Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman...
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Descriptive Complexity
Neil Immerman
College of Computer and Information SciencesUniversity of Massachusetts, Amherst
Amherst, MA, USA
people.cs.umass.edu/˜immerman
Neil Immerman Descriptive Complexity
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P =
∞⋃k=1
DTIME[nk ]
P is a goodmathematical
wrapper for “trulyfeasible”.
“truly feasible” isthe informal set ofproblems we can
solve exactly on allreasonably sized
instances.
r.e. completeco-r.e. complete
r.e.co-r.e.Recursive
NP completeco-NP complete
NPco-NPNP ∩ co-NP
P completeP
“truly
feasible”
FO(CFL)
FO(REGULAR)
FO
Neil Immerman Descriptive Complexity
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P =
∞⋃k=1
DTIME[nk ]
P is a goodmathematical
wrapper for “trulyfeasible”.
“truly feasible” isthe informal set ofproblems we can
solve exactly on allreasonably sized
instances.
r.e. completeco-r.e. complete
r.e.co-r.e.Recursive
NP completeco-NP complete
NPco-NPNP ∩ co-NP
P completeP
“truly
feasible”
FO(CFL)
FO(REGULAR)
FO
Neil Immerman Descriptive Complexity
![Page 4: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/4.jpg)
P =
∞⋃k=1
DTIME[nk ]
P is a goodmathematical
wrapper for “trulyfeasible”.
“truly feasible” isthe informal set ofproblems we can
solve exactly on allreasonably sized
instances.
r.e. completeco-r.e. complete
r.e.co-r.e.Recursive
NP completeco-NP complete
NPco-NPNP ∩ co-NP
P completeP
“truly
feasible”
FO(CFL)
FO(REGULAR)
FO
Neil Immerman Descriptive Complexity
![Page 5: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/5.jpg)
NTIME[t(n)]: a mathematical fiction
input w , |w | = n
N accepts w
if at least
one of the 2t(n)
paths accepts.
2s
0
00000000
0000000000000
1
t(n)
t(n)
b1 b2 b3 · · · bt(n)
Neil Immerman Descriptive Complexity
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NP =∞⋃
k=1
NTIME[nk ]
Many optimizationproblems we want
to solve are NPcomplete.
SAT, TSP,3-COLOR,
CLIQUE, . . .
As descisonproblems, all NP
complete problemsare isomorphic.
r.e. completeco-r.e. complete
r.e.co-r.e.Recursive
NP completeSAT
co-NP completeSAT
NPco-NPNP ∩ co-NP
P completeP
“truly
feasible”
FO(CFL)
FO(REGULAR)
FO
Neil Immerman Descriptive Complexity
![Page 7: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/7.jpg)
NP =∞⋃
k=1
NTIME[nk ]
Many optimizationproblems we want
to solve are NPcomplete.
SAT, TSP,3-COLOR,
CLIQUE, . . .
As descisonproblems, all NP
complete problemsare isomorphic.
r.e. completeco-r.e. complete
r.e.co-r.e.Recursive
NP completeSAT
co-NP completeSAT
NPco-NPNP ∩ co-NP
P completeP
“truly
feasible”
FO(CFL)
FO(REGULAR)
FO
Neil Immerman Descriptive Complexity
![Page 8: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/8.jpg)
NP =∞⋃
k=1
NTIME[nk ]
Many optimizationproblems we want
to solve are NPcomplete.
SAT, TSP,3-COLOR,
CLIQUE, . . .
As descisonproblems, all NP
complete problemsare isomorphic.
r.e. completeco-r.e. complete
r.e.co-r.e.Recursive
NP completeSAT
co-NP completeSAT
NPco-NPNP ∩ co-NP
P completeP
“truly
feasible”
FO(CFL)
FO(REGULAR)
FO
Neil Immerman Descriptive Complexity
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NP =∞⋃
k=1
NTIME[nk ]
Many optimizationproblems we want
to solve are NPcomplete.
SAT, TSP,3-COLOR,
CLIQUE, . . .
As descisonproblems, all NP
complete problemsare isomorphic.
r.e. completeco-r.e. complete
r.e.co-r.e.Recursive
PSPACE
NP completeSAT
co-NP completeSAT
NPco-NPNP ∩ co-NP
P completeP
“truly
feasible”
FO(CFL)
FO(REGULAR)
FO
Neil Immerman Descriptive Complexity
![Page 10: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/10.jpg)
NP =∞⋃
k=1
NTIME[nk ]
Many optimizationproblems we want
to solve are NPcomplete.
SAT, TSP,3-COLOR,
CLIQUE, . . .
As descisonproblems, all NP
complete problemsare isomorphic.
r.e. completeco-r.e. complete
r.e.co-r.e.Recursive
EXPTIME
PSPACE
NP completeSAT
co-NP completeSAT
NPco-NPNP ∩ co-NP
P completeP
“truly
feasible”
FO(CFL)
FO(REGULAR)
FO
Neil Immerman Descriptive Complexity
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Descriptive Complexity
Queryq1 q2 · · · qn
7→ Computation 7→ Answera1 a2 · · · ai · · · am
· · · S · · ·
Restrict attention to the complexity of computing individual bitsof the output, i.e., decision problems.
How hard is it to check if input has property S ?
How rich a language do we need to express property S?
There is a constructive isomorphism between these twoapproaches.
Neil Immerman Descriptive Complexity
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Descriptive Complexity
Queryq1 q2 · · · qn
7→ Computation 7→ Answera1 a2 · · · ai · · · am
· · · S · · ·
Restrict attention to the complexity of computing individual bitsof the output, i.e., decision problems.
How hard is it to check if input has property S ?
How rich a language do we need to express property S?
There is a constructive isomorphism between these twoapproaches.
Neil Immerman Descriptive Complexity
![Page 13: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/13.jpg)
Descriptive Complexity
Queryq1 q2 · · · qn
7→ Computation 7→ Answera1 a2 · · · ai · · · am
· · · S · · ·
Restrict attention to the complexity of computing individual bitsof the output, i.e., decision problems.
How hard is it to check if input has property S ?
How rich a language do we need to express property S?
There is a constructive isomorphism between these twoapproaches.
Neil Immerman Descriptive Complexity
![Page 14: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/14.jpg)
Descriptive Complexity
Queryq1 q2 · · · qn
7→ Computation 7→ Answera1 a2 · · · ai · · · am
· · · S · · ·
Restrict attention to the complexity of computing individual bitsof the output, i.e., decision problems.
How hard is it to check if input has property S ?
How rich a language do we need to express property S?
There is a constructive isomorphism between these twoapproaches.
Neil Immerman Descriptive Complexity
![Page 15: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/15.jpg)
Descriptive Complexity
Queryq1 q2 · · · qn
7→ Computation 7→ Answera1 a2 · · · ai · · · am
· · · S · · ·
Restrict attention to the complexity of computing individual bitsof the output, i.e., decision problems.
How hard is it to check if input has property S ?
How rich a language do we need to express property S?
There is a constructive isomorphism between these twoapproaches.
Neil Immerman Descriptive Complexity
![Page 16: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/16.jpg)
Think of the Input as a Finite Logical Structure
Graph G = (v1, . . . , vn,≤, E , s, t)
Σg = (E2, s, t)
r rr r
rrr
r*
HHHHj
*
HHHHj
HHHHj
*s t
Binary String Aw = (p1, . . . ,p8,≤ ,S)S = p2,p5,p7,p8
Σs = (S1) w = 01001011
Neil Immerman Descriptive Complexity
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First-Order Logic
input symbols: from Σvariables: x , y , z, . . .
boolean connectives: ∧,∨,¬quantifiers: ∀, ∃
numeric symbols: =,≤,+,×,min,max
α ≡ ∀x∃y(E(x , y)) ∈ L(Σg)
β ≡ ∃x∀y(x ≤ y ∧ S(x)) ∈ L(Σs)
β ≡ S(min) ∈ L(Σs)
In this setting, with the structure of interest being the finiteinput, FO is a weak, low-level complexity class.
Neil Immerman Descriptive Complexity
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First-Order Logic
input symbols: from Σvariables: x , y , z, . . .
boolean connectives: ∧,∨,¬quantifiers: ∀, ∃
numeric symbols: =,≤,+,×,min,max
α ≡ ∀x∃y(E(x , y)) ∈ L(Σg)
β ≡ ∃x∀y(x ≤ y ∧ S(x)) ∈ L(Σs)
β ≡ S(min) ∈ L(Σs)
In this setting, with the structure of interest being the finiteinput, FO is a weak, low-level complexity class.
