Time Complexity
CS 133 : Automata Theory and Computability Intractability
Nestine Hope S. Hernandez Algorithms and Complexity Laboratory Department of Computer Science University of the Philippines, Diliman
[email protected]
Day 19
Time Complexity Intractability
Time Complexity The Class P The Class NP
Time Complexity The number of steps that an algorithm uses on a particular input may depend on several parameters.
Time Complexity The number of steps that an algorithm uses on a particular input may depend on several parameters. For simplicity, we compute the running time of an algorithm purely as a function of the length of the string representing the input and don't consider any other parameters.
Time Complexity The number of steps that an algorithm uses on a particular input may depend on several parameters. For simplicity, we compute the running time of an algorithm purely as a function of the length of the string representing the input and don't consider any other parameters.
De nition Let
M
be a deterministic Turing machine that halts on all inputs.
The running time or time complexity of where
f (n)
M
f : N → N, M uses on any M , we say that M
is the function
is the maximum number of steps that
n. If f (n) is the running time of f (n) and that M is an f (n) time Turing machine. we use n to represent the length of the input.
input of length runs in time Customarily
Time Complexity Big-O Notation Because the exact running time of an algorithm often is a complex expression, we usually just estimate it.
Time Complexity Big-O Notation Because the exact running time of an algorithm often is a complex expression, we usually just estimate it. We consider only the highest order term of the expression for the running time of the algorithm, disregarding both the coe cient of that term and any lower order terms, because the highest order term dominates the other terms on large inputs.
Time Complexity Big-O Notation Because the exact running time of an algorithm often is a complex expression, we usually just estimate it. We consider only the highest order term of the expression for the running time of the algorithm, disregarding both the coe cient of that term and any lower order terms, because the highest order term dominates the other terms on large inputs. Example:
f1 (n) = 6n3 + 2n2 + 20n + 45
Time Complexity Big-O Notation Because the exact running time of an algorithm often is a complex expression, we usually just estimate it. We consider only the highest order term of the expression for the running time of the algorithm, disregarding both the coe cient of that term and any lower order terms, because the highest order term dominates the other terms on large inputs. Example:
f1 (n) = 6n3 + 2n2 + 20n + 45 ⇒ f1 (n) = O(n3 ), the asymptotic
notation or big-O notation
Time Complexity Big-O Notation Because the exact running time of an algorithm often is a complex expression, we usually just estimate it. We consider only the highest order term of the expression for the running time of the algorithm, disregarding both the coe cient of that term and any lower order terms, because the highest order term dominates the other terms on large inputs. Example:
f1 (n) = 6n3 + 2n2 + 20n + 45 ⇒ f1 (n) = O(n3 ), the asymptotic
notation or big-O notation
g be functions f, g : N → R+ . Say that f (n) = O(g(n)) if positive integers c for every integer n ≥ n0 , f (n) ≤ c g(n). Let
f
and
and
n0
exist such that
Time Complexity Big-O Notation Because the exact running time of an algorithm often is a complex expression, we usually just estimate it. We consider only the highest order term of the expression for the running time of the algorithm, disregarding both the coe cient of that term and any lower order terms, because the highest order term dominates the other terms on large inputs. Example:
f1 (n) = 6n3 + 2n2 + 20n + 45 ⇒ f1 (n) = O(n3 ), the asymptotic
notation or big-O notation
g be functions f, g : N → R+ . Say that f (n) = O(g(n)) if positive integers c for every integer n ≥ n0 , f (n) ≤ c g(n). Let
f
and
f2 (n) = 3n log2 n + 5nlog2 log2 n + 2
and
n0
exist such that
