2011/5/6
optimization problem
Decision problem: 輸出是yes/no (1/0) (可不可以 找到答案)
Optimization problem: 輸出是最好的解 (可以的 答案中找出最好的那個)
例子: Shortest path v.s. Path
NP‐Complete只適用於decision problems
Decision problem v.s.
optimization problem
怎麼將optimization problem轉換成decision problem呢?
對要optimize的值設定一個bound
將Shortest path轉換成decision problem:
給定graph G, vertices u & v, integer k, 有沒有從u 到v的路徑使用少於k個edge?
用Reduction證明”一樣難”
假設以下兩樣存在
則B也沒有polynomial‐time algorithm可以解
使用反證法,假設有
A: no polynomial‐time algorithm exist Polynomial‐time reduction algorithm:
instances of A instances of B
B: polynomial‐time algorithm Polynomial‐time reduction algorithm:
instances of A instances of B
A: polynomial‐time algorithm
矛盾. 所以B也沒有.
易解”?
Θ 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
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
Polynomials are closed under addition, multiplication, and composition.
Abstract problem
I: instances of
problem S: solutions
Q: Abstract problem (binary relation)
Decision problem:
S={0,1}
Example: PATH
, , , 1
0
if a shortest path from u to v has at most k edges
otherwise
Encoding
A set S of abstract objects
The set of binary strings
encoding:
mapping Polygons, numbers,
graphs, functions, ordered pairs, programs, …
Some more definitions
Concrete problem: 一個problem的instance是 binary string的set, 則稱為concrete problem.
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 的algorithm可以解此
problem
P的正式定義
The complexity class P:
The set of concrete decision problems
that are polynomial‐time solvable
Abstract problem轉換成 encoding problem
I: instances of problem
S: solutions Q: Abstract problem
(binary relation)
Decision problem:
S={0,1}
e(I): {0,1}*
: → 0,1 ∗
e(Q): Concrete problem
Encoding和花的時間有關嗎?
有!
極端的例子: unary
input: integer k, running time: Θ
假設encoding是unary: 11111…1111
則在這樣的case下
input length: n running time: Θ
可是如果以正常的binary encoding表示
input lennth: n log 1running time: Θ 2
Encoding決定是Θ or Θ 2 !!
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例子(unary), 大部 分的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.
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
: 0,1 ∗
: → 0,1 ∗ : → 0,1 ∗
: 0,1 ∗
如果有 和 是polynomial‐time computable, 則 和 為
polynomially related.
I: instances of
problem S: solutions
Q: Abstract problem
(binary relation) Decision problem:
S={0,1}
: 0,1 ∗ : → 0,1 ∗
: Concrete problem : → 0,1 ∗
: Concrete problem
polynomially related
if:
Then: ∈ if and only if ∈
因為是對稱的, 所以只需要證明一個方向.
假設 可以在 時間內解決(for some constant k)
假設對每個problem instance i, 轉換成 需花 (for some constant c),
則解決 (input為 ) 先花 轉換成
再解決 (input為 ), 花
c, k都是constant, 因此為polynomial time
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 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 , (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: Σ∗
Concatenation of :
: ∈ ∈
Closure or Kleene Star:
∗ ⋃ ⋃ ⋃ ⋃…
:concatenation 自己k次
應用formal language framework…
Decision problem Q的set of instances為Σ∗
Σ 0,1
Q的主要特性可以想成是會產生答案為1(yes)的這些
instances
因此可以把Q重新定義為一個language over Σ 0,1 ,
而 ∈ Σ∗: 1
例子: PATH= , , , : , is an undirected graph, , ∈ , 0 is an integer, and 從u到v在G裡面 有一條路徑含有最多k條edge
input x, A outputs 1
An algorithm A rejects a string ∈ 0,1 ∗ if, given input x, A outputs 0
The language accepted by an algorithm A is the set of strings ∈ 0,1 ∗: 1
注意: L is accepted by A, 不一定表示 ∉ 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 for a constant k and any length‐n string ∈ .
: PATH
The language PATH can be accepted in polynomial time. 這句話什麼意思?
如果 , , , 可以找到解 (從u到v在G裡面有一條 路徑含有最多k條edge), 則可以在polynomial time裡 面說”有 (1, yes)”!
