[PPT] - From Independence to Expansion and Back Again Tobias Christiani, PowerPoint Presentation

SLIDE 1

From Independence to Expansion and Back Again

1

From Independence to Expansion and Back Again

Tobias Christiani, Rasmus Pagh

IT University of Copenhagen

Mikkel Thorup

University of Copenhagen

SLIDE 2

From Independence to Expansion and Back Again

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Introduction Topic of this talk:

Upper bounds on the space-time tradeoff of k-independent functions

in the word RAM model

SLIDE 3

From Independence to Expansion and Back Again

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Introduction Topic of this talk:

Upper bounds on the space-time tradeoff of k-independent functions

in the word RAM model Definition A family of functions F from [u] to [r] is k-independent if for every set of k distinct keys x1, x2, . . . , xk ∈ [u] and k values y1, y2, . . . , yk ∈ [r] we have that Pr

f2F[f(x1) = y1, f(x2) = y2, . . . , f(xk) = yk] = rk

SLIDE 4

From Independence to Expansion and Back Again

2

Introduction Topic of this talk:

Upper bounds on the space-time tradeoff of k-independent functions

in the word RAM model Definition A family of functions F from [u] to [r] is k-independent if for every set of k distinct keys x1, x2, . . . , xk ∈ [u] and k values y1, y2, . . . , yk ∈ [r] we have that Pr

f2F[f(x1) = y1, f(x2) = y2, . . . , f(xk) = yk] = rk

Example of a k-independent function:

f(x) =

k1

X

i=0

aixi mod p

SLIDE 5

From Independence to Expansion and Back Again

2

Introduction Topic of this talk:

Upper bounds on the space-time tradeoff of k-independent functions

in the word RAM model Definition A family of functions F from [u] to [r] is k-independent if for every set of k distinct keys x1, x2, . . . , xk ∈ [u] and k values y1, y2, . . . , yk ∈ [r] we have that Pr

f2F[f(x1) = y1, f(x2) = y2, . . . , f(xk) = yk] = rk

Tradeoff:

– Space used to represent f ∈ F – Time used to evaluate f ∈ F

Example of a k-independent function:

f(x) =

k1

X

i=0

aixi mod p

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From Independence to Expansion and Back Again

3

Space-time tradeoff: Lower and upper bounds Theorem [Siegel’89] A data structure for representing a k-independent function f : [u] → [r] with evaluation time t < k must use at least ku1/t words of space

Lower bound

Domain Memory . . . . . . . . . t Probes f(x) x

SLIDE 7

From Independence to Expansion and Back Again

3

Space-time tradeoff: Lower and upper bounds Theorem [Siegel’89] A data structure for representing a k-independent function f : [u] → [r] with evaluation time t < k must use at least ku1/t words of space Randomized data structure for representing a k-independent function f : [u] → [r] with a space usage of O(ku1/tt) and evaluation time O(t log t)

Main result (vanilla version) Lower bound

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From Independence to Expansion and Back Again

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Space-time tradeoff: Lower and upper bounds Theorem [Siegel’89] A data structure for representing a k-independent function f : [u] → [r] with evaluation time t < k must use at least ku1/t words of space Randomized data structure for representing a k-independent function f : [u] → [r] with a space usage of O(ku1/tt) and evaluation time O(t log t)

Main result (vanilla version)

Reference Space Time Polynomials [Joffe’74] O(k) O(k) Graph powering [Siegel’89] O(ktu1/t) O(t)t Recursive tabulation [Thorup’13] O(poly k + u1/t) O(tlog t)

Lower bound Previous results

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From Independence to Expansion and Back Again

4

From expansion to independence U V . . . d

Constructions of k-independent families of functions based on

bipartite expander graphs

Neighbor function Γ : U → V d

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From Independence to Expansion and Back Again

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From expansion to independence U V . . . d Definition A bipartite graph Γ is k-unique if for every S ⊆ U with |S| ≤ k there exists v ∈ V with exactly one neighbor in S

Constructions of k-independent families of functions based on

bipartite expander graphs

Neighbor function Γ : U → V d

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From Independence to Expansion and Back Again

