Quiz2 45mins. Personal Number: Problem 1. (20pts) Here is an Table of Perl Regular Ex

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1 Long Quiz2 45mins Nme: Personl Numer: Prolem. (20pts) Here is n Tle of Perl Regulr Ex Chrcter Description. single chrcter \s whitespce chrcter (spce, t, newline) \S non-whitespce chrcter \d digit (0-9) \D non-digit \w word chrcter (-z, A-Z, 0-9, _) \W non-word chrcter [eiou] mtches single chrcter in the given set [^eiou] mtches single chrcter outside the given set (foo r z) mtches ny of the lterntives specified Quntifiers cn e used to specify how mny of the previous thing you wnt to mtch on, where "thing" mens either literl chrcter, one of the metchrcters listed ove, or group of chrcters or metchrcters in prentheses. Chrcter Description * zero or more of the previous thing + one or more of the previous thing? zero or one of the previous thing {3} mtches exctly 3 of the previous thing {3,6} mtches etween 3 nd 6 of the previous thing {3,} mtches 3 or more of the previous thing The ^ metchrcter mtches the eginning of the string nd the $ metsymol mtches the end of the string.

2 Plese write down R.E. (most of them will using the symol in the tle except the lst one ) to mtch following ptterns: ) three digits, ech followed y white spce chrcter (eg 3 4 5) /(\d\s){3}/ 2) mtches string in which every odd- numered letter is ( eg cdf ) /(.)+/ 3) string strts with one or more digits /^\d+/ 4) string tht ends with one or more digits /^d+/ 5) nothing in the string /^$/ Prolem 2. (20pts) Consider the following grmmr G: S " XY X " X X Y " Y Y ) Give leftmost derivtion of. S XY XY XY

3 Y Y Y 2) Build the derivtion tree for the derivtion in prt (). Y X Y X Y X Y 3) Wht is L(G)? L(G) = ( + )* ( + ) *

4 Prolem 3. (20pts) Given the following term: (λx.λz.x(λu.λy.y)(xzz))((λu.λv.u(uv))(λw.λz.wz))(λx.λy.yx) reduce this term s much s possile. Ans: Notice the nswer could e represented in different wys, so it is not the only symolic solution. Firstly, Identify first level lists (λx.λz.x(λu.λy.y)(xzz)) ((λu.λv.u(uv))(λw.λz.wz)) (λx.λy.yx) Secondly, reduce in ech list = (λx.λz.xzz) (λv. (λw.λz.wz) ((λw.λz.wz)v) ) (λx.λy.yx) = (λx.λz.xzz) (λv. (λw.λz.wz) (λz.vz) ) (λx.λy.yx) Thirdly, reduce the 2 nd list = (λx.λz.xzz) (λw.λz.wz) (λz. (λx.λy.yx)z) = (λx.λz.xzz) (λw.λz.wz) (λz.λy.yz ) = (λw.λz.wz) (λz.λy.yz ) (λz.λy.yz ) Prolem 4. (20pts) Below is C lnguge progrm #include<stdio.h> #include<conio.h> int fctoril(int); int fctoril (int i) { int f; if(i==) return ; else f = i* fctoril (i-);

5 } return f; void min() { int x; printf("enter ny numer to clculte fctoril :"); scnf("%d",&x); printf("\nfctoril : %d", fctoril (x)); } It is C progrm clculting fctoril numer, so plese do : ) Since Fortrn77 does not support recursive clcultion, plese write n equivlent Fortrn77 progrm. Here re the keyword you cn use : progrm, Integer, red, if, elseif, else, endif, stop, end, function, return, write. progrm min integer x, nswer red(*,*) x nswer = fctoril(x) write(*,*) nswer stop end INTEGER FUNCTION fctoril(n) INTEGER n, i fctoril = DO i = 2,n fctoril = fctoril * i END DO END FUNCTION 2) Write equivlent MIPS progrm nd show how stck chnges

6 slti $t0, $0, 2 # if i < 2 (i.e i == ) eq $t0, $zero, cont # if i >= 2 go to cont ddi $v0, $zero, # else mke the resturn vlue jr $r cont: # OPERATION : sve into stck min: ddi $sp, $sp, - 8 sw $r, 0($sp) sw $0, 4($sp) # mke spce in the stck # sve the return ddress # sve the rgument vlue # OPERATION 2: compute fct(n - ) ddi $0, $0, - jl fct # OPERATION 3: restore from stck lw $r, 0($sp) # get old return ddress from stck lw $0, 4($sp) # get old rgument vlue from stck ddi $sp, $sp, 8 # return stck pointer to originl vlue, # thus ersing ll vlues # OPERATION 4: finlly n * fct(n - ) mult $v0, $0 # multiply n * fi(n - ) mflo $v0 # gets the result of the multipliction from # the low register jr $r

7 3) Sketch, riefly nd informlly, how PDA for L = {x = y : x, y {0, } re unequl strings} would work. ANS: Recll in Lecture 7 Construct PDA ccepting Ide Exmple 3 { x" {, } # ( x) < # ( x) 2# ( x)}. To hve # ( x) < # ( x), we must hve which cncel two ' s. Let the first do the jo. nd Solution, / s, /, /, z / z, z / z, / ', / ', /, /, /, /, / ',' /, z / z, / ' z,', /, z / z, /, z / z z,', /, s /, z / z, /

8 Now we remove the prt to cncel 2 s, we hve : Solution, / s, /, /, z / z, z / z, / ', / ', /, /, /, /, / ',' /, z / z, / ' z,', /, z / z, /, z / z z,', /, s /, z / z, / Prolem 5. (20pts ) The mount of usle spce on DVD devote the sme mount to dt(2048ytes per sector). Clculte the size of Doule- Sided, 2,295,072 sectors DVD. Size = 2048ytes per sector * 2,295,072 sectors * Doule- Sided = * 2 = = 8.5 G