝑙 is a monomial

Define

• 𝑞𝑙|𝑦: restriction of 𝑙-th monomial to variables 𝑦[1], … , 𝑦[𝑒]
• 𝑞𝑙|𝑧: restriction of 𝑙-th monomial to variables 𝑧[1], … , 𝑧[𝑒]
• (empty product = 1)

“Inner product”

• 𝑄 = 𝑞1 𝑦 ⋅ 𝑞1 𝑧 + ⋯ + 𝑞𝑛 𝑦 ⋅ 𝑞𝑛 𝑧

Reduction to matrix multiplication

𝑦1 ⋮ 𝑦𝑜 × 𝑜0.1 monomials 𝑧1 … 𝑧𝑜 𝑜0.1 monomials

Entry 𝑗, 𝑙 : Evaluation of 𝑞𝑙|𝑦𝑗 Entry 𝑙, 𝑘 : Evaluation of 𝑞𝑙|𝑧𝑘

Result matrix 𝑆[𝑗, 𝑘]: Evaluation of 𝑄 under 𝑦𝑗 = 𝑦𝑗[1], … , 𝑦𝑗[𝑒] and 𝑧𝑘 = 𝑧𝑘[1], … , 𝑧𝑘[𝑒]

Tool 3: Union Bound and Chernoff Bound

Union Bound: Pr 𝐵 ∪ 𝐶 ≤ Pr 𝐵 + Pr 𝐶 (Variant of) Chernoff Bound: Let 𝑌1, … , 𝑌𝑙 be independent 0/1-valued random variables such that 0 < 𝐹 𝑌𝑗 < 1. Then, the random variable 𝑌 = 𝑗=1

𝑙

𝑌𝑗 satisfies: Pr 𝑌 < 1 − 𝜀 𝐹[𝑌] ≤ 𝑓−𝜀2𝐹[𝑌]/2 for every 0 ≤ 𝜀 ≤ 1

Solving the Problem

Min-plus matrix product

We give an algorithm for the following problem:

• Given: 𝑜 × 𝑒 integer matrix 𝐵 and 𝑒 × 𝑜 integer matrix 𝐶
• Output: 𝑜 × 𝑜 matrix 𝐷 such that

𝐷 𝑗, 𝑘 = min

𝑙∈{1,…,𝑒}(𝐵 𝑗, 𝑙 + 𝐶 𝑙, 𝑘 )

Matrix multiplication in min-plus semiring:

• min is addition
• + is multiplication
• 0 is 1-element
• ∞ is 0-element

𝑙∗ such that 𝐵 𝑗, 𝑙∗ + 𝐶 𝑙∗, 𝑘 = min

𝑙∈{1,…,𝑒}(𝐵 𝑗, 𝑙 + 𝐶 𝑙, 𝑘 ) is a witness for 𝑗, 𝑘

Tripartite graph for min-plus product

⋮ ⋮ ⋮ 𝑜 𝑜 𝑒

𝑥 𝑗, 𝑙 = 𝐵[𝑗, 𝑙] 𝑥 𝑙, 𝑘 = 𝐶[𝑙, 𝑘] (no edge if weight is ∞)

𝐷 𝑗, 𝑘 = min

1≤𝑙≤𝑒(𝐵 𝑗, 𝑙 + 𝐶 𝑙, 𝑘 )

• 1. Min-plus product = APSP in tripartite graph
• 2. If 𝐵 = 𝐶 = 𝐻: 𝐻 × 𝐻 = matrix of 2-hop distances

APSP and min-plus product are “equivalent”

In general: 𝐻𝑗 = matrix distances using exactly 𝑗 hops Distance matrix 𝐸: 𝐸 = 𝐽 + 𝐻 + 𝐻2 + ⋯ + 𝐻𝑜−1 = 𝐻 + 𝐽 𝑜−1

+ is entry-wise minimum Identity matrix 𝐽: 0 at diagonal, ∞ otherwise

Repeated squaring: Compute 𝐻 + 𝐽 2, 𝐻 + 𝐽 4, 𝐻 + 𝐽 8, …,

⇒ 𝑃(log 𝑜) min-plus products for distances, shortest paths through witnesses

Even stronger relationship known: Theorem: APSP on 𝑜 nodes can be solved in time 𝑃(𝑈 𝑜 ) if and only min-plus product on 𝑜 × 𝑜 matrices can be solved in time 𝑃(𝑈 𝑜 ).

