computational complexity
Post on 25-Apr-2022
4 Views
Preview:
TRANSCRIPT
Computational Complexity
Lecture 3in which we come across
Diagonalization and Time-hierarchies(But first some more of NP-completeness)
1
NP-Complete Languages
2
NP-Complete Languages
A language L1 is NP-complete if L1 is in NP and any NP language L can be reduced to L1 (Karp reduction: polynomial time many-one reduction)
2
NP-Complete Languages
A language L1 is NP-complete if L1 is in NP and any NP language L can be reduced to L1 (Karp reduction: polynomial time many-one reduction)
Trivial example: L1 = TMSAT
2
NP-Complete Languages
A language L1 is NP-complete if L1 is in NP and any NP language L can be reduced to L1 (Karp reduction: polynomial time many-one reduction)
Trivial example: L1 = TMSAT
First interesting example: L1 = CKT-SAT
2
NP-Complete Languages
A language L1 is NP-complete if L1 is in NP and any NP language L can be reduced to L1 (Karp reduction: polynomial time many-one reduction)
Trivial example: L1 = TMSAT
First interesting example: L1 = CKT-SAT
Convert x into a circuit C s.t. C is satisfiable iff x in L
2
NP-Complete Languages
A language L1 is NP-complete if L1 is in NP and any NP language L can be reduced to L1 (Karp reduction: polynomial time many-one reduction)
Trivial example: L1 = TMSAT
First interesting example: L1 = CKT-SAT
Convert x into a circuit C s.t. C is satisfiable iff x in L
More examples, bootstrapping from CKT-SAT
2
NP-Complete Languages
A language L1 is NP-complete if L1 is in NP and any NP language L can be reduced to L1 (Karp reduction: polynomial time many-one reduction)
Trivial example: L1 = TMSAT
First interesting example: L1 = CKT-SAT
Convert x into a circuit C s.t. C is satisfiable iff x in L
More examples, bootstrapping from CKT-SAT
If L ≤p L1 and L1 ≤p L2, then L ≤p L2
2
CKT-SAT ≤p SAT
3
CKT-SAT ≤p SATSAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form)
3
CKT-SAT ≤p SATSAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form)
Converting a circuit to a collection of clauses:
3
CKT-SAT ≤p SATSAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form)
Converting a circuit to a collection of clauses:
For each wire (connected component), add a variable
3
CKT-SAT ≤p SATSAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form)
Converting a circuit to a collection of clauses:
For each wire (connected component), add a variable
Add output variable as a clause. And for each gate, add a clause involving variables for wires connected to the gate:
3
CKT-SAT ≤p SATSAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form)
Converting a circuit to a collection of clauses:
For each wire (connected component), add a variable
Add output variable as a clause. And for each gate, add a clause involving variables for wires connected to the gate:
e.g. : (z⇒x), (z⇒y), (¬z ⇒ ¬x ∨ ¬y). i.e., (¬z ∨ x), (¬z ∨ y), (z ∨ ¬x ∨ ¬y).
AND
xy z
3
CKT-SAT ≤p SATSAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form)
Converting a circuit to a collection of clauses:
For each wire (connected component), add a variable
Add output variable as a clause. And for each gate, add a clause involving variables for wires connected to the gate:
e.g. : (z⇒x), (z⇒y), (¬z ⇒ ¬x ∨ ¬y). i.e., (¬z ∨ x), (¬z ∨ y), (z ∨ ¬x ∨ ¬y).
and : (z⇒x ∨ y), (¬z⇒¬x), (¬z⇒¬y).
AND
xy z
OR
xy z
3
SAT ≤p 3SAT
4
SAT ≤p 3SATPrevious reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT ≤p 3SAT.
4
SAT ≤p 3SATPrevious reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT ≤p 3SAT.
More directly:
4
SAT ≤p 3SATPrevious reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT ≤p 3SAT.
More directly:
(a ∨ b ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ y), (¬y ∨ d ∨ e)
4
SAT ≤p 3SATPrevious reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT ≤p 3SAT.
