But it is often needed to recognize this language Example: Programming language syntax have matching brackets, not regular.
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- Jonah Cross
- 7 years ago
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1 Big picture All languages Decidable Turing machines NP P Context-free Context-free grammars, push-down automata Regular Automata, non-deterministic automata, regular expressions
2 Recall: Theorem: L := {0 n 1 n : n 0} is not regular But it is often needed to recognize this language Example: Programming language syntax have matching brackets, not regular. Next: Introduce context-free languages
3 Why study context-free languages Practice with more powerful model Programming languages: Syntax of C++, Java, etc. is specified by context-free grammar Other reasons: human language has structures that can be modeled as context-free language English is not a regular language
4 Example: Context-free grammar G, Σ = {0,1} S 0 S 1 S ε Two substitution rules (a.k.a. productions) Variables = {S}, Terminals = {0,1} Derivation of 0011 in grammar: S 0S1 00S L(G) = {0 n 1 n : n 0}
5 Example: Context-free grammar G, Σ = {0,1} S A S B A 0 A 1 A ε B 1 B 0 B ε L(G) = L(A) U L(B) = {0 n 1 n : n 0} U {1 n 0 n : n 0} Next: A convention to write this more compactly
6 Example: Context-free grammar G, Σ = {0,1} S A B A 0 A 1 ε B 1 B 0 ε Convention: Write A w w' for A w and A w'
7 Definition: A context-free grammar (CFG) G is a 4 tuple (V, Σ, R, S) where V is a finite set of variables Σ is a finite set of terminals (V Σ = ) R is a finite set of rules, where each rule is A w S V is the start variable A V, w (V U Σ)*
8 Example The language L = {a m b n : m > n} is described by the CFG G = (V, Σ, R, S) where: V = {S, T} Σ = {a, b} R = { S as at T atb ε } Derive aaab: S?
9 Example The language L = {a m b n : m > n} is described by the CFG G = (V, Σ, R, S) where: V = {S, T} Σ = {a, b} R = { S as at T atb ε } Derive aaab: S as?
10 Example The language L = {a m b n : m > n} is described by the CFG G = (V, Σ, R, S) where: V = {S, T} Σ = {a, b} R = { S as at T atb ε } Derive aaab: S as aat?
11 Example The language L = {a m b n : m > n} is described by the CFG G = (V, Σ, R, S) where: V = {S, T} Σ = {a, b} R = { S as at T atb ε } Derive aaab: S as aat aaatb?
12 Example The language L = {a m b n : m > n} is described by the CFG G = (V, Σ, R, S) where: V = {S, T} Σ = {a, b} R = { S as at T atb ε } Derive aaab: S as aat aaatb aaab
13 Definition: Let G = (V, Σ, R, S) be a CFG we write uav uwv and say uav yields uwv if A w is a rule We say u derives v, written u * v, if u = v, or u 1, u 2,, u k k 1 : u u 1 u 2 u k = v The language of the grammar is L(G) = {w : S * w}
14 Definition: A language L is context-free if CFG G : L(G) = L
15 Example: = {0,1,#} Give a CFG for L = { x#y : x,y in {0,1}* x y } G = S BL S RB L BL A R RB A A BAB # B 0 1 Remark: B *? To understand, explain what each piece does!
