By Boolos, Burgess, Jeffrey

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Unless otherwise stated, it is to be understood that a machine starts in its lowest-numbered state. The machine we have been considering halts when it is in state qs scanning S1, for there is no table entry or arrow or quadruple telling it what to do in such a case. A virtue of the flow chart as a way of representing the machine program is that if the starting state is indicated somehow (for example, if it is understood that the leftmost node represents the starting state unless there is an indication to the contrary), then we can dispense with the names of the states: It doesn't matter what you call them.

We could in principle start them all going in state 1 with input k and await developments. Some machines will halt at once, with score 0. As time passes, one or another of the other machines may halt; then we can check whether or not it has halted in standard position. If not, its score is 0; if so, its score can be determined simply by counting the number of strokes in a row on the tape. If this number is less than or equal to the score of some k-state machine that stopped earlier, we can ignore it.

We may think of this transformation as a matter of subtracting each member of the diagonal sequence from 1: we write the antidiagonal sequence as This sequence can be relied upon not to appear as a row in Figure 2-1, for if it did appear-say, as the mth row-we should have But the mth of these equations cannot hold. [Proof: s,(m) must be zero or one. If zero, the rnth equation says that 0 = 1. 1 Then the antidiagonal sequence differs from every row of our array, and so the antidiagonal set differs from every set in our list L.