More directly:
(a ∨ b ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ y), (¬y ∨ d ∨ e)
Reduction needs 3SAT
4
SAT ≤p 3SATPrevious reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT ≤p 3SAT.
More directly:
(a ∨ b ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ y), (¬y ∨ d ∨ e)
Reduction needs 3SAT
2SAT is in fact in P! [Exercise]
4
SAT ≤p 3SATPrevious reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT ≤p 3SAT.
More directly:
(a ∨ b ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ y), (¬y ∨ d ∨ e)
Reduction needs 3SAT
2SAT is in fact in P! [Exercise]
Reduction not parsimonious (can you make it? [Exercise])
4
3SAT ≤p CLIQUE
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
(x ∨ ¬y ∨ ¬z)
(w ∨ x ∨ ¬z)
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
(x ∨ ¬y ∨ ¬z)
(w ∨ x ∨ ¬z)
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
(x ∨ ¬y ∨ ¬z)
(w ∨ x ∨ ¬z)
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
(x ∨ ¬y ∨ ¬z)
(w ∨ x ∨ ¬z)
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)00*0
01*101*0
10*0
10*111*0
11*1
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
vertices: each clause’s satisfying assignments (for its variables)
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)00*0
01*101*0
10*0
10*111*0
11*1
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
vertices: each clause’s satisfying assignments (for its variables)
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)00*0
01*101*0
10*0
10*111*0
11*1
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
vertices: each clause’s satisfying assignments (for its variables)
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)00*0
01*101*0
10*0
10*111*0
11*1
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
vertices: each clause’s satisfying assignments (for its variables)
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)00*0
01*101*0
10*0
10*111*0
11*1
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
vertices: each clause’s satisfying assignments (for its variables)
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)00*0
01*101*0
10*0
10*111*0
11*1
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
vertices: each clause’s satisfying assignments (for its variables)
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)00*0
01*101*0
10*0
10*111*0
11*1
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
vertices: each clause’s satisfying assignments (for its variables)
edges between consistent assignments
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)00*0
01*101*0
10*0
10*111*0
11*1
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
vertices: each clause’s satisfying assignments (for its variables)
edges between consistent assignments
m-clique iff all m clauses satisfiable
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)00*0
01*101*0
10*0
10*111*0
11*1
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
vertices: each clause’s satisfying assignments (for its variables)
edges between consistent assignments
m-clique iff all m clauses satisfiable
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)00*0
01*101*0
10*0
10*111*0
11*1
0*1*1*1*
1*0*
(w ∨ y)
5
3SAT ≤p CLIQUECLIQUE: Does graph G have a clique of size m?
Clauses → Graph
vertices: each clause’s satisfying assignments (for its variables)
edges between consistent assignments
m-clique iff all m clauses satisfiable
(x ∨ ¬y ∨ ¬z) *000
*001*010
*100
*101*110
*111
(w ∨ x ∨ ¬z)00*0
01*101*0
10*0
10*111*0
11*1
0*1*1*1*
1*0*
(w ∨ y)
1*1**11011*0----1110sat assignment
3-Clique
5
INDEP-SET and VERTEX-COVER
6
INDEP-SET and VERTEX-COVER
CLIQUE ≤p INDEP-SET
6
INDEP-SET and VERTEX-COVER
CLIQUE ≤p INDEP-SET
G has an m-clique iff Gc has an m-independent-set
6
INDEP-SET and VERTEX-COVER
CLIQUE ≤p INDEP-SET
G has an m-clique iff Gc has an m-independent-set
INDEP-SET ≤p VERTEX-COVER
6
INDEP-SET and VERTEX-COVER
CLIQUE ≤p INDEP-SET
G has an m-clique iff Gc has an m-independent-set
INDEP-SET ≤p VERTEX-COVER
G has an m-indep-set iff G has an (n-m)-vertex-cover
6
NP, P, co-NP and NPC
P
NP coNP
NPC coNPC
7
NP, P, co-NP and NPCWe say class X is “closed under polynomial reductions” if (L1 ≤p L2 and L2 in class X) implies L1 in X
P
NP coNP
NPC coNPC
7
NP, P, co-NP and NPCWe say class X is “closed under polynomial reductions” if (L1 ≤p L2 and L2 in class X) implies L1 in X
e.g. P, NP are closed under polynomial reductions
P
NP coNP
NPC coNPC
7
NP, P, co-NP and NPCWe say class X is “closed under polynomial reductions” if (L1 ≤p L2 and L2 in class X) implies L1 in X
e.g. P, NP are closed under polynomial reductions
So is co-NP (If X is closed, so is co-X. Why?)
