1 Structural induction, finite automata, regular expressions
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1 Discrete Structures Prelim 2 smple uestions s CS2800 Questions selected for spring Structurl induction, finite utomt, regulr expressions 1. We define set S of functions from Z to Z inductively s follows: Rule 1. For ny n Z, the trnsltion (or offset) function t n : x x + n is in S. Rule 2. For ny k 0 Z, the scling function r k : x kx is in S. Rule 3. If f nd g re elements of S, then the composition f g S. Rule 4. If f S nd f hs right inverse g, then g is lso in S. In other words, S consists of functions tht trnslte nd scle integers, nd compositions nd right inverses thereof. Note: This semester, we mde bigger distinction between the elements of n inductively defined set nd the mening of n inductively defined set. We probbly would hve phrsed this uestion s follows: Let S be given by s S ::= t n r k s 1 s 2 rinv s nd inductively, let the function defined by s (written F s : Z) be given by the rules F tn (x) ::= x + n, F rk (x) ::= ks, F s1 s 2 (x) ::= F s1 F s2 nd let F rinv s ::= g where g is right inverse of F s. () [1 point] Show tht the function f : x 3x + 17 is in S. By rule 1, the function t 17 : x x + 17 is in S, nd by rule 2, r 3 : x 3x is in S. By rule 3, therefore, t 17 r 3 : x 3x + 17 is in S. (b) Use structurl induction to prove tht for ll f S, f is injective. You my use without proof the fct tht the composition of injective functions is injective. We must show tht ll functions formed with ech of the rules re injective. Let P (s) be the sttement s is injective. P (t k ) holds, becuse t k hs two sided inverse t k, nd is therefore injective. P (r k ) holds, becuse we reuired tht k 0. Therefore, if kx 1 = kx 2, we cn cncel k to find x 1 = x 2. P (f g) holds, ssuming P (f) nd P (g), becuse the composition of injections is n injection. If g is the right inverse of f, then P (g) holds, becuse g hs left-inverse (nmely f) nd is therefore injective. (c) Give surjection φ from S to Z (proof of surjectivity not necessry). Remember tht this surjection must mp function to n integer, nd for every integer there must be function tht mps to it. 1
2 Let φ(s) ::= s(0). This is surjection, becuse t n (0) = 0 + n = n, so for ny n there exists s S (nmely t n ) with φ(s) = n. 2. Seeking to void the limittions of regulr expressions, student designs set S of super expressions. Just s with regulr expressions, ech super expression s S mtches set of strings, denoted by L(s). The set S nd the function L re defined inductively s follows: A chrcter clss is denoted [ n ] (where the i re single chrcters, i.e. elements of Σ). It mtches single chrcter which is ny of the i. For exmple, [012] mtches 0, 1, nd 2, but does not mtch 12. Formlly, [ n ] S nd L([ n ]) = { 1, 2,..., n }. If s S, positive repetition of s is denoted s+. It mtches one or more strings, ech of which is mtched by s. For exmple, [45]+ mtches 4, 54, nd 555 but does not mtch. Formlly, s+ S nd L(s+) = {x 1 x 2 x n n 1 nd x i L(s)}. If s 1 S nd s 2 S then the difference of s 1 nd s 2 is denoted by s 1 \ s 2. It mtches ll strings tht s 1 mtches but s 2 does not. For exmple, ([123]+)\[123] mtches 1322 