4

From expansion to independence U V Definition A bipartite graph Γ is k-unique if for every S ⊆ U with |S| ≤ k there exists v ∈ V with exactly one neighbor in S

Constructions of k-independent families of functions based on

bipartite expander graphs x1 x2 x3

Neighbor function Γ : U → V d

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From Independence to Expansion and Back Again

4

From expansion to independence U V Definition A bipartite graph Γ is k-unique if for every S ⊆ U with |S| ≤ k there exists v ∈ V with exactly one neighbor in S

Constructions of k-independent families of functions based on

bipartite expander graphs x1 x2 x3 v1 v2 v3 v4 v5

Neighbor function Γ : U → V d

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From Independence to Expansion and Back Again

4

From expansion to independence U V Definition A bipartite graph Γ is k-unique if for every S ⊆ U with |S| ≤ k there exists v ∈ V with exactly one neighbor in S

Constructions of k-independent families of functions based on

bipartite expander graphs x1 x2 x3 v1 v2 v3 v4 v5

Neighbor function Γ : U → V d

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From Independence to Expansion and Back Again

4

From expansion to independence U V Lemma [Siegel’89] Let Γ : U → V d be k-unique and h : V → [r] be a random function. Then f(x) = P

i h(Γ(x)i) mod r defines a

k-independent family of functions Definition A bipartite graph Γ is k-unique if for every S ⊆ U with |S| ≤ k there exists v ∈ V with exactly one neighbor in S

Constructions of k-independent families of functions based on

bipartite expander graphs x1 x2 x3 v1 v2 v3 v4 v5

Neighbor function Γ : U → V d

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From Independence to Expansion and Back Again

4

From expansion to independence U V Lemma [Siegel’89] Let Γ : U → V d be k-unique and h : V → [r] be a random function. Then f(x) = P

i h(Γ(x)i) mod r defines a

k-independent family of functions Definition A bipartite graph Γ is k-unique if for every S ⊆ U with |S| ≤ k there exists v ∈ V with exactly one neighbor in S

Constructions of k-independent families of functions based on

bipartite expander graphs x1 x2 x3 v1 v2 v3 v4 v5 f(x3) = h(v2) + h(v4) + h(v5) mod r

Neighbor function Γ : U → V d
Peeling argument example:

SLIDE 16

From Independence to Expansion and Back Again

4

From expansion to independence U V Lemma [Siegel’89] Let Γ : U → V d be k-unique and h : V → [r] be a random function. Then f(x) = P

i h(Γ(x)i) mod r defines a

k-independent family of functions Definition A bipartite graph Γ is k-unique if for every S ⊆ U with |S| ≤ k there exists v ∈ V with exactly one neighbor in S

Constructions of k-independent families of functions based on

bipartite expander graphs x1 x2 v1 v2 v3 v4 v5 f(x3) = h(v2) + h(v4) + h(v5) mod r

Neighbor function Γ : U → V d
Peeling argument example:

SLIDE 17

From Independence to Expansion and Back Again

4

From expansion to independence U V Lemma [Siegel’89] Let Γ : U → V d be k-unique and h : V → [r] be a random function. Then f(x) = P

i h(Γ(x)i) mod r defines a

k-independent family of functions Definition A bipartite graph Γ is k-unique if for every S ⊆ U with |S| ≤ k there exists v ∈ V with exactly one neighbor in S

Constructions of k-independent families of functions based on

bipartite expander graphs x1 x2 v1 v2 v3 v4 v5 f(x3) = h(v2) + h(v4) + h(v5) mod r

Neighbor function Γ : U → V d
Peeling argument example:

SLIDE 18

From Independence to Expansion and Back Again

4

From expansion to independence U V Lemma [Siegel’89] Let Γ : U → V d be k-unique and h : V → [r] be a random function. Then f(x) = P

i h(Γ(x)i) mod r defines a

k-independent family of functions Definition A bipartite graph Γ is k-unique if for every S ⊆ U with |S| ≤ k there exists v ∈ V with exactly one neighbor in S

Constructions of k-independent families of functions based on

bipartite expander graphs x2 v1 v2 v3 v4 v5 f(x3) = h(v2) + h(v4) + h(v5) mod r

Neighbor function Γ : U → V d
Peeling argument example:

SLIDE 19

From Independence to Expansion and Back Again

4

From expansion to independence U V Lemma [Siegel’89] Let Γ : U → V d be k-unique and h : V → [r] be a random function. Then f(x) = P

i h(Γ(x)i) mod r defines a

k-independent family of functions Definition A bipartite graph Γ is k-unique if for every S ⊆ U with |S| ≤ k there exists v ∈ V with exactly one neighbor in S

Constructions of k-independent families of functions based on

bipartite expander graphs x2 v1 v2 v3 v4 v5 f(x3) = h(v2) + h(v4) + h(v5) mod r

Neighbor function Γ : U → V d
Peeling argument example:

SLIDE 20

From Independence to Expansion and Back Again

4

From expansion to independence U V Lemma [Siegel’89] Let Γ : U → V d be k-unique and h : V → [r] be a random function. Then f(x) = P

i h(Γ(x)i) mod r defines a

k-independent family of functions Definition A bipartite graph Γ is k-unique if for every S ⊆ U with |S| ≤ k there exists v ∈ V with exactly one neighbor in S

Constructions of k-independent families of functions based on

bipartite expander graphs v1 v2 v3 v4 v5 f(x3) = h(v2) + h(v4) + h(v5) mod r

Neighbor function Γ : U → V d
Peeling argument example:

SLIDE 21

From Independence to Expansion and Back Again

5

Existence of optimal k-unique graphs k-unique Γ : U → V d

. . . . . . . . . d

U V

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From Independence to Expansion and Back Again

5

Existence of optimal k-unique graphs k-unique Γ : U → V d

. . . . . . . . . d

U V k-independent family F f(x) = P

i h(Γ(x)i) mod r

Time d Space |V | h : V → [r] is random

SLIDE 23

From Independence to Expansion and Back Again

5

Existence of optimal k-unique graphs k-unique Γ : U → V d

. . . . . . . . . d

U V k-independent family F f(x) = P

i h(Γ(x)i) mod r

Time d Space |V | h : V → [r] is random Lemma [Siegel’89] Most graphs Γ give an optimal space-time tradeoff for k-independent hashing

SLIDE 24

From Independence to Expansion and Back Again

5

Existence of optimal k-unique graphs k-unique Γ : U → V d

. . . . . . . . . d

U V k-independent family F f(x) = P

i h(Γ(x)i) mod r

Time d Space |V | h : V → [r] is random Lemma [Siegel’89] Most graphs Γ give an optimal space-time tradeoff for k-independent hashing

Explicit k-unique graphs with optimal parameters are not known

SLIDE 25

From Independence to Expansion and Back Again

5

Existence of optimal k-unique graphs k-unique Γ : U → V d

. . . . . . . . . d

U V k-independent family F f(x) = P

i h(Γ(x)i) mod r

Time d Space |V | h : V → [r] is random Lemma [Siegel’89] Most graphs Γ give an optimal space-time tradeoff for k-independent hashing

Explicit k-unique graphs with optimal parameters are not known
Storing a random Γ defeats the purpose of k-independent hashing

SLIDE 26

From Independence to Expansion and Back Again

5

Existence of optimal k-unique graphs k-unique Γ : U → V d

. . . . . . . . . d

U V k-independent family F f(x) = P

i h(Γ(x)i) mod r

Time d Space |V | h : V → [r] is random Lemma [Siegel’89] Most graphs Γ give an optimal space-time tradeoff for k-independent hashing

Explicit k-unique graphs with optimal parameters are not known
Storing a random Γ defeats the purpose of k-independent hashing
Verifying that a given Γ is k-unique is infeasible

SLIDE 27

From Independence to Expansion and Back Again

6

From independence to expansion Set of all functions Γ : U → V d

SLIDE 28

From Independence to Expansion and Back Again

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From independence to expansion Set of all functions Γ : U → V d Optimally k-unique

SLIDE 29

From Independence to Expansion and Back Again

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From independence to expansion Set of all functions Γ : U → V d Optimally k-unique k-independent family

