{"id":"3b898c27-44a0-4427-ba63-aa45d881d845","arxiv_id":"2505.09809","paper_version":2,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":6.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"The density of colour-alternating 6-cycles in any red/blue edge-coloured graph is at most (1/2)^6, matching the random colouring.","lead":"This paper proves that a uniformly random red/blue edge-colouring of a large complete graph maximizes the number of colour-alternating 6-cycles. It settles the first open case of a recently posed problem on semi-inducibility in edge-coloured graphs.","discovery_kind":"new_application","skeptic_critique":{"model":"deepseek-v4-flash","headline":"The proof's key inequality (2.6) needs A to be positive semidefinite, and the paper asserts this without proof or certificate; if A has a negative eigenvalue, the argument fails.","rationale":"The reader's weakest assumption identifies exactly the load-bearing gap: the PSD assertion for A is indispensable for (2.6), the only inequality that converts the large coefficient computation into an upper bound on tinj(C_A^6, G). The coefficient arithmetic spot-checks reported by the reader support the cancellation, so this is a verifiability concern rather than a detected falsehood. The quasirandomness remark at the end is also underproved, but it is not needed for Theorem 1.1. Supplying exact eigenvalues or a semidefinite certificate, together with a script reproducing the tbip tables, would close the gap. Since the reader already assigned CONDITIONAL with moderate confidence, no verdict change is needed.","tokens_in":21508,"tokens_out":28469,"duration_ms":290492,"concrete_test":"Compute the eigenvalues of the rational matrix A exactly, for example with sympy.Matrix.eigenvals or an exact rational LDL^T / Cholesky factorization. The claim survives this check if and only if the minimum eigenvalue is 0 (with the all-ones vector as kernel) and all other eigenvalues are nonnegative; any negative eigenvalue invalidates inequality (2.6).","verdict_should_be":"UNCHANGED","load_bearing_attack":"The central step in the proof of Theorem 2.1 is the assertion, in Section 2, that the 8x8 matrix A is positive semidefinite. This is what justifies adding the quadratic forms x^T A x and y^T A y to the right-hand side of (2.6) and then showing that the resulting expression has coefficient (1/2)^6 for every d(J_l, G). If A had a negative eigenvalue, (2.6) would not be an upper bound: a negative quadratic term could make the displayed expression smaller than tinj(C_A^6, G), so the subsequent coefficient computation would prove nothing. The all-ones vector is visibly in the kernel because all row sums are zero, but the remaining eigenvalues are not shown. The long tables of tinj(R_i · R_j, G) and tinj(B_i · B_j, G) are also asserted after 'computing explicitly', but those can be checked entry-by-entry once the tables are generated; the PSD property is the one unconditional inequality that cannot be recovered by arithmetic cancellation. This is not a claim that the theorem is false; it is a missing certificate for a load-bearing computational step.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper proves that in every red/blue edge-coloured graph G, the homomorphism density t(C^A_6, G) of the colour-alternating 6-cycle is at most (1/2)^6, with equality asymptotically for uniformly random colourings. The proof follows the flag algebra method: it reduces to injective homomorphism densities, partitions the copies of K_{3,3} into 26 isomorphism types, derives an expression for tinj(C^A_6, G) as a weighted combination of the densities d(J_ℓ, G), introduces 16 rooted flags, and proves the desired upper bound by adding two positive semidefinite quadratic forms based on an 8×8 matrix A. The proof concludes by computing, for each of the 26 types J_ℓ, that the coefficient of d(J_ℓ, G) is (1/2)^6. The final paragraph outlines an argument that extremality forces quasirandomness.","tokens_in":21750,"tokens_out":4120,"duration_ms":42075,"significance":"If the proof is correct, this solves the first open case of a problem of Basit, Granet, Horsley, Kündgen and Staden on semi-inducibility of alternating cycles, showing that random colourings are asymptotically extremal for C^A_6. The paper has two notable strengths: the reduction from homomorphism densities to injective densities via the blow-up argument is clean, and the flag algebra certificate is unusually small and presented in human-readable form, sparing the reader the large semidefinite programs typical of such proofs. The paper also correctly credits independent work of Balogh, Lidický, Mubayi and Pfender. However, the main result rests on two computational assertions that are not actually proved: the positive semidefiniteness of the matrix A and the correctness of the long tables of tbip values. Both are load-bearing, so the current manuscript is not yet a complete proof.","major_comments":[{"comment":"The claim 'The matrix A is positive semi-definite (i.e. all of its eigenvalues are non-negative)' is the only justification for the inequalities x^T A x ≥ 0 and y^T A y ≥ 0 that are used in (2.6). This is a load-bearing step: if A had any negative eigenvalue, the expression in (2.6) would not be an upper bound for tinj(C^A_6, G). The paper does not provide a proof or certificate for this assertion. I verified that all row sums are zero, so the all-ones vector is in the kernel, but the remaining eigenvalues are not shown. Please provide a verifiable certificate, such as an exact Cholesky factorization A = LL^T with rational entries, a list of all eigenvalues with algebraic or numerical verification, or a sum-of-squares decomposition of the quadratic form. Without this, the central inequality (2.6) is unsupported.","section":"Section 2, definition of A"},{"comment":"The displayed formulas for tinj(R_i · R_j, G) and tinj(B_i · B_j, G) (36 + 36 identities) are introduced with 'By computing the quantities tbip(...) explicitly ... we obtain the following:' and no further derivation. These formulas are used as input to the coefficient calculations that prove every coefficient of d(J_ℓ, G) equals (1/2)^6, so if any one of them is wrong the proof fails. The computation is in principle checkable by direct enumeration, but as written it is a bare assertion. Please provide either a human-readable derivation (e.g., the generating table of tbip values for each pair (i,j) and each J_ℓ) or a short computer script that verifies all of them, to make the proof independently checkable.","section":"Section 2, tbip tables"},{"comment":"The arithmetic showing that each coefficient of d(J_ℓ, G) in (2.6) equals (1/2)^6 is long and not machine-verified in the manuscript. While the displayed sums are explicit, the sheer number of terms makes this a possible source of transcription errors. In particular, the line for J11 contains the garbled fragment 'J11: 1/12 1/6+ +' which appears to be a typesetting mistake; the actual computation is interleaved with the J9 and J12 computations. Please audit the coefficient listings and include the global calculation in a more transparent form (e.g., a table of the final coefficient for each J_ℓ, or an appendix with the complete arithmetic) so that the conclusion that (2.6) evaluates to (1/2)^6 + o(1) can be checked without manual re-entry of every term.","section":"Section 2, coefficient calculations"}],"minor_comments":[{"comment":"The abstract says 'the number of colour alternating 6-cycles', but the paper actually studies homomorphism densities (and, via Theorem 2.1, injective homomorphism densities). Please clarify the phrasing to reflect the formal definition used in the paper.","section":"Abstract"},{"comment":"The quasirandomness characterization is stated only as a brief summary. Since it also relies on the unproved assertion that the kernel of A is spanned by the all-ones vector, please either prove it in full or clearly label it as a conjecture/remark. As written, it is a stronger claim than Theorem 1.1 and would require the same computational certificate.","section":"Final paragraph"},{"comment":"Reference [1] is to a preprint arXiv:2501.09842v1; reference [2] is to another preprint. Please add any published or more stable versions if they have appeared, and in Remark 1.2 the independent work of Balogh, Lidický, Mubayi and Pfender is mentioned without a bibliographic entry; a citation after their work becomes public would be appropriate.","section":"References"}],"recommendation":"major_revision","confidential_remarks":"The main obstacle is the missing PSD certificate for A. The proof structure is otherwise coherent, and the paper is a good candidate for acceptance once the missing computational certificates are supplied. The independent work of Balogh et al. may make this a race-to-the-bottom competitive area; I would encourage the editor to give the authors a reasonable but not unlimited time to provide the certificates, since the result appears correct but the written proof is currently incomplete. No concerns about novelty or citation fairness beyond the missing bibliographic entry for the independent group."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"One thing to know: the paper proves the first open case of Basit et al.'s semi-inducibility problem, showing t(C_A_6, G) ≤ (1/2)^6, tight for a random red/blue clique. The argument is a standard flag algebra proof, but the key inequality depends on an 8x8 matrix A being positive semidefinite, and the paper merely asserts this. If that assertion is wrong, the proof collapses. I don't think it is wrong, but the authors need to show their work.\n\nThe result is new and correctly positioned: Theorem 1.1 does not follow from the earlier cases solved in [1] and [2], and the authors are transparent that Balogh, Lidický, Mubayi and Pfender independently have the same theorem. The proof structure is clean. The reduction to injective homomorphisms via the blow-up argument is standard and correct. The decomposition of K3,3 colorings into 26 isomorphism types and the use of 16 rooted flags with missing edges is a nice simplification; it avoids the much larger K6 flag set and makes the coefficient arithmetic human-checkable in principle. I spot-checked several of the displayed coefficient computations and they cancel to (1/2)^6 as claimed.\n\nThe soft spot is exactly what the stress-test note flags. In Section 2, the step from (2.5) to (2.6) adds x^T A x and y^T A y to the right-hand side, justified solely by 'A is positive semi-definite.' That is the load-bearing unconditional inequality of the proof. No eigenvalues, no Cholesky factor, no certificate are provided. The all-ones vector is in the kernel by construction, but the remaining eigenvalues are not shown. The long tables of tbip(R_i·R_j, J_l) and tbip(B_i·B_j, J_l) are also asserted after 'computing explicitly,' though those are finite enumerations and can be checked entry-by-entry. The PSD claim cannot be recovered from the displayed arithmetic alone; it is a separate computational fact. The quasirandomness remark at the end is stated without proof, but it is clearly labeled as a remark and not needed for the theorem.\n\nI would not desk-reject this. The theorem is significant for extremal graph theory, the presentation is honest, and the method is appropriate. But I would send it to a referee who knows flag algebras and ask for the PSD certificate—eigenvalues or a rational Cholesky decomposition—and ideally the script that generated the tbip tables. With that supplied, I would accept. As it stands, it is a conditional accept rather than a clean one.","headline":"Proves the first open case of the semi-inducibility problem for alternating 6-cycles with a clean flag algebra argument, but the key inequality depends on an uncertified positive semidefiniteness claim.","tokens_in":22256,"tokens_out":2571,"would_cite":true,"duration_ms":24945,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["05C35","05C15","05D40"],"pacs":[],"model":"deepseek-v4-flash","headline":"Every edge-coloured graph has at most $(1/2)^6$ alternating 6-cycles per 6-vertex set, with equality approached only by random colourings.","keywords":["alternating cycles","edge-coloured graphs","homomorphism density","semi-inducibility problem","flag algebras","quasirandomness","extremal combinatorics","red-blue colourings"],"falsifier":"Compute the eigenvalues of the $8\\times 8$ matrix $A$ printed in Section 2; if any eigenvalue is negative, the inequality (2.6) is not justified and the proof collapses. Alternatively, run a computer search over edge-coloured cliques on 8–10 vertices for one whose alternating-6-cycle homomorphism density exceeds $1/64$.","tokens_in":21294,"feed_emoji":"","tokens_out":4734,"duration_ms":47197,"temperature":0.7,"pith_summary":"The paper proves that in any red/blue edge-coloured graph, the number of colour-alternating 6-cycles, counted as homomorphic copies and normalized per 6-tuple of vertices, is at most $(1/2)^6 = 1/64$. This bound is asymptotically tight, because a uniformly random red/blue colouring of a large clique achieves it. The result solves the first open case of the semi-inducibility problem for alternating cycles.","feed_headline":"Random colourings win the alternating-6-cycle maximum","feed_subtitle":"A flag-algebra proof shows no red-blue graph packs more alternating 6-cycles than a uniformly random clique.","key_machinery":"The proof reduces the problem to injective homomorphisms, partitions the $512$ red/blue colourings of $K_{3,3}$ into the $26$ isomorphism classes $J_1,\\ldots,J_{26}$, and expresses the alternating-6-cycle density as a linear combination of the class densities $d(J_\\ell,G)$. The core object is an $8\\times 8$ matrix $A$ together with sixteen rooted flags $R_1,\\ldots,R_8$ and $B_1,\\ldots,B_8$; because $A$ is asserted positive semidefinite, the quadratic forms $\\vec{x}_{u,v}^T A \\vec{x}_{u,v}$ and $\\vec{y}_{u,v}^T A \\vec{y}_{u,v}$ are nonnegative. Adding these nonnegative terms to the expression for $t_{\\mathrm{inj}}(C_6^A,G)$ and double-counting the resulting products of flag densities yields exactly $(1/2)^6\\sum_{\\ell=1}^{26} d(J_\\ell,G) + o(1)$.","core_discovery":"The central claim is Theorem 1.1: every edge-coloured graph $G$ satisfies $t(C_6^A, G) \\leq (1/2)^6$. Since the uniformly random colouring of a large clique has alternating-6-cycle density $(1/2)^6 + o(1)$, this identifies random colouring as the asymptotic extremal configuration. The proof further shows that a sequence of edge-coloured cliques approaches density $(1/2)^6$ if and only if the sequence is quasirandom, meaning every edge-coloured pattern appears with the frequency expected from a random colouring.","pith_inferences":["Because extremality implies quasirandomness, the alternating 6-cycle is a forcing object: a single subgraph density constraint forces the entire colouring to be random, a phenomenon that may hold for every alternating cycle of length $2 \\bmod 4$.","A short rational Cholesky decomposition of the $8\\times 8$ matrix $A$ would turn the asserted positive semidefiniteness into a human-checkable certificate, making the proof fully verifiable without computer algebra.","The use of $K_{3,3}$ rather than $K_6$ suggests a general template: for alternating $2k$-cycles, one might work with $K_{k,k}$ colourings and rooted flags with missing edges, which could keep the flag count manageable for $k=5$ or $6$.","If the suggested bound $t(C_{4k+2}^A, G) \\leq (1/2)^{4k+2}$ holds, then random colourings simultaneously extremize all alternating cycles of length $2 \\bmod 4$, strengthening the case for a general semi-inducibility theorem."],"forward_implications":["Every edge-coloured graph has alternating-6-cycle homomorphism density at most $(1/2)^6$, so no red/blue pattern can beat a random colouring.","The extremal sequences are exactly the quasirandom ones: approaching the maximum forces the red graph to be quasirandom.","This solves the first open case of the semi-inducibility problem for alternating cycles.","The paper suggests, but does not prove, the general bound $t(C_{4k+2}^A, G) \\leq (1/2)^{4k+2}$ for all $k \\geq 1$.","The unusually small flag set (16 flags rather than 40) hints that flag-algebra proofs with missing edges and bipartite ambient graphs may apply to larger alternating cycles."],"supporting_citations":[{"why":"Introduces the semi-inducibility problem, defines the class of alternating cycles, and poses the open case solved in this paper.","marker":"[1]"},{"why":"Provides the companion entropy method for alternating paths of odd length, situating the present method and result within the same programme.","marker":"[2]"},{"why":"Supplies the flag-algebra machinery used to reduce the extremal problem to a finite double-counting calculation.","marker":"[3]"}],"fun_headline_variants":["Random colourings max out alternating 6-cycles","Flag algebra proves random colourings maximize alternating 6-cycles","Quasirandom colourings are extremal for alternating 6-cycles","Alternating 6-cycles: random colouring is the asymptotic max"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The 8-by-8 matrix $A$ is asserted, without proof or certificate, to be positive semidefinite, and the entire upper-bound inequality (2.6) depends on that assertion.","fun_headline_variants_meta":{"raw":{"variants":["Random colourings max out alternating 6-cycles","Flag algebra proves random colourings maximize alternating 6-cycles","Quasirandom colourings are extremal for alternating 6-cycles","Alternating 6-cycles: random colouring is the asymptotic max"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.00187,"raw_usage":{"total_tokens":7222,"prompt_tokens":708,"completion_tokens":6514,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":324,"completion_tokens_details":{"reasoning_tokens":6440}},"tokens_in":324,"tokens_out":6514,"duration_ms":48724,"temperature":1.0,"reasoning_tokens":6440,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-15T21:24:16.574427+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Compute the eigenvalues of the $8\\times 8$ matrix $A$ printed in Section 2; if any eigenvalue is negative, the inequality (2.6) is not justified and the proof collapses. Alternatively, run a computer search over edge-coloured cliques on 8–10 vertices for one whose alternating-6-cycle homomorphism density exceeds $1/64$.","supporting_citations":[{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"Supplies the flag-algebra machinery used to reduce the extremal problem to a finite double-counting calculation."}],"review_version":1}