btree: add basic in memory b-tree with integers
continuous-integration/drone/push Build is passing Details

This commit is contained in:
Antoine Bartuccio 2019-07-30 17:26:36 +02:00
parent 4f361f7d55
commit 33657122f9
Signed by: klmp200
GPG Key ID: E7245548C53F904B
4 changed files with 670 additions and 0 deletions

13
.drone.yml Normal file
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pipeline:
build:
image: golang:1.12
group: build
commands:
- go get -v -d ./...
- go build .
test:
image: golang:1.12
group: test
commands:
- go get -v -d ./...
- go test -v ./...

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btree/btree.go Normal file
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package btree
import (
"fmt"
)
// Tree is the tree itself
type Tree struct {
root *Node // Pointer to the Node root
t int // Minimum degree
}
// Node is a Node of a Btree
type Node struct {
numberOfKeys int // The number of keys really stored
t int // The value of t dependes upon disk blok size
isLeaf bool
keys []int
children []*Node
}
// Constructors
// NewBtree creates a new btree
func NewBtree(t int) *Tree {
return &Tree{
root: nil,
t: t,
}
}
func newNode(t int, isLeaf bool) *Node {
return &Node{
numberOfKeys: 0,
t: t,
isLeaf: isLeaf,
keys: make([]int, 2*t-1),
children: make([]*Node, 2*t),
}
}
// Tree methods
// Traverse the tree
func (t *Tree) Traverse() {
if t.root != nil {
t.root.traverse()
}
}
// Search k in the tree
func (t *Tree) Search(k int) *Node {
if t.root == nil {
return nil
}
return t.root.search(k)
}
// Remove k in the tree
func (t *Tree) Remove(k int) error {
if t.root == nil {
return fmt.Errorf("The tree is empty")
}
err := t.root.remove(k)
// If the root node has 0 keys, makes its first child as the new root
// If it has no child, set root as nil
if t.root.numberOfKeys == 0 {
if t.root.isLeaf {
t.root = nil
return err
}
t.root = t.root.children[0]
}
return err
}
// Insert k in the tree
func (t *Tree) Insert(k int) {
// If the tree is empty
if t.root == nil {
t.root = newNode(t.t, true)
t.root.keys[0] = k
t.root.numberOfKeys = 1
return
}
// If the tree is not empty
if !t.root.isFull() {
// If root is not full, insert in non full root
t.root.insertNonFull(k)
return
}
// If the root is full, then the tree grows in height
s := newNode(t.t, false)
// Make the old root as a child of the new root
s.children[0] = t.root
// Split the old root and move 1 key to the new root
s.splitChild(0, t.root)
// The new root has two children now.
// We decide which of the two children is going to have the new key
i := 0
if s.keys[0] < k {
i++
}
s.children[i].insertNonFull(k)
// Change root
t.root = s
}
// Node methods
// traverse all nodes in a subtree rooted with this node
func (n *Node) traverse() {
// There are n entries and n+1 children, treverse trough n keys and n first children
for i := 0; i < n.numberOfKeys; i++ {
// If this is not a leaf, then traverse the subtree before printing the key
if !n.isLeaf {
n.children[i].traverse()
}
fmt.Printf(" %d", n.keys[i])
}
// Print the subtree rooted with the last child
if !n.isLeaf {
n.children[n.numberOfKeys].traverse()
}
}
// search k in the subtree rooted with this node
func (n *Node) search(k int) *Node {
// Find the first entry greater than or equal to k
i := 0
for i < n.numberOfKeys && k > n.keys[i] {
i++
}
// If theh found key is equal to k, return this node
if n.keys[i] == k {
return n
}
// If the key is not found here and this is a leaf node
if n.isLeaf {
return nil
}
// Go to the approipriate child
return n.children[i].search(k)
}
func (n *Node) isFull() bool {
return n.numberOfKeys == 2*n.t-1
}
func (n *Node) insertNonFull(k int) {
// Initialize the index as the index of the rightmost element
i := n.numberOfKeys - 1