Step 1: : Divide into subproblems

Overall algorithm

1. Set 𝑒 = 2 log 𝑜/100 2. Divide problem into

𝑜 𝑒 subproblems

3. Solve each subproblem in time 𝑃 𝑜2poly log 𝑜 4. Merge solutions in time 𝑃 𝑜3/𝑒 Total time: 𝑃 𝑜3 𝑒 poly log 𝑜 = 𝑜3/2Ω log 𝑜 1/2 For every 𝑙 = 1, … ,

𝑜 𝑒:

• Compute product 𝐷𝑙 of 𝐵𝑙 and 𝐶𝑙

(𝑜 × 𝑒 matrix with 𝑒 × 𝑜 matrix) Return: min 𝐷1, … , 𝐷𝑜/𝑒 (entry-wise minimum) 𝐵 𝐶

𝐵1 𝐵𝑜/𝑒 𝐶1 𝐶𝑜/𝑒

𝑒

Subproblem

We solve the following subproblem:

• Given: 𝑜 × 𝑒 matrix 𝐵 and 𝑒 × 𝑜 matrix 𝐶
• Output: 𝑜 × 𝑜 matrix 𝑋 of witnesses such that

𝑋 𝑗, 𝑘 = arg min

𝑙∈{1,…,𝑒}

(𝐵 𝑗, 𝑙 + 𝐶 𝑙, 𝑘 ) From witnesses in 𝑋 we can easily reconstruct values of min-plus product min

𝑙∈{1,…,𝑒} (𝐵 𝑗, 𝑙 + 𝐶 𝑙, 𝑘 ) in time 𝑃 𝑜2

Step 2: : Preprocess input of subproblem

Enforce unique minimum

For every entry 𝑗, 𝑙 of 𝐵: 𝐵∗ 𝑗, 𝑙 ∶= 𝐵 𝑗, 𝑙 ⋅ 𝑜 + 1 + 𝑙 For every entry 𝑙, 𝑘 of 𝐶 𝐶∗ 𝑙, 𝑘 ∶= 𝐶 𝑙, 𝑘 ⋅ 𝑜 + 1 Fix some pair 𝑗, 𝑘 and define 𝑙∗ as smallest 𝑙′ ∈ {1, … , 𝑒} such that 𝐵 𝑗, 𝑙′ + 𝐶 𝑙′, 𝑘 = min

𝑙∈{1,…,𝑒} 𝐵 𝑗, 𝑙 + 𝐶 𝑙, 𝑘

⇒ Work with 𝐵∗ and 𝐶∗ instead of 𝐵 and 𝐶 to ensure unique minima Running time: 𝑃(log 𝑜) additions per entry (add to itself for 𝑃(log 𝑜) times) ⇒ 𝑃 𝑜𝑒 log 𝑜

