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<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8" />
<title>Turing Machine with 2D Tape</title>
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<optgroup label="Deterministic, Two-Way, Counter">
<option value="axbxcx-det">{ω ∈ (a ∪ b ∪ c)* | aˣbˣcˣ, x ≥ 0}</option>
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<h3>Turing Machine with Two-Dimensional Tape</h3>
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Path-local Decision:
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<ul class="nav nav-tabs" id="remarks-tab" role="tablist">
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<span class="small">2DTM‑2D‑Counter</span>
</button>
</li>
<li class="nav-item" role="presentation">
<button class="nav-link tab" id="tab-2" data-bs-toggle="tab" data-bs-target="#remarks-2" type="button" role="tab" aria-controls="tab-2" aria-selected="false">
<span class="small">2NTM‑2D‑Counter</span>
</button>
</li>
</ul>
<div id="remarks" class="tab-content">
<div id="remarks-1" class="tab-pane fade show active"> <br />
A two-way deterministic counter Turing machine with a two-dimensional tape (<span class="small">2DTM‑2D‑Counter</span>)
is equivalent in power to a two-way deterministic free-tape Turing machine with a one-dimensional tape
(<span class="small">Ordinary‑TM</span>). <br /> <br />
<div class="accordion" id="remarks-1-head">
<div class="accordion-item">
<h2 class="accordion-header">
<button class="accordion-button collapsed" type="button" data-bs-toggle="collapse" data-bs-target="#remarks-1-given" aria-expanded="false" aria-controls="remarks-1-given">Given</button>
</h2>
<div id="remarks-1-given" class="accordion-collapse collapse" aria-labelledby="remarks-1-given">
<div class="accordion-body">
<ul class = "indented-list">
<li class="indented-list-item">
<i>M</i><sub><span class="small">Ordinary‑TM</span></sub> = (<i>Q</i><sub><span class="small">Ordinary‑TM</span></sub>,
<i>T</i><sub><span class="small">Ordinary‑TM</span></sub>, <i>S</i><sub><span class="small">Ordinary‑TM</span></sub>,
<i>P</i><sub><span class="small">Ordinary‑TM</span></sub>,
<i>q</i><sub><i>I</i><sub><span class="small">Ordinary‑TM</span></sub></sub>)
</li>
<li class="indented-list-item">
<i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub> = (<i>Q</i><sub><span class="small">2DTM‑2D‑Counter</span></sub>,
<i>T</i><sub><span class="small">2DTM‑2D‑Counter</span></sub>, <i>S</i><sub><span class="small">2DTM‑2D‑Counter</span></sub>,
<i>P</i><sub><span class="small">2DTM‑2D‑Counter</span></sub>,
<i>q</i><sub><i>I</i><sub><span class="small">2DTM‑2D‑Counter</span></sub></sub>)
</li>
<li class="indented-list-item">
Counter = 𝜏
</li>
<li class="indented-list-item">
The convention follows Denning et al. (1978) but adds the input alphabet <i>S</i> ⊆ <i>T</i> to the definition.
</li>
<li class="indented-list-item">
We impose the restriction that
<i>T</i><sub><span class="small">2DTM‑2D‑Counter</span></sub> = <i>S</i><sub><span class="small">2DTM‑2D‑Counter</span></sub> ∪ {𝜏}.
</li>
<li class="indented-list-item">
We define <i>P</i><sub><span class="small">2DTM‑2D‑Counter</span></sub> such that it includes four (4) forms:
<ul class="indented-list">
<li class="indented-list-item">
<i>q</i> ] L(<i>t</i>/<i>t'</i>, <i>q'</i>), <i>q</i> ] R(<i>t</i>/<i>t'</i>, <i>q'</i>), <i>q</i> ] U(<i>t</i>/<i>t'</i>, <i>q'</i>),
and <i>q</i> ] D(<i>t</i>/<i>t'</i>, <i>q'</i>)
</li>
<li class="indented-list-item">
<i>t</i> ∈ <i>T</i><sub><span class="small">2DTM‑2D‑Counter</span></sub> ∪ {#}
</li>
<li class="indented-list-item">
However, we impose the restriction that <i>t'</i> ∈ {<i>t</i>, 𝜏}.
