I'm confused about the Claim 1.6 in Computational Complexity A mordern approach (AB), which is :Claim 1.6 Define a single-tape Turing machine to be a TM that has only  one read-write tape, that is used as input, work, and output tape.For   every $f : \{0,1\}^* \to \{0,1\}$ and time-constructible $T : \mathbb N \to \mathbb N$, if $f$ is computable in time $T(n)$ by a TM $M$  using k tapes, then it is computable in time $5kT(n)^2$ by a  single-tape TM $\widetilde M$.PROOF SKETCH: Again the idea is simple: The TM $\widetilde M$ encodes  k tapes of M on a single tape by using locations $1,k+1,2k+1,...$ to  encode the first tape, locations $2,k+2,2k+2,...$ to encode the second  tape etc. For every symbol $a$ in M's alphabet, $\widetilde M$ will  contain both the symbol $a$ and the symbol $\hat a$. In the encoding  of each tape, exactly one symbol will be of the ^type, indicating  that the corresponding head of $M$ is positioned in that location.  $\widetilde M$ will not touch the first $n+1$ locations of its tape  (where the input is located) but rather start by taking $O(n^2)$ steps  to copy the input bit by bit into the rest of the tape, while encoding  it in the above way.To simulate one step of $M$, the machine $\widetilde M$ makes two  sweeps of its work tape: First it sweeps the tape in the left-to-right  direction and records to its register the $k$ symbols that are marked  by ^. Then $\widetilde M$ uses $M$'s transition function to  determine the new state, symbols, and  head movements and sweeps the  tape back in the right-to-left direction to update the encoding  accordingly. Clearly, $\widetilde M$ will have the same output as $M$.  Also, since on $n$-length inputs $M$ never reaches more than location  $T(n)$ of any of its tapes, $\widetilde M$ will never need to reach  more than location $2n+kT(n) \lt (k+2)T(n)$ of its work tape,meaning  that for each of the at most $T(n)$ steps of $M$, $\widetilde M$  performs at most $5 \cdot k \cdot T(n)$ work (sweeping back and forth  requires about $4 \cdot k \cdot T(n)$ steps, and some additional steps  may be needed for updating head movement and book keeping).My questions and guesses are shown below:$k$ includes input-tape, work-tapes,and output-tape,right?Why $\widetilde M$ will not touch the first $n+1$ locations of its tape after copy operations?About inequation $2n+kT(n) \lt (k+2)T(n)$, the $2n$ in it means we need introduce two new characters to indicate start and end of every k-length inputs, right?the time of sweeping tape in the left-to-right direction or right-to-left direction is $2n+kT(n)$ at most, to simulate one step of $M$, $\widetilde M$ should need $2 \cdot (2n+kT(n)$, or $2kT(n)$ after deleting $2n$, but rather than $4 \cdot k \cdot T(n)$. I don't know why?Any reply will be appriciated!
Transformation from $k$-tape Turing Machine in time $T(n)$ to single-tape Turing Machine in time $5kT(n)^2$ at most
complexity theory;turing machines;simulation