Neil Immerman Descriptive Complexity
![Page 19: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/19.jpg)
Second-Order Logic: FO plus Relation Variables
Fagin’s Theorem: NP = SO∃
Φ3color ≡ ∃R1 G1 B1 ∀ x y ((R(x) ∨G(x) ∨ B(x)) ∧ (E(x , y)→(¬(R(x) ∧ R(y)) ∧ ¬(G(x) ∧G(y)) ∧ ¬(B(x) ∧ B(y)))))
a
ds
b
c
g
t
f
e
a
s
b
c
g
d t
f
e
Neil Immerman Descriptive Complexity
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Second-Order Logic: FO plus Relation Variables
Fagin’s Theorem: NP = SO∃
Φ3color ≡ ∃R1 G1 B1 ∀ x y ((R(x) ∨G(x) ∨ B(x)) ∧ (E(x , y)→(¬(R(x) ∧ R(y)) ∧ ¬(G(x) ∧G(y)) ∧ ¬(B(x) ∧ B(y)))))
a
ds
b
c
g
t
f
e
a
s
b
c
g
d t
f
e
Neil Immerman Descriptive Complexity
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r.e. completeco-r.e. complete
r.e.co-r.e.Recursive
EXPTIME
PSPACE
PTIME Hierarchy SO NP completeSAT
co-NP completeSAT
SO∃ NPSO∀co-NPNP ∩ co-NP
P completeP
“truly
feasible”
FO(CFL)
FO(REGULAR)
FO
Neil Immerman Descriptive Complexity
![Page 22: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/22.jpg)
Addition is First-Order
Q+ : STRUC[ΣAB]→ STRUC[Σs]
A a1 a2 . . . an−1 anB + b1 b2 . . . bn−1 bn
S s1 s2 . . . sn−1 sn
C(i) ≡ (∃j > i)(
A(j) ∧ B(j) ∧
(∀k .j > k > i)(A(k) ∨ B(k)))
Q+(i) ≡ A(i) ⊕ B(i) ⊕ C(i)
Neil Immerman Descriptive Complexity
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Addition is First-Order
Q+ : STRUC[ΣAB]→ STRUC[Σs]
A a1 a2 . . . an−1 anB + b1 b2 . . . bn−1 bn
S s1 s2 . . . sn−1 sn
C(i) ≡ (∃j > i)(
A(j) ∧ B(j) ∧
(∀k .j > k > i)(A(k) ∨ B(k)))
Q+(i) ≡ A(i) ⊕ B(i) ⊕ C(i)
Neil Immerman Descriptive Complexity
![Page 24: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/24.jpg)
Addition is First-Order
Q+ : STRUC[ΣAB]→ STRUC[Σs]
A a1 a2 . . . an−1 anB + b1 b2 . . . bn−1 bn
S s1 s2 . . . sn−1 sn
C(i) ≡ (∃j > i)(
A(j) ∧ B(j) ∧
(∀k .j > k > i)(A(k) ∨ B(k)))
Q+(i) ≡ A(i) ⊕ B(i) ⊕ C(i)
Neil Immerman Descriptive Complexity
![Page 25: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/25.jpg)
Parallel Machines:
Quantifiers are Parallel
CRAM[t(n)] = CRCW-PRAM-TIME[t(n)]-HARD[nO(1)]
Assume array A[x ] : x = 1, . . . , r in memory.
∀x(A(x)) ≡ write(1); proc pi : if (A[i] = 0) then write(0)
PP
PP
PP
PP
PP PP
Memory
Global
r
2
3
1
4
5
10
Neil Immerman Descriptive Complexity
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Parallel Machines: Quantifiers are Parallel
CRAM[t(n)] = CRCW-PRAM-TIME[t(n)]-HARD[nO(1)]
Assume array A[x ] : x = 1, . . . , r in memory.
∀x(A(x)) ≡ write(1); proc pi : if (A[i] = 0) then write(0)
PP
PP
PP
PP
PP PP
Memory
Global
r
2
3
1
4
5
10
Neil Immerman Descriptive Complexity
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Parallel Machines: Quantifiers are Parallel
CRAM[t(n)] = CRCW-PRAM-TIME[t(n)]-HARD[nO(1)]
Assume array A[x ] : x = 1, . . . , r in memory.
∀x(A(x)) ≡ write(1);
proc pi : if (A[i] = 0) then write(0)
PP
PP
PP
PP
PP PP
Memory
Global
r
2
3
1
4
5
1
0
Neil Immerman Descriptive Complexity
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Parallel Machines: Quantifiers are Parallel
CRAM[t(n)] = CRCW-PRAM-TIME[t(n)]-HARD[nO(1)]
Assume array A[x ] : x = 1, . . . , r in memory.
∀x(A(x)) ≡ write(1); proc pi : if (A[i] = 0) then write(0)
PP
PP
PP
PP
PP PP
Memory
Global
r
2
3
1
4
5
1
0
Neil Immerman Descriptive Complexity
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FO
=
CRAM[1]
=
AC0
=
Logarithmic-TimeHierarchy
Arithmetic Hierarchy FO(N) r.e. completeHalt
co-r.e. completeHalt
r.e. FO∃(N)co-r.e. FO∀(N)Recursive
EXPTIME
PSPACE
PTIME Hierarchy SO NP completeSAT
co-NP completeSAT
NP SO∃co-NP SO∀NP ∩ co-NP
P completeP
“truly
feasible”
FO(CFL)
FO(REGULAR)
CRAM[1] AC0FO LOGTIME Hierarchy
Neil Immerman Descriptive Complexity
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Inductive Definitions and Least Fixed Point
E?(x , y)def= x = y ∨ E(x , y) ∨ ∃z(E?(x , z) ∧ E?(z, y))
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
ϕGtc : binRel(G) → binRel(G) is a monotone operator
G ∈ REACH ⇔ G |= (LFPϕtc)(s, t) E? = (LFPϕtc)
REACH =
G, s, t∣∣ s ?→ t
REACH 6∈ FO
r rr r
rrr
r*
HHHHj
*
HHHHj
HHHHj
*s t
Neil Immerman Descriptive Complexity
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Inductive Definitions and Least Fixed Point
E?(x , y)def= x = y ∨ E(x , y) ∨ ∃z(E?(x , z) ∧ E?(z, y))
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
ϕGtc : binRel(G) → binRel(G) is a monotone operator
G ∈ REACH ⇔ G |= (LFPϕtc)(s, t) E? = (LFPϕtc)
REACH =
G, s, t∣∣ s ?→ t
REACH 6∈ FO
r rr r
rrr
r*
HHHHj
*
HHHHj
HHHHj
*s t
Neil Immerman Descriptive Complexity
![Page 32: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/32.jpg)
Inductive Definitions and Least Fixed Point
E?(x , y)def= x = y ∨ E(x , y) ∨ ∃z(E?(x , z) ∧ E?(z, y))
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
ϕGtc : binRel(G) → binRel(G) is a monotone operator
G ∈ REACH ⇔ G |= (LFPϕtc)(s, t) E? = (LFPϕtc)
REACH =
G, s, t∣∣ s ?→ t
REACH 6∈ FO
r rr r
rrr
r*
HHHHj
*
HHHHj
HHHHj
*s t
Neil Immerman Descriptive Complexity
![Page 33: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/33.jpg)
Inductive Definitions and Least Fixed Point
E?(x , y)def= x = y ∨ E(x , y) ∨ ∃z(E?(x , z) ∧ E?(z, y))
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
ϕGtc : binRel(G) → binRel(G) is a monotone operator
G ∈ REACH ⇔ G |= (LFPϕtc)(s, t) E? = (LFPϕtc)
REACH =
G, s, t∣∣ s ?→ t
REACH 6∈ FO
r rr r
rrr
r*
HHHHj
*
HHHHj
HHHHj
*s t
Neil Immerman Descriptive Complexity
![Page 34: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/34.jpg)
Inductive Definitions and Least Fixed Point
E?(x , y)def= x = y ∨ E(x , y) ∨ ∃z(E?(x , z) ∧ E?(z, y))
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
ϕGtc : binRel(G) → binRel(G) is a monotone operator
G ∈ REACH ⇔ G |= (LFPϕtc)(s, t) E? = (LFPϕtc)
REACH =
G, s, t∣∣ s ?→ t
REACH 6∈ FO
r rr r
rrr
r*
HHHHj
*
HHHHj
HHHHj
*s t
Neil Immerman Descriptive Complexity
![Page 35: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/35.jpg)
Inductive Definitions and Least Fixed Point
E?(x , y)def= x = y ∨ E(x , y) ∨ ∃z(E?(x , z) ∧ E?(z, y))
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
ϕGtc : binRel(G) → binRel(G) is a monotone operator
G ∈ REACH ⇔ G |= (LFPϕtc)(s, t)
E? = (LFPϕtc)
REACH =
G, s, t∣∣ s ?→ t
REACH 6∈ FO
r rr r
rrr
r*
HHHHj
*
HHHHj
HHHHj
*s t
Neil Immerman Descriptive Complexity
![Page 36: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/36.jpg)
Inductive Definitions and Least Fixed Point
E?(x , y)def= x = y ∨ E(x , y) ∨ ∃z(E?(x , z) ∧ E?(z, y))
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
ϕGtc : binRel(G) → binRel(G) is a monotone operator
G ∈ REACH ⇔ G |= (LFPϕtc)(s, t) E? = (LFPϕtc)
REACH =
G, s, t∣∣ s ?→ t
REACH 6∈ FO
r rr r
rrr
r*
HHHHj
*
HHHHj
HHHHj
*s t
Neil Immerman Descriptive Complexity
![Page 37: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/37.jpg)
Tarski-Knaster Theorem
Thm. If ϕ : Relk (G)→ Relk (G) is monotone, then LFP(ϕ)exists and can be computed in P.
