Double-star reconstruction (A3) #
AUDIT-NOTES A3 and paper Section sec:doublestar, in the binary case T = 2.
The analytic content of the reconstruction is proved here in full. Everything rests on one
identity about a Navascués–Wolfe witness at order two, gTwist_law: after relabelling the copy
indices of each source by an arbitrary permutation, the t rows read on the diagonal are still
t independent copies of the target. From it follow, in order,
blockMarg_union_of_sourceDisjoint— source-disjoint blocks of vertices are independent under the target (step (i) of the paper proof, and the pair-source form of AUDIT-NOTES A3(i)), and its iterateblockMarg_biUnion_of_sourceDisjoint;centreLeaf_mass— conditioning on the event that each leaf's two copies list the prescribed array, the joint law of the two centre observations is the conditional law of the two centres given the leaf values (steps (ii)–(iii), the finite-order step);DSStruct.ctrFactor_mul— the component identity that the reconstruction needs;gCompatible_of_localDecoder— a model whose responses are deterministic local readings of a decoder, packaged so that no dependent latent types appear;DSStruct.gCompatible_of_dsStruct— the reconstruction (step (iv)).
What is not proved here is the purely graph-theoretic step exists_dsStruct: that a
double-star forest carries the combinatorial data DSStruct. That step, exists_dsStruct,
and the theorem doubleStar_terminates assembled from it, live in
InflationGraphOpen/DoubleStar.lean; everything in this file is proved.
Relabelling the copy indices #
The whole finite-order content of the double-star reconstruction is a single identity about a
Navascués–Wolfe witness: after relabelling the copy indices of each source by a permutation
π, the t diagonal rows are still t independent copies of the target. Sections (i)–(iii)
of the paper proof are all read off from it.
Relabelling copied observations composes.
Relabelling assignments composes.
The identity relabelling.
Relabelling by π is a bijection of assignments, with inverse the relabelling by
π⁻¹.
Equations
- TriangleInflation.Graph.gRelabelEquiv π = { toFun := TriangleInflation.Graph.gRelabel π, invFun := TriangleInflation.Graph.gRelabel fun (e : Γ.Edge) => (π e)⁻¹, left_inv := ⋯, right_inv := ⋯ }
Instances For
A symmetric witness has the same pushforward along F and along F ∘ gRelabel π.
The rows read by a table after relabelling the copy indices of each source: row r reads,
at every vertex, the copy π_e r of each incident source e. For π = 1 these are the t
diagonal rows.
Equations
- TriangleInflation.Graph.gTwist π ω r v = ω ⟨v, fun (e : ↥(Γ.inc v)) => (π ↑e) r⟩
Instances For
The twisted-row law. For a symmetric witness whose diagonal law is the t-fold
tensor power of the target, the t rows read after any relabelling of the copy indices are
again t independent copies of the target. This is the only property of the witness that the
double-star reconstruction uses.
Reading an expectation off a pushforward #
Expectations only see the pushforward.
The expectation of a function of the twisted rows.
A vertex whose incident sources are all relabelled so that row r becomes row s reads,
in the twisted row r, exactly what it reads in the diagonal row s.
The two-row identity at order two. Twisting by π and reading the two rows gives a
pair of independent draws from the target.
Marginals of the target on blocks of vertices #
The marginal of a target on a set of vertices, presented as a function of a full outcome
vector; it depends on w only through the coordinates in A.
Equations
Instances For
Source-disjoint blocks of the target are independent. If no source touches both A
and B then the joint marginal of the target on A ∪ B is the product of its marginals on
A and on B. This is step (i) of the paper proof, and it needs only order two.
The iterated form of blockMarg_union_of_sourceDisjoint: pairwise source-disjoint blocks
are mutually independent under the target.
The conditioning event and the conditional law of the centres #
The relabelling that moves the copy index j e of each source to the index 0 (and
back).
Equations
- TriangleInflation.Graph.swapTwist j e = Equiv.swap 0 (j e)
Instances For
The other of the two copy indices.
Instances For
Two-row block masses. The probability that the first twisted row matches w₀ on A₀
and the second matches w₁ on A₁ is the product of the two target marginals: this is the
one computation the reconstruction performs.
The conditional marginal of the centres (step (iii) of the paper proof).
Fix a copy index j e for every source. Condition a Navascués–Wolfe witness at order two on
the event E that each leaf's two copies read a prescribed array x. Then the joint mass of
E together with the event that the vertices of C — in practice the two centres — read, on
the copy j of each of their leaf sources and the copy 0 of every other source, the values
w, is the corresponding two-block marginal of the target. Dividing by the same identity with
C = ∅ turns this into the conditional law of the centres given the leaf values, which is the
only fact about the witness the reconstruction uses.