F

SLIDE 30

From Independence to Expansion and Back Again

6

From independence to expansion Set of all functions Γ : U → V d Optimally k-unique k-independent family

F

SLIDE 31

From Independence to Expansion and Back Again

6

From independence to expansion Lemma A k-independent function Γ : U → V d with |U| = u, |V | = O(ku1/tt) and d = O(t) is k-unique with high probability. Set of all functions Γ : U → V d Optimally k-unique k-independent family

F

SLIDE 32

From Independence to Expansion and Back Again

6

From independence to expansion Lemma A k-independent function Γ : U → V d with |U| = u, |V | = O(ku1/tt) and d = O(t) is k-unique with high probability. Set of all functions Γ : U → V d Optimally k-unique k-independent family

Γ F Γ ∈ F

k-unique k-independent family k-unique whp.

F

SLIDE 33

From Independence to Expansion and Back Again

7

Overview of technique

. . . . . .

k-unique

SLIDE 34

From Independence to Expansion and Back Again

7

Overview of technique

. . . . . . . . . . . . . . . . . . Increase domain

k-unique k-unique

SLIDE 35

From Independence to Expansion and Back Again

7

Overview of technique

. . . . . . . . . . . . . . . . . .

F

Increase domain Define family

k-unique k-unique k-independent

SLIDE 36

From Independence to Expansion and Back Again

7

Overview of technique

. . . . . . . . . . . . . . . . . . . . . . . .

F

Increase domain Sample Γ ∈ F Define family

k-unique k-unique k-independent k-unique whp.

SLIDE 37

From Independence to Expansion and Back Again

8

A randomized recursive construction of a k-unique function

View x ∈ [u] as a string of characters from an alphabet Σ

Γ(ax) = L

i h(a, Γ(x)i),

a ∈ Σ, ax ∈ Σ∗ h : Σ × V → V d is random

SLIDE 38

From Independence to Expansion and Back Again

8

A randomized recursive construction of a k-unique function

View x ∈ [u] as a string of characters from an alphabet Σ

Γ(ax) = L

i h(a, Γ(x)i),

a ∈ Σ, ax ∈ Σ∗

. . . . . .

h : Σ × V → V d is random

SLIDE 39

From Independence to Expansion and Back Again

8

A randomized recursive construction of a k-unique function

View x ∈ [u] as a string of characters from an alphabet Σ

Γ(ax) = L

i h(a, Γ(x)i),

a ∈ Σ, ax ∈ Σ∗

. . . |Σ| a . . . . . . . . . . . .

h : Σ × V → V d is random

SLIDE 40

From Independence to Expansion and Back Again

8

A randomized recursive construction of a k-unique function

View x ∈ [u] as a string of characters from an alphabet Σ

Γ(ax) = L

i h(a, Γ(x)i),

a ∈ Σ, ax ∈ Σ∗

. . . |Σ| a x . . . . . . . . . . . .

h : Σ × V → V d is random

Γ(x)

SLIDE 41

From Independence to Expansion and Back Again

8

A randomized recursive construction of a k-unique function

View x ∈ [u] as a string of characters from an alphabet Σ

Γ(ax) = L

i h(a, Γ(x)i),

a ∈ Σ, ax ∈ Σ∗

. . . |Σ| a x . . . . . . . . .

h : Σ × V → V d is random

⊕ ⊕

01..0 00..1 10..1

=

11..0

Γ(ax) Γ(x)

SLIDE 42

From Independence to Expansion and Back Again

8

A randomized recursive construction of a k-unique function Lemma Γ is k-unique over Σj ⇒ Γ is k-independent over Σj+1 ⇒ Γ is k-unique over Σj+1 whp.

