[PPT] - Quantum query complexity and the adversary bound Part II: Learning PowerPoint Presentation

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Quantum query complexity and the adversary bound Part II: Learning graphs

Alexander Belov University of Latvia

22nd EWSCS, 5-10 March 2017, Palmse

SLIDE 2

Learning graphs

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

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SLIDE 3

Dual Adversary

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

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Recall the dual adversary bound minimise

max

z

j∈[n]

Xj[ [z, z] ]

subject to

j:xj=yj

Xj[ [x, y] ] = 1

whenever f(x) = f(y);

Xj is a p.s.d. D × D matrix

for all j ∈ [n],

SLIDE 4

Dual Adversary

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

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Recall the dual adversary bound minimise

max

z

j∈[n]

Xj[ [z, z] ]

subject to

j:xj=yj

Xj[ [x, y] ] = 1

whenever f(x) = f(y);

Xj is a p.s.d. D × D matrix

for all j ∈ [n],

How do we ensure the feasibility condition?

In general this is difficult,
but there is a way for functions with short certificates.

SLIDE 5

Certificate Structure

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

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Function

f : [q]n ⊇ D → {0, 1}

For x ∈ f −1(1), write out:

Mx =

S ⊆ [n] | xS is enough to deduce f(x) = 1
.

The set of all Mx is the certificate structure of f. (Interested in inclusion-wise minimal Mx only.)

SLIDE 6

Example

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

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Element Distinctness Problem Function f : [q]n → {0, 1}: there are two equal elements.

1147 1417 1471

∅

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3

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4117

4171 4711

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1
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❏ ❏ ❏ ❏ 2 ✴ ✴

3

✎✎

4

t

t t t t

✴

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❏

❏ ❏ ❏ ❏

✕✕
▲

▲ ▲ ▲ ▲ ♣♣♣♣♣♣

❇

❇ ❇

✜✜
❁

❁ ❁ rrrrr

✮

✮ t t t t t

✎✎
✎✎

✮ ✮

❇

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t

t t t t t t t t t

❁

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▲

▲ ▲ ▲ ▲

✎

✎ rrrrr

✜✜
❏

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✴✴

♣♣♣♣♣♣

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✴ ✴ ❏ ❏ ❏ ❏ ❏

∅
1
❏

❏ ❏ ❏ ❏ 2 ✴ ✴

3

✎✎

4

t

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∅
1
❏

❏ ❏ ❏ ❏ 2 ✴ ✴

3

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4

t

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SLIDE 7

Another Example

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

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Collision Problem Distinguish between two cases: Negative: each symbol in the input string is unique; or Positive: each symbol has exactly two appearances. E.g., negative input: 2746 and three variants of positive inputs:

1144 1414 1441

∅

1
❏

❏ ❏ ❏ ❏ 2 ✴ ✴

3

✎✎

4

t

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✴

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❏ ❏ ❏ ❏

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▲

▲ ▲ ▲ ▲ ♣♣♣♣♣♣

❇

❇ ❇

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1
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3

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4

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1
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3

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4

t

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✕✕
▲

▲ ▲ ▲ ▲ ♣♣♣♣♣♣

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✕✕

✴ ✴ ❏ ❏ ❏ ❏ ❏

SLIDE 8

Idea

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

7 / 35

How do we ensure the feasibility condition?

j:xj=yj

Xj[ [x, y] ] = 1

whenever f(x) = f(y); Element distinctness on positive input x = 122.

❄

❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ⑧ ⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

∅

1 2 3

SLIDE 9

Idea

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

7 / 35

How do we ensure the feasibility condition?

j:xj=yj

Xj[ [x, y] ] = 1

whenever f(x) = f(y); Element distinctness on positive input x = 122. Define flow pe(x) of value 1 from ∅ to Mx.

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧

⑧

⑧ ⑧ ⑧ ⑧

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

∅

1 2 3

SLIDE 10

Idea

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

7 / 35

How do we ensure the feasibility condition?

j:xj=yj

Xj[ [x, y] ] = 1

whenever f(x) = f(y); Element distinctness on positive input x = 122.

y = 456 y = 123

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧

⑧

⑧ ⑧ ⑧ ⑧

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

∅

1 2 3

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧

⑧

⑧ ⑧ ⑧ ⑧

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

∅

1 2 3

SLIDE 11

Idea

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

7 / 35

How do we ensure the feasibility condition?

j:xj=yj

Xj[ [x, y] ] = 1

whenever f(x) = f(y); Element distinctness on positive input x = 122.

