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| --- | ||
| title: "CS135-lecture-20210318" | ||
| # date: 2021-03-18T13:03:19-07:00 | ||
| draft: false | ||
| bookToc: true | ||
| tags: ["context-free language", "pumping lemma"] | ||
| --- | ||
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| ## Pumping lemma for CFL | ||
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| PL for RL | ||
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| Given {{<k>}} RL, L {{</k>}} | ||
| there exists {{<k>}} p {{</k>}} such | ||
| that if {{<k>}} w \in L {{</k>}} and {{<k>}} |w| \geq p {{</k>}} | ||
| then there exists {{<k>}} w = xyz {{</k>}} where | ||
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| - {{<k>}} |y| > 0 {{</k>}} | ||
| - {{<k>}} |xy| \leq p {{</k>}} | ||
| - {{<k>}} xy^iz \in L {{</k>}} | ||
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| We consume {{<k>}} w {{</k>}} and we get to a state | ||
| that is the first repeated state. | ||
| At some point we come back to it, and then to the accept. | ||
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|  | ||
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| Since {{<k>}} w {{</k>}} is {{<k>}} p {{</k>}} or longer, it must visit one state more than once. | ||
| We can identify this state as the one we get repeated. | ||
| The part that gets us there the first time {{<k>}} x {{</k>}}, | ||
| the part that gets us there the second time {{<k>}} y {{</k>}}, | ||
| and the part that goes to the accept state {{<k>}} z {{</k>}}. | ||
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| If {{<k>}} x {{</k>}} gets us to the repeated state, then we can go around {{<k>}} y {{</k>}} as many times we want (or omit). | ||
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| ## Proof template | ||
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| Given context-free language | ||
| {{<k>}} L {{</k>}}, there exists {{<k>}} p {{</k>}} such | ||
| that if {{<k>}} w \in L {{</k>}} and {{<k>}} |w| \geq p {{</k>}}, | ||
| then there exists a {{<k>}} w = uvxyz {{</k>}} where | ||
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| - {{<k>}} |vy| > 0 {{</k>}} | ||
| - {{<k>}} |vxy| \leq p {{</k>}} | ||
| - {{<k>}} uv^ixy^iz \in L {{</k>}} | ||
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| Main use: we look for strings that are in the language and at least {{<k>}} p {{</k>}} long, and then apply the pumping lemma to show | ||
| that the language is not context-free. | ||
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| To prove: | ||
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| - Assume {{<k>}} L {{</k>}} is | ||
| context-free | ||
| - Pick {{<k>}} w \in L {{</k>}} with {{<k>}} |w| \geq p {{</k>}} | ||
| - Argue that the pumping lemma says that there exists {{<k>}} w = uvxyz {{</k>}} with | ||
| - {{<k>}} |vy| > 0 {{</k>}} | ||
| - {{<k>}} |vyx| \leq p {{</k>}} | ||
| - {{<k>}} uv^ixy^i \in L {{</k>}} | ||
| - Argue that either if we pump down ({{<k>}} i = 0 {{</k>}} gives us {{<k>}} uxz {{</k>}}, | ||
| or if we pump up {{<k>}} i > 0 {{</k>}} we get {{<k>}} uvvxyyz {{</k>}} | ||
| and show that {{<k>}} e \not \in L {{</k>}} | ||
| - Pumping lemma says string is in {{<k>}} L {{</k>}}, contradiction, so {{<k>}} L {{</k>}} is not | ||
| context-free. | ||
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| ### Example | ||
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| Consider {{<k>}} L = \{a^n b^n c^n : n \geq 0\} {{</k>}} | ||
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| This starts by consuming {{<k>}} a {{</k>}}'s and pushing them onto the stack. | ||
| Then when we start seeing {{<k>}} b {{</k>}}'s we pop the {{<k>}} a {{</k>}}'s off. | ||
| When we reach the {{<k>}} c {{</k>}}'s we don't have a way to remember how many {{<k>}} a {{</k>}}'s or {{<k>}} b {{</k>}}'s we saw. | ||
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| {{< hint info >}} | ||
| Note: A Turning machine can do this because it has 2 stacks. | ||
| {{< /hint >}} | ||
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| So lets prove its not context-free using the pumping lemma. | ||
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| The string we pick must be length {{<k>}} p {{</k>}} and exist in {{<k>}} L {{</k>}}: | ||
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| {{<k display>}} | ||
| \begin{aligned} | ||
| w = a^p b^p c^p | ||
| \end{aligned} | ||
| {{</k>}} | ||
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| {{<k>}} w {{</k>}} exists in the language and is at least {{<k>}} p {{</k>}} long (it is in fact {{<k>}} 3p {{</k>}} long). | ||
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| So lets assume that this language is context-free, and apply the pumping lemma. | ||
