CS 241 Week 4 Tutorial Solutions

Transcription

1 CS 4 Week 4 Tutoril Solutions Writing n Assemler, Prt & Regulr Lnguges Prt Winter 8 Assemling instrutions utomtilly. slt $d, $s, $t. Solution: $d, $s, nd $t ll fit in -it signed integers sine they re 5-it unsigned ints, so we n keep them unhnged. So we n do the following: int s, t, d //ssume these re initilized ppropritely //Begin with the opode for slt, whih is int slt = xa // slt = slt (s << ) // ss sss slt = slt (t << 6) // ss ssst tttt slt = slt (d << ) // ss ssst tttt dddd d We ould lso mke sure tht register integers only ontin 5-it integers y first tking s & xf nd similr for t nd d, ut this isn t neessry sine our snner lredy heks this.. eq $s, $t, $i, where i is n immedite vlue (INT or HEXINT token). Solution: Proeeding similr to the ove: int s, t, i //ssume these re initilized ppropritely //Begin with the opode for eq, whih is //However, we need to mke sure it s on the left 6 //its rther thn the right 6 its sine eq is immedite-formt! eq = x // //Now dd s, t in the sme wy s efore eq = eq (s << ) // ss sss eq = eq (t << 6) // ss ssst tttt //Finlly, dd i. We need to mke sure to ler its most signifint its first! eq = eq (i & xffff) // ss ssst tttt iiii iiii iiii iiii Note tht we need to mke sure to zero out the high its of i efore dding it to the instrution! In wht ses n we get undesirle ehviour if we forget? How ould we design our ode to mximize ode reuse etween vrious instrutions? Solution: hve generi funtions for ny register or immedite-formt instrution whih simply ept s, t, d, f or s, t, i, o respetively s rguments.

2 Regulr Lnguge Solutions. {, }{}{, }{, }{}{, }. {, } {}{}{}{}{}{}{, } DFA Exerise Solutions. We wnt to find DFA for the lnguge of strings over,, with n n nd n even numer of s. Our DFA should ept strings suh s, nd while rejeting strings like, nd the empty string. Generlly, good wy to pproh DFA prolems is to think out wht sttes you will need to differentite strings tht re not in the lnguge from strings tht re in the lnguge. One you figure out the set of sttes, filling in the trnsitions is often strightforwrd. For exmple, we might hve stte lelled one, even, whih we enter whenever we hve onsumed n nd n even numer of s. It should e n epting stte, sine these re preisely the strings we wnt to ept. To differentite etween these strings nd ones not in the lnguge, we might need to rete sttes representing other possiilities. Nmely, one, odd, no, even, no, odd, > one, odd, > one, even. None of these sttes should e epting. Note tht we do not need to expliitly rete the sttes > one, odd nd > one, even, euse one we red more thn one there is no wy tht the string n possily e epted: we re essentilly in n error stte. strt no, even no, odd one, even one, odd. Following the dvie from the previous prolem, we n e in four possile onfigurtions: evenness nd oddness of nd. Reding new or simply toggles the evenness of tht letter nd hs no effet on the other letter. As result we get:

3 strt even, even even, odd odd, even odd, odd. The DFA is identil to tht of the previous prolem exept tht every stte trnsitions to itself on. 4. In this prolem, we wnt DFA tht reognizes inry strings ending in. For exmple,,, nd should e epted, nd strings suh s nd nything else not ending in should e rejeted. We proeed one gin y figuring out wht sttes re neessry to reognize this lnguge. Sometimes when you re not sure wht sttes you will need, it is helpful to pik simple exmple string from the lnguge nd figure out wht sttes you will need to reognize just tht prtiulr string. For exmple, it is ler tht the sttes nd trnsitions shown elow re required to reognize : strt ε Think out wht these sttes orrespond to. If you re in the stte tht mens the string you hve red ends in, nd you must red to otin string in the lnguge. If you re in the stte, the string you hve red ends in nd you only hve to red to get string in the lnguge. The sttes trk how muh of the terminting suffix we hve seen, nd how muh we still need to see to omplete the suffix. If the entire suffix is red, we will e in the epting stte. Otherwise, we will e in one of the intermedite sttes leding up to it. To omplete the DFA, we just need to fill in the missing trnsitions one y one. When we re in the initil stte, we still need to see the hrters in order to reh the epting stte. If we next red, we hven t seen the first hrter of tht suffix, so we loop on. When we re in the stte, our string ends in nd we still wnt to see. If we red, our string ends in nd we still wnt to see. This mens we must loop on. When we re in the stte, our string ends in nd we still wnt to see. If we red, our string ends in, whih mens we must see the entire suffix next to reh the epting stte. So we return to the initil stte on. When we re in the stte, out string ends in nd we still wnt to see. If we red, our string ends in, whih mens we need to see to reh the epting stte. So we should go to the stte on.