proof: Monotone means, for all R ⊆ S, ϕ(R) ⊆ ϕ(S).
Let I0 def= ∅; Ir+1 def
= ϕ(Ir ) Thus, ∅ = I0 ⊆ I1 ⊆ · · · ⊆ I t .
Let t be min such that I t = I t+1. Note that t ≤ nk wheren = |V G|. ϕ(I t ) = I t , so I t is a fixed point of ϕ.
Suppose ϕ(F ) = F . By induction on r , for all r , Ir ⊆ F .
base case: I0 = ∅ ⊆ F .
inductive case: Assume I j ⊆ F
By monotonicity, ϕ(I j) ⊆ ϕ(F ), i.e., I j+1 ⊆ F .
Thus I t ⊆ F and I t = LFP(ϕ).
Neil Immerman Descriptive Complexity
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Tarski-Knaster Theorem
Thm. If ϕ : Relk (G)→ Relk (G) is monotone, then LFP(ϕ)exists and can be computed in P.
proof: Monotone means, for all R ⊆ S, ϕ(R) ⊆ ϕ(S).
Let I0 def= ∅; Ir+1 def
= ϕ(Ir ) Thus, ∅ = I0 ⊆ I1 ⊆ · · · ⊆ I t .
Let t be min such that I t = I t+1. Note that t ≤ nk wheren = |V G|. ϕ(I t ) = I t , so I t is a fixed point of ϕ.
Suppose ϕ(F ) = F . By induction on r , for all r , Ir ⊆ F .
base case: I0 = ∅ ⊆ F .
inductive case: Assume I j ⊆ F
By monotonicity, ϕ(I j) ⊆ ϕ(F ), i.e., I j+1 ⊆ F .
Thus I t ⊆ F and I t = LFP(ϕ).
Neil Immerman Descriptive Complexity
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Tarski-Knaster Theorem
Thm. If ϕ : Relk (G)→ Relk (G) is monotone, then LFP(ϕ)exists and can be computed in P.
proof: Monotone means, for all R ⊆ S, ϕ(R) ⊆ ϕ(S).
Let I0 def= ∅; Ir+1 def
= ϕ(Ir )
Thus, ∅ = I0 ⊆ I1 ⊆ · · · ⊆ I t .
Let t be min such that I t = I t+1. Note that t ≤ nk wheren = |V G|. ϕ(I t ) = I t , so I t is a fixed point of ϕ.
Suppose ϕ(F ) = F . By induction on r , for all r , Ir ⊆ F .
base case: I0 = ∅ ⊆ F .
inductive case: Assume I j ⊆ F
By monotonicity, ϕ(I j) ⊆ ϕ(F ), i.e., I j+1 ⊆ F .
Thus I t ⊆ F and I t = LFP(ϕ).
Neil Immerman Descriptive Complexity
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Tarski-Knaster Theorem
Thm. If ϕ : Relk (G)→ Relk (G) is monotone, then LFP(ϕ)exists and can be computed in P.
proof: Monotone means, for all R ⊆ S, ϕ(R) ⊆ ϕ(S).
Let I0 def= ∅; Ir+1 def
= ϕ(Ir ) Thus, ∅ = I0 ⊆ I1 ⊆ · · · ⊆ I t .
Let t be min such that I t = I t+1. Note that t ≤ nk wheren = |V G|. ϕ(I t ) = I t , so I t is a fixed point of ϕ.
Suppose ϕ(F ) = F . By induction on r , for all r , Ir ⊆ F .
base case: I0 = ∅ ⊆ F .
inductive case: Assume I j ⊆ F
By monotonicity, ϕ(I j) ⊆ ϕ(F ), i.e., I j+1 ⊆ F .
Thus I t ⊆ F and I t = LFP(ϕ).
Neil Immerman Descriptive Complexity
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Tarski-Knaster Theorem
Thm. If ϕ : Relk (G)→ Relk (G) is monotone, then LFP(ϕ)exists and can be computed in P.
proof: Monotone means, for all R ⊆ S, ϕ(R) ⊆ ϕ(S).
Let I0 def= ∅; Ir+1 def
= ϕ(Ir ) Thus, ∅ = I0 ⊆ I1 ⊆ · · · ⊆ I t .
Let t be min such that I t = I t+1. Note that t ≤ nk wheren = |V G|.
ϕ(I t ) = I t , so I t is a fixed point of ϕ.
Suppose ϕ(F ) = F . By induction on r , for all r , Ir ⊆ F .
base case: I0 = ∅ ⊆ F .
inductive case: Assume I j ⊆ F
By monotonicity, ϕ(I j) ⊆ ϕ(F ), i.e., I j+1 ⊆ F .
Thus I t ⊆ F and I t = LFP(ϕ).
Neil Immerman Descriptive Complexity
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Tarski-Knaster Theorem
Thm. If ϕ : Relk (G)→ Relk (G) is monotone, then LFP(ϕ)exists and can be computed in P.
proof: Monotone means, for all R ⊆ S, ϕ(R) ⊆ ϕ(S).
Let I0 def= ∅; Ir+1 def
= ϕ(Ir ) Thus, ∅ = I0 ⊆ I1 ⊆ · · · ⊆ I t .
Let t be min such that I t = I t+1. Note that t ≤ nk wheren = |V G|. ϕ(I t ) = I t , so I t is a fixed point of ϕ.
Suppose ϕ(F ) = F . By induction on r , for all r , Ir ⊆ F .
base case: I0 = ∅ ⊆ F .
inductive case: Assume I j ⊆ F
By monotonicity, ϕ(I j) ⊆ ϕ(F ), i.e., I j+1 ⊆ F .
Thus I t ⊆ F and I t = LFP(ϕ).
Neil Immerman Descriptive Complexity
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Tarski-Knaster Theorem
Thm. If ϕ : Relk (G)→ Relk (G) is monotone, then LFP(ϕ)exists and can be computed in P.
proof: Monotone means, for all R ⊆ S, ϕ(R) ⊆ ϕ(S).
Let I0 def= ∅; Ir+1 def
= ϕ(Ir ) Thus, ∅ = I0 ⊆ I1 ⊆ · · · ⊆ I t .
Let t be min such that I t = I t+1. Note that t ≤ nk wheren = |V G|. ϕ(I t ) = I t , so I t is a fixed point of ϕ.
Suppose ϕ(F ) = F .
By induction on r , for all r , Ir ⊆ F .
base case: I0 = ∅ ⊆ F .
inductive case: Assume I j ⊆ F
By monotonicity, ϕ(I j) ⊆ ϕ(F ), i.e., I j+1 ⊆ F .
Thus I t ⊆ F and I t = LFP(ϕ).
Neil Immerman Descriptive Complexity
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Tarski-Knaster Theorem
Thm. If ϕ : Relk (G)→ Relk (G) is monotone, then LFP(ϕ)exists and can be computed in P.
proof: Monotone means, for all R ⊆ S, ϕ(R) ⊆ ϕ(S).
Let I0 def= ∅; Ir+1 def
= ϕ(Ir ) Thus, ∅ = I0 ⊆ I1 ⊆ · · · ⊆ I t .
Let t be min such that I t = I t+1. Note that t ≤ nk wheren = |V G|. ϕ(I t ) = I t , so I t is a fixed point of ϕ.
Suppose ϕ(F ) = F . By induction on r , for all r , Ir ⊆ F .
base case: I0 = ∅ ⊆ F .
inductive case: Assume I j ⊆ F
By monotonicity, ϕ(I j) ⊆ ϕ(F ), i.e., I j+1 ⊆ F .
Thus I t ⊆ F and I t = LFP(ϕ).
Neil Immerman Descriptive Complexity
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Tarski-Knaster Theorem
Thm. If ϕ : Relk (G)→ Relk (G) is monotone, then LFP(ϕ)exists and can be computed in P.
proof: Monotone means, for all R ⊆ S, ϕ(R) ⊆ ϕ(S).
Let I0 def= ∅; Ir+1 def
= ϕ(Ir ) Thus, ∅ = I0 ⊆ I1 ⊆ · · · ⊆ I t .
Let t be min such that I t = I t+1. Note that t ≤ nk wheren = |V G|. ϕ(I t ) = I t , so I t is a fixed point of ϕ.
Suppose ϕ(F ) = F . By induction on r , for all r , Ir ⊆ F .
base case: I0 = ∅ ⊆ F .
inductive case: Assume I j ⊆ F
By monotonicity, ϕ(I j) ⊆ ϕ(F ), i.e., I j+1 ⊆ F .
Thus I t ⊆ F and I t = LFP(ϕ).
Neil Immerman Descriptive Complexity
![Page 46: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/46.jpg)
Tarski-Knaster Theorem
Thm. If ϕ : Relk (G)→ Relk (G) is monotone, then LFP(ϕ)exists and can be computed in P.
proof: Monotone means, for all R ⊆ S, ϕ(R) ⊆ ϕ(S).
Let I0 def= ∅; Ir+1 def
= ϕ(Ir ) Thus, ∅ = I0 ⊆ I1 ⊆ · · · ⊆ I t .
Let t be min such that I t = I t+1. Note that t ≤ nk wheren = |V G|. ϕ(I t ) = I t , so I t is a fixed point of ϕ.
Suppose ϕ(F ) = F . By induction on r , for all r , Ir ⊆ F .
base case: I0 = ∅ ⊆ F .
inductive case: Assume I j ⊆ F
By monotonicity, ϕ(I j) ⊆ ϕ(F ), i.e., I j+1 ⊆ F .
Thus I t ⊆ F and I t = LFP(ϕ).
Neil Immerman Descriptive Complexity
![Page 47: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/47.jpg)
Tarski-Knaster Theorem
Thm. If ϕ : Relk (G)→ Relk (G) is monotone, then LFP(ϕ)exists and can be computed in P.
proof: Monotone means, for all R ⊆ S, ϕ(R) ⊆ ϕ(S).
Let I0 def= ∅; Ir+1 def
= ϕ(Ir ) Thus, ∅ = I0 ⊆ I1 ⊆ · · · ⊆ I t .
Let t be min such that I t = I t+1. Note that t ≤ nk wheren = |V G|. ϕ(I t ) = I t , so I t is a fixed point of ϕ.
Suppose ϕ(F ) = F . By induction on r , for all r , Ir ⊆ F .
base case: I0 = ∅ ⊆ F .
inductive case: Assume I j ⊆ F
By monotonicity, ϕ(I j) ⊆ ϕ(F ), i.e., I j+1 ⊆ F .
Thus I t ⊆ F and I t = LFP(ϕ).
Neil Immerman Descriptive Complexity
![Page 48: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/48.jpg)
Tarski-Knaster Theorem
Thm. If ϕ : Relk (G)→ Relk (G) is monotone, then LFP(ϕ)exists and can be computed in P.
proof: Monotone means, for all R ⊆ S, ϕ(R) ⊆ ϕ(S).
Let I0 def= ∅; Ir+1 def
= ϕ(Ir ) Thus, ∅ = I0 ⊆ I1 ⊆ · · · ⊆ I t .
Let t be min such that I t = I t+1. Note that t ≤ nk wheren = |V G|. ϕ(I t ) = I t , so I t is a fixed point of ϕ.
Suppose ϕ(F ) = F . By induction on r , for all r , Ir ⊆ F .
base case: I0 = ∅ ⊆ F .
inductive case: Assume I j ⊆ F
By monotonicity, ϕ(I j) ⊆ ϕ(F ), i.e., I j+1 ⊆ F .
Thus I t ⊆ F and I t = LFP(ϕ).
Neil Immerman Descriptive Complexity
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Inductive Definition of Transitive Closure
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
I1 = ϕGtc(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 1
I2 = (ϕGtc)2(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2
I3 = (ϕGtc)3(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 4
... =...
...
Ir = (ϕGtc)r (∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2r−1... =
......
(ϕGtc)d1+log ne(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ n
LFP(ϕtc) = ϕd1+log netc (∅); REACH ∈ IND[log n]
Next we will show that IND[t(n)] = FO[t(n)].
Neil Immerman Descriptive Complexity
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Inductive Definition of Transitive Closure
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
I1 = ϕGtc(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 1
I2 = (ϕGtc)2(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2
I3 = (ϕGtc)3(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 4
... =...
...
Ir = (ϕGtc)r (∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2r−1... =
......
(ϕGtc)d1+log ne(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ n
LFP(ϕtc) = ϕd1+log netc (∅); REACH ∈ IND[log n]
Next we will show that IND[t(n)] = FO[t(n)].
Neil Immerman Descriptive Complexity
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Inductive Definition of Transitive Closure
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
I1 = ϕGtc(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 1
I2 = (ϕGtc)2(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2
I3 = (ϕGtc)3(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 4
... =...
...
Ir = (ϕGtc)r (∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2r−1... =
......
(ϕGtc)d1+log ne(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ n
LFP(ϕtc) = ϕd1+log netc (∅); REACH ∈ IND[log n]
Next we will show that IND[t(n)] = FO[t(n)].
Neil Immerman Descriptive Complexity
![Page 52: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/52.jpg)
Inductive Definition of Transitive Closure
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
I1 = ϕGtc(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 1
I2 = (ϕGtc)2(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2
I3 = (ϕGtc)3(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 4
... =...
...
Ir = (ϕGtc)r (∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2r−1... =
......
(ϕGtc)d1+log ne(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ n
LFP(ϕtc) = ϕd1+log netc (∅); REACH ∈ IND[log n]
Next we will show that IND[t(n)] = FO[t(n)].
Neil Immerman Descriptive Complexity
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Inductive Definition of Transitive Closure
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
I1 = ϕGtc(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 1
I2 = (ϕGtc)2(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2
I3 = (ϕGtc)3(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 4
... =...
...
Ir = (ϕGtc)r (∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2r−1... =
......
(ϕGtc)d1+log ne(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ n
LFP(ϕtc) = ϕd1+log netc (∅); REACH ∈ IND[log n]
Next we will show that IND[t(n)] = FO[t(n)].
Neil Immerman Descriptive Complexity
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Inductive Definition of Transitive Closure
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
I1 = ϕGtc(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 1
I2 = (ϕGtc)2(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2
I3 = (ϕGtc)3(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 4
... =...
...
Ir = (ϕGtc)r (∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2r−1... =
......
(ϕGtc)d1+log ne(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ n
LFP(ϕtc) = ϕd1+log netc (∅); REACH ∈ IND[log n]
Next we will show that IND[t(n)] = FO[t(n)].
Neil Immerman Descriptive Complexity
![Page 55: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/55.jpg)
Inductive Definition of Transitive Closure
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
I1 = ϕGtc(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 1
I2 = (ϕGtc)2(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2
I3 = (ϕGtc)3(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 4
... =...
...
Ir = (ϕGtc)r (∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2r−1... =
......
(ϕGtc)d1+log ne(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ n
LFP(ϕtc) = ϕd1+log netc (∅); REACH ∈ IND[log n]
Next we will show that IND[t(n)] = FO[t(n)].
Neil Immerman Descriptive Complexity
![Page 56: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/56.jpg)
Inductive Definition of Transitive Closure
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z(R(x , z) ∧ R(z, y))
I1 = ϕGtc(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 1
I2 = (ϕGtc)2(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2
I3 = (ϕGtc)3(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 4
... =...
...
Ir = (ϕGtc)r (∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ 2r−1... =
......