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| Possible divisions for {{<k>}} w {{</k>}}: | ||
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| Since {{<k>}} w = uvxyz {{</k>}}, we don't know exactly where they are however | ||
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| We do know that {{<k>}} 0 < |vxy| \leq p {{</k>}} | ||
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| Its possible that {{<k>}} vxy {{</k>}}: | ||
| - appear in the region of just {{<k>}} a {{</k>}}'s, since there are at least {{<k>}} p {{</k>}} {{<k>}} a {{</k>}}'s. | ||
| So {{<k>}} vxy = {{</k>}} all {{<k>}} a {{</k>}}'s. | ||
| - appear in the {{<k>}} a {{</k>}}'s and {{<k>}} b {{</k>}}'s. | ||
| - any other combination of 2, but not all 3 | ||
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|  | ||
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| So, {{<k>}} uvvxyyz {{</k>}} will increase 1 or 2 types of characters, but not all 3. | ||
| Thus it is out of balance, and the string is not in {{<k>}} L {{</k>}}. | ||
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| {{<k display>}} | ||
| \begin{aligned} | ||
| uvvxyyz \not \in L | ||
| \end{aligned} | ||
| {{</k>}} | ||
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| Therefore, the language {{<k>}} L = \{a^n b^n c^n : n \geq 0\} {{</k>}} | ||
| is not context-free. | ||
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| ## Why the pumping lemma is true | ||
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| If {{<k>}} L {{</k>}} is | ||
| context-free, then there is a context-free grammar for it. | ||
| So, every string {{<k>}} w \in L {{</k>}} has a parse tree: | ||
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| where the fringe is {{<k>}} w {{</k>}} and the interior is non-terminals. | ||
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|  | ||
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| Observe, | ||
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| - The longer that {{<k>}} w {{</k>}} is, the taller the tree | ||
| - The taller the tree, the more non-terminals in the longest root-to-leaf path | ||
| - Once the root-to-leaf path is long enough, some non-terminal must repeat in it | ||
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| There are a finite number of productions in the grammar, and so as the tree gets longer and longer until there must be a repeat. | ||
| This is where the pumping lemma comes in. | ||
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| As an abstraction: | ||
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| Some part of {{<k>}} w {{</k>}} will be derived from {{<k>}} A {{</k>}}. | ||
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|  | ||
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| This tells us that | ||
| - {{<k>}} x \in L(A) {{</k>}}, and | ||
| - {{<k>}} vxy \in L(A) {{</k>}} | ||
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| So each time that {{<k>}} A {{</k>}} is reached, we can replace it with {{<k>}} x {{</k>}} or {{<k>}} vxy {{</k>}}, | ||
| we can pump the {{<k>}} v {{</k>}} and {{<k>}} y {{</k>}} as many times we want. | ||
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| {{< hint info >}} | ||
| Note: This is similar to the pumping lemma for regular languages because it also uses the pigeon hole principle. | ||
| {{< /hint >}} | ||
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,36 @@ | ||
| --- | ||
| title: "CS137-lecture-20210318" | ||
| # date: 2021-03-18T19:04:36-07:00 | ||
| draft: false | ||
| bookToc: true | ||
| tags: ["sequential circuits", "finite state machine"] | ||
| --- | ||
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| # FSM cont. | ||
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| ## Design rules | ||
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| - If we cannot determine function(s) of combinational circuit(s) in advance: | ||
| 1. Model FSM as FSD | ||
| - May need to design bit-slice 1st | ||
| 2. Determine number of flip flops (dependent on the number of states you need) | ||
| 3. Convert the FSD to truth table | ||
| 4. Find minimal expressions for next state variable(s) and output(s) | ||
| 5. Draw the complete circuit with flip-flops | ||
| - Otherwise | ||
| - Use bit-serial design with known modules | ||
| - Or, bit-parallel design with known modules | ||
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| ### Example | ||
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| Lets design a Moore state machine that accepts the string {{<k>}} 101 {{</k>}}. | ||
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| As a box diagram: | ||
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|  | ||
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| Step 1, create a finite state diagram: | ||
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