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This a very high level review post that I am making for myself and other people taking CS Theory. |
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If you want to lean about the theory behind the content in this blog post I recommed looking else where. |
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This post will cover how to solve typical problems relating to topics covered by my second CS Theory exam. |
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# Myhill-Nerode Theorem |
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## Definition |
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L is regular if and only if it has a finite index. The index is the maximum number of elements thar are pairwise distibguishable. |
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Two strings are said to be pairwise distinguishable if you can append something to both of the strings and it makes one string |
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accepted by the language and the other string non-accepting. |
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The size of an index set X equals the number of equivalence classes it has. Each element in the language is accepted by only |
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one equivalence class. |
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## Problem Approach |
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Prove that language L is regular. |
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1) Define a set X which is infinite in size - this doesn;t necesarrily need to be in the language. |
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2) Make a general argument that show that each element in X is pairwise distinguishable. |
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Pick any two elements x, y in X and show that if you append z to them one is accepted by the language and |
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the other is not in the language. |
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## Example |
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Prove the following language is non-regular: |
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$$ |
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L={ww^r | w \in {0,1}^*} |
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$$ |
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answer: |
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1) |
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$$ |
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X = {(01)^i | i \geq 0} |
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$$ |
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Pick any 2 elements of X and show pairwise distinguishable |
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$$ |
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x = (01)^i, y = (01)^j | i \neq j |
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$$ |
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suppose we pick |
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$$ |
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z = (10)^i\\ |
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xz \in L\\ |
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yz \notin L |
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$$ |
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# DFA minimization algorithm |
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Types of Problems: |
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- Prove DFA is minimal |
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- Minimize the DFA |
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The argument for DFA minimization comes from the Myhill-Nerode theorem. Given |
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a DFA, if you can form a set of strings which represent each state and they are all |
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pairwise distinguishable, then the DFA is minimal with that many states. |
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## Prove DFA is minimal |
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For these types of problems you simply construct a table and show that each state is pairwise distinguishable. |
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To show pairwise distinguishably you have to show that there exists a string where if appended to one element |
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makes it accepted by the language but pushes the other string out of the language. |
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### Example |
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Prove the following DFA is minimal. |
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![DFA Example](media/CSTHEORY/DFAMinimalProof.png) |
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Find a set of strings which represent the minimal path to each state in the DFA. |
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$$ |
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X = \{\epsilon, b, bb, ba\} |
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$$ |
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Show that each state is pairwise distinguishable. |
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![DFA Example](media/CSTHEORY/DFAMinimalTable.png) |
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## Minimize the DFA |
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To use this concept of being indistinguishable to minimize a DFA, you can use a table to keep track which |
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states are distinguishable from each other. The states which are not indistinguishable can |
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be combined. To solve one of these problems you start by creating a table which compares each of the |
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states in the DFA. You then go through and mark the states which are indistinguishable -- start with |
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the ones with different accepting statuses. Then you continue marking off states where if you transition with |
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a symbol on the DFA you are distinguishable and the other state is non-distinguishable according to the table. |
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### Example |
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Minify the Following DFA: |
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![DFA Example](media/CSTHEORY/DFAMinification.png) |
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After marking the states with different accepting criteria as being distinguishable you get this table: |
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![Half Complete Table](media/CSTHEORY/MinificationTable.svg) |
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After looping through all pairs and marking them on the table if there exists symbol which results in one state |
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to be distinguishable and one to be indistinguishable you get this table: |
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![Min TableTable](media/CSTHEORY/MinTable2.svg) |
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According to the table you are able to combine {D, A, B}, {C, F}, and {E, G}. |
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Minimal DFA: |
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![Min DFA](media/CSTHEORY/MinimalDFA.svg) |
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# Pumping lemma for regular languages |
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The pumping lemma cannot prove that a language is regular, however, you can use it |
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to show that some languages are non-regular. This theory gets at the idea that if |
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a regular language is long enough/infinite, it will have a state somewhere which is |
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repeated on the path that accepts the string. |
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The accepted strings can be divided into three parts: |
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- Symbols leading up to the loop |
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- Symbols which complete a loop and come back to start of loop |
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- Symbols at the end of the string |
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![Min DFA](media/CSTHEORY/PumpingLemmaTheory.svg) |
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To Show that a language L is not regular using pumping lemma: |
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- Proof by Contradiction |
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- Assume L is regular |
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- Choose a representative string S which is just barely in the language and is represented in terms of p. |
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- Express S = xyz such that |xy| < p and y > 0 |
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- Show that you can pump y some amount of times such that it is not in the language. |
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- This contradicts the pumping lemma. |
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- The assumption that L is regular is wrong. |
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- L must not be regular. |
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## Example |
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Show that the following language is non-regular. |
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$$ |
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{0^n1^n | n \geq 0} |
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$$ |
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Proof by contradiction |
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Assume that L is regular. |
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Let p be the pumping length associated with L |
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$$ |
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S = o^p1^p |
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$$ |
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S is valid since |
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$$ |
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|s| \geq p, S \in L |
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$$ |
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For any valid decomposition |
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S = xyz |
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such that |xy| <= p and |y| > 0 |
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Consider: |
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$$ |
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xy^2z |
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$$ |
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By the pumping lemma this should be in the language but it is not. Therefore our assumption that the |
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language is regular is false. |
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![Min DFA](media/CSTHEORY/PumpingLemmaExample.svg) |
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# Context-free grammars, closure properties for CFLs |
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The context-free grammars are a superset of the regular languages. This means that CFG's can represent |
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some non-regular languages and every regular language is also a CFL. Contest-free Languages are defined by Context-Free Grammars and accepted using |
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Pushdown Automata machines. |
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Context Free Grammars are Represented using: |
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- **Terminals** = Set of symbols in that language |
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- **Variables** = Set of symbols representing categories |
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- **Start Symbol** = Variable which you start with- written on top |
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- **Substitution Rules** = Set of rules that recursively define the language. |
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## Example 1 |
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Grammar G: |
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$$ |
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A \rightarrow 0A1 \\ |
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A \rightarrow B \\ |
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B \rightarrow \# \\ |
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$$ |
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This grammar describes the following language: |
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$$ |
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L = \{0^k\#1^k | k \geq 0\} |
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$$ |
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## Example 2 |
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Give CFG for non-Palindromes |
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$$ |
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S \rightarrow aXb | bXa | aSa | bSb | ab | ba \\ |
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X \rightarrow aX | bX | a | b \\ |
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$$ |
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In this example, the S rule states in that recursive state until something that is not a palindrome is found. |
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Once you exit the S state, you can finish by appending anything to the middle of the string. |
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## Example 3 |
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Give CFG for the following language: |
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$$ |
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\{a^ib^jc^kd^l | i+k = j + l\} |
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$$ |
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$$ |
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S \rightarrow aSd | XYZ \\ |
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X \rightarrow aXb | \epsilon\\ |
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Y \rightarrow bYc | \epsilon\\ |
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Z \rightarrow cZd | \epsilon |
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$$ |
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# Parse trees, ambiguity |
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# Chomsky Normal Form |
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# Pushdown automata |
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# Construction to convert CFG to a PDA |
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