Time Complexity Big-O Notation Because the exact running time of an algorithm often is a complex expression, we usually just estimate it. We consider only the highest order term of the expression for the running time of the algorithm, disregarding both the coe cient of that term and any lower order terms, because the highest order term dominates the other terms on large inputs. Example:
f1 (n) = 6n3 + 2n2 + 20n + 45 ⇒ f1 (n) = O(n3 ), the asymptotic
notation or big-O notation
g be functions f, g : N → R+ . Say that f (n) = O(g(n)) if positive integers c for every integer n ≥ n0 , f (n) ≤ c g(n). Let
f
and
and
n0
exist such that
f2 (n) = 3n log2 n + 5nlog2 log2 n + 2 ⇒ f2 (n) = O(n log n) f3 (n) = O(n2 ) + O(n)
Time Complexity Big-O Notation Because the exact running time of an algorithm often is a complex expression, we usually just estimate it. We consider only the highest order term of the expression for the running time of the algorithm, disregarding both the coe cient of that term and any lower order terms, because the highest order term dominates the other terms on large inputs. Example:
f1 (n) = 6n3 + 2n2 + 20n + 45 ⇒ f1 (n) = O(n3 ), the asymptotic
notation or big-O notation
g be functions f, g : N → R+ . Say that f (n) = O(g(n)) if positive integers c for every integer n ≥ n0 , f (n) ≤ c g(n). Let
f
and
and
n0
exist such that
f2 (n) = 3n log2 n + 5nlog2 log2 n + 2 ⇒ f2 (n) = O(n log n) f3 (n) = O(n2 ) + O(n) ⇒ f3 (n) = O(n2 )
Time Complexity Analyzing Algorithms
k k
A = {0 1 | k ≥ 0}
Time Complexity Analyzing Algorithms
k k
A = {0 1 | k ≥ 0} M1 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat if both 0s and 1s remain on the tape:
B
Scan across the tape, crossing o a single 0 and a single 1.
3. If 0s still remain after all the 1s have been crossed o, or if 1s still remain after all the 0s have been crossed o, reject. Otherwise, if neither 0s nor 1s remain on the tape, accept.
Time Complexity Analyzing Algorithms
k k
A = {0 1 | k ≥ 0} M1 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat if both 0s and 1s remain on the tape:
B
Scan across the tape, crossing o a single 0 and a single 1.
3. If 0s still remain after all the 1s have been crossed o, or if 1s still remain after all the 0s have been crossed o, reject. Otherwise, if neither 0s nor 1s remain on the tape, accept. stage 1:
Time Complexity Analyzing Algorithms
k k
A = {0 1 | k ≥ 0} M1 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat if both 0s and 1s remain on the tape:
B
Scan across the tape, crossing o a single 0 and a single 1.
3. If 0s still remain after all the 1s have been crossed o, or if 1s still remain after all the 0s have been crossed o, reject. Otherwise, if neither 0s nor 1s remain on the tape, accept. stage 1:
2n = O(n)
Time Complexity Analyzing Algorithms
k k
A = {0 1 | k ≥ 0} M1 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat if both 0s and 1s remain on the tape:
B
Scan across the tape, crossing o a single 0 and a single 1.
3. If 0s still remain after all the 1s have been crossed o, or if 1s still remain after all the 0s have been crossed o, reject. Otherwise, if neither 0s nor 1s remain on the tape, accept. stage 1: stage 2:
2n = O(n)
Time Complexity Analyzing Algorithms
k k
A = {0 1 | k ≥ 0} M1 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat if both 0s and 1s remain on the tape:
B
Scan across the tape, crossing o a single 0 and a single 1.
3. If 0s still remain after all the 1s have been crossed o, or if 1s still remain after all the 0s have been crossed o, reject. Otherwise, if neither 0s nor 1s remain on the tape, accept. stage 1: stage 2:
2n = O(n) n 2 2 O(n) = O(n )
Time Complexity Analyzing Algorithms
k k
A = {0 1 | k ≥ 0} M1 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat if both 0s and 1s remain on the tape:
B
Scan across the tape, crossing o a single 0 and a single 1.
3. If 0s still remain after all the 1s have been crossed o, or if 1s still remain after all the 0s have been crossed o, reject. Otherwise, if neither 0s nor 1s remain on the tape, accept. stage 1: stage 2: stage 3:
2n = O(n) n 2 2 O(n) = O(n )
Time Complexity Analyzing Algorithms
k k
A = {0 1 | k ≥ 0} M1 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat if both 0s and 1s remain on the tape:
B
Scan across the tape, crossing o a single 0 and a single 1.