Claim: 𝑙∗ is unique minimum of 𝐵∗ 𝑗, 𝑙 + 𝐶∗ 𝑙, 𝑘 over 𝑙∗ ∈ {1, … , 𝑒}

Proof of Cla laim im: : 𝑙∗ is unique minimum of 𝐵∗ 𝑗, 𝑙 + 𝐶∗ 𝑙, 𝑘

• ver 𝑙∗ ∈ {1, … , 𝑒}

Let 𝑙 ≠ 𝑙∗. We show that 𝐵∗ 𝑗, 𝑙 + 𝐶∗ 𝑙, 𝑘 > 𝐵∗ 𝑗, 𝑙∗ + 𝐶∗ 𝑙, 𝑘

• r equivalently

(1) 𝐵 𝑗, 𝑙 + 𝐶 𝑙, 𝑘 ⋅ 𝑜 + 1 + 𝑙 > 𝐵 𝑗, 𝑙∗ + 𝐶 𝑙∗, 𝑘 ⋅ 𝑜 + 1 + 𝑙∗ Case 1: 𝐵 𝑗, 𝑙 + 𝐶 𝑙, 𝑘 = 𝐵 𝑗, 𝑙∗ + 𝐶 𝑙, 𝑘 Then 𝑙∗ < 𝑙 because 𝑙∗ is smallest index assuming min value (1) follows immediately Case 2: 𝐵 𝑗, 𝑙 + 𝐶 𝑙, 𝑘 > 𝐵 𝑗, 𝑙∗ + 𝐶 𝑙, 𝑘 ⇒ 𝐵 𝑗, 𝑙 + 𝐶 𝑙, 𝑘 ≥ 𝐵 𝑗, 𝑙∗ + 𝐶 𝑙, 𝑘 + 1 (integers!) ⇒ 𝐵 𝑗, 𝑙 + 𝐶 𝑙, 𝑘 ⋅ 𝑜 + 1 + 𝑙 ≥ 𝐵 𝑗, 𝑙∗ + 𝐶 𝑙∗, 𝑘 ⋅ 𝑜 + 1 + 𝑙 + 𝑜 + 1 > 𝐵 𝑗, 𝑙∗ + 𝐶 𝑙∗, 𝑘 ⋅ 𝑜 + 1 + 𝑙∗

Fredman’s trick: Get rid of weights

Construct 𝑜 × 𝑒2 matrix 𝐵′ and 𝑒2 × 𝑜 matrix 𝐶′

• 𝐵′ 𝑗, 𝑙, ℓ

≔ 𝐵∗ 𝑗, 𝑙 − 𝐵∗ 𝑗, ℓ

• 𝐶′

𝑙, ℓ , 𝑘 ≔ 𝐶∗ ℓ, 𝑘 − 𝐶∗ 𝑙, 𝑘 Idea: Compare alternatives 𝑙 and ℓ without taking sums Observation: 𝐵′ 𝑗, 𝑙, ℓ ≤ 𝐶′ 𝑙, ℓ , 𝑘 ⇔ 𝐵∗ 𝑗, 𝑙 + 𝐶∗ 𝑙, 𝑘 ≤ 𝐵∗ 𝑗, ℓ + 𝐶∗ ℓ, 𝑘

Fredman’s trick continued

For every pair 𝑙, ℓ sort set 𝑇𝑙,ℓ ≔ {𝐵′ 𝑗, 𝑙, ℓ , 𝐶′ 𝑙, ℓ , 𝑗 ∣ 𝑗 = 1, … , 𝑜} Breaking ties:

• Precedence of 𝐵′-entries over 𝐶′-entries
• Otherwise arbitrarily

Properties: 1. Entries of 𝐵′′ and 𝐶′′ from {1, … , 2𝑜} 2. Comparisons preserved:

𝐵′ 𝑗, 𝑙, ℓ ≤ 𝐶′ 𝑙, ℓ , 𝑘 iff 𝐵′′ 𝑗, 𝑙, ℓ ≤ 𝐶′′ 𝑙, ℓ , 𝑘

3. For every 𝑗, 𝑘 there is unique 𝑙∗ such that for all ℓ: 𝐵′′ 𝑗, 𝑙∗, ℓ ≤ 𝐶′′ 𝑙∗, ℓ , 𝑘

Define matrices 𝐵′′ and 𝐶′′:

• 𝐵′′ 𝑗, 𝑙, ℓ

= rank(𝐵′ 𝑗, 𝑙, ℓ ; 𝑇𝑙,ℓ)

• 𝐶′′

𝑙, ℓ , 𝑘 = rank(𝐶′ 𝑙, ℓ , 𝑘 ; 𝑇𝑙,ℓ) (replace each value by rank in 𝑇𝑙,ℓ) ⇒ Every entry needs 1 + log 𝑜 bits

(no weight dependence!)

Footnote on running time: 𝐵′ and 𝐶′ do not need to be computed explicitly. No subtractions necessary!

