# Turing Machine for comparing, copying, and operating

If one wants to design a Turing Machine for a function such as this:

Where $x>0,y>0$ and are both integers represented in unary, so an example movement in this TM on the read-write head would be like:

But to draw the transition graph for a TM for this function, that's where I'm getting lost. I'm not sure how one would start at $q_0$ and end at $q_f$ for a TM like this. Maybe someone can help me understand this better.

EDIT:

Someone with great math skills gave an amazing answer below. I had independently come up with my own (possible) solution at the same time. It was a headache-inducing few hours of playing with the nodes in Tikz/Latex to get everything to basically just not overlap over each other too much. I say 'possible' because it was quite long, and it would be very easy for me to have missed some steps or made mistakes in various places.

Here's what I did for the conditional and the case that $x \leq y$ --

And here's what I did for the other case --

The solution is split between 2 images because it’s too big to show one page. The 1st part has the conditional decision process that determines whether (A) x>y, (B) x

NOTE: Due to the large size and number of nodes/edges in this graph, I wrote a program in Latex/MathJax/Tikz to allow me to illustrate this better (clearer and easier) than anything I could do with my typically messy handwriting. However, there may be 1 or 2 transitions that have an edge passing close to a value, I didn’t see many. However, I noticed 1 that might be difficult to read in the 1st part, the edge going from q6 to q8 has the value ‘blank,blank,R’.

1st part:

The conditional works by turning all the 1’s into x’s, then turning each x back into a 1 only if there is a corresponding 1 on the other side of the 0. So this compares the x-value on the left side of the 0 with the y-value on the right side of the 0.

The section of this Turing machine that deals with whether B or C occurred, works by going to the middle and erasing everything to the left of the 0 (erasing the x-value), then erasing the 0, then moves all the way to the right and writes a 0, then goes back to the leftmost 1 so that it points at the beginning.

2nd part:

This part of this Turing Machine starts where q13 left off in the last part of the graph. This part works by copying the y-value and then adds that to the x-value by removing any 0’s remaining between the 1’s. The result is then the final state for x+2y.

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Do you have Turing machines for multiplication, addition, and conditionals involving inequalities already, or are you asking for help constructing these as well? Because if you have the simpler machines, the machine that computes the indicated function can be pieced together from them. – portin.daniel Nov 30 '13 at 20:16
I have some transition graphs, but they are very basic and won't work here. I get the delta transitions, but the graph is different, and I'm not sure how the RW head would be moving around in this case. – stackuser Nov 30 '13 at 20:38
+1 for the Tikz. I love the automata package, though I think the default loops and arrowheads are a bit too small. – Rick Decker Dec 2 '13 at 3:40

Suppose that we have Turing machines $M_1$ and $M_2$ that compute the number-theoretic functions $f_1$ and $f_2$, respectively. Then we can construct the Turing machine $M_2 \circ M_1$ that computes the number-theoretic function $f_2 \circ f_1$, assuming appropriate restrictions on the domain and range of $f_1$ and $f_2$. This is done by first computing $f_1(n)$ with machine $M_1$, then computing $f_2(f_1(n))$ using $M_2$. So we see that computing the composition of number-theoretic functions corresponds to performing sequential computations of Turning machines. I'll sketch your Turing machine below.

Let's stipulate that a Turing machine that computes a number-theoretic function $n \mapsto f(n)$ accepts $u(n)$ as input and terminates with $u(f(n))$ printed at the beginning of the tape with the tape head at initial position, where $u(n)$ denotes the representation of $n$ in unary. A Turing machine that computes the number theoretic function $(n,m) \mapsto f(n,m)$ is similarly defined. The first problem is to find a Turing machine that computes $(n,m) \mapsto n + 2m$ and $(n,m) \mapsto m$. The first Turing machine can be constructed from Turing machines that compute $(n,m) \mapsto n + m$ and $m \mapsto 2m$, which in turn can be constructed from Turing machines that compute $(n,m) \mapsto nm$ and $n \mapsto 2$.

