plt.imshow(plt.imread('./res/fig18_3.png'))

"""
Explanation: 18 B-Trees
B-Trees are balanced search trees, similar to red-black trees, but they are better at minizing disk I/O operations.
We use the number of pages read or written as a approximation of the total time spent accessing the disk.
A B-tree node is usually as large as a whole disk page, and this size limits the number of children a B-tree node can have.
End of explanation
"""


plt.figure(figsize=(15,10))
plt.imshow(plt.imread('./res/fig18_4.png'))

"""
Explanation: 18.1 Definition of B-trees
Properties

node attributes:     
$x.n$: the number of keys.     
$x.\text{keys}$: $x.\text{key}_1 \leq x.\text{key}_2 \leq \dotsb \leq x.\text{key}_n$.    

$x.leaf$: True if it is, False otherwise.


Each internal node has $x.n+1$ points $x.c_1, \dotsc, x.c_{x.n+1}$ to its children.


$x.c_i.keys \leq x.key_i \leq x.c_{i+1}.keys$


All leaves have the same depth $h$.


minimum degree of the B-tree $T$: a fixed integer $t \geq 2$.    

Every node other than the root must have at least $t-1$ keys.     
Every node may contain at most $2t-1$ keys. FULL.

The simplest B-tree occurs when $t=2$: 2-3-4 tree.
The worst-case height of B-tree
If $n \geq 1$, then for any $n$-key B-tree $T$ of height $h$ and minimum degree $t \geq 2$.  
$$h \leq \log_t \frac{n+1}{2}$$
Proof:
The root of a B-tree $T$ contains at least one key, and all other nodes contain at least $t-1$ keys.
End of explanation
"""


#todo: code

"""
Explanation: the number $n$ of keys statisfies the inequality:
$$ n \geq 1 + (t-1) \sum_{i=1}^h 2t^{i-1} = 2t^h -1$$
B-trees save a factor of about $\lg t$ over red-black trees in the number of nodes examined for most tree operations.
%maybe: exercise
18.2 Basic operations on B-trees
two conventions:   


The root of the B-tree is always in main memory.


Any nodes that are passed as parameters must already have had a Disk-Read operation performed on them.


Searching a B-tree
We make a multiway branching decision according to the number of the node's children. $O(t lg_t n) = O(t) \times O(h)$.
End of explanation
"""


#todo: code

"""
Explanation: Creating an empty B-tree
$O(1)$ time to allcate one disk page.
End of explanation
"""


plt.imshow(plt.imread('./res/fig18_5.png'))

"""
Explanation: Inserting a key into a B-tree
As we trave down the tree, we split each full node we meet, in order to assure that the parent is always not full whenever we want to split a full node.
(a) B-TREE-SPLIT-CHILD(X,I)
End of explanation
"""


plt.imshow(plt.imread('./res/fig18_6.png'))

"""
Explanation: If $T$ is full tree, then create a new root node, and split them. (Height increase 1).
(b)
```
B-TREE-INSERT(T,K)
    if T is FULL:
        S = B-TREE-CREATE()
        S.children = T
        R = B-TREE-SPLIT-CHILD(S)
    else
        R = T
B-TREE-INSERT-NONFULL(R,K)

```
End of explanation
"""


#todo: exercises

"""
Explanation: (c)
```
B-TREE-INSERT-NONFULL(T,K)
    if T is LEAF
        FIND_AND_INSERT(T,K)
    else
       C = FIND_CHILD(T,K)
       if C is FULL
           N = B-TREE-SPLIT-CHILD(C)
           R = FIND_CHILD(N,K)
       else
           R = T
   B-TREE-INSERT-NONFULL(R, K)

```
End of explanation
"""


plt.figure(figsize=(10,20))
plt.subplot(2,1,1)
plt.imshow(plt.imread('./res/fig18_8.png'))
plt.subplot(2,1,2)
plt.imshow(plt.imread('./res/fig18_9.png'))

#todo: exercise

"""
Explanation: 18.3 Deleting a key from a B-tree

When going down, if the node's child has only $t-1$ keys, move its key down to the child.     

if the node is root, then delete the root, use its child as new root. (Height -1).


If the node to delete is a leaf node, then directly remove it.


If the node to delete is a internal node, then     

If the key's left child has more than $t-1$ keys, move up the lastest key.    
If the key's right child has more than $t-1$ keys, move up the first key.     
otherwise, the sum of left and right child's keys is less than $2t-1$, merge them as a new node.

Base on the three priciples, we summary all possible cases as follows:
End of explanation
"""


