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Semi-supervised graph propagation and pseudo-labeling limits

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

AlgorithmAlgorithm: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Finite-round confidence-threshold pseudo-labeling

Definition

Let q≥2, let C={1,…,q}, and let (X,d) be a finite metric space, meaning d:X×X→[0,∞) is symmetric, d(x,y)=0 iff x=y, and d(x,z)≤d(x,y)+d(y,z) for all x,y,z. The input is a nonempty ordered list S0=((xi,yi))i=1m of labeled examples with yi∈C, a finite ordered set U0⊆X∖{x1,…,xm} of unlabeled examples, an integer k with 1≤k≤m, a confidence threshold τ∈[1/q,1], a pseudo-label weight λ>0, and a round cap T∈N0. A true labeled record has weight 1; a pseudo-labeled record has weight λ.

Maintain an ordered training list Rt, initially S0, a remaining unlabeled set Ut, initially U0, and an initially empty pseudo-label log. To predict at x∈Ut, order all records (xi,yi,wi)∈Rt by increasing (d(x,xi),i), where i is their stable record index, and take the first k records Nt(x). For c∈C, put Vt,c(x)=∑(xi,yi,wi)∈Nt(x)wi1{yi=c}. Let y^t(x) be the smallest class index attaining max⁡cVt,c(x), and define confidence κt(x)=max⁡cVt,c(x)/∑cVt,c(x). The denominator is positive because k≥1 and every record weight is positive.

If U0=∅, set status no-unlabeled-inputs and skip the round loop. Otherwise, for rounds t=0,…,T−1, compute all predictions from the same current list Rt. Let At={x∈Ut:κt(x)≥τ}. If At=∅, set status no-additions and leave the loop. Otherwise append one pseudo-labeled record for each x∈At, in the supplied order on U0, using its predicted class and weight λ; append (t,x,y^t(x),λ) to the pseudo-label log, and remove all points in At from Ut. A pseudo-label is never revised. If the updated Ut=∅, set status all-labelled and leave the loop. If all T rounds finish without an earlier stop and points remain in Ut, set status round-cap; this includes the case T=0.

For any query x∈X, define the returned classifier by the same nearest-record ordering and weighted vote using the final training list, with the smallest-index vote tie rule. In every case return that classifier, the pseudo-label log, the remaining unlabeled set, and the status. The procedure makes no population-risk or pseudo-label-correctness guarantee.

Definition references

Chapelle, Schölkopf, and Zien, Semi-Supervised Learning, Chapter 1 §1.3.2: iterative prediction-and-retraining context; the finite weighted k-nearest-neighbor procedure above is specified locally.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

The real Cholesky recurrence computes the positive-diagonal factor

Statement

Let n≥1 and let M∈Rn×n be symmetric positive definite, meaning xTMx>0 for every nonzero x∈Rn. Initialize R as the zero n×n matrix. For j=1,…,n in order, set pj=Mjj−∑k<jRjk2,Rjj=pj, and then, for i=j+1,…,n in order, set Rij=Mij−∑k<jRikRjkRjj. Every pivot pj is positive, so the recurrence is well-defined and returns a lower-triangular matrix R with positive diagonal satisfying M=RRT.

Statement references

Hermitian positive-definite matrices and Cholesky factorisation A = LL* with positive diagonal: over the real numbers, positive definiteness means xTMx>0 for every nonzero vector, and the Cholesky factor has the form RRT with R lower triangular and positive diagonal.

Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T: a linear map with finite-dimensional domain satisfies dimension equals nullity plus rank; in particular, a square map with zero kernel is invertible.

David Bindel, CS 4220: Numerical Analysis, “Cholesky factorization”: complete block-recursion discussion; it leaves positivity of the Schur complement as an exercise, which the proof below supplies.

Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, §2.9.3: symmetric positive-definite matrices, the positive-diagonal Cholesky factor, and the LDLᵀ square-root construction.

Facts & Assumptions

Given: A real symmetric positive-definite matrix M∈Rn×n with n≥1. Prove the displayed recurrence is defined at every stage and returns its positive-diagonal Cholesky factor.

[A1]

M is real symmetric and xTMx>0 for every nonzero x∈Rn.

[F1]

Hermitian positive-definite matrices and Cholesky factorisation A = LL* with positive diagonal: over R, positive definiteness means symmetry and xTMx>0 for every nonzero vector; a Cholesky factor is lower triangular with positive diagonal and satisfies M=RRT.