P
NP coNP
NPC coNPC
7
NP, P, co-NP and NPCWe say class X is “closed under polynomial reductions” if (L1 ≤p L2 and L2 in class X) implies L1 in X
e.g. P, NP are closed under polynomial reductions
So is co-NP (If X is closed, so is co-X. Why?)
If any NPC language is in P, then NP = P
P
NP coNP
NPC coNPC
7
NP, P, co-NP and NPCWe say class X is “closed under polynomial reductions” if (L1 ≤p L2 and L2 in class X) implies L1 in X
e.g. P, NP are closed under polynomial reductions
So is co-NP (If X is closed, so is co-X. Why?)
If any NPC language is in P, then NP = P
If any NPC language is in co-NP, then NP = co-NPP
NP coNP
NPC coNPC
7
NP, P, co-NP and NPCWe say class X is “closed under polynomial reductions” if (L1 ≤p L2 and L2 in class X) implies L1 in X
e.g. P, NP are closed under polynomial reductions
So is co-NP (If X is closed, so is co-X. Why?)
If any NPC language is in P, then NP = P
If any NPC language is in co-NP, then NP = co-NP
Note: X ⊆ co-X ⇒ X = co-X (Why?)P
NP coNP
NPC coNPC
7
NP, P, co-NP and NPCWe say class X is “closed under polynomial reductions” if (L1 ≤p L2 and L2 in class X) implies L1 in X
e.g. P, NP are closed under polynomial reductions
So is co-NP (If X is closed, so is co-X. Why?)
If any NPC language is in P, then NP = P
If any NPC language is in co-NP, then NP = co-NP
Note: X ⊆ co-X ⇒ X = co-X (Why?)
L is NP-complete iff Lc is co-NP-complete (Why?)
P
NP coNP
NPC coNPC
7
NP, P, co-NP and NPCWe say class X is “closed under polynomial reductions” if (L1 ≤p L2 and L2 in class X) implies L1 in X
e.g. P, NP are closed under polynomial reductions
So is co-NP (If X is closed, so is co-X. Why?)
If any NPC language is in P, then NP = P
If any NPC language is in co-NP, then NP = co-NP
Note: X ⊆ co-X ⇒ X = co-X (Why?)
L is NP-complete iff Lc is co-NP-complete (Why?)
co-NP complete = co-(NP-complete)
P
NP coNP
NPC coNPC
7
Separating Classes
8
Separating Classes
How to prove a set X strictly bigger than Y
8
Separating Classes
How to prove a set X strictly bigger than Y
Show an element not in Y, but in X? For us, not in Y may often be difficult to prove for (familiar) elements
8
Separating Classes
How to prove a set X strictly bigger than Y
Show an element not in Y, but in X? For us, not in Y may often be difficult to prove for (familiar) elements
Count? What if both infinite?!
8
Separating Classes
How to prove a set X strictly bigger than Y
Show an element not in Y, but in X? For us, not in Y may often be difficult to prove for (familiar) elements
Count? What if both infinite?!
Comparing infinite sets: diagonalization!
8
Cantor’s Diagonal Slash
9
Cantor’s Diagonal SlashAre real numbers (say in the range [0,1) ) countable?
