426.lean`

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Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1

2 declarations, 1 contain sorry, 0 declared as axioms

Erdős Problem 426

References:

erdosproblems.com/426
[Er76b] Erdős, P., Problems and results in graph theory and combinatorial analysis. Proceedings of the Fifth British Combinatorial Conference (1976), 169-192.
[EnEr72] Entringer, R. C. and Erdős, Paul, On the number of unique subgraphs of a graph. J. Combinatorial Theory Ser. B (1972), 112-115.
[HaSc73] Harary, Frank and Schwenk, Allen J., On the number of unique subgraphs. J. Combinatorial Theory Ser. B (1973), 156-160.
[Br75] Brouwer, A. E., Note: "On the number of unique subgraphs of a graph" (J. Combinatorial Theory Ser. B 13 (1972), 112-115) by R. C. Entringer and P. Erdős. J. Combinatorial Theory Ser. B (1975), 184-185.
[BrCh24] Bradač, D. and Christoph, M., Unique subgraphs are rare. arXiv:2410.16233 (2024).

Imports (1)

FormalConjecturesUtil

theorem Erdos426.isUniqueSubgraph_bot_bot Sanity check: the empty graph ⊥ is a unique subgraph of itself line 46 source
Source documentation
Sanity check: the empty graph ⊥ is a unique subgraph of itself. Its only subgraph is ⊥ (everything ≤ ⊥ equals ⊥), which is isomorphic to ⊥ via the identity.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: SimpleGraph.IsUniqueSubgraph, Bot.bot, SimpleGraph
Proof uses: Exists.intro, And, LE.le, Nonempty, SimpleGraph.Iso, Eq, And.intro, le_refl, Nonempty.intro, SimpleGraph.Iso.refl, Iff.mp, le_bot_iff
17 instances, coercions and auxiliary constants left out.

Tactic trace: 3 steps
Open to load the checked before-and-after goal states.
```
theorem isUniqueSubgraph_bot_bot {V : Type*} : IsUniqueSubgraph (⊥ : SimpleGraph V) ⊥ := by
…
```
theorem Erdos426.erdos_426 We say $H$ is a unique subgraph of $G$ if there is exactly one way to find $H$ as a subgraph (not necessarily induced) of $G$ sorry line 71 source
Source documentation
We say $H$ is a unique subgraph of $G$ if there is exactly one way to find $H$ as a subgraph (not necessarily induced) of $G$. Is there a graph on $n$ vertices with $$\gg \frac{2^{\binom{n}{2}}}{n!}$$ many distinct unique subgraphs?
Bradač and Christoph [BrCh24] have proved the answer is no: if $f(n)$ is the maximum number of unique subgraphs in a graph on $n$ vertices then $$f(n) = o\left(\frac{2^{\binom{n}{2}}}{n!}\right).$$
The $\gg$ below is read as: some constant $c>0$ works for arbitrarily large $n$. The negation of the proposition on the right is then exactly $f(n) = o(2^{\binom{n}{2}}/n!)$, the form in which Bradač and Christoph [BrCh24] resolved the problem.
The linked file states the resolution in that negated form, as Tendsto fSeq atTop (nhds 0). It counts the isomorphism classes occurring as unique subgraphs, whereas uniqueSubgraphCount counts their representatives $G\leq H$; uniqueness forces exactly one representative per class, so the two counts agree.

depends on sorry what this means
Statement uses: Iff, False, Exists, Real, And, LT.lt, OfNat.ofNat, Filter.Frequently, Nat, SimpleGraph, Fin, LE.le, HMul.hMul, HDiv.hDiv, HPow.hPow, Nat.choose, Nat.factorial, SimpleGraph.uniqueSubgraphCount, Filter.atTop
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
40 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_426 : answer(False) ↔
    ∃ c : ℝ, 0 < c ∧ ∃ᶠ (n : ℕ) in atTop, ∃ H : SimpleGraph (Fin n),
      c * ((2 : ℝ) ^ n.choose 2 / n.factorial) ≤ (uniqueSubgraphCount H : ℝ) := by
…
```
Mentions: c