{"id":"13ade574-2aca-4ad6-93ef-9b6388d0c508","arxiv_id":"2505.16808","paper_version":1,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":8.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"A planar graph construction using signed Wenger gadgets disproves the conjecture that every planar graph has fractional arboricity at most 2.","lead":"Researchers construct planar graphs from signed-graph gadgets to show that the fractional arboricity of planar graphs can exceed 2, refuting a 2020 conjecture. The same constructions push lower bounds on the fractional balanced chromatic number of planar signed graphs to 83/41.","discovery_kind":"new_application","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Lemma 3's positive-triangle mini-gadget is asserted, not proved; Theorem 6 collapses if the assertion fails.","rationale":"The reader identified Lemma 3 as the weakest assumption, and the stress-test concurs. Lemma 4 is a careful case analysis conditional on the triangle property; the negative-triangle enforcement is rigorous by the (K4,-) counting argument. The positive-triangle mini-gadget, however, is left as a figure plus a one-line assertion. Because the entire counterexample to Conjecture 1 depends on forcing the two positive triangles ux1x2 and vx3x4 to have the triangle property, an unverified mini-gadget is a genuine load-bearing gap. The paper contains additional minor defects (the sign reversal in Theorem 13's 'In other words' line and an inconsistent variable in Theorem 16's counting inequality), but these do not affect Theorem 6; they would be corrected in revision. Since the reader's CONDITIONAL verdict already reflects this uncertainty and the central construction is plausible, no verdict change is recommended.","tokens_in":12690,"tokens_out":44407,"duration_ms":346540,"concrete_test":"Obtain from the authors the explicit signed graph of Figure 1 (all vertices, edges, signs, including any vertex placed inside the inner negative triangle). Then verify the key assertion for k=1 and k=2 by exhaustive search (or ILP/SAT) over balanced (2k,k)-colorings: no color appears on all three outer vertices. Independently re-derive the general counting argument: if a color c is on u_1,u_2,u_3, the negative triangles (u_{j-1},u_{j+1},u'_j) force c off each u'_j; then the (K4,-) completion of u'_1u'_2u'_3 forces c to appear twice on the added vertex, impossible. If either the k=2 search or the counting argument fails, Lemma 3 is false and Theorem 6 is unproved.","verdict_should_be":"UNCHANGED","load_bearing_attack":"Theorem 6 rests on Lemma 4, which requires the seven specified triangles of W' to have the 'triangle property' in every balanced (2k,k)-coloring. The negative-triangle part of Lemma 3 is sound: adding a vertex to form (K4,-) forces each color to appear exactly twice on the four vertices, so every color appears at least once on the original triangle. The positive-triangle part, however, is not demonstrated. The proof says only: 'if a color c appears in all three of the u_i, then it can appear in none of the u'_j which is not possible.' The graph of Figure 1 is not described by an edge/sign list, the phrase 'which is not possible' depends on the inner negative triangle u'_1u'_2u'_3 having the triangle property, and that property is only 'assumed.' If the actual mini-gadget does not force c off all u'_j, then in Lemma 4 the step 'by the triangle property on ux1x2 we have x2 not in A' (the text has a missing overline) is unjustified, and the whole (2k,k)-coloring obstruction may not hold. This is the single most load-bearing step for the paper's main counterexample; the 83/41 sign error and Theorem 16 variable inconsistency are separate and do not rescue this gap.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper studies fractional balanced colorings of planar signed graphs and fractional arboricity of planar graphs. Its central construction is a planar signed simple graph whose fractional balanced chromatic number is claimed to be strictly larger than 2, which would refute the Bonamy–Kardoš–Kelly–Postle conjecture that every planar graph has fractional arboricity at most 2. The paper further claims, by iterating the construction, that the supremum of the fractional balanced chromatic number over planar signed simple graphs is at least 83/41, and that a specific planar graph W1 has fractional arboricity exactly 2 + 2/25. The arguments rest on a Wenger-type gadget and a 'triangle property' enforcement lemma, together with explicit colorings given in several tables.","tokens_in":12881,"tokens_out":21208,"duration_ms":148584,"significance":"If the main claims are correct, the paper settles a notable open conjecture in the negative and provides the best known lower bounds both for fractional balanced chromatic numbers of planar signed graphs and for fractional arboricity of planar graphs. The paper is constructive: it gives explicit colorings in Tables 1–4, including a (172,85)-coloring and a (52,25)-coloring, and it identifies concrete graphs rather than relying on non-effective arguments. These are genuine strengths. However, the central construction depends on a lemma whose proof is not fully specified, and the derivation of the 83/41 lower bound contains a sign error. Because the main theorem is load-bearing on these points, the paper cannot be accepted in its current form, but the identified issues appear repairable within the scope of the manuscript.","major_comments":[{"comment":"The enforcement of the triangle property for positive facial triangles is asserted, not proved. The proof states that 'if a color c appears in all three of the u_i, then it can appear in none of the u′_j which is not possible,' but it does not specify the signs of the edges between the outer triangle u1u2u3 and the inner triangle u′1u′2u′3, nor does it justify why c cannot appear on any u′_j. A complete argument requires that the three triangles u_{j+1}u_{j+2}u′_j (indices mod 3) be negative, so that a color appearing on all three outer vertices and on one inner vertex would form a negative monochromatic triangle. In addition, the inner negative triangle u′1u′2u′3 and the three new facial negative triangles created by the mini-gadget must themselves be given the triangle property by the (K4,−) construction, which is not stated. Since Lemma 4 subsequently uses the triangle property on the positive triangles ux1x2 and vx3x4, this gap is load-bearing for Theorem 6.","section":"§2, Lemma 3 (Figure 1)"},{"comment":"The derivation of p/q ≥ 83/41 contains a sign error. Corollary 12 gives m_{p,q} ≤ p − 3q + 21m_{p,q}; rearranging yields m_{p,q} ≥ (3q − p)/20, not m_{p,q} ≤ (3q − p)/20 as written. Only after this correction does the combination with m_{p,q} ≤ 2p − 4q (Lemma 10) give p/q ≥ 83/41. The theorem statement is therefore plausible, but the proof as printed is invalid.","section":"§3, Theorem 13"}],"minor_comments":[{"comment":"There is a typo: 'minimum total wight' should be 'weight'.","section":"§1, Abstract"},{"comment":"The sentence 'In coloring of Table 2 for every edge the number of common colors on the end points of each edge is 14' should refer to Table 3, not Table 2.","section":"§4, proof of Theorem 15"},{"comment":"In the third base-case tuple for operation [1], the values (a12,a13,a23,a14,a24,a34) = (14,14,14,14,14,13) are inconsistent with the listed third possibility (a12,a13,a23) = (14,13,13) and with the counting identity 2Σa_ij + Σb_i = 4q. Presumably the tuple should be (14,13,13,14,14,13).","section":"§4, proof of Theorem 15"},{"comment":"The list of triangles required to satisfy the triangle property omits wx1x5, which is among the seven triangles enumerated in Remark 5.","section":"§4, Theorem 14"},{"comment":"The mini-gadget of Figure 1 would be much easier to verify if the signs of all edges were given explicitly in an adjacency list; the current figure alone is insufficient to support the argument in Lemma 3.","section":"§2, Figure 1"}],"recommendation":"major_revision","confidential_remarks":"The paper addresses a significant conjecture and the constructions are plausible, but the referee cannot verify the main theorem without a substantial expansion of Lemma 3. The sign error in Theorem 13 is easy to fix, but the missing verification of the positive-triangle mini-gadget is load-bearing. The later explicit colorings (Tables 1–4) give some confidence that the gadget is valid, so I recommend major revision rather than rejection. The paper fits the journal's scope well."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Short version: this is a serious paper with a genuinely new counterexample to a plausible conjecture, but the proof of the key lemma is sketched in a way that I can't fully verify from the text. It should be refereed, not desk-rejected, but the authors need to fill a real gap.\n\nWhat's new: the construction using Wenger's gadget to get a planar signed graph with χ_fb > 2, the iterative lower bound to 83/41, and the explicit (52,25)-coloring showing fractional arboricity 2+2/25 for W1. The paper is self-contained and the main idea is nice. The list of maximal balanced sets in Lemma 8 is checkable, and the coloring tables are concrete.\n\nWhere I'm nervous: Lemma 3 is central. The negative-triangle part (adding a vertex to make (K4,−)) is fine. The positive-triangle part is one sentence: 'if a color c appears in all three u_i, then it can appear in none of the u'_j which is not possible.' That 'which is not possible' depends on the inner negative triangle having the triangle property, but that property is only asserted. The figure isn't described by an edge/sign list, so I can't check whether the mini-gadget actually forces this. Without this, Lemma 4 and Theorem 6 don't go through. This is not a minor typo; it's a proof obligation.\n\nThe sign error in Theorem 13 is minor: from m ≤ p−3q+21m you get m ≥ (3q−p)/20, not ≤. The paper then continues as if the correct inequality were there, so the intended argument is recoverable. Theorem 16's proof has a notational tangle with a and a_uv, but the inequalities work when you sort out which variable is which.\n\nBottom line: the central claim is plausible and the construction is likely correct, but the paper as written is not convincing enough for me to cite as a disproof until Lemma 3 is proved in detail. I'd send it to a knowledgeable referee with a request to verify the mini-gadget, and I'd tell the authors to fix the sign error. It deserves peer review.","headline":"A likely correct counterexample to the Bonamy–Kardos–Kelly–Postle conjecture, but the proof of the key mini-gadget lemma is too sketchy to certify.","tokens_in":13476,"tokens_out":5855,"would_cite":false,"duration_ms":41405,"reading_group":"yes","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["05C15","05C10","05C22"],"pacs":[],"model":"deepseek-v4-flash","headline":"A planar signed simple graph exists whose fractional balanced chromatic number is strictly larger than 2.","keywords":["fractional balanced chromatic number","signed graphs","planar graphs","fractional arboricity","balanced coloring","fragment gadget","arboricity conjecture","graph coloring"],"falsifier":"Search exhaustively for a balanced $(2,1)$-coloring of the 64-vertex signed planar graph obtained by identifying the three $z$-vertices in the construction; if any assignment of two colors to the vertices has each color class free of negative cycles, then the claimed nonexistence of $(2k,k)$-colorings fails at $k=1$ and the main theorem collapses.","tokens_in":12443,"feed_emoji":"🎨","tokens_out":9043,"duration_ms":66586,"temperature":0.7,"pith_summary":"For a signed graph, a balanced set is a vertex set inducing no negative cycle, and a fractional balanced coloring places nonnegative weights on balanced sets so each vertex receives total weight at least 1; the minimum total weight is the fractional balanced chromatic number. This paper constructs a planar signed simple graph whose fractional balanced chromatic number is strictly larger than 2. Since that number is always at most the fractional arboricity of the underlying graph, the same construction refutes the conjecture that every planar graph has fractional arboricity at most 2. Iterating the gadget pushes the lower bound on the supremum for planar signed simple graphs to $83/41$, with a concrete first member at $2+2/85$.","feed_headline":"Planar signed graph needs more than 2 fractional colors","feed_subtitle":"Construction refutes the conjecture that every planar graph has fractional arboricity at most 2.","key_machinery":"The load-bearing object is a signed planar gadget built from the classical fragment construction, together with the triangle property. A facial triangle has the triangle property if, in every balanced $(2k,k)$-coloring, a negative triangle receives each color at least once and a positive triangle receives each color at most twice. Lemma 3 claims that any facial triangle of a plane signed graph can be completed by adding a vertex inside the face, and a mini-gadget for positive triangles, so that the triangle property holds, and Lemma 4 then forces the distinguished endpoints $u$ and $v$ of the gadget to share no color in any balanced $(2k,k)$-coloring. That no-common-color conclusion is what makes the global construction exceed two colors.","core_discovery":"The paper's central result is that a balanced $(2k,k)$-coloring cannot exist for a certain finite planar signed graph, so its fractional balanced chromatic number exceeds 2. The proof's key step shows that in the gadget $<span class=\"math\">$\\widehat{W}''$</span> the distinguished endpoints $u$ and $v$ must receive disjoint color sets in every such coloring; placing copies of the gadget on the three edges of a triangle then demands $3k$ colors from a palette of $2k$, a contradiction. Because balanced color classes are more restrictive than acyclic ones, this also yields a planar graph with fractional arboricity above 2, refuting the conjecture of $<span class=\"math\">$[2]$</span>. The paper then computes the exact value $2+2/85$ for the first iterated graph, proves a limiting lower bound of $83/41$, and exhibits an explicit planar graph with fractional arboricity $2+2/25$.","pith_inferences":["The exact supremum of the fractional balanced chromatic number for planar signed simple graphs is not determined here; it lies somewhere between $83/41$ and the general upper bound $5/2$.","Because the same gadget drives both parameters, the unsigned counterpart has a gap as well: $a_f(W_1)=2+2/25$ is a concrete planar graph exceeding 2, but the supremum of planar fractional arboricity could be larger.","A direct computer verification of Lemma 3 on the 64-vertex graph for small $k$ would be a natural independent test of whether the