Proof that uniform circuit families can efficiently simulate a Turing Machine

Question

Can someone explain (or provide a reference for) how to show that uniform circuit families can efficiently simulate Turing machines? I have only seen them discussed in terms of specific complexity classes (e.g., $\mathbf{P}$ or $\mathbf{NC}$). I would like to see how uniform circuit families is a strong enough model for universal, efficient computation.

Answer 1

Here is the basic idea. We'll take P as an example.

Given a Turing machine running in polynomial time, we construct, for each input length, a layered circuit in which each circuit represents the current configuration of the Turing machine:

• For each cell, the contents of the cell.
• For each cell, whether the head is there.
• The current state of the machine.

The number of layers is the same as the polynomial upper bound on the running time, and the width of each layer (number of cells) is the maximum space used by the machine.

The initial layer is initialized directly from the input. Each other layer is constructed from the preceding layer (using the convention that once the machine halts, its configuration remains static). Finally, the output is extracted from the last layer. The construction is uniform, as the description hopefully makes clear.

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This shows that you can convert Turing machines running in time $T(n)$ and space $S(n)$ to uniform circuits of size $O(T(n)S(n))$. This can likely be improved.

Answer 2

Let us prove the following theorem:

A language $L$ is decidable if there is a Turing machine $T$ that on input $n$ outputs a circuit for $L \cap \{0,1\}^n$.

$\Longrightarrow$ Suppose that $L$ is decidable. Given $n$, we can compute $L \cap \{0,1\}^n$. We can then construct a circuit for the corresponding Boolean function.

$\Longleftarrow$ Suppose that there is a Turing machine $T$ than on input $n$ outputs a circuit for $L \cap \{0,1\}^n$. To decide $L$, given an input $x$, run $T(|x|)$ to get a circuit for $L \cap \{0,1\}^{|x|}$, and run it on $x$ to determine whether $x \in L$.