View x ∈ [u] as a string of characters from an alphabet Σ

Γ(ax) = L

i h(a, Γ(x)i),

a ∈ Σ, ax ∈ Σ∗

. . . |Σ| a x . . . . . . . . .

h : Σ × V → V d is random

⊕ ⊕

01..0 00..1 10..1

=

11..0

Γ(ax) Γ(x)

SLIDE 43

From Independence to Expansion and Back Again

8

A randomized recursive construction of a k-unique function Lemma Γ is k-unique over Σj ⇒ Γ is k-independent over Σj+1 ⇒ Γ is k-unique over Σj+1 whp. |Σ| = u1/2t |V | = O(ku1/2tt) d = O(t)

View x ∈ [u] as a string of characters from an alphabet Σ

Γ(ax) = L

i h(a, Γ(x)i),

a ∈ Σ, ax ∈ Σ∗

. . . |Σ|

Parameterizing for k-independence with

domain size u and tradeoff parameter t a x . . . . . . . . .

h : Σ × V → V d is random

⊕ ⊕

01..0 00..1 10..1

=

11..0

Γ(ax)

Space O(ku1/tt2). Time O(t2).

Γ(x)

SLIDE 44

From Independence to Expansion and Back Again

9

A stronger expansion property Definition A bipartite graph Γ is k-majority-unique if for every S ⊆ U with |S| ≤ k there exists x ∈ S such that the majority of vertices in Γ({x}) have exactly one neighbor in S U V

. . . d

SLIDE 45

From Independence to Expansion and Back Again

9

A stronger expansion property Definition A bipartite graph Γ is k-majority-unique if for every S ⊆ U with |S| ≤ k there exists x ∈ S such that the majority of vertices in Γ({x}) have exactly one neighbor in S U V

SLIDE 46

From Independence to Expansion and Back Again

9

A stronger expansion property Definition A bipartite graph Γ is k-majority-unique if for every S ⊆ U with |S| ≤ k there exists x ∈ S such that the majority of vertices in Γ({x}) have exactly one neighbor in S U V

SLIDE 47

From Independence to Expansion and Back Again

9

A stronger expansion property Definition A bipartite graph Γ is k-majority-unique if for every S ⊆ U with |S| ≤ k there exists x ∈ S such that the majority of vertices in Γ({x}) have exactly one neighbor in S U V

SLIDE 48

From Independence to Expansion and Back Again

10

A graph product based on component-wise concatenation Lemma Let Γ be k-majority-unique over an alphabet Σ. Then the function defined by Γ(x1x2)i = Γ(x1)iΓ(x2)i is k-unique over Σ2.

SLIDE 49

From Independence to Expansion and Back Again

10

A graph product based on component-wise concatenation Lemma Let Γ be k-majority-unique over an alphabet Σ. Then the function defined by Γ(x1x2)i = Γ(x1)iΓ(x2)i is k-unique over Σ2.

S

Proof:

Goal: Find a key x0

1x0 2 ∈ S ⊆ Σ2 that has a unique neighbor

SLIDE 50

From Independence to Expansion and Back Again

10

A graph product based on component-wise concatenation Lemma Let Γ be k-majority-unique over an alphabet Σ. Then the function defined by Γ(x1x2)i = Γ(x1)iΓ(x2)i is k-unique over Σ2.

S

Proof:

Goal: Find a key x0

1x0 2 ∈ S ⊆ Σ2 that has a unique neighbor

There exists x0

1 ∈ {x1 | x1x2 ∈ S} with > d/2 unique neighbors

Γ(x0

1)

SLIDE 51

From Independence to Expansion and Back Again

10

A graph product based on component-wise concatenation Lemma Let Γ be k-majority-unique over an alphabet Σ. Then the function defined by Γ(x1x2)i = Γ(x1)iΓ(x2)i is k-unique over Σ2.

S

x0

1a

x0

1b x0 1c

Proof:

Goal: Find a key x0

1x0 2 ∈ S ⊆ Σ2 that has a unique neighbor

There exists x0

1 ∈ {x1 | x1x2 ∈ S} with > d/2 unique neighbors

Γ(x0

1)

SLIDE 52

From Independence to Expansion and Back Again

10

A graph product based on component-wise concatenation Lemma Let Γ be k-majority-unique over an alphabet Σ. Then the function defined by Γ(x1x2)i = Γ(x1)iΓ(x2)i is k-unique over Σ2.