9
Cantor’s Diagonal SlashAre real numbers (say in the range [0,1) ) countable?
Suppose they were: consider enumerating them along with their binary representations in a table
9
Cantor’s Diagonal SlashAre real numbers (say in the range [0,1) ) countable?
Suppose they were: consider enumerating them along with their binary representations in a table
Ri
R1 =
R2 =
R3 =
R4 =
R5 =
R6 =
R7 =
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 19
Cantor’s Diagonal SlashAre real numbers (say in the range [0,1) ) countable?
Suppose they were: consider enumerating them along with their binary representations in a table
Consider the real number corresponding to the “flipped diagonal”
Ri
R1 =
R2 =
R3 =
R4 =
R5 =
R6 =
R7 =
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 19
Cantor’s Diagonal SlashAre real numbers (say in the range [0,1) ) countable?
Suppose they were: consider enumerating them along with their binary representations in a table
Consider the real number corresponding to the “flipped diagonal”
Ri
R1 =
R2 =
R3 =
R4 =
R5 =
R6 =
R7 =
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 19
Cantor’s Diagonal SlashAre real numbers (say in the range [0,1) ) countable?
Suppose they were: consider enumerating them along with their binary representations in a table
Consider the real number corresponding to the “flipped diagonal”
Doesn’t appear in this table!
Ri
R1 =
R2 =
R3 =
R4 =
R5 =
R6 =
R7 =
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 19
Cantor’s Diagonal SlashAre real numbers (say in the range [0,1) ) countable?
Suppose they were: consider enumerating them along with their binary representations in a table
Consider the real number corresponding to the “flipped diagonal”
Doesn’t appear in this table!
Ri
R1 =
R2 =
R3 =
R4 =
R5 =
R6 =
R7 =
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 1
This table
can’t
have all re
als
9
Undecidable Languages
LM1 =
LM2 =
LM3 =
LM4 =
LM5 =
LM6 =
LM7 =
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 110
Undecidable LanguagesLanguages, like real numbers, can be represented as infinite bit-vectors LM1 =
LM2 =
LM3 =
LM4 =
LM5 =
LM6 =
LM7 =
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 110
Undecidable LanguagesLanguages, like real numbers, can be represented as infinite bit-vectors
TMs can be enumerated!
LM1 =
LM2 =
LM3 =
LM4 =
LM5 =
LM6 =
LM7 =
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 110
Undecidable LanguagesLanguages, like real numbers, can be represented as infinite bit-vectors
TMs can be enumerated!
Table of languages recognized by the TMs
LM1 =
LM2 =
LM3 =
LM4 =
LM5 =
LM6 =
LM7 =
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 110
Undecidable LanguagesLanguages, like real numbers, can be represented as infinite bit-vectors
TMs can be enumerated!
Table of languages recognized by the TMs
L = ”diagonal language”
LM1 =
LM2 =
LM3 =
LM4 =
LM5 =
LM6 =
LM7 =
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 110
Undecidable LanguagesLanguages, like real numbers, can be represented as infinite bit-vectors
TMs can be enumerated!
Table of languages recognized by the TMs
L = ”diagonal language”
Lc does not appear as a row in this table. Hence not recognizable!
LM1 =
LM2 =
LM3 =
LM4 =
LM5 =
LM6 =
LM7 =
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 110
Undecidable LanguagesLanguages, like real numbers, can be represented as infinite bit-vectors
TMs can be enumerated!
Table of languages recognized by the TMs
L = ”diagonal language”
Lc does not appear as a row in this table. Hence not recognizable!