𝑃 𝑜𝑒2 log 𝑜 ≤ 𝑃 𝑜2

Step 3: : Design circuit for subproblem

Circuit for min-plus product

Circuit with 0/1 as inputs Gates:

• Boolean functions: AND, OR
• XOR (i.e., sum modulo 2)

Circuit only outputs 1 bit! ⇒ Compute result bit-per-bit For every pair 𝑗, 𝑘 and every 𝑐 ∈ {1, … , log 𝑜}:

Design circuit 𝐷𝑐 𝐵′′ 𝑗,∗ , 𝐶′′[∗, 𝑘] computing 𝑐-th bit of unique 𝑙∗ for which 𝐵′′ 𝑗, 𝑙∗, ℓ ≤ 𝐶′′ 𝑙∗, ℓ , 𝑘 for all ℓ Input: Each bit of 𝑗-th row of 𝐵′′ and 𝑘-th colum of 𝐶′′

Structure of circuit

Goal: For every 𝑗, 𝑘, compute 𝑙∗ s.t. ∀ℓ: 𝐵′′ 𝑗, 𝑙∗, ℓ ≤ 𝐶′′ 𝑙∗, ℓ , 𝑘 𝐷𝑐 𝐵′′ 𝑗,∗ , 𝐶′′[∗, 𝑘] =

𝑙∈ 1,…,𝑒 , 𝑐th bit of 𝑙 is 1 ℓ=1 𝑒

𝐵′′ 𝑗, 𝑙, ℓ ≤ 𝐶′′ 𝑙, ℓ , 𝑘

Claim: 𝐷𝑐 ⋅,⋅ = 𝑐-th bit of 𝑙∗ for which ∀ℓ: 𝐵′′ 𝑗, 𝑙∗, ℓ

≤ 𝐶′′ 𝑙∗, ℓ , 𝑘 Proof:

• Big AND returns 1 if and only if 𝑙 = 𝑙∗

(uniqueness of minimum)

• If 𝑐-th bit of 𝑙∗ is 1: Big OR includes 𝑙∗ and thus returns 1
• If 𝑐-th bit of 𝑙∗ is 0: Big OR does not include 𝑙∗ and thus returns 0

1 iff comparison true (to be specified)

Step 4: : Represent circuit by polynomial

Outer OR

𝐷𝑐 𝐵′′ 𝑗,∗ , 𝐶′′[∗, 𝑘] =

𝑙∈ 1,…,𝑒 , 𝑐th bit of 𝑙 is 1 ℓ=1 𝑒

𝐵′′ 𝑗, 𝑙, ℓ ≤ 𝐶′′ 𝑙, ℓ , 𝑘 May be replaced by ⨁ due to uniqueness: AND outputs 1 for exactly one 𝑙

Polynomial for outer circuit

Fixing 𝑗, 𝑘, and 𝑙, we want to replace the following circuit by a polynomial: Apply Razborov-Smolensky with 𝑞 = 3 + log 𝑒:

• Error probability for specific 𝑙: ≤

1 2𝑞 = 1 8𝑒

• Error probability for all 𝑙: ≤ 𝑒 ⋅

1 8𝑒 = 1 8

(union bound)

ℓ=1 𝑒

𝐵′′ 𝑗, 𝑙, ℓ ≤ 𝐶′′ 𝑙, ℓ , 𝑘

𝑦=1 𝑞

1 ⊕

ℓ=1 𝑒

𝑠

𝑦,ℓ ⋅ 𝑀𝐹𝑅𝑙,ℓ 𝐵′′ 𝑗,∗ , 𝐶′′[∗, 𝑘] ⊕ 1

=: 𝑀𝐹𝑅𝑙,ℓ ⋅,⋅

Less-or-equal-circuit for two numbers 𝑏 and 𝑐

𝑀𝐹𝑅 𝑏, 𝑐 =

𝑗=1 𝑢

(1 ⊕ 𝑏𝑗 ⊕ 𝑐𝑗) ∨

𝑗=1 𝑢

1 ⊕ 𝑏𝑗 ∧ 𝑐𝑗 ∧

𝑘=1 𝑗−1

1 ⊕ 𝑏𝑘 ⊕ 𝑐

𝑘

= 1 iff 𝑏 = 𝑐 = 1 iff

• First 𝑗 − 1 bits of 𝑏 and 𝑐 equal,
• 𝑗-th bit of 𝑏 = 0, and
• 𝑗-th bit of 𝑐 = 1

