====== PCP is undecidable ======

In this section, we will prove that Post's Correspondence Problem is undecidable. Compared with the other proofs of this Chapter, PCP undecidability will be more involved. Basically, the proof is a two-step reduction from the **halting problem**.

===== Outline of the proof =====

The core idea of the proof is to construct a set of pairs $math[\alpha_i, \beta_i], such that **adding a pair to a partial matching** corresponds to a **transition of the Turing Machine**.

Informally, a word $math[\beta_i] will describe a **Turing Machine configuration**, i.e. the current state, the head position, and the current word on the tape. 

The bottom partial matching will always be longer than the upper one. 

A complete matching will represent **a sequence of configurations** of the Turing Machine, starting from the initial state, and ending in a final (accepting) state.

In order to make the proof work, we need a few technical fixes:

  * we need to restrict PCP in order to force **the selection of the first tile**, for any possible matching 
  * we need to add some (minor) restrictions to the Turing Machines which are the input of the halting problem.

We shall describe these restrictions below:

===== Modifications of the Turing Machine =====

  * for this proof, we consider that Turing Machines have only two final states $math[F=\{s_{yes},s_{no}\}], which model the output of $math[0/1]. It is straightforward how one can take an arbitrary Turing Machine an turn it into a //yes/no//-Turing Machine.
  * We require that the execution of $math[M] on $math[w] never moves the head to the left of the first symbol of the word. Making sure this happens is a little more technically involved, however algorithmically possible. We shall skip these details.
  * We require that the word $ w \neq \epsilon $, hence we do not allow simulations of the empty word. To accommodate for this, we can easily add a new symbol to the alphabet, which is used specifically for encoding the empty word.


===== The problem FTPCP =====
$def[FTPCP]
Let $math[\alpha_1, \ldots, \alpha_n$ and $\beta_1, \ldots, \beta_n] be sequences of words over a fixed alphabet. There exists a finite sequence $math[a_1a_2 \ldots a_k], with $math[a_1 = 1] and $math[a_i = 1, \ldots, n] such that:

$math[\alpha_{a_1}\alpha_{a_2}\ldots \alpha_{a_k} = \beta_{a_1}\beta_{a_2}\ldots \beta_{a_k}]
$end

===== FTPCP is undecidable =====

We show $math[f_h\leq_T] FTPCP.

Let $math[M] and $math[w=c_1\ldots c_n] be an input of the halting problem. We construct a matrix of $math[\alpha_i / \beta_i] pairs such that a match exists for FTPCP iff $math[M(w)] halts.

**Step 1 (the first pair)**:

$math[\left(\begin{array}{l} \alpha_1 \\ \beta_1 \end{array}\right) = \left(\begin{array}{l} \# \\ \# s_0c_1\ldots c_n \# \end{array}\right)] 
  * The first tile must correspond to the initial configuration of $math[M]. Note that the alphabet of PCP is $math[\Sigma\cup K] - it also contains the states of $math[M] as symbols.
  * Also, the position of the state symbol is used to **represent the head position**. In the initial state, this points to the first symbol of the input word, namely $math[c_1].

**Example**

Suppose we have $math[\Sigma = \{\#,0,1\}] and the input word $math[w=100]. Then, any matching of FTPCP must start as follows:

$math[\left(\begin{array}{l} \# \\ \# s_0100 \# \end{array}\right)] 


** Step 2 (//moving right// transitions)**:

For every pair of symbols $math[c,c'\in \Sigma] and states $math[s,s' \in K] where $math[s\neq s_{no}] such that $math[\delta(s,c)=(s',c',R)], construct the tile:

$math[ \left(\begin{array}{l} sc \\ c's' \end{array}\right)]
  * when a matching with $math[sc] is performed, this indicates that the current state is $math[s], and the head position indicates $math[c].

** Example **
Suppose in our machine we have the transition $math[\delta(s_0,1)=(s_1,0,R)]. Then we can continue our matching as follows:

$math[\left(\begin{array}{l} \#s_01 \\ \# s_0100 \# 0s_1 \end{array}\right)] 

Note that on the bottom part, we have partially constructed the next configuration of the Turing Machine. The current state is now $math[s_1] and the head position indicates $math[0]. We need the following ingredients to obtain the complete configuration description in the bottom part:

** Step 3 (completing configurations) **

For every $math[c\in \Sigma], construct the tile:
$math[ \left(\begin{array}{l} c \\ c \end{array}\right)]

** Example **

We can now continue our matching by selecting tiles from Step 3. We obtain:

$math[\left(\begin{array}{l} \#s_0100\# \\ \# s_0100 \# 0s_100\# \end{array}\right)] 

** Step 4 (//moving left// transitions) **

For every $math[c,c',c''\in \Sigma] and every $math[s,s'\in K] where $math[s\neq s_{no}] such that: $math[\delta(s,c) = (s',c',L)], add tiles:

$math[ \left(\begin{array}{l} c''sc \\ s'c''c' \end{array}\right)]

  * Unlike the previous step, where we add one tile per transition, here we add $math[|\Sigma|] tiles. The intuition is as follows: for each //left// transition, and **each possible //previous// symbol** $math[c''], we add the corresponding transition.

