Theory of computation 学习笔记（Monash FIT2014）
声明
笔记内容来源于 monash university FIT2014 课程内容
仅供学习交流
Conjunctive Normal Form
在布尔逻辑中，如果一个公式是子句的合取，那么它是合取范式(CNF)的。
即，多个表达式的交集（∩）。
例： (B ∨ C ∨ T) ∧ (¬B ∨ ¬C) ∧ (¬B ∨ ¬T) ∧ (¬C ∨ ¬T)
predicate logic
In mathematical logic, a predicate is commonly understood to be a Boolean-valued function .
P: X→ {true, false}, called the predicate on X.
例：∀X : computer(X) ⇐⇒ automatic(X) ∧ programmable(X) ∧ storedProgram(X) ∧ generalPurpose(X)
这句表达式的意义为：对于任意计算机， 当且仅当它同时满足，是自动机，可被运行，可存储运行，能运算所有可计算问题，它被称为计算机。
induction
数学归纳法
证明步骤：
(i) Inductive basis:
(ii) Our inductive hypothesis is that …… is true for n. We need to use this to show that n can be replaced by n + 1 in this inequality
So, we’ve shown that, if the claimed inequality holds for n, then it holds for n + 1
(iii) By the Principle of Mathematical Induction, the claimed inequality must hold for all n.
即：
- 先列出base case， 例如对于n>=1的问题， n=1就是base case。
- 假设n=k时，成立
用1和2证明k+1时仍成立 - 得出结论，原式正确。
Regular expression
证明一个语言不是正则：
Pumping Lemma :
Then for all words w in L with more than N letters,
there exist strings x, y, z, with y ≠ ε, such that
◦ w = xyz
◦ length(x) + length(y) £ N
◦ for all i ≥ 0, xyiz is in L.
例：（证明语言DOG非正则）
Assume, by way of contradition, that DOG is regular. Then there is a FA that recognises it. Let N be the number of states in such an FA. Let w be the string grN (woof) N . By the Pumping Lemma, w can be divided up into three parts, w = xyz, such that y is nonempty, |xy| ≤ N, and xyi z ∈ DOG for all i ≥ 0. The requirement that |xy| ≤ N forces y to fall within the first part, grN , of w. Consider the string xyyz. If y contains g, then xyyz has two gs, so xyyz 6∈ DOG, since every string in DOG has exactly one g. If y contains no g, then it’s all-r, so repetition of y creates at least one extra r (since y is nonempty), so the number of rs is greater than the number of woofs, which violates the definition of DOG. So xyyz 6∈ DOG, which contradicts the conclusion of the Pumping Lemma. So our initial assumption, that DOG is regular, must be incorrect. So DOG is not regular.
Automata
自动机原理讲解见：
Finite Automaton (FA)
Every string traces a unique path in the automaton
即：对于每个state（点），需要分配所有输入类型的输出。比如：a,b 进入state3，那么state3一定有a和b的指出。
解题技巧：对于复杂情况，简化成该问题的反面，取补集。
accept->not accept and not accept-> accept
一个FA的反面：讲所有final改成nonfinal，nonfinal改为final。
Nondeterministic Finite Automata (NFA)
待补充
Kleene’s Theorem
Regular expression, FA, NFA, GNFA 都可以互相转换。
互相转换
Regular->NFA
按如下图例转换
NFA->FA
从start state开始， 找一步之内能接受对应值的state，将这些state作为新集合，继续寻找。直到没有新的集合出现。
空，不作为计数路径，可以直接跳过：
GNFA->regular
简化FA
图例待补充
Convert this into an equivalent FA with the minimum possible number of states：
Final state 作为不同的颜色标记，查看其他行的颜色类型是否一样，不一样的state作为新的颜色。直到不可再分。
正则语言的闭合性
Closure properties of regular languages
complement, union, intersection,concatenation
即正则语言的，补集，并集，交集，级联，仍是正则语言。
context free
Derivation of context free
Context free -> PushDown
S前的直接读，
S后的先放进stack，然后放S（S stack在顶层），S处理完后最后读取。
NFA->CFG
Chomsky Normal Form
Nonterminal -> Nonterminal Nonterminal
Nonterminal -> terminal
Cocke-Younger-Kasami (CYK) algorithm
Pumping Lemma for CFG
Church-Turing Thesis
Any function which can defined by an algorithm can be represented by a Turing Machine.
Evidence:
• different approaches to computability end up in agreement
• long experience, that algorithms can be implemented as programs, and therefore on Turing machines
• no known counterexamples, i.e., no algorithms which seem to be unimplementable
Decidable
Recursively enumerable languages
