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A Complexity Class Based on ComparatorCircuits
Stephen Cook
Joint work with Dai Tri Man Le and Yuval Filmus and Yuli Ye
Department of Computer ScienceUniversity of Toronto
Canada
Based Partly on a paper in CSL 2011
1 / 17
Outline of the talk
1 Comparator circuits and the class CCI Interesting natural complete problems: stable marriage, lex-first
maximal matching, comparator circuit value problem. . .
2 Example Reductions
3 Universal Circuits and applications (new)
4 Oracle separations (new)
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Comparator Circuits
Originally invented for sorting, e.g.,I Batcher’s O(log2 n)-depth sorting
networks (’68)I Ajtai-Komlos-Szemeredi (AKS)O(log n)-depth sorting networks (’83)
Can also be considered as booleancircuits.
Comparator gatep x • p ∧ q
q y H p ∨ q
Example
1 w0 • 0 • 0 01 w1 • 0 N 11 w2 10 w3 H 1 • 00 w4 H 1 10 w5 H 0 0
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Comparator Circuit Value (Ccv) Problem (decision)
Given a comparator circuit with specifiedBoolean inputs, determine the outputvalue of a designated wire.
1 w0 • •1 w1 • N1 w2
0 w3 H • ?0 w4 H0 w5 H
Complexity classes
1 CCSubr =
decision problems log-space many-one-reducible to Ccv
I [Subramanian’s PhD thesis ’90], [Mayr-Subramanian ’92]
2 CC =
decision problems AC0 many-one-reducible to Ccv
I Complete problems: stable marriage, lex-first maximal matching. . .
3 CC∗ =
decision problems AC0 Turing-reducible to Ccv
I Needed when developing a Cook-Nguyen style theory for CCI The function class FCC∗ is closed under compostion
NC1 ⊆ NL ⊆ CC ⊆ CCSubr ⊆ CC∗ ⊆ P
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Comparator Circuit Value (Ccv) Problem (decision)
Given a comparator circuit with specifiedBoolean inputs, determine the outputvalue of a designated wire.
1 w0 • •1 w1 • N1 w2
0 w3 H • ?0 w4 H0 w5 H
Complexity classes
1 CCSubr =
decision problems log-space many-one-reducible to Ccv
I [Subramanian’s PhD thesis ’90], [Mayr-Subramanian ’92]
2 CC =
decision problems AC0 many-one-reducible to Ccv
I Complete problems: stable marriage, lex-first maximal matching. . .
3 CC∗ =
decision problems AC0 Turing-reducible to Ccv
I Needed when developing a Cook-Nguyen style theory for CCI The function class FCC∗ is closed under compostion
NC1 ⊆ NL ⊆ CC ⊆ CCSubr ⊆ CC∗ ⊆ P
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Comparator Circuit Value (Ccv) Problem (decision)
Given a comparator circuit with specifiedBoolean inputs, determine the outputvalue of a designated wire.
1 w0 • •1 w1 • N1 w2
0 w3 H • ?0 w4 H0 w5 H
Complexity classes
1 CCSubr =
decision problems log-space many-one-reducible to Ccv
I [Subramanian’s PhD thesis ’90], [Mayr-Subramanian ’92]
2 CC =
decision problems AC0 many-one-reducible to Ccv
I Complete problems: stable marriage, lex-first maximal matching. . .
3 CC∗ =
decision problems AC0 Turing-reducible to Ccv
I Needed when developing a Cook-Nguyen style theory for CCI The function class FCC∗ is closed under compostion
NC1 ⊆ NL ⊆ CC ⊆ CCSubr ⊆ CC∗ ⊆ P
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Stable Marriage Problem (search version) (Gale-Shapley ’62)
Given n men and n women together with their preference lists
Find a stable marriage between men and women, i.e.,1 a perfect matching2 satisfies the stability condition: no two people of the opposite sex like
each other more than their current partners
Preference lists
Men:a x y
b y x
Women:x a b
y a b
a
b
x
y
stable marriage
a
b
x
y
unstable marriage
Stable Marriage Problem (decision version)
Is a given pair of (m,w) in the man-optimal (woman-optimal) stablemarriage?