LM1 =
LM2 =
LM3 =
LM4 =
LM5 =
LM6 =
LM7 =
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 1
This table ca
n’t have
all language
s
10
Diagonalization to Separate Classes
Diagonalization can separate the class of decidable languages (from the class of all languages)
Plan: Use similar techniques to separate complexity classes
11
DTIME Hierarchy
12
DTIME HierarchyFix a TM model (one-tape, binary alphabet)
12
DTIME HierarchyFix a TM model (one-tape, binary alphabet)
DTIME(T) = class of languages that can be decided in O(T(n)) time, by such a TM
12
DTIME HierarchyFix a TM model (one-tape, binary alphabet)
DTIME(T) = class of languages that can be decided in O(T(n)) time, by such a TM
Theorem: DTIME(nc) ⊊ DTIME(nc+1) for all c ≥ 1
12
DTIME HierarchyFix a TM model (one-tape, binary alphabet)
DTIME(T) = class of languages that can be decided in O(T(n)) time, by such a TM
Theorem: DTIME(nc) ⊊ DTIME(nc+1) for all c ≥ 1
More generally DTIME(T) ⊊ DTIME(T’) if T, T’ “nice” (and ≥ n)
and T(n)log(T(n)) = o(T’(n))
12
DTIME HierarchyFix a TM model (one-tape, binary alphabet)
DTIME(T) = class of languages that can be decided in O(T(n)) time, by such a TM
Theorem: DTIME(nc) ⊊ DTIME(nc+1) for all c ≥ 1
More generally DTIME(T) ⊊ DTIME(T’) if T, T’ “nice” (and ≥ n)
and T(n)log(T(n)) = o(T’(n))
Consequences, for e.g., P ⊊ EXP
12
DTIME HierarchyFix a TM model (one-tape, binary alphabet)
DTIME(T) = class of languages that can be decided in O(T(n)) time, by such a TM
Theorem: DTIME(nc) ⊊ DTIME(nc+1) for all c ≥ 1
More generally DTIME(T) ⊊ DTIME(T’) if T, T’ “nice” (and ≥ n)
and T(n)log(T(n)) = o(T’(n))
Consequences, for e.g., P ⊊ EXP
P ⊆ DTIME(2n) ⊊ DTIME(22n) ⊆ EXP
12
DTIME Hierarchy: Proof
13
Mi be an enumeration of TMs, each TM appearing infinitely often
DTIME Hierarchy: Proof
13
Mi be an enumeration of TMs, each TM appearing infinitely often
Consider table(i,j) = UTM|T’ (Mi,j), where T log T = o(T’)
DTIME Hierarchy: Proof
Mi
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 1
j
13
Mi be an enumeration of TMs, each TM appearing infinitely often
Consider table(i,j) = UTM|T’ (Mi,j), where T log T = o(T’)
DTIME Hierarchy: Proof
Mi
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 1
j
T’ large and nice enough to allow
simulation
13
Mi be an enumeration of TMs, each TM appearing infinitely often
Consider table(i,j) = UTM|T’ (Mi,j), where T log T = o(T’)
DTIME Hierarchy: Proof
Mi
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 1
j
Think DTIME(T) ⊆ rows
T’ large and nice enough to allow
simulation
13
Mi be an enumeration of TMs, each TM appearing infinitely often
Consider table(i,j) = UTM|T’ (Mi,j), where T log T = o(T’)
DTIME Hierarchy: Proof
Mi
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 1
j
Think DTIME(T) ⊆ rows
T’ large and nice enough to allow
simulation
13
Mi be an enumeration of TMs, each TM appearing infinitely often
Consider table(i,j) = UTM|T’ (Mi,j), where T log T = o(T’)
Let L’ = inverted diagonal.
DTIME Hierarchy: Proof
Mi
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 1
j
Think DTIME(T) ⊆ rows
T’ large and nice enough to allow
simulation
13
L’ in DTIME(T’)
Mi be an enumeration of TMs, each TM appearing infinitely often
Consider table(i,j) = UTM|T’ (Mi,j), where T log T = o(T’)
Let L’ = inverted diagonal.
DTIME Hierarchy: Proof
Mi
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 1
j
Think DTIME(T) ⊆ rows
T’ large and nice enough to allow
simulation
13
L’ in DTIME(T’)
On input i, run UTM|T’ (Mi,i), modified to invert output
Mi be an enumeration of TMs, each TM appearing infinitely often
Consider table(i,j) = UTM|T’ (Mi,j), where T log T = o(T’)
Let L’ = inverted diagonal.
