Completeness NP-

Completeness

Jennya 2013/10/31

為什麼Polynomial time就是

”容易解”, ”可解的”?

 (1)

Θ(𝑛¹⁰⁰)雖然是polynomial time, 但實務上這麼高次的多項式並 不常見 通常如果找到一個polynomial-time algorithm, 比較快的方法很 快也會被找到

 (2)

通常使用不同的computation model(之後自動機會教到, 現在可 以想像是單CPU v.s. 多CPU的機器), 某model可用polynomial- time解的問題在另外一個model也可用polynomial-time解

 (3)

Polynomials are closed under addition, multiplication, and composition.

Abstract problem

I

problem instances

S

problem solutions Q: Abstract problem

(binary relation) Example: PATH

𝑖 = 𝐺, 𝑢, 𝑣, 𝑘

𝑃𝐴𝑇𝐻 𝑖 = 1 if a shortest path from u to v has at most k edges 𝑃𝐴𝑇𝐻 𝑖 = 0 otherwise

Decision problem:

S={0,1}

Encoding

A set S of abstract objects

The set of binary strings

encoding:

mapping Polygons, numbers,

graphs, functions, ordered pairs, programs, …

Abstract problem轉換成 concrete problem

We can use encodings to map abstract problems to concrete problems

Instance set I solution set S

{0,1}

Q

Abstract problem Abstract Decision Problem Q

𝑒: 𝐼 → 0,1 ^∗

e(I): {0,1}* solution set S

{0,1}

e(Q)

Concrete problem Concrete Decision Problem e(Q)

 Concrete problem:

instance set = the set of binary strings

 「An algorithm solves a concrete problem in O(T(n))」

一個problem的instance長度為n (i的長度, 即為binary string長度)

而此algorithm可在O(T(n))時間產生解

 「A concrete problem is polynomial-time solvable」

有一個𝑂 𝑛^𝑘 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑘的algorithm可以解此 problem

Concrete problem

P的正式定義

The complexity class P:

The set of concrete decision problems

that are polynomial-time solvable

Encoding和花的時間有關嗎?

 有! 極端的例子: unary

 input: integer k

running time: Θ 𝑘

 Unary encoding: 11111…1111 input length n

 running time: Θ(𝑘) = Θ(𝑛)

 binary encoding:

input length n = log 𝑘 + 1

running time:Θ(𝑘) = Θ(2^𝑛)

 Encoding決定是Θ(𝑛) or Θ(2^𝑛)!!

k個

Encoding和花的時間有關嗎?

 然而如果我們不考慮這麼極端的encoding方式 (unary), 其他的encoding都不會影響到一個問題是 否可以在polynomial time解決.

 例: 使用三進位數和二進位數是沒有差別的, 因為

我們可以在polynomial time裡面將三進位數轉換 成二進位數.

polynomial-time computable function

f: {0,1}*{0,1}*

input output

如果f花polynomial time可以把任何input轉成 output, 則稱為polynomial-time computable

Polynomially related

I: instances of problem

𝑒₁: 0,1 ^∗

𝑒₁: 𝐼 → 0,1 ^∗ 𝑒₂: 𝐼 → 0,1 ^∗

𝑒₂: 0,1 ^∗ 𝑓₁₂ 𝑒₁ 𝑖 = 𝑒₂(𝑖)

𝑓₂₁ 𝑒₂ 𝑖 = 𝑒₁(𝑖)

如果有𝑓₁₂和𝑓₂₁是polynomial-time computable, 則𝑒₁和𝑒₂為 polynomially related.

I: instances of

problem S: solutions

Q: Abstract problem

(binary relation) Decision problem:

S={0,1}

𝑒₁: 0,1 ^∗ 𝑒₁: 𝐼 → 0,1 ^∗

𝑒₁(𝑄): Concrete problem 𝑒₂: 0,1 ^∗

𝑒₂: 𝐼 → 0,1 ^∗

𝑒₂(𝑄): Concrete problem

polynomially related

if:

Then: 𝑒₁(𝑄) ∈ 𝑃 if and only if 𝑒₂(𝑄) ∈ 𝑃

 Proof:

 假設𝑒₁(𝑄)可以在𝑂(𝑛^𝑘)時間內解決(for some constant k)

 假設 對每個problem instance i, 𝑒₂(𝑖)轉換成𝑒₁(𝑖)需花 𝑂(𝑛^𝑐)(for some constant c), 𝑛 = 𝑒₂ 𝑖

 則解決𝑒₂ 𝑄 (input為𝑒₂(𝑖)) 先花𝑂(𝑛^𝑐)轉換成𝑒₁(𝑖)

 𝑒₁ 𝑖 = 𝑂(𝑛^𝑐)

 再解決𝑒₁(𝑄) (input為𝑒₁(𝑖)), 花𝑂 𝑒₁ 𝑖 ^𝑘 = 𝑂 𝑛^𝑐𝑘

 c, k都是constant, 因此為polynomial time

 因為是對稱的, 所以只需要證明一個方向.