In all cases, the value of the cut is 1! y = 456 y = 123

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧

⑧

⑧ ⑧ ⑧ ⑧

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

∅

1 2 3

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧

⑧

⑧ ⑧ ⑧ ⑧

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

∅

1 2 3

SLIDE 12

Construction

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

8 / 35

For each arc e from S to S ∪ {j}, we define a block-diagonal matrix

Xe

j =

α

Yα,

where the sum is over all assignments α on S. Each Yα is defined as ψψ∗, where (we is the weight of e):

ψ[ [z] ] =      pe(z)/√we, f(z) = 1, and z satisfies α; √we, f(z) = 0, and z satisfies α; 0,

therwise.

Finally, we define

Xj =

e loads j

Xe

j .

SLIDE 13

Feasibility

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

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j:xj=yj

Xj[ [x, y] ] = 1

whenever f(x) = f(y)

Claim. Left-hand side is the value of the cut, and Xj 0.

Xj =

e loads j

Xe

j ,

Xe

j =

α

Yα, Yα = ψψ∗ ψ[ [z] ] =      pe(z)/√we, f(z) = 1, and z satisfies α; √we, f(z) = 0, and z satisfies α; 0,

therwise.

SLIDE 14

Objective value

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

10 / 35

minimise

max

z

j∈[n]

Xj[ [z, z] ]

Claim. The objective value is
j∈[n]

Xj[ [z, z] ] =

e we,

if f(z) = 0;

e

pe(z)2 we ,

if f(z) = 1.

Xj =

e loads j

Xe

j ,

Xe

j =

α

Yα, Yα = ψψ∗ ψ[ [z] ] =      pe(z)/√we, f(z) = 1, and z satisfies α; √we, f(z) = 0, and z satisfies α; 0,

therwise.

SLIDE 15

Summary

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

11 / 35

First, for each arc e from S to S ∪ {j},

define its weight we.

❄

❄ ❄ ❄ ❄ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄ ⑧ ⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄ ⑧ ⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

∅ 1 2 3

SLIDE 16

Summary

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

11 / 35

First, for each arc e from S to S ∪ {j},

define its weight we.

Next, for each Mx in the certificate structure,

define flow pe(x) of value 1 from ∅ to Mx.

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧

⑧

⑧ ⑧ ⑧ ⑧

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

∅

1 2 3

SLIDE 17

Summary

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

11 / 35

First, for each arc e from S to S ∪ {j},

define its weight we.

Next, for each Mx in the certificate structure,

define flow pe(x) of value 1 from ∅ to Mx. The complexity of the learning graph is

max

e

we, max

x∈f−1(1)

pe(x)2 we

.
❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧

⑧

⑧ ⑧ ⑧ ⑧

❄

❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

⑧

⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄

∅

1 2 3

SLIDE 18

Example: OR function

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

12 / 35

For each arc from ∅ to {j}, its weight is we =

1 √n.

···
❖❖❖❖❖❖❖❖❖❖

❄ ❄ ❄ ❄ ❄ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦

∅ 1 2 n

SLIDE 19

Example: OR function

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

12 / 35

For each arc from ∅ to {j}, its weight is we =

1 √n.

For each Mx, the flow goes to a ∈ Mx.
···
❖❖❖❖❖❖❖❖❖❖
❄

❄ ❄ ❄ ❄ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦

∅ 1 2 n

SLIDE 20

Example: OR function

Learning graphs Dual Adversary Certificate Structure Examples Idea Construction Feasibility Objective value Summary OR function Symmetry Element Distinctness Triangle Detection

12 / 35

For each arc from ∅ to {j}, its weight is we =

1 √n.

For each Mx, the flow goes to a ∈ Mx.

The complexity of the learning graph is

max

e

we, max

x∈f−1(1)

pe(x)2 we

= max
n · 1

√n, 1 1/√n

= √n.
···
❖❖❖❖❖❖❖❖❖❖
❄

❄ ❄ ❄ ❄ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦

∅ 1 2 n

SLIDE 21

Symmetry

Learning graphs Symmetry Alternative Description Transitions Theorem OR function Element Distinctness Triangle Detection

13 / 35

SLIDE 22

Alternative Description

Learning graphs Symmetry Alternative Description Transitions Theorem OR function Element Distinctness Triangle Detection

14 / 35

Randomized procedure for loading values of variables.
For each positive input: the goal is to load a 1-certificate.