Models with deterministic local responses #
The reconstruction builds a model whose sources are independent and whose responses are
deterministic functions of the incident sources. It is convenient to package such a model as a
single decoder dec from source values to outcomes, subject to the locality requirement that
the outcome at v depend only on the sources incident to v.
A deterministic local decoder gives a model. If every source e carries an
independent law ν e and the outcome at each vertex v is a function dec of the source
values that depends only on the sources incident to v, then the pushforward of the product
law along dec is compatible.
A sum over a function space of a product of per-coordinate weights is the product of the coordinate sums.
Reconstruction: the source-disjoint fibre case #
A double-star forest whose components are single sources — a perfect matching — is already
covered by the machinery above: the source of a component carries the outcomes of its two
endpoints. This is the degenerate case p = q = 0 of the paper's Theorem, and it is recorded
here because it exercises the whole pipeline (twisted rows, block independence, deterministic
local decoder) end to end.
Reconstruction when the vertex fibres of the sources are source-disjoint. Choose for
each vertex v an incident source edgeAt v. If the vertex sets {v | edgeAt v = e} are
pairwise source-disjoint, every target passing the order-two Navascués–Wolfe test is
compatible: the source e carries the joint outcome of the vertices that read it, which by
source-disjoint block independence is exactly the right marginal.
The combinatorics of a double-star forest #
The combinatorial data of a double-star forest. Every vertex is a leaf or a centre;
a leaf carries a single source joining it to its centre; a centre has a partner centre, and the
two share the centre source of their component. root v names the centre source of the
component of v, so the components are the fibres of root.
Is the vertex a leaf?
For a leaf, its unique source; for a centre, the centre source of its component.
For a leaf, its centre; for a centre, its partner.
The centre source of the component of a vertex.
A vertex whose
edgeAtis the given source.
Instances For
The latent alphabet used by the reconstruction: a response table together with a bit. The same type serves every source; the source law is what distinguishes leaf sources, which carry the bit, from centre sources, which carry the table.
Instances For
The table that unsupported arguments are sent to.
Equations
- TriangleInflation.Graph.dfltTable Γ x✝¹ x✝ = false
Instances For
The one-vertex marginal of the target.
Equations
- TriangleInflation.Graph.vMarg P v β = TriangleInflation.Graph.blockMarg {v} P (Function.const Γ.V β)
Instances For
The array of two symbols listed at each leaf: both symbols when both are supported, and otherwise the supported symbol twice.
Equations
Instances For
A copy index at which the leaf ℓ reads the symbol β.
Equations
- TriangleInflation.Graph.pick P ℓ β = if TriangleInflation.Graph.xarr P ℓ 0 = β then 0 else 1
Instances For
The decoder of the reconstructed model: a leaf copies its source, a centre evaluates its half of the centre table on the values of its leaf sources.
Equations
Instances For
The response table read off a witness table: on the leaf values ξ it returns the
observations that read, at each leaf source, the copy showing ξ.
Equations
- D.tableOf P ω ξ = TriangleInflation.Graph.gTwist (TriangleInflation.Graph.swapTwist (D.jOf P ξ)) ω 0
Instances For
The conditioning event of the component with centre source y: every leaf of that
component has its two copies reading the prescribed array.
Equations
- D.evt P y ω = ∀ ℓ ∈ D.leaves y, ∀ (m : Fin 2), TriangleInflation.Graph.readDiag ω m ℓ = TriangleInflation.Graph.xarr P ℓ m
Instances For
Equations
- D.instDecidableEvt P y ω = id inferInstance
One-vertex marginals and the listed arrays #
Any set of leaves is a union of source-disjoint singletons, so its marginal is the product of the one-vertex marginals.
The mass of the conditioning event, evaluated by the conditional-marginal identity.
The conditional mass that the reconstructed centre source assigns to the outcomes w at
the two centres of the component y.
Equations
- One or more equations did not get rendered due to their size.
Instances For
The component identity. The centre factor times the leaf marginals of a component is the marginal of the target on the whole component. This is step (iii)–(iv) of the paper proof in the form the reconstruction needs.
At a centre, the reconstructed table evaluated on the local leaf values reads the twisted observation used by the conditional-marginal identity.
The per-source factor of the reconstructed model at a centre source.
The double-star reconstruction. A target passing the order-two Navascués–Wolfe test on a pair graph carrying a double-star structure is compatible.