The Turing machine in .NET Creation Data Matrix barcode in .NET The Turing machine

How to generate, print barcode using .NET, Java sdk library control with example project source code free download:
7.2 The Turing machine generate, create ecc200 none in .net projects International Standard Serial Numbers Table 7.1 Example of acti Data Matrix ECC200 for .NET on table for Turing machine.

Input state s1 s2 s3 s4 s5 Change of cell contents 0 1 0 1 0 0 0 1 0 1 Move tape (R, right; L, left) R R R R L Output state s2 s3 s4 s5 s1. Table 7.2 Tape changes du ring Turing machine computation, using the action table de ned in Table 7.1.

The input tape is blank, representing a string of 0 bits or 0000000 . . .

The initial position of the head is on the leftmost zero bit. At each step, the position of the head on the tape is shown by the bold, underlined bit. The machine halts at step 6, leaving the codeword 1101100 .

. . on the output tape.

Step 1 2 3 4 5 6 Halt Input state s1 s2 s3 s4 s5 s1 Tape contents 0 0 0 0 0 0 0 ...

1 0 0 0 0 0 0 ...

1 1 0 0 0 0 0 ...

1 1 0 0 0 0 0 ...

1 1 0 1 0 0 0 ...

1 1 0 1 1 0 0 ...

Output state s2 s3 s4 s5 s1 . the possibility of return Data Matrix barcode for .NET ing to a state previously used as an input (e.g.

, s1 ; 0 1; R; s1 ). If there is no output state corresponding to the instruction, the machine halts. The same happens if the cell reading is not a case covered by the action table given a new input state.

To summarize, the tape and machine keep on moving, as long as the combination of initial state and cell reading have a matching de nition in the action table. The nal output of the program is the bit sequence written on the tape, which is left at the point where the machine halted. To illustrate how the TM works, consider next a few program examples including the basic operations of addition, subtraction, multiplication, and division.

Example 7.1 This is a TM program that creates the 5-bit sequence 11011 out of an initially blank tape with equivalent sequence 00000 . .

. Table 7.1 shows the action table instructions and Table 7.

2 shows the step-by-step implementation on the tape. It is assumed that the rst initial state is s1 and that the head is located at the leftmost cell, as underlined. As Table 7.

2 illustrates, each of the computation steps is characterized by a left-to-right tape move (or equivalently, right-to-left head move, as viewed from the tape) with a possible change in the cell contents. It is seen that the machine halts at step 6, because there is no instruction in the action table concerning an input state s1 with a cell reading of 1..

Algorithmic entropy and Kolmogorov complexity Table 7.3 Example of acti gs1 datamatrix barcode for .NET on table for Turing machine de ning the addition of two numbers in the unary system.

Input state s1 s2 s3 s4 Change of cell contents 1 1 0 0 1 1 0 0 1 0 0 1 Move tape R R R L L R Output state s2 s3 s2 s5 s4 s2. Example 7.2 This is a TM program that adds two integers. Such an operation requires one to use the unary system.

In the unary system, the only symbol is 1 (for instance) and an integer number n is represented by a string of n consecutive 1s. The decimal numbers 3 and 7 are, thus, represented in unary as 111 and 1111111, respectively. Adding unary numbers is just a matter of concatenating the two strings, namely, 3 + 7 111 + 1111111 = 1111111111 10.

Because we are manipulating two different numbers, we need an extra blank symbol, 0, to use as a delimiter to show where each of the numbers ends. Thus, the two numbers, 2 and 3, must be noted as the unary string 1101110. We use this string as the TM input tape.

The action table and the step-by-step implementation of its instructions are shown in Table 7.3 and 7.4, respectively.

It is seen that the TM halts at step 14. The output tape then contains the string 1111100 5, which is the expected result. It is left to the reader as an exercise to show that the program also works with other integer choices, meaning that this is a true adding machine! We note that the program only requires four states with two symbols, which illustrates its simplicity.

The output state s5 , which is unde ned, forces the TM to halt. Analyzing the tape moves line by line in Table 7.4, we can better understand how the algorithm works.

r Basically, each time the head sees a 1 in the cell, the only action is to look at the next cell to the right, and so on, until a 0 input is hit (here, at step 3). The head then inspects the next cell with input state s3 . r If the result is another 0, then the output state is s , meaning that the machine must 5 halt at the next step, the output tape remaining as it is.

This is because reading two successive 0 means that either (a) the second operand is equal to zero, or (b) that the addition is completed. r If the result is a 1, then the program ips the two cell values from 01 to 10, and the same inspection as before is resumed, until another 0 is hit (here, at step 12). If the sequence 00 is identi ed from there, the TM halts, meaning that the addition is completed.

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