[F2]

Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T: for a linear map on a finite-dimensional space, dimension equals kernel dimension plus image dimension; hence zero kernel implies full-dimensional image and invertibility.

Proof

1.1givenA1F1F2inductionalgebra

Induct on j. At j=1, testing e1 in [A1] gives p1=M11>0. For j>1, suppose columns 1,…,j−1 have been computed with positive diagonal and let T=R{1,…,j−1},{1,…,j−1} and A=M{1,…,j−1},{1,…,j−1}; the induction hypothesis gives A=TTT. The block A is positive definite because extending any nonzero x∈Rj−1 by zeros gives a nonzero vector to which [A1] applies. Its kernel is therefore zero; [F2] makes A invertible. The lower-triangular matrix T is invertible as well, since solving Tx=0 row by row with its positive diagonal forces every coordinate of x to be zero. Put m=(M1j,…,Mj−1,j)T and u=(Rj1,…,Rj,j−1)T. The already computed recurrence at each column i<j gives Mij=∑k<iRikRjk+RiiRji, so Tu=m. Since A=TTT, this implies mTA−1m=uTu. The leading j×j block P of M is positive definite by the same zero-extension argument. For the nonzero vector (−A−1m,1), its quadratic form under P is Mjj−mTA−1m=Mjj−uTu=pj>0. Thus Rjj=pj is positive and every division defining Rij is by a positive number. The recurrence is well-defined through column j, and the same formulas give the factorization identity on the leading j×j block.

2.1step 1.1F1algebra∎

For every j, the pivot equation gives Mjj=∑k<jRjk2+Rjj2=(RRT)jj. For every i>j, rearranging the recurrence gives Mij=∑k<jRikRjk+RijRjj=(RRT)ij. These identities fill the diagonal and lower triangle; symmetry of M and RRT fills the upper triangle. The initialized matrix remains lower triangular, and each diagonal entry is positive by step 1.1, so it is the required Cholesky factor.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Graph-harmonic label extension on a finite weighted graph

Definition

Let n≥1, V={1,…,n}, and let W=(wij)∈Rn×n be symmetric, entrywise nonnegative, and zero-diagonal. Put di=∑jwij, D=diag⁡(d1,…,dn), and L=D−W. The support graph has edge {i,j} exactly when wij>0; two vertices lie in the same connected component exactly when a finite path of support edges joins them; connected components include isolated singleton vertices. Let B⊆V be nonempty, put U=V∖B, and prescribe yB∈RB. Call the boundary anchored when every connected component meeting U also meets B. For f∈RV with f∣B=yB, define the Dirichlet graph energy E(f)=12fTLf. A function is harmonic on U when (Lf)u=0 for every u∈U. A graph-harmonic label extension is any minimizer of E over the affine set of functions with the prescribed boundary values; the definition itself makes no uniqueness assertion.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

The unlabeled marginal alone does not identify labels

Statement

Let X be a finite nonempty feature set with its discrete sigma-algebra, and let μ:X→[0,1] satisfy ∑x∈Xμ(x)=1. Let S⊊X be a nonempty set of observed labeled inputs, and choose x⋆∈X∖S with q=μ(x⋆)>0. Let h0,h1:X→{0,1} agree on X∖{x⋆} and differ at x⋆. For j∈{0,1}, let Pj be the joint law assigning mass μ(x) to (x,hj(x)). Both laws have the same feature marginal and the same observed labels on S. Let H={g:X→{0,1}}, and let Q:H→[0,1] satisfy ∑g∈HQ(g)=1, so it is a probability vector; take the same Q under both laws, as a predictor rule using their common information must do. For the zero-one population risk RPj(g)=∑x∈Xμ(x)1{g(x)≠hj(x)}, the expected risks satisfy max⁡j∈{0,1}∑g∈HQ(g)RPj(g)≥q2.

Statement references

Supervised prediction problems: a prediction problem supplies a joint law, measurable predictors, and a measurable loss.

Population and empirical risk: population risk is the integral of the loss under the joint law; for these finite discrete laws it is the displayed finite sum.

Chapelle, Schölkopf, and Zien, Semi-Supervised Learning, Chapter 1 §§1.3.2–1.3.3

Facts & Assumptions

Given: A finite discrete feature set, a probability mass function μ, a nonempty observed-input set S, an unseen atom x⋆∉S with positive mass q, two binary label functions that differ only at x⋆, and a common randomized predictor distribution Q. Prove the worst of the two expected zero-one population risks is at least q/2.

