Reducing Pcp (Post's correspondence problem) to mPcp Recently I have been studying Post's correspondence problem ($Pcp$), and I have stumbled upon a problem where I need to find a reduction from $Pcp$ to a modified version, $mPcp$. This modified version requires any matching sequence of "dominoes" $p_{i_1}, p_{i_2}, ..., p_{i_m}$ to start with the first "domino", $p_1$.

I am very well aware of the fact that there  exists  a  match  for
an  instance $I=\langle p_1, ..., p_n\rangle$ of $Pcp$  if  and  only  if  there  exists  a match  for  one  of  the  instances $I_1, I_2, ..., I_n$ of $mPcp$ where $I_j=\langle p_j, p_{j+1}, ..., p_n, p_1, ..., p_{j-1}\rangle$ for $j=1, ..., n$. In other words, if and only if $I$ is a yes-instance for $Pcp$, there is a way to rotate the list of "dominoes" in a cyclic manner such that the starting piece in the found match for $Pcp$ is put in the first position.

However, it is not clear to me how this observation could lead to a mapping from an instance $I$ of $Pcp$ to the "correct" instance $I_j$ of $mPcp$ that can be carried out in a finite amount of steps for any $I$, thus serving as a suitable reduction.

Could anybody point me in the right direction? Thank you very much.
http://infolab.stanford.edu/~ullman/ialc/slides/slides14.pdf

To summarise what is said in the attached pdf:

Let the dominos be (u1,v1), (u2,v2),...(un,vn).
(* is some symbol not already in the alphabet) For all the u's, put a * after them, and for all the v's, put a star before them, like such:
(u1*,v1), (u2,v2),...(un,*vn)

Now in the MPCP if we wanna start with the first term, we put a * before u1 like such:
(u1,v1), (u2,v2),...(un,*vn)

So now you the first domino MUST start, as it is the only one which has the first symbol matching. If there is a solution to this, it must start with the first domino. 

Now to make the endings match, we add a domino:
(@,*@)
where @ is anything not in the alphabet.

Now MPCP is yes iff PCP is yes.
Put the *s before (or after) each symbol, not the words. So for d=(01101,001) add (0*1*1*0*1*,*0*0*1) but no (01101*,*001). If d is a starting domino add  (*0*1*1*0*1*,*0*0*1), too.