Conjectura
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← problemsCB005openMathematics/ Combinatorics

The lonely runner conjecture

Maintainer — open

Runners at distinct speeds on a circular track: is each of them, at some moment, far from all the others?

Motivation

Wills 1967, Cusick 1973, named by Goddyn in 1998. Proved for up to seven runners; open for eight.

The combinatorial statement is equivalent to a question about view-obstruction in geometry and to one about chromatic numbers of distance graphs, which is unusual — three unrelated-looking areas stuck at the same place.

Wills 1967; Cusick 1973; named by Goddyn 1998. Known for n ≤ 7.

Lean API
import Mathlib.Data.Finset.Card
import Mathlib.Data.Rat.Defs
import Mathlib.Algebra.Order.Ring.Rat

namespace Conjectura.CB005

/-- Runners with distinct constant speeds start together on a circular track of
circumference one. Is every runner, at some moment, at distance at least `1/n` from
all the others? Proved for up to seven runners. -/
def goal : Prop :=
  ∀ (n : ℕ) (v : Fin n → ℚ), 0 < n → Function.Injective v →
    ∀ i : Fin n, ∃ t : ℚ, ∀ j : Fin n, j ≠ i →
      ∃ k : ℤ, 1 / (n : ℚ) ≤ |(v i - v j) * t - (k : ℚ)|

end Conjectura.CB005
Definition1
Related work0
    For your AI
    Download as .md
    I am proving a theorem in Lean 4 and submitting it to Conjectura.
    
    ## Problem CB005 — The lonely runner conjecture
    
    Runners at distinct speeds on a circular track: is each of them, at some moment, far from all the others?
    
    ## Environment (fixed — do not assume anything newer)
    
    - Lean toolchain: `leanprover/lean4:v4.33.0-rc1`
    - Mathlib: `v4.33.0-rc1`
    
    If a lemma you want does not exist in that Mathlib, prove it inline instead of
    importing something newer.
    
    ## The exact statement I must prove
    
    ```lean
    theorem solution : ∀ (n : ℕ) (v : Fin n → ℚ), 0 < n → Function.Injective v →
        ∀ i : Fin n, ∃ t : ℚ, ∀ j : Fin n, j ≠ i → ∃ k : ℤ, 1 / (n : ℚ) ≤ |(v i - v j) * t - k| := by
      sorry
    ```
    
    ## The file I submit
    
    ```lean
    import Conjectura.Problems.CB005.Statement
    
    namespace Submission
    
    theorem solution : Conjectura.CB005.goal := by
      sorry
    
    end Submission
    
    ```
    
    ## The Lean definitions of every term in this problem
    
    These are the actual definitions your proof will be checked against. Do not
    substitute your own version of any of them.
    
    ### Injective
    
    class
    
    ```lean
    Function.Injective
    ```
    Defined in Mathlib.
    
    ## Rules — submissions violating these are rejected automatically
    
    1. **Do not change the name or type of `solution`.** It must satisfy the
       statement above exactly.
    2. **Do not redefine or shadow anything from the problem's Statement module.**
       Declaring your own `goal`, or redefining a name it depends on, produces a
       proof of a *different* statement and is rejected. This is the single most
       common rejection.
    3. **No `sorry`** anywhere, including in helper lemmas. It surfaces as the
       axiom `sorryAx` and is detected transitively through imports.
    4. **No `native_decide`** — it surfaces as `Lean.ofReduceBool` and is not
       accepted, because it trusts compiled code rather than the kernel.
    5. Only these axioms are permitted: `propext`, `Classical.choice`,
       `Quot.sound`.
    6. Follow Mathlib style: hypotheses left of the colon, explicit types,
       `snake_case` theorem names, `UpperCamelCase` types.
    
    ## What I want from you
    
    Here is my argument in informal mathematics:
    
    > [PASTE YOUR PROOF SKETCH HERE]
    
    Turn it into Lean 4 that compiles under the environment above and satisfies the
    statement exactly. Where you are unsure a lemma exists in this Mathlib version,
    say so explicitly rather than guessing a name.

    Submissions are not open yet

    Conjectura is in beta. You can read every statement, every definition and the Lean behind them, and download the exact files the checker uses — but proofs are not being accepted yet.

    The reason is a deliberate order of operations. Accepting a proof means running a stranger’s code and standing behind a verdict, and no statement here yet carries a researcher’s name. A machine-checked answer to a question nobody has vouched for is worth very little, so the vouching comes first.

    The English write-ups are also switched off during the beta. Nothing on this page is generated by a model.

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