3. If 0s still remain after all the 1s have been crossed o, or if 1s still remain after all the 0s have been crossed o, reject. Otherwise, if neither 0s nor 1s remain on the tape, accept. stage 1: stage 2: stage 3:
2n = O(n) n 2 2 O(n) = O(n ) O(n)
Time Complexity Analyzing Algorithms
k k
A = {0 1 | k ≥ 0} M1 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat if both 0s and 1s remain on the tape:
B
Scan across the tape, crossing o a single 0 and a single 1.
3. If 0s still remain after all the 1s have been crossed o, or if 1s still remain after all the 0s have been crossed o, reject. Otherwise, if neither 0s nor 1s remain on the tape, accept.
2n = O(n) n 2 2 O(n) = O(n ) stage 3: O(n) Thus the total time of M1 on an O(n) + O(n2 ) + O(n) = O(n2 ). stage 1: stage 2:
input of length
n
is
Time Complexity
De nition Let
t : N → R+
be a function.
De ne the time complexity class,
T IM E(t(n)), to be the collection of O(t(n)) time Turing machine.
all languages that are decidable by an
Time Complexity
De nition Let
t : N → R+
be a function.
De ne the time complexity class,
T IM E(t(n)), to be the collection of O(t(n)) time Turing machine.
all languages that are decidable by an
A ∈ T IM E(n2 ) because M1 decides A T IM E(n ) contains all languages that can be
Hence, in the preceding slide, in time
2
O(n ) and O(n2 ).
decided on
2
Time Complexity
De nition Let
t : N → R+
be a function.
De ne the time complexity class,
T IM E(t(n)), to be the collection of O(t(n)) time Turing machine.
all languages that are decidable by an
A ∈ T IM E(n2 ) because M1 decides A T IM E(n ) contains all languages that can be
Hence, in the preceding slide, in time
2
O(n ) and O(n2 ).
2
decided on
Is there a machine that decides
A
asymptotically more quickly?
Time Complexity
M2 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat as long as some 0s and some 1s remain on the tape:
B
Scan across the tape, checking whether the total number of 0s
and 1s remaining is even or odd. If it is odd, reject.
B
Scan again across the tape, crossing o every other 0 starting
with the rst 0, and then crossing o every other 1 starting with the rst 1. 3. If no 0s and no 1s remain on the tape, accept. Otherwise, reject.
Time Complexity
M2 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat as long as some 0s and some 1s remain on the tape:
B
Scan across the tape, checking whether the total number of 0s
and 1s remaining is even or odd. If it is odd, reject.
B
Scan again across the tape, crossing o every other 0 starting
with the rst 0, and then crossing o every other 1 starting with the rst 1. 3. If no 0s and no 1s remain on the tape, accept. Otherwise, reject. stage 1:
Time Complexity
M2 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat as long as some 0s and some 1s remain on the tape:
B
Scan across the tape, checking whether the total number of 0s
and 1s remaining is even or odd. If it is odd, reject.
B
Scan again across the tape, crossing o every other 0 starting
with the rst 0, and then crossing o every other 1 starting with the rst 1. 3. If no 0s and no 1s remain on the tape, accept. Otherwise, reject. stage 1:
O(n)
Time Complexity
M2 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat as long as some 0s and some 1s remain on the tape:
B
Scan across the tape, checking whether the total number of 0s
and 1s remaining is even or odd. If it is odd, reject.
B
Scan again across the tape, crossing o every other 0 starting
with the rst 0, and then crossing o every other 1 starting with the rst 1. 3. If no 0s and no 1s remain on the tape, accept. Otherwise, reject. stage 1: stage 2:
O(n)
Time Complexity
M2 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat as long as some 0s and some 1s remain on the tape:
B
Scan across the tape, checking whether the total number of 0s
and 1s remaining is even or odd. If it is odd, reject.
