反向多重索引

Question

我试图从这个 paper, which has also a smaller version here 中理解倒置多指数。为此，我构建了一个玩具示例并希望有人验证 or/and 与我分享 his/her 意见。

例子：

Assume we have N = 6 points in M = 4 dimensions. We use two blocks to
create sub-vecrtors. Let the points be these:

p0 = (0, 0, 1, 1)
p1 = (2, 2, 3, 3)
p2 = (4, 4, 5, 5)
p3 = (6, 6, 7, 7)
p4 = (8, 8, 9, 9)
p5 = (10, 10, 11, 11) // p5^1 = (10, 10), which is appended in D^1 etc.

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We run k-means twice, once for D^1 and once for D^2, by requiring 3
centroids and we get:
U = { u1 = (1, 1), u2 = (5, 5), u3 = (9, 9) }
V = { v1 = (2, 2), v2 = (6, 6), v3 = (10, 10) }

Now we have to assign the points to the Wi,j. All we be empty, except:

W1,1 = (u1, v1): p0, p1
W2,2 = (u2, v2): p2, p3
W3,3 = (u3, v3): p4, p5

Note: We store the PQ-copressed version with every point ( for example
instead of storing just p0, store [p0, (u1, v1)] ), which will be used
during reranking.
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Assume T = 4 (which sets L = 2 for the sake of the example) and the query
is: (5, 5, 0, 0). Posing the query...


q^1 VS U
i   u_α(i) r
1   u2     0
2   u1    32 <-- |d1( (5,5), (1,1) )|^2 = (5 - 1)^2 + (5 - 1)^2 = 32

q^2 VS V
j   v_β(j) s
1   v1     8
2   v2    72

Invoke the Multi Sequence Algorithm which will output:
W2,1 --> [u2 v1] --> empty
W1,1 --> [u1 v1] --> p0, p1
W2,2 --> [u2 v2] --> p4, p5
// stop when the points that lie in the Ws returned so far are >= T

So the candidate list is {p0, p1, p2, p3}
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Rerank the candidate list, by computing the distance between the query and
the PQ-compressed representation (sum the distances of the subvectors).

请问正确吗？

Answer 1

是的，这似乎是正确的，正如 Babenko 本人在电子邮件对话中提到的那样。