** Example **

Continuing the previous example, suppose we have transition $math[\delta(s_1,0) = (s_2,1,L)], We extend the matching by selecting the appropriate tile with $math[c''=0]:

$math[\left(\begin{array}{l} \#s_0100\#0s_10 \\ \# s_0100 \# 0s_100\#s_201 \end{array}\right)] 

and by completing the configuration, we get:

$math[\left(\begin{array}{l} \#s_0100\#0s_100\# \\ \# s_0100 \# 0s_100\#s_2010\# \end{array}\right)] 

** Step 5: (//hold transitions//) **

Can you figure this step by yourself?


** Step 6: (//completing a match//) **

For every $math[c\in\Sigma\setminus\{\#\}], we add tiles:

$math[ \left(\begin{array}{l} cs_{yes} \\ s_{yes} \end{array}\right)], $math[ \left(\begin{array}{l} s_{yes}c \\ s_{yes} \end{array}\right)] 

  * these pairs of tiles //consume// the rest of the input (from both the left and right-side of the head), when the //yes//-state is reached.

** Example **

Suppose that, in our previous example $math[s_2 = s_{yes}]. Then we can attempt to complete the matching as follows:

$math[\left(\begin{array}{l} \#s_0100\#0s_100\# \\ \# s_0100 \# 0s_100\#s_{yes}010\# \end{array}\right)] 

$math[\left(\begin{array}{l} \#s_0100\#0s_100\# s_{yes}0 \\ \# s_0100 \# 0s_100\#s_{yes}010\# s_{yes} \end{array}\right)] 

$math[\left(\begin{array}{l} \#s_0100\#0s_100\# s_{yes}010\# \\ \# s_0100 \# 0s_100\#s_{yes}010\# s_{yes}10\# \end{array}\right)] 

$math[\left(\begin{array}{l} \#s_0100\#0s_100\# s_{yes}010\# s_{yes}1 \\ \# s_0100 \# 0s_100\#s_{yes}010\# s_{yes}10\# s_{yes} \end{array}\right)] 

$math[\left(\begin{array}{l} \#s_0100\#0s_100\# s_{yes}010\# s_{yes}10\# \\ \# s_0100 \# 0s_100\#s_{yes}010\# s_{yes}10\# s_{yes}0\# \end{array}\right)] 

$math[\left(\begin{array}{l} \#s_0100\#0s_100\# s_{yes}010\# s_{yes}10\# s_{yes}0 \\ \# s_0100 \# 0s_100\#s_{yes}010\# s_{yes}10\# s_{yes}0\# s_{yes} \end{array}\right)] 

$math[\left(\begin{array}{l} \#s_0100\#0s_100\# s_{yes}010\# s_{yes}10\# s_{yes}0\# \\ \# s_0100 \# 0s_100\#s_{yes}010\# s_{yes}10\# s_{yes}0\# s_{yes}\# \end{array}\right)] 

** Step 7: (final touches)**

What more is needed ?

{{## the tile syes diez diez / diez ##}}

===== PCP is undecidable =====
We prove PCP is undecidable by showing FTPCP $math[\leq_T] PCP.
Let $math[ \left\{\left(\begin{array}{l} \alpha_i \\ \beta_i \end{array}\right)\right\}_{ 1 \leq i \leq n}] be an instance of FTPCP. To build an instance of PCP, we need the following ingredients.

Consider an arbitrary word $math[w=c_1\ldots c_k] over alphabet $math[\Sigma], and the symbol $math[*] which does not occur in $math[\Sigma]. We define the words:
  * $math[\lhd w = *c_1*c_2* \ldots *c_k]
  * $math[w \rhd = c_1*c_2* \ldots *c_k*]
  * $math[\lhd w \rhd = *c_1*c_2* \ldots *c_k*]
by inserting $math[*] to the left, right and left & right of every symbol in $math[w].

Using our word constructions, we can construct the following instance of PCP:
$math[ \left\{\left(\begin{array}{l} \lhd \alpha_i \\ \lhd \beta_i \rhd \end{array}\right)\right\} \cup \left\{\left(\begin{array}{l} \lhd \alpha_i \\ \beta_i \rhd \end{array}\right)\right\}_{ 1 \leq i \leq n} \cup \left\{\left(\begin{array}{l} \lhd @ \\ @ \end{array}\right)\right\}]

where $math[*] and $math[@] are symbols outside of the alphabet of the PCP instance.

We prove the construction is correct. Direction "$math[\Rightarrow]". Suppose there is a matching of FTPCP starting with the first tile and continuing with $math[a_2,\ldots,a_k]. Then choose the pair $math[\left(\begin{array}{l} \lhd \alpha_1 \\ \lhd \beta_1 \rhd \end{array}\right)], and continue with the pairs $math[a_2=\left(\begin{array}{l} \lhd \alpha_{a_2} \\ \beta_{a_2} \rhd \end{array}\right)] up to $math[a_k], which will constitute a matching.

Direction "$math[\Leftarrow]". Suppose we have a matching in our constructed PCP instance. Since $math[*] can only appear between other symbols from $math[\Sigma], it follows that any matching can only start with the pair $math[\left(\begin{array}{l} \lhd \alpha_1 \\ \lhd \beta_1 \rhd \end{array}\right)], and must continue with pairs $math[\left(\begin{array}{l} \lhd \alpha_i \\ \beta_i \rhd \end{array}\right)], from which we can immediately construct a matching in FTPCP.