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Defective correspondence coloring of planar graphs

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Pith's one-line read The paper constructs a planar graph that is 1-defective 3-correspondable but not 4-correspondable, and shows outerplanar graphs need exactly three defects for 2-correspondence coloring.

desk verdict Main theorem rests on a false lemma; the other two results look plausible, but the paper's headline claim is not proven as written. read the letter →

arxiv 2411.15336 v1 pith:57YWEJJK submitted 2024-11-22 math.CO

James Anderson This is my paper · ORCID
classification math.CO MSC 05C1505C10
keywords defectivecoloringcorrespondenceDP-coloringplanargraphsouterplanarlistrelaxed
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

This paper studies defective correspondence coloring, a generalization of list coloring in which each edge carries its own matching between color lists and each vertex may share its color with up to d neighbors. Its central result is a planar graph that is 1-defective 3-correspondable, meaning every triple of lists with edge-specific matchings admits a coloring where each vertex matches at most one neighbor, yet is not 4-correspondable, meaning some four-list cover forces a proper-coloring obstruction. This extends to correspondence coloring a separation previously established for list coloring. The paper also builds a planar graph that is not 3-defective 3-correspondable, tightening known bounds, and proves all outerplanar graphs are 3-defective 2-correspondable with three defects best possible. Together these results narrow which pairs of defect allowance and color count can hope to color every planar graph.

What carries the argument

The load-bearing object is the gadget T, formed by two copies R1 and R2 of K4-minus-an-edge sharing a central vertex z, with endpoint colors for u and v fixed by the partial coloring. A correspondence cover of each half is classified as twisted when it contains a specific 6-cycle of conflicts and wedged when it contains a smaller conflict pattern; the lemmas use these two patterns to decide whether a 1-defective coloring with zero defect on z exists. The counting lemma then shows that for any 3-fold cover, only six of the nine (u,v) color pairs are bad, so with four copies of T there is a pair that is good for at least two copies. That pair is colored through the extension lemmas, and the four-copy graph is the configuration that makes the counting work; the non-4-correspondability direction is carried by a permutation-based 4-fold cover that blocks every proper coloring.

What would settle it

Enumerate all 3-fold correspondence covers of T(4) up to isomorphism and check each for a 1-defective coloring; if any cover has none, the positive half of the main theorem fails, and similarly, finding a proper coloring of the specific 4-fold cover constructed in Section 9 would refute the non-4-correspondability claim.

Watch

Extended reading notes

Core claim

The main construction is the graph T(4): four copies of a twelve-vertex gadget T glued by identifying the x-vertices and identifying the y-vertices. The paper proves that every 3-fold correspondence cover of T(4) has a 1-defective coloring, while a carefully chosen 4-fold cover has no proper coloring at all. The proof works by classifying the two halves of T, each half being K4 with one edge deleted, as twisted or wedged depending on which conflict cycles appear in the cover, then showing that among the nine possible color pairs for the shared endpoints u and v, at most six pairs can be bad in a sense that forces an extra defect. Counting over the four copies leaves a good pair that can be colored with zero defect on u and v, while the 4-fold obstruction is built from a permutation construction on the four copies. The paper also establishes that some planar graph fails 3-defective 3-correspondability, and that outerplanar graphs need exactly three defects for two-list correspondence coloring.

Load-bearing premise

The positive half of the main construction depends on Lemma 5.4(i), which asserts that every assignment of conflict matchings to the four-vertex half-gadget R, with one vertex given one color and three vertices given two colors, admits a 1-defective coloring in which the special vertex c is left with zero defect.

Editorial extensions

If this is right

  • The main construction answers a question asked for list coloring in the correspondence setting: 1-defective 3-correspondability does not imply 4-correspondability for planar graphs.
  • Combining Theorem 1.3 with known decomposition results leaves the smallest d for which every planar graph is d-defective 3-correspondable between 4 and 6.
  • The graph witnessing Theorem 1.3 is 4-correspondable, so a planar graph can be properly 4-correspondable and still fail to be 3-defective 3-correspondable.
  • Every outerplanar graph is 3-defective 2-correspondable, and the fan-based construction shows that two defects do not suffice.
  • The same framework leaves open whether the separating graph could be chosen not 4-choosable, which would strengthen the break with list coloring.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • The twisted/wedged classification is checkable by brute force: for the small cover types that appear after fixing u and v, exhaustive enumeration would independently confirm the counting lemmas and could be reused for larger gadgets.
  • If the T(4) construction is as flexible as it appears, similar identified-copy gadgets should produce separations for other defect and list-size pairs, such as d-defective 3-correspondability versus d+1-correspondability for small d.
  • The outerplanar theorem suggests that on graphs of bounded treewidth, defective correspondence coloring may track defective list coloring with the defect threshold shifted by one; series-parallel graphs would be a natural test class.
  • Because the non-4-correspondability cover is built from permutations on the four copies, varying the permutation family could probe whether the threshold is sharp, for example whether T(3) is already 1-defective 3-correspondable.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Referee Report

