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TURING MA 4.4. TURING MACHINES Alan Turing introduced a new mathematic Cty, al Mode] . Cal] Machine during the year of 1936. It is mostly used to define lat e "Rua Bes My mee a integer ns. compute integer functio’ h The basic model has a finite control, an input tape that is divig f . ed; cells, and a tape head that scans one cell of the tape at a time, tin, Mareperar re Tape head Finite control Fig.4.13 Turing Machine Model Vv Each cell can hold one of a finite number of tape symbols, Y All other tape symbols extending infinitely to the left and rat hold a special symbol called blank. v Initially the tape head is pointing the leftmost cell that hols the input. ; In one move, the turin I ‘ ne & machine depending upon the symbol sca! by the tape ‘head and the state of finite control. * Itchanges its State. wg wht ‘i ing * It Prints a symbol! on the tape cell scanned, replaci"s Was written there, * Moves its head One cell, to its left or right. y Scanned by CamScanner properties of CFL and Turing Machines 44 Definition ~— A Turing Machine M is a 7-tuple M=(Q, 5, 5; do: B, F) where Q- finite set of states 5 — finite set of input symbols r- finite set of tape symbols 8- transition function mapping the states of finite and tape symbols to states, tape symbols and aioe the head. ement of ie, QX PF Q*P*{L, BY qo = Qis the initial state F cQis the set of final states BE j- is the blank symbol. 4.4.2 Instantancous Description for Turing Machines The ID of a turing machine is defined in terms of the entire input string and the current state. Definition : An ID ofa Turing Machine is a string «By, where B is the present state of M, the entire input string is split as ay, the first symbol of y is the current symbol ‘a’ under R/w head and y has all the subsequent symbols of the input string and the string a is the substring of the input string formed by all the symbols to the left of a. £g. Construct the ID. . R/w head Fig 4.14 Instantaneous Description Scanned by CamScanner a—_ Suppose 8(q, Xi) = (P; Ys R)], and the input string to be processed is X1 Xq-- + Xq ae ane Xp Qi Xn XXQ Xi) YP Kigy ee Xy 4.4.4 Language ofa TM The language accepted by M denoted L(M) is the set of those words in 5* that cause M to enter a final state. L(M) = {w | w in £* and qow a pct, for some p in F and 4), Oy £ FY [1]Design a TM to accept the language L = {0" I" | n 21} [Apr/May-2005] Solution : Given a finite sequence of 0’s and 1’s on its tape. The turing machine is designed using the following way. (i) M replaces the leftmost 0: by x, moves right to the leftmost 1, replacing it by y. : (ii) Then M moves left to find the rightmost x, and moves one cell right to the leftmost 0 and repeats the cycle. (iii) While searching for a 1, if a blank is encountered, then M halts without accepting. (iv) After changing a1 toa y, if M finds no more 0’s, then M checks that no more 1’s remain, accepting the string else not. Assume the set of states Q = {dg 45 Gy I> Ig} r= 0,1} fF =(0. 1, x, y, B} F= {qq} let oq be the initial state and at state Qo» it replaces the leftmost 0 by x, and changes it to q,. Atq,, M searches right for 1’s, skipping over 0’s and y’s. Scanned by CamScanner . . a If M finds a 1, it changes it to y, entering state qp. From gy itseagee ts : Catch : 8 le ty, an x and moves right to change the state to Qo: At qo; if y is encountered, it goes to state q; and check s remain. If the y’s are followed by a B, state 4, is entered and th ta Oy, And for all others, M rejects. ” "ep 0 | x y B qo (q,.x.R) a ~ (q3, y, R) q) (q,, 9, R) (q,, y, L) = (q,. ys R) ~ —& 1G OL) = @xR @yl) - q _ _ _— 3 (43.