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loki/pkg/util/loser/tree.go

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// Loser tree, from https://en.wikipedia.org/wiki/K-way_merge_algorithm#Tournament_Tree
package loser
type Sequence interface {
Next() bool // Advances and returns true if there is a value at this new position.
}
func New[E any, S Sequence](sequences []S, maxVal E, at func(S) E, less func(E, E) bool, close func(S)) *Tree[E, S] {
nSequences := len(sequences)
t := Tree[E, S]{
maxVal: maxVal,
at: at,
less: less,
close: close,
nodes: make([]node[E, S], nSequences*2),
}
for i, s := range sequences {
t.nodes[i+nSequences].items = s
t.moveNext(i + nSequences) // Must call Next on each item so that At() has a value.
}
if nSequences > 0 {
t.nodes[0].index = -1 // flag to be initialized on first call to Next().
}
return &t
}
// Call the close function on all sequences that are still open.
func (t *Tree[E, S]) Close() {
for _, e := range t.nodes[len(t.nodes)/2 : len(t.nodes)] {
if e.index == -1 {
continue
}
t.close(e.items)
}
}
// A loser tree is a binary tree laid out such that nodes N and N+1 have parent N/2.
// We store M leaf nodes in positions M...2M-1, and M-1 internal nodes in positions 1..M-1.
// Node 0 is a special node, containing the winner of the contest.
type Tree[E any, S Sequence] struct {
maxVal E
at func(S) E
less func(E, E) bool
close func(S) // Called when Next() returns false.
nodes []node[E, S]
}
type node[E any, S Sequence] struct {
index int // This is the loser for all nodes except the 0th, where it is the winner.
value E // Value copied from the loser node, or winner for node 0.
items S // Only populated for leaf nodes.
}
func (t *Tree[E, S]) moveNext(index int) bool {
n := &t.nodes[index]
if n.items.Next() {
n.value = t.at(n.items)
return true
}
t.close(n.items) // Next() returned false; close it and mark as finished.
n.value = t.maxVal
n.index = -1
return false
}
func (t *Tree[E, S]) Winner() S {
return t.nodes[t.nodes[0].index].items
}
func (t *Tree[E, S]) Next() bool {
if len(t.nodes) == 0 {
return false
}
if t.nodes[0].index == -1 { // If tree has not been initialized yet, do that.
t.initialize()
return t.nodes[t.nodes[0].index].index != -1
}
if t.nodes[t.nodes[0].index].index == -1 { // already exhausted
return false
}
if t.moveNext(t.nodes[0].index) {
t.replayGames(t.nodes[0].index)
} else {
t.sequenceEnded(t.nodes[0].index)
}
return t.nodes[t.nodes[0].index].index != -1
}
func (t *Tree[E, S]) initialize() {
winners := make([]int, len(t.nodes))
// Initialize leaf nodes as winners to start.
for i := len(t.nodes) / 2; i < len(t.nodes); i++ {
winners[i] = i
}
for i := len(t.nodes) - 2; i > 0; i -= 2 {
// At each stage the winners play each other, and we record the loser in the node.
loser, winner := t.playGame(winners[i], winners[i+1])
p := parent(i)
t.nodes[p].index = loser
t.nodes[p].value = t.nodes[loser].value
winners[p] = winner
}
t.nodes[0].index = winners[1]
t.nodes[0].value = t.nodes[winners[1]].value
}
// Starting at pos, which is a winner, re-consider all values up to the root.
func (t *Tree[E, S]) replayGames(pos int) {
n := parent(pos)
for n != 0 {
// If n.value < pos.value then pos loses.
// If they are equal, pos wins because n could be a sequence that ended, with value maxval.
if t.less(t.nodes[n].value, t.nodes[pos].value) {
loser := pos
// Record pos as the loser here, and the old loser is the new winner.
pos = t.nodes[n].index
t.nodes[n].index = loser
t.nodes[n].value = t.nodes[loser].value
}
n = parent(n)
}
// pos is now the winner; store it in node 0.
t.nodes[0].index = pos
t.nodes[0].value = t.nodes[pos].value
}
func (t *Tree[E, S]) sequenceEnded(pos int) {
// Find the first active sequence which used to lose to it.
n := parent(pos)
for n != 0 && t.nodes[t.nodes[n].index].index == -1 {
n = parent(n)
}
if n == 0 {
// There are no active sequences left
t.nodes[0].index = pos
t.nodes[0].value = t.maxVal
return
}
// Record pos as the loser here, and the old loser is the new winner.
loser := pos
winner := t.nodes[n].index
t.nodes[n].index = loser
t.nodes[n].value = t.nodes[loser].value
t.replayGames(winner)
}
func (t *Tree[E, S]) playGame(a, b int) (loser, winner int) {
if t.less(t.nodes[a].value, t.nodes[b].value) {
return b, a
}
return a, b
}
func parent(i int) int { return i / 2 }
// Add a new sequence to the merge set
func (t *Tree[E, S]) Push(sequence S) {
// First, see if we can replace one that was previously finished.
for newPos := len(t.nodes) / 2; newPos < len(t.nodes); newPos++ {
if t.nodes[newPos].index == -1 {
t.nodes[newPos].index = newPos
t.nodes[newPos].items = sequence
t.moveNext(newPos)
t.nodes[0].index = -1 // flag for re-initialize on next call to Next()
return
}
}
// We need to expand the tree. Pick the next biggest power of 2 to amortise resizing cost.
size := 1
//nolint: revive
for ; size <= len(t.nodes)/2; size *= 2 {
}
newPos := size + len(t.nodes)/2
newNodes := make([]node[E, S], size*2)
// Copy data over and fix up the indexes.
for i, n := range t.nodes[len(t.nodes)/2:] {
newNodes[i+size] = n
newNodes[i+size].index = i + size
}
t.nodes = newNodes
t.nodes[newPos].index = newPos
t.nodes[newPos].items = sequence
// Mark all the empty nodes we have added as finished.
for i := newPos + 1; i < len(t.nodes); i++ {
t.nodes[i].index = -1
t.nodes[i].value = t.maxVal
}
t.moveNext(newPos)
t.nodes[0].index = -1 // flag for re-initialize on next call to Next()
}