The proof of $P\subseteq P_{\\poly}$, Let $M$ is a Turing machine with $T(n)$ is running time and goal here is to design a boolean circuit of size $O(T(n))$ (for more detail see Arora and Barak page no -105 ).
The idea is that you can compute configuration $c_i$ from configuration $c_{i−1}$ by examining the content of 3 adjacent cells (i.e. by constant size circuit).
First Case : Let us consider that snapshot is the configuration of Turing machine i.e. requires $O(T(n))$ size.
Number of variables $x_{i,j}$, where $i$ represents a row in table and $j$ represent the colum( cook -levin reduction proof) are $T(n) \times T(n)$ many. Now when we combine the constant size circuits of all variables then size of the resultant circuit is going to be $T(n) \times T(n)$ ,which is polynomial but my goal was to design a circuit of size $O(T(n))$.
Second case : Let us consider that snapshot is machine's state and symbols read by all heads. let us consider they are $z_1,z_2 \cdots z_{T(n)}$ and size of each $z_i$ is constant string. To compute a snapshot $z_{i+1}$for this computation I need $z_i$, all old head positions and current head symbol .This all are constant size string, so I need only constant size circuit to encode the computation (current head position only depends upon the three symbol in the table in cook - levin theorem, or In other words, it is a map say $M$: $\tau^3 \mapsto \tau$, $\tau$ is a tape alphabet. So the total size of the circuit is $O(T(n))$
Question : To design a boolean circuit of size $O(T(n))$ for Turing machine $M$ with running time $T(n)$?