B
Scan again across the tape, crossing o every other 0 starting
with the rst 0, and then crossing o every other 1 starting with the rst 1. 3. If no 0s and no 1s remain on the tape, accept. Otherwise, reject. stage 1: stage 2:
O(n) (1 + log2 n)O(n) = O(n log n)
Time Complexity
M2 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat as long as some 0s and some 1s remain on the tape:
B
Scan across the tape, checking whether the total number of 0s
and 1s remaining is even or odd. If it is odd, reject.
B
Scan again across the tape, crossing o every other 0 starting
with the rst 0, and then crossing o every other 1 starting with the rst 1. 3. If no 0s and no 1s remain on the tape, accept. Otherwise, reject. stage 1: stage 2: stage 3:
O(n) (1 + log2 n)O(n) = O(n log n)
Time Complexity
M2 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat as long as some 0s and some 1s remain on the tape:
B
Scan across the tape, checking whether the total number of 0s
and 1s remaining is even or odd. If it is odd, reject.
B
Scan again across the tape, crossing o every other 0 starting
with the rst 0, and then crossing o every other 1 starting with the rst 1. 3. If no 0s and no 1s remain on the tape, accept. Otherwise, reject. stage 1: stage 2: stage 3:
O(n) (1 + log2 n)O(n) = O(n log n) O(n)
Time Complexity
M2 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat as long as some 0s and some 1s remain on the tape:
B
Scan across the tape, checking whether the total number of 0s
and 1s remaining is even or odd. If it is odd, reject.
B
Scan again across the tape, crossing o every other 0 starting
with the rst 0, and then crossing o every other 1 starting with the rst 1. 3. If no 0s and no 1s remain on the tape, accept. Otherwise, reject.
O(n) (1 + log2 n)O(n) = O(n log n) stage 3: O(n) Thus the total time of M1 on an input of O(n) + O(n log n) + O(n) = O(n log n). stage 1:
stage 2:
length
n
is
Time Complexity
M2 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Repeat as long as some 0s and some 1s remain on the tape:
B
Scan across the tape, checking whether the total number of 0s
and 1s remaining is even or odd. If it is odd, reject.
B
Scan again across the tape, crossing o every other 0 starting
with the rst 0, and then crossing o every other 1 starting with the rst 1. 3. If no 0s and no 1s remain on the tape, accept. Otherwise, reject.
O(n) (1 + log2 n)O(n) = O(n log n) stage 3: O(n) Thus the total time of M1 on an input of O(n) + O(n log n) + O(n) = O(n log n). stage 1:
stage 2:
Hence,
A ∈ T IM E(n log n)
length
n
is
Time Complexity
M3 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Scan across the 0s on tape 1 until the rst 1. At the same time, copy the 0s onto tape 2. 3. Scan across the 1s on tape 1 until the end of the input. For each 1 read on tape 1, cross o a 0 on tape 2. If all 0s are crossed o before all the 1s are read, reject. 4. If all the 0s have now been crossed o, accept. if any 0s remain, reject.
Time Complexity
M3 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Scan across the 0s on tape 1 until the rst 1. At the same time, copy the 0s onto tape 2. 3. Scan across the 1s on tape 1 until the end of the input. For each 1 read on tape 1, cross o a 0 on tape 2. If all 0s are crossed o before all the 1s are read, reject. 4. If all the 0s have now been crossed o, accept. if any 0s remain, reject.
A O(n).
Hence, the time complexity of and on a 2-tape TM it is
on a single-tape TM is
O(n log n)
Time Complexity
M3 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Scan across the 0s on tape 1 until the rst 1. At the same time, copy the 0s onto tape 2. 3. Scan across the 1s on tape 1 until the end of the input. For each 1 read on tape 1, cross o a 0 on tape 2. If all 0s are crossed o before all the 1s are read, reject. 4. If all the 0s have now been crossed o, accept. if any 0s remain, reject.
A O(n).
Hence, the time complexity of and on a 2-tape TM it is
on a single-tape TM is
O(n log n)
In computability theory, the Church-Turing thesis implies that all reasonable models of computation are equivalent, that is, they all decide the same class of languages.
