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planner: fix column evaluator can not detect input's column-ref and t…
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,85 @@ | ||
| // Copyright 2024 PingCAP, Inc. | ||
| // | ||
| // Licensed under the Apache License, Version 2.0 (the "License"); | ||
| // you may not use this file except in compliance with the License. | ||
| // You may obtain a copy of the License at | ||
| // | ||
| // http://www.apache.org/licenses/LICENSE-2.0 | ||
| // | ||
| // Unless required by applicable law or agreed to in writing, software | ||
| // distributed under the License is distributed on an "AS IS" BASIS, | ||
| // WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. | ||
| // See the License for the specific language governing permissions and | ||
| // limitations under the License. | ||
|
|
||
| package disjointset | ||
|
|
||
| // Set is the universal implementation of a disjoint set. | ||
| // It's designed for sparse cases or non-integer types. | ||
| // If you are dealing with continuous integers, you should use SimpleIntSet to avoid the cost of a hash map. | ||
| // We hash the original value to an integer index and then apply the core disjoint set algorithm. | ||
| // Time complexity: the union operation has an inverse Ackermann function time complexity, which is very close to O(1). | ||
| type Set[T comparable] struct { | ||
| parent []int | ||
| val2Idx map[T]int | ||
| idx2Val map[int]T | ||
| tailIdx int | ||
| } | ||
|
|
||
| // NewSet creates a disjoint set. | ||
| func NewSet[T comparable](size int) *Set[T] { | ||
| return &Set[T]{ | ||
| parent: make([]int, 0, size), | ||
| val2Idx: make(map[T]int, size), | ||
| idx2Val: make(map[int]T, size), | ||
| tailIdx: 0, | ||
| } | ||
| } | ||
|
|
||
| func (s *Set[T]) findRootOriginalVal(a T) int { | ||
| idx, ok := s.val2Idx[a] | ||
| if !ok { | ||
| s.parent = append(s.parent, s.tailIdx) | ||
| s.val2Idx[a] = s.tailIdx | ||
| s.tailIdx++ | ||
| s.idx2Val[s.tailIdx-1] = a | ||
| return s.tailIdx - 1 | ||
| } | ||
| return s.findRootInternal(idx) | ||
| } | ||
|
|
||
| // findRoot is an internal implementation. Call it inside findRootOriginalVal. | ||
| func (s *Set[T]) findRootInternal(a int) int { | ||
| if s.parent[a] != a { | ||
| // Path compression, which leads the time complexity to the inverse Ackermann function. | ||
| s.parent[a] = s.findRootInternal(s.parent[a]) | ||
| } | ||
| return s.parent[a] | ||
| } | ||
|
|
||
| // InSameGroup checks whether a and b are in the same group. | ||
| func (s *Set[T]) InSameGroup(a, b T) bool { | ||
| return s.findRootOriginalVal(a) == s.findRootOriginalVal(b) | ||
| } | ||
|
|
||
| // Union joins two sets in the disjoint set. | ||
| func (s *Set[T]) Union(a, b T) { | ||
| rootA := s.findRootOriginalVal(a) | ||
| rootB := s.findRootOriginalVal(b) | ||
| // take b as successor, respect the rootA as the root of the new set. | ||
| if rootA != rootB { | ||
| s.parent[rootB] = rootA | ||
| } | ||
| } | ||
|
|
||
| // FindRoot finds the root of the set that contains a. | ||
| func (s *Set[T]) FindRoot(a T) int { | ||
| // if a is not in the set, assign a new index to it. | ||
| return s.findRootOriginalVal(a) | ||
| } | ||
|
|
||
| // FindVal finds the value of the set corresponding to the index. | ||
| func (s *Set[T]) FindVal(idx int) (T, bool) { | ||
| v, ok := s.idx2Val[s.findRootInternal(idx)] | ||
| return v, ok | ||
| } |