[A1]

The feature set X is finite and discrete, μ is a probability mass function, x⋆∉S has mass q>0, and h0,h1 agree away from x⋆ and differ there.

[F1]

Supervised prediction problems: a supervised problem has measurable instance and label spaces, a probability law on their product, measurable predictors, and a nonnegative measurable loss. Every map on a finite discrete feature set is measurable.

[F2]

Population and empirical risk: RP(g) is the integral of the loss ℓ(g(x),y) under the joint law P; on the finite laws here with zero-one loss it equals the sum in the Statement.

Proof

1.1givenA1F1F2

For each j, assigning mass μ(x) to (x,hj(x)) defines a probability law because the masses are nonnegative and sum to one. The feature marginal is μ for both laws, and h0=h1 on S, so the observed labeled input and unlabeled marginal are identical. Finiteness and the discrete sigma-algebra make every predictor and zero-one loss measurable, so [F2] gives the risk formula in the Statement.

2.1givenA1step 1.1algebra

Fix g∈H. At each x≠x⋆, the two risk summands agree because h0(x)=h1(x). At x⋆, the binary labels differ, so exactly one of the two indicators 1{g(x⋆)≠h0(x⋆)} and 1{g(x⋆)≠h1(x⋆)} equals one. Therefore RP0(g)+RP1(g)≥q.

3.1givenstep 2.1algebra∎

Multiply the inequality in step 2.1 by Q(g)≥0 and sum over the finite class H. Since ∑gQ(g)=1, the two expected risks sum to at least q, so at least one is at least q/2. This proves the claimed maximum.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Anchored graph-harmonic extensions are unique and obey a maximum principle

Statement

Let n≥1, V={1,…,n}, and let W=(wij)∈Rn×n be symmetric, entrywise nonnegative, and zero-diagonal. Put di=∑jwij, D=diag⁡(d1,…,dn), and L=D−W. The support graph has edge {i,j} exactly when wij>0. Two vertices lie in the same connected component exactly when a finite path of support edges joins them. Let B⊆V be nonempty, U=V∖B, and prescribe yB∈RB. For index sets S,T⊆V, let LST denote the submatrix with rows in S and columns in T, in increasing index order. Assume every support-graph component meeting U also meets B. Then the energy E(f)=12fTLf, minimized over f∣B=yB, has exactly one minimizer. If U=∅, it is f=yB. If U≠∅, the principal block LUU is real symmetric positive definite, meaning vTLUUv>0 for every nonzero v∈RU, and the minimizer is fU=−LUU−1LUByB. Each u∈U has du>0 and fu=∑jwujfj/du; hence min⁡b∈Byb≤fu≤max⁡b∈Byb. For C≥2, if each boundary value is one-hot (one coordinate equals 1 and all others equal 0) in RC, the coordinatewise energy minimizers have every row in the probability simplex ΔC={p∈[0,1]C:∑c=1Cpc=1}.

Facts & Assumptions

Given: A finite symmetric nonnegative zero-diagonal weight matrix W, its support graph and Laplacian L, a nonempty boundary B with prescribed values yB, and the assumption that every support component meeting U=V∖B meets B. Prove the energy minimizer is unique, satisfies the block formula and maximum principle, and has the stated one-hot multiclass consequence.

[F1]

Finite weighted graph Laplacian: di=∑jwij, D=diag⁡(di), L=D−W, and the support graph has edge {i,j} exactly when wij>0.

[F2]

Graph-harmonic label extension on a finite weighted graph: the admissible extensions satisfy f∣B=yB, their energy is E(f)=12fTLf, and an anchored boundary means each support component meeting U meets B.

[F3]

Quadratic form and kernel of a weighted graph Laplacian: for every g∈RV, gTLg=12∑i,jwij(gi−gj)2; this quantity vanishes exactly when g is constant on each support component.

[F4]

Hermitian positive-definite matrices and Cholesky factorisation A = LL* with positive diagonal: over the real numbers, a Cholesky factorization of M means M=RRT with R lower triangular and positive diagonal; this definition does not assert that a factorization exists.

[F5]

A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique: every finite real symmetric positive-definite matrix has a unique Cholesky factorization with positive diagonal.

