(Solved) : Give Sequence Configurations Specified Machine Enters Started Indicated Input String Examp Q36597682 . . .
Give the sequence of configurations that the specified machinebelow enters when started on the indicated input string:
Example 3.9 for a and b:
Example 3.8 for c and d:
2. Give the sequence of configurations that the specified machine below enters when started on the indicated input string: a. b. c. d. # on Mi (see Turing machine M, from Exam ple 3.9 in Sipser). 101 10#10100 on Mi (see Turing machine Mi from Example 3.9 in Sipser). 00 on M2 (see Turing machine M2 from Example 3.8 in Sipser). 00000000 on M2 (see Turing machine M2 from Example 3.8 in Sipser). EXAMPLE 3.9 The following is a formal description of Mi-Q. T. 91.4 pt, freject), the Turing machine that we informally described (page 167) for deciding the larn guage B-(w#w| w {0.1}*) .We describe 6 with a state diagram (see the following figure) .The start, accept, and reject states are q1, laccept, and (reject, respectively 4i 0 0,1 R d8 74 accept 16 0,1,x-L Чт 0,1 L FIGURE 3.10 State diagram for Turing machine Mi In Figure 3.10, which depicts the state diagram of TM M1, you will find the label 0,1R on the transition going from qs to itself. That label means that the machine stays in qs and moves to the right when it reads a O or a 1 in state q3. It doesn’t change the symbol on the tape Stage 1 is implemented by states q1 through qr, and stage 2 by the remaining states. To simplify the figure, we dont show the reject state or the transitions going to the reject state. Those transitions occur implicitly whenever a state lacks an outgoing transition for a particular symbol. Thus because in state qs, no outgoing arrow with a # is present, if a # occurs under the head when the machine is in state qs, it goes to state greject. For completeness, we say that the head moves right in each of these transitions to the reject state. Here we describe a Turing machine (TM) M2 that decides A o2″ n0, the language consisting of all strings of Os whose length is a power of 2. M2- “On input string w: Sweep left to right across the tape, crossing off every other 0. 2. 1. If in stage 1 the tape contained a single 0, accept. 3. If in stage 1 the tape contained more than a single 0 and the number of Os was odd, reject. 4. Return the head to the left-hand end of the tape. 5. Go to stage 1.” Each iteration of stage 1 cuts the number of Os in half. As the machine sweeps across the tape in stage 1, it keeps track of whether the number of Os seen is even or odd. If that number is odd and greater than 1, the original number of Os in the input could not have been a power of 2. Therefore, the machine rejects in this instance. However, if the number of Os seen is 1, the original number must have been a power of 2. So in this case, the machine accepts. Now we give the formal description of M2-(QΣΤ. δ.qifbccept, greject) Σ (0), and . We describe 6 with a state diagram (see Figure 3.8) . The start, accept, and reject states are qi, Qaccept, and greject, respectively 0-L x-R 91 93 0-x,R 0+x,R reject daccept 94 FIGURE 3.8 State diagram for Turing machine M In this state diagram, the label 0–+u,R appears on the transition from qı to q2 This label signifies that when in state qı with the head reading 0, the machine goes to state q2, writes u, and moves the head to the right. In other words, δ(Y1 ,0) = (g2.u.R). For clarity we use the shorthand O→R in the transition from a3 to q4, to mean that the machine moves to the right when reading 0 in state q3 but doesn’t alter the tape, so δ(q3:0) (94,0,R) This machine begins by writing a blank symbol over the leftmost 0 on the tape so that it can find the left-hand end of the tape in stage 4. Whereas we would normally use a more suggestive symbol such as # for the left-hand end delimiter, we use a blank here to keep the tape alphabet, and hence the state diagram, small. Example 3.11 gives another method of finding the left-hand end of the tape. Next we give a sample run of this machine on input 0000. The starting con- figuration is q10000. The sequence of configurations the machine enters appears as follows; read down the columns and left to right. 10000 Lg2000 uxg300 uq5x0xu 5uxOxu g2x0xu uxq20xu Xq5xxU u5 xxxL Xq2xxu Uxxg2XU Uxxxg2u UXxxudaccept uxq50xu Show transcribed image text
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