combinatorics / Graph theory (induced trees / graph invariants)

WOWII Conjecture 72: Two induced trees pin down tree($ G $)

For a connected graph $ G $, let $ t= $ tree($ G $) (order of a largest induced tree), $ A= $ average eccentricity, and $ L= $ maximum independence number of a neighbourhood. Then $ \lceil (A+L)/3 \rceil \le t. $ (The evenly-divided reading of the conjecture holds; a stronger reading that divides only $ L $ by three is false.)

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combinatoricsJul 23, 2026Significance 5/100Registry: site confirmed

WOWII Conjecture 72: Two induced trees pin down tree($ G $)

Prior state unknownproved

The evenly-divided reading of WOWII Conjecture 72 holds: $ \lceil(A + L)/3\rceil \le t, $ where $ t = $ tree($ G $) (order of a largest induced tree), $ A = $ average eccentricity and $ L = $ maximum neighbourhood independence number. A stronger reading that divides only $ L $ by three is false. The argument rests on two elementary observations (a diametral path is chordless and therefore induces a t…

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For a connected graph $ G $, let $ t= $ tree($ G $) (order of a largest induced tree), $ A= $ average eccentricity, and $ L= $ maximum independence number of a neighbourhood. Then $ \lceil (A+L)/3 \rceil \le t. $ (The evenly-divided reading of the conjecture holds; a stronger reading that divides only $ L $ by three is false.)

The evenly-divided reading of WOWII Conjecture 72 holds: $ \lceil(A + L)/3\rceil \le t, $ where $ t = $ tree($ G $) (order of a largest induced tree), $ A = $ average eccentricity and $ L = $ maximum neighbourhood independence number. A stronger reading that divides only $ L $ by three is false. The argument rests on two elementary observations (a diametral path is chordless and therefore induces a tree on $ D+1 $ vertices; a maximum independent set in a neighbourhood induces a star on $ L+1 $ vertices). The original conjecture’s precise wording is not yet pinned in a public formal repository, so statement fidelity remains to be audited.

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