// If this is a leaf node
if n.isLeaf {
// Finds the location of the new key to be inserted
// Moves all greater keys to one place ahead
for i >= 0 && n.keys[i] > k {
n.keys[i+1] = n.keys[i]
i--
}
// Insert the new key at the found location
n.keys[i+1] = k
n.numberOfKeys++
return
}
// If this is not a leaf
// Finds the child wich is going to have the new key
for i >= 0 && n.keys[i] > k {
i--
}
// Check if the found chird is full
if n.children[i+1].isFull() {
// If the child is full, then split it
n.splitChild(i+1, n.children[i+1])
// After the split, the middle key of children[i] goes up and
// children[i] is splitted into two
// See which of those two is going to have the new key
if n.keys[i+1] < k {
i++
}
}
n.children[i+1].insertNonFull(k)
}
func (n *Node) splitChild(i int, y *Node) {
// Create a new node that will store (t-1) keys of y
z := newNode(y.t, y.isLeaf)
z.numberOfKeys = n.t - 1
// TODO: make optimisations
// Copy the last (t-1) keys of y to z
for j := 0; j < n.t-1; j++ {
z.keys[j] = y.keys[j+n.t]
}
// Copy the last t children of y to z
if !y.isLeaf {
for j := 0; j < n.t; j++ {
z.children[j] = y.children[j+n.t]
}
}
// Reduce the number of keys in y
y.numberOfKeys = n.t - 1
// Since this node is going to have a new child, create space for it
for j := n.numberOfKeys; j >= i+1; j-- {
n.children[j+1] = n.children[j]
}
// Link the new child to this node
n.children[i+1] = z
// A key of y will move to this node
// Find the location of the new key and move all greater keys ahead
for j := n.numberOfKeys - 1; j >= i; j-- {
n.keys[j+1] = n.keys[j]
}
// Copy the middle key of y to this node
n.keys[i] = y.keys[n.t-1]
// Increment the count of keys in this node
n.numberOfKeys++
}
// findKey returns the index of the first key that is greater than or equal to k
func (n *Node) findKey(k int) int {
index := 0
for index < n.numberOfKeys && n.keys[index] < k {
index++
}
return index
}
// remove the key k from the sub-tree rooted with this node
func (n *Node) remove(k int) error {
index := n.findKey(k)
// The key to be removed is in this node
if index < n.numberOfKeys && n.keys[index] == k {
if n.isLeaf {
return n.removeFromLeaf(index)
}
return n.removeFromNonLeaf(index)
}
// If this is a leaf, the key is not in the tree
if n.isLeaf {
return fmt.Errorf("The key %d does not exist in the tree", k)
}
isInLastChild := false
if index == n.numberOfKeys {
isInLastChild = true
}
// If the child where is the key has less than t keys, wi fill it
if n.children[index].numberOfKeys < n.t {
n.fill(index)
}
// If the last child has been merged, it must be merged with the previous
// child and so we recurse on the (index-1)th child.
if isInLastChild && index > n.numberOfKeys {
return n.children[index-1].remove(k)
}
// We recurse on the (index)th child which now has at least t keys
return n.children[index].remove(k)
}
// removeFromLeaf the index-th key from this node which is a leaf node
func (n *Node) removeFromLeaf(index int) error {
// Move all the keys after the index-th position one place backward
for i := index + 1; i < n.numberOfKeys; i++ {
n.keys[i-1] = n.keys[i]
}
n.numberOfKeys--
return nil
}
// removeFromNonLeaf the index-th key from this node which is not a leaf node
func (n *Node) removeFromNonLeaf(index int) error {
k := n.keys[index]
// If the child that precedes k has at least t keys,
// find the predecessor of k in the subtree and replace k with it
// Recursively delete the predecessor in the child
if n.children[index].numberOfKeys >= n.t {
pred := n.getPred(index)
n.keys[index] = pred
return n.children[index].remove(pred)
}
// If the child has less than t keys, examine children[index+1]
// If it has at least t keys, find the successor of k in this subtree