S

x0

1a

x0

1b x0 1c

Proof:

Goal: Find a key x0

1x0 2 ∈ S ⊆ Σ2 that has a unique neighbor

There exists x0

1 ∈ {x1 | x1x2 ∈ S} with > d/2 unique neighbors

There exists x0

2 ∈ {x2 | x0 1x2 ∈ S} with > d/2 unique neighbors

Γ(x0

2)

SLIDE 53

From Independence to Expansion and Back Again

10

A graph product based on component-wise concatenation Lemma Let Γ be k-majority-unique over an alphabet Σ. Then the function defined by Γ(x1x2)i = Γ(x1)iΓ(x2)i is k-unique over Σ2.

S

x0

1a

x0

1b x0 1c

Proof:

Goal: Find a key x0

1x0 2 ∈ S ⊆ Σ2 that has a unique neighbor

There exists x0

1 ∈ {x1 | x1x2 ∈ S} with > d/2 unique neighbors

There exists x0

2 ∈ {x2 | x0 1x2 ∈ S} with > d/2 unique neighbors

Γ(x0

2)

Γ(x0

1)

SLIDE 54

From Independence to Expansion and Back Again

10

A graph product based on component-wise concatenation Lemma Let Γ be k-majority-unique over an alphabet Σ. Then the function defined by Γ(x1x2)i = Γ(x1)iΓ(x2)i is k-unique over Σ2.

S

x0

1a

x0

1b x0 1c

Proof:

Goal: Find a key x0

1x0 2 ∈ S ⊆ Σ2 that has a unique neighbor

There exists x0

1 ∈ {x1 | x1x2 ∈ S} with > d/2 unique neighbors

There exists x0

2 ∈ {x2 | x0 1x2 ∈ S} with > d/2 unique neighbors

Γ(x0

2)

Γ(x0

1)

SLIDE 55

From Independence to Expansion and Back Again

10

A graph product based on component-wise concatenation Lemma Let Γ be k-majority-unique over an alphabet Σ. Then the function defined by Γ(x1x2)i = Γ(x1)iΓ(x2)i is k-unique over Σ2.

S

x0

1a

x0

1b x0 1c

Proof:

Goal: Find a key x0

1x0 2 ∈ S ⊆ Σ2 that has a unique neighbor

There exists x0

1 ∈ {x1 | x1x2 ∈ S} with > d/2 unique neighbors

There exists x0

2 ∈ {x2 | x0 1x2 ∈ S} with > d/2 unique neighbors

Γ(x0

2)

Γ(x0

1)

SLIDE 56

From Independence to Expansion and Back Again

10

A graph product based on component-wise concatenation Lemma Let Γ be k-majority-unique over an alphabet Σ. Then the function defined by Γ(x1x2)i = Γ(x1)iΓ(x2)i is k-unique over Σ2.

S

x0

1a

x0

1b x0 1c

Proof:

Goal: Find a key x0

1x0 2 ∈ S ⊆ Σ2 that has a unique neighbor

There exists x0

1 ∈ {x1 | x1x2 ∈ S} with > d/2 unique neighbors

There exists x0

2 ∈ {x2 | x0 1x2 ∈ S} with > d/2 unique neighbors

Γ(x0

2)

Γ(x0

1)

SLIDE 57

From Independence to Expansion and Back Again

10

A graph product based on component-wise concatenation Lemma Let Γ be k-majority-unique over an alphabet Σ. Then the function defined by Γ(x1x2)i = Γ(x1)iΓ(x2)i is k-unique over Σ2.

S

x0

1a

x0

1b x0 1c

Proof:

Goal: Find a key x0

1x0 2 ∈ S ⊆ Σ2 that has a unique neighbor

There exists x0

1 ∈ {x1 | x1x2 ∈ S} with > d/2 unique neighbors

There exists x0

2 ∈ {x2 | x0 1x2 ∈ S} with > d/2 unique neighbors

Γ(x0

2)

Γ(x0

1)

SLIDE 58

From Independence to Expansion and Back Again

10

A graph product based on component-wise concatenation Lemma Let Γ be k-majority-unique over an alphabet Σ. Then the function defined by Γ(x1x2)i = Γ(x1)iΓ(x2)i is k-unique over Σ2.