DTIME Hierarchy: Proof
Mi
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 1
j
Think DTIME(T) ⊆ rows
T’ large and nice enough to allow
simulation
13
L’ in DTIME(T’)
On input i, run UTM|T’ (Mi,i), modified to invert output
L’ not in DTIME(T)
Mi be an enumeration of TMs, each TM appearing infinitely often
Consider table(i,j) = UTM|T’ (Mi,j), where T log T = o(T’)
Let L’ = inverted diagonal.
DTIME Hierarchy: Proof
Mi
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 1
j
Think DTIME(T) ⊆ rows
T’ large and nice enough to allow
simulation
13
L’ in DTIME(T’)
On input i, run UTM|T’ (Mi,i), modified to invert output
L’ not in DTIME(T)
If M accepts L’ in time T, then for sufficiently large i s.t. Mi=M, UTM can finish simulating Mi(i). Then table(i,i)=L’(i)!
Mi be an enumeration of TMs, each TM appearing infinitely often
Consider table(i,j) = UTM|T’ (Mi,j), where T log T = o(T’)
Let L’ = inverted diagonal.
DTIME Hierarchy: Proof
Mi
1 0 0 1 0 0 0 0 1
0 0 1 0 1 0 0 1 1
1 1 1 1 1 1 1 0 0
1 1 0 1 0 1 0 1 1
1 1 0 0 0 0 1 0 0
0 0 0 0 0 0 1 1 0
0 1 0 1 0 1 0 1 1
j
Think DTIME(T) ⊆ rows
T’ large and nice enough to allow
simulation
13
NTIME Hierarchy
14
NTIME Hierarchy
Finer hierarchy
14
NTIME Hierarchy
Finer hierarchy
NTIME(T) ⊊ NTIME(T’) if T(n)=o(T’(n)), and T, T’ nice
14
NTIME Hierarchy
Finer hierarchy
NTIME(T) ⊊ NTIME(T’) if T(n)=o(T’(n)), and T, T’ nice
Because a more sophisticated Universal NTM has less overhead
14
NTIME Hierarchy
Finer hierarchy
NTIME(T) ⊊ NTIME(T’) if T(n)=o(T’(n)), and T, T’ nice
Because a more sophisticated Universal NTM has less overhead
Diagonalization is more complicated
14
NTIME Hierarchy
Finer hierarchy
NTIME(T) ⊊ NTIME(T’) if T(n)=o(T’(n)), and T, T’ nice
Because a more sophisticated Universal NTM has less overhead
Diagonalization is more complicated
Issue: NTIME(T’) enough to simulate NTIME(T), but not to simulate co-NTIME(T)!
14
NTIME Hierarchy
Finer hierarchy
NTIME(T) ⊊ NTIME(T’) if T(n)=o(T’(n)), and T, T’ nice
Because a more sophisticated Universal NTM has less overhead
Diagonalization is more complicated
Issue: NTIME(T’) enough to simulate NTIME(T), but not to simulate co-NTIME(T)!