nd 11 but not 1, 2, or 3. Formlly, s 1 \ s 2 S nd L(s 1 \ s 2 ) = L(s 1 ) \ L(s 2 ). If s 1 S nd s 2 S then the conctention of s 1 nd s 2 is denoted by s 1 s 2. It mtches ny string tht is formed by conctenting string mtched by s 1 with string mtched by s 2. For exmple [012][34] mtches 03 nd 24 but not 0 or 32. Formlly, s 1 s 2 S nd L(s 1 s 2 ) = {xy x L(s 1 ) nd y L(s 2 )}. () Let Σ = {0, 1, 2}. Let s be the super expression ([01][12])+ \ ([012][012]). Write (norml) regulr expression r such tht L(r) = L(s). To explin the nswer, I ll refer to string mtching [01][12] s unit. We must mtch strings contining two or more units. A unit is mtched by the (norml) regulr expression (0 + 1)(1 + 2). Therefore, two or more units is mtched by (0 + 1)(1 + 2)(0 + 1)(1 + 2)((0 + 1)(1 + 2)). (b) Prove by structurl induction tht every super expression s S hs n euivlent regulr expression. You my use without proof the fct tht the union, intersection, nd complement of regulr lnguges re regulr. Let P (s) be the sttement there exists r RE such tht L(r) = L(s). We must show P ([ 1... n ]), P (s + ) (ssuming P (s)), P (s 1 \ s 2 ) (ssuming P (s 1 ) nd P (s 2 )), nd P (s 1 s 2 ) (gin, ssuming P (s 1 ) d P (s 2 )) P ([ 1... n ]) holds, becuse L([ 1... n ]) = L( n ). P (s 1 \ s 2 ) holds; we ssume inductively tht s 1 nd s 2 re regulr. We know tht L(s 1 \ s 2 ) = L(s 1 ) \ L(s 2 ) = L(s 1 ) L(s 2 ), which is regulr, becuse the intersection nd complement of regulr lnguges is regulr. Finlly, we must show tht L(s 1 s 2 ) is regulr. We inductively ssume tht s 1 nd s 2 re regulr. Therefore, there exist r 1 nd r 2 with L(r 1 ) = L(s 1 ) nd L(r 2 ) = L(s 2 ). By definition, this shows tht L(r 1 r 2 ) = L(s 1 s 2 ), so s 1 s 2 is regulr. 3. Given DFAs M 1 = (Q 1, Σ, δ 1, 01, F 1 ) nd M 2 = (Q 2, Σ, δ 2, 02, F 2 ), we cn construct mchine M 12 with L(M 12 ) = L(M 1 ) L(M 2 ) s follows: Let Q = Q 1 Q 2 = the set of ll ordered pirs ( 1, 2 ), where 1 Q 1 nd 2 Q 2. Let 0 Q = ( 01, 02 ). 2
3 Let F = F 1 F 2 = {( 1, 2 ) 1 F 1 nd 2 F 2 }. Let δ 12 (( 1, 2 ), ) = (δ 1 ( 1, ), δ 2 ( 2, )). Let M 12 = (Q, Σ, δ 12, 0, F ). Use structurl induction to prove tht for ll x Σ, δ 12 (( 1, 2 ), x) = ( δ1 ( 1, x), δ ) 2 ( 2, x). The min chllenge here is wding through the nottionl jungle to understnd wht the problem ctully sys. Once you ve done this, the proof is short nd strightforwrd. Here it is in ll its glory: We will prove the result by structurl induction on x, s suggested. Both the set of strings Σ nd the extended trnsition function δ re defined recursively (see the definitions t the end). The bse cse for x is the empty string. We cn simply red off the corresponding line in the definition of δ, which tells us tht δ 1 ( 1, ) = 1, δ 2 ( 2, ) = 2, nd δ 12 (( 1, 2 ), ) = ( 1, 2 ). Hence δ 12 (( 1, 2 ), ) = ( 1, 2 ) = ( δ1 ( 1, ), δ ) 2 ( 2, ), so the sttement is true in the bse cse. Now ssume the sttement is true for some string x, nd consider the next lrger string x. Agin reding off the pproprite line in the definition of δ, we know tht δ 1 ( 1, x) = δ 1 ( δ 1 ( 1, x), ), nd δ 2 ( 2, x) = δ 2 ( δ 2 ( 2, x), ) Wht is δ 12 (( 1, 2 ), x)? Well, we lso hve δ 12 (( 1, 2 ), x) = δ 12 ( δ 12 (( 1, 2 ), x), ) (definition of δ 12 ) = δ 12 (( δ1 ( 1, x), δ ) ) 2 ( 2, x), (inductive hypothesis) ( ) )) = δ 1 ( δ1 ( 1, x),, δ 2 ( δ2 ( 2, x), (definition of δ 12 ) = ( δ1 ( 1, x), δ ) 2 ( 2, x) (definition of δ 1, δ 2 ) This proves the sttement for ll strings x Σ by (structurl) induction. 