Storage–repair tradeoff for (6,5,5) exact-repair codes

For exact-repair regenerating codes with (n,k,d)=(6,5,5), what is the optimal tradeoff between per-node storage and repair bandwidth? Any k nodes must recover the data, and a failed node must be reproduced exactly from d helpers; nonlinear codes are allowed.

Why it matters and what counts as progress

Why it matters: A complete converse identifies the storage and repair costs that no exact-repair code, including nonlinear codes, can beat in this model.

What counts as progress: Prove converse bounds matching attainable layered-code and space-sharing points, or improve constructions outside a known converse range.

How the problem developed

3 Dec 2013 · Partial result [1]
Chao Tian Characterized the exact storage-repair tradeoff for the smaller (4,3,3) regenerating-code case using an information-theoretic converse.
2015 · Partial result [2]
Chao Tian Characterized the (5,4,4) rate region, extending the exact-repair tradeoff results to a larger parameter set.
20 Oct 2026 · Tradeoff theorem [3]
Chao Tian Reported a complete converse for the (6,5,5) exact-repair storage-repair-bandwidth region without assuming linear codes; layered codes attain the bounds.

AI contribution

Exact Storage-Repair-Bandwidth Tradeoffs for (6,5,5) and (7,6,6) Exact-Repair Regenerating Codes

Chao Tian

The manuscript gives converse bounds that completely characterize the storage/repair-bandwidth tradeoffs of exact-repair regenerating codes for (n,k,d)=(6,5,5) and (7,6,6), without assuming linear codes. Previously known layered codes with space sharing attain the bounds.

AI involvementAI-led

ImportanceNotable

Technical verificationPartial

Human understandingFull