Given a function and its complexity, what is the complexity of its circuit?

For example, given an element-wise function $f$, with input $x\in\{0,1\}^{p\times n}$, the complexity $T(f(x))=O(n)$, and that all numbers are represented using $p$ binary digits. Suppose that we also know the complexity of each individual operations inside $f$. For example, multiplication takes $O(m^{1.585})$ using Karatsuba algorithm, here $m$ is the number of digits of the product. Similarly, suppose that we know all the complexity of arithmetic operations like addition, subtraction, division, square root etc. In addition, suppose that we also know the complexity of all the basic boolean cells, e.g. Full Adder, Subtractor, Multiplier etc.

What is the complexity, in terms of number of boolean gates, of its equivalent boolean circuit? Can we derive a formula containing $n,p,m$?

If so, what is the complexity of a boolean circuits of a Cholesky decomposition, with input $X\in\{0,1\}^{p\times n\times n}$?

If you can compute a function on $N$-bit inputs in time $T(N)$ with code that doesn't use random-access memory (only straight-line code; no arrays or pointers or memory), then there exists a combinational circuit with $O(T(N) \cdot (\log N)^2)$ gates that computes the function. That's because each basic operation you can do in code -- add, multiply, compare, etc. -- can be done in $O((\log N)^2)$ gates.

If you can compute the function on $N$-bit inputs in time $T(N)$ with code, and the code does use random-access memory, there exists a sequential circuit with $O(T(N) \cdot (\log N)^2)$ gates that computes the function in $O(T(N) \log N)$ time steps. Or, you can do it with a combinational circuit with $O(T(N)^2 \cdot (\log N)^3)$ gates -- though this might not be tight.

Technical details:

• I am making some assumptions about the model of computation for the code -- e.g., a RAM machine, where each cell is $O(\log N)$ bits wide.

• The exponent 2 can be reduced to the exponent for multiplication (though this might not be very useful in practice).

• It may be helpful to note that any algorithm that runs in time $T(N)$ can be converted to an algorithm that runs in time $O(T(N) \log N)$ and that uses only $O(T(N) \cdot \log N)$ cells of memory, by using a balanced binary tree data structure to build a sparse array that is used to simulate its memory.

• Great answer, any chance providing some related materials to read, especially about that each basic operation can be done in $O((\log N)^2)$ and that transforming the algorithm to a sequential circuit ? Why $O(T(N)log(N))$ time steps? – xtt Mar 31 '17 at 7:57
• @Jason, for basic operations, start with en.wikipedia.org/wiki/…. You can also look at the different basic operations; there's stuff on specific operations at, e.g., en.wikipedia.org/wiki/Multiplication_algorithm, en.wikipedia.org/wiki/Division_algorithm, etc. For the second paragraph, that's more involved. Basically you construct a circuit that is a small processor, with a random-access memory large enough to store $O(T(N) \cdot (\log N)^2)$ bits; then you count the number of gates needed to do that. – D.W. Mar 31 '17 at 20:36

There is no mapping from algorithmic complexity to circuit size. A counterexample: consider the language of strings of all ones, $1^n$, for only those $n$ such that the $n$-th Turing Machine rejects the string $1^n$ within $2^{2^n}$ time. Then this function requires $\Omega\left(2^{2^n}\right)$ time to compute on any machine, but the circuit for it only has size $O(n)$, because you can "hard-wire" the answer into the circuit, which then only checks that all the inputs are $1$.

D.W. provides a construction from an algorithm to a circuit. However, note that this gives only an upper bound, which is not necessarily tight. In particular, no function requires circuits of size larger than $O\left(\frac{2^n}{n}\right)$, because that is the size of a circuit which simply enumerates that function's truth table for inputs of size $n$.

Consequently, if an algorithm takes $O(2^{2^n})$ time, then you can use D.W.'s construction for a circuit of size $O(2^{2^n}\log(n)^2)$, but it is immediate that this bound is not tight; there is also a circuit of size $O(\frac{2^n}{n})$, and there may be a much smaller circuit for it, as in the example above. What we really want to know when we ask what is the size of this function's circuit is a lower bound: a minimum size for any circuit.

Circuit size lower bounds is a particularly enigmatic part of computational complexity about which very little is known. For example, we do not know whether or not SAT can be solved with $O(n)$-size circuits, and we have not ruled out that each problem in NEXP can be solved with polynomial-size circuits.