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Since subtraction is often required, a TM for complementing and incrementing a binary number is interesting. Here are the instructions for such a machine: 1 2 3 4 5 6 7 8 (1, (1, (1, (2, (2, (3, (3, (3, 0, 1, , 0, 1, 1, 0, , 1, 0, , 1, 0, 1, 0, , 1, Right ) 1, Right ) 2, Left ) 3, Right ) 2, Left ) 3, Right ) 3, Right ) halt, Stationary)
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Instructions 1 and 2 are the same as for the simpler TM which complemented the bits on the tape. Instruction 3 will apply when the TM has complemented all the bits and encountered the blank on the right end of the tape. When that happens, the machine will go into state 2 and move left. If the machine is in state 2 and encounters a 0, instruction 4 will cause the 0 to be replaced by a 1, the machine to enter state 3, and move right. Once the machine is in state 3, instructions 6 and 7 will cause the machine to move right without further changing the contents of the tape. When the machine finally encounters the blank on the right again, instruction 8 will cause the machine to halt. If the machine is in state 2 and encounters a 1, instruction 5 will cause the 1 to be replaced by a 0, the machine to stay in state 2, and move left again. This will continue in such manner until the TM encounters a 0, in which case instruction 4 will apply, as described in the previous paragraph. Using the binary number 2 as the example again, the TM will create the following contents on the tape as it executes: 0 1 1 1 1 1 0 0 1 1 0 1 0 0 0 original tape complementing complete after executing instruction 5 after executing instruction 4 halted after executing instruction 8
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This TM works for many inputs, but not all. Suppose the original input tape were all zeros: 0 0 0 original tape
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After the complementing is complete, and all the 0s become 1s, the TM will back up over the tape repeatedly executing instruction 5. That is, it will back up changing each 1 to 0. In this case, however, the TM will never encounter a 0, where instruction 4 would put the TM into state 3 and start the TM moving toward the end of the tape and a proper halt. Instead, the TM will ultimately encounter the first symbol on the tape, and instruction 5 will command it to move again left. Since the machine can go no further in that direction, the machine crashes. Likewise, the TM will crash if one of the symbols on the tape is something other than 1 or 0. There are no instructions in this TM for handling any other symbol, so an input tape such as this will also cause the TM to crash: 0 3 0 original tape
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Another way a TM can fail is by getting into an infinite loop. If instruction 7 above specified a move to the left instead of the right, certain input tapes containing only 1s and 0s would cause the TM to enter an endless loop, moving back and forth endlessly between two adjacent cells on the tape. Algorithms can be specified as TMs and, like all algorithms, TMs must be tested for correctness, given expected inputs.
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CHURCH TURING THESIS The Turing machine is thought to be a very general model of computation. In 1936, logician Alonzo Church advanced the thesis that any algorithmic procedure to manipulate symbols, conducted by humans or any machine, can be conducted by some TM. It is not possible to prove this proposition rigorously, for the notion of an algorithm is not specified mathematically. However, the Church Turing thesis has been widely tested, and is now accepted as true. One would not want to write a TM for a complex task like designing a set of stairs for a staircase, but it could be done. The significance of having such a model of computation is that the model has been used to show that some tasks cannot be accomplished with a TM. If the Church Turing thesis is true, then tasks for which a TM cannot be successful are tasks which simply have no algorithmic solution. UNSOLVABLE PROBLEMS It would be very useful to have a way of quickly knowing whether any particular program, when provided with any particular set of inputs, will execute to completion and halt, or instead continue endlessly. In computer science, this is known as the halting problem. Given a program, and a set of inputs, will the program execute to completion or not Is there some algorithm one can apply that will, for any program and any set of inputs, determine whether the program will run to completion or not One might suggest simply running the program, providing the particular inputs, and seeing whether the program halts or not. If the program were to run to completion and halt, you would know that it halts. However, if the program were to continue to run, you would never know whether the program would continue forever, or halt eventually. What is needed is an algorithm for inspecting the program, an algorithm which will tell us whether the program will eventually halt, given a particular set of inputs. If there is such an algorithm for inspecting a program, there is a TM to implement it. Unfortunately however, the halting problem has been shown to be an unsolvable problem, and the proof that there is no solution is a proof by contradiction. We begin by assuming there is, indeed, a TM that implements a solution to the halting problem. We will call this TM 'H', for it solves the big halting problem. The input to H must include both the program under test p, and the input to the program i. In pseudocode, we call H like this: H(p, i) We assume that H must itself halt, and that the output from H must be true or false the program under test must be found either to halt, or not to halt. Whatever H does, it does not rely on simply running the program under test, because H itself must always halt in a reasonable time. Now suppose that we create another TM called NotH that takes a symbolic argument that will include the encoding of a program, p. NotH will call H, passing the code for p as both the program p and the input data i to be tested. (TMs can be linked this way, but the details are not important to this discussion.) NotH will return true if H fails to halt under these conditions, and will loop forever if H does halt. In pseudocode NotH looks like this: NotH(p) if(H(p, p) is false) return true else while(true) {} //loop forever endNotH Now suppose we test NotH itself with this approach. That is, suppose we pass the code for NotH itself to NotH. We will refer to the code for NotH as 'nh', and we can ask, Does the program NotH halt when it is run with its own code as input Saying this another way, does NotH(nh) halt If NotH(nh) halts, this can only be because H(nh,nh) reports that NotH does not halt. On the other hand, if NotH(nh) does not halt, this can only be because H(nh,nh) reports that NotH does halt. These are obviously contradictions.
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