The space of the Tree Evaluation read machine #
On the encoding of an instance of value length ℓ and height H, the space budget of every state
that Cslib.TreeEval.Reader.machine reaches is O((ℓ + H + 1) · log (ℓ + 2)): residues
below the prime p ≤ 4ℓ + 2 have O(log (ℓ + 2)) bits. The registers and the sums hold ℓ
of them, the stack holds at most H frames with a residue each, and the counters have
O(ℓ + log (H + 2)) bits.
Main definitions #
Cslib.TreeEval.Reader.spaceBound: the bound(ℓ + H + 1) · (|ℓ| + 1), with|ℓ|the length ofℓin binary.
Main results #
Cslib.TreeEval.Reader.stateBudget_le: the budget of every state reached isO(spaceBound ℓ H).
References #
- J. Cook and I. Mertz, Tree Evaluation is in Space O(log n · log log n)
- O. Goldreich, On the Cook–Mertz Tree Evaluation procedure, Algorithm 1, which the machine runs.
The space bound (ℓ + H + 1) · (|ℓ| + 1).
Instances For
The bounds on the components of a state that satisfies the invariant.
- rounds (fr : Iter.Frame) : fr ∈ s.stack → (Turing.MultiTapeTM.encNat fr.round).length ≤ (Turing.MultiTapeTM.encNat ℓ).length + 3
- n₂ : (Turing.MultiTapeTM.encNat s.ctl.toTuple.2.2.1).length ≤ (Turing.MultiTapeTM.encNat ℓ).length + 3
Instances For
The budget of a state within the space bound.