S

x0

1a

x0

1b x0 1c

Proof:

Goal: Find a key x0

1x0 2 ∈ S ⊆ Σ2 that has a unique neighbor

There exists x0

1 ∈ {x1 | x1x2 ∈ S} with > d/2 unique neighbors

There exists x0

2 ∈ {x2 | x0 1x2 ∈ S} with > d/2 unique neighbors

Γ(x0

2)

Γ(x0

1)

SLIDE 59

From Independence to Expansion and Back Again

11

A divide-and-conquer recursion

View x ∈ [u] as a string of two characters and recurse on each

Γ(x1x2) = L

i h(Γ(x1)i, Γ(x2)i)

SLIDE 60

From Independence to Expansion and Back Again

11

A divide-and-conquer recursion

View x ∈ [u] as a string of two characters and recurse on each

Γ(x1x2) = L

i h(Γ(x1)i, Γ(x2)i) Universe Degree Γ(x) u d

SLIDE 61

From Independence to Expansion and Back Again

11

A divide-and-conquer recursion

View x ∈ [u] as a string of two characters and recurse on each

Γ(x1x2) = L

i h(Γ(x1)i, Γ(x2)i) Universe Degree Γ(x) Γ(x1) Γ(x2) u d u1/2 d/2

SLIDE 62

From Independence to Expansion and Back Again

11

A divide-and-conquer recursion

View x ∈ [u] as a string of two characters and recurse on each

Γ(x1x2) = L

i h(Γ(x1)i, Γ(x2)i) Γ(x1,1) Γ(x1,2) Γ(x2,1) Γ(x2,2) Universe Degree Γ(x) Γ(x1) Γ(x2) u d u1/2 d/2 u1/4 d/4

SLIDE 63

From Independence to Expansion and Back Again

11

A divide-and-conquer recursion

View x ∈ [u] as a string of two characters and recurse on each

Γ(x1x2) = L

i h(Γ(x1)i, Γ(x2)i) Γ(x1,1) Γ(x1,2) Γ(x2,1) Γ(x2,2) Universe Degree Γ(x) Γ(x1) Γ(x2) u d u1/2 d/2 u1/4 d/4 . . . . . . . . .

SLIDE 64

From Independence to Expansion and Back Again

11

A divide-and-conquer recursion

View x ∈ [u] as a string of two characters and recurse on each

Γ(x1x2) = L

i h(Γ(x1)i, Γ(x2)i) Γ(x1,1) Γ(x1,2) Γ(x2,1) Γ(x2,2) Problems:

By the lower bound h must have a domain of size at least k2

– Use the first recursion to implement h Universe Degree Γ(x) Γ(x1) Γ(x2) u d u1/2 d/2 u1/4 d/4 . . . . . . . . .

SLIDE 65

From Independence to Expansion and Back Again

11

A divide-and-conquer recursion

View x ∈ [u] as a string of two characters and recurse on each

Γ(x1x2) = L

i h(Γ(x1)i, Γ(x2)i) Γ(x1,1) Γ(x1,2) Γ(x2,1) Γ(x2,2) Problems:

By the lower bound h must have a domain of size at least k2

– Use the first recursion to implement h Universe Degree Γ(x) Γ(x1) Γ(x2) u d u1/2 d/2 u1/4 d/4 . . . . . . . . . Problems:

By the lower bound h must have a domain of size at least k2

– Use the first recursion to implement h

Low space usage, high degree: representing Γ(x) in few words

– Graph products that take the structure of S into account

SLIDE 66

From Independence to Expansion and Back Again

12

Summary Result:

Near-optimal space-time tradeoff for k-independent functions

SLIDE 67

From Independence to Expansion and Back Again

12

Summary Result:

Near-optimal space-time tradeoff for k-independent functions

Technique:

Graph products and alternating between expansion and independence

F

SLIDE 68

From Independence to Expansion and Back Again

12

Summary Result:

Near-optimal space-time tradeoff for k-independent functions

Technique:

Graph products and alternating between expansion and independence

Open questions:

Optimal expanders without k-independence

F

SLIDE 69

From Independence to Expansion and Back Again

12

Summary Result:

Near-optimal space-time tradeoff for k-independent functions

Technique:

Graph products and alternating between expansion and independence

Open questions:

Optimal expanders without k-independence