2 major / 3 minor

Summary. The paper studies defective correspondence coloring of planar graphs. It claims three main results: (i) there is a planar graph that is not 3-defective 3-correspondable (Theorem 1.3); (ii) there is a planar graph that is 1-defective 3-correspondable but not 4-correspondable, extending a recent result of Ma, Xu, and Zhu from list coloring to correspondence coloring (Theorem 1.4); and (iii) every outerplanar graph is 3-defective 2-correspondable, with 3 defects best possible (Theorem 1.6). The proof of Theorem 1.4 is built around a gadget T(4) formed from four copies of a graph T identified at two vertices, with Section 5 setting up the reduction, Sections 6-8 proving the 1-defective 3-correspondability direction, and Section 9 proving the failure of 4-correspondability. The outerplanar and Theorem 1.3 arguments are independent of the T(4) construction. The main defect of the manuscript is that a central lemma used to combine colorings across the two halves of T is false as stated, and the proof of Theorem 1.4 collapses as a result.

Significance. If Theorem 1.4 were correct, it would be a genuine separation result: a planar graph that is 1-defective 3-correspondable but not 4-correspondable would extend the Ma-Xu-Zhu list-coloring example to the correspondence setting, where matchings may vary per edge. Theorems 1.3 and 1.6 are also interesting contributions to the defective correspondence coloring literature, and the non-4-correspondability construction in Section 9 is explicit and appears self-contained. However, the 1-defective 3-correspondability direction of the main theorem rests on Lemma 5.4(i), which admits a concrete counterexample. The manuscript is not a reliable proof of Theorem 1.4 as written, although the other two main theorems may well survive independently.

major comments (2)
  1. [§5.1, Lemma 5.4(i)] Lemma 5.4(i) is false as stated. Consider the correspondence cover of R with L(a)={a1}, L(b)={b1,b2}, L(c)={c1,c2}, L(d)={d1,d2} and edges a1b1, a1d1, b1c1, b2c2, c1d2, c2d1, b1d2, b2d1. All matchings in this cover are maximal. Since φ(a)=a1 is forced, if φ(c)=c1 then def(c)=0 forces φ(b)=b2 and φ(d)=d1, giving def(d)=2 because d1 is adjacent to both a1 and b2. If φ(c)=c2 then def(c)=0 forces φ(b)=b1 and φ(d)=d2, giving def(b)=2 because b1 is adjacent to both a1 and d2. Hence no 1-defective H-coloring has def(c)=0, contradicting the lemma. The proof of the lemma fails in Case 2, where the proposed assignment φ(b)=b1, φ(c)=c2, φ(d)=d2 creates exactly the second defective vertex in this cover.
  2. [§7–§8, dependence on Lemma 5.4(i)] The invalidity of Lemma 5.4(i) is load-bearing for the proof of Theorem 1.4, not a peripheral gap. The concatenation arguments in Lemmas 7.2, 7.3, and 7.7 use Lemma 5.4(i) precisely to obtain a coloring of one half of T with zero defect at the shared vertex z; this is exactly the false conclusion def(c)=0 when z plays the role of c. Lemma 7.7, Case 2 is representative: the proof chooses φ(z)=z1 with def φ(z)=0 and then colors the other half with only a 1-defective coloring, so if the first half could only guarantee def φ(z)≤1, the total defect at z could become 2. The counterexample from my first comment is a cover of R that is neither wedged nor twisted, so it can occur in a good cover of T\{u,v} in the case ℓ(x)=1, ℓ(z)=2, ℓ(y)=2, which is automatically good under Definition 5.6. Consequently Lemma 7.1, Lemma 5.9, and Theorem 5.11, and therefore the 1-defective 3-correspondability half of Theorem 1.4, are not established. Section 9's proof of non-4-correspondability appears independent of this lemma, but the two halves of Theorem 1.4 together are unsupported as written.
minor comments (3)
  1. [§2, Definition 2.1] In the displayed definition of E(H) for a list cover, the text reads "c= d}}" with a doubled closing brace and an unspaced equality; this should be "c = d" and a single closing brace.
  2. [§8, Lemma 8.7] In the final paragraph of the proof of Lemma 8.7, the sentence "if ψ is the coloring induced by ... then ψ is a 1-defective H coloring φ of T" mixes the names ψ and φ; the last symbol should be ψ.
  3. [§4, Theorem 4.1 proof, Case 2] In Case 2 of the proof of Theorem 4.1, the line "with ψ2(v) ⩽ 1 and ψ2(z) ⩽ 2" should read "with def ψ2(v) ⩽ 1 and def ψ2(z) ⩽ 2" for clarity.