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Stable Marriage Problem (search version) (Gale-Shapley ’62)
Given n men and n women together with their preference lists
Find a stable marriage between men and women, i.e.,1 a perfect matching2 satisfies the stability condition: no two people of the opposite sex like
each other more than their current partners
Preference lists
Men:a x y
b y x
Women:x a b
y a b
a
b
x
y
stable marriage
a
b
x
y
unstable marriage
Stable Marriage Problem (decision version)
Is a given pair of (m,w) in the man-optimal (woman-optimal) stablemarriage?
6 / 17
Stable Marriage Problem (search version) (Gale-Shapley ’62)
Given n men and n women together with their preference lists
Find a stable marriage between men and women, i.e.,1 a perfect matching2 satisfies the stability condition: no two people of the opposite sex like
each other more than their current partners
Preference lists
Men:a x y
b y x
Women:x a b
y a b
a
b
x
y
stable marriage
a
b
x
y
unstable marriage
Stable Marriage Problem (decision version)
Is a given pair of (m,w) in the man-optimal (woman-optimal) stablemarriage?
6 / 17
Stable Marriage Problem (search version) (Gale-Shapley ’62)
Given n men and n women together with their preference lists
Find a stable marriage between men and women, i.e.,1 a perfect matching2 satisfies the stability condition: no two people of the opposite sex like
each other more than their current partners
Preference lists
Men:a x y
b y x
Women:x a b
y a b
a
b
x
y
stable marriage
a
b
x
y
unstable marriage
Stable Marriage Problem (decision version)
Is a given pair of (m,w) in the man-optimal (woman-optimal) stablemarriage?
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Lex-first maximal matching problem
Lex-first maximal matching
Let G be a bipartite graph.
Successively match the bottom nodes x , y , z , . . . to the least availabletop node
a b c
x y z w
Lex-first maximal matching problem (decision)
Is a given node v (resp. edge u, v) in the lex-first maximal matching ofG ?
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Lex-first maximal matching problem
Lex-first maximal matching
Let G be a bipartite graph.
Successively match the bottom nodes x , y , z , . . . to the least availabletop node
a b c
x y z w
Lex-first maximal matching problem (decision)
Is a given node v (resp. edge u, v) in the lex-first maximal matching ofG ?
7 / 17
Lex-first maximal matching problem
Lex-first maximal matching
Let G be a bipartite graph.
Successively match the bottom nodes x , y , z , . . . to the least availabletop node
a b c
x y z w
Lex-first maximal matching problem (decision)
Is a given node v (resp. edge u, v) in the lex-first maximal matching ofG ?
7 / 17
Lex-first maximal matching problem
Lex-first maximal matching
Let G be a bipartite graph.
Successively match the bottom nodes x , y , z , . . . to the least availabletop node
a b c
x y z w
Lex-first maximal matching problem (decision)
Is a given node v (resp. edge u, v) in the lex-first maximal matching ofG ?
7 / 17
Lex-first maximal matching problem
Lex-first maximal matching
Let G be a bipartite graph.
Successively match the bottom nodes x , y , z , . . . to the least availabletop node
a b c
x y z w
Lex-first maximal matching problem (decision)
Is a given node v (resp. edge u, v) in the lex-first maximal matching ofG ?
7 / 17
Reducing node lex-first maximal matching to Ccv
a b c d
x y z
1 x • • • • 01 y • • • • 01 z • • • • 00 a H H 10 b H H 10 c H H 10 d H 0
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Theorem
NL ⊆ CC
Proof.
Show that stCONN ≤AC0 Ccv.
Note
Easier to show NC1 ⊆ CC.
Proof.
A Boolean function f (~x) is in NC1 iff it is computed by a uniform polysizelog-depth family of binary tree circuits with input (many copies of) xi and¬xi .