</li>
</ul>
</li>
</ul>
</div>
</div>
</div>
<div class="accordion-item">
<h2 class="accordion-header">
<button class="accordion-button collapsed" type="button" data-bs-toggle="collapse" data-bs-target="#remarks-1-proof1" aria-expanded="false" aria-controls="remarks-1-proof1">
Proof Sketch for <i>M</i><sub><span class="small">Ordinary-TM</span></sub> ⊆
<i>M</i><sub><span class="small">2DTM-2D-Counter</span></sub>
</button>
</h2>
<div id="remarks-1-proof1" class="accordion-collapse collapse" aria-labelledby="remarks-1-proof1">
<div class="accordion-body">
<ul class="indented-list">
<li class="indented-list-item">
Our key idea is to map each symbol in <i>T</i><sub><span class="small">Ordinary‑TM</span></sub>
to a unique "vertical" unary encoding terminated by #.
</li>
<li class="indented-list-item">
Albeit appealing, it is incorrect to simply map the <i>i</i><sup>th</sup>
symbol in <i>T</i><sub><span class="small">Ordinary‑TM</span></sub> (which we denote as <i>t</i><sub><i>i</i></sub>)
to a vertical string of <i>i</i> 𝜏's.
<ul class="indented-list">
<li class="indented-list-item">
Reason: This does not allow us to simulate the instruction <i>q</i> ] 𝛘(<i>t</i><sub><i>k</i></sub>/<i>t</i><sub><i>j</i></sub>,
<i>q'</i>) if <i>j</i> < <i>k</i>.
</li>
<li class="indented-list-item">
The unary encoding of the replacement is shorter than that of the
existing symbol, but <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub>
is not capable of changing 𝜏's to #'s.
</li>
</ul>
</li>
<li class="indented-list-item">
To account for this, we perform a fixed-length "vertical" unary mapping for each symbol in <i>T</i><sub><span class="small">Ordinary‑TM</span></sub>,
as well as for #, <span class="small">Boundary</span> and <span class="small">Void</span>.
</li>
<li class="indented-list-item">
More formally:
<ul class="indented-list">
<li class="indented-list-item">
We set the length of each unary mapping to <i>N</i> = |<i>T</i><sub><span class="small">Ordinary‑TM</span></sub>| + 3.
</li>
<li class="indented-list-item">
We map the <i>i</i><sup>th</sup> symbol in <i>T</i><sub><span class="small">Ordinary‑TM</span></sub> to
<i>i</i> 𝜏's followed by (<i>N</i> - <i>i</i>) #'s.
</li>
<li class="indented-list-item">
We map # to (<i>N</i> - 2) 𝜏's
followed by two #'s. This accounts for instructions where the replacement symbol is #.
</li>
<li class="indented-list-item">
We map <span class="small">Void</span> to (<i>N</i> - 1) 𝜏's followed by one #.
</li>
<li class="indented-list-item">
We map <span class="small">Boundary</span> to <i>N</i> 𝜏's.
</li>
</ul>
</li>
<li class="indented-list-item">
We can, thus, simulate the instruction
<i>q</i> ] 𝛘(<i>t</i><sub><i>k</i></sub>/<i>t</i><sub><i>j</i></sub>) by changing the unary encoding of <i>t</i><sub><i>k</i></sub>
to that of <span class="small">Void</span>, then writing the unary encoding of <i>t</i><sub><i>j</i></sub>.
</li>
<li class="indented-list-item">
Note that the machine can encounter several <span class="small">Void</span>'s before the unary encoding of
the symbol of interest. As a result, rewinding back to the top row of <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub>'s tape
is not straightforward. To address this, we pad the top row of the tape with <span class="small">Boundary</span>.