May be replaced by XOR: at most one of inner expressions is true

Polynomial for LEQ circuit

Apply Razborov/Smolensky with 𝑟 = 3 + 2 log 𝑒 + log(𝑢 + 1): Additional trick: For every entry 𝑏 of 𝐵′′ and every entry 𝑐 of 𝐶′′: Precompute XOR of 𝑏𝑗’s and XOR of 𝑐𝑗’s: additional time 𝑃 𝑜𝑒2𝑢𝑟 ≤ 𝑃 𝑜2 Introduce new variables for these combinations for later evaluation New form:

𝑀𝐹𝑅 𝑏, 𝑐 =

𝑗=1 𝑢

(1 ⊕ 𝑏𝑗 ⊕ 𝑐𝑗) ⊕

𝑗=1 𝑢

1 ⊕ 𝑏𝑗 ∧ 𝑐𝑗 ∧

𝑘=1 𝑗−1

1 ⊕ 𝑏𝑘 ⊕ 𝑐

𝑘

𝑢+1 𝑟 ≤𝑢

"2 ⊕ gates" 𝑀𝐹𝑅′ 𝑏, 𝑐 =

𝑢+1 𝑟

("2 ⊕ gates")

at most one 𝑏𝑗, at most one 𝑐𝑗, at most one constant

Polynomial for LEQ circuit cont’d

Expansion (distributive law): → polynomial over 𝐺2 with

• degree ≤ 𝑟
• #monomials: 𝑛 ≤ 𝑢 + 1 ⋅ 3𝑟 monomials

Error probability: For each application of Raz/Smol: Error prob. ≤

1 2𝑟

By union bound:

• For comparing a fixed pair (𝑏, 𝑐): error probability ≤

𝑢+1 2𝑟

• For all 𝑒2 comparisons: error probability ≤

𝑒2(𝑢+1) 2𝑟

≤

𝑒2 𝑢+1 23+2 log 𝑒+log 𝑢+1 = 1 8

𝑀𝐹𝑅′ 𝑏, 𝑐 =

𝑢+1 𝑟

("2 ⊕ gates")

Final polynomial

Apply distributive law: #monomials bounded by

𝑁 ≤ 𝑒 ⋅ 𝑒 + 1 𝑛

𝑞 = 𝑒 ⋅

𝑒 + 1 𝑛

2+log 𝑒

Error probability: ≤

1 8 + 1 8 = 1 4

𝑄𝑐 𝐵′′ 𝑗,∗ , 𝐶′′[∗, 𝑘] =

𝑙=1,…,𝑒 𝑐th bit of 𝑙 is 1 𝑦=1 𝑞

1 ⊕

ℓ=1 𝑒

𝑠

𝑦,ℓ ⋅ (𝑀𝐹𝑅𝑙,ℓ ′

𝐵′′ 𝑗,∗ , 𝐶′′[∗, 𝑘] ⊕ 1)

XOR with 𝑛 ≤ 𝑢 + 1 ⋅ 3𝑟 monomials XOR with ≤ (𝑒 + 1)𝑛 monomials

The calculation

#monomials 𝑁 ≤ 𝑒 ⋅

𝑒 + 1 𝑛

𝑞 = 𝑒 ⋅

𝑒 + 1 𝑛

3+log 𝑒

= 𝑒 ⋅ 𝑒 + 1 ⋅ 𝑢 + 1 ⋅ 3𝑟 3+log 𝑒 = 𝑒 ⋅ 𝑒 + 1 ⋅ 𝑢 + 1 ⋅ 33+2 log 𝑒+log 𝑢+1

3+log 𝑒

Claim: 𝑁 ≤ 𝑜0.1 Taking logarithms:

log 𝑁 ≤ log 𝑒 + 3 + log 𝑒 log 𝑒 + 1 + log 𝑢 + 1 + 3 + 2 log 𝑒 + log 𝑢 + 1 ⋅ log 3 ≤ log 𝑒 + 3 + log 𝑒 log 𝑒 + 1 + log 𝑒 + 1 + 3 + 2 log 𝑒 + log 𝑒 + 1 ⋅ 2 ≤ log 𝑒 + 3 log 𝑒 + log 𝑒 2 log 𝑒 + 2 log 𝑒 + 3 log 𝑒 + 2 log 𝑒 + 2 log 𝑒 ⋅ 2 = log 𝑒 + 4 log 𝑒 4 log 𝑒 + 7 log 𝑒 ⋅ 2 = log 𝑒 + 76 log2 𝑒 ≤ 100 log2 𝑒 = 100 log 𝑜 100

2

≤ 0.1 log 𝑜 𝑒 ≥ 𝑢

𝑒 = 2 log 𝑜/100 𝑞 = 3 + log 𝑒 𝑟 = 3 + 2 log 𝑒 + log(𝑢 + 1)

Step 5: : Fast evaluation of polynomial

Fast evaluation of polynomial

For every 𝑐 ∈ 1, … , log 𝑜 : Generate probabilistic polynomial 𝑄𝑐 with the following properties

• 𝑄

𝑐 is XOR of 𝑁 ≤ 𝑜0.1 monomials

• Variables of 𝑄

𝑐 can partitioned into two subsets 𝑌 and 𝑍

• For every pair 𝑗, 𝑘: if
• variables of 𝑌 evaluated according to 𝑗-th row of 𝐵′′ and
• variables of 𝑍 evaluated according to 𝑘-th column of 𝐶′′,
• then 𝑄𝑐 returns 𝑐-th bit of arg min

𝑙∈{1,…,𝑒}

(𝐵′′ 𝑗, 𝑙, ℓ ≤ 𝐶′′ 𝑙, ℓ , 𝑘 ) with probability ≥

3 4

⇒ (Fast Evaluation Lemma): Can evaluate 𝑄𝑐 for all 𝑜2 pairs 𝑗, 𝑘 in time 𝑃(𝑜2poly log 𝑜 ) Result matrix 𝑆𝑐 with entries 𝑆𝑐 𝑗, 𝑘

Step 6: : Amplify success probability

Majority amplification

For all pairs 𝑗, 𝑘 and every 𝑐 ∈ 1, … , log 𝑜 : 𝑆𝑐 𝑗, 𝑘 = 𝐷𝑐 𝐵′′ 𝑗,∗ , 𝐶′′ ∗, 𝑘 with probability ≥

3 4

Repeat evaluation with 𝑠 = 18 log 𝑜 different random polynomials Define 𝑋

𝑐[𝑗, 𝑘] as majority output of all 𝑠 evaluations

Fix pair 𝑗, 𝑘 and 𝑐 ∈ 1, … , log 𝑜

𝑌: Random variable counting how often 𝑆𝑐[𝑗, 𝑘] and 𝐷𝑐(𝑗, 𝑘 ) agree over all 𝑠 trials Pr 𝑋

𝑐 𝑗, 𝑘 ≠ 𝐷𝑐(𝑗, 𝑘) ≤ Pr 𝑌 < 𝑠

2 𝐹 𝑌 ≥ 3 ⋅ 𝑠 4

…still 𝑃(𝑜2poly log 𝑜 )

Bounding success probability

Bound error probability using tail bound:

Pr 𝑁𝑐 𝑗, 𝑘 ≠ 𝐷(𝑗, 𝑘, 𝑐) ≤ Pr 𝑌 < 𝑠 2 ≤ Pr 𝑌 < 4 6 E 𝑌 = Pr 𝑌 < 1 − 1 3 E 𝑌 ≤ 𝑓

− 2 3

2

E 𝑌 2 = 𝑓 −4E 𝑌 18 ≤ 𝑓 −3𝑠 18 = 𝑓−3 log 𝑜 ≤ 2−4 log 𝑜 = 𝑜−4

Majority needs to be correct for all 𝑜2 pairs 𝑗, 𝑘 and log 𝑒 bit positions 𝑐 in all

𝑜 𝑒 instances of the algorithm:

Union bound: Pr ∃𝑗, 𝑘, 𝑐: 𝑁𝑐 𝑗, 𝑘 ≠ 𝐷𝑐 𝑗, 𝑘 in some instance ≤ 𝑜3 log 𝑒 𝑒 ⋅ 𝑜−4 ≤ 1 𝑜

Chernoff: Pr 𝑌 < 1 − 𝜀 𝐹[𝑌] ≤ 𝑓−𝜀2𝐹[𝑌]/2

Questions?