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<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8" />
<title>k-Queue Automaton</title>
<link rel="icon" type="image/x-icon" href="assets/automata-icon.png" />
<link href="style/colors.css" rel="stylesheet" />
<link href="style/style.css" rel="stylesheet" />
<link href="https://cdn.jsdelivr.net/npm/bootstrap@5.2.1/dist/css/bootstrap.min.css" rel="stylesheet" />
<script src="https://cdn.jsdelivr.net/npm/bootstrap@5.2.1/dist/js/bootstrap.bundle.min.js"></script>
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<link rel="stylesheet" href="https://cdn.jsdelivr.net/npm/bootstrap-icons@1.10.2/font/bootstrap-icons.css" />
<script src="scripts/array.js"></script>
</head>
<body>
<nav class="navbar navbar-expand-lg navbar-dark">
<div class="container-fluid">
<a class="navbar-brand">
<img id="automata-icon" src="assets/automata-icon.png" class="d-inline-block" />
AUTOMATA THEORY
</a>
<button class="navbar-toggler" type="button" data-bs-toggle="collapse" data-bs-target="#menu" aria-controls="menu" aria-expanded="false" aria-label="Toggle navigation">
<span class="navbar-toggler-icon"></span>
</button>
<div class="collapse navbar-collapse" id="menu">
<ul class="navbar-nav me-auto mb-2 mb-lg-0">
<li class="nav-item">
<a class="nav-link" aria-disabled="true" href="index.html" tabindex="-1">Finite Automaton</a>
</li>
<li class="nav-item">
<a class="nav-link" tabaria-current="page" tabindex="-1" href="turing-2d.html">Turing Machine with 2D Tape</a>
</li>
<li class="nav-item">
<a class="nav-link active" href="#" aria-disabled="true">
<span>
<i>k</i>-Queue Automaton
</span>
</a>
</li>
</ul>
<div class="d-flex">
<div id="view-github" class="d-flex">
<a class="nav-link" target="_blank" href="https://github.com/memgonzales/turing-machine-variants"><img id="github-icon" src="assets/github.png" class="d-inline-block" /></a>
</div>
<div id="view-github" class="d-flex" style="padding-top: 1.5px">
<a class="nav-link" target="_blank" href="https://github.com/memgonzales/turing-machine-variants">View on GitHub</a>
</div>
</div>
</div>
</div>
</nav>
<div class="container-fluid">
<div class="row">
<div id="program-container" class="col-sm-5">
<label for="test-cases" class="form-label">Sample Machines</label>
<div class="input-group">
<select id="test-cases" class="form-select">
<optgroup label="Custom">
<option value="instructions">Instructions</option>
<option value="custom">Blank Machine</option>
</optgroup>
<optgroup label="Deterministic, One-Way, Counter, Multi-Queue">
<option value="axbxcx-det">{ω ∈ (a ∪ b ∪ c)* | aˣbˣcˣ, x ≥ 0}</option>
</optgroup>
<optgroup label="Nondeterministic, One-Way, Counter, Multi-Queue">
<option value="ax-axbxcx-non">{ω ∈ (a ∪ b ∪ c)* | aʸbˣcˣ, x, y ≥ 0}</option>
</optgroup>
<optgroup label="Deterministic, Two-Way, Free Stack/Tape, One-Queue">
<option value="wwr">{ω ∈ (H ∪ A)* | ωωᴿ}</option>
</optgroup>
<optgroup label="Nondeterministic, Two-Way, Free Stack/Tape, One-Queue">
<option value="ww">{ω ∈ (H ∪ A)* | ωω}</option>
</optgroup>
</select>
<button id="save" type="button" class="btn btn-outline-primary"><i class="bi bi-download"></i></button>
</div>
<br />
<div id="editor"></div>
<br />
<div class="d-grid"><input id="input-string" type="text" class="form-control" placeholder="Input String" /></div>
<br />
<div class="d-grid">
<button id="run" type="button" class="btn btn-primary">Run Machine</button>
</div>
</div>
<div id="simulation-container" class="col-sm-7">
<h3>
<i>k</i>-Queue Automaton
</h3>
<div class="row">
<div id="simulation" class="table-responsive">
<div id="simulation-config">
<br />
<label for="config" class="form-label">Path</label>
<select id="config" class="form-select"></select>
</div>
</div>
</div>
<div id="simulation-controls" class="row">
<br />
<div class="col-sm-12 text-center unselectable">
<button id="prev" type="button" class="btn btn-primary" disabled="true">< Prev</button>
<span class="align-middle">
Step
<input id="step-number" type="number" value="1" min="1" />
of
<span id="total-steps"></span>
</span>
<button id="next" type="button" class="btn btn-primary">Next ></button>
</div>
<br />
<br />
<div>
<h5>
<span id="final-decision-emoji"></span>
Final Decision:
<span id="final-decision"></span>
</h5>
<p id="final-decision-sub"></p>
<p id="config-decision-header">
<b>
Path-local Decision:
<span id="config-decision"></span>
</b>
<br />
<span id="config-decision-sub"></span>
</p>
</div>
<br />
<hr />
</div>
<br />
<ul class="nav nav-tabs" id="remarks-tab" role="tablist">
<li class="nav-item" role="presentation">
<button class="nav-link active tab" id="tab-1" data-bs-toggle="tab" data-bs-target="#remarks-1" type="button" role="tab" aria-controls="tab-1" aria-selected="true">
<span class="small">Multi‑Queue‑1D/NQA‑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">1‑Queue‑2DQA‑Free-Tape</span>
</button>
</li>
<li class="nav-item" role="presentation">
<button class="nav-link tab" id="tab-3" data-bs-toggle="tab" data-bs-target="#remarks-3" type="button" role="tab" aria-controls="tab-3" aria-selected="false">
<span class="small">1‑Queue‑2NQA‑Free-Tape</span>
</button>
</li>
</ul>
<div id="remarks" class="tab-content">
<div id="remarks-1" class="tab-pane fade show active">
<br />
Both deterministic and nondeterministic variants of a one-way multi-queue counter automaton
can accept all regular languages.