¥5R) q,,B,R) q4 TT ~ _ - Eg: (i) q0011 xq 011 Fx0q,11 Fxq,0y1 Fq,x0yl Xqo0yl Fxxq,y1 Pay Pxxqyy Pxq,xyy F TADYY PXXY Gy Pxxyyq, i FxxyyB qu. Accepted. (ii) q,0 4 1] xq, 11 Cq,xyl Pr xqoyl PF xyq,! rejected, Scanned by CamScanner : Theory of ~ i ntering state q Fron LC i : = : male qo. Fr n it se, ty IfM finds a 1, it changes it to y, e 2 5 Goo it Sergi ‘ an x and moves right to change the state to qo: iy At qo. ify is encountered, it goes to st 0° remain. If the y’s are followed by And for all others, M rejects. ate 43 and Check . 8 that ho 1 4B, state 44'S entered ang then ag, | Ny “Dtey, 0 | X y B Mo | (4,,.x,R) ~ - (43, y, R) _ 4] %R) yb) 7 @pyR % | (q,, 0, L) ~ q3 C - (dp, x, R) (4), y, L) - (43, y, R) (44, B, R) (i) q,0011 Fxq,011 Fx0q,11 xq, 0y1 Fa,x0y1 xq,0yl Pxxqyl Pxxyq) 1 Pxxqnyy Pxqgxyy XXUoYY Pxxy qay Pxxyyqs PxxyyB q4- Accepted, _ Gi) got F xq. 11 Paqyxyl p XY F xyqy 1 Tejected, Scanned by CamScanner yly Fig. 4.15 Transition diagram for 01", M=(QE, 1,8, 4B, (ay) where Q= {Mos Gy> Ip» 432 a} x = {0,1} r=(0, 1,x, y, B} Qy = Initial state q4 = Final state 8 is given in the table. L: The set of strings with an 2) Design a Turing Machine for the language equal number of 0’s and I's. Solution : iin either 0 oF lL. Assume that the input string may start with 101, 0110 001.9" “ meet ef ant ane , jpothor the input syne (i) Change all 0’s to x’s and all V's pole . blank gyunbol . myveinany ~ ee if it finds the aes m ue wong ic at state do: At Me" vnigh (> ety eT | symbol as ; Scanned by CamScanner -ayqer out Us UoAIB St. Q areys jeury =°b uy = Ob 9yeys fqn x1 oa4 {1‘0}=3 (5b th hb lbp lb by = fe) EXCIIy ({5b} “a pg 4 ‘3 0) =NWL orb Bud Scanned by CamScanner et, ty, Nags e Ohedy mtaedtinte tat accepts the language L = 4 ihe, , hy | | Lahey > SC ownstiietion ts similar to the design of 0&2, YW We have to replace each a by x and b by y and ¢ by z repeat lh: © Ally the Turing Machine is at state Qo: At qos if it finds g ae Steph, & dy v's and moves right with state q, At q, if it finds D's, reply J . NT dy y's and moves right with state q,. At State q,, if it finds o', rp ’ > F Tl it by z, and enters state q; by moving left. At 3, if it finds the lefing X. by skipping z, b, y, a, then it goes to state qo. Repeat the Proves At qo, it finds y. a b c x y Zz B B GpXR = - — @yyR) = - % /@paR) @,y,R — (@pyR) = - a, } - (46 R) (G21) = — — (q,zR) q; | As a,L) (q3, b, L) — (2R) @yy,L) = - % | - - 7 ~ Gy YR) (qq. 2 R) (Gy BA Eg: (i) qgaabbeo EF xqyabbec xaq, bce E xayq,beo Xaybq,cc EF xayq,bzo xaq,ybze Xqjaybze Q3Xaybze f- Xqgaybze xq ybze Fr xxyq,bze f- XXYYqoze 0 A Xxyyzqje KR Xxyyq,2Z fF *XYqny2z A XXq3yyzz I Xq3xyyzz qoyye2 XNyquy2z xxyyqyzz YqZ *xyyZzq.Bo xxyyzz2Bq, Accepted, Scanned by CamScanner ——__—p eM Construct a Turing machine that performs addition op erat; io (Nov Solution : Oe The function is defined as [(xty) =x + y. x is given by 0X y is given by 0” The input is placed on the tape as OX10Y. The sum of thes are performed by replacing the intermediate | by 0 and replacing blank symbol. C ly Ww i the li, The transition table is given as follows : é: 0 1 B qo | (4,0R) (q,,0,R) 9 - q, | (q,,0,R) = (q,; B, L) 4, | (q3,B,R) - _ q, - - _ x=3, y=2, f(xty) (i) qy03102 => q)900100 oF 0q,00100 fF 00q,0100 fF 000q,100 fF 0000q,00 00000q,0 fF 00000q,B 00000q,0B fF 00000Bq, Transition diagram; 010 + Start Scanned by CamScanner O16 Br © old (%)
0} gio’ MB, 6} op syges ye bo SIIB, Fig. 4.21 (m—n) Proper Subtraction _v) Turing Machine, M= (Q,.5, b. 8, dos B, {q,}) ‘where Q = {qos qd: q>> G3; Wy Is 46) == {0,1}: | Scanned by CamScanner 
