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dab8dff9c2
We keep wanting this, and it's shardwidth-dependent code, and we keep rewriting it. It should be in the shardwidth package.
70 lines
2.5 KiB
Go
70 lines
2.5 KiB
Go
// Copyright 2021 Molecula Corp.
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//
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// Licensed under the Apache License, Version 2.0 (the "License");
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// you may not use this file except in compliance with the License.
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// You may obtain a copy of the License at
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//
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// http://www.apache.org/licenses/LICENSE-2.0
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//
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// Unless required by applicable law or agreed to in writing, software
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// distributed under the License is distributed on an "AS IS" BASIS,
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// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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// See the License for the specific language governing permissions and
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// limitations under the License.
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package shardwidth
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import (
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"math/bits"
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)
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// FindNextShard returns the index of the first item which is not in the
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// same shard as i. The index it returns may be equal to the length of the
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// haystack, indicatincg that the rest of the list is in the same shard.
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func FindNextShard(i int, haystack []uint64) int {
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// compute the last thing that's in the same shard as haystack[i].
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if i >= len(haystack) {
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return i
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}
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// current shard:
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shard := (haystack[i] >> Exponent)
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// last value in shard:
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shardEnd := ((shard + 1) << Exponent) - 1
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j := i
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// We want to do a binary search of the haystack. For any length of
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// haystack, its topmost bit gives us a reasonable halfway point; it may
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// not actually be halfway, but the number of steps it'll take to search
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// it will be the same as if it were. sort.Search has interface overhead
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// and makes us sad.
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for incr := 1 << (bits.Len64(uint64(len(haystack) - i))); incr > 0; incr >>= 1 {
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if j+incr < len(haystack) {
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if haystack[j+incr] <= shardEnd {
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j += incr
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}
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}
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}
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// we've found the last item that is in the same shard as i, so...
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return j + 1
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}
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// FindShards finds the shards in a given haystack
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func FindShards(haystack []uint64) (shards []uint64, endIndexes []int) {
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if len(haystack) == 0 {
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return nil, nil
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}
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index := 0
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// the steady state of this loop is that shards contains the current
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// shard, but not its ending index; each time we find a new ending
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// index, we record that index as the end for the current shard, and
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// the new shard, until we reach the end and append len(haystack)
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// as the last index.
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shards = []uint64{haystack[index] >> Exponent}
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index = FindNextShard(index, haystack)
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for index < len(haystack) {
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shards = append(shards, haystack[index]>>Exponent)
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endIndexes = append(endIndexes, index)
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index = FindNextShard(index, haystack)
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}
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endIndexes = append(endIndexes, index)
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return shards, endIndexes
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}
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