</li>
<li class="indented-list-item">
Consequently, the first step in converting <i>M</i><sub><span class="small">Ordinary‑TM</span></sub> to
to <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub> is to translate each symbol in the
input string to the unary encoding of <span class="small">Boundary</span> followed by the unary encoding of the symbol.
</li>
<li class="indented-list-item">
To illustrate, suppose 𝜏 = 1 and <i>T</i><sub><span class="small">Ordinary‑TM</span></sub> = {a, b}. <br /> <br />
Suppose the initial snapshot of <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>'s tape
is as follows: <br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center head-color">#</td>
<td class="text-center">a</td>
<td class="text-center">b</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
</table>
The equivalent snapshot of <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub>'s tape is
shown below. The unary encodings
of <span class="small">Boundary</span> are highlighted in light blue,
while the unary encodings of the symbols in the input string are highlighted in yellow. <br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center head-color">#</td>
<td class="text-center blue-color">1</td>
<td class="text-center blue-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center blue-color">1</td>
<td class="text-center blue-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center blue-color">1</td>
<td class="text-center blue-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center blue-color">1</td>
<td class="text-center blue-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center blue-color">1</td>
<td class="text-center blue-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center yellow-color">1</td>
<td class="text-center yellow-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center yellow-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
</table>
Suppose we want to perform the instruction <i>q</i> ] R(a/b, <i>q'</i>).
The equivalent snapshot of <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub>'s tape is
shown below.
The unary encoding of the newly written b is highlighted in yellow. <br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">#</td>
<td class="text-center head-color">1</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center yellow-color">1</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center yellow-color">1</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">...</td>
<td class="text-center">...</td>
<td class="text-center">...</td>
<td class="text-center">...</td>
<td class="text-center">...</td>
</tr>
</table>
Suppose we want to perform the instruction <i>q'</i> ] R(b/a, <i>q''</i>).
The equivalent snapshot of <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub>'s tape
is shown below. The unary encoding of <span class="small">Void</span> is highlighted in light blue,
while the unary encoding of the newly written a is highlighted in yellow. <br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center head-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center blue-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center blue-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center blue-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center blue-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center blue-color">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center yellow-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center yellow-color">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">...</td>
<td class="text-center">...</td>
<td class="text-center">...</td>
<td class="text-center">...</td>
<td class="text-center">...</td>
</tr>
</table>
</li>
</ul>
</div>
</div>
</div>
<div class="accordion-item">
<h2 class="accordion-header">
<button class="accordion-button collapsed" type="button" data-bs-toggle="collapse" data-bs-target="#remarks-1-proof2" aria-expanded="false" aria-controls="remarks-1-proof2">
Proof Sketch for <i>M</i><sub><span class="small">2DTM-2D-Counter</span></sub> ⊆
<i>M</i><sub><span class="small">Ordinary-TM</span></sub>
</button>
</h2>
<div id="remarks-1-proof2" class="accordion-collapse collapse" aria-labelledby="remarks-1-proof2">
<div class="accordion-body">
<ul class="indented-list">
<li class="indented-list-item">
Our key idea is to establish a bijection between the one-dimensional tape of <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>
and the two-dimensional tape of <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub>.
This can be done by placing the rows of <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub> next to each other
and using a marker ⬤ ∉ <i>T</i><sub><span class="small">2DTM-2D-Counter</span></sub> to demarcate them.