4. () Drw finite utomton (DFA, NFA or -NFA) with lphbet {0, 1} to recognize the lnguge {x {0, 1} x contins the substring 010} An NFA is probbly the esiest to construct. 0, 1 0, 1 0 strt (b) Drw finite utomton (DFA, NFA or -NFA) with lphbet {, b} to recognize the sme lnguge s the regulr expression (b b). 3
4 We cn blindly pply the regex -NFA construction from the proof of Kleene s Theorem to get something like this: b strt 0 1 f b Of course this is bit messy, more compressed version might look like this: 11 b strt 1 b Prove tht L = {0 n 10 n n N} is not regulr. Suppose tht this lnguge is ccepted by some deterministic finite utomton with N sttes. Consider the string x = 0 n 10 n. Since x is in the lnguge nd x N, by the Pumping Lemm, there exist strings u, v, nd w such tht x = uvw, v 1, uv N, nd M ccepts uv i w for ll i > 0. Since uv N, it must be the cse tht uv is string of 0 s, nd tht w contins the 1 in 0 N 10 N. Thus, if i > 1, uv i w hs more thn N 0s to the left of the 1 nd only N 0s to the right of the 1, nd thus is not in the lnguge. This contrdicts the ssumption tht the lnguge is ccepted by M (since M ccepts string not in the lnguge). 6. Let r be regulr expression. Show tht there exists regulr expression r with L(r ) = L(r) (the complement of L(r)). If your proof involves the construction of regulr expression or utomton, you must prove tht the lnguge of the regulr expression/utomton is wht you clim it is (using the 4
5 definitions). First, note tht L(r) is lnguge, which mens tht it s set of strings. So its complement consists of ll the strings in L(r) tht re not in Σ. (Mny of you took complement to men the result of switching the 0s nd 1s in the strings in L(r). This is not wht complement mens in this context; moreover, this pproch fils if Σ {0, 1}. Unfortuntely, you typiclly got 0, 1, or 2 out of 6 if you did this, depending on wht else you did.) By Kleene s theorem, there is DFA M = (Q, Σ, δ, 0, F ) with L(M) = L(R). We cn form new utomton M by flipping the ccept nd reject sttes of M: M = (Q, Σ, δ, 0, Q \ F ). I clim L(M ) = L(M). Indeed, x L(M ) ˆδ( 0, x) Q \ F ˆδ( 0, x) / F x / L(M) (You typiclly lost 1 point if you didn t give proof.) Applying Kleene s theorem gin tells us tht since there is DFA with L(M ) = L(r) there must be regulr expression r with L(r ) = L(r). 7. Build deterministic finite utomton tht recognizes the set of strings of 0 s nd 1 s, tht only contin single 0 (nd ny number of 1 s). Describe the set of strings tht led to ech stte. 0 strt ,1 The strings leding to i contin i 0 s. 8. Given string x, we cn define the chrcter doubling of x to be x with every chrcter doubled: for exmple cd(bc) = bbcc. Formlly, cd() =, nd cd(x) = cd(x). We cn then define the chrcter doubling of lnguge L to be the set of ll strings formed by doubling the chrcters of strings in L; formlly cd(l) = {cd(x) x L}. Given DFA M = (Q, Σ, δ, 0, F ), we cn construct new DFA M cd tht recognizes