22c: 21 Data Structures

1. 22c: 21 Data Structures Lecture on 03/13/2009

2. Outline • Infix to Postfix Conversion with Stack • B-Trees

3. Infix to Postfix

4. Why do we need the stack? a + b ab + a + b a b + +

5. Example Infix String : a+b*c-d Postfix String : abc*+d- b*c -> bc* a+bc* -> abc*+ abc*+-d -> abc*+d-

6. Algorithm • Initially the Stack is empty and our Postfix string has no characters. • Now, the first character scanned is 'a'. 'a' is added to the Postfix string. • The next character scanned is '+'. It being an operator, it is pushed to the stack.

7. a +

8. Algorithm • Next character scanned is 'b' which will be placed in the Postfix string. • Next character is '*' which is an operator. • Now, the top element of the stack is '+' which has lower precedence than '*', so '*' will be pushed to the stack.

9. a b * +

10. Algorithm • The next character is 'c' which is placed in the Postfix string. • Next character scanned is '-'. The topmost character in the stack is '*' which has a higher precedence than '-'.

11. Algorithm • Thus '*' will be popped out from the stack and added to the Postfix string. • Even now the stack is not empty. • Now the topmost element of the stack is '+' which has equal priority to '-'. So pop the '+' from the stack and add it to the Postfix string. • The '-' will be pushed to the stack.

12. a b c + * -

13. Algorithm • Next character is 'd' which is added to Postfix string. • Now all characters have been scanned so we must pop the remaining elements from the stack and add it to the Postfix string.

14. - d a a b c + *

15. Algorithm • scan the Infix string from left to right. • Initialize an empty stack. • If the scanned character is an operand, add it to the Postfix string. • If the scanned character is an operator and if the stack is empty push the character to stack.

16. Algorithm • If the scanned character is an operator and the stack is not empty, compare the precedence of the character with the element on top of the stack (topStack). • If topStack has higher precedence over the scanned character Pop the stack else Push the scanned character to stack.

17. Algorithm • Repeat this step as long as stack is not empty and topStack has precedence over the character. • Repeat this step till all the characters are scanned.

18. Algorithm • (After all characters are scanned, we have to add any character that the stack may have to the Postfix string.) • If stack is not empty add topStack to Postfix string and Pop the stack. • Repeat this step as long as stack is not empty. • Return the Postfix string.

19. B-Trees • In computer science, a B-tree is a tree data structure that keeps data sorted and allows searches, insertions, and deletions in logarithmic time.

20. B -Tree • Unlike self-balancing binary search trees, it is optimized for systems that read and write large blocks of data. It is most commonly used in databases and file systems. • Example?

21. Secondary Storage Access • Binary Search Trees • AVL Trees • M-ary Search Trees • B- Trees Lesser the height of the tree, quicker is an element access

22. 72 8 48 92 97 78 18 51 83 26 54 35 B-tree of order 5 41 66 87 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 31 32 35 36 37 38 39 66 68 69 70 72 73 74 76 78 79 81 83 84 85 41 42 44 46 48 49 50 51 52 53 54 56 58 59 87 89 90 92 93 95 97 98 99

23. Properties • Data items are stored at the leaves • The non leaf nodes store up to M-1 keys • The root is either a leaf or has between two and M children

24. Properties • All non leaf nodes (except the root) have at least M/2 up to M children. • All leaves are at the same depth and have at least L/2 up to L data items, for some L.

25. Properties • Each node represents a disk block • So we choose M and L on the basis of the size of the items that are being stored

26. Inserting an Element • Search if it already exists (no duplicates allowed)

27. 72 8 48 92 97 78 18 51 83 26 54 35 Insert 57 41 66 87 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 31 32 35 36 37 38 39 66 68 69 70 72 73 74 76 78 79 81 83 84 85 41 42 44 46 48 49 50 51 52 53 54 56 57 58 59 87 89 90 92 93 95 97 98 99

28. Insert 57 • We had to rearrange the data in the leaf. • Cost of doing this is negligible compared to a disk access.

29. Insert 55 • Leaf is already full • Since we now have L+1 items we split them into two leaves • Distribute data evenly between leaves • Update parent

30. 72 92 8 78 18 97 83 26 35 Insert 55 41 66 87 48 51 54 57 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 31 32 35 36 37 38 39 66 68 69 70 72 73 74 76 78 79 81 83 84 85 41 42 44 46 48 49 50 51 52 53 54 55 56 57 58 59 87 89 90 92 93 95 97 98 99

31. Splitting • Splitting is time consuming, but it is a rare occurrence. • For every split, there are roughly L/2 non splits.

32. Insert 40 • The leaf is full, so we need to split. • But there is no place to add an extra key. • So, we need to add an extra child under root. • But, root cannot have more than M=5 children!

33. Insert 40 • Hence, the solution is to split parent. • Then, update all the values

34. Insert 40 26 41 66 87 8 18 35 38 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 31 32 35 36 37 38 39 40

35. Increasing height • When a non leaf node is split, its parent gains a child. • What if the parent is already full? • We continue splitting nodes up the tree till no splitting is required. • If we split the root, we have two roots, so add another single root at the top.

36. Deletion • Find the item, and then remove it • What if the leaf already had minimum number of elements? • Adopt a neighbor item if the neighbor is not itself at its minimum • Otherwise, combine neighbors.

37. Delete 99 26 41 66 87 8 18 35 38 72 78 83 92 97 87 89 90 92 93 95 97 98 99 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 31 32 35 36 37 38 39 40 66 68 69 70 72 73 74 76 78 79 81 83 84 85

38. Delete 99 26 41 66 87 8 18 35 38 72 78 87 92 83 84 85 87 89 90 92 93 95 97 98 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 31 32 35 36 37 38 39 40 66 68 69 70 72 73 74 76 78 79 81

39. Questions?