For OR function: (x is a positive input, and {a} is a 1-certificate) I: Load a

We start in the empty set ∅.
We divide in a number of stages.
On each stage we load a number of variables: a transition.
Length of the transition: number of loaded variables.

SLIDE 23

Transitions

Learning graphs Symmetry Alternative Description Transitions Theorem OR function Element Distinctness Triangle Detection

15 / 35

I: Load a

Define the set of transitions as the union over all inputs:

I: From ∅ to {j} for all j ∈ [n]

···
❖❖❖❖❖❖❖❖❖❖

❄ ❄ ❄ ❄ ❄ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦

∅ 1 2 n

SLIDE 24

Theorem

Learning graphs Symmetry Alternative Description Transitions Theorem OR function Element Distinctness Triangle Detection

16 / 35

Symmetry Assumption:

On each stage, the length of each transition is the same.
On each stage, the number of taken transitions is the same for

all inputs, all taken with the same probability. Then, there exists a learning graph with complexity

i

Li

Ti,

where Length

Li:

Number of variables loaded on the stage i Speciality

Ti:

Number of transitions
n the stage i
/
Number of transitions

used for one input

SLIDE 25

OR function

Learning graphs Symmetry Alternative Description Transitions Theorem OR function Element Distinctness Triangle Detection

17 / 35

Symmetry Assumption:

On each stage, the length of each transition is the same.
On each stage, the number of taken transitions is the same for

all inputs, all taken with the same probability.

···
❖❖❖❖❖❖❖❖❖❖

❄ ❄ ❄ ❄ ❄ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦

∅ 1 2 n

···
❖❖❖❖❖❖❖❖❖❖
❄

❄ ❄ ❄ ❄ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦

∅ 1 2 n

Transitions Used Length Speciality I: From ∅ to {j}

j = a 1 n

Complexity:

i Li

√Ti = √n.

SLIDE 26

Element Distinctness

Learning graphs Symmetry Element Distinctness Formulation Na¨ ıve learning graph Learning graph Idea More generality Triangle Detection

18 / 35

SLIDE 27

Formulation

Learning graphs Symmetry Element Distinctness Formulation Na¨ ıve learning graph Learning graph Idea More generality Triangle Detection

19 / 35

Given x1, . . . , xn ∈ [q], detect whether there exist a = b such that xa = xb. Certificate: {a, b}. Goal: To load two specific elements a and b.

SLIDE 28

Na¨ ıve learning graph

Learning graphs Symmetry Element Distinctness Formulation Na¨ ıve learning graph Learning graph Idea More generality Triangle Detection

20 / 35

I: Load a II: Load b Transitions Used Length Speciality I: From ∅ to {i}

i = a 1 n

II: From {i} to {i, j}

i = a, j = b 1 n2

Complexity:

i Li

√Ti = n.

SLIDE 29

Na¨ ıve learning graph

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Transitions Used Length Speciality I: From ∅ to {i}

i = a 1 n

II: From {i} to {i, j}

i = a, j = b 1 n2

The second stage is a bottleneck. Can we improve length? Can we improve speciality?

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Learning graph

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I: Load r elements not from {a, b} II: Load a III: Load b Transitions Used Length Speciality I: From ∅ to S of r elements

a, b / ∈ S r 1

II: From S to S ∪{j} for |S| =

r and j / ∈ S a, b / ∈ S, j = a

1

n

III: From S to S ∪{j} for |S| =

r + 1 and j / ∈ S a ∈ S, j = b

1

n2/r

Complexity:

i

Li

Ti = O
r + √n + n/√r
= O(n2/3)

when r = n2/3.

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Idea

Learning graphs Symmetry Element Distinctness Formulation Na¨ ıve learning graph Learning graph Idea More generality Triangle Detection

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I: Load r elements not from {a, b} II: Load a III: Load b Main idea: Before loading b, a is hidden among the r previously loaded elements. Where does a man hide a leaf? In the forest. But what does he do if there is no forest?.. He grows a forest to hide it in. Gilbert Keith Chesterton

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More generality

Learning graphs Symmetry Element Distinctness Formulation Na¨ ıve learning graph Learning graph Idea More generality Triangle Detection

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A similar algorithm solves any problem with 1-certificate complexity

k = O(1).