</li>
<li class="indented-list-item">
To illustrate, suppose we have the following snapshot of <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub>'s tape:
<br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center head-color">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
<tr>
<td class="text-center">...</td>
<td class="text-center">...</td>
<td class="text-center">...</td>
<td class="text-center">...</td>
<td class="text-center">...</td>
<td class="text-center">...</td>
</tr>
</table>
</li>
<li class="indented-list-item">
The equivalent snapshot of <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>'s tape is
as follows:
<br /> <br />
<div class="table-responsive">
<table class="table table-bordered">
<tr>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center head-color">1</td>
<td class="text-center">⬤</td>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">1</td>
<td class="text-center">⬤</td>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">⬤</td>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center">1</td>
<td class="text-center">⬤</td>
<td class="text-center">#</td>
<td class="text-center">⬤</td>
<td class="text-center">...</td>
</tr>
</table>
</div>
</li>
<li class="indented-list-item">
Simulating the left and right movements of <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub> is
straightforward, except for two cases:
<ul class="indented-list">
<li class="indented-list-item">
If the tape head attempts to move (right) to a cell that has not yet been previously visited,
<i>M</i><sub><span class="small">Ordinary‑TM</span></sub> has to make room for it by shifting
all the contents starting from the nearest ⬤ up to the rightmost ⬤, one unit to the right.
</li>
<li class="indented-list-item">
The tape head attempts to move past the leftmost cell in a row. This is illegal.
</li>
</ul>
</li>
<li class="indented-list-item">
Simulating the up and down movements of <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub> requires
moving the tape head of <i>M</i><sub><span class="small">Ordinary‑TM</span></sub> to the left (for up) or to the right
(for down) until the demarcator ⬤ is reached, signaling that the tape head is already "entering" the target row.
From there, we continue moving the tape head in the same direction
until the target column is reached.
<ul class="indented-list">
<li class="indented-list-item">
This simulation entails keeping track of how many times the tape head has already moved. This information can be
written on a separate
tape. A two-tape (or, in general, a multi-tape) Turing machine is equivalent in power to a single-tape Turing machine
(a sketch of the conversion is provided as an addendum).
</li>
<li class="indented-list-item">
If the target cell has not yet been previously visited, <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>
has to make room for it by shifting all the contents starting from the right of the newly vacated cell up to the rightmost ⬤,
one unit to the right.
</li>
<li class="indented-list-item">
Attempting to move past the topmost row is illegal.
</li>
</ul>
</li>
</ul>
</div>
</div>
</div>
<div class="accordion-item">
<h2 class="accordion-header">
<button class="accordion-button collapsed" type="button" data-bs-toggle="collapse" data-bs-target="#remarks-1-proof3" aria-expanded="false" aria-controls="remarks-1-proof3">
Addendum: Simulating a Multi-Tape Using a Single-Tape Turing Machine
</button>
</h2>
<div id="remarks-1-proof3" class="accordion-collapse collapse" aria-labelledby="remarks-1-proof3">
<div class="accordion-body">
<ul class="indented-list">
<li class="indented-list-item">
Let <i>M</i><sub><span class="small">Multi‑Tape‑TM</span></sub> be a multi-tape Turing machine
and <i>M</i><sub><span class="small">Ordinary‑TM</span></sub> be an ordinary (single-tape) Turing machine.
</li>
<li class="indented-list-item">
Our key idea is to place the rows of the multi-tape machine next to each other and using a marker ⬤ that is
not in the tape alphabet of <i>M</i><sub><span class="small">Multi‑Tape‑TM</span></sub> to demarcate them.
</li>
<li class="indented-list-item">
Meanwhile, to keep track of the position of the tape heads, we introduce an accented version of each symbol
in the tape alphabet of <i>M</i><sub><span class="small">Multi‑Tape‑TM</span></sub>.
This accented version is used to indicate that a tape head is pointing to that cell.
<ul class="indented-list">
<li class="indented-list-item">
In this regard, the tape alphabet of
<i>M</i><sub><span class="small">Ordinary‑TM</span></sub> is the union of the tape alphabet of
<i>M</i><sub><span class="small">Multi‑Tape‑TM</span></sub> and their accented versions.
