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I am trying to attack TAOCP once again, given the sheer literal heaviness of the volumes I have trouble committing to it seriously. In TAOCP 1 Knuth writes, page 8, basic concepts::

Let $A$ be a finite set of letters. Let $A^*$ be the set of all strings in $A$ (the set of all ordered sequences $x_1$ $x_2$ ... $x_n$ where $n \ge 0$ and $x_j$ is in $A$ for $1 \le j \le n$). The idea is to encode the states of the computation so that they are represented by strings of $A^*$ . Now let $N$ be a non-negative integer and Q (the state) be the set of all $(\sigma, j)$, where $\sigma$ is in $A^*$ and j is an integer $0 \le j \le N$; let $I$ (the input) be the subset of Q with $j=0$ and let $\Omega$ (the output) be the subset with $j = N$. If $\theta$ and $\sigma$ are strings in $A^*$, we say that $\theta$ occurs in $\sigma$ if $\sigma$ has the form $\alpha \theta \omega$ for strings $\alpha$ and $\omega$. To complete our definition, let $f$ be a function of the following type, defined by the strings $\theta_j$, $\phi_j$ and the integers $a_j$, $b_j$ for $0 \le j \le N$:

  • $f((\sigma, j)) = (\sigma, a_j)$ if $\theta_j$ does not occur in $\sigma$
  • $f((\sigma, j)) = (\alpha \psi_j \omega, b_j)$ if $\alpha$ is the shortest possible string for which $\sigma = \alpha \theta_j \omega$
  • $f((\sigma,N)) = (\sigma, N)$

Not being a computer scientist, I have trouble grasping the whole passage. I kind of get the idea that is behind a system of opcodes, but I haven't progressed effectively in understanding. I think that the main problem is tat I don't know how to read it effectively.

Would it be possible to explain the passage above so that I can understand it, and give me a strategy in order to get in the logic in interpreting these statements?

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  • $\begingroup$ Then you should not include your commentary in the alleged quote, confusing anybody who does not have the book handy. -.- Hope my answer helps... $\endgroup$ – Raphael May 10 '12 at 13:41
  • $\begingroup$ @Raphael: the quote is verbatim from the book. I just added explanation in parentheses of the symbols for I and omega $\endgroup$ – Stefano Borini May 10 '12 at 14:18
  • $\begingroup$ @SteanoBorini: But it is not "explanation", it is wrong. I see how you can read the original text to come to the same conclusion you did, but it is still not helpful. If you say you cite something and add commentary, please mark it as such so people can take it with a grain of salt. $\endgroup$ – Raphael May 10 '12 at 14:27
  • $\begingroup$ There is context missing here: which computation and which states? $\endgroup$ – reinierpost Feb 12 '14 at 12:46
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We are missing some context so I have no idea what point Knuth is trying to make, but here's how to interpret a Turing machine this way. Perhaps it will help you understand what's going on. In general, a good way of getting a grip on a concept is playing with it. In the case of programming paradigms, that means writing a program. In this case, I will show how to write any program.

Suppose the tape of the Turing machine has symbols $\{0,1,\epsilon\}$ (where $\epsilon$ stands for "empty"), and add one more symbol which represents the location of the head $H$. Your states are going to be pairs of the form $(q,\alpha)$, where $q$ is a state of the Turing machine, and $\alpha \in \{0,\ldots,14\}$. We also identify $(F,0)$ with $N$ for any final state.

On (non-empty) input $x$, your starting point will be $(Hx,(s,0))$, where $s$ is the starting state. The difficult part is to encode states. Suppose that at state $q$, upon reading input $x$, you replace it with $a(q,x)$, move in direction $D(q,x) \in \{L,R\}$, and switch to state $\sigma(q,x)$. For the $\theta$s, we have $$ \begin{align*} \theta_{q,0} &= 0H0, & \theta_{q,1} &= 0H1, & \theta_{q,2} &= 0H\epsilon, \\ \theta_{q,3} &= 1H0, & \theta_{q,4} &= 1H1, & \theta_{q,5} &= 1H\epsilon, \\ \theta_{q,6} &= \epsilon H0 & \theta_{q,7} &= \epsilon H1, & \theta_{q,8} &= \epsilon H\epsilon, \\ \theta_{q,9} &= H0, & \theta_{q,10} &= H1, & \theta_{q,11} &= H\epsilon, \\ \theta_{q,12} &= 0H, & \theta_{q,13} &= 1H, & \theta_{q,14} &= \epsilon H. \end{align*} $$ For the $a$s, we have $a_{q,i} = (q,i+1)$ for $i < 14$, and $a_{q,14} = (q,14)$, though we should never really get that far. For the $b$s, we have $$ \begin{align*} &b_{q,0} = b_{q,3} = b_{q,6} = b_{q,9} = (\sigma(q,0),0), \\ &b_{q,1} = b_{q,4} = b_{q,7} = b_{q,10} = (\sigma(q,1),0), \\ &b_{q,2} = b_{q,5} = b_{q,8} = b_{q,11} = b_{q,12} = b_{q,13} = b_{q,14} = (\sigma(q,\epsilon),0). \end{align*} $$ Now it remains to determine the $\psi$s. Let $a_0 = a(q,0)$. If $D(q,0) = L$ then $$ \begin{align*} \psi_{q,0} &= H0a_0, & \psi_{q,3} &= H1a_0, & \psi_{q,6} &= \psi_{q,9} = H\epsilon a_0. \end{align*} $$ If $D(q,0) = R$ then $$ \begin{align*} \psi_{q,0} &= 0a_0H, & \psi_{q,3} &= 1a_0H, & \psi_{q,6} &= \epsilon a_0 H, & \psi_{q,9} &= a_0H\epsilon. \end{align*} $$ Next, let $a_1 = a(q,1)$. If $D(q,1) = L$ then $$ \begin{align*} \psi_{q,1} &= H0a_1, & \psi_{q,4} &= H1a_1, & \psi_{q,7} &= \psi_{q,10} = H\epsilon a_1. \end{align*} $$ If $D(q,1) = R$ then $$ \begin{align*} \psi_{q,1} &= 0a_1H, & \psi_{q,4} &= 1a_1H, & \psi_{q,7} &= \epsilon a_1 H, & \psi_{q,10} &= a_1 H\epsilon. \end{align*} $$ Finally, let $a_\epsilon = a(q,\epsilon)$. If $D(q,\epsilon) = L$ then $$ \begin{align*} \psi_{q,2} &= H0a_\epsilon, & \psi_{q,5} &= H1a_\epsilon, & \psi_{q,8} &= \psi_{q,11} = H\epsilon a_\epsilon, \\ \psi_{q,12} &= H0a_\epsilon, & \psi_{q,13} &= H1a_\epsilon, &\psi_{q,14} &= H\epsilon a_\epsilon. \end{align*} $$ If $D(q,\epsilon) = R$ then $$ \begin{align*} \psi_{q,2} &= 0a_\epsilon H, & \psi_{q,5} &= 1a_\epsilon H, & \psi_{q,8} &= \epsilon a_\epsilon H, & \psi_{q,11} &= a_\epsilon H\epsilon, \\ \psi_{q,12} &= 0a_\epsilon H, & \psi_{q,13} &= 1a_\epsilon H, & \psi_{q,14} &= \epsilon a_\epsilon H. \end{align*} $$

Now apply $f$ repeatedly until you get stuck. If you follow the construction, you will see that we have simulated the running of the Turing machine.

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  • $\begingroup$ understood: nothing. Not your fault. Thank you anyway :( $\endgroup$ – Stefano Borini May 5 '12 at 0:24
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    $\begingroup$ "We are missing some context." It's: we should have some precise description of what we mean by a 'method of computation'; here's one given by A.A. Markov; there are other equivalent ones, such as Turing machines. $\endgroup$ – rgrig May 6 '12 at 1:46
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Let us break it down bit by bit. First of all, remember what Knuth wrote on page 7:

Let us formally define a computational method to be a quadruple $(Q,I,\Omega,f)$, in which $Q$ is a set containing subsets $I$ and $\Omega$, and $f$ is a function from $Q$ into itself. [...] The four quantities $Q$, $I$, $\Omega$, $f$ are intended to represent repectively the state of the computation, the input, the output, and the computational rule.

This is the outline. You have to read "represent" as "contain"; $Q$ is going to contain states (some of which are in $I$, some are in $\Omega$) and $f$ is going to be a transition function between states; think of it as a program.

Let $A$ be a finite set of letters. Let $A^*$ be the set of all strings in $A$ (the set of all ordered sequences $x_1$ $x_2$ ... $x_n$ where $n \ge 0$ and $x_j$ is in $A$ for $1 \le j \le n$).

This is just a reiteration of what $A^*$ is. See also here.

The idea is to encode the states of the computation so that they are represented by strings of $A^*$.

This is probably the key sentence. We are talking about computations, that is execution sequences of some (programming language) statements which manipulate some state, which can be thought of as values in memory cells, or valuations of variables. Knuth says here that he wants to encode these states in an abstract way, namely as word over some alphabet.

