COMP 103 Hashing

2 RECAP-TODAY RECAP Bitmaps are a fast way to implement Sets of integers, characters, etc TODAY  Hashing is a similar idea  Detecting collisions  Dealing with collisions:  Buckets/chaining  Probing

3 O(1) Sets with big values? ✔ We need a way to compute an array index for an object: add(“Show me the book that refers to all the books that don’t reference themselves”) “Hashing”: the number is the “hash code” of object N ✔✗✔✔✗✗✗✗✗✗✗ ⋯⋯ ✗ Hash function 581

4 O(1) Sets with big values?  But there are too many possible sentences!!  Suppose the hash function always produces a number between (say) 0 and 10,000… ⇒ some sentences must end up with the same number!! ⇒ “Collision” N ✔✗✔✔✗✗✗✗✗✗✗ ⋯⋯ ✔✗ HASH “ Show me the book th… ” “ The barber shaves everyone who doesn’t shave themselves ” HASH

5 Detecting collisions  Store the item in the array, instead of just a boolean  When asked about an item, or adding one  need to check if there is a different item there already.  Questions: 1. How do you compute the hash code? 2. What do you do about collisions? N ⋯⋯ “ Show me the book th… ” “ The barber shaves everyone… ” HASH

6 A HashSet private E[ ] data ; public boolean contains(E value) { int hash = Math.abs(value.hashCode() % data.length); if (data[hash] == null) return false; else if (data[hash].equals(value)) return true; else //Collision !!! } public boolean add(E value) { int hash = Math.abs(value.hashCode() % data.length); if (data[hash] == null) { data[hash] = value; size++; return true; } else if (data[hash].equals(value)) return false; else //Collision !!! } Cost is independent of number of items in Set Cost is determined by cost of hashCode() must be consistent

7 Computing Hash Codes Wish list:  Should produce an integer  Should distribute the hash codes evenly through the range. minimises collisions  Should be fast to compute  Should take account of all components of the object  Must be consistent with equals() !!! two items that are equal must have the same hash value. Can we avoid clashes altogether? That would be perfect!

8 A Simple Hash Function for Strings  We could add up the codes of all the characters: private int hash(String value) { int hash = 0; for (int i = 0; i < value.length(); i++) hash += value.charAt(i); return hash; } So why is this not very good?

9 Example: Hashing course codes  Course codes: four letters and three digits 418 ← DEAF ← DEAF102 DEAF201 ⋮ 429 ← BBSC201 MDIA ← ECHI410 MDIA102 MDIA ← ECHI303 JAPA111 JAPA201 MDIA202 MDIA220 MDIA ← ARCH101 ASIA101 BBSC231 BBSC303 BBSC321 CHEM201 ECHI403 ECHI412 JAPA112 JAPA211 JAPA301 MDIA203 MDIA302 MDIA320 ⋮ 450 ← ANTH412 ARCH389 ARTH111 BIOL228 BIOL327 BIOL372 CHEM489 COML304 COML403 COML421 COMP102 COMP201 CRIM313 CRIM421 DESN215 DESN233 ECON328 ECON409 ECON418 ECON508 EDUC449 EDUC458 EDUC548 EDUC557 ENGL228 ENGL408 ENGL426 ENGL435 ENGL444 ENGL453 FREN124 FREN331 FREN403 FREN412 GEOL362 GEOL407 GERM214 GERM403 GERM412 INFO213 INFO312 INFO402 ITAL206 ITAL215 LALS501 LATI404 LING224 LING323 LING404 MAOR102 MARK304 MARK403 MATH206 MATH314 MATH323 MATH431 MOFI403 PHIL104 PHIL203 PHIL302 PHIL320 PHIL401 PHIL410 RELI321 RELI411 SAMO101 ⋮ ie. a lot of collisions!

10 Better Hash Functions  Make the contribution of each character depend on its position: private int hash(String course) { int k = 257; int hash = 0; for (int i = 0; i < course.length(); i ++ ) hash = hash*k + course.charAt(i); return hash; } k 6 x c 0 + k 5 x c 1 + k 4 x c 2 + k 3 x c 3 + k 2 x c 4 + k 1 x c 5 + c 6 (it is best to use a prime number for the constant)

11 Perfect Hash Functions  Perfect hash function gives no collisions for a given data set!  Example - for VUW courses (in ): private int hash(String course) { int hash = 0; for (int i = 0; i < course.length(); i++) hash = (hash * 51 + course.charAt(i)) % 72201; return hash; }  Building a perfect hash function:  very difficult  very specific to a particular set of possible values  only useful in very specialised circumstances

12 Designing Hash Functions  Integers:  use the value itself (mod table size)  Strings:  use a standard method  Objects with several fields, Arrays, Lists  combine the hash values of the individual fields/cells each multiplied by a position (like String)  Sets (where order doesn’t matter)  ?  Look up a text book  This is a specialised and well studied topic!

13 Java and hashCode  All objects have a hashCode method and an equals method, so:  you can call equals on any object  and you can put any object into a HashSet, HashMap, …  Many predefined objects (eg String) have good equals and hashCode methods defined.  The default equals method:  are the references identical? ie, equals is ==  if this is not what you want, define your own equals method  The default hashCode (eg. for user defined classes):  returns an integer based on the reference (pointer value)  So… if you redefine equals, you should redefine hashCode too!

14 Dealing with Collisions  Two approaches:  Use a collection at each place (“buckets” or “chaining”)  Look for an empty place in the hashtable (“probing” or “open addressing”) N ⋯⋯ “ The barber shaves everyone… ” HASH “ Show me the book th… ” HASH “ Show me the book th… ” ✔ ✘

15 Collisions: chaining / buckets  Store a Set in each cell: hash value → which set  Performance?  if the array is of size k, each subset will be about 1/k th of size().  cost ≈ cost(hashCode) + cost (subset) ant fox hen dog bee kea cow elk owl pig sow tui ape bat bug cat eel gnu jay nit ray yak cod roe What kind of Sets ? This is what Java's HashMap does. If the sets get too big.... just rehash!

16 Linear Probing Hash value tells us where to start looking.  if value.hashCode() → p start at index p if cell is used, try p+1, p+2, p+3 … wrap round to 0 at the end of the array: “the division method” hash = (name[0]+name[1])% Sam Steve StigStu Sven Sun (3) (2) (5) (4) (2)