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B D D D D in .NET Make qr codes in .NET B D D D D

1 B D D D D generate, create qrcode none with .net projects Platform SDK (a ) A loop. Fig. 16.22 Illustration of computations. (b) Rejection of a tape. 16.7 Two-way recognizers In the next sectio QR Code for .NET n we shall prove that two-way machines are as powerful as one-way machines with respect to the classes of tapes that they can recognize. For some computations, however, it is convenient to use two-way recognizers since they may require fewer states than the equivalent one-way recognizers.

However, for the ability of a two-way machine to reverse direction and reread its tape, we pay in terms of an increased computation time. Example Consider the two-way machine shown in Table 16.2, which accepts a tape if and only if it contains at least three 1 s and at least two 0 s.

The starting and accepting states are A and G, respectively. Some typical computations are shown in Fig. 16.

23. The operation of the machine can be summarized as follows. Initially the machine is in state A and the head is scanning the left-end marker.

The head then proceeds to the right to determine whether the tape contains at least three 1 s. If the tape contains two or fewer 1 s, it is rejected; if it contains three 1 s then the head reverses its direction and moves left until it again reaches the left-end marker. The machine then proceeds to the right to determine whether the tape contains two or more 0 s.

If it does, the machine enters state G and will eventually accept the tape; otherwise the tape will be rejected.. Table 16.2 A two-w ay machine A B C D E F G A, R 0 A, R B, R C, R D, L F, R G, R G, R 1 B, R C, R D, L D, L E, R F, R G, R. E, R 0 C C 1 A D E 0 B D E 1 B D F (a) Rejecting a tape. (b) Accepting a tape. Fig. 16.23 Example of computations. The minimal one-wa .net framework qr barcode y machine that is equivalent to the two-way machine in Table 16.2 has 12 states.

This larger number of states is necessary because of the way in which a one-way machine operates. Any one-way machine that recognizes the above set of tapes must examine the tapes for the proper. Finite-state recognizers number of 0 s and QR Code for .NET 1 s simultaneously. This can be done, for example, by the use of two separate counters, one for the 1 s and the other for the 0 s.

The state of the machine in such a case is the composite state of the two counters. Consequently, the number of states required to perform the above computation is proportional to the product of the numbers of states required to test the tapes for the number of 0 s and the number of 1 s separately. The two-way machine in this example tests the tapes rst for the appropriate number of 1 s and then for the appropriate number of 0 s.

Thus, the number of states is proportional to the sum of the numbers of states required to test the tapes for the two requirements separately.. Conversion to one-way recognizers We now turn to pro visual .net Denso QR Bar Code ving that two-way machines can recognize sets of tapes (or strings) if and only if they are regular sets. Speci cally, we shall show that for every given two-way machine there is an equivalent one-way machine that recognizes the same set of tapes.

Since the details of the construction procedure do not add signi cantly to its understanding, we shall con ne our discussion to sketching the main ideas of the proof. Since a one-way machine makes as many moves as there are symbols on the tape while a two-way machine can make moves by reversing direction, the one-way machine cannot keep track of all the moves of the two-way machine or simulate them. It is, therefore, necessary to isolate the signi cant information gained by a two-way machine on moving to the left from the particular sequence of moves.

Consider an initial segment at the left of the input tape, and suppose that the head is scanning the rightmost square of this segment. The only way in which this segment can in uence the future behavior of the two-way machine is via the state which the machine is in when (and if) it leaves this segment. Thus, when a two-way machine backs up and reexamines a segment of the tape, the state Si in which the machine reenters the segment and the corresponding state Si which the machine would be in if it left the segment are the only two factors of signi cance in predicting the future behavior of the machine.

A two-way machine having n states can be in any of these states when it scans the rightmost square of the initial segment. Two cases must be considered. First, the machine may never leave the segment but oscillate within it.

Second, the machine will ultimately leave the segment on the right in one of its n states. Thus, a reentry into a segment may have n + 1 outcomes, that is, leaving the segment in one of the n states or not leaving it. Consequently the effect of the segment on the computation can be determined by specifying, for each state Si in which the machine might reenter the segment, which of the n + 1 outcomes would indeed result.

Such a speci cation is accomplished by means of a crossing function (or crossing table), denoted C(S)..
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