(ϕGtc)d1+log ne(∅) =
(a,b) ∈ V G × V G
∣∣ dist(a,b) ≤ n
LFP(ϕtc) = ϕd1+log netc (∅); REACH ∈ IND[log n]
Next we will show that IND[t(n)] = FO[t(n)].Neil Immerman Descriptive Complexity
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ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z (R(x , z)∧R(z, y))
1. Dummy universal quantification for base case:
ϕtc(R, x , y) ≡ (∀z.M1)(∃z)(R(x , z) ∧ R(z, y))
M1 ≡ ¬(x = y ∨ E(x , y))
2. Using ∀, replace two occurrences of R with one:
ϕtc(R, x , y) ≡ (∀z.M1)(∃z)(∀uv .M2)R(u, v)
M2 ≡ (u = x ∧ v = z) ∨ (u = z ∧ v = y)
3. Requantify x and y .
M3 ≡ (x = u ∧ y = v)
ϕtc(R, x , y) ≡ [ (∀z.M1)(∃z)(∀uv .M2)(∃xy .M3) ] R(x , y)
Every FO inductive definition is equivalent to a quantifier block.
Neil Immerman Descriptive Complexity
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ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z (R(x , z)∧R(z, y))
1. Dummy universal quantification for base case:
ϕtc(R, x , y) ≡ (∀z.M1)(∃z)(R(x , z) ∧ R(z, y))
M1 ≡ ¬(x = y ∨ E(x , y))
2. Using ∀, replace two occurrences of R with one:
ϕtc(R, x , y) ≡ (∀z.M1)(∃z)(∀uv .M2)R(u, v)
M2 ≡ (u = x ∧ v = z) ∨ (u = z ∧ v = y)
3. Requantify x and y .
M3 ≡ (x = u ∧ y = v)
ϕtc(R, x , y) ≡ [ (∀z.M1)(∃z)(∀uv .M2)(∃xy .M3) ] R(x , y)
Every FO inductive definition is equivalent to a quantifier block.
Neil Immerman Descriptive Complexity
![Page 59: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/59.jpg)
ϕtc(R, x , y) ≡ x = y ∨ E(x , y) ∨ ∃z (R(x , z)∧R(z, y))
1. Dummy universal quantification for base case:
ϕtc(R, x , y) ≡ (∀z.M1)(∃z)(R(x , z) ∧ R(z, y))
M1 ≡ ¬(x = y ∨ E(x , y))
2. Using ∀, replace two occurrences of R with one:
ϕtc(R, x , y) ≡ (∀z.M1)(∃z)(∀uv .M2)R(u, v)
M2 ≡ (u = x ∧ v = z) ∨ (u = z ∧ v = y)
3. Requantify x and y .
M3 ≡ (x = u ∧ y = v)
ϕtc(R, x , y) ≡ [ (∀z.M1)(∃z)(∀uv .M2)(∃xy .M3) ] R(x , y)
Every FO inductive definition is equivalent to a quantifier block.
Neil Immerman Descriptive Complexity
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QBtc ≡ [(∀z.M1)(∃z)(∀uv .M2)(∀xy .M3)]
ϕtc(R, x , y) ≡ [(∀z.M1)(∃z)(∀uv .M2)(∃xy .M3)]R(x , y)
ϕtc(R, x , y) ≡ [QBtc]R(x , y)
ϕrtc(∅) ≡ [QBtc]r (false)
Thus, for any structure A ∈ STRUC[Σg],
A ∈ REACH ⇔ A |= (LFPϕtc)(s, t)
⇔ A |= ([QBtc]d1+log ||A||e false)(s, t)
Neil Immerman Descriptive Complexity
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QBtc ≡ [(∀z.M1)(∃z)(∀uv .M2)(∀xy .M3)]
ϕtc(R, x , y) ≡ [(∀z.M1)(∃z)(∀uv .M2)(∃xy .M3)]R(x , y)
ϕtc(R, x , y) ≡ [QBtc]R(x , y)
ϕrtc(∅) ≡ [QBtc]r (false)
Thus, for any structure A ∈ STRUC[Σg],
A ∈ REACH ⇔ A |= (LFPϕtc)(s, t)
⇔ A |= ([QBtc]d1+log ||A||e false)(s, t)
Neil Immerman Descriptive Complexity
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QBtc ≡ [(∀z.M1)(∃z)(∀uv .M2)(∀xy .M3)]
ϕtc(R, x , y) ≡ [(∀z.M1)(∃z)(∀uv .M2)(∃xy .M3)]R(x , y)
ϕtc(R, x , y) ≡ [QBtc]R(x , y)
ϕrtc(∅) ≡ [QBtc]r (false)
Thus, for any structure A ∈ STRUC[Σg],
A ∈ REACH ⇔ A |= (LFPϕtc)(s, t)
⇔ A |= ([QBtc]d1+log ||A||e false)(s, t)
Neil Immerman Descriptive Complexity
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QBtc ≡ [(∀z.M1)(∃z)(∀uv .M2)(∀xy .M3)]
ϕtc(R, x , y) ≡ [(∀z.M1)(∃z)(∀uv .M2)(∃xy .M3)]R(x , y)
ϕtc(R, x , y) ≡ [QBtc]R(x , y)
ϕrtc(∅) ≡ [QBtc]r (false)
Thus, for any structure A ∈ STRUC[Σg],
A ∈ REACH ⇔ A |= (LFPϕtc)(s, t)
⇔ A |= ([QBtc]d1+log ||A||e false)(s, t)
Neil Immerman Descriptive Complexity
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CRAM[t(n)] = concurrent parallel random access machine;polynomial hardware, parallel time O(t(n))
IND[t(n)] = first-order, depth t(n) inductive definitions
FO[t(n)] = t(n) repetitions of a block of restricted quantifiers:
QB = [(Q1x1.M1) · · · (Qkxk .Mk )]; Mi quantifier-free
ϕn = [QB][QB] · · · [QB]︸ ︷︷ ︸t(n)
M0
Neil Immerman Descriptive Complexity
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parallel time = inductive depth = QB iterationThm. For all constructible, polynomially bounded t(n),
CRAM[t(n)] = IND[t(n)] = FO[t(n)]
proof idea: CRAM[t(n)] ⊇ FO[t(n)]: For QB with k variables,keep in memory current value of formula on all possibleassignments, using nk bits of global memory.Simulate each next quantifier in constant parallel time.
CRAM[t(n)] ⊆ FO[t(n)]: Inductively define new state of everybit of every register of every processor in terms of this globalstate at the previous time step.
Thm. For all t(n), even beyond polynomial,
CRAM[t(n)] = FO[t(n)]
Neil Immerman Descriptive Complexity
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parallel time = inductive depth = QB iterationThm. For all constructible, polynomially bounded t(n),
CRAM[t(n)] = IND[t(n)] = FO[t(n)]
proof idea: CRAM[t(n)] ⊇ FO[t(n)]: For QB with k variables,keep in memory current value of formula on all possibleassignments, using nk bits of global memory.
Simulate each next quantifier in constant parallel time.
CRAM[t(n)] ⊆ FO[t(n)]: Inductively define new state of everybit of every register of every processor in terms of this globalstate at the previous time step.
Thm. For all t(n), even beyond polynomial,
CRAM[t(n)] = FO[t(n)]
Neil Immerman Descriptive Complexity
![Page 67: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/67.jpg)
parallel time = inductive depth = QB iterationThm. For all constructible, polynomially bounded t(n),
CRAM[t(n)] = IND[t(n)] = FO[t(n)]
proof idea: CRAM[t(n)] ⊇ FO[t(n)]: For QB with k variables,keep in memory current value of formula on all possibleassignments, using nk bits of global memory.Simulate each next quantifier in constant parallel time.
CRAM[t(n)] ⊆ FO[t(n)]: Inductively define new state of everybit of every register of every processor in terms of this globalstate at the previous time step.
Thm. For all t(n), even beyond polynomial,
CRAM[t(n)] = FO[t(n)]
Neil Immerman Descriptive Complexity
![Page 68: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/68.jpg)
parallel time = inductive depth = QB iterationThm. For all constructible, polynomially bounded t(n),
CRAM[t(n)] = IND[t(n)] = FO[t(n)]
proof idea: CRAM[t(n)] ⊇ FO[t(n)]: For QB with k variables,keep in memory current value of formula on all possibleassignments, using nk bits of global memory.Simulate each next quantifier in constant parallel time.
CRAM[t(n)] ⊆ FO[t(n)]: Inductively define new state of everybit of every register of every processor in terms of this globalstate at the previous time step.
Thm. For all t(n), even beyond polynomial,
CRAM[t(n)] = FO[t(n)]
Neil Immerman Descriptive Complexity
![Page 69: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/69.jpg)
parallel time = inductive depth = QB iterationThm. For all constructible, polynomially bounded t(n),
CRAM[t(n)] = IND[t(n)] = FO[t(n)]
proof idea: CRAM[t(n)] ⊇ FO[t(n)]: For QB with k variables,keep in memory current value of formula on all possibleassignments, using nk bits of global memory.Simulate each next quantifier in constant parallel time.