</li>
</ul>
</li>
<li class="indented-list-item">
To illustrate, suppose we have the following snapshot of <i>M</i><sub><span class="small">Multi‑Tape‑TM</span></sub>'s
tapes:
<br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">#</td>
<td class="text-center">a</td>
<td class="text-center">a</td>
<td class="text-center">e</td>
<td class="text-center head-color">e</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
</table>
<table class="table table-bordered">
<tr>
<td class="text-center">#</td>
<td class="text-center head-color">a</td>
<td class="text-center">#</td>
<td class="text-center">e</td>
<td class="text-center">e</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
</table>
<table class="table table-bordered">
<tr>
<td class="text-center">#</td>
<td class="text-center">a</td>
<td class="text-center">#</td>
<td class="text-center head-color">a</td>
<td class="text-center">#</td>
<td class="text-center">#</td>
<td class="text-center">...</td>
</tr>
</table>
</li>
<li class="indented-list-item">
The equivalent snapshot of <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>'s tape is
as follows:
<br /> <br />
<div class="table-responsive">
<table class="table table-bordered">
<tr>
<td class="text-center head-color">#</td>
<td class="text-center">a</td>
<td class="text-center">a</td>
<td class="text-center">e</td>
<td class="text-center">ê</td>
<td class="text-center">⬤</td>
<td class="text-center">#</td>
<td class="text-center">â</td>
<td class="text-center">#</td>
<td class="text-center">e</td>
<td class="text-center">e</td>
<td class="text-center">⬤</td>
<td class="text-center">#</td>
<td class="text-center">a</td>
<td class="text-center">#</td>
<td class="text-center">â</td>
<td class="text-center">⬤</td>
<td class="text-center">#</td>
<td class="text-center">⬤</td>
<td class="text-center">...</td>
</tr>
</table>
</div>
</li>
<li class="indented-list-item">
Simulating the left and right movements of <i>M</i><sub><span class="small">Multi‑Tape‑TM</span></sub>
is analogous to the simulation presented in the conversion of <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub>
to <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>.
</li>
</ul>
</div>
</div>
</div>
</div>
</div>
<div id="remarks-2" class="tab-pane fade"> <br />
A two-way nondeterministic counter Turing machine with a two-dimensional tape (<span class="small">2NTM-2D-Counter</span>) is equivalent in power
to a two-way deterministic free-tape Turing machine with a one-dimensional tape (<span class="small">Ordinary-TM</span>). <br /> <br />
<div class="accordion" id="remarks-2-head">
<div class="accordion-item">
<h2 class="accordion-header">
<button class="accordion-button collapsed" type="button" data-bs-toggle="collapse" data-bs-target="#remarks-2-given1" aria-expanded="false" aria-controls="remarks-1-given">
Given
</button>
</h2>
<div id="remarks-2-given1" class="accordion-collapse collapse" aria-labelledby="remarks-2-given1">
<div class="accordion-body">
<ul class="indented-list">
<li class="indented-list-item">
Let <i>M</i><sub><span class="small">2NTM‑2D‑Free‑Tape</span></sub> be a two-way nondeterministic free-tape Turing machine with a two-dimensional tape.
</li>
<li class="indented-list-item">
Let <i>M</i><sub><span class="small">2DTM‑Multi-2D‑Free‑Tape</span></sub> be a two-way deterministic free-tape Turing machine with multiple two-dimensional tapes.
</li>
</ul>
</div>
</div>
</div>
<div class="accordion-item">
<h2 class="accordion-header">
<button class="accordion-button collapsed" type="button" data-bs-toggle="collapse" data-bs-target="#remarks-2-given" aria-expanded="false" aria-controls="remarks-1-given">
Proof Sketch for <i>M</i><sub><span class="small">2NTM‑2D‑Counter</span></sub> ⊆ <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>
</button>
</h2>
<div id="remarks-2-given" class="accordion-collapse collapse" aria-labelledby="remarks-2-given">
<div class="accordion-body">
<ul class="indented-list">
<li class="indented-list-item">
We follow this line of thought:
<ul class="indented-list">
<li class="indented-list-item">
Clearly, <i>M</i><sub><span class="small">2NTM‑2D‑Counter</span></sub> ⊆ <i>M</i><sub><span class="small">2NTM‑2D‑Free‑Tape</span></sub>.