cd(l(m)) by dding new stte to the middle of every trnsition from on chrcter : 1 2 becomes () Formlly describe the components (Q cd, Σ cd, δ cd, 0cd, F cd ) of M cd in terms of the components of M. Be sure to describe δ cd on ll inputs (you my need to dd one or more dditionl sttes). Q cd = Q { Q, A} {X} δ cd : (, ) ; δ cd : (, ) δ(, ); δ cd : (, b) X if b, nd δ cd : (X, ) X. The remining components re unchnged: Σ cd = Σ, 0cd = 0, nd F cd = F. (b) Use structurl induction on x to prove tht for ll x, δ( 0, x) = δ cd ( 0cd, cd(x)). 5
6 Let P (x) be the sttement tht δ( 0, x) = δ cd ( 0, cd(x)). I will prove x, P (x) by structurl induction. To show P (), note tht δ( 0, ) = 0. Moreover, cd() =, so δ cd ( 0, cd()) = δ cd ( 0, ) = 0 = δ( 0, ), s reuired. To show P (x), we ssume the inductive hypothesis P (x). we compute: δ cd ( 0, cd(x)) = δ cd ( 0, cd(x)) by definition of cd = δ cd (δ cd ( δ cd ( 0, cd(x)), ), ) by definition of δ cd = δ cd (δ cd ( δ( 0, x), ), ) by definition of δ cd = δ cd (, ) ( δ(0,x)) by definition of δ cd = δ( δ( 0, x), ) by definition of δ cd = δ( 0, x) by definition of δ 9. We cn lso define the string doubling of x to be xx. For exmple, sd(bc) = bcbc. Show tht the set of regulr lnguges is not closed under string doubling. In other words, give regulr lnguge L nd prove tht sd(l) = {sd(x) x L} is not regulr. You cn use ny theorem proved in clss to help prove this result. Let L = 0 1. Clerly L is regulr. Moreover, sd(l) = {0 n 10 n 1 n N}. This lnguge is not regulr. To see this, ssume for the ske of contrdiction tht it is. Then there exists some nturl number m s in the pumping lemm. Let x = 0 m 10 m 1. Clerly x sd(l), nd x m, so we cn split x into u, v, nd w, s in the pumping lemm. We know tht uv n, so v cn only contin 0 s. Then x = uv 2 w contins more 0 s before the first 1 thn fter, nd thus x / sd(l). But the pumping lemm sys tht x sd(l); this is contrdiction, nd thus sd(l) is not regulr. 10. Hppy Ct hs been shown the following proof, nd hs promptly turned into Grumpy Ct. Briefly but clerly identify the error which hs induced grumpiness. To prove: The lnguge of the regulr expression 0 1 is, in fct, not DFA-recognizble. Proof. Let L be the lnguge of 0 1. Assume there is some DFA M with n sttes tht recognizes L. Let x = 0 n Clerly, x L nd x n. Therefore ccording to the Pumping Lemm, we cn split x into three prts u, v nd w, such tht uv n, v 1, nd uv i w L for ll nturl numbers i. Let v = n. Since uv n, it must be the cse tht u =, nd v = 0 n 1 1. Then uv 2 w = 0 n 1 10 n 1 11, which is clerly not in L. This contrdicts our ssumption tht there is DFA which recognizes the lnguge. The pumping lemm sys there exists some v, but we hve chosen specific v. It my be tht the pumping lemm gives some other v (such s 0). 2 Probbility problems 1. () Give the definition of vrince in terms of expecttion. 6
