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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31
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Every fixed multitape Turing machine has an equivalent one-tape simulation

Statement

Fix k1. For every deterministic k-tape Turing machine M, there is a deterministic one-tape Turing machine SM that, on every input, has the same accept, reject, diverge, and output behavior as M.

Facts & Assumptions

Given: A fixed integer k1 and a deterministic k-tape Turing machine M.

[L1]

A deterministic k-tape Turing machine has k right-infinite tapes, k head positions, and one simultaneous move on each tape at each step, by Multitape and nondeterministic Turing machine variants.

[F1]

For this theorem, extend the standard terminology to a deterministic multitape machine by using its accept, reject, and divergence behavior for recognition and decision, and by treating its first tape as the output tape for partial-function computation. These are exactly the clauses of Languages recognized and decided by a Turing machine and Partial functions computed by a machine under fixed encodings, with "tape" replaced by "first tape" in the output clause.

[L2]

Stay-put bookkeeping may be removed without changing accept, reject, divergence, or output behavior, by Stay-put moves can be eliminated from fixed-tape Turing-machine models.

Proof

technique · direct
1.1

Enlarge the simulator's finite tape alphabet by fresh delimiters and marked copies of the simulated symbols. On its ordinary input, SM first makes finitely many sweeps to write k blocks separated by #_1,\dots,#_{k-1}. A block records the simulated tape prefix from cell 0 through at least both its head and its rightmost nonblank cell, and exactly one recorded symbol is marked to indicate the head. The first block initially records the input (or one marked blank for empty input), and every other block initially consists of one marked blank. Thus initialization terminates and the encoding invariant represents the entire finite-support contents and head position of every simulated tape.

L1givenconstruct
2.1

In one simulation cycle, the one-tape machine SM first scans once from left to right, reads the k marked symbols, and stores in its finite control the unique transition of M determined by the current state and those symbols.

L1step 1.1construct
3.1

The machine then makes finitely many return sweeps, rewrites each old marked symbol, and places its new mark at the simulated successor head position. A left move from simulated cell 0 leaves the mark at that cell, matching the library's clamped left boundary. If a right move leaves the recorded prefix, SM shifts the delimiter and the entire encoded suffix following that block one cell right (or extends the last block into the blank suffix) and inserts a marked blank cell. These operations preserve all later blocks and restore the invariant from step 1.1.

L1step 2.1construct
4.1

After the passes from steps 2.1 and 3.1, the encoding represents precisely the next k-tape configuration of M. If the simulated state is halting, SM postpones entering its own halting state while it copies the first block cell-for-cell, with the mark removed and internal blank cells preserved, to tape cells beginning at 0, and erases every later cell, delimiter, and auxiliary symbol. It then enters the corresponding halting state. Hence its final tape is exactly the simulated first tape, not merely a compaction of its nonblank symbols. Induction on simulated steps now proves identical acceptance, rejection, divergence, and first-tape output on every original input.

step 1.1step 2.1step 3.1construct
4.2

If one chooses a local implementation of step 3.1 that temporarily uses stay-put head motions during block maintenance, [L2] converts that implementation to an equivalent one-tape machine using only the library's standard left/right move convention.

L2step 3.1
5.1

Hence the matching behavior in step 4.1 means, by [F1], that SM recognizes and decides the same languages and computes the same partial functions as M.

F1step 4.1step 4.2

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources