Inductive Representation Learning on Large Graphs

• p. 4 Algorithm 1, lines 7–9

  hᵏᵥ ← hᵏᵥ/‖hᵏᵥ‖₂

• p. 9 Theorem 1

  are final output values generated by Algorithm 1

• p. 18 Theorem 1 proof, normalization argument

  having aggregators prepend a unit value to all output representations

Only the authors can say what is right here.

Judge's note: I checked Algorithm 1 on p.4: line 7 sits inside the k-loop, and line 9 sets z_v ← h^K_v, so the final layer is normalized. Theorem 1 is stated on p.9 and restated on p.18 with a scalar z_v ∈ ℝ. The proof on p.18 says it ignores 'issues related to vector normalization'. Its 'prepend a unit value' remark only lets later layers recover intermediate normalization constants; it does nothing for the final scalar output. I checked the counterexample myself. In the diamond graph with edges 12, 13, 14, 23, 24, vertex 1 has 3 neighbours with 2 edges among them, so c_1 = 2·2/(3·2) = 2/3. A normalized scalar is ±1, which is at least 1/3 away from 2/3. The features can be chosen distinct and separated, so the theorem's assumptions hold. The theorem is false as printed unless the final normalization is dropped or a separate readout is added.

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• p. 17 Lemma 3 proof, final paragraph

  this single positive value can be mapped to an indicator vector

• p. 4 Algorithm 1, lines 4–5

  CONCAT

• p. 18 Theorem 1 proof, architecture assumptions

  all non-linearities are rectified linear units

Only the authors can say what is right here.

Judge's note: I checked Algorithm 1 lines 4–5 (p.4), the Lemma 3 proof (p.17) and the architecture assumptions (p.18). The proof applies f_θ to x_v of the node itself, but the pooling aggregator applies its MLP to the h_u of neighbours. I reworked the counterexample. Take a star with centre feature 0 and leaves 1, 2, 3. Every leaf has the same single neighbour, so every leaf gets the same pooled vector q. Each leaf's h¹ is then normalize(ReLU(a·x + r)), with every coordinate affine in x. The three leaves are pairwise in each other's 2-hop neighbourhoods, so they need distinct one-hot vectors. Leaf 2's active coordinate j then needs L_j(2) > 0 with L_j(1) ≤ 0 and L_j(3) ≤ 0, but affinity gives L_j(2) = (L_j(1) + L_j(3))/2 ≤ 0. Normalization keeps zero coordinates at zero. The features meet the separation assumption (C = 1/2). So the encoding claimed in Lemma 3 is not available as printed. This is a different defect from the A⁴ issue in C029.

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• p. 17 Proof of Lemma 3, last paragraph

  equivalently each MLP corresponds to a different max-pooling dimension in equation 3 of the main text

• p. 6 Eq. (3)

  ∀ui ∈N(v)

Only the authors can say what is right here.

Judge's note: I checked Eq. 3 on p.6 and Algorithm 1 on p.4. The pooling runs over u∈N(v), and N(v) is drawn from {u : (u,v)∈E}. v's own features enter only through σ(W^1·CONCAT(h^0_v, h^1_N(v))), a single layer to which Lemmas 1–2 are not applied. Under the proof's construction, pooled dimension c = max over u∈N(v) of ReLU(f_c(x_u)), which is positive whenever any neighbour has colour c. A node with neighbours of two colours therefore gets a multi-hot vector, not v's one-hot vector, and normalisation does not change that. Lemma 3, as stated (pooling at depth k=1 yielding h^1_v ∈ E_I), is left without a valid proof. A repair such as an extra iteration may exist, but only the authors can say which architecture they intended.

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• p. 7 Section 4.1, Reddit data

  We sampled 50 large communities

• p. 13 Appendix B, Reddit data

  selected the communities with ranks [11,50] (inclusive)

• p. 7 Section 4.1, Reddit data

  50 large communities and built a post-to-post graph

Only the authors can say what is right here.

Judge's note: Read page 7 (Section 4.1, Reddit data) and page 13 (Appendix B, Reddit data) on the page images. Ranks 11 to 50 inclusive are 40 communities, and the appendix adds that the largest communities were omitted. Nothing on either page reconciles this with 50 communities sampled. The community is the node label, so the number of classes is left unclear; only the authors can say which figure is right.

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• p. 16 Appendix E, proof of Lemma 1, property 1

  g_v has a unique maximum of 3|D|ε − 2ε > 2|D|ε at d_v = 0

• p. 17 Proof of Lemma 1, inequality chain

  > 2(|D|)ε − 2(|D|)ε

• p. 16 Lemma 1, Equation (7)

  g(x) > ϵ

• p. 16 Equation (8) and property 1

  3|D|ϵ − 2ϵ > 2|D|ϵ

Only the authors can say what is right here.

Judge's note: I worked through both parts on the page images of pp.16–17. (a) 3|D|ε − 2ε > 2|D|ε holds only when |D| > 2. For D = {v} and x = x_v, Eq. (8) gives g = 3ε − 2ε = ε, which breaks the strict g > ε of Eq. (7); singletons are allowed by 'any non-empty finite subset'. The last two lines of the chain read 2|D|ε − 2|D|ε > 0, i.e. 0 > 0. The chain only ever aims at g > 0, never at g > ε. For |D| ≥ 2 the direct bound g > 3|D|ε − 2ε − 2(|D|−1)ε = |D|ε ≥ 2ε still gives the result, and Lemma 3 only uses the sign (Eq. 9). On its own, part (a) would be a step-level defect (S3). (b) An entry of A⁴ is positive only if there is a walk of length exactly 4. For the path a–b–c, the only off-diagonal edge of A⁴ is a–c, so χ = 2. But a, b and c all lie in each other's 2-hop neighbourhoods, and Lemma 3's conclusion needs three distinct one-hot vectors, which is impossible in 2 dimensions. Lemma 3 is therefore false as printed under its own definition of G⁴. It would hold under a distance-≤4 power-graph reading. Severity S1 because of part (b).

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• p. 18 Proof of Theorem 1, before Eq. 11

  containing only v and it’s immediate neighbors

• p. 18 Appendix E, proof of Theorem 1, text before Eq. (11)

  Thus, we have that b⊤c is the number of edges in the subgraph containing only v and it’s immediate neighbors

• p. 18 Appendix E, Eq. (11)-(12)

  2(b⊤c −dv)

• p. 18 Proof

  For clarity and we ignore issues related to vector normalization

Only the authors can say what is right here.

Judge's note: Definitions from the p.18 image: h^3_v = h^1_v ⊕ A_v ⊕ (Σ_u h^1_u ⊕ A_u), b = h^3_v[m:2m] = A_v, and c = h^3_v[3m:4m] = Σ_{u∈N(v)} A_u. Then b⊤c = Σ_{u∈N(v)} |N(u)∩N(v)| = 2T, where T is the number of edges among v's neighbours, not d_v+T. Triangle check: b⊤c = 2, so the LHS of Eq. 11 is 2(2−2)/2 = 0 while c_v = 1. Star with two leaves: b⊤c = 0, so the LHS is −2 while c_v = 0. Adding self-loops gives 2T+2d_v, which does not rescue it. A continuous function of h^3_v that gives c_v still exists (b⊤c/(d_v(d_v−1))), so Theorem 1's conclusion is not in question, but the formula is wrong as printed. The 'it's' → 'its' slip in the same sentence is a minor copyedit, and the other grammar slips are covered in C043.

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• p. 18 Theorem 1 proof, definitions of b and c

  the sum of the adjacency rows of v’s neighbors

• p. 18 Equations (11)–(12)

  2(b⊤c − dv)

Only the authors can say what is right here.

Judge's note: I reworked this on p.18. b⊤c = Σ_{u∈N(v)} |N(u) ∩ N(v)|, which counts each edge between two neighbours twice. For a triangle at v: b = (0,1,1), c = (2,1,1), b⊤c = 2 and d_v = 2, so Eq. (11) gives 2(2−2)/(2·1) = 0, while the true coefficient is 1. The right-hand side of Eq. (11) is correct; only the expression in b and c is wrong. A corrected expression, b⊤c/(d_v(d_v−1)), is still continuous, so Theorem 1's approach survives. The displayed formula is wrong as printed.