<ul class="indented-list">
<li class="indented-list-item">
Although their stack alphabet is limited, they still have auxiliary memory, making them more powerful than finite-state accepters.
</li>
</ul>
Both deterministic and nondeterministic variants of a one-way multi-queue counter automaton
can recognize some context-free languages but fail to recognize some.
<ul class="indented-list">
<li class="indented-list-item">
The context-free languages that they can recognize include:
<ul class="indented-list">
<li class="indented-list-item">
{ω ∈ (a ∪ b)* | aˣbˣ, x ≥ 0}
</li>
<li class="indented-list-item">
{ω ∈ (a ∪ b)* | N(a) = N(b)}
</li>
</ul>
</li>
<li class="indented-list-item">
The context-free languages that they cannot recognize include:
<ul class="indented-list">
<li class="indented-list-item">
{ω ∈ (H ∪ A)* | ωωᴿ}
</li>
</ul>
</li>
</ul>
Both deterministic and nondeterministic variants of a one-way deterministic multi-queue counter queue automaton
can recognize some (i.e., but not all) context-sensitive languages.
<ul class="indented-list">
<li class="indented-list-item">
The context-sensitive languages that they can recognize include:
<ul class="indented-list">
<li class="indented-list-item">
{ω ∈ (a ∪ b ∪ c)* | aˣbˣcˣ, x ≥ 0}
</li>
</ul>
</li>
</ul>
</div>
<div id="remarks-2" class="tab-pane fade">
<br />
A two-way deterministic free-tape single-queue automaton (<span class="small">1‑Queue‑2DQA‑Free-Tape</span>)
is equivalent in power to a Turing machine (<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-given" aria-expanded="false" aria-controls="remarks-2-given">
Proof Sketch for <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>
⊆ <i>M</i><sub><span class="small">1‑Queue‑2DQA‑Free-Tape</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 prove the tighter result that a Turing machine can be simulated with a one-way deterministic
free-tape single-queue automaton.
</li>
<li class="indented-list-item">
Our key idea is to establish a bijection between the contents of the tape of <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>
and the queue of <i>M</i><sub><span class="small">1‑Queue‑2DQA‑Free-Tape</span></sub>
via the following strategy:
<ul class="indented-list">
<li class="indented-list-item">
Use a special symbol ⬤ that is not part of the tape alphabet to mark the start of the simulated tape.
</li>
<li class="indented-list-item">
The symbol to which the tape head of <i>M</i><sub><span class="small">Ordinary‑TM</span></sub> points
is the front of the queue. Symbols on the right of the tape head are in front of ⬤, while symbols on
the left of the tape head are behind ⬤.
</li>
<li class="indented-list-item">
To illustrate, suppose we have the following snapshot of
<i>M</i><sub><span class="small">Ordinary‑TM</span></sub>'s
tape:
<br /> <br />
<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">b</td>
<td class="text-center">c</td>
<td class="text-center">...</td>
</tr>
</table>
</li>
<li class="indented-list-item">
The equivalent snapshot of the queue is as follows:
<br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">b</td>
<td class="text-center">c</td>
<td class="text-center">...</td>
<td class="text-center">⬤</td>
<td class="text-center">#</td>
<td class="text-center">a</td>
<td class="text-center">...</td>
</tr>
</table>
</li>
</ul>
</li>
<li class="indented-list-item">
Simulating the <i>M</i><sub><span class="small">Ordinary‑TM</span></sub> instruction "read symbol d, write symbol b, and move right"
is straightforward:
<ul class="indented-list">
<li class="indented-list-item">
Dequeue symbol d, then enqueue symbol b.