Time Complexity
M3 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Scan across the 0s on tape 1 until the rst 1. At the same time, copy the 0s onto tape 2. 3. Scan across the 1s on tape 1 until the end of the input. For each 1 read on tape 1, cross o a 0 on tape 2. If all 0s are crossed o before all the 1s are read, reject. 4. If all the 0s have now been crossed o, accept. if any 0s remain, reject.
A O(n).
Hence, the time complexity of and on a 2-tape TM it is
on a single-tape TM is
O(n log n)
In computability theory, the Church-Turing thesis implies that all reasonable models of computation are equivalent, that is, they all decide the same class of languages. In complexity theory, the choice of model aects the time complexity of languages.
Time Complexity
M3 =
On input string
w:
1. Scan across the tape and reject if a 0 is found to the right of a 1. 2. Scan across the 0s on tape 1 until the rst 1. At the same time, copy the 0s onto tape 2. 3. Scan across the 1s on tape 1 until the end of the input. For each 1 read on tape 1, cross o a 0 on tape 2. If all 0s are crossed o before all the 1s are read, reject. 4. If all the 0s have now been crossed o, accept. if any 0s remain, reject.
A O(n).
Hence, the time complexity of and on a 2-tape TM it is
on a single-tape TM is
O(n log n)
In computability theory, the Church-Turing thesis implies that all reasonable models of computation are equivalent, that is, they all decide the same class of languages. In complexity theory, the choice of model aects the time complexity of languages. In complexity theory, we classify computational problems according to their time complexity. But with which model do we measure time?
Time Complexity Complexity Relationships among models
Theorem
Let t(n) be a function, where t(n) ≥ n. Then every t(n) time multitape Turing machine has an equivalent O(t2 (n)) time single-tape TM.
Time Complexity Complexity Relationships among models
Theorem
Let t(n) be a function, where t(n) ≥ n. Then every t(n) time multitape Turing machine has an equivalent O(t2 (n)) time single-tape TM.
Theorem
Let t(n) be a function, where t(n) ≥ n. Then every t(n) time nondeterministic single-tape Turing machine has an equivalent 2O(t(n)) time deterministic single-tape TM.
Time Complexity Complexity Relationships among models
Theorem
Let t(n) be a function, where t(n) ≥ n. Then every t(n) time multitape Turing machine has an equivalent O(t2 (n)) time single-tape TM.
Theorem
Let t(n) be a function, where t(n) ≥ n. Then every t(n) time nondeterministic single-tape Turing machine has an equivalent 2O(t(n)) time deterministic single-tape TM. The theorems above demonstrated at most a square or polynomial dierence between the time complexity measures on deterministic single-tape and multitape TMs
Time Complexity Complexity Relationships among models
Theorem
Let t(n) be a function, where t(n) ≥ n. Then every t(n) time multitape Turing machine has an equivalent O(t2 (n)) time single-tape TM.
Theorem
Let t(n) be a function, where t(n) ≥ n. Then every t(n) time nondeterministic single-tape Turing machine has an equivalent 2O(t(n)) time deterministic single-tape TM. The theorems above demonstrated at most a square or polynomial dierence between the time complexity measures on deterministic single-tape and multitape TMs and at most an exponential dierence between the time complexity of problems on deterministic and nondeterministic TMs.
Time Complexity
From here on we focus on aspects of time complexity theory that are unaected by polynomial dierences in running time. We consider such dierences to be insigni cant and ignore them. The decision to disregard polynomial dierences doesn't imply that we consider such dierences unimportant. On the contrary, we certainly do consider the dierence between time
n
and time
n3
to be an important one.
But some questions that do not depend on polynomial dierences are important too.
Time Complexity
De nition P
is the class of language that are decidable in polynomial time on a
deterministic single-tape TM.
P=
[ k
T IM E(nk )
Time Complexity
De nition P
is the class of language that are decidable in polynomial time on a
deterministic single-tape TM.
P=
[
T IM E(nk )
k
1.
P
is invariant for all models of computation that are
polynomially equivalent to the deterministic single-tape TM, and 2.
P
roughly corresponds to the class of problems that are
realistically solvable on a computer.
