Questions, not proof records

Open-problem statements, with their sources attached.

Each statement record keeps the mathematical question, a dated status source, accessible references, and any pinned Lean formulation separate from proof verification.

“Open” is a dated source assertion. In these pinned sources, sorry marks an admitted statement, not a proof. Source indexing does not mean the formulation has been independently built or certified by Therefore.
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Source labels openWikipedia · Combinatorics

Babai–Seress Conjectures on the Diameter of Finite Groups

Babai–Seress Conjecture (Conjecture 1.5)**: There exists an absolute constant CC such that the diameter of the alternating group AnA_n satisfies diam(An)nC.\operatorname{diam}(A_n) \leq n^C. Reference: L. Babai and Á. Seress, On the diameter of permutation groups, European Journal of Combinatorics 13 (1992), Conjecture 1.5

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Combinatorics

Babai–Seress Conjectures on the Diameter of Finite Groups

Babai–Seress Conjecture (Conjecture 1.7)**: There exists an absolute constant CC such that every finite simple non-abelian group GG satisfies diam(G)(logG)C.\operatorname{diam}(G) \leq (\log |G|)^C. Reference: L. Babai and Á. Seress, On the diameter of permutation groups, European Journal of Combinatorics 13 (1992), Conjecture 1.7

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Combinatorics

Graceful Tree Conjecture (Ringel–Kotzig conjecture)

Every tree admits a graceful labeling.

A graceful labeling of a tree TT with mm edges is an injective map f:V{0,,m}f : V \to \{0, \dots, m\} such that the multiset of absolute differences f(u)f(v)|f(u) - f(v)| over edges {u,v}\{u,v\} of TT equals {1,,m}\{1, \dots, m\}.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Combinatorics

Komlós conjecture

The Komlós conjecture*

There exists a universal constant K>0K > 0 such that for all n,mNn, m \in \mathbb{N} and all vectors v1,,vnRmv_1, \dots, v_n \in \mathbb{R}^m with vi21\|v_i\|_2 \le 1 (encoded here as jvij21\sum_j v_{ij}^2 \le 1), there exist signs εi{1,+1}\varepsilon_i \in \{-1, +1\} such that iεiviK\left\|\sum_i \varepsilon_i v_i\right\|_\infty \le K, i.e. iεivijK\left|\sum_i \varepsilon_i v_{ij}\right| \le K for every coordinate jj.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Combinatorics

Magic Squares

Does there exist a 3×33 \times 3 semi-magic square whose entries are all distinct positive integer cubes? A square is semi-magic if all rows and columns sum to the same total.

More precisely, we seek a 3×33 \times 3 matrix with entries aija_{ij} such that each aij=nij3a_{ij} = n_{ij}^3 for some positive integer nijn_{ij}, all nine cubes are distinct, and all row sums and column sums are equal. Reference:* Semi-Magic Square of Cubes

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Combinatorics

Pebbling number conjecture

The pebbling number conjecture: the pebbling number of a Cartesian product of connected graphs is at most equal to the product of the pebbling numbers of the factors. See Asplund, Hurlbert, and Kenter.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Combinatorics

Ramsey numbers

The open problem: determine the Ramsey number R(5,5)R(5,5).

It is known that 43R(5,5)4643 \le R(5,5) \le 46.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Combinatorics

Sidorenko's conjecture (1993)

Sidorenko's conjecture (1993).*

For every finite bipartite simple graph HH and every finite simple graph GG: t(H,G)t(K2,G)e(H)t(H, G) \ge t(K_2, G)^{e(H)}, where K2K_2 denotes the single-edge graph on 2 vertices (i.e. completeGraph (Fin 2)).

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Combinatorics

Snake in the box

For dimension 99, the length of the longest snake in the box is not known. This is currently the smallest dimension where this question is open.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Combinatorics

Sparse Ruler

Wichmann's conjecture on optimal rulers.* Every optimal ruler with more than 1313 segments is a Wichmann ruler W(r,s)W(r, s) (up to reflection, i.e. reversing the segment list). Posed by Wichmann [Wi63]; the finitely many known exceptions all have at most 1313 segments (lengths 1,13,17,23,581, 13, 17, 23, 58), and no further exceptions are known up to length 213213.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Combinatorics

Steiner Systems

Construct an S(t,k,n)S(t, k, n)-Steiner system with n>k>t>5n > k > t > 5, t<10t < 10, and n<200n < 200.