// Replace k by its successor and recursively delete the successor in the subtree
if n.children[index+1].numberOfKeys >= n.t {
succ := n.getSucc(index)
n.keys[index] = succ
return n.children[index+1].remove(succ)
}
// Merge k and all of children[index+1] int children[index]
// Free children[index+1] and recursively delete k from children[index]
n.merge(index)
return n.children[index].remove(k)
}
// getPred returns the predecessor of keys[index]
func (n *Node) getPred(index int) int {
// Keep moving to the rightmost node until we reach a leaf
current := n.children[index]
for !current.isLeaf {
current = current.children[current.numberOfKeys]
}
// Return the last key of the leaf
return current.keys[current.numberOfKeys-1]
}
// getSucc returns the successor of keys[index]
func (n *Node) getSucc(index int) int {
// Keep moving to the leftmost node starting from children[index+1] until we reach a leaf
current := n.children[index+1]
for !current.isLeaf {
current = current.children[0]
}
// Return the first key of the leaf
return current.keys[0]
}
// fill child children[index] which has less than t-1 keys
func (n *Node) fill(index int) {
// If the previous child has more than t-1 keys, borrow a key from that child
if index != 0 && n.children[index-1].numberOfKeys >= n.t {
n.borrowFromPrev(index)
return
}
// If the next child has more than t-1 keys, borrow a key from that child
if index != n.numberOfKeys && n.children[index+1].numberOfKeys >= n.t {
n.borrowFromNext(index)
return
}
// Merge children[index] with its sibling
if index != n.numberOfKeys {
n.merge(index)
return
}
// If this is the last child, merge with the previous sibling
n.merge(index - 1)
}
// borrowFromPrev takes a key from children[index+1] and insert it in children[index]
func (n *Node) borrowFromPrev(index int) {
child := n.children[index]
sibling := n.children[index-1]
// Moves all keys in children[index] one step ahead
for i := child.numberOfKeys - 1; i >= 0; i-- {
child.keys[i+1] = child.keys[i]
}
// If the child is not a leaf, move all its child pointers one step ahead
if !child.isLeaf {
for i := child.numberOfKeys; i >= 0; i-- {
child.children[i+1] = child.children[i]
}
}
// Sets child's first key equal to keys[index-1] from the current node
child.keys[0] = n.keys[index-1]
// Moves sibling's last child as children[index]'s first child
if !child.isLeaf {
child.children[0] = sibling.children[sibling.numberOfKeys]
}
// Moves the key from the sibling to the parent
n.keys[index-1] = sibling.keys[sibling.numberOfKeys-1]
child.numberOfKeys++
sibling.numberOfKeys--
}
// borrowFromNext takes a key from children[index+1] and insert it in children[index]
func (n *Node) borrowFromNext(index int) {
child := n.children[index]
sibling := n.children[index+1]
// keys[index] is inserted as the last key in children[index]
child.keys[child.numberOfKeys] = n.keys[index]
// Sibling's first child is inserted as the last child into children[index]
if !child.isLeaf {
child.children[child.numberOfKeys+1] = sibling.children[0]
}
// The first key from sibling is inserted into keys[index]
n.keys[index] = sibling.keys[0]
// TODO: optimize loops here
// Moving all keys in sibling one step behind
for i := 1; i < sibling.numberOfKeys; i++ {
sibling.keys[i-1] = sibling.keys[i]
}
// Moving the child pointers one step behind
if !sibling.isLeaf {
for i := 1; i <= sibling.numberOfKeys; i++ {
sibling.children[i-1] = sibling.children[i]
}
}
child.numberOfKeys++
sibling.numberOfKeys--
}
// merge children[index] with children[index+1]
func (n *Node) merge(index int) {
child := n.children[index]
sibling := n.children[index+1]
// Pulls a key from the current node ande insert it into the (t-1)th position
child.keys[n.t-1] = n.keys[index]