4 When we re in the epting stte, our string ends in the desired suffix. If we red, it ends in, whih mens we must see next to reh the epting stte, so we trnsition to stte. If we red, it ends in, so we trnsition to stte. We otin this DFA, whih epts the desired lnguge: strt ε 5. The DFA is identil to tht of the previous question, exept tht the finl stte hs trnsition to itself on or. 6. We wnt to find DFA tht reognizes strings over the lphet,,, whih re integers with digit sum of, nd my hve leding zeros. For exmple, should e in the lnguge sine + + =, nd should e in the lnguge sine =, ut nd should not e in the lnguge sine their digit sums re nd 6, respetively. One gin, the est wy to think out this prolem is to think out wht sttes we ould use to distinguish strings in the lnguge from those not in the lnguge. Sine we wnt to ept strings with digit sum of, it mke sense to hve our sttes trk the digit sum. We n hve n epting stte lled, whih represents tht the sum of the digits we hve red so fr is. Then we n hve non-epting sttes lled, nd orresponding to eh of those digit sums. We lso wnt to rejet ll strings with digit sum lrger thn ; we ould dd nother non-epting stte lled something like > to represent this. The trnsition funtion n then e defined very esily. If we re in stte x nd the symol i is next in the input, then the digit sum of the integer we hve seen so fr is x nd the new digit sum fter reding i will e x+i. Thus we should hve trnsitions from stte x on symol i to stte x+i if x+i is less thn or equl to, or to stte > if x+i is greter thn. For exmple, we would hve trnsition from stte on the symol to stte, sine + =, nd trnsition from stte on symol to stte > sine + >. Filling in ll the trnsitions in this wy, we get the following DFA:,,,,,, strt > Notie tht the > stte is not only non-epting, ut it loops k to itself on every symol. Insted of drwing this stte, we ould use the impliit error stte onvention disussed in lss. If trnsition is not drwn on the DFA digrm, we n ssume it impliitly goes to non-epting error stte tht loops k to itself on every symol, nd thus if we enounter n undefined trnsition when trying to reognize string, the input will e rejeted. Using this onvention lets us remove the > stte from the digrm nd mke it simpler (though we would still hve to list the error stte in forml desription of this DFA). 4

5 strt 7. We wnt to find DFA tht reognizes strings over the lphet,, whih end in nd hve n even numer of s. Sine itself hs one, whih is n odd numer of s, there must e t lest one efore. So the smllest string in our lnguge is. We n uild DFA to ept this string: strt even s odd s end end end Now, like for prolem, we simply need to fill in the missing trnsitions: From the initil stte if we see or it doesn t hnge the numer of s in the string so we loop. From the odd s stte if we see n we go k to the even s stte, if we see the numer of s don t hnge so we loop. From the end stte if we see we loop euse we still ould e reding the first letter of, if we see we nnot e in so we return to the odd s stte. From the end stte if we see we return to the even s stte, if we see n we return to the odd s stte. From the end stte if we see or we return to the even s stte, nd if we see n we return to the odd s stte., strt even s odd s end end end,.. NFA Exerise Solutions. Given DFA M = {Σ, Q, q, A, δ}, define the NFA N = {Sigm, Q, q, A, δ } where δ (q, ) = {δ(q, )}. 5

6 . () The DFA for this prolem ws omplited, ut notie how muh simpler the NFA is! We strt with the sme si struture s the DFA: list of sttes representing the sustring to e reognized. However, fter tht point in the DFA we needed to figure out how to reover from reding the wrong thing nywhere in the wrong ple, wheres in the NFA we n simply wit s long s possile on the initil stte y looping on ny input., strt ε () We n pproh this prolem in the sme si wy s the previous question, ut with two rnhing endings:, strt ε () It turns out tht for prefixes NFAs don t help us over DFAs, sine there s no wy to reover if we red the wrong thing. The solution here is oth n NFA nd DFA., strt ε 6