16 Example: = {0,1,#} Give a CFG for L = { x#y : x,y in {0,1}* x y } G = S BL S RB L BL A R RB A A BAB # Remark: A *? B 0 1 Remark: B * 0, B * 1
17 Example: = {0,1,#} Give a CFG for L = { x#y : x,y in {0,1}* x y } G = S BL S RB L BL A R RB A Remark: R *? A BAB # Remark: A * x#y : x = y B 0 1 Remark: B * 0, B * 1
18 Example: = {0,1,#} Give a CFG for L = { x#y : x,y in {0,1}* x y } G = S BL S RB L BL A Remark: L *? R RB A Remark: R * x#y : x y A BAB # Remark: A * x#y : x = y B 0 1 Remark: B * 0, B * 1
19 Example: = {0,1,#} Give a CFG for L = { x#y : x,y in {0,1}* x y } G = S BL S RB Remark: RB *? L BL A Remark: L * x#y : x y R RB A Remark: R * x#y : x y A BAB # Remark: A * x#y : x = y B 0 1 Remark: B * 0, B * 1
20 Example: = {0,1,#} Give a CFG for L = { x#y : x,y in {0,1}* x y } G = S BL Remark: BL *? S RB Remark: RB * x#y : x < y L BL A Remark: L * x#y : x y R RB A Remark: R * x#y : x y A BAB # Remark: A * x#y : x = y B 0 1 Remark: B * 0, B * 1
21 Example: = {0,1,#} Give a CFG for L = { x#y : x,y in {0,1}* x y } G = S BL Remark: BL * x#y : x > y S RB Remark: RB * x#y : x < y L BL A Remark: L * x#y : x y R RB A Remark: R * x#y : x y A BAB # Remark: A * x#y : x = y B 0 1 Remark: B * 0, B * 1 L(G) = L
22 Example: CFG for expressions in programming languages Task: recognize strings like x (1 + 0) S S+S S x S ( S ) 0 1 S S + S 0 + S 0 + S + S S S x S x S x (S) x (S + S) x (1 + S) x (1 + 0)
23 We have seen: CFG, definition, and examples Next: Ambiguity
24 Ambiguity: Some string may have multiple derivations in a CFG Ambiguity is a problem for compilers: Compilers use derivation to give meaning to strings. Example: meaning of 1+0x0 * is its value, 1 N If there are two different derivations, the value may not be well defined.
25 Example: The string 1+0x0 has two derivations in S S+S S x S ( S ) 0 1 One derivation: S S+S 1+S 1+SxS 1+0xS 1+0x0 Another derivation: S?
26 Example: The string 1+0x0 has two derivations in S S+S S x S ( S ) 0 1 One derivation: S S+S 1+S 1+SxS 1+0xS 1+0x0 Another derivation: S SxS?
27 Example: The string 1+0x0 has two derivations in S S+S S x S ( S ) 0 1 One derivation: S S+S 1+S 1+SxS 1+0xS 1+0x0 Another derivation: S SxS Sx0?
28 Example: The string 1+0x0 has two derivations in S S+S S x S ( S ) 0 1 One derivation: S S+S 1+S 1+SxS 1+0xS 1+0x0 Another derivation: S SxS Sx0 S+Sx0?
29 Example: The string 1+0x0 has two derivations in S S+S S x S ( S ) 0 1 One derivation: S S+S 1+S 1+SxS 1+0xS 1+0x0 Another derivation: S SxS Sx0 S+Sx0 S+0x0?
30 Example: The string 1+0x0 has two derivations in S S+S S x S ( S ) 0 1 One derivation: S S+S 1+S 1+SxS 1+0xS 1+0x0 Another derivation: S SxS Sx0 S+Sx0 S+0x0 1+0x0
31 We now want to define CFG with no ambiguity Definition: A derivation is leftmost if at every step the leftmost variable is expanded Example: the 1 st previous derivation was leftmost S S+S 1+S 1+SxS 1+0xS 1+0x0 Definition: A CFG G is un-ambiguous if no string has two different leftmost derivations.
32 Example The CFG S S+S S x S ( S ) 0 1 is ambiguous because 1+0x0 has two distinct leftmost derivations One leftmost derivation: S S+S 1+S 1+SxS 1+0xS 1+0x0 Another leftmost derivation: S SxS S+SxS 1+SxS 1+0xS 1+0x0
33 Example Instead of using CFG S S+S S x S ( S ) 0 1 we may use un-ambiguous grammar S S + T T T T x F F F 0 1 (S) Unique leftmost derivation of 1+0x0: S?
34 Example Instead of using CFG S S+S S x S ( S ) 0 1 we may use un-ambiguous grammar S S + T T T T x F F F 0 1 (S) Unique leftmost derivation of 1+0x0: S S + T?
35 Example Instead of using CFG S S+S S x S ( S ) 0 1 we may use un-ambiguous grammar S S + T T T T x F F F 0 1 (S) Unique leftmost derivation of 1+0x0: S S + T T + T?
36 Example Instead of using CFG S S+S S x S ( S ) 0 1 we may use un-ambiguous grammar S S + T T T T x F F F 0 1 (S) Unique leftmost derivation of 1+0x0: S S + T T + T F + T?
37 Example Instead of using CFG S S+S S x S ( S ) 0 1 we may use un-ambiguous grammar S S + T T T T x F F F 0 1 (S) Unique leftmost derivation of 1+0x0: S S + T T + T F + T 1 + T?
