require TM output at sq. 1; definition of disciplined machine (issue #…

…191)

content/turing-machines/machines-computations/combining-machines.tex

@@ -21,15 +21,47 @@
 creating several simple Turing machines and combining them into one
 machine that can solve the problem. This point is especially important
 when tackling the Halting Problem in the next section.
+
+How do we combine Turing machines $M = \tuple{Q, \Sigma, q_0, \delta}$
+and~$M' = \tuple{Q', \Sigma', q_0', \delta'}$?  We now use the
+configuration of the tape after $M$~has halted as the input
+configuration of a run of machine~$M'$.  To get a single Turing
+machine $M \frown M'$ that does this, do the following:
+\begin{enumerate}
+    \item Renumber (or relabel) all the states~$Q'$ of~$M'$ so that $M$
+    and~$M'$ have no states in common ($Q \cap Q' = \emptyset$).
+    \item The states of $M \frown M'$ are $Q \cup Q'$.
+    \item The tape alphabet is $\Sigma \cup \Sigma'$.
+    \item The start state is~$q_0$.
+    \item The transition function is the function $\delta''$ given by:
+    \[\delta''(q,\sigma) =
+    \begin{cases}
+      \delta(q,\sigma) & \text{if $q \in Q$}\\
+      \delta'(q,\sigma) & \text{if $q \in Q'$}\\
+      \tuple{q_0', \sigma, \TMstay} & \text{if $q \in Q$ and
+      $\delta(q,\sigma)$ undefined}
+    \end{cases}\]
+\end{enumerate}
+The resulting machine uses the instructions of~$M$ when it is in a
+state $q \in Q$, the instructions of~$M'$ when it is in a state~$q \in
+Q'$. When it is in a state $q \in Q$ and is scanning a symbol~$\sigma$
+for which $M$ has no transition (i.e., $M$ would have halted), it
+enters the start state of~$M'$ (and leaves the tape contents and head
+position as it is).
+
+Note that unless the machine~$M$ is disciplined, we don't know where
+the tape head is when $M$~halts, so the halting configuration of~$M$
+need not have the head scanning square~$1$. When combining machines,
+it's important to keep this in mind.
 \end{explain}
 
 \begin{ex}
-\emph{Combining Machines:} Design a machine that computes the function
-$f(m,n) = 2(m+n)$.
-
-In order to build this machine, we can combine two machines we are already
-familiar with: the addition machine, and the doubler. We begin by drawing 
-a state diagram for the addition machine.
+\emph{Combining Machines:} We'll design a machine which, when started
+on input consisting of two blocks of~$\TMstroke$'s of length $n$
+and~$m$, halts with a single block of $2(m+n)$ $\TMstroke$'s on the
+tape. In order to build this machine, we can combine two machines we
+are already familiar with: the addition machine, and the doubler. We
+begin by drawing a state diagram for the addition machine.
 \[
 \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=2.8cm,
                     semithick]
@@ -39,18 +71,18 @@
   \node[state]         (B) [right of=A] {$q_1$};
   \node[state]         (C) [right of=B] {$q_2$};
 
-  \path (A) edge node {\TMtrans{\TMblank}{\TMstroke}{\TMright}} (B)
+  \path (A) edge node {\TMtrans{\TMblank}{\TMstroke}{\TMstay}} (B)
             edge [loop above] node {\TMtrans{\TMstroke}{\TMstroke}{\TMright}} (B)
         (B) edge [loop above] node {\TMtrans{\TMstroke}{\TMstroke}{\TMright}} (B)
             edge node {\TMtrans{\TMblank}{\TMblank}{\TMleft}} (C)
         (C) edge [loop above] node {\TMtrans{\TMstroke}{\TMblank}{\TMstay}} (C);
 \end{tikzpicture}
 \]
-Instead of halting at state $q_2$, we want to continue operation in
-order to double the output. Recall that the doubler machine erases
-the first stroke in the input and writes two strokes in a separate
-output. Let's add an instruction to make sure the tape head is reading 
-the first stroke of the output of the addition machine.
+Instead of halting in state~$q_2$, we want to continue operation in
+order to double the output. Recall that the doubler machine erases the
+first stroke in the input and writes two strokes in a separate output.
+Let's add an instruction to make sure the tape head is reading the
+first stroke of the output of the addition machine.
 \[
 \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=2.8cm,
                     semithick]
@@ -62,7 +94,7 @@
   \node[state]         (D) [below left of=C] {$q_3$};
   \node[state]         (E) [below left of=D] {$q_4$};
 
-  \path (A) edge node {\TMtrans{\TMblank}{\TMstroke}{\TMright}} (B)
+  \path (A) edge node {\TMtrans{\TMblank}{\TMstroke}{\TMstay}} (B)
             edge [loop above] node {\TMtrans{\TMstroke}{\TMstroke}{\TMright}} (B)
         (B) edge [loop above] node {\TMtrans{\TMstroke}{\TMstroke}{\TMright}} (B)
             edge node {\TMtrans{\TMblank}{\TMblank}{\TMleft}} (C)
@@ -72,10 +104,10 @@
 \end{tikzpicture}
 \]
 It is now easy to double the input---all we have to do is connect the
-doubler machine onto state $q_4$. This requires renaming the states
-of the doubler machine so that they start at $q_4$ instead of
-$q_0$---this way we don't end up with two starting states. The final
-diagram should look as in \olref{fig:combined}.
+doubler machine onto state~$q_4$. This requires renaming the states of
+the doubler machine so that they start at~$q_4$ instead
+of~$q_0$---this way we don't end up with two starting states. The
+final diagram should look as in \olref{fig:combined}.
 \begin{figure}
 \[
 \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=2.8cm,
@@ -92,7 +124,7 @@
   \node[state]         (5) [left of=4]  {$q_8$};
   \node[state]         (6) [left of=5]  {$q_9$};
 
-  \path (A) edge node {\TMtrans{\TMblank}{\TMstroke}{\TMright}} (B)
+  \path (A) edge node {\TMtrans{\TMblank}{\TMstroke}{\TMstay}} (B)
             edge [loop above] node {\TMtrans{\TMstroke}{\TMstroke}{\TMright}} (B)
         (B) edge [loop above] node {\TMtrans{\TMstroke}{\TMstroke}{\TMright}} (B)
             edge node {\TMtrans{\TMblank}{\TMblank}{\TMleft}} (C)
@@ -117,4 +149,23 @@
 \end{figure}
 \end{ex}
 
+\begin{prop}
+    If $M$ and $M'$ are disciplined and compute the functions $f\colon
+    \Nat^k \to \Nat$ and $f'\colon \Nat \to \Nat$, respectively, then
+    $M \frown M'$ is disciplined and computes~$\comp{f}{f'}$.
+\end{prop}
+
+\begin{proof}
+    Since $M$ is disciplined, when it halts with
+    output~$f(n_1,\dots,n_k) = m$, the head is scanning square~$1$. If
+    we now enter the start state of~$M'$, the machine will halt with
+    output $f(m)$, again scanning square~$1$. The other conditions of
+    \olref[dis]{defn:disciplined} are also satisfied.
+\end{proof}
+
+\begin{prob}
+    Give a disciplined Turing machine computing $f(x) = x+2$ by taking
+    the machine~$M$ from \cref{tur:mac:dis:prob:disc-succ} and
+    construct $M \frown M$.
+\end{prob}
 \end{document}

content/turing-machines/machines-computations/configuration.tex

@@ -18,8 +18,8 @@
 \emph{configurations}---and a configuration in turn is a sequence of
 symbols (corresponding to the contents of the tape at a given point in
 the computation), a number indicating the position of the read/write
-head, and a state.  Using these, we can define what the Turing machine
-$M$ computes on a given input.
+head, and a state. Using these, we can define what the Turing
+machine~$M$ computes on a given input.
 \end{explain}
 
 \begin{defn}[Configuration]
@@ -56,15 +56,14 @@
 \end{defn}
 
 \begin{explain}
-The~$\frown$ symbol is for \emph{concatenation}---we want to
-ensure that there are no blanks between the left end marker and
-the beginning of the input.
+The~$\frown$ symbol is for \emph{concatenation}---the input string
+begins immediately to the left end marker.
 \end{explain}
 
 \begin{defn}
 We say that a configuration $\tuple{C, m, q}$ \emph{yields the
-  configuration $\tuple{C', m', q'}$ in one step} (according to~$M$),
-  iff
+configuration $\tuple{C', m', q'}$ in one step} (according to~$M$),
+iff
 \begin{enumerate}
 \item the $m$-th symbol of $C$ is $\sigma$,
 \item the instruction set of $M$ specifies $\delta(q, \sigma) =
@@ -82,24 +81,24 @@
 \end{enumerate}
 \end{defn}
 
-\begin{defn}
+\begin{defn}\ollabel{defn:run-output}
 A \emph{run of $M$ on input~$I$} is a sequence $C_i$ of configurations
 of $M$, where $C_0$ is the initial configuration of $M$ for input~$I$,
 and each $C_i$ yields $C_{i+1}$ in one step.
 
 We say that $M$ \emph{halts on input $I$ after $k$ steps} if $C_k =
-\tuple{C, m, q}$, the $m$th symbol of $C$ is~$\sigma$, and $\delta(q,
+\tuple{C, m, q}$, the $m$th symbol of~$C$ is~$\sigma$, and $\delta(q,
 \sigma)$ is undefined.  In that case, the \emph{output} of~$M$ for
-input $I$ is~$O$, where $O$ is a string of symbols not beginning or
-ending in~$\TMblank$ such that $C = \TMendtape \concat \TMblank^i
-\concat O \concat \TMblank^j$ for some~$i, j \in \Nat$.
+input~$I$ is~$O$, where $O$ is a string of symbols not ending
+in~$\TMblank$ such that $C = \TMendtape \concat O \concat \TMblank^j$
+for some~$i, j \in \Nat$.
 \end{defn}
 
 \begin{explain}
-According to this definition, the output~$O$ of~$M$ always begins and
+According to this definition, the output~$O$ of~$M$ always 
 ends in a symbol other than~$\TMblank$, or, if at time~$k$ the entire
 tape is filled with~$\TMblank$ (except for the leftmost~$\TMendtape$),
-$O$ is the empty string.
+$O$~is the empty string.
 \end{explain}
 
 \end{document}

content/turing-machines/machines-computations/disciplined-machines.tex

@@ -0,0 +1,90 @@
+% Part: turing-machines
+% Chapter: machines-computations
+% Section: well-behaved-machines
+
+\documentclass[../../../include/open-logic-section]{subfiles}
+
+\begin{document}
+
+\olfileid{tur}{mac}{dis}
+\olsection{Disciplined Machines}
+
+\begin{explain}
+In section \olref[hal]{sec}, we considered Turing machines that have a
+single, designated halting state~$h$---such machines are guaranteed to
+halt, if they halt at all, in state~$h$.  In this way, machines with a
+single halting state are more ``disciplined'' than we allow Turing
+machines in general to be.  There are other restrictions we might
+impose on the behavior of Turing machines.  For instance, we also have
+not prohibited Turing machines from ever erasing the tape-end marker
+on square~$0$, or to attempt to move left from square~$0$. (Our
+definition states that the head simply stays on square~$0$ in this
+case; other definitions have the machine halt.) It is likewise
+sometimes desirable to be able to assume that a Turing machine, if
+it halts at all, halts on square~$1$.
+\end{explain}
+
+\begin{defn}\ollabel{defn:disciplined}
+A Turing machine~$M$ is \emph{disciplined} iff
+\begin{enumerate}
+    \item it has a designated single halting state~$h$,
+    \item it halts, if it halts at all, while scanning square~$1$,
+    \item it never erases the $\TMendtape$ symbol on square~$0$, and
+    \item it never attempts to move left from square~$0$.
+\end{enumerate}
+\end{defn}
+
+\begin{explain}
+We have already discussed that any Turing machine can be changed into
+one with the same behavor but with a designated halting state. This is
+done simply by adding a new state~$h$, and adding an instruction
+$\delta(q, \sigma) = \tuple{h, \sigma, N}$ for any pair
+$\tuple{q,\sigma}$ where the original $\delta$~is undefined.  It is
+true, although tedious to prove, that any Turing machine~$M$ can be
+turned into a disciplined Turing machine~$M'$ which halts on the same
+inputs and produces the same output. For instance, if the Turing
+machine halts and is not on square~$1$, we can add some instructions
+to make the head move left until it finds the tape-end marker, then
+move one square to the right, then halt.  We'll leave you to think
+about how the other conditions can be dealt with.
+\end{explain}
+
+\begin{ex}
+    In \olref{fig:adder-disc}, we turn the addition machine
+    from \olref[una]{ex:adder} into a disciplined machine.
+    \begin{figure}\[
+\begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=2.8cm,
+                    semithick]
+  \tikzstyle{every state}=[fill=none,draw=black,text=black]
+
+  \node[initial,state]         (A)              {$q_0$};
+  \node[state]         (B) [right of=A] {$q_1$};
+  \node[state]         (C) [below right of=B] {$q_2$};
+  \node[state]         (D) [below left of=C] {$q_3$};
+  \node[state]         (H) [left of=D] {$h$};
+
+  \path (A) edge node {\TMtrans{\TMblank}{\TMstroke}{\TMstay}} (B)
+           edge [loop below] node {\TMtrans{\TMstroke}{\TMstroke}{\TMright}} (B)
+        (B) edge [loop below] node {\TMtrans{\TMstroke}{\TMstroke}{\TMright}} (B)
+            edge node {\TMtrans{\TMblank}{\TMblank}{\TMleft}} (C)
+        (C) edge node {\TMtrans{\TMstroke}{\TMblank}{\TMleft}} (D)
+        (D) edge [loop above] node {\TMtrans{\TMstroke}{\TMstroke}{\TMleft}} (D)
+        edge node {\TMtrans{\TMendtape}{\TMendtape}{\TMright}} (H);
+\end{tikzpicture}
+\]\caption{A disciplined addition machine}
+\ollabel{fig:adder-disc}
+\end{figure}
+\end{ex}
+
+\begin{prop}\ollabel{prop:disciplined} For evevery Turing machine~$M$,
+there is a disciplined Turing machine~$M'$ which halts with output~$O$
+if $M$~halts with output~$O$, and does not halt if $M$~does not halt.
+In particular, and function $f\colon\Nat^n \to \Nat$ computable by a
+Turing machine is computable by a disciplined Turing machine.
+\end{prop}
+
+\begin{prob}\label{tur:mac:dis:prob:disc-succ}
+Give a disciplined machine that computes $f(x) = x+1$.
+\end{prob}
+
+\end{document}

content/turing-machines/machines-computations/halting-states.tex

@@ -6,27 +6,27 @@
 
 \begin{document}
 
-\olfileid{tur}{tur}{tur}
+\olfileid{tur}{mac}{hal}
 \olsection{Halting States}
 
 \begin{explain}
 Although we have defined our machines to halt only when there
 is no instruction to carry out, common representations of Turing
-machines have a dedicated \emph{halting state}, $h$, such that
+machines have a dedicated \emph{halting state}~$h$, such that
 $h \in Q$.
 
 The idea behind a halting state is simple: when the machine has
-finished operation (it is ready to accept input, or has finished writing
-the output), it goes into a state $h$ where it halts. Some
-machines have two halting states, one that accepts input and one
-that rejects input. 
+finished operation (it is ready to accept input, or has finished
+writing the output), it goes into a state~$h$ where it halts. Some
+machines have two halting states, one that accepts input and one that
+rejects input.
 \end{explain}
 
 \begin{ex}\emph{Halting States}.
 To elucidate this concept, let us begin with an alteration of the
-even machine. Instead of having the machine halt in state $q_0$
+even machine. Instead of having the machine halt in state~$q_0$
 if the input is even, we can add an instruction to send the machine
-into a halt state.
+into a halting state.
 \[
 \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=2.8cm,
                     semithick]

content/turing-machines/machines-computations/machines-computations.tex

@@ -19,6 +19,8 @@
 
 \olimport{halting-states}
 
+\olimport{disciplined-machines}
+
 \olimport{combining-machines}
 
 \olimport{variants}

content/turing-machines/machines-computations/representing-tms.tex

@@ -6,7 +6,7 @@
 
 \begin{document}
 
-\olfileid{tms}{tms}{rep}
+\olfileid{tur}{mac}{rep}
 \olsection{Representing Turing Machines}
 
 \begin{explain}
@@ -195,7 +195,7 @@
 \multicolumn{1}{l|}{\phantom{\TMtrans{\TMstroke}{q_1}{\TMright}}} &  \\ \cline{1-4}
 \end{tabular}
 \]
-As we can see, the machine halts when scanning a blank in state~$q_0$.
+As we can see, the machine halts when scanning a~$\TMblank$ in state~$q_0$.
 \end{ex}
 
 \begin{explain}
@@ -256,11 +256,11 @@
 
 \begin{prob}
 Choose an arbitary input and trace through the configurations of the
-doubler machine in \olref[tms][tms][rep]{ex:doubler}.
+doubler machine in \olref[tur][mac][rep]{ex:doubler}.
 \end{prob}
 
 \begin{prob}
-The double machine in \olref[tms][tms][rep]{ex:doubler} writes its
+The double machine in \olref[tur][mac][rep]{ex:doubler} writes its
 output to the right of the input.  Come up with a new method for
 solving the doubler problem which generates its output immediately to
 the right of the end-of-tape marker. Build a machine that executes

content/turing-machines/machines-computations/turing-machines.tex

@@ -41,7 +41,7 @@
 machine programs to tell when they're ``in danger'' of running off the
 tape. We could assume that this symbol is never overwritten, i.e.,
 that $\delta(q,\TMendtape) = \tuple{q', \TMendtape, x}$ if
-$\delta(q,\TMendtape)$ is defined. Some textbook do this, we do not.
+$\delta(q,\TMendtape)$ is defined. Some textbooks do this, we do not.
 You can simply be careful when constructing your Turing machine that
 it nevery overwrites~$\TMendtape$.  Moreover, there are cases where
 allowing such overwriting provides some convenient flexibility.