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  Post machine simulates Turing machine, Equivalence of TMs, PMs and Markov Algorithms

Post machine simulates Turing machine:

Assume that the TM M encompass the alphabet A = {0, 1, <, > }, where the angular brackets  are employed to delimit the finite part of the tape which initially includes the input and later gets expanded to delimit the part of the tape which has been visited so far. Assume that the M’s transition function be of form f: Q x A -> Q x A x {L, R}.

We sketch the design of Post machine P which simulates M utilizing the alphabet B = {0, 1, <, >, #}. The fundamental idea is straight-forward, the details are complicated and will be skipped.

Let consider a configuration where M is in the state q and presently reads the (bold) symbol x of the tape:

M: < B y x z C>, with y, x, z ∈ A and B, C ∈ A*. That is, we focus attention on what occurs to x and to its two neighbor symbols y to left and z to the right, and merely carry all along the remaining parts of the tape, B and C.

In this configuration, M performs either a move-right transition q, x -> q’ x’ R or move-left transition q, x -> q’ x’ L

Post machine P’s state space comprises a subspace that is in 1-to-1 correspondence with the state space Q of TM M. Devoid of danger of confusion; we call this subspace Q and select its states with similar label as the corresponding state of M. When referring to a state q, the context will make it apparent whether we signify the state qM of M or qP of P. P’s state space comprises additional states, to be introduced as required, as in general a single transition of M needs a sequence of transition s of P.M’s configuration: state q,  tape < B y x z C > is modeled as P’s configuration: state q, tape x z C > < B y

The encoding is practically forced, as P should inspect the similar symbol x as M does, however the only symbol it can examine is at the head of queue. Since M executes a transition q, x -> q’ x’ R or q, x -> q’ x’ L, P should be programmed to convert its queue in a corresponding manner. First try, that doesn’t quite work.

The easy case, that is, a move-right transition q, x -> q’ x’ R:

M: q, < B y x z C > becomes q’, < B y x’ z C >
P: q, x z C > < B y becomes q’, z C > < B y x’

The single P transition attains accurately the needed transformation.

The cumbersome case, a move-left transition q, x -> q’ x’ L:

M: q, < B y x z C > becomes q’, < B y x’ z C>
P: q, x z C > < B y should become q’, y x’ z C > < B

P’s transformation needs a complete rotation of queue in order to get the tail symbol to the head of queue. The rotation is possible when there is a distinguished symbol # which tells us when to stop. Therefore, the tentative encoding employed so far fails for a move-left transition and should be corrected.

Accurate encoding of M’s configuration in P’s configuration: Place a marker # two slots to left of the scanned symbol x. ‘To the left’ is interpreted in a circular manner, as when head and tail were glued altogether, and therefore # appears as next-to-last symbol in the queue.

M’s configuration:  state q, tape < B y x z C > is modeled as

P’s configuration: state q, tape x z C > < B # y

Simulating the move-right transition q, x -> q’ x’ R:

M: q, < B y x z C > becomes q’, < B y x’ z C >
P: q, x z C > < B # y becomes q’, z C > < B y # x’

Unluckily, this former ‘easy case’ has become more complex, as the consecutive pair # y should be permuted to y #. This can be attained by using auxiliary states and complete rotation.

Simulating the move-left transition q, x -> q’ x’ L, focus on two neighboring symbols to the left of scanned symbol x:

M: q, < B y z x C > becomes q’, < B y z x’ C >
P: q, x C > < B y # z becomes q’, z x’ C > < B # y

Again, the consecutive pair # y should be permuted to y #, attained by using auxiliary states and complete rotation.

Two more cases should be considered, if M’s read or write head is placed at either end of tape visited so far, on a symbol <or>, and probably extends the visited part. In sum up, the design of a PM which simulates a random TM is theoretically straightforward once one mastered the acrobatics of the P’s queue rotations.

Therefore, we have completed cycle TM ≥ MA ≥ PM ≥ TM which verifies the computational equivalence of such three universal models of the computation.

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