[F6]

Forward and backward substitution are correct, unique, and quadratic in scalar operations: forward substitution solves a lower-triangular system with nonzero diagonal uniquely, and backward substitution solves an upper-triangular system with nonzero diagonal uniquely.

[F7]

Cholesky solves Hermitian positive-definite systems and has about half the factorisation cost of LU: with that factorization, forward and backward triangular substitution solve every right-hand side in finitely many scalar operations.

Proof

1.1givenF1F2algebra

If U=∅, then B=V and the affine set of extensions contains only f=yB. It is therefore the unique minimizer; the maximum and minimum assertions over U are vacuous, and the one-hot boundary rows are in ΔC.

1.2givenF1F3algebra

Suppose U≠∅ and put M=LUU. For nonzero v∈RU, extend v by zero on B to v~∈RV. By [F3], vTMv=v~TLv~=12∑i,jwij(v~i−v~j)2. If this sum were zero, every positive-weight edge would have equal endpoint values, so v~ would be constant on each support component. Every component meeting U meets B, where v~=0, forcing v=0, a contradiction. Thus M is positive definite; it is symmetric as a principal block of the symmetric matrix L.

2.1step 1.2F5F6F7F4

By [F5], M has a Cholesky factorization M=RRT with positive diagonal. By [F6] and [F7], solving Rv=−LUByB and then RTz=v gives a unique vector z=−M−1LUByB in finitely many scalar operations.

3.1step 1.2step 2.1F2algebra

Let fB=yB and fU=z. For any h∈RU, block expansion and Mz+LUByB=0 give E(f+h)−E(f)=hT(Mz+LUByB)+12hTMh=12hTMh. This is positive for h≠0 by step 1.2, so f is the unique minimizer and has the stated block formula.

3.2step 2.1F1F2algebra

The equation solved in step 2.1 gives (Lf)u=0 for every u∈U. If du=0, then u is an isolated support component contained in U, contradicting anchoring; hence du>0 and fu=∑jwujfj/du, a weighted average of neighboring values.

4.1givenF1step 3.2choose

Let m=max⁡i∈Vfi. If m>max⁡b∈Byb, then no maximizing vertex lies in B, so choose a maximizing vertex u0∈U. At u0, the weighted-average equation in step 3.2 and fj≤m force fj=m for each neighbor with positive weight. Choose a shortest finite path in its support component to B; every interior vertex lies in U, so applying the same argument successively forces the boundary value at the end of the path to equal m, a contradiction. Therefore max⁡i∈Vfi≤max⁡b∈Byb.

5.1step 4.1algebra

Apply step 4.1 to −f, whose boundary values are −yB and whose unlabeled coordinates satisfy the same weighted-average equations. This yields min⁡b∈Byb≤fu for every u∈U, completing the maximum-principle bounds.

6.1step 1.2step 2.1step 4.1step 5.1F1algebra∎

For one-hot boundary data, let f(c) be the minimizer for coordinate c and put g=∑c=1Cf(c). By step 2.1 each coordinate satisfies (Lf(c))u=0, and by steps 4.1 and 5.1 each fu(c) is between 0 and 1. Linearity gives (Lg)u=0 on U, while the one-hot boundary data give g∣B=1. The constant-one function also has these properties because every row sum of L=D−W is zero. The difference h=g−1 therefore has h∣B=0 and LUUhU=0; positive definiteness from step 1.2 forces hU=0. Thus g is identically 1, so every unlabeled row lies in ΔC.

AlgorithmAlgorithm: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Finite graph-harmonic label propagation with a residual status

Definition

Inputs are n≥1, V={1,…,n}, a matrix W∈Rn×n, a nonempty labeled subset B⊆V, a boundary-label matrix YB∈R∣B∣×r with r≥1, an operation cap K∈N0 per label coordinate, and a residual tolerance ε≥0. Use exact real arithmetic in the mathematical specification. Reject with status invalid-input unless W is symmetric, entrywise nonnegative, and zero-diagonal. Set U=V∖B, di=∑jwij, D=diag⁡(d1,…,dn), and L=D−W. The support graph has edge {i,j} exactly when wij>0; two vertices are in the same connected component exactly when a finite path of support edges joins them. Use increasing vertex order for every submatrix, boundary vector, and returned output coordinate.

Inspect every connected component in increasing vertex order. If some component meeting U misses B, return status unanchored-component and the unanchored components before forming a solve. If U=∅, return YB, an empty unlabeled output, and status no-unlabeled-vertices. Otherwise set M=LUU.