</li>
<li class="indented-list-item">
We attempt to show that <i>M</i><sub><span class="small">2NTM‑2D‑Free‑Tape</span></sub>
⊆ <i>M</i><sub><span class="small">2DTM‑Multi-2D‑Free‑Tape</span></sub> by simulating the former
using three (two-dimensional) tapes.
</li>
<li class="indented-list-item">
Each (two-dimensional) tape of <i>M</i><sub><span class="small">2DTM‑Multi-2D‑Free‑Tape</span></sub>
can be converted to a (one-dimensional) tape of <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>.
</li>
<li class="indented-list-item">
A multi-tape Turing machine is equivalent in power to an ordinary (single-tape) Turing machine; a sketch of the
conversion is provided as an addendum.
</li>
</ul>
</li>
<li class="indented-list-item">
Therefore, it suffices to prove that <i>M</i><sub><span class="small">2NTM‑2D‑Free‑Tape</span></sub>
can be converted to <i>M</i><sub><span class="small">2DTM‑Multi-2D‑Free‑Tape</span></sub> using three tapes:
<ul class="indented-list">
<li class="indented-list-item">
The first tape serves as an immutable storage for the input string.
</li>
<li class="indented-list-item">
The second tape carries out the actual simulation.
</li>
<li class="indented-list-item">
The third tape marks the address of the configuration executed by <i>M</i><sub><span class="small">Multi‑Tape‑TM</span></sub>
in the computational tree of <i>M</i><sub><span class="small">2NTM‑2D‑Free‑Tape</span></sub>.
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<li class="indented-list-item">
The nodes of the computational tree of <i>M</i><sub><span class="small">2NTM‑2D‑Free‑Tape</span></sub>
should be visited in a breadth-first fashion. Visiting them in a depth-first fashion risks trapping the machine
in a nonterminating branch.
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</ul>
</li>
</ul>
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</div>
</div>
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<h2 class="accordion-header">
<button class="accordion-button collapsed" type="button" data-bs-toggle="collapse" data-bs-target="#remarks-2-proof1" aria-expanded="false" aria-controls="remarks-1-given">
Proof Sketch for <i>M</i><sub><span class="small">Ordinary‑TM</span></sub> ⊆ <i>M</i><sub><span class="small">2NTM‑2D‑Counter</span></sub>
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</h2>
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<li class="indented-list-item">
Since <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub> ⊆ <i>M</i><sub><span class="small">2NTM‑2D‑Counter</span></sub> and
we have already established that <i>M</i><sub><span class="small">2DTM‑2D‑Counter</span></sub> = <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>,
it follows that <i>M</i><sub><span class="small">Ordinary‑TM</span></sub> ⊆ <i>M</i><sub><span class="small">2NTM‑2D‑Counter</span></sub>.
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</ul>
</div>
</div>
</div>
<div class="accordion-item">
<h2 class="accordion-header">
<button class="accordion-button collapsed" type="button" data-bs-toggle="collapse" data-bs-target="#remarks-1-proof3" aria-expanded="false" aria-controls="remarks-1-proof3">
Addendum: Simulating a Multi-Tape Using a Single-Tape Turing Machine
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<li class="indented-list-item">
Let <i>M</i><sub><span class="small">Multi‑Tape‑TM</span></sub> be a multi-tape Turing machine
and <i>M</i><sub><span class="small">Ordinary‑TM</span></sub> be an ordinary (single-tape) Turing machine.
</li>
<li class="indented-list-item">
Our key idea is to place the rows of the multi-tape machine next to each other and using a marker ⬤ that is
not in the tape alphabet of <i>M</i><sub><span class="small">Multi‑Tape‑TM</span></sub> to demarcate them.
</li>