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity: the proof is a self-contained constructive argument with no fitted inputs and no load-bearing self-citations.

full rationale

The paper's claims are proved by explicit combinatorial constructions: it builds correspondence covers and derives defective colorability or non-colorability from local matching conditions. There is no parameter fitted to the target result, no definition that presupposes the conclusion, and no 'prediction' that is forced by the way an input was constructed. The only self-citation, reference [1], is used for expository text on list/correspondence coloring, not as a load-bearing premise. Background results such as the degeneracy-correspondence bound from Bernshteyn and Lee are external and used as standard lemmas; they do not assume the paper's conclusions. The reader's proposed counterexample to Lemma 5.4(i) concerns the mathematical validity of a lemma, not circularity: an incorrect proof step is a correctness problem, not a reduction of the theorem to its own inputs. Accordingly, no circular step meeting the quoted-evidence standard was found.

Assumptions & free parameters 0 free parameters · 3 assumptions · 0 invented entities

No empirical or fitted parameters are used; construction sizes such as 63, 7, 9, 12, and 4 are exact constants rather than tuned values. The central proof relies on standard correspondence coloring definitions, the degeneracy-correspondence fact, and the false Lemma 5.4(i). No new particles, forces, or external entities are postulated.

assumptions (3)
  • standard math A d-degenerate graph is (d+1)-correspondable (Lemma 2.2, cited to Bernshteyn and Lee).
    Used to show the graph in Theorem 1.3 is 4-correspondable and to connect degeneracy with correspondence coloring.
  • domain assumption Every planar graph is 0-def 5-correspondable, adapting Thomassen's 5-choosability to correspondence coloring.
    Background for the question of which pairs (d,k) work for all planar graphs; not load-bearing for the new constructions.
  • ad hoc to paper Lemma 5.4(i): every cover of R with list sizes 1,2,2,2 admits a 1-defective coloring with zero defect at c.
    Assumed in the proof of Theorem 1.4 and used in multiple lemmas. This assumption is false, which invalidates the proof of Theorem 1.4.

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Pith. "Pith review of Defective correspondence coloring of planar graphs." pith.science (2026). https://pith.science/paper/57YWEJJK

@misc{pith2026241115336,
  author       = {Pith},
  title        = {Pith review of: Defective correspondence coloring of planar graphs},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/57YWEJJK}},
  note         = {Machine review of arXiv:2411.15336}
}
abstract

Defective coloring (also known as relaxed or improper coloring) is a generalization of proper coloring defined as follows: for $d \in \mathbb{N}$, a coloring of a graph is $d$-defective if every vertex is colored the same as at most $d$ of its neighbors. We investigate defective coloring of planar graphs in the context of correspondence coloring, a generalization of list coloring introduced by Dvo\v{r}\'ak and Postle. First we show there exists a planar graph that is not $3$-defective $3$-correspondable, strengthening a recent result of Cho, Choi, Kim, Park, Shan, and Zhu. Then we construct a planar graph that is $1$-defective $3$-correspondable but not $4$-correspondable, thereby extending a recent result of Ma, Xu, and Zhu from list coloring to correspondence coloring. Finally we show all outerplanar graphs are $3$-defective $2$-correspondence colorable, with 3 defects being best possible.

Figures

Figures reproduced from arXiv: 2411.15336 by the authors.