Time Complexity Examples of Problems in
P
Time Complexity Examples of Problems in
P AT H = {< G, s, t > | G is a directed graph
P
that has a directed path from
s
to
t.}
Time Complexity Examples of Problems in
P AT H = {< G, s, t > | G is a directed graph
P
that has a directed path from
P AT H ∈ P
s
to
t.}
Time Complexity Examples of Problems in
P AT H = {< G, s, t > | G is a directed graph
P
that has a directed path from
s
to
t.}
P AT H ∈ P M = On and t:
input
< G, s, t >
1. Place a mark on node
where
G
is a directed graph with nodes
s
s.
2. Repeat the until no addtl nodes are marked:
B
Scan all the edges of
marked node 3. If
t
a
G.
If an edge
to an unmarked node
(a, b) is found going b, mark node b.
is marked, accept. Otherwise, reject.
from a
Time Complexity Examples of Problems in
P
Time Complexity Examples of Problems in
RELP RIM E = {< x, y > | x
and
y
P
are relatively prime.}
Time Complexity Examples of Problems in
RELP RIM E = {< x, y > | x
and
y
P
are relatively prime.}
RELP RIM E ∈ P
Time Complexity Examples of Problems in
RELP RIM E = {< x, y > | x
and
y
P
are relatively prime.}
RELP RIM E ∈ P E=
On input
< x, y >
1. Repeat until
B
Exchange
2. Output
x.
where
y = 0: B x and y .
x
and
Assign
y
are natural numbers in binary:
x ← x mod y .
Time Complexity Examples of Problems in
RELP RIM E = {< x, y > | x
and
y
P
are relatively prime.}
RELP RIM E ∈ P E=
< x, y >
On input
1. Repeat until
B
Exchange
2. Output
R=
where
y = 0: B x and y .
x
and
Assign
y
are natural numbers in binary:
x ← x mod y .
x.
On input
1. Run E on
< x, y >
where
x
and
y
are natural numbers in binary:
< x, y >.
2. If the result is 1, accept. Otherwise, reject.
Time Complexity
There are problems, including many interesting and useful ones, wherein polynomial time algorithms that solve them aren't known to exist.
Time Complexity
There are problems, including many interesting and useful ones, wherein polynomial time algorithms that solve them aren't known to exist. Why have we been unsuccessful in nding polynomial time algorithms for these problems?
Time Complexity
There are problems, including many interesting and useful ones, wherein polynomial time algorithms that solve them aren't known to exist. Why have we been unsuccessful in nding polynomial time algorithms for these problems? We don't know.
Time Complexity
There are problems, including many interesting and useful ones, wherein polynomial time algorithms that solve them aren't known to exist. Why have we been unsuccessful in nding polynomial time algorithms for these problems? We don't know. Perhaps these problems have, as yet undiscovered, polynomial time algorithms that rest on unknown principles.
Time Complexity
There are problems, including many interesting and useful ones, wherein polynomial time algorithms that solve them aren't known to exist. Why have we been unsuccessful in nding polynomial time algorithms for these problems? We don't know. Perhaps these problems have, as yet undiscovered, polynomial time algorithms that rest on unknown principles. Or possibly some of these problems simply cannot be solved in polynomial time.
Time Complexity
Time Complexity
HAM P AT H = {< G, s, t > | G is a directed graph with a
Hamiltonian path from
s
to
t.}
Time Complexity
HAM P AT H = {< G, s, t > | G is a directed graph with a
Hamiltonian path from
s
to
t.}
We can easily obtain an exponential time algorithm for this problem by modifying the algorithm for
P AT H .
We need only add a check to
verify that the potential path is Hamiltonian.
Time Complexity
HAM P AT H = {< G, s, t > | G is a directed graph with a
Hamiltonian path from
s
to
t.}
We can easily obtain an exponential time algorithm for this problem by modifying the algorithm for
P AT H .
We need only add a check to
verify that the potential path is Hamiltonian. No one knows whether
HAM P AT H
is solvable in polynomial time.
Time Complexity
HAM P AT H = {< G, s, t > | G is a directed graph with a
Hamiltonian path from
s
to
t.}
We can easily obtain an exponential time algorithm for this problem by modifying the algorithm for
P AT H .