38 Example Instead of using CFG S S+S S x S ( S ) 0 1 we may use un-ambiguous grammar S S + T T T T x F F F 0 1 (S) Unique leftmost derivation of 1+0x0: S S + T T + T F + T 1 + T 1 + T x F?
39 Example Instead of using CFG S S+S S x S ( S ) 0 1 we may use un-ambiguous grammar S S + T T T T x F F F 0 1 (S) Unique leftmost derivation of 1+0x0: S S + T T + T F + T 1 + T 1 + T x F x F x 0
40 Actual Java specification grammar snippet Cumbersome but un-ambiguous MultiplicativeExpression: UnaryExpression MultiplicativeExpression * UnaryExpression MultiplicativeExpression / UnaryExpression MultiplicativeExpression % UnaryExpression AdditiveExpression: MultiplicativeExpression AdditiveExpression + MultiplicativeExpression AdditiveExpression - MultiplicativeExpression
41 Next: understand power of context-free languages Study closure under not, U, o, * Recall from regular langues: If A, B are regular then not A is regular? A U B is regular? A o B is regular? A* is regular?
42 Next: understand power of context-free languages Study closure under not, U, o, * Recall from regular langues: If A, B are regular then not A regular A U B regular A o B regular A* regular
43 Suppose A, B are context-free: A = L(G A ) for CFG G A =(V A, Σ, R A, S A ) B = L(G B ) for CFG G B =(V B, Σ, R B, S B ) What about A U B S? A o B A*
44 Suppose A, B are context-free: A = L(G A ) for CFG G A =(V A, Σ, R A, S A ) B = L(G B ) for CFG G B =(V B, Σ, R B, S B ) What about A U B S S A S B Context-free A o B S? A*
45 Suppose A, B are context-free: A = L(G A ) for CFG G A =(V A, Σ, R A, S A ) B = L(G B ) for CFG G B =(V B, Σ, R B, S B ) What about A U B S S A S B Context-free A o B S S A S B Context-free A* S?
46 Suppose A, B are context-free: A = L(G A ) for CFG G A =(V A, Σ, R A, S A ) B = L(G B ) for CFG G B =(V B, Σ, R B, S B ) What about A U B S S A S B Context-free A o B S S A S B Context-free A* S SS A ε Context-free Above all context-free! In general, (not A) is NOT context-free
47 Suppose A, B are context-free: A = L(G A ) for CFG G A =(V A, Σ, R A, S A ) B = L(G B ) for CFG G B =(V B, Σ, R B, S B ) What about A U B S S A S B Context-free A o B S S A S B Context-free A* S SS A ε Context-free Above also shows regular context-free Context-free languages contain regular languages
48 Example: Context Free UNION = {0,1,#} Give a CFG for L = { x#y : x,y in {0,1}* x y OR x = y R } y R is the reverse of y: 001 R = R = R = 1
49 Example: Context Free UNION = {0,1,#} Give a CFG for L = { x#y : x,y in {0,1}* x y OR x = y R } Write L = L 1 U L 2, where L 1 = { x#y : x y } L 2 = { x#y : x = y R }
50 Example: Context Free UNION = {0,1,#} Give a CFG for L = { x#y : x,y in {0,1}* x y OR x = y R } Write L = L 1 U L 2, where L 1 = { x#y : x y } L 2 = { x#y : x = y R } G 1 = S 1 BL RB L BL A R RB A A BAB # B 0 1 Remark: L * x#y : x y Remark: R * x#y : x y Remark: A * x#y : x = y
51 Example: Context Free UNION = {0,1,#} Give a CFG for L = { x#y : x,y in {0,1}* x y OR x = y R } Write L = L 1 U L 2, where L 1 = { x#y : x y } L 2 = { x#y : x = y R } G 1 = S 1 BL RB L BL A R RB A A BAB # B 0 1 G 2 = S 2 0S 2 0 1S 2 1 #
52 Example: Context Free UNION = {0,1,#} Give a CFG for L = { x#y : x,y in {0,1}* x y OR x = y R } Write L = L 1 U L 2, where L 1 = { x#y : x y } L 2 = { x#y : x = y R } G 1 = S 1 BL RB L BL A R RB A A BAB # B 0 1 G 2 = S 2 0S 2 0 1S 2 1 # Let G = S S 1 S 2 Then, L(G 1 ) = L 1 & L(G 2 ) = L 2 L(G) = L 1 U L 2 = L
53 Example: Context Free CONCATENATION Give a CFG for L = { 0 m 1 m 0 n 1 n : m even and n odd}
54 Example: Context Free CONCATENATION Give a CFG for L = { 0 m 1 m 0 n 1 n : m even and n odd} Write L = L 1 o L 2, where L 1 = { 0 m 1 m : m even} L 2 = { 0 n 1 n : n odd}
55 Example: Context Free CONCATENATION Give a CFG for L = { 0 m 1 m 0 n 1 n : m even and n odd} Write L = L 1 o L 2, where L 1 = { 0 m 1 m : m even} L 2 = { 0 n 1 n : n odd} G 1 = S 1 00S 1 11 ε
56 Example: Context Free CONCATENATION Give a CFG for L = { 0 m 1 m 0 n 1 n : m even and n odd} Write L = L 1 o L 2, where L 1 = { 0 m 1 m : m even} L 2 = { 0 n 1 n : n odd} G 1 = S 1 00S 1 11 ε G 2 = S 2 00S
57 Example: Context Free CONCATENATION Give a CFG for L = { 0 m 1 m 0 n 1 n : m even and n odd} Write L = L 1 o L 2, where L 1 = { 0 m 1 m : m even} L 2 = { 0 n 1 n : n odd} G 1 = S 1 00S 1 11 ε G 2 = S 2 00S Let G = S S 1 S 2 Then, L(G 1 ) = L 1 & L(G 2 ) = L 2 L(G) = L 1 o L 2 = L
58 Example: Context Free STAR Give a CFG for L = { w in {0,1}* : w = w 1 w 2 w k, k 0 where each w i is a palindrome } A string w is a palindrome if w = w R That is, w reads the same forwards and backwards Example: 00100, 1001, and 0 are palindromes; 0011, 01 are not
59 Example: Context Free STAR Give a CFG for L = { w in {0,1}* : w = w 1 w 2 w k, k 0 where each w i is a palindrome } Write L = L 1 *, where L 1 = {w : w is a palindrome} Note: In fact, L = {0,1}*, but we will not use that.
60 Example: Context Free STAR Give a CFG for L = { w in {0,1}* : w = w 1 w 2 w k, k 0 where each w i is a palindrome } Write L = L 1 *, where L 1 = {w : w is a palindrome} G 1 = S 1 0S 1 0 1S ε
61 Example: Context Free STAR Give a CFG for L = { w in {0,1}* : w = w 1 w 2 w k, k 0 where each w i is a palindrome } Write L = L 1 *, where L 1 = {w : w is a palindrome} G 1 = S 1 0S 1 0 1S ε Let G = S SS 1 ε. Then, L(G 1 ) = L 1 L(G) = L 1 * = L.
62 Beyond regular A string and its reversal with C in middle: S 0S0 1S1 C Example: S * 0001C1000 More generally, to get strings of the form A k C B k use rules: S A S B C
63 Example: Σ = {0,1,#}, w R is reverse of w L = {w # x : w R is a substring of x} Useful to rewrite L as:
64 Example: Σ = {0,1,#}, w R is reverse of w L = {w # x : w R is a substring of x} = { w#x w R y : w,x,y {0,1}* } G := S CB C 0C0 1C1 #B B 0 B 1 B ε Remark: B *?
65 Example: Σ = {0,1,#}, w R is reverse of w L = {w # x : w R is a substring of x} = { w#x w R y : w,x,y {0,1}* } G := S CB C 0C0 1C1 #B Remark: C *? B 0 B 1 B ε Remark: B * {0,1}*
66 Example: Σ = {0,1,#}, w R is reverse of w L = {w # x : w R is a substring of x} = { w#x w R y : w,x,y {0,1}* } G := S CB C 0C0 1C1 #B Remark: C * w#{0,1}*w R B 0 B 1 B ε Remark: B * {0,1}* L(G) = L
67 CFG vs. automata CFG non-deterministic pushdown automata (PDA) A PDA is simply an NFA with a stack. x, y z q 1 q 2 This means: read x from the input; pop y off the stack; push z onto the stack Any of x,y,z may be ε.
