1. Hashing & HashMaps
CS-2851 Dr. Mark L. Hornick

2. Let’s review the worst-case performance characteristics of previously covered data structures
ArrayList – JCF class
get()
add()
contains()
SortedArrayList (uses binary searching)
LinkedList – JCF class
BinaryTree
CS-2851 Dr. Mark L. Hornick

3. Let’s review the worst-case performance characteristics of previously covered data structures
ArrayList
get() – O(k); constant-time access
add() – O(n); due to shifting
contains() – O(n); sequential search
SortedArrayList
get()
add()
contains()
LinkedList
BinaryTree
CS-2851 Dr. Mark L. Hornick

4. Let’s review the worst-case performance characteristics of previously covered data structures
ArrayList
get() – O(k); constant-time access
add() – O(n); due to shifting
contains() – O(n); sequential search
SortedArrayList
add() – O(n+log n)->O(n); due to need to figure out first where to add; then the need to shift elements to the right
contains() – O(log n); search based on the Splitting Rule (binary search)
LinkedList
get()
add()
contains()
BinaryTree
CS-2851 Dr. Mark L. Hornick

5. Let’s review the worst-case performance characteristics of previously covered data structures
ArrayList
get() – O(k); constant-time access
add() – O(n); due to shifting
contains() – O(n); sequential search
SortedArrayList
add() – O(n+log n)->O(n); due to need to figure out first where to add; then the need to shift elements to the right
contains() – O(log n); search based on the Splitting Rule (binary search)
LinkedList
get() – O(n); sequential access
add() – O(k); once where to add has been determined
really O(n), because that’s how it takes to find the location to insert
BinaryTree
get()
add()
contains()
CS-2851 Dr. Mark L. Hornick

6. Let’s review the worst-case performance characteristics of previously covered data structures
ArrayList
get() – O(k); constant-time access
add() – O(n); due to shifting
contains() – O(n); sequential search
SortedArrayList
add() – O(n+log n)->O(n); due to need to figure out first where to add; then the need to shift elements to the right
contains() – O(log n); search based on the Splitting Rule (binary search)
LinkedList
get() – O(n); sequential access
add() – O(k); once where to add has been determined
really O(n), because that’s how it takes to find the location to insert
BinaryTree
get() – not supported due to lack of indexing (but do we always need it?)
add() – O(log n); due to sorting built into the tree structure
contains() – O(log n); due to sorting built into the tree structure
What about memory usage??
CS-2851 Dr. Mark L. Hornick

7. Is there anything faster at everything?
CS-2851 Dr. Mark L. Hornick

8. Map definition
A map is a collection in which each Entry element has two parts
a unique key part
a value part (which may not be unique)
Each unique key “maps” to a corresponding value
Example: Morse code map – each character maps to a (unique) sequence of dots and dashes
Example: a map of Students, in which each key is the (unique) student ID, and each (non-unique?) value is a reference to the Student object itself
Example: a phonebook, where each number (each key) maps to a person
Entry
key
value
CS-2851 Dr. Mark L. Hornick

9. What is a Key?
A key is just something that uniquely identifies a particular instance of an value/object
A key can be a number, a string, or an object, so long as it is unique
If two values/objects have the same key, then they are (theoretically) equal
Only one ID per MSOE student, so if the ID’s match, it must (by definition) be the same student
If the equals() method comparing two keys returns true, then the objects are equal too, by definition
CS-2851 Dr. Mark L. Hornick

10. What if an object doesn’t possess a specific unique attribute?
Scenario: pretend MSOE ID’s didn’t exist
Can any of the attributes of a student, taken together, be unique?
…even though any individual attribute may not exhibit this uniqueness?
Exercise
CS-2851 Dr. Mark L. Hornick

11. A key can be generated from a unique combination of non-unique attributes
All of an object’s attributes can be used to generate the key
That is, the object itself is the key
Or the key can be generated from just a subset of an object’s attributes
Provided that subset is unique
CS-2851 Dr. Mark L. Hornick