只要encoding都是”合理的” (“簡要的”)表示方式, 一個問題的複雜度 (能否在polynomial time裡面解掉)不會被encoding影響.

A Formal-language Framework

 An alphabet Σ: a finite set of symbols

 A language 𝐿 over Σ: 使用Σ裡面的symbol組合而 成的字串 (不一定包含全部可能的字串)

 Ex: Σ = 0,1 , 𝐿 (over Σ)={10,11,101,111, … }

 empty string: ϵ

 empty language: ∅

 Σ^∗: the language with all strings over Σ

Operations on languages

 Union

 Intersection

 Complement: 𝐿 = Σ^∗ − 𝐿

 Concatenation of 𝐿₁𝐿₂:

𝐿 = 𝑥₁𝑥₂: 𝑥₁ ∈ 𝐿₁ 𝑎𝑛𝑑 𝑥₂ ∈ 𝐿₂

 Closure (Kleene Star):

𝐿^∗ = 𝜖 ⋃𝐿 ⋃𝐿² ⋃𝐿³⋃…

𝐿^𝑘:concatenation 自己k次

應用formal language framework…

 We can view

「a decision problem Q 」as

「a language L over Σ = 0,1 」

=> 𝐿 = {𝑥 ∈ Σ^∗: 𝑄 𝑥 = 1}

 Q的instance set為Σ^∗

 Q = 能夠產生答案為1(yes)的這些instances

 i.e. PATH problem的language：

Accepts and Rejects

 An algorithm A accepts a string 𝑥 ∈ 0,1 ^∗ if, given input x, the algorithm’s output A(x) = 1

 An algorithm A rejects a string 𝑥 ∈ 0,1 ^∗ if, given input x, the algorithm’s output A(x) = 0

 The language accepted by an algorithm A is the set of strings 𝐿 = {𝑥 ∈ 0,1 ^∗: 𝐴 𝑥 = 1}

 注意: L is accepted by A, 不一定表示𝑥 ∉ 𝐿會被A reject! (ex. 無窮迴圈)

 A language is decided by an algorithm A if every binary string in L is accepted by A and every binary string not in L is rejected by A

 A language is accepted in polynomial time if it is accepted by A and if A accepts x in time 𝑂(𝑛^𝑘) for a constant k and any length-n string 𝑥 ∈ 𝐿.

使用formal-language framework 定義complexity class P

 可以用「a set of languages」定義 「complexity class」

如何決定是不是在這個class(set)中：

由「決定一個string x是否屬於L」的 algorithm的 running time而定

 使用這個方式, 我們可以重新定義P這個complexity class:

𝑃 = {𝐿 ⊆ 0,1 ^∗:

there exists an algorithm A that decides L in polynomial time}

 Theorem:

𝑃 = {𝐿: 𝐿 is accepted by a polynomial time algorithm}.



 Proof:

 The class of languages decided by polynomial-time

algorithms是the class of languages accepted by polynomial- time algorithms的subset.

 所以我們只需要證如果L is accepted by a polynomial-time algorithm, 它也可以decided by a polynomial-time algorithm.

𝑃 = {𝐿 ⊆ 0,1 ^∗:

there exists an algorithm A that 𝐝𝐞𝐜𝐢𝐝𝐞𝐬 L in polynomial time}

𝑃 = {𝐿 ⊆ 0,1 ^∗:

there exists an algorithm A that 𝐚𝐜𝐜𝐩𝐞𝐭𝐬 L in polynomial time}

 假設L是被某polynomial-time algorithm A accept.

 我們要利用A做成一個algorithm A’可以decides L.

 因為A accepts L in 𝑂(𝑛^𝑘) for some constant k, 所 以我們也可以說 A accepts L 最多花𝑐𝑛^𝑘個steps for a constant c

 對任何input x, A’ 利用A, 先執行𝑐𝑛^𝑘個steps. 如 果這時候A accept x了, A’就accept x. 如果A還沒 accept x, A’就reject x.

 A’使用A的overhead不會超過一個polynomial factor, 所以A’是一個可以decide L的polynomial time algorithm.