Let a1, . . . , ak be a 1-certificate. I: Load r elements not from {a1, a2, . . . , ak} II.1: Load a1 . . . II.k: Load ak

Complexity is O(nk/(k+1)).

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Triangle Detection

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SLIDE 34

Settings

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Given xi,j ∈ {0, 1}, with 1 ≤ i < j ≤ n, detect whether there exist 1 ≤ a < b < c ≤ n such that

xa,b = xa,c = xb,c = 1.

NB: the number of input variables is Θ(n2).

c
a
b
⑧

⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄ ❚ ❚ ❚ ❚ ❚ ❚ ❚ ❚ ❚ ❚ ❚ ❚ ❚ ♦♦♦♦♦♦♦♦♦♦♦♦♦♦♦♦♦♦♦♦ ✎ ✎ ✎ ✎ ✎ ✎ ✎ ✎ ✎ ❄ ❄ ❄ ❄ ❄ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ⑧ ⑧ ⑧ ⑧ ⑧ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴

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Settings

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Given xi,j ∈ {0, 1}, with 1 ≤ i < j ≤ n, detect whether there exist 1 ≤ a < b < c ≤ n such that

xa,b = xa,c = xb,c = 1.

We can use the learning graph of the last section with

complexity

(n2)3/4 = n3/2.

But we can do better.

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Learning graph

Learning graphs Symmetry Element Distinctness Triangle Detection Settings Learning graph Complexity

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c
a
b
In the beginning nothing is loaded.

Continue as follows...

SLIDE 37

Learning graph

Learning graphs Symmetry Element Distinctness Triangle Detection Settings Learning graph Complexity

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c
a
b
❖

❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ❄ ❄ ❄ ❄ ❄ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ I: Take disjoint A, B ⊆ [n] \ {a, b, c} of sizes n4/7 and

n5/7, and load all edges between A and B

Length:

|A||B| = n9/7

Speciality:

1

Complexity:

n9/7

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Learning graph

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c
a
b
❖

❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ❄ ❄ ❄ ❄ ❄ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ II: Add a to A and load all edges between a and B Length:

|B| = n5/7

Speciality:

n

Complexity:

n17/14

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Learning graph

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c
a
b
❖

❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ❄ ❄ ❄ ❄ ❄ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ♦♦♦♦♦♦♦♦♦ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ III: Add b to B and load all edges between b and A Length:

|A| = n4/7

Speciality:

n2/|A| = n10/7

Complexity:

n9/7

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Learning graph

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c
a
b
❖

❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ❄ ❄ ❄ ❄ ❄ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ♦♦♦♦♦♦♦♦♦ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ IV: Load ℓ = n3/7 edges connecting c to vertices in B, but b Length:

ℓ = n3/7

Speciality:

n3/(|A||B|) = n3/(n4/7n5/7) = n12/7

Complexity:

n9/7

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Learning graph

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c
a
b
❖

❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ❄ ❄ ❄ ❄ ❄ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ♦♦♦♦♦♦♦♦♦ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ V: Load edge bc Length:

1

Speciality:

n3/|A| = n3/n4/7 = n17/7

Complexity:

n17/14

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Learning graph

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c
a
b
❖

❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ✴ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ♦ ❄ ❄ ❄ ❄ ❄ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❖ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ❄ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ✹ ♦♦♦♦♦♦♦♦♦ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ⑧ ❄ ❄ ❄ ❄ ❄ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ ✯ VI: Load edge ac Length:

1

Speciality:

n3/ℓ = n18/7

Complexity:

n9/7

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Complexity

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I: Take disjoint A, B ⊆ [n] \ {a, b, c} of sizes n4/7 and

n5/7 and load all edges between A and B

II: Add a to A and load all edges between a and B III: Add b to B and load all edges between b and A IV: Load ℓ = n3/7 edges connecting c to elements in B, but

b

V: Load edge bc VI: Load edge ac Stage I II III IV V VI Length

n9/7 n5/7 n4/7 n3/7

1 1 Speciality 1

n n10/7 n12/7 n17/7 n18/7

Complexity

n9/7 n17/14 n9/7 n9/7 n17/14 n9/7

Total complexity: O(n9/7).

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Thank you!