CRAM[t(n)] ⊆ FO[t(n)]: Inductively define new state of everybit of every register of every processor in terms of this globalstate at the previous time step.
Thm. For all t(n), even beyond polynomial,
CRAM[t(n)] = FO[t(n)]
Neil Immerman Descriptive Complexity
![Page 70: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/70.jpg)
For t(n) poly bdd,
CRAM[t(n)]
=
IND[t(n)]
=
FO[t(n)]
Arithmetic Hierarchy FO(N) r.e. completeHalt
co-r.e. completeHalt
FO∃(N)r.e.co-r.e.
FO∀(N)Recursive
EXPTIME
FO[2nO(1)
]CRAM[2n
O(1)
] PSPACE
PTIME Hierarchy SO NP completeSAT
co-NP completeSAT
NPco-NPNP ∩ co-NP
P completeCRAM[nO(1)] P
FO[nO(1)]
FO(LFP)
FO[logO(1) n] CRAM[logO(1) n] NC“truly
feasible”FO[log n] CRAM[log n] AC1
FO(CFL)
FO(REGULAR)
CRAM[1] AC0FO LOGTIME Hierarchy
Neil Immerman Descriptive Complexity
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Remember that
for all t(n),
CRAM[t(n)]
=
FO[t(n)]
Arithmetic Hierarchy FO(N) r.e. completeHalt
co-r.e. completeHalt
FO∃(N)r.e.
FO∀(N)co-r.e.
Recursive
EXPTIME
FO[2nO(1)
] PSPACE
PTIME Hierarchy SO NP completeSAT
co-NP completeSAT
SO∃ NPSO∀co-NPNP ∩ co-NP
P completeHorn-SAT P
FO[nO(1)]
FO(LFP)
FO[logO(1) n] NC“truly
feasible”FO[log n] AC1
FO(CFL)
FO(REGULAR)
AC0FO LOGTIME Hierarchy
Neil Immerman Descriptive Complexity
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Number of Variables Determines Amount of Hardware
Thm. For k = 1,2, . . . , DSPACE[nk ] = VAR[k + 1]
Since variables range over a universe of size n, a constantnumber of variables can specify a polynomial number of gates.
The proof is just a more detailed look at CRAM[t(n)] = FO[t(n)].
A bounded number, k , of variables, is k log n bits andcorresponds to nk gates, i.e., polynomially much hardware.
A second-order variable of arity r is nr bits, corresponding to2nr
gates.
Neil Immerman Descriptive Complexity
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Number of Variables Determines Amount of Hardware
Thm. For k = 1,2, . . . , DSPACE[nk ] = VAR[k + 1]
Since variables range over a universe of size n, a constantnumber of variables can specify a polynomial number of gates.
The proof is just a more detailed look at CRAM[t(n)] = FO[t(n)].
A bounded number, k , of variables, is k log n bits andcorresponds to nk gates, i.e., polynomially much hardware.
A second-order variable of arity r is nr bits, corresponding to2nr
gates.
Neil Immerman Descriptive Complexity
![Page 74: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/74.jpg)
Number of Variables Determines Amount of Hardware
Thm. For k = 1,2, . . . , DSPACE[nk ] = VAR[k + 1]
Since variables range over a universe of size n, a constantnumber of variables can specify a polynomial number of gates.
The proof is just a more detailed look at CRAM[t(n)] = FO[t(n)].
A bounded number, k , of variables, is k log n bits andcorresponds to nk gates, i.e., polynomially much hardware.
A second-order variable of arity r is nr bits, corresponding to2nr
gates.
Neil Immerman Descriptive Complexity
![Page 75: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/75.jpg)
Number of Variables Determines Amount of Hardware
Thm. For k = 1,2, . . . , DSPACE[nk ] = VAR[k + 1]
Since variables range over a universe of size n, a constantnumber of variables can specify a polynomial number of gates.
The proof is just a more detailed look at CRAM[t(n)] = FO[t(n)].
A bounded number, k , of variables, is k log n bits andcorresponds to nk gates, i.e., polynomially much hardware.
A second-order variable of arity r is nr bits, corresponding to2nr
gates.
Neil Immerman Descriptive Complexity
![Page 76: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/76.jpg)
Number of Variables Determines Amount of Hardware
Thm. For k = 1,2, . . . , DSPACE[nk ] = VAR[k + 1]
Since variables range over a universe of size n, a constantnumber of variables can specify a polynomial number of gates.
The proof is just a more detailed look at CRAM[t(n)] = FO[t(n)].
A bounded number, k , of variables, is k log n bits andcorresponds to nk gates, i.e., polynomially much hardware.
A second-order variable of arity r is nr bits, corresponding to2nr
gates.
Neil Immerman Descriptive Complexity
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SO: Parallel Machines with Exponential HardwareGiven ϕ with n variables and m clauses, is ϕ ∈ 3-SAT?
With r = m2n processors, recognize 3-SAT in constant time!
Let S be the first n bits of our processor number.If processors S1, . . .Sm notice that truth assignment S makesall m clauses of ϕ true, then ϕ ∈ 3-SAT, so S1 writes a 1.
PP
PP
PP
PP
PP PP
Memory
Global
r
2
3
1
4
5
0
1
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SO: Parallel Machines with Exponential HardwareGiven ϕ with n variables and m clauses, is ϕ ∈ 3-SAT?
With r = m2n processors, recognize 3-SAT in constant time!
Let S be the first n bits of our processor number.If processors S1, . . .Sm notice that truth assignment S makesall m clauses of ϕ true, then ϕ ∈ 3-SAT, so S1 writes a 1.
PP
PP
PP
PP
PP PP
Memory
Global
r
2
3
1
4
5
0
1
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SO: Parallel Machines with Exponential HardwareGiven ϕ with n variables and m clauses, is ϕ ∈ 3-SAT?
With r = m2n processors, recognize 3-SAT in constant time!
Let S be the first n bits of our processor number.
If processors S1, . . .Sm notice that truth assignment S makesall m clauses of ϕ true, then ϕ ∈ 3-SAT, so S1 writes a 1.
PP
PP
PP
PP
PP PP
Memory
Global
r
2
3
1
4
5
0
1
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SO: Parallel Machines with Exponential HardwareGiven ϕ with n variables and m clauses, is ϕ ∈ 3-SAT?
With r = m2n processors, recognize 3-SAT in constant time!
Let S be the first n bits of our processor number.If processors S1, . . .Sm notice that truth assignment S makesall m clauses of ϕ true, then ϕ ∈ 3-SAT,
so S1 writes a 1.
PP
PP
PP
PP
PP PP
Memory
Global
r
2
3
1
4
5
0
1
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SO: Parallel Machines with Exponential HardwareGiven ϕ with n variables and m clauses, is ϕ ∈ 3-SAT?
With r = m2n processors, recognize 3-SAT in constant time!
Let S be the first n bits of our processor number.If processors S1, . . .Sm notice that truth assignment S makesall m clauses of ϕ true, then ϕ ∈ 3-SAT, so S1 writes a 1.
PP
PP
PP
PP
PP PP
Memory
Global
r
2
3
1
4
5
0
1
Neil Immerman Descriptive Complexity
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SO: Parallel Machines with Exponential Hardware
Thm. SO[t(n)] = CRAM[t(n)]-HARD[2nO(1)] .
proof: SO[t(n)] is like FO[t(n)] but using a quantifier blockcontaining both first-order and second-order quantifiers.The proof is similar to FO[t(n)] = CRAM[t(n)].
Cor.
SO = PTIME Hierarchy = CRAM[1]-HARD[2nO(1)]
SO[nO(1)] = PSPACE = CRAM[nO(1)]-HARD[2nO(1)]
SO[2nO(1)] = EXPTIME = CRAM[2nO(1)
]-HARD[2nO(1)]
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SO: Parallel Machines with Exponential Hardware
Thm. SO[t(n)] = CRAM[t(n)]-HARD[2nO(1)] .
proof: SO[t(n)] is like FO[t(n)] but using a quantifier blockcontaining both first-order and second-order quantifiers.The proof is similar to FO[t(n)] = CRAM[t(n)].
Cor.
SO = PTIME Hierarchy = CRAM[1]-HARD[2nO(1)]
SO[nO(1)] = PSPACE = CRAM[nO(1)]-HARD[2nO(1)]
SO[2nO(1)] = EXPTIME = CRAM[2nO(1)
]-HARD[2nO(1)]
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SO: Parallel Machines with Exponential Hardware
Thm. SO[t(n)] = CRAM[t(n)]-HARD[2nO(1)] .
proof: SO[t(n)] is like FO[t(n)] but using a quantifier blockcontaining both first-order and second-order quantifiers.The proof is similar to FO[t(n)] = CRAM[t(n)].