The second problem can be divided into two parts. First, we need a Turing machine that computes $(n,m) \mapsto 1$ if $n \leq m$ and $(n,m) \mapsto 0$ if $n > m$. Let's agree to call this function $le$. Then, we need a Turing machine that runs a Turing machine $M_1$ if its input is $u(1)$ and a Turing machine $M_2$ if its input is $u(0)$. If we piece together these two machines appropriately, we can construct a Turing machine that computes $(n,m) \mapsto n + 2m$ if $le(n,m) = 1$ and $(n,m) \mapsto m$ if $le(n,m) = 0$. Note that you will need to either make clever use of the constant and projection functions, or else save the input $u(n,m)$ somewhere, in order to run $M_1$ and $M_2$ on $u(n,m)$. If you need some help constructing any of the particular machines, please leave a comment. In any case, hopefully this post will give you some ideas on how to go about constructing more complicated Turing machines.

Edit. Here's how I would go about solving the second part of the problem, namely constructing a machine $M_1$ that computes the function $(n,m) \mapsto (n,m,\mathbf{1})$ if $n \leq m$ and $(n,m) \mapsto (n,m,\mathbf{0})$ if $n > m$, where $\mathbf{1}$ and $\mathbf{0}$ are special symbols that indicate a positive or negative answer to the question whether $n \leq m$. Because $n$ and $m$ are represented in unary, we know that $n \leq m$ if and only if $\operatorname{len} u(n) \leq \operatorname{len} u(m)$. Determining whether this is the case is the job of the inner loop, consisting of states $q_1$, $q_2$, $q_3$, $q_4$ and $q_5$. The inner loop compares the first nonzero symbol of $u(n)$ with the first nonzero symbol of $u(m)$, marking each considered symbol with $0$. It does this until either $u(n)$ or $u(m)$ have no nonzero symbols. If we run out of nonzero symbols in $u(n)$ to compare with nonzero symbols in $u(m)$, then we know that $\operatorname{len} u(n) \leq \operatorname{len} u(m)$. But if we run out of nonzero symbols in $u(m)$ to compare with nonzero symbols in $u(n)$, then we know that $\operatorname{len}u(m) < \operatorname{len} u(n)$. The transition to state $q_y$ indicates that the former possibility occurs, while the transition to the state $q_n$ indicates the latter. After transitioning to either $q_y$ or $q_n$, we print either $\mathbf{1}$ or $\mathbf{0}$ at the end of the tape, and convert all the $0$ symbols back to $1$, returning to the start.

Here are three examples of runs in $M_1$ on inputs $u(2,2)$, $u(2,3)$ and $u(3,2)$. We see that $M_1$ outputs $u(2,2,\mathbf{1})$ in the first case, $u(2,3,\mathbf{1})$ in the second, and $u(3,2,\mathbf{0})$ in the third.

\begin{align} \underline{\text{B}}111\text{B}111\text{B} & \vdash^1 \text{B}\underline{1}11\text{B}111\text{B} && q_0 \rightarrow^1 q_1 \\ & \vdash^1 \text{B}0\underline{1}1\text{B}111\text{B} && q_1 \rightarrow^1 q_2 \\ & \vdash^* \text{B}011\text{B}\underline{1}11\text{B} && q_2 \rightarrow^* q_3 \\ & \vdash^1 \text{B}011\underline{\text{B}}011\text{B} && q_3 \rightarrow^1 q_4 \\ & \vdash^* \text{B}0\underline{1}1\text{B}011\text{B} && q_4 \rightarrow^* q_1 \\ & \vdash^1 \text{B}00\underline{1}\text{B}011\text{B} && q_1 \rightarrow^1 q_2 \\ & \vdash^* \text{B}001\text{B}0\underline{1}1\text{B} && q_2 \rightarrow^* q_3 \\ & \vdash^1 \text{B}001\text{B}\underline{0}01\text{B} && q_3 \rightarrow^1 q_4 \\ & \vdash^* \text{B}00\underline{1}\text{B}001\text{B} && q_4 \rightarrow^* q_1 \\ & \vdash^1 \text{B}000\underline{\text{B}}001\text{B} && q_2 \rightarrow^1 q_2 \\ & \vdash^* \text{B}000\text{B}00\underline{1}\text{B} && q_2 \rightarrow^* q_3 \\ & \vdash^1 \text{B}000\text{B}0\underline{0}0\text{B} && q_3 \rightarrow^1 q_4 \\ & \vdash^* \text{B}000\underline{\text{B}}000\text{B} && q_4 \rightarrow^* q_1 \\ & \vdash^1 \text{B}000\text{B}\underline{0}00\text{B} && q_1 \rightarrow^1 q_y \\ & \vdash^* \underline{\text{B}}111\text{B}111\text{B}1\text{B} && q_y \rightarrow^* q_f \\ \end{align}