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• p. 4 Section 3.1, paragraph after Algorithm 1 (two occurrences)

  with the aggregated neighborhood vector, h^{k−1}_{N(v)}

• p. 4 Algorithm 1, lines 4-5

  h^k_{N(v)} ← AGGREGATE_k({h^{k−1}_u, ∀u ∈ N(v)})

• p. 4 Section 3.1, paragraph after Algorithm 1

  with the aggregated neighborhood vector, h

• p. 4 Algorithm 1, line 4

  AGGREGATEk({h

Judge's note: Checked on the page 4 image. Algorithm 1 lines 4 and 5 use superscript k on h_{N(v)}; the explanatory paragraph prints superscript k−1 in both places. Section 3.3 on page 5 also uses h^k_{N(v)}. The prose explicitly describes the algorithm's steps, so this is a notation inconsistency whose fix is to use k in the prose.

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• p. 5 Section 3.1, Neighborhood definition, last sentence

  S1 · S2 ≤ 500 (see Section 4.4 for details)

• p. 8 Section headings

  4.3 Runtime and parameter sensitivity

• p. 5 Section 3.1, Neighborhood definition (end)

  Section 4.4 for details)

• p. 6 Section 4, Experimental set-up

  4.4 for sensitivity analyses)

Judge's note: On the page 8 image, Section 4.3 is 'Runtime and parameter sensitivity' and holds the discussion of K and neighbourhood sample size (Figure 2.B). Section 4.4 is 'Summary comparison between the different aggregator architectures' (Wilcoxon tests) and says nothing about K or sample sizes. Both pointers should read Section 4.3.

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• p. 10 Reference [1]

  arXiv preprint , 2016.

• p. 6 Experimental set-up

  Since, the “convo-

Only the authors can say what is right here.

Judge's note: The page 10 image shows "arXiv preprint , 2016." with no identifier, whereas reference [6] carries one (arXiv:1710.10568). The authors need to supply the missing number. The "Since, the" comma is visible on the page 6 image and is S4 at most.

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• p. 10 Reference [13]

  Identity mappings in deep residual networks. In EACV, 2016.

• p. 10 References, entry [13]

  In EACV,

• p. 10 Reference [13]

  Identity mappings in deep residual networks. In EACV,

Only the authors can say what is right here.

Judge's note: Rests on outside knowledge: that this paper appeared at ECCV 2016 and that "EACV" is not a real venue. The page 10 image does print "In EACV, 2016.", so it is not an extraction artefact.

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• p. 6 Equation 3

  AGGREGATE^pool_k = max({σ(W_pool h^k_{u_i} + b), ∀u_i ∈ N(v)})

• p. 4 Algorithm 1, line 4

  AGGREGATE_k({h^{k−1}_u, ∀u ∈ N(v)})

Only the authors can say what is right here.

Judge's note: The page 6 image shows superscript k on h_{u_i} in Eq. 3. Algorithm 1 line 4 (page 4) and the mean aggregator text and Eq. 2 (page 5) use h^{k−1}_u as the aggregator input at step k; h^k_u is not yet computed at that point. Probably should be h^{k−1}_{u_i}; the authors should confirm.

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• p. 5 Section 3.3, Mean aggregator paragraph, after Eq. 2

  the convolutional aggregator does concatenate the node’s previous layer

Judge's note: Read on the page 5 image. The first half of the sentence says the convolutional aggregator does not perform the concatenation, and Eq. 2 has a MEAN over the union with no CONCAT. The 'i.e.' clause then says it 'does concatenate', contradicting both. The intended wording is evidently 'does not concatenate'.

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• p. 16 Proof of Lemma 1, property 1

  gv has a unique maximum of 3|D|ϵ −2ϵ > 2|D|ϵ at dv = 0.

• p. 17 Proof of Lemma 1, inequality chain

  > 2(|D|)ϵ −2(|D|)ϵ

Only the authors can say what is right here.

Judge's note: I checked p.16 (Eq. 8, property 1) and the chain on p.17. 3|D|ε−2ε > 2|D|ε is equivalent to |D|>2. For |D|=2 the two sides are equal, and for |D|=1 the left side is smaller. The chain's step '> gv(x) − 2|D|ε > 2|D|ε − 2|D|ε' uses this property. The conclusion g>0 can still be reached directly: g > (3|D|−2)ε − 2(|D|−1)ε = |D|ε > 0. So the step is faulty but the stated result stands, which makes this S3.

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