We need only add a check to
verify that the potential path is Hamiltonian. No one knows whether
HAM P AT H
But the HAMPATH problem is
is solvable in polynomial time.
polynomial veri able.
Time Complexity
HAM P AT H = {< G, s, t > | G is a directed graph with a
Hamiltonian path from
s
to
t.}
We can easily obtain an exponential time algorithm for this problem by modifying the algorithm for
P AT H .
We need only add a check to
verify that the potential path is Hamiltonian. No one knows whether
HAM P AT H
But the HAMPATH problem is
is solvable in polynomial time.
polynomial veri able.
That is, even though we don't know of a fast way to determine whether a graph contains a Hamiltonian path, if such a path were discovered somehow, we could easily convince someone else of its existence, simply by presenting it.
Time Complexity
De nition A veri er for a language A is an algorithm V, where
A = {w | V
accepts
< w, c >
for some string
c}. w,
so
a polynomial time veri er runs in polynomial time in the length of
w.
We measure the time of a veri er only in terms of the length of
A language A is polynomially veri able if it has a polynomial time veri er.
Time Complexity
De nition A veri er for a language A is an algorithm V, where
A = {w | V
accepts
< w, c >
for some string
c}. w,
so
a polynomial time veri er runs in polynomial time in the length of
w.
We measure the time of a veri er only in terms of the length of
A language A is polynomially veri able if it has a polynomial time veri er.
De nition NP
is the class of languages that have polynomial time veri ers.
Time Complexity
Theorem
A language is in NP i it is decided by some nondeterministic polynomial time TM.
Time Complexity
Theorem
A language is in NP i it is decided by some nondeterministic polynomial time TM. The following nondeterministic TM decides the
HAM P AT H
problem
in nondeterministic polynomial time.
N1 = On and t:
input
< G, s, t >
1. Write a list of nodes in
G.
m
where
numbers
G
is a directed graph with nodes
p1 , · · · , pm ,
where
m
is the number of
Each number in the list is nondeterministically
selected to be between
1
and
m.
2. Check for repetitions in the list. If any are found, reject. 3. Check whether
s = p1
t = pm .
i between 1 and m − 1, G. If any are not, reject.
4. For each edge of
and
passed, so accept.
if either fail, reject.
check whether
s
(pi , pi+1 )
is an
Otherwise, all tests have been
Time Complexity
De nition N T IM E(t(n)) = {L|L is a language
decided by a
Corollary NP =
S
k
N T IM E(nk )
O(t(n))
time nondeterministic TM}.
Time Complexity Examples of Problems in NP
CLIQU E = {< G, k > |G
is an undirected graph with a
k -clique}
Time Complexity Examples of Problems in NP
CLIQU E = {< G, k > |G
CLIQUE is in NP.
is an undirected graph with a
k -clique}
Time Complexity Examples of Problems in NP
CLIQU E = {< G, k > |G
CLIQUE is in NP.
is an undirected graph with a
k -clique}
Time Complexity Examples of Problems in NP
CLIQU E = {< G, k > |G
CLIQUE is in NP.
is an undirected graph with a
k -clique}
Time Complexity Examples of Problems in NP
SU BSET SU M = {< S, t > |S = {x1P , · · · , xk } and {y1 , · · · , yl } ⊆ {x1 , · · · , xk }, we have yi = t}.
SU BSET SU M
is in NP.
for some
Time Complexity Examples of Problems in NP
SU BSET SU M = {< S, t > |S = {x1P , · · · , xk } and {y1 , · · · , yl } ⊆ {x1 , · · · , xk }, we have yi = t}.
SU BSET SU M
is in NP.
for some
Time Complexity Examples of Problems in NP
SU BSET SU M = {< S, t > |S = {x1P , · · · , xk } and {y1 , · · · , yl } ⊆ {x1 , · · · , xk }, we have yi = t}.
SU BSET SU M
is in NP.
for some
Time Complexity
P = NP ?
Time Complexity
Questions? See you next meeting!