Cor.
SO = PTIME Hierarchy = CRAM[1]-HARD[2nO(1)]
SO[nO(1)] = PSPACE = CRAM[nO(1)]-HARD[2nO(1)]
SO[2nO(1)] = EXPTIME = CRAM[2nO(1)
]-HARD[2nO(1)]
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SO: Parallel Machines with Exponential Hardware
Thm. SO[t(n)] = CRAM[t(n)]-HARD[2nO(1)] .
proof: SO[t(n)] is like FO[t(n)] but using a quantifier blockcontaining both first-order and second-order quantifiers.The proof is similar to FO[t(n)] = CRAM[t(n)].
Cor.
SO = PTIME Hierarchy = CRAM[1]-HARD[2nO(1)]
SO[nO(1)] = PSPACE = CRAM[nO(1)]-HARD[2nO(1)]
SO[2nO(1)] = EXPTIME = CRAM[2nO(1)
]-HARD[2nO(1)]
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SO: Parallel Machines with Exponential Hardware
Thm. SO[t(n)] = CRAM[t(n)]-HARD[2nO(1)] .
proof: SO[t(n)] is like FO[t(n)] but using a quantifier blockcontaining both first-order and second-order quantifiers.The proof is similar to FO[t(n)] = CRAM[t(n)].
Cor.
SO = PTIME Hierarchy = CRAM[1]-HARD[2nO(1)]
SO[nO(1)] = PSPACE = CRAM[nO(1)]-HARD[2nO(1)]
SO[2nO(1)] = EXPTIME = CRAM[2nO(1)
]-HARD[2nO(1)]
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Parallel Time versus Amount of Hardware
PSPACE = FO[2nO(1)] = CRAM[2nO(1)
]-HARD[nO(1)]
= SO[nO(1)] = CRAM[nO(1)]-HARD[2nO(1)]
I We would love to understand this tradeoff.
I Is there such a thing as an inherently sequential problem?,i.e., is NC 6= P?
I Same tradeoff as number of variables vs. number ofiterations of a quantifier block.
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Parallel Time versus Amount of Hardware
PSPACE = FO[2nO(1)] = CRAM[2nO(1)
]-HARD[nO(1)]
= SO[nO(1)] = CRAM[nO(1)]-HARD[2nO(1)]
I We would love to understand this tradeoff.
I Is there such a thing as an inherently sequential problem?,i.e., is NC 6= P?
I Same tradeoff as number of variables vs. number ofiterations of a quantifier block.
Neil Immerman Descriptive Complexity
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Parallel Time versus Amount of Hardware
PSPACE = FO[2nO(1)] = CRAM[2nO(1)
]-HARD[nO(1)]
= SO[nO(1)] = CRAM[nO(1)]-HARD[2nO(1)]
I We would love to understand this tradeoff.
I Is there such a thing as an inherently sequential problem?,i.e., is NC 6= P?
I Same tradeoff as number of variables vs. number ofiterations of a quantifier block.
Neil Immerman Descriptive Complexity
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Parallel Time versus Amount of Hardware
PSPACE = FO[2nO(1)] = CRAM[2nO(1)
]-HARD[nO(1)]
= SO[nO(1)] = CRAM[nO(1)]-HARD[2nO(1)]
I We would love to understand this tradeoff.
I Is there such a thing as an inherently sequential problem?,i.e., is NC 6= P?
I Same tradeoff as number of variables vs. number ofiterations of a quantifier block.
Neil Immerman Descriptive Complexity
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SO[t(n)]
=
CRAM[t(n)]-
HARD-[2nO(1)]
Arithmetic Hierarchy FO(N) r.e. completeHalt
co-r.e. completeHalt
FO∃(N)r.e.
FO∀(N)co-r.e.
Recursive
SO[2nO(1)
] EXPTIME
FO[2nO(1)
] SO[nO(1)] PSPACE
PTIME Hierarchy SO NP completeSAT
co-NP completeSAT
SO∃ NPSO∀co-NPNP ∩ co-NP
P completeHorn-SAT P
FO[nO(1)]
FO(LFP)
FO[logO(1) n] NC“truly
feasible”FO[log n] AC1
FO(CFL)
FO(REGULAR)
AC0FO LOGTIME Hierarchy
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Recent Breakthroughs in Descriptive Complexity
Theorem [Ben Rossman] Any first-order formula with anynumeric relations (≤,+,×, . . .) that means “I have a clique ofsize k ” must have at least k/4 variables.
Creative new proof idea using Hastad’s Switching Lemma givesthe essentially optimal bound.
This lower bound is for a fixed formula, if it were for a sequenceof polynomially-sized formulas, i.e., a fixed-point formula, itwould follow that CLIQUE 6∈ P and thus P 6= NP.
Best previous bounds:
I k variables necessary and sufficient without ordering orother numeric relations [I 1980].
I Nothing was known with ordering except for the trivial factthat 2 variables are not enough.
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Recent Breakthroughs in Descriptive Complexity
Theorem [Martin Grohe] Fixed-Point Logic with Countingcaptures Polynomial Time on all classes of graphs withexcluded minors.
Grohe proves that for every class of graphs with excludedminors, there is a constant k such that two graphs of the classare isomorphic iff they agree on all k -variable formulas infixed-point logic with counting.
Using Ehrenfeucht-Fraısse games, this can be checked inpolynomial time, (O(nk (log n))). In the same time we can give acanonical description of the isomorphism type of any graph inthe class. Thus every class of graphs with excluded minorsadmits the same general polynomial time canonizationalgorithm: we’re isomorphic iff we agree on all formulas in Ckand in particular, you are isomorphic to me iff your Ck canonicaldescription is equal to mine.
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Thm. REACH is complete for NL = NSPACE[log n].
Proof: Let A ∈ NL, A = L(N), uses c log n bits of worktape.Input w , n = |w |
w 7→ CompGraph(N,w) = (V ,E , s, t)
V =
ID = 〈q,h,p〉∣∣ q ∈ States(N),h ≤ n, |p| ≤ cdlog ne
E =
(ID1, ID2)
∣∣ ID1(w) −→N
ID2(w)
s = initial ID
t = accepting ID
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NSPACE[log n] Turing Machine
2
read-only input w
worktape p
n
c(log n)
h
q
1
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Claim. w ∈ L(N) ⇔ CompGraph(N,w) ∈ REACH
v
s t
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Cor: NL ⊆ P
Proof: REACH ∈ P
P is closed under (logspace) reductions.
i.e., (B ∈ P ∧ A ≤ B) ⇒ A ∈ P
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NSPACE vs. DSPACE
Prop.NSPACE[s(n)] ⊆ NTIME[2O(s(n))] ⊆ DSPACE[2O(s(n))]
We can do much better!
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Savitch’s TheoremREACH ∈ DSPACE(log n)2
s
p
t
proof:
G ∈ REACH ⇔ G |= PATHn(s, t)PATH1(x , y) ≡ x = y ∨ E(x , y)
PATH2d (x , y) ≡ ∃z (PATHd (x , z) ∧ PATHd (z, y))
Sn(d) = space to check paths of dist. d in n-nodegraphs
Sn(n) = log n + Sn(n/2)
= O((log n)2)
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Savitch’s Theorem
DSPACE[s(n)] ⊆ NSPACE[n] ⊆ DSPACE[(s(n))2]
proof: Let A ∈ NSPACE[s(n)]; A = L(N)
w ∈ A ⇔ CompGraph(N,w) ∈ REACH
|w | = n; |CompGraph(N,w)| = 2O(s(n))
Testing if CompGraph(N,w) ∈ REACH takes space,
(log(|CompGraph(N,w)|))2 = (log(2O(s(n))))2
= O((s(n))2)
From w build CompGraph(N,w) in DSPACE[s(n)].
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REACH ∈ NL
proof: Fix G, let Nd =∣∣v
∣∣ distance(s, v) ≤ d∣∣
Claim: The following problems are in NL:1. dist(x ,d): distance(s, x) ≤ d2. NDIST(x ,d ; m): if m = Nd then ¬dist(x ,d)
proof:1. Guess the path of length ≤ d from s to x .2. Guess m vertices, v 6= x , with dist(v ,d).
c := 0;for v := 1 to n do // nondeterministically
( dist(v ,d) && v 6= x ; c + + ) ||( no-op )
if (c == m) then ACCEPT
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Claim. We can compute Nd in NL.
proof: By induction on d .
Base case: N0 = 1
Inductive step: Suppose we have Nd .
