combinatorics / Discrepancy theory; vector balancing

Prescribed-order Euclidean prefix discrepancy O(sqrt d), independent of length (Bansal-Jiang-Meka-Singla-Sinha Conjecture 6.3)

Theorem 1.1: an absolute constant CC such that every finite sequence in the Euclidean unit ball of Rd\mathbb R^d, in prescribed order, has one signing with every prefix of norm at most CdC\sqrt d, independent of NN. This removes the log⁡N\sqrt{\log N} term from Banaszczyk's bound and is optimal up to the constant. It is existential: no efficient or online algorithm is given, so the algorithmic and online versions of the question remain open.

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combinatoricsSep 24, 2026Significance 22/100Registry: lean checked

Prescribed-order Euclidean prefix discrepancy O(sqrt d), independent of length (Bansal-Jiang-Meka-Singla-Sinha Conjecture 6.3)

Prior state unknown→proved

Theorem 1.1: an absolute constant CC such that every finite sequence in the Euclidean unit ball of Rd\mathbb R^d, in prescribed order, has one signing with every prefix of norm at most CdC\sqrt d, independent of NN. This removes the log⁡N\sqrt{\log N} term from Banaszczyk's bound and is optimal up to the constant. It is existential: no efficient or online algorithm is given, so the algorithmic and online versions of…

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Theorem 1.1: an absolute constant CC such that every finite sequence in the Euclidean unit ball of Rd\mathbb R^d, in prescribed order, has one signing with every prefix of norm at most CdC\sqrt d, independent of NN. This removes the log⁡N\sqrt{\log N} term from Banaszczyk's bound and is optimal up to the constant. It is existential: no efficient or online algorithm is given, so the algorithmic and online versions of the question remain open.

Theorem 1.1: an absolute constant CC such that every finite sequence in the Euclidean unit ball of Rd\mathbb R^d, in prescribed order, has one signing with every prefix of norm at most CdC\sqrt d, independent of NN. This removes the log⁡N\sqrt{\log N} term from Banaszczyk's bound and is optimal up to the constant. It is existential: no efficient or online algorithm is given, so the algorithmic and online versions of the question remain open.

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