7 I would lso ccept Vr(X) = E(X 2 ) E 2 (X). Vr(X) = E((X E(X)) 2 ) (b) Let X nd Y be rndom vribles with E(X) = E(Y ) = 0. Prove tht Vr(X+Y ) = Vr(X)+Vr(Y ). Mke (nd clerly stte) dditionl ssumptions if necessry. We must reuire tht X nd Y re independent. Therefore E(XY ) = 0. Then we hve Vr(X + Y ) = E ( (X + Y E(X + Y )) 2) by definition = E ( (X + Y ) 2) since E(X + Y ) = E(X) + E(Y ) = 0 = E(X 2 + 2XY + Y 2 ) rithmetic = E(X 2 ) + 2E(XY ) + E(Y 2 ) expecttion of the sum is sum of expecttions = E(X 2 ) + E(Y 2 ) since E(XY ) = 0 (independence) = Vr(X) + Vr(Y ) by definition nd fct tht E(X) = E(Y ) = 0 2. () The verge humn height is 5 feet nd 4 inches, nd the vrince is 2 sured inches. How lrge smple must I tke so tht my estimte of the verge will (with 90% probbility) be correct to within hlf inch? Let H denote the height rndom vrible, nd let E be the estimted verge. The lw of lrge numbers sttes tht P( E E(H) ) σ 2 (H)/n 2. Plugging in hlf inch for epsilon nd solving for n, we see tht if n 2/( ) tht the probbility of being incorrect is no lrger thn 0.1. (b) A certin high school is divided into two tems: 35% of the students re beliebers, nd the remining 65% re directioners. 90% of the songs on belieber s plylist will be Justin Bieber songs, while the other 10% will be by One Direction. Directioners re bit more brod-minded: 80% of their songs will be One Direction songs, while the remining 20% will be Justin Bieber songs. A student is selected t rndom, nd rndom song is selected from their plylist. It turns out to be Bby by Justin Bieber. Wht is the probbility tht the student ws directioner? Let D be the event tht the student is directioner, nd let B be the event tht the student is belieber. Let S D be the event tht the selected song ws One Direction song, nd S B be the event tht the selected song ws Justin Bieber song. We re given P (B) = 35% P (D) = 65% P (S D D) = 80% P (S D B) = 10% P (S B D) = 20% P (S B B) = 90% 7
8 We wish to find P (D S B ). The correct nswer is given by Byes rule: P (D S B ) = P (S B D)P (D) P (S B ) = P (S B D)P (D) P (S B D)P (D) + P (S B B)P (B) = Suppose tht coin hs probbility.6 of lnding heds. You flip it 100 times. The coin flips re ll mutully independent. () Wht is the expected number of heds? Let X i be the outcome of the ith coin toss; X i = 1 if the ith coin toss lnds heds nd 0 otherwise. The totl number of heds is Y = X X 100. We re interested in E(Y ). By linerity, E(Y ) = E(X 1 ) + + E(X 100 ) = 100(.6) = 60. (b) Wht upper bound does Mrkov s Theorem give for the probbility tht the number of heds is t lest 80? Mrkov s Theorem sys tht Pr(Y 80) E(Y )/80 = 60/80 = 3/4. (c) Wht is the vrince of the number of heds for single toss? Clculte the vrince using either of the euivlent definitions of vrince. Let X i be n indictor vrible tht is 1 is the ith toss is heds, nd 0 otherwise. We re given P (X i = 1) = 0.6. The expecttion of X i is = 0.6. Note tht Xi 2 = X i, so E(Xi 2) = 0.6 s well. Therefore V r(x i) = E(X i ) 2 E(X i ) 2 = = (d) Wht is the vrince of the number of heds for 100 tosses? You my use the fct tht if X 1,..., X n re mutully independent, then Vr( X i ) = Vr(X i ); you don t need to prove this. V r(y ) = V r(x 1 )+ +V r(x 100 ). Since V r(x i ) =.24 for i = 1,..., 100, V r(y ) = = 24. (e) Wht upper bound does Chebyshev s Theorem give for the probbility tht the number of heds is either less thn 40 or greter thn 80? By Chebyshev s Theorem. Pr( Y E(Y ) 20) Vr(Y )/400. Since E(Y ) = 60, Pr(Y 80 Y 40) 24/400 =.06. 8
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