Week 9 - Monday
What did we talk about last time?
Practiced with red-black trees
AVL trees
Balanced add
Takes a potentially unbalanced tree and turns itinto a balanced tree
Step 1
Turn the tree into a degenerate tree (backbone orvine) by right rotating any nodes with left children
Step 2
Turn the degenerate tree into a balanced tree bydoing sequences of left rotations starting at the root
The analysis is not obvious, but it takes O(n)total rotations
How much time does it take to insert n itemswith:
Red-black or AVL tree
Balanced insertion method
Unbalanced insertion + DSW rebalance
How much space does it take to insert n itemswith:
Red-black or AVL tree
Balanced insertion method
Unbalanced insertion + DSW rebalance
Balanced binary trees ensure that accessingany element takes O(log n) time
But, maybe only a few elements are accessedrepeatedly
In this case, it may make more sense torestructure the tree so that these elementsare closer to the root
1.Single rotation
Rotate a child about its parent if an element in achild is accessed, unless it's the root
2.Moving to the root
Repeat the child-parent rotation until the elementbeing accessed is the root
A concept called splaying takes the rotate to the root idea further,doing special rotations depending on the relationship of the childR with its parent Q and grandparent P
Cases:
1.Node R's parent is the root
Do a single rotation to make R the root
2.Node R is the left child of Q and Q is the left child of P(respectively right and right)
Rotate Q around P
Rotate R around Q
3.Node R is the left child of Q and Q is the right child of P(respectively right and left)
Rotate R around Q
Rotate R around P
We are not going deeply into self-organizingtrees
It's an interesting concept, and rigorousmathematical analysis shows that splay treesshould perform well, in terms of Big O bounds
In practice, however, they usually do not
An AVL tree or a red-black tree is generally thebetter choice
If you come upon some special problem whereyou repeatedly access a particular element mostof the time, only then should you consider splaytrees
Student Lecture
We can define a symbol table ADT with a few essentialoperations:
put(Key key, Value value)
▪Put the key-value pair into the table
get(Key key):
▪Retrieve the value associated with key
delete(Key key)
▪Remove the value associated with key
contains(Key key)
▪See if the table contains a key
isEmpty()
size()
It's also useful to be able to iterate over all keys
We have been talking a lot about trees andother ways to keep ordered symbol tables
Ordered symbol tables are great, but we maynot always need that ordering
Keeping an unordered symbol table mightallow us to improve our running time
Balanced binary search trees give us:
O(log n) time to find a key
O(log n) time to do insertions and deletions
Can we do better?
What about:
O(1) time to find a key
O(1) to do an insertion or a deletion
We make a huge array, so big that we’ll havemore spaces in the array than we expect datavalues
We use a hashing function that maps keys toindexes in the array
Using the hashing function, we know whereto put each key but also where to look for aparticular key
Let’s make a hash table to store integer keys
Our hash table will be 13 elements long
Our hashing function will be simply moddingeach value by 13
Insert these keys: 3, 19, 7, 104, 89
0
1
2
3
4
5
6
7
8
9
10
11
12
104
3
19
7
89
0
1
2
3
4
5
6
7
8
9
10
11
12
Find these keys:
19
YES!
88
NO!
16
NO!
104
3
19
7
89
0
1
2
3
4
5
6
7
8
9
10
11
12
We are using a hash table for a space/timetradeoff
Lots of space means we can get down to O(1)
How much space do we need?
How do we pick a good hashing function?
What happens if two values collide (map tothe same location)
We want a function that will map data tobuckets in our hash table
Important characteristics:
Efficient: It must be quick to execute
Deterministic: The same data must always mapto the same bucket
Uniform: Data should be mapped evenly acrossall buckets
We want a function h(k) that computes a hashfor every key k
The simplest way of guaranteeing that we hashonly into legal locations is by setting h(k) to be:
h(k) = k mod N where N is the size of the hashtable
To avoid crowding the low indexes, N should beprime
If it is not feasible for N to be prime, we can addanother step using a prime p > N:
h(k) = (k mod p) mod N
Pros
Simple
Fast
Easy to do
Good if you know nothing about the data
Cons
Prime numbers are involved (What’s the nearestprime to the size you want?)
Uses no information about the data
If the data is strangely structured (multiples of p, forexample) it could all hash to the same location
Break the key into parts and combine thoseparts
Shift folding puts the parts together withouttransformations
SSN: 123-45-6789 is broken up and summed 123 +456 + 789 = 1,368, then modded by N, probably
Boundary folding puts the parts togetherreversing every other part of the key
SSN: 123-45-6789 is broken up and summed 123 +654 + 789 = 1,566, then modded by N, probably
Pros
Relatively Simple and Fast
Mixes up the data more than division
Points out a way to turn strings or other non-integer data into an integer that can be hashed
Transforms the numbers so that patterns in thedata are likely to be removed
Cons
Primes are still involved
Uses no special information about the data
Square the key, then take the “middle”numbers out of the result
Example: key = 3,121 then 3,1212 = 9,740,641and the hash value is 406
One nice thing about this method is that wecan make the table size be a power of 2
Then, we can take the log2 N middle bits outof the squared value using bitwise shifts andmasking
Pros
Randomizes the data a lot
Fast when implemented correctly
Primes are not necessary
Cons
Uses no special information about the data
Remove part of the key, especially if it isuseless
Example:
Many SSN numbers for Indianapolis residentsbegin with 313
Removing the first 3 digits will, therefore, notreduce the randomness very much, provided thatyou are looking at a list of SSN’s for Indianapolisresidents
Pros
Uses information about the key
Can be efficient and easy to implement
Cons
Requires special knowledge
Careless extraction of digits can give poor hashingperformance
Change the number to a different base
Then, treat the base as if it were still base 10and use the division method
Example: 345 is 423 in base 9
If N = 100, we could take the mod and put 345in location 23
Pros
If many numbers have similar final digits or valuesmod N (or p), they can be randomized by thismethod
Cons
Choice of base can be difficult
Effects are unpredictable
Not as quick as many of the other methods
Values that didn’t collide before might now collide
Determine if a string has any duplicatecharacters
Weak!
Okay, but do it in O(m) time where m is thelength of the string
Collisions
Chaining implementation of hash tables
Keep working on Project 3!