\begin{align} \underline{\text{B}}111\text{B}1111\text{B} & \vdash^1 \text{B}\underline{1}11\text{B}1111\text{B} && q_0 \rightarrow^1 q_1 \\ & \vdash^1 \text{B}0\underline{1}1\text{B}1111\text{B} && q_1 \rightarrow^1 q_2 \\ & \vdash^* \text{B}011\text{B}\underline{1}111\text{B} && q_2 \rightarrow^* q_3 \\ & \vdash^1 \text{B}011\underline{\text{B}}0111\text{B} && q_3 \rightarrow^1 q_4 \\ & \vdash^* \text{B}0\underline{1}1\text{B}0111\text{B} && q_4 \rightarrow^* q_1 \\ & \vdash^1 \text{B}00\underline{1}\text{B}0111\text{B} && q_1 \rightarrow^1 q_2 \\ & \vdash^* \text{B}001\text{B}0\underline{1}11\text{B} && q_2 \rightarrow^* q_3 \\ & \vdash^1 \text{B}001\text{B}\underline{0}011\text{B} && q_3 \rightarrow^1 q_4 \\ & \vdash^* \text{B}00\underline{1}\text{B}0011\text{B} && q_4 \rightarrow^* q_1 \\ & \vdash^1 \text{B}000\underline{\text{B}}0011\text{B} && q_1 \rightarrow^1 q_2 \\ & \vdash^* \text{B}000\text{B}00\underline{1}1\text{B} && q_2 \rightarrow^* q_3 \\ & \vdash^1 \text{B}000\text{B}0\underline{0}01\text{B} && q_3 \rightarrow^1 q_4 \\ & \vdash^* \text{B}000\underline{\text{B}}0001\text{B} && q_4 \rightarrow^* q_1 \\ & \vdash^* \text{B}000\text{B}\underline{0}001\text{B} && q_1 \rightarrow^1 q_y \\ & \vdash^* \underline{\text{B}}000\text{B}0001\text{B}\mathbf{1}\text{B} && q_y \rightarrow^* q_f \\ \end{align}

\begin{align} \underline{\text{B}}1111\text{B}111\text{B} & \vdash^1 \text{B}\underline{1}111\text{B}111\text{B} && q_0 \rightarrow^1 q_1 \\ & \vdash^1 \text{B}0\underline{1}11\text{B}111\text{B} && q_1 \rightarrow^1 q_2 \\ & \vdash^* \text{B}0111\text{B}\underline{1}11\text{B} && q_1 \rightarrow^* q_3 \\ & \vdash^1 \text{B}0111\underline{\text{B}}011\text{B} && q_3 \rightarrow_1 q_4 \\ & \vdash^* \text{B}0\underline{1}11\text{B}011\text{B} && q_4 \rightarrow^* q_1 \\ & \vdash^1 \text{B}00\underline{1}1\text{B}011\text{B} && q_1 \rightarrow_1 q_2 \\ & \vdash^* \text{B}0011\text{B}0\underline{1}1\text{B} && q_2 \rightarrow^* q_3 \\ & \vdash^1 \text{B}0011\text{B}\underline{0}01\text{B} && q_3 \rightarrow^1 q_4 \\ & \vdash^* \text{B}00\underline{1}1\text{B}001\text{B} && q_4 \rightarrow^* q_1 \\ & \vdash^1 \text{B}000\underline{1}\text{B}001\text{B} && q_1 \rightarrow^1 q_2 \\ & \vdash^* \text{B}0001\text{B}00\underline{1}\text{B} && q_2 \rightarrow^* q_3 \\ & \vdash^1 \text{B}0001\text{B}0\underline{0}0\text{B} && q_2 \rightarrow_1 q_4 \\ & \vdash^* \text{B}000\underline{0}\text{B}000\text{B} && q_4 \rightarrow^* q_1 \\ & \vdash^1 \text{B}0000\underline{\text{B}}000\text{B} && q_1 \rightarrow^1 q_2 \\ & \vdash^* \text{B}0000\text{B}000\underline{\text{B}} && q_2 \rightarrow^* q_3 \\ & \vdash^1 \text{B}0000\text{B}000\text{B}\underline{\text{B}} && q_3 \rightarrow^1 q_n \\ & \vdash^* \underline{\text{B}}0000\text{B}000\text{B}\mathbf{0}\text{B} && q_n \rightarrow^* q_f \\ \end{align}