68 Example: PDA for L = {0 n 1 n : n 0} 0, ε 0 1, 0 ε q 1 q 2 q 3 q 0 ε, ε $ ε, ε ε ε, $ ε The $ is a special symbol to recognize end of stack Idea: q 1 : read and push 0s onto stack until no more q 2 : read 1s and match with 0s popped from stack
69 Unlike the case for regular automata, non-deterministic PDA are strictly more powerful than deterministic PDA. Compilers must work with deterministic PDA, an important subclass of context-free languages
70 Non-context-free languages Intuition: If L involves regular expressions and/or nested matchings then probably context-free. If not, probably not. { 0 n 1 n : n 0 } CF : nested {w w : w Σ*} not CF: not nested {0 n 1 n 2 n : n 0} not CF: not nested
71 Non-context-free languages There is a pumping lemma for context-free languages. Similar to the one for regular, but simultaneously pump string in two parts: w = u v i x y i z
72 Context-free pumping lemma: L is CF language p 0 w L, w p u,v,x,y,z : w= uvxyz, vy > 0, vxy p i 0 : uv i xy i z L
73 Context-free pumping lemma: L is CF language Proof idea: Let G be CFG : L(G) = L p 0 w L, w p u,v,x,y,z : w= uvxyz, vy > 0, vxy p i 0 : uv i xy i z L If w L is very long, derivation repeats a variable V (like repeat states in regular P.L.) vxy = piece of w that V derives: V * vxy Because V repeated once, can repeat it again
74 Context-free pumping lemma: L is CF language p 0 A w L, w p u,v,x,y,z : w= uvxyz, vy > 0, vxy p i 0 : uv i xy i z L Useful to prove L NOT context-free. Use contrapositive: L context-free language A same as (not A) L not context-free
75 Context-free pumping lemma (contrapositive) p 0 not A w L, w p u,v,x,y,z : w = uvxyz, vy > 0, vxy p i 0 : uv i xy i z L L not context-free To prove L not context-free it is enough to prove not A Not A is the stuff in the box.
76 Context-free pumping lemma (contrapositive) p 0 w L, w p u,v,x,y,z : w = uvxyz, vy > 0, vxy p i 0 : uv i xy i z L Adversary picks p, You pick w L of length p, L not context-free Adversary decomposes w = uvxyz, vy > 0, vxy p You pick i 0 Finally, you win if uv i xy i z L
77 Theorem: L := {a n b n c n : n 0} is not context-free. Proof: p 0 Adversary moves p w L, w p You move w := a p b p c p u,v,x,y,z : w = uvxyz, Adversary moves u,v,x,y,z vy > 0, vxy p You move i := 2 i 0 : uv i xy i z L You must show uvvxyyz L: vy misses at least one symbol in = {a,b,c} since?
78 Theorem: L := {a n b n c n : n 0} is not context-free. Proof: p 0 Adversary moves p w L, w p You move w := a p b p c p u,v,x,y,z : w = uvxyz, Adversary moves u,v,x,y,z vy > 0, vxy p You move i := 2 i 0 : uv i xy i z L You must show uvvxyyz L: vy misses at least one symbol in = {a,b,c} since between as and cs there are p bs, and vy p so uvvxyyz????
79 Theorem: L := {a n b n c n : n 0} is not context-free. Proof: p 0 Adversary moves p w L, w p You move w := a p b p c p u,v,x,y,z : w = uvxyz, Adversary moves u,v,x,y,z vy > 0, vxy p You move i := 2 i 0 : uv i xy i z L You must show uvvxyyz L: vy misses at least one symbol in = {a,b,c} since between as and cs there are p bs, and vy p so uvvxyyz has too few of that symbol, so L DONE
80 Theorem: L := {a i b j c k : 0 i j k } is not context-free. Proof: p 0 Adversary moves p w L, w p You move w := a p b p c p u,v,x,y,z : w = uvxyz, Adversary moves u,v,x,y,z vy > 0, vxy p i 0 : uv i xy i z L So far, same as {a n b n c n : n 0}. But now we need a few cases. Our choice of i depends on u,v,x,y,z
81 Theorem: L := {a i b j c k : 0 i j k } is not context-free. Proof (cont.): You have w = a p b p c p, with w = uvxyz, vy > 0, vxy p. You must pick i 0 and show uv i xy i z L. If no a's in vy:?
82 Theorem: L := {a i b j c k : 0 i j k } is not context-free. Proof (cont.): You have w = a p b p c p, with w = uvxyz, vy > 0, vxy p. You must pick i 0 and show uv i xy i z L. If no a's in vy: uv 0 xy 0 z has fewer b's or c's than a's. If no c's in vy:?
83 Theorem: L := {a i b j c k : 0 i j k } is not context-free. Proof (cont.): You have w = a p b p c p, with w = uvxyz, vy > 0, vxy p. You must pick i 0 and show uv i xy i z L. If no a's in vy: uv 0 xy 0 z has fewer b's or c's than a's. If no c's in vy: uv 2 xy 2 z has more a's or b's than c's. If no b's in vy:?
84 Theorem: L := {a i b j c k : 0 i j k } is not context-free. Proof (cont.): You have w = a p b p c p, with w = uvxyz, vy > 0, vxy p. You must pick i 0 and show uv i xy i z L. If no a's in vy: uv 0 xy 0 z has fewer b's or c's than a's. If no c's in vy: uv 2 xy 2 z has more a's or b's than c's. If no b's in vy: You fall in a previous case, since vxy p DONE
85 Theorem: L := {s s : s {0,1}* } is not context-free. Proof: p 0 Adversary moves p w L, w p You move w := 0 p 1 p 0 p 1 p u,v,x,y,z : w = uvxyz, vy > 0, vxy p i 0 : uv i xy i z L Note: To prove L not regular we moved w = 0 p 1 0 p 1 That move does not work for context-free!
86 Theorem: L := {s s : s {0,1}* } is not context-free. Proof: p 0 Adversary moves p w L, w p You move w := 0 p 1 p 0 p 1 p u,v,x,y,z : w = uvxyz, Adversary moves u,v,x,y,z vy > 0, vxy p Three cases: i 0 : uv i xy i z L vxy in 1 st half of w:?
87 Theorem: L := {s s : s {0,1}* } is not context-free. Proof: p 0 Adversary moves p w L, w p You move w := 0 p 1 p 0 p 1 p u,v,x,y,z : w = uvxyz, Adversary moves u,v,x,y,z vy > 0, vxy p Three cases: i 0 : uv i xy i z L vxy in 1 st half of w: 2 nd half of uv 2 xy 2 z starts with 1, but uv 2 xy 2 z still starts with 0. vxy in 2 nd half of w:?
88 Theorem: L := {s s : s {0,1}* } is not context-free. Proof: p 0 Adversary moves p w L, w p You move w := 0 p 1 p 0 p 1 p u,v,x,y,z : w = uvxyz, Adversary moves u,v,x,y,z vy > 0, vxy p Three cases: i 0 : uv i xy i z L vxy in 1 st half of w: 2 nd half of uv 2 xy 2 z starts with 1, but uv 2 xy 2 z still starts with 0. vxy in 2 nd half of w: 1 st half of uv 2 xy 2 z ends with 0, but uv 2 xy 2 z still ends with 1. vxy touches midpoint:?
89 Theorem: L := {s s : s {0,1}* } is not context-free. Proof: p 0 Adversary moves p w L, w p You move w := 0 p 1 p 0 p 1 p u,v,x,y,z : w = uvxyz, Adversary moves u,v,x,y,z vy > 0, vxy p Three cases: i 0 : uv i xy i z L vxy in 1 st half of w: 2 nd half of uv 2 xy 2 z starts with 1, but uv 2 xy 2 z still starts with 0. vxy in 2 nd half of w: 1 st half of uv 2 xy 2 z ends with 0, but uv 2 xy 2 z still ends with 1. vxy touches midpoint: uv 0 xy 0 z = 0 p 1 i 0 j 1 p with either i < p or j < p. DONE
90 L := { w {a,b}* : w has same number of a and b} Grammar for L??
91 L := { w {a,b}* : w has same number of a and b} Grammar for L S ε SS asb bsa Not clear why this works. It requires a proof.
92 Proofs by induction Let P(n) be any claim To prove n 0, P(n) is true it suffices to prove Base case: P(0) is true Induction step: n : ( ( i < n, P(i) ) => P(n) ) Induction hypothesis You can replace 0 by any fixed value
93 Example: P(n) = i=0 n i = n(n+1)/2 Claim: n 0, P(n) Proof by induction: Base case: P(0) 0 = 0(1)/2 = 0 is true Induction step: n : ( ( i < n, P(i) ) => P(n) ) i=0 n i =??