12. OK, so what role do keys play in making a faster data structure?
What if each unique key corresponded to a unique index within an array of Entries?
Maps to
key
index
Entry
key
value
CS-2851 Dr. Mark L. Hornick

13. Hash definition
A hash is a transformation of a key into a numeric value that maps to the index of an array (or table)
This is done in two steps:
generate a numeric hashcode from the key (which is not necessarily numeric)
If the key is already numeric and unique (like an ID), then the key can be used as the hashcode
transform the hashcode into an array index
Key
hashcode
index
Winter 2005
CS-2851 Dr. Mark L. Hornick

14. HashMap definition
A HashMap<E> is an array-based collection of Entry<E> elements
a value part (which could be anything)
a unique key part (somehow derived from value)
Each Entry is at a specific index in the array, where the index is determined from the hashcode of the key
Example: a map of Students, in which each key is the (unique) student ID, and each (non-unique?) value is a reference to the Student object itself
Entry<E>
key
E value
CS-2851 Dr. Mark L. Hornick

15. How do you generate a hashcode?
In Java, all classes have a built-in hashCode() method defined in the Object class
Key
hashcode
CS-2851 Dr. Mark L. Hornick

16. Classes that don’t override hashCode() inherit the Object class’s hashCode() method
Which returns the memory address of the object
Is this a repeatable hashcode??? No!
Mem addr
Object
hashcode
CS-2851 Dr. Mark L. Hornick

17. A given key should always generate the same hashcode
So that the hashcode computation can be repeated at any time, and always result in the same value
…and therefore, the same index
Q: If keys are unique, does this guarantee the hashcode generated from the keys are also unique??
Key
hashcode
index
CS-2851 Dr. Mark L. Hornick

18. Exercise Generate a hashcode from a String of characters
What approach should you use??
CS-2851 Dr. Mark L. Hornick

19. How do you generate a hashcode?
In Java, many classes override Objects hashcode() method in order to generate unique hashcodes
Integer class
Integer’s hashCode( ) method simply returns the underlying int value
String class
Look at the javadoc for String.hashCode
Key
hashcode
CS-2851 Dr. Mark L. Hornick

20. Writing your own hashCode()
A key should uniquely identify an object
Hashcodes generated from keys should be as unique as possible
to avoid collisions
Depending on the hashcode algorithm, different keys can generate the same hashcode
Key
hashcode
index
CS-2851 Dr. Mark L. Hornick

21. How do you transform a hashcode into an array index?
Assume you have an array with length=1024
An array index in the range 0…1023 can be computed as follows using modulo arithmetic:
int index = hashCode( )% 1024;
The resulting index=933
CS-2851 Dr. Mark L. Hornick

22. More hashing examples (for a table 1024 in length)
indexes to 933
indexes to 500
indexes to 234
CS-2851 Dr. Mark L. Hornick

23. Exercise table size null 3 … xxx Anne xxx yyy yyy Susan zzz Ed zzz
xxx
yyy
zzz
1023 3
null …
xxx Anne
yyy Susan
zzz Ed
What are the index values xxx, yyy, and zzz?
CS-2851 Dr. Mark L. Hornick

24. Hashing can result in Collisions
indexes to 933
indexes to 500
indexes to 234
also indexes to 933
When two different keys yield the same index (even from different hashcodes), that is called a collision
Keys that yield the same index are called synonyms
Special handling is required
CS-2851 Dr. Mark L. Hornick

25. Hashing is inefficient when there are a lot of collisions
Ideally, we want the hashing algorithm to generate indices “sprinkled” randomly throughout the underlying table
The Uniform Hashing Assumption assumes
Each key is equally likely to hash to any one of the table addresses, independently of where the other keys have hashed
CS-2851 Dr. Mark L. Hornick

26. Even if this assumption is true, collisions still occur
This is due to the finite set of indices in a table
The bigger the table, the less likely a collision is to occur
But tables cannot be made infinitely large
An infinite number of keys cannot be mapped into a finite set of indices
So collision handlers have to be implemented
Winter 2005
CS-2851 Dr. Mark L. Hornick