In fact,
T(n+1) = o(
T’(n))
14
NTIME Hierarchy
15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
Think NTIME(T) ⊆ rows15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
Think NTIME(T) ⊆ rows15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Think NTIME(T) ⊆ rows15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
f(1) f(2) f(3) f(4) f(5)
Think NTIME(T) ⊆ rows15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Let L be the “diagonal” language
f(1) f(2) f(3) f(4) f(5)
Think NTIME(T) ⊆ rows15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Let L be the “diagonal” language
f(1) f(2) f(3) f(4) f(5)
L
Think NTIME(T) ⊆ rows15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Let L be the “diagonal” language
f(1) f(2) f(3) f(4) f(5)
L
L’
Think NTIME(T) ⊆ rows15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Let L be the “diagonal” language
L’(j)=L(j+1)f(1) f(2) f(3) f(4) f(5)
L
L’
Think NTIME(T) ⊆ rows15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Let L be the “diagonal” language
L’(j)=L(j+1)f(1) f(2) f(3) f(4) f(5)
L
L’
Think NTIME(T) ⊆ rows15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Let L be the “diagonal” language
L’(j)=L(j+1)
except if j=f(i), then L’(j) = 1 - L(f(i-1)+1)
f(1) f(2) f(3) f(4) f(5)
L
L’
Think NTIME(T) ⊆ rows15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Let L be the “diagonal” language
L’(j)=L(j+1)
except if j=f(i), then L’(j) = 1 - L(f(i-1)+1)
f(1) f(2) f(3) f(4) f(5)
L
L’
Think NTIME(T) ⊆ rows15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Let L be the “diagonal” language
L’(j)=L(j+1)
except if j=f(i), then L’(j) = 1 - L(f(i-1)+1)
L’ not in NTIME(T), but is in NTIME(T’)
f(1) f(2) f(3) f(4) f(5)
L
L’
Think NTIME(T) ⊆ rows15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Let L be the “diagonal” language
L’(j)=L(j+1)
except if j=f(i), then L’(j) = 1 - L(f(i-1)+1)
L’ not in NTIME(T), but is in NTIME(T’)
f(1) f(2) f(3) f(4) f(5)
L
L’
Think NTIME(T) ⊆ rows
Flip, Diagonal
15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Let L be the “diagonal” language
L’(j)=L(j+1)
except if j=f(i), then L’(j) = 1 - L(f(i-1)+1)
L’ not in NTIME(T), but is in NTIME(T’)
f(1) f(2) f(3) f(4) f(5)
L
L’
Think NTIME(T) ⊆ rows
Flip, Diagonal
15
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Let L be the “diagonal” language
L’(j)=L(j+1)
except if j=f(i), then L’(j) = 1 - L(f(i-1)+1)
L’ not in NTIME(T), but is in NTIME(T’)
f(1) f(2) f(3) f(4) f(5)
L
L’
Think NTIME(T) ⊆ rows
Flip, Diagonal
15
Delay, R
apid
thicken
ing
NTIME Hierarchy“Delayed flip” on a “rapidly thickening diagonal”
f(i+1)=exp(f(i))
Let L be the “diagonal” language
L’(j)=L(j+1)
except if j=f(i), then L’(j) = 1 - L(f(i-1)+1)
L’ not in NTIME(T), but is in NTIME(T’)
f(1) f(2) f(3) f(4) f(5)
L
L’
Think NTIME(T) ⊆ rows
Flip, Diagonal
15
Time Hierarchy
16
Time Hierarchy
Within DTIME and NTIME fine gradation
16
Time Hierarchy
Within DTIME and NTIME fine gradation
In particular P ⊊ EXP, NP ⊊ NEXP
16
Time Hierarchy
Within DTIME and NTIME fine gradation
In particular P ⊊ EXP, NP ⊊ NEXP
Tells nothing across DTIME and NTIME
16
Time Hierarchy
Within DTIME and NTIME fine gradation
In particular P ⊊ EXP, NP ⊊ NEXP
Tells nothing across DTIME and NTIME
P and NP?
16
Time Hierarchy
Within DTIME and NTIME fine gradation
In particular P ⊊ EXP, NP ⊊ NEXP
Tells nothing across DTIME and NTIME
P and NP?
Just diagonalization won’t help (next lecture)
16
Today
DTIME Hierarchy
DTIME(T) ⊊ DTIME(T’) if T log T = o(T’)
NTIME Hierarchy
NTIME(T) ⊊ NTIME(T’) if T = o(T’)
Using diagonalization
17
Next Lecture
Another application of diagonalization
Ladner’s Theorem: If P≠NP, NP language which is neither in P nor NP-complete
Limits of Diagonalization
Starting Space Complexity
18
top related