add partitioning #3
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instead of 1 big heap we now split he heap in n (n=6) heaps to and assign elements to to one of the heaps based on the hash. This way we reduce switching on heavyhitters and reduce merge errors
mschoch
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Mar 23, 2022
mschoch
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mschoch
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So, I spotted a few mechanical things noted here. And my understanding is that you have replaced the single map/heap with N partitioned map/heaps, where the hash of the key (which you computed anyway) is used to choose the partition. I will wait for your explanation to review again with that in mind.
topk.go
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| var ( | ||
| err error | ||
| ) | ||
| for i := 0; i < 6; i++ { |
topk.go
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| if len(s.k.elts) < s.n { | ||
| // there is free space | ||
| if len(s.p[i].elts) < s.n { | ||
| // there is free sp[i]ace |
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Search/Replace error or some deep joke I don't get?
| idx2, ok2 := other.p[i].m[k] | ||
| xhash := reduce(metro.Hash64Str(k, 0), len(s.alphas)) | ||
| min1 := other.alphas[xhash] | ||
| min2 := other.alphas[xhash] |
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min1 and min2 are always assigned the same value, and they appear to only be read inside this scope. Is there some reason we need both?
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instead of 1 big heap we now split he heap in n (n=6) heaps to and assign elements to to one of the heaps based on the hash. This way we reduce switching on heavy hitters and reduce merge errors