可以嗎? 可以! 用BFS找出最短路徑, 然後看有沒有比 k大 (只需polynomial time)
假設我們設計這個algorithm發現比k大的話就無窮 迴圈, 這樣的話就沒有 decided in polynomial time.
定義complexity class P
Complexity class: a set of languages, 是不是在其 中由algorithm的complexity measure決定(ex.
running time), 而此algorithm是決定一個string x 是否屬於L.
使用這個方式, 我們可以重新定義P這個
complexity class:
⊆ 0,1 ∗:
there exists an algorithm A that decides L in polynomial time
: 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.
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.
for a constant c
對任何input x, A’ 利用A, 先執行 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.
Algorithms that accepts/rejects/decides L
告訴你某個instance是不 是有解(在L裡面)
Algorithms that verify L with a certificate
給你一個certificate (可能 是答案), 可以讓你檢查某 個instance是不是有解(在 L裡面)
給 , , , 和一條path p, 我們可以檢查path是 不是真的是在G中uv的path, 且長度是不是不 超過k. 此p是一個certificate, 用來幫助algorithm 看此instance是不是屬於PATH.
對PATH來說其實沒有太大差別, 因為本來就可以在polynomial time decide PATH. 但是對於其他問題可能有差別!
G=(V,E) is a simple cycle that contains each vertex in V.
Not all graph is Hamiltonian (找不到Hamiltonian cycle)
HAM‐CYCLE={ : G is a Hamiltonian graph}
Hamiltonian cycles
暴力法?
假設使用adjacency matrix, n= 是G的
encoding的長度 (也就是 ,m是G中vertex數)
檢查所有的vertex permutation需要Ω !
Ω ! Ω 2 , not for any constant k.
目前還找不到polynomial time algorithm to decide/accept HAM‐CYCLE
Verify會簡單一點嗎?
會!
假設告訴你某一個graph G是Hamiltonian, 然後 告訴你一個vertex的序列(certificate)可以組成 Hamiltonian cycle.
則我們可以在polynomial time裡面檢查:
這個vertex序列是不是真的是G裡面的vertex的 permutation
vertex序列的相鄰vertices之間是不是在G中有那 個edge
Verification Algorithm
Verification Algorithm: Algorithm A with two arguments:
Ordinary input string x
Binary string y (certificate)
A verifies an input string x if there exists a certificate y such that A(x,y)=1.
The language verified by a verification algorithm A is
∈ 0,1 ∗: there exists ∈ 0,1 ∗ such that , 1
如果x在L裡面, 則一定找得到y.
如果x不在L裡面, 則一定找不到y.
The complexity class NP
Complexity class NP: the class of languages that
can be verified by a polynomial‐time algorithm. A language L belongs to NP if and only if there exist a two‐input polynomial‐time algorithm A and a constant c such that
L= ∈ 0,1 ∗: there exists a certificate with such that , 1 .
Then algorithm A verifies L in polynomial time.
What is in NP?
HAM‐CYCLE∈NP
if L ∈ , then L ∈ . Why?
可以做出一個algorithm是完全不甩certificate的, 就可以模擬出verification algorithm的效果
意思就是說 ⊆ .
但P=NP or not? (尚未得知)
Complexity class co‐NP
class NP is closed under complement? (尚未得知)
意思就是說 ∈ 的話, ∈ 否?
co‐NP: all languages that satisfies ∈
NP‐Complete languages
“The hardest languages in NP”
If NP‐P is nonempty, then these in NP‐Complete are in NP‐P (such as HAM‐CYCLE)
Reducibility 解一個破全部, 一箭千雕
Reducibility
如果Q可以reduce成Q’, 則表示任何一個Q的 instance都可以”換句話說”變成Q’的一個
instance
一元一次方程式: ax+b=0可以視為一元二次方程
式的特例: 0 0, 解出來可以得到對
應的一元一次方程式解.
如果一個問題Q可以reduce成另外一個問題Q’, 則Q不會比Q’難解.
Reduction
is polynomial‐time reducible to , (寫成 ) if there exists a polynomial‐time
computable function f: 0,1 ∗ → 0,1 ∗ such that for all ∈ 0,1 ∗, ∈ if and only if ∈ .
f: reduction function
用來計算f 的polynomial‐time algorithm F:
