![David Luebke 6 02/10/17
Direct Addressing
● Suppose:
■ The range of keys is 0..m-1
■ Keys are distinct
● The idea:
■ Set up an array T[0..m-1] in which
○ T[i] = x if x∈ T and key[x] = i
○ T[i] = NULL otherwise
■ This is called a direct-address table
○ Operations take O(1) time!
○ So what’s the problem?](https://image.slidesharecdn.com/hashtablesh3-170210091749/85/Hash-table-6-320.jpg)
![David Luebke 8 02/10/17
Review: Direct Addressing
● Suppose:
■ The range of keys is 0..m-1
■ Keys are distinct
● The idea:
■ Set up an array T[0..m-1] in which
○ T[i] = x if x∈ T and key[x] = i
○ T[i] = NULL otherwise