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Polynomial-time computations have logspace-uniform circuit families

Statement

For every fixed deterministic Turing machine M that halts in polynomial time, there is a polynomial-size logspace-uniform Boolean circuit family (Cn) such that Cn(x)=1 exactly when M accepts x{0,1}n.

Facts & Assumptions

Given: a fixed deterministic multitape machine M and a fixed integer k1 such that B(n)=(n+2)k+k bounds its running time on every input of length n.

[L2]

Logspace uniformity requires logspace procedures for the indexed circuit's size, gate types, predecessor queries, and designated output (A logspace-uniform Boolean circuit family).

Proof

technique · computation tableau
1.1

For 0tB=B(n), encode each configuration by one-hot state bits, head-position bits Ht,r,j for each tape r, and symbol bits Xt,r,j,a. Use positions 0jn+B+1; no head can leave this interval within B steps. There are exactly n input nodes labeled x1,,xn; the initial row uses these, their negations, and constants for the input, blank cells, initial heads, and start state.

givenconstruct
2.1

For each tape and symbol, compute the scanned-symbol bit Rt,r,a=j(Ht,r,jXt,r,j,a) with a binary OR chain. The fixed transition table applied to the state and these scanned-symbol bits gives constant-size Boolean formulas for the next state, write symbols, and movement directions. At a halting state define this circuit update to leave the entire configuration unchanged. This is an extension of the tableau update only; no transition from a halted machine is assumed.

step 1.1construct
3.1

At each cell, retain its old symbol unless its head bit is one, in which case use the computed write symbol. A new head bit is the OR of the old adjacent head bits conjoined with their computed movement directions; in a halted configuration keep its old head bit. At position zero also retain a head commanded to move left, as required by the right-infinite tape convention. Treat other out-of-range head bits as zero. These are constant-size formulas once the scanned-symbol bits are available. Induction on t gives the genuine configuration until halting and its unchanged copy thereafter. Therefore the accepting-state bit at time B equals one exactly when M accepts x.

step 2.1induction
4.1

Each layer uses O(n+B) gates: the tape count, alphabet, and state set are fixed, each scanned-symbol OR chain is linear, and each cell update has constant size. There are B layers, so the size is O(B(n+B)), hence polynomial. All formulas use binary AND/OR and unary NOT. Designate the final accepting-state node as output.

step 3.1algebra
5.1

Number nodes by initial-row order and then by time layer, with scanned-symbol chains before transition formulas and cell updates. Pad fixed formula blocks with unused constant nodes if necessary so that each block length is a fixed constant or fixed multiple of n+B. A node is addressed by its layer, block, position, and constant-size formula slot; all are O(log(n+2))-bit integers. Fixed-degree arithmetic computes block offsets, adjacent positions, OR-chain predecessors, and the initial input label from these indices. Counting n on unary input and evaluating the fixed polynomial B(n) also use logarithmic space. These procedures answer SIZE, TYPE, predecessor, and output queries and emit the topological encoding. Thus [L2] gives logspace uniformity, including n=0 where the initial row has only constants.

L2step 4.1construct

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