94 Example: P(n) = i=0 n i = n(n+1)/2 Claim: n 0, P(n) Proof by induction: Base case: P(0) 0 = 0(1)/2 = 0 is true Induction step: n : ( ( i < n, P(i) ) => P(n) ) i=0 n i = i=0 n-1 i + n = (n-1)n/2 + n = n(n+1)/2
95 L := { w {a,b}* : w has same number of a and b} S ε SS asb bsa Claim: For any w {a,b}*, S * w if and only if w L Proof of only if : Suppose S * w. Must show w L. This fact is self-evident. We show a proof by induction nevertheless, as a warm-up for the other direction, which is not self-evident.
96 L := { w {a,b}* : w has same number of a and b} S ε SS asb bsa Claim: For any w {a,b}*, S * w if and only if w L Proof of only if : Suppose S * w. Must show w L. Let P(n) = any w Base case (n=1):?? {S,a,b}* such that S * w in has same number of a and b. n steps
97 L := { w {a,b}* : w has same number of a and b} S ε SS asb bsa Claim: For any w {a,b}*, S * w if and only if w L Proof of only if : Suppose S * w. Must show w L. Let P(n) = any w {S,a,b}* such that S * w in has same number of a and b. n steps Base case (n=1): ε, SS, asb, bsa have same number. Induction step: Suppose S * w' w By induction hypothesis,?? where S * w' in n-1 steps.
98 L := { w {a,b}* : w has same number of a and b} S ε SS asb bsa Claim: For any w {a,b}*, S * w if and only if w L Proof of only if : Suppose S * w. Must show w L. Let P(n) = any w {S,a,b}* such that S * w in has same number of a and b. n steps Base case (n=1): ε, SS, asb, bsa have same number. Induction step: Suppose S * w' w where S * w' in n-1 steps. By induction hypothesis, w' has same number of a, b. Since any rule adds same number of a and b, w has too.
99 L := { w {a,b}* : w has same number of a and b} S ε SS asb bsa Claim: For any w {a,b}*, S * w if and only if w L Proof of if : Suppose w L. Must show S * w Let P(n) = w {S,a,b}*, w = n, S * w. Base case: w = ε. Use rule??
100 L := { w {a,b}* : w has same number of a and b} S ε SS asb bsa Claim: For any w {a,b}*, S * w if and only if w L Proof of if : Suppose w L. Must show S * w Let P(n) = w {S,a,b}*, w = n, S * w. Base case: w = ε. Use rule S ε Induction step: Let w = n. This step is more complicated, and is the creative step of this proof.
101 S ε SS asb bsa Induction step: Let w = n. Define c i := number of a - number of b in w 1 w 2 w i c 0 = 0 c n =??
102 S ε SS asb bsa Induction step: Let w = n. Define c i := number of a - number of b in w 1 w 2 w i c 0 = 0 c n = 0 If 0 < i < n : ci = 0 then w =??
103 S ε SS asb bsa Induction step: Let w = n. Define c i := number of a - number of b in w 1 w 2 w i c 0 = 0 c n = 0 If 0 < i < n : ci = 0 then w = w' w'', where w', w'' L, and w' <n, w'' < n By induction hypothesis. S * w', S * w''. Hence S SS * w' S w' w'' = w
104 S ε SS asb bsa Induction step: Let w = n. Define c i := number of a - number of b in w 1 w 2 w i c 0 = 0 c n = 0 If 0 < i < n : ci > 0 then w =??
105 S ε SS asb bsa Induction step: Let w = n. Define c i := number of a - number of b in w 1 w 2 w i c 0 = 0 c n = 0 If 0 < i < n : ci > 0 then w = a w' b, where w' L and w' < n By induction hypothesis. S * w' Hence S asb * a w' b = w
106 S ε SS asb bsa Induction step: Let w = n. Define c i := number of a - number of b in w 1 w 2 w i c 0 = 0 c n = 0 If 0 < i < n : ci < 0 then w =??
107 S ε SS asb bsa Induction step: Let w = n. Define c i := number of a - number of b in w 1 w 2 w i c 0 = 0 c n = 0 If 0 < i < n : ci < 0 then w = b w' a, where w' L and w' < n By induction hypothesis. S * w' Hence S bsa * b w' a = w
108 S ε SS asb bsa Induction step: Let w = n. Define c i := number of a - number of b in w 1 w 2 w i c 0 = 0 c n = 0 These three cover all cases, because two consecutive c i differ by 1. So the c i cannot change sign without going through 0 DONE
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