// Copies the keys from children[index+1] to children[index] at the end
for i := 0; i < sibling.numberOfKeys; i++ {
child.keys[i+n.t] = sibling.keys[i]
}
// Copies the child pointers from C[index+1] to children[index]
if !child.isLeaf {
for i := 0; i <= sibling.numberOfKeys; i++ {
child.children[i+n.t] = sibling.children[i]
}
}
// Moves all keys after index in the current node one step before
// to fill the gap created by moving keys[index] to children[index]
for i := index + 1; i < n.numberOfKeys; i++ {
n.keys[i-1] = n.keys[i]
}
// Moves the child pointer after (index+1) in the current node one step before
// This action marks sibling for deletion by the GC
for i := index + 2; i <= n.numberOfKeys; i++ {
n.children[i-1] = n.children[i]
}
// Updates the key count of child and the current node
child.numberOfKeys += sibling.numberOfKeys + 1
n.numberOfKeys--
}

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btree/btree_test.go Normal file
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package btree
import (
"fmt"
"testing"
)
func TestTree_Search(t *testing.T) {
tree := NewBtree(3)
tree.Insert(6)
k := 6
if tree.Search(k) == nil {
t.Errorf("Not present %d", k)
}
k = 15
if tree.Search(k) != nil {
t.Errorf("Present %d", k)
}
}
func TestTree_Remove(t *testing.T) {
tree := NewBtree(3)
toInsert := []int{10, 20, 5, 6, 12, 30, 7, 17}
for _, k := range toInsert {
tree.Insert(k)
}
tree.Remove(toInsert[0])
for _, k := range toInsert[1:] {
if tree.Search(k) == nil {
t.Errorf("Not present %d", k)
}
}
if tree.Search(toInsert[0]) != nil {
t.Errorf("Present %d", toInsert[0])
}
}
func TestTree_Insert(t *testing.T) {
tree := NewBtree(3)
toInsert := []int{10, 20, 5, 6, 12, 30, 7, 17}
// Test before insertion
for _, k := range toInsert {
if tree.Search(k) != nil {
t.Errorf("Present %d", k)
}
}
for _, k := range toInsert {
tree.Insert(k)
}
// Test after insertion
for _, k := range toInsert {
if tree.Search(k) == nil {
t.Errorf("Not present %d", k)
}
}
}
/*
func TestTree_Traverse(t *testing.T) {
tree := NewBtree(3)
toInsert := []int{1, 3, 7, 10, 11, 13, 14, 15, 18, 16, 19, 24, 25, 26, 21, 4, 5, 20, 22, 2, 17, 12, 6}
for _, k := range toInsert {
tree.Insert(k)
}
tree.Traverse()
fmt.Println("\n 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 18 19 20 21 22 24 25 26")
}
*/
func ExampleTree_Traverse() {
tree := NewBtree(3)
toInsert := []int{1, 3, 7, 10, 11, 13, 14, 15, 18, 16, 19, 24, 25, 26, 21, 4, 5, 20, 22, 2, 17, 12, 6}
for _, k := range toInsert {
tree.Insert(k)
}
tree.Traverse()
fmt.Println("")
tree.Remove(6)
tree.Traverse()
fmt.Println("")
tree.Remove(13)
tree.Traverse()
fmt.Println("")
tree.Remove(7)
tree.Traverse()
fmt.Println("")
tree.Remove(4)
tree.Traverse()
fmt.Println("")
tree.Remove(2)
tree.Traverse()
fmt.Println("")
tree.Remove(16)
tree.Traverse()
// Output:
// 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 18 19 20 21 22 24 25 26
// 1 2 3 4 5 7 10 11 12 13 14 15 16 17 18 19 20 21 22 24 25 26
// 1 2 3 4 5 7 10 11 12 14 15 16 17 18 19 20 21 22 24 25 26
// 1 2 3 4 5 10 11 12 14 15 16 17 18 19 20 21 22 24 25 26
// 1 2 3 5 10 11 12 14 15 16 17 18 19 20 21 22 24 25 26
// 1 3 5 10 11 12 14 15 16 17 18 19 20 21 22 24 25 26
// 1 3 5 10 11 12 14 15 17 18 19 20 21 22 24 25 26
}

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package main
import (
"fmt"
"git.klmp200.net/klmp200/kvs/btree"
)
func main() {
t := btree.NewBtree(3)
t.Insert(10)
t.Insert(20)
t.Insert(5)
t.Insert(6)
t.Insert(12)
t.Insert(30)
t.Insert(7)
t.Insert(17)
fmt.Print("Traversal of the constructed tree is")
t.Traverse()
k := 6
if t.Search(k) != nil {
fmt.Print("\nPresent")
} else {
fmt.Print("\nNot Present")
}
k = 15
if t.Search(k) != nil {
fmt.Print("\nPresent")
} else {
fmt.Print("\nNot Present")
}
}