Figure 1
Figure 1. Graph T in the proof of Theorem 1.3. Definition 2.3. Let G1 and G2 be isomorphic (say by σ) and let H1 = (L1, H1) and H2 = (L2, H2) be correspondence covers of G1 and G2 respectively. Then we say H1 and H2 are isomorphic, denoted H1 ∼= H2, if there exists an isomorphism ψ from H1 to H2 that preserves lists, i.e., for all v ∈ G1 and c ∈ H1 it follows c ∈ L1(v) ⇐⇒ ψ(c) ∈ L2(σ(v)). Definition 2.4. Let G be a graph and … view at source ↗
Figure 2
Figure 2. Graph H; a cover for graph T in the proof of Theorem 1.3. argument shows x3 and y3 are not the same color. But then it is easy to see that z3 will conflict with at least two of x2, y2, x3, y3, and thus def(z3) ⩾ 4, a contradiction. ■ We are now ready to prove Theorem 1.3. Theorem (1.3). There exists a planar graph that is not 3-defective 3-correspondable. Proof. Let G consist of 63 copies of T with all the copies of… view at source ↗
Figure 3
Figure 3. G and correspondence cover H in proof of Theorem 1.6. Consider the fan of length 12, i.e., a path z1, z2, . . . , z12 with a vertex u adjacent to each zi . Now for each edge zizi+1, i ∈ [11], add a vertex yi adjacent to zi and zi+1. Call the resulting graph G (see Fig 3i). Clearly G is outerplanar. Let H be a correspondence cover for T as follows: for each i ∈ [11], let Myizi+1 be the matching with 1 corresponding t… view at source ↗
Figures from the paper (9 more)
Figure 4
Figure 4. Figure 4: Graphs R and T. 5.1. Twisted and Wedged Definition 5.1. Let H = (L, H) be a correspondence cover for R with L(a) = {a1}, L(b) = {b1, b2}, L(c) = {c1}, L(d) = {d1, d2}. Then we say H is (1) twisted if a1b1d2c1b2d1a1 is a cycle in H, (2) wedged if a1b1, a1d1, and at leas…
Figure 5
Figure 5. Figure 5: Twisted and wedged covers for R. Proof. By applying an appropriate isomorphism to H, we may assume without loss of gener￾ality that H = (L, H) is such that L(a) = {a1}, L(b) = {b1, b2}, L(c) = {c1}, L(d) = {d1, d2}. By Remark 2.5, all matchings are maximal, and thus we…
Figure 6
Figure 6. Figure 6: A wedged cover for R with |L(c)| = 2. Note that although we are defining wedged in both Definition 5.1 and Definition 5.3, the first is for covers H of R with |L(c)| = 1, and the second for covers H of R with |L(c)| = 2. It should be clear from context how the definiti…
Figure 7
Figure 7. Figure 7: 1-Bad cover H as in Lemma 8.1 (some edges and vertices not shown). 8. Proof of Lemma 5.10 The proof of Lemma 5.10 follows from the series of lemmas below, which cover all the cases when Hφ is bad. For convenience, throughout this section assume H is such that L(x) = {x…
Figure 8
Figure 8. Figure 8: 2-bad covers as in Lemmas 8.2 and 8.3 respectively (some edges and vertices not shown). ■ Lemma 8.3. If H is 2(ii)-bad (i.e., ℓφ(x) = 1, ℓφ(z) = 1, ℓφ(y) = 2, and Hφ(R1) and Hφ(R2) are wedged), then φ can be extended to 1-defective H-colorings ψ, ρ of T such that defψ …
Figure 9
Figure 9. Figure 9: 4-bad cover H as in Lemma 8.6 (some edges and vertices not shown). Lemma 8.6. If H is 4-bad (i.e., ℓφ(x) = 1, ℓφ(z) = 2, ℓφ(y) = 1, and there exists distinct zi , zj ∈ L(z) such that H(R1) \ {zi} and H(R2) \ {zj} are twisted), then φ can be extended to 1-defective H-co…
Figure 10
Figure 10. Figure 10: 5-bad covers H as in Lemma 8.7 (some edges and vertices not shown). constructed similarly, taking x3 in place of x1. Thus we may assume Hφ(R1) is not twisted. By symmetry, we may assume Hφ(R2) is not twisted. Thus we may assume Hφ(R1) and Hφ(R2) are not twisted, and t…
Figure 11
Figure 11. Figure 11: H as in the last case of the proof of Lemma 8.7 (some edges and vertices not shown). 2 1 3 4 2 1 3 4 1 2 3 4 1 2 3 4 2 1 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 [PITH_FULL_IMAGE:figures/full_fig_p020_11.png]
Figure 12
Figure 12. Figure 12: H1,2,3;4(1, 1) (edges in identity matchings not shown). 9. Proof of Theorem 5.12 We construct a 4-fold correspondence cover H of T(4) such that T(4) is not H-colorable. We start by constructing covers for T. Let L := V (T) × [4]. For variables α, β ∈ [4], let H1,2,3;4…

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