1. c := 0;2. for v := 1 to n do // nondeterministically3. ( dist(v ,d + 1); c + +) ||4. (∀z (NDIST(z,d ; Nd ) ∨ (z 6= v ∧ ¬E(z, v))))5. 6. Nd+1 := c
G ∈ REACH ⇔ NDIST(t ,n; Nn)
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Thm. NSPACE[s(n)] = co-NSPACE[s(n)].
proof: Let A ∈ NSPACE[s(n)]; A = L(N)
w ∈ A ⇔ CompGraph(N,w) ∈ REACH
|w | = n; |CompGraph(N,w)| = 2O(s(n))
Testing if CompGraph(N,w) ∈ REACH takes space,
log(|CompGraph(N,w)|) = log(2O(s(n)))
= O(s(n))
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What We Know
I Diagonalization: more of the same resource gives usmore:
DTIME[n] ⊂6= DTIME[n2],
same for DSPACE, NTIME, NSPACE, . . .
I Natural Complexity Classes have Natural CompleteProblems
SAT for NP, CVAL for P, QSAT for PSPACE, . . .
I Major Missing Idea: concept of work or conservation ofenergy in computation, i.e,
in order to solve SAT or other hard problem we must do acertain amount of computational work.
Neil Immerman Descriptive Complexity
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What We Know
I Diagonalization: more of the same resource gives usmore:
DTIME[n] ⊂6= DTIME[n2],
same for DSPACE, NTIME, NSPACE, . . .
I Natural Complexity Classes have Natural CompleteProblems
SAT for NP, CVAL for P, QSAT for PSPACE, . . .
I Major Missing Idea: concept of work or conservation ofenergy in computation, i.e,
in order to solve SAT or other hard problem we must do acertain amount of computational work.
Neil Immerman Descriptive Complexity
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What We Know
I Diagonalization: more of the same resource gives usmore:
DTIME[n] ⊂6= DTIME[n2],
same for DSPACE, NTIME, NSPACE, . . .
I Natural Complexity Classes have Natural CompleteProblems
SAT for NP, CVAL for P, QSAT for PSPACE, . . .
I Major Missing Idea: concept of work or conservation ofenergy in computation, i.e,
in order to solve SAT or other hard problem we must do acertain amount of computational work.
Neil Immerman Descriptive Complexity
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Strong Lower Bounds on FO[t(n)] for small t(n)
I [Sipser]: strict first-order alternation hierarchy: FO.
I [Beame-Hastad]: hierarchy remains strict up toFO[log n/ log log n].
I NC1 ⊆ FO[log n/ log log n] and this is tight.
I Does REACH require FO[log n]? This would implyNC1 6= NL.
Neil Immerman Descriptive Complexity
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Strong Lower Bounds on FO[t(n)] for small t(n)
I [Sipser]: strict first-order alternation hierarchy: FO.
I [Beame-Hastad]: hierarchy remains strict up toFO[log n/ log log n].
I NC1 ⊆ FO[log n/ log log n] and this is tight.
I Does REACH require FO[log n]? This would implyNC1 6= NL.
Neil Immerman Descriptive Complexity
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Strong Lower Bounds on FO[t(n)] for small t(n)
I [Sipser]: strict first-order alternation hierarchy: FO.
I [Beame-Hastad]: hierarchy remains strict up toFO[log n/ log log n].
I NC1 ⊆ FO[log n/ log log n] and this is tight.
I Does REACH require FO[log n]? This would implyNC1 6= NL.
Neil Immerman Descriptive Complexity
![Page 110: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/110.jpg)
Strong Lower Bounds on FO[t(n)] for small t(n)
I [Sipser]: strict first-order alternation hierarchy: FO.
I [Beame-Hastad]: hierarchy remains strict up toFO[log n/ log log n].
I NC1 ⊆ FO[log n/ log log n] and this is tight.
I Does REACH require FO[log n]? This would implyNC1 6= NL.
Neil Immerman Descriptive Complexity
![Page 111: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/111.jpg)
Does It Matter? How important is P 6= NP?
I Much is known about approximation, e.g., some NPcomplete problems, e.g., Knapsack, Euclidean TSP, can beapproximated as closely as we want, others, e.g., Clique,can’t be.
I We conjecture that SAT requires DTIME[Ω(2εn)] for someε > 0, but no one has yet proved that it requires more thanDTIME[n].
I Basic trade-offs are not understood, e.g., trade-off betweentime and number of processors. Are any problemsinherently sequential? How can we best use mulitcores?
I SAT solvers are impressive new general purpose problemsolvers, e.g., used in model checking, AI planning, codesynthesis. How good are current SAT solvers? How muchcan they be improved?
Neil Immerman Descriptive Complexity
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Does It Matter? How important is P 6= NP?
I Much is known about approximation, e.g., some NPcomplete problems, e.g., Knapsack, Euclidean TSP, can beapproximated as closely as we want, others, e.g., Clique,can’t be.
I We conjecture that SAT requires DTIME[Ω(2εn)] for someε > 0, but no one has yet proved that it requires more thanDTIME[n].
I Basic trade-offs are not understood, e.g., trade-off betweentime and number of processors. Are any problemsinherently sequential? How can we best use mulitcores?
I SAT solvers are impressive new general purpose problemsolvers, e.g., used in model checking, AI planning, codesynthesis. How good are current SAT solvers? How muchcan they be improved?
Neil Immerman Descriptive Complexity
![Page 113: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/113.jpg)
Does It Matter? How important is P 6= NP?
I Much is known about approximation, e.g., some NPcomplete problems, e.g., Knapsack, Euclidean TSP, can beapproximated as closely as we want, others, e.g., Clique,can’t be.
I We conjecture that SAT requires DTIME[Ω(2εn)] for someε > 0, but no one has yet proved that it requires more thanDTIME[n].
I Basic trade-offs are not understood, e.g., trade-off betweentime and number of processors. Are any problemsinherently sequential? How can we best use mulitcores?
I SAT solvers are impressive new general purpose problemsolvers, e.g., used in model checking, AI planning, codesynthesis. How good are current SAT solvers? How muchcan they be improved?
Neil Immerman Descriptive Complexity
![Page 114: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/114.jpg)
Does It Matter? How important is P 6= NP?
I Much is known about approximation, e.g., some NPcomplete problems, e.g., Knapsack, Euclidean TSP, can beapproximated as closely as we want, others, e.g., Clique,can’t be.
I We conjecture that SAT requires DTIME[Ω(2εn)] for someε > 0, but no one has yet proved that it requires more thanDTIME[n].
I Basic trade-offs are not understood, e.g., trade-off betweentime and number of processors. Are any problemsinherently sequential? How can we best use mulitcores?
I SAT solvers are impressive new general purpose problemsolvers, e.g., used in model checking, AI planning, codesynthesis. How good are current SAT solvers? How muchcan they be improved?
Neil Immerman Descriptive Complexity
![Page 115: Neil Immerman - UMass Amherstimmerman/pub/DescriptiveComplexityLongSurvey.pdfNeil Immerman Descriptive Complexity. Descriptive Complexity Query q1 q2 qn 7!Computation 7! Answer a1](https://reader033.fdocuments.us/reader033/viewer/2022041911/5e67632f6e42cc033b34cebf/html5/thumbnails/115.jpg)
Descriptive Complexity
Fact: For constructible t(n), FO[t(n)] = CRAM[t(n)]
Fact: For k = 1,2, . . . , VAR[k + 1] = DSPACE[nk ]
The complexity of computing a query is closely tied to thecomplexity of describing the query.
P = NP ⇔ FO(LFP) = SO
ThC0 = NP ⇔ FO(MAJ) = SO
P = PSPACE ⇔ FO(LFP) = SO(TC)
Neil Immerman Descriptive Complexity
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Arithmetic Hierarchy FO(N) r.e. completeHalt
co-r.e. completeHalt
FO∃(N)r.e.
FO∀(N)co-r.e.
Recursive
Primitive Recursive
SO[2nO(1)
] EXPTIME
QSAT PSPACE completeFO[2n
O(1)
] SO[nO(1)] PSPACE
PTIME Hierarchy SO NP completeSAT
co-NP completeSAT
SO∃ NPSO∀co-NPNP ∩ co-NP
P completeHorn-SAT P
FO[nO(1)]
FO(LFP)
FO[logO(1) n] NC“truly
feasible”FO[log n] AC1
sAC1FO(CFL)
NL2SAT NL comp.FO(TC)
2COLOR L comp. LFO(DTC)
NC1FO(REGULAR)
ThC0FO(COUNT)
AC0FO LOGTIME Hierarchy
Neil Immerman Descriptive Complexity