No example of a Steiner system with t>5t > 5 is known, despite a 2014 existence theorem by Keevash showing that such systems must exist for sufficiently large nn. Reference:* Large Steiner Systems

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWikipedia · Combinatorics

Union-closed sets conjecture

For every finite union-closed family of sets, other than the family containing only the empty set, there exists an element that belongs to at least half of the sets in the family.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWritten on the Wall II · Combinatorics

Written on the Wall II - Conjecture 100

WOWII Conjecture 100 (status O):

For a simple connected graph G, α(G) ≤ ⌈(max_v l(v) + 0.5 · diam(Gᶜ)) / 2⌉ where α(G) = G.indepNum is the independence number, max_v l(v) is the maximum over all vertices of the independence number of the neighbourhood (in G), and diam(Gᶜ) is the diameter of the complement Gᶜ. Note:* length(Ḡ) in DeLaVina's original is interpreted here as the diameter of the complement. The hypothesis hGc : Gᶜ.Connected is added so that diam(Gᶜ) is finite (otherwise Gᶜ.ediam = ⊤ and Gᶜ.ediam.toNat collapses silently to 0); see the module docstring above.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWritten on the Wall II · Combinatorics

Written on the Wall II - Conjecture 133

WOWII Conjecture 133:

For a simple connected graph GG, path(G)rad(G)+avgvl(v)cC4(G)\operatorname{path}(G) \ge \operatorname{rad}(G) + \lfloor \mathrm{avg}_v\, l(v) \rfloor^{cC_4(G)}, where path(G)\operatorname{path}(G) is the path number of the graph (number of vertices of a largest induced path), rad(G)\operatorname{rad}(G) is the radius (minimum eccentricity, as a natural number), avgvl(v)=l(G)\mathrm{avg}_v\, l(v) = l(G) is the average independence number of vertex neighbourhoods, and cC4(G)cC_4(G) is the C4C_4-free characteristic function (1 if GG is C4C_4-free, not necessarily induced, and 0 otherwise).

We read DeLaVina's bracket notation [average of λ(v)] in the source as the floor (a standard Graffiti.pc convention), hence ⌊l G⌋ in Lean.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWritten on the Wall II · Combinatorics

Written on the Wall II - Conjecture 141

WOWII Conjecture 141

For a simple connected graph G, tree(G) ≥ ⌊girth(G) / 2⌋ - 1 + max_v l(v) where tree(G) is the number of vertices of a largest induced tree subgraph, girth(G) is the length of the shortest cycle (0 if acyclic), and l(v) = indepNeighbors G v is the independence number of the neighbourhood of v.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWritten on the Wall II · Combinatorics

Written on the Wall II - Conjecture 142

WOWII Conjecture 142:

For a simple connected graph GG, tree(G)(2/3)girth(G)+ecc(B)\mathrm{tree}(G) \ge (2/3) \cdot \mathrm{girth}(G) + \mathrm{ecc}(B) where tree(G)\mathrm{tree}(G) is the largest induced tree size, girth(G)\mathrm{girth}(G) is the length of the shortest cycle (00 if acyclic), BB is the set of boundary vertices (those of maximum eccentricity), and ecc(B)\mathrm{ecc}(B) is the eccentricity of the set BB.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWritten on the Wall II · Combinatorics

Written on the Wall II - Conjecture 144

WOWII Conjecture 144

For a simple connected graph GG, tree(G)girth(G)1+ecc(Centers)\mathrm{tree}(G) \ge \mathrm{girth}(G) - 1 + \mathrm{ecc}(\mathrm{Centers}) where tree(G)\mathrm{tree}(G) is the largest induced tree size, girth(G)\mathrm{girth}(G) is the length of the shortest cycle (00 if acyclic), Centers=G.center\mathrm{Centers} = G.\mathrm{center} is the set of vertices with minimum eccentricity (the center of GG), and ecc(Centers)\mathrm{ecc}(\mathrm{Centers}) is the eccentricity of the center set , the maximum distance from any non-center vertex to the nearest center vertex.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWritten on the Wall II · Combinatorics

Written on the Wall II - Conjecture 145

WOWII Conjecture 145

For a simple connected graph GG, tree(G)2ecc(B)/λmin(G)\mathrm{tree}(G) \ge 2 \cdot \mathrm{ecc}(B) / \lambda_{\min}(\overline{G}) where tree(G)\mathrm{tree}(G) is the number of vertices in a largest induced subtree, ecc(B)\mathrm{ecc}(B) is the eccentricity of the boundary vertices (eccSet and boundaryVertices), and λmin(G)\lambda_{\min}(\overline{G}) is the minimum local independence number of the complement graph.

We state the inequality in the form tree(G)lMin(G)2ecc(B)\mathrm{tree}(G) \cdot \mathrm{lMin}(\overline{G}) \ge 2 \cdot \mathrm{ecc}(B) to avoid division.

Source checked Jul 26, 20261 pinned Lean statementInspect problem
Source labels openWritten on the Wall II · Combinatorics

Written on the Wall II - Conjecture 146

WOWII Conjecture 146

For a simple connected graph GG, tree(G)2ecc(B)/rad(G2)\mathrm{tree}(G) \ge 2 \cdot \mathrm{ecc}(B) / \mathrm{rad}(G^2) where tree(G)\mathrm{tree}(G) is the number of vertices in a largest induced subtree, ecc(B)\mathrm{ecc}(B) is the eccentricity of the boundary vertices of GG (eccSet and maxEccentricityVertices), and rad(G2)\mathrm{rad}(G^2) is the radius of the square graph of GG.

We state the inequality in the form tree(G)rad(G2)2ecc(B)\mathrm{tree}(G) \cdot \mathrm{rad}(G^2) \ge 2 \cdot \mathrm{ecc}(B) to avoid division.

Source checked Jul 26, 20261 pinned Lean statementInspect problem