Example: Consider a program that uses (at most) $k$ variables, each of which stores an integer. That is, a state is given by the tuple of values $(x_1, \dots, x_k)$ where $x_k$ is the (current) value of the $k$-th variable. In order to encode states of this form in a formal language, we can choose $A = \{0,1,\#\}$ with $\#$ a separator. Now model such a state by $\#\overline{x_1}\#\cdots\#\overline{x_k}\#$ where $\overline{x_i}$ is the binary encoding of $x_i$.

Specifically, $(3,5,0)$ would be $\#11\#101\#0\#$.

Now let $N$ be a non-negative integer and Q be the set of all $(\sigma, j)$, where $\sigma$ is in $A^*$ and j is an integer $0 \le j \le N$; let $I$ be the subset of Q with $j=0$ and let $\Omega$ be the subset with $j = N$.

You misquoted there (bad Stefano!); the parentheses are not in the original text, and they were misleading (see above). Knuth defines $Q$ here as the set of all possible states ($\sigma \in A^*$) at all possible places in the computation ($j$ can be understood as program counter). Therefore, $Q$ contains all statement-indexed states any computation of the algorithm given by $f$ can assume. By definition, we start with program counter $0$ and end in $N$, thus states indexed $0$ are input states and those indexed $N$ are output states.

If $\theta$ and $\sigma$ are strings in $A^*$, we say that $\theta$ occurs in $\sigma$ if $\sigma$ has the form $\alpha \theta \omega$ for strings $\alpha$ and $\omega$.

I hope that this is clear; it is just a (re)definition of substrings.

To complete our definition, let $f$ be a function of the following type, defined by the strings $\theta_j$, $\phi_j$ and the integers $a_j$, $b_j$ for $0 \le j \le N$:

  • $f((\sigma, j)) = (\sigma, a_j)$ if $\theta_j$ does not occur in $\sigma$
  • $f((\sigma, j)) = (\alpha \psi_j \omega, b_j)$ if $\alpha$ is the shortest possible string for which $\sigma = \alpha \theta_j \omega$
  • $f((\sigma,N)) = (\sigma, N)$

This is a a small programming language; if you fix $\theta_j, \psi_j, a_j, b_j$, you have a program. On program counter $j$, $f$ replaces the left-most occurrence $\theta_j$ in the state with $\psi_j$ and goes to statement $b_j$. If there is no $\theta_j$ in the current state, it goes to statement $a_j$. The program loops if statement $N$ is reached, modelling termination.

On the upper half of page 8, there is a more concrete example of a "program" $f$. Keep in mind that Knuth is going to use assembly language later on; this informs how he looks at programs (atomic statements connected by jumps).

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    $\begingroup$ Now I got a bit better understanding of what is going on. However, two things are still not clear and I would really appreciate if you could expand your answer. First, θj,ψj,aj,bj - what are these strings and numbers? What do they represent? If I understand correctly, aj and bj represent the step number or command counter for state j+1. But I am not sure what θj,ψj strings mean. Can you explain what do you mean by " if you fix θj,ψj,aj,bj, you have a program"? Or rather, how would I fix it for some example? $\endgroup$ – Georgy Bolyuba Jun 11 '12 at 19:38
  • $\begingroup$ @GeorgyBolyuba: You are right about $a_j$ and $b_j$. The program's state is a string $\sigma$ and a "program counter" $j$. $\theta_j$ and $\psi_j$ are used to modify that state (see second case of $f$). They can have all kinds of shapes; it really depends on how you encode state as a string. See the book for an example. $\endgroup$ – Raphael Jun 11 '12 at 23:24
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That text describes the following (Python) pseudocode:

subs = a list of string pairs  
As = a list of integers  
Bs = a list of integers

def f(state, pc):  
  if pc == N: return (state, pc)  
  if state.find(subs[pc][0]) != -1:  
    return (state.replace(subs[pc][0],subs[pc][1],1), Bs[pc])  
  else:  
    return (state,As[pc])  

The function f is presumably going to be applied repeatedly.

The last three bullet points is all you really need once you understand the notations. All that comes before is a bit analogous to explaining how Python works before giving the Python code.

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  • $\begingroup$ Ah ok, it's a Turing machine. $\endgroup$ – Stefano Borini May 5 '12 at 22:16
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    $\begingroup$ Rather, it is a different model of computation with the same power as a Turing machine. $\endgroup$ – Yuval Filmus May 6 '12 at 1:31
  • $\begingroup$ Well, three lines below your quote Knuth says that this is equivalent to Turing machines, so presumably you already knew this when you asked. I thought you were asking for help with the notation. Now I have no idea what is it that you wanted to ask. $\endgroup$ – rgrig May 6 '12 at 1:49

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