triangle-completion step is as robust as stated.","If the triangle property can be enforced on other face types, the same construction pattern could be adapted to signed graphs on higher-genus surfaces or to other families with similar balancing constraints."],"forward_implications":["Conjecture 1 is false: there is a planar graph with $a_f(G)>2$, and the construction can be made concrete with $a_f(W_1)=2+2/25$.","The fractional balanced chromatic number of planar signed simple graphs has no upper bound of 2; its supremum is at least $2+1/41$.","In the iterative family, the first nontrivial member has exact fractional balanced chromatic number $2+2/85$, so the obstruction appears already in a finite explicit graph.","No graph built from a negative triangle by the two allowed operations needs more than $83/41$ colors in the fractional sense, so the bound $83/41$ is the ceiling for this construction method."],"supporting_citations":[{"why":"states the conjecture that every planar graph has fractional arboricity at most 2, which the construction refutes.","marker":"[2]"},{"why":"supplies the fragment gadget whose signed version is the core of the construction.","marker":"[12]"},{"why":"provides the planar acyclic 5-coloring result used to bound fractional arboricity by 2.5 for planar graphs.","marker":"[3]"},{"why":"introduces balanced colorings of signed graphs, the basis of the parameter studied here.","marker":"[13]"},{"why":"refutes a related balanced-partition conjecture with a fragment-based planar example, motivating the approach.","marker":"[6]"},{"why":"introduces the fractional balanced chromatic number and its Hadwiger-style context.","marker":"[5]"},{"why":"defines fractional balanced coloring of signed graphs and the parameter used throughout the paper.","marker":"[7]"}],"fun_headline_variants":["Planar signed graph beats fractional chromatic bound 2","Fractional balanced chromatic number breaks 2 for planar signed graphs","Refuting conjecture: planar arboricity exceeds 2","Planar signed graphs require fractional colors >2"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The proof depends on the claim that every facial triangle can be equipped with a small completion that forces the triangle property in every balanced $(2k,k)$-coloring; if that enforcement ever fails, the conclusion that the gadget's distinguished endpoints share no color no longer follows.","fun_headline_variants_meta":{"raw":{"variants":["Planar signed graph beats fractional chromatic bound 2","Fractional balanced chromatic number breaks 2 for planar signed graphs","Refuting conjecture: planar arboricity exceeds 2","Planar signed graphs require fractional colors >2"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000557,"raw_usage":{"total_tokens":2664,"prompt_tokens":974,"completion_tokens":1690,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":590,"completion_tokens_details":{"reasoning_tokens":1625}},"tokens_in":590,"tokens_out":1690,"duration_ms":9485,"temperature":1.0,"reasoning_tokens":1625,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-07T14:54:33.362069+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Search exhaustively for a balanced $(2,1)$-coloring of the 64-vertex signed planar graph obtained by identifying the three $z$-vertices in the construction; if any assignment of two colors to the vertices has each color class free of negative cycles, then the claimed nonexistence of $(2k,k)$-colorings fails at $k=1$ and the main theorem collapses.","supporting_citations":[{"cited_title":"Fractional vertex- arboricity of planar graphs, 2020","cited_arxiv_id":null,"evidence_quote":"states the conjecture that every planar graph has fractional arboricity at most 2, which the construction refutes."},{"cited_title":"Note on a paper of B","cited_arxiv_id":null,"evidence_quote":"supplies the fragment gadget whose signed version is the core of the construction."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"provides the planar acyclic 5-coloring result used to bound fractional arboricity by 2.5 for planar graphs."},{"cited_title":"Balanced decompositions of a signed graph.J","cited_arxiv_id":null,"evidence_quote":"introduces balanced colorings of signed graphs, the basis of the parameter studied here."},{"cited_title":"On the 4-color theorem for signed graphs","cited_arxiv_id":null,"evidence_quote":"refutes a related balanced-partition conjecture with a fragment-based planar example, motivating the approach."},{"cited_title":"Balanced-chromatic number and hadwiger-like conjectures.Priprint, 2024+","cited_arxiv_id":null,"evidence_quote":"introduces the fractional balanced chromatic number and its Hadwiger-style context."},{"cited_title":"Fractional balanced colouring of signed graphs.Priprint, 2025+","cited_arxiv_id":null,"evidence_quote":"defines fractional balanced coloring of signed graphs and the parameter used throughout the paper."}],"review_version":1}