The comparator circuit uses one comparator gate for each AND or ORgate in the tree, and ignores one of the two outputs.
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The Function Class FCC
Let f : 0, 1∗ → 0, 1∗The bit graph of f satisfies
BITf (x , i)↔ f (x)i = 1
Definef ∈ FCC↔ f is p-bounded and BITf ∈ CC
Recall BITf ∈ CC iff BITf ≤AC0 Ccv
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Claim
We need universal cirucits to show that FCC is closed under composition.
Assume that the functions in question are p-bounded.
Roughly speaking, f ∈ FCC iff there are AC0 functions CIRf and INf suchthat the comparator circuit CIRf (x) on input IN(x) computes f (x).
Assume f and g are in FCC.
We want AC0 functions CIRf g and INf g such that CIRf g (x) on inputINf g (x) computes f (g(x)).
We know CIRf (g(x)) on input INf (g(x)) computes f (g(x)) but we needuniversal circuits to show CIRf g and INf g are AC0 functions.
In other words, we need universal circuits to convert descriptions ofdescriptions to descriptions.
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Universal Comparator Circuits
Below is a universal comparator gadget.
1 If b = 0 then x ′ = x and y ′ = y .
2 If b = 1 then x ′ = x ∨ y and y ′ = x ∧ y .
b • • • 0
x H x ′
y H • H y ′
b H 1
To simulate one arbitrary comparator in a circuit with m wires we put inm(m − 1) gadgets in a row, for the m(m − 1) possible comparators.A universal circuit UNIV(m, n) for circuits with at most m wires and ngates requires m′ wires and n′ gates, where
m′ = 2m(m − 1)n + m n′ = 4m(m − 1)n
12 / 17
Universal Comparator Circuits
Below is a universal comparator gadget.
1 If b = 0 then x ′ = x and y ′ = y .
2 If b = 1 then x ′ = x ∨ y and y ′ = x ∧ y .
b • • • 0
x H x ′
y H • H y ′
b H 1
To simulate one arbitrary comparator in a circuit with m wires we put inm(m − 1) gadgets in a row, for the m(m − 1) possible comparators.A universal circuit UNIV(m, n) for circuits with at most m wires and ngates requires m′ wires and n′ gates, where
m′ = 2m(m − 1)n + m n′ = 4m(m − 1)n
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Applications of Universal Circuits
FCC is closed under composition
CC is closed under AC0-Turing reductions
CC can be characterized using uniform cirucit families of comparatorcircuits, analgous to other circuit-based complexity classes such asAC0,NC1,TC1, . . ..
CC = CCSubr = CC∗
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From now on we allow NOT gates in comparator circuits.
It is easy to see that this does not change CC.
To simulate a comparator circuit with NOT gates, use ‘Double-rail logic’and DeMorgan’s laws.
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Oracle separationsWe use length-preserving functions f : 0, 1∗ → 0, 1∗ as oracles.
f = fnn∈N fn : 0, 1n → 0, 1n
Insert an oracle-gate computing fn into a comparator circuit by selectingany n wires as inputs and any n wires as outputs.
Theorem (ACN2007)
Any oracle circuit Cn(fn) with ∧,∨,¬, fn gates (arbitrary fanin and fanout)which computes f d
n , d ≥ 1, requires the oracle depth of Cn to be at least d.
Corollary
CC(f ) 6⊆ NC(f )
Proof.
Let gn = f d(n), where d(n) = (log n)Ω(1) and d(n) = nO(1). Theng ∈ CC(f ) but g 6∈ NC(f ). (Obviously CC(f ) ⊆ P(f ))
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Work in ProgressUse the superior fanout of NC circuits to show that
NC(f ) 6⊆ CC(f )
Flip Path in a comparator circuit.
Lemma
For each comparator circuit C and for each bit setting of inputs x1, . . . , xmand for each i , if xi is flipped, then there is a unique ‘flip-path’ in C fromxi to some output wire such that each wire on the path is flipped.
Proof.
This is true for each comparator gate.
p x • p ∧ q
q y H p ∨ q
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