For each column c=1,…,r in order, start a fresh counter at zero and form bc=−LUB(YB):,c, evaluating the matrix-vector product in increasing vertex order. Recompute a fresh lower-triangular Cholesky factor R of M for this column by the recurrence: initialize R to zero; in increasing order j=1,…,∣U∣, compute pj=Mjj−∑k<jRjk2, return status factorization-failure if pj≤0, set Rjj=pj, and for i>j set Rij=(Mij−∑k<jRikRjk)/Rjj. Recompute R for each column so that the factorization cost is part of that column's cap.

Count each real addition, subtraction, multiplication, division, comparison, and square root as one scalar operation. Evaluate every finite sum left-to-right in increasing index order, with an empty sum equal to zero. For each coordinate of bc, implement the unary negation as one subtraction from zero and include that subtraction in the count. Input validation, degree/Laplacian construction, component scans, loop-control comparisons, and budget checks are outside the per-column count. The count for column c includes forming bc, its Cholesky recurrence, both triangular substitutions, and its residual calculation. Before each counted operation, if it would make the counter exceed K, return status budget-exhausted with outputs and residuals for only the fully completed columns before c. If a Cholesky pivot satisfies pj≤0, return status factorization-failure with outputs and residuals for only the fully completed columns before c.

Solve Rvc=bc by vc,i=((bc)i−∑j<iRijvc,j)/Rii in increasing i, and solve RTzc=vc by zc,i=(vc,i−∑j>iRjizc,j)/Rii in decreasing i, using the same counter and sum convention. Compute ec,i=∑j=1∣U∣Mijzc,j−(bc)i in increasing i, then ρc=(∑i=1∣U∣ec,i2)1/2, again using the same counter. If ρc>ε, return status residual-failure with candidate coordinates and residuals through column c. If all columns pass, return YU=(z1,…,zr), the residuals, and status residual-passed.

Definition references

Graph-harmonic label extension on a finite weighted graph: fixes the finite support graph, boundary data, componentwise anchor condition, and energy.

Anchored graph-harmonic extensions are unique and obey a maximum principle: proves that the anchored principal block is positive definite and gives the exact harmonic solution.

The real Cholesky recurrence computes the positive-diagonal factor: proves the displayed deterministic recurrence returns the real positive-diagonal Cholesky factor for the anchored positive-definite block.

Forward and backward substitution are correct, unique, and quadratic in scalar operations: supplies the unique triangular substitutions used to solve each column.

Zhu, Ghahramani, and Lafferty, Semi-Supervised Learning Using Gaussian Fields and Harmonic Functions, §2

Chapelle, Schölkopf, and Zien, Semi-Supervised Learning, Chapter 11 §§11.2–11.3

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Structural assumptions behind semi-supervised gains

Remarks

Here semi-supervised input means labeled examples together with unlabeled examples or their feature marginal PX. The finite harmonic extension uses the supplied graph weights to prefer scores that vary little across large-weight edges; it does not identify the ground-truth conditional label law P(Y∣X). A cluster assumption posits that points in the same cluster are likely to share a class; low-density separation posits that a classifier boundary lies in a region of small PX-mass; graph smoothness favors scores that vary little across edges with large supplied weights. These are additional modeling assumptions relating PX and P(Y∣X), not consequences of observing PX. The pseudo-label procedure is also a training heuristic. This batch states no general population-risk improvement, consistency, minimax, calibration, or graph-limit theorem. The exact harmonic maximum principle concerns the supplied boundary labels, not unknown true labels.

Remarks references

Finite-round confidence-threshold pseudo-labeling: the finite pseudo-label procedure returns a fixed training assignment and makes no risk guarantee.

Finite graph-harmonic label propagation with a residual status: the finite graph solver returns score candidates and residual statuses; only zero-residual exact solutions are the harmonic extension, and neither certifies true labels.

The unlabeled marginal alone does not identify labels: identical feature marginals and observed labels can leave positive-mass labels unidentified.

Zhu, Ghahramani, and Lafferty, Semi-Supervised Learning Using Gaussian Fields and Harmonic Functions, §2

Chapelle, Schölkopf, and Zien, Semi-Supervised Learning, Chapter 1 §1.2.2 and §§1.3.2–1.3.3, and Chapter 11 §11.6

4 · Examples, counterexamples and false statements

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