Next, we construct a machine $M_2$ that accepts as input a representation of $(n,m,\alpha)$ where $\alpha \in \{\mathbf{1},\mathbf{0}\}$, erases the input $\alpha$, and then returns to initial position. However, if $\alpha$ was $\mathbf{1}$, then $M_2$ returns to initial position ready to perform computations in some Turing machine $M_3$, and if $\alpha$ was $\mathbf{0}$, then $M_2$ returns to initial position ready to perform computations in some Turing machine $M_4$.

In practice, we would combine the operations of $M_1$ and $M_2$, so as not to waste space when drawing diagrams. However, having macro operations that can be easily combined is quite useful. If you ran the machine $M_2$ on the output from $M_1$, then you would be ready to perform computations in some Turing machine $M_3$ if $n \leq m$ and, conversely, in some other machine if $m < n$. I've belabored the point by making these machines separate in order to demonstrate the logic behind them. It's up to you to make the machines more efficient, and to design them to the task at hand, which is presumably what your instructor wants. I hope this helps!

Edit. For the first part of the problem, consider the fact that you're representing natural numbers in unary. This means that you can convert arithmetic operations on numbers to relatively simple string operations. For instance, consider the function $(n,m) \mapsto n + m$. Assuming you have tape $\text{B}11\text{B}111\text{B}$ representing an input of $(1,2)$, you want to construct the tape $\text{B}1111\text{B}$ representing $3$ as output. The idea will be to move the string representation of the second number over, concatenating it with the first string, ten adjusting the length of the concatenated string by $1$. To compute the function $n \mapsto 2n$, you merely need to add $n$ to itself, which can be done using the machine computing the addition function.

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Hi, that's a great explanation for how the TM would overall look. Like I said, I'm lost specifically with regards to how the transition graph would look for this thing. I'm afraid to ask you for more help, I don't want you to be angry with me after the 3 paragraphs you wrote. However, as I said, I'm not finding any material to help me understand how the transition graph will be formed. If you have the patience for it, maybe you can consider helping me start with the conditional and then, next step after each value goes it's own way. – stackuser Nov 30 '13 at 22:08
I can add something more later, or perhaps someone else can give some help. But I won't be able to for several hours. So good luck! – portin.daniel Nov 30 '13 at 23:43
I added examples of machines that perform some of the computations in the second part of your problem in order to demonstrate the logic behind them. In practice, you will need to combine their functionality in order to produce something fitting the task at hand. – portin.daniel Dec 2 '13 at 0:35
I also added a note about arithmetic operations on natural numbers represented in unary to help with the first part of the problem. – portin.daniel Dec 2 '13 at 0:40
Please don't be angry but I had actually had a solution since this morning and spent the last few hours coding it with Tikz/Latex. Posted it in the question... I'm not sure what your $B$ is that you keep using in your transitions. Are you just literally writing the value $B$ in each box, or does it have some other meaning? – stackuser Dec 2 '13 at 3:22