</li>
<li class="indented-list-item">
To illustrate, consider these tape and queue snapshots before executing this instruction:
<br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">a</td>
<td class="text-center">b</td>
<td class="text-center">c</td>
<td class="text-center head-color">d</td>
<td class="text-center">e</td>
<td class="text-center">...</td>
</tr>
</table>
<table class="table table-bordered">
<tr>
<td class="text-center">d</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">b</td>
<td class="text-center">c</td>
</tr>
</table>
</li>
<li class="indented-list-item">
After reading symbol d, writing symbol b, and moving right, the tape snapshot is as follows:
<br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">a</td>
<td class="text-center">b</td>
<td class="text-center">c</td>
<td class="text-center">b</td>
<td class="text-center head-color">e</td>
<td class="text-center">...</td>
</tr>
</table>
</li>
<li class="indented-list-item">
After dequeueing symbol b and enqueueing symbol b (following our proposed simulation), the queue snapshot is as follows:
<br /> <br />
<table class="table table-bordered">
<tr>
<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">b</td>
<td class="text-center">c</td>
<td class="text-center">b</td>
</tr>
</table>
</li>
</ul>
</li>
<li class="indented-list-item">
Simulating the <i>M</i><sub><span class="small">Ordinary‑TM</span></sub> instruction "read symbol d, write symbol b, and move left"
is more involved:
<ul class="indented-list">
<li class="indented-list-item">
Introduce the primitive operation <span class="small">Cyclic-Shift</span> whose goal
is to transfer a symbol at the back of the queue to the front:
<ul class="indented-list">
<li class="indented-list-item">
Suppose the queue snapshot is as follows:
<br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">e</td>
<td class="text-center">f</td>
<td class="text-center">⬤</td>
<td class="text-center">a</td>
</tr>
</table>
</li>
<li class="indented-list-item">
Enqueue a special symbol ☆ that is not part of the tape alphabet.
<br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">e</td>
<td class="text-center">f</td>
<td class="text-center">⬤</td>
<td class="text-center">a</td>
<td class="text-center">☆</td>
</tr>
</table>
</li>
<li class="indented-list-item">
Replace each symbol <i>x</i> with a new symbol <i>xy</i> where <i>y</i> is the symbol
immediately behind <i>x</i> in the queue.
<ul class="indented-list">
<li class="indented-list-item">
This can be done by using a special state <i>q</i><sub><i>x</i></sub>
to remember that we dequeued <i>x</i>. When the machine encounters <i>y</i> from this state, it will
enqueue <i>xy</i>.
</li>
<li class="indented-list-item">
We keep doing this until we scan the marker ☆, at which point we enqueue ☆ then the symbol that
used to be in front of it.
</li>
<li class="indented-list-item">
In other words, we are using the states as some form of "memory," which works
since the tape alphabet is finite.
</li>
</ul>
<br />
<table class="table table-bordered">
<tr>
<td class="text-center">ef</td>
<td class="text-center">f⬤</td>
<td class="text-center">⬤a</td>
<td class="text-center">☆</td>
<td class="text-center">a</td>
</tr>
</table>
</li>
<li class="indented-list-item">
Afterwards, dequeue the symbols and enqueue only their first characters
until the marker ☆ is scanned.
<br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">☆</td>
<td class="text-center">a</td>
<td class="text-center">e</td>
<td class="text-center">f</td>
<td class="text-center">⬤</td>
</tr>
</table>
</li>
<li class="indented-list-item">
Dequeue ☆ to complete the <span class="small">Cyclic-Shift</span> operation.
<br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">a</td>
<td class="text-center">e</td>
<td class="text-center">f</td>
<td class="text-center">⬤</td>
</tr>
</table>
</li>
</ul>
</li>
<li class="indented-list-item">
The instruction "read symbol d, write symbol b, and move left" can then be simulated as follows:
<ul class="indented-list">
<li class="indented-list-item">
Dequeue d, enqueue b, and perform two <span class="small">Cyclic-Shift</span> operations.
</li>
<li class="indented-list-item">
To illustrate, consider these tape and queue snapshots before executing this instruction:
<br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">a</td>
<td class="text-center">b</td>
<td class="text-center">c</td>
<td class="text-center head-color">d</td>
<td class="text-center">e</td>
<td class="text-center">...</td>
</tr>
</table>
<table class="table table-bordered">
<tr>
<td class="text-center">d</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">b</td>
<td class="text-center">c</td>
</tr>
</table>
</li>
<li class="indented-list-item">
After reading symbol d, writing symbol b, and moving left, the tape snapshot is as follows:
<br /> <br />
<table class="table table-bordered">
<tr>
<td class="text-center">a</td>
<td class="text-center">b</td>
<td class="text-center head-color">c</td>
<td class="text-center">b</td>
<td class="text-center">e</td>
<td class="text-center">...</td>
</tr>
</table>
</li>
<li class="indented-list-item">
After dequeueing d, enqueueing b, and performing two <span class="small">Cyclic-Shift</span> operations,
the queue snapshot is as follows:
<br /> <br />
<table class="table table-bordered">
<tr>
<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">b</td>
<td class="text-center">c</td>
</tr>
</table>
<table class="table table-bordered">
<tr>
<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">b</td>
<td class="text-center">c</td>
<td class="text-center">b</td>
</tr>
</table>
<table class="table table-bordered">
<tr>
<td class="text-center">b</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">b</td>
<td class="text-center">c</td>
</tr>
</table>
<table class="table table-bordered">
<tr>
<td class="text-center">c</td>
<td class="text-center">b</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">b</td>
</tr>
</table>
</li>
</ul>
</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-2-given1" aria-expanded="false" aria-controls="remarks-2-given1">
Proof Sketch for <i>M</i><sub><span class="small">1‑Queue‑2DQA‑Free-Tape</span></sub>
⊆ <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>
</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">
Our key idea is to simulate the queue automaton using two 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 is for simulating the queue operations.
<ul class="indented-list">
<li class="indented-list-item">
This can be done in a variety of ways. Perhaps, th simplest way to simulate
the dequeueing of
a symbol is to to lazily mark it as deleted using a symbol that is not part of the
queue alphabet.
</li>
</ul>
</li>
</ul>
</li>
<li class="indented-list-item">
Since a multi-tape Turing machine is equivalent in power to an ordinary (single-tape) Turing machine,
our proof is completed. You may refer to the addendum in our
<a href = "https://turing-machine-variants.vercel.app/turing-2d.html" target = "_blank" class = "link-secondary">Turing machine page</a> for the sketch of the conversion
from multi-tape to single-tape Turing machine.
</li>
</ul>
</div>
</div>
</div>
</div>
</div>
<div id="remarks-3" class="tab-pane fade">
<br />
A two-way nondeterministic free-tape single-queue automaton (<span class="small">1‑Queue‑2NQA‑Free-Tape</span>)
is equivalent in power to a Turing machine (<span class="small">Ordinary‑TM</span>). <br /> <br />
<div class="accordion" id="remarks-3-head">
<div class="accordion-item">
<h2 class="accordion-header">
<button class="accordion-button collapsed" type="button" data-bs-toggle="collapse" data-bs-target="#remarks-3-given1" aria-expanded="false" aria-controls="remarks-3-given1">
Proof Sketch for
<i>M</i><sub><span class="small">1‑Queue‑2NQA‑Free-Tape</span></sub>
⊆ <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>
</button>
</h2>
<div id="remarks-3-given1" class="accordion-collapse collapse" aria-labelledby="remarks-3-given1">
<div class="accordion-body">
<ul class="indented-list">
<li class="indented-list-item">
Our key idea is to simulate the automaton 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 is for simulating the queue operations.
</li>
<li class="indented-list-item">
The third tape marks the address of the configuration executed by the multi-tape Turing machine
in the computational tree of <i>M</i><sub><span class="small">1‑Queue‑2NQA‑Free-Tape</span></sub>.
</li>
<li class="indented-list-item">
The nodes of the computational tree of <i>M</i><sub><span class="small">1‑Queue‑2NQA‑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.
</li>
</ul>
</li>
<li class="indented-list-item">
Since a multi-tape Turing machine is equivalent in power to an ordinary (single-tape) Turing machine,
our sketch is completed. You may refer to the addendum in our
<a href = "https://turing-machine-variants.vercel.app/turing-2d.html" target = "_blank" class = "link-secondary">Turing machine page</a> for the sketch of the conversion
from multi-tape to single-tape Turing machine.
</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-3-given" aria-expanded="false" aria-controls="remarks-3-given">
Proof Sketch for <i>M</i><sub><span class="small">Ordinary‑TM</span></sub>
⊆ <i>M</i><sub><span class="small">1‑Queue‑2NQA‑Free-Tape</span></sub>
</button>
</h2>
<div id="remarks-3-given" class="accordion-collapse collapse" aria-labelledby="remarks-3-given">
<div class="accordion-body">
<ul class="indented-list">
<li class="indented-list-item">
Since <i>M</i><sub><span class="small">1‑Queue‑2DQA‑Free-Tape</span></sub>
⊆ <i>M</i><sub><span class="small">1‑Queue‑2NQA‑Free-Tape</span></sub>
and we have already established that <i>M</i><sub><span class="small">1‑Queue‑2DQA‑Free-Tape</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">1‑Queue‑2DQA‑Free-Tape</span></sub>
</li>
</ul>
</div>
</div>
</div>
</div>
</div>
</div>
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