FA-99521 / Vector retrieval ranking / Open access
Vector retrieval: Per-shard top results are concatenated without global ranking · case 01
Per-shard top results are concatenated without global ranking.
ROOT CAUSE
Each shard’s local order is concatenated and treated as globally ranked.
THE FAILURE
Each shard’s local order is concatenated and treated as globally ranked.
Unsuccessful approach: Sorting ascending fixes merge order in the wrong score direction.
Case contract
Shard lists contain [id,score], unique IDs; globally return top k by score then ID.
Why this case matters
An offline vector retrieval toy model isolates one ranking contract from embedding generation, storage, and serving.
1 / The failure
Exit 1"""Failure Map reference implementation. Python standard library only."""
import json
N = 1
observations = []
def solve(shards, k):
return [r[0] for shard in shards for r in shard][:k]
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('global winner',solve([[['a',N]], [['b',N+4]]],1),['b'])
check('tie across shards',solve([[['z',N]], [['a',N]]],2),['a','z'])
check('third shard',solve([[['x',1]],[['y',N+2]],[['z',N+3]]],2),['z','y'])
check('zero limit',solve([[['a',N]]],0),[])
check('empty shards',solve([[],[]],N),[])
check('reordered shards',solve([[['b',N+1]],[['a',N+2]]],2),['a','b'])
print(json.dumps({"observations": observations, "passed": all(x["passed"] for x in observations)}, ensure_ascii=False))
raise SystemExit(0 if all(x["passed"] for x in observations) else 1)
| Boundary fixture | Actual | Expected | Outcome |
|---|---|---|---|
| global winner | ['a'] | ['b'] | Failed |
| tie across shards | ['z', 'a'] | ['a', 'z'] | Failed |
| third shard | ['x', 'y'] | ['z', 'y'] | Failed |
| zero limit | [] | [] | Passed |
| empty shards | [] | [] | Passed |
| reordered shards | ['b', 'a'] | ['a', 'b'] | Failed |
SHA-256 / 6ec991e5969b05c268d9d8d0c98be79387c5908cd319c705d16205d84fe90b9f
2 / The unsuccessful fix
Exit 1"""Failure Map reference implementation. Python standard library only."""
import json
N = 1
observations = []
def solve(shards, k):
return [r[0] for r in sorted([r for shard in shards for r in shard],key=lambda r:r[1])[:k]]
def check(label, actual, expected):
observations.append({"check": label, "actual": actual, "expected": expected, "passed": actual == expected})
check('global winner',solve([[['a',N]], [['b',N+4]]],1),['b'])
check('tie across shards',solve([[['z',N]], [['a',N]]],2),['a','z'])
check('third shard',solve([[['x',1]],[['y',N+2]],[['z',N+3]]],2),['z','y'])
check('zero limit',solve([[['a',N]]],0),[])
check('empty shards',solve([[],[]],N),[])
check('reordered shards',solve([[['b',N+1]],[['a',N+2]]],2),['a','b'])
print(json.dumps({"observations": observations, "passed": all(x["passed"] for x in observations)}, ensure_ascii=False))
raise SystemExit(0 if all(x["passed"] for x in observations) else 1)
| Boundary fixture | Actual | Expected | Outcome |
|---|---|---|---|
| global winner | ['a'] | ['b'] | Failed |
| tie across shards | ['z', 'a'] | ['a', 'z'] | Failed |
| third shard | ['x', 'y'] | ['z', 'y'] | Failed |
| zero limit | [] | [] | Passed |
| empty shards | [] | [] | Passed |
| reordered shards | ['b', 'a'] | ['a', 'b'] | Failed |
SHA-256 / 948d230f7c51df7e9725d5173556c9cab36f7e67fa83f258fd8555596e0e4789
HELD IN THE MEMBER ARCHIVE
The verified repair and its recorded checks are member-only.
This mechanism has 6 recorded checks per implementation. The open-access tier publishes the failure and the unsuccessful fix; the repaired source that passes every check, and the observations that prove it, are available to members.
Every case sharing this mechanism uses the same contract and the same repair, so this one record is held back for all of them.
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Sign in to the archive ↗Verification & scope
Inputs are small exact toy vectors and scores. No production index, ANN recall, embedding quality, or latency claim is implied. This reproducer isolates one failure mechanism. Results cover the supplied fixtures. Variants within a family share a test contract and should remain grouped when constructing evaluation splits. Related mechanisms with a shared evaluation_group must also remain together; these controlled models are not independent production incidents.
Observations recorded using Python 3.12.14 at 2026-09-30T16:48:25.348868+00:00.
Case digest / de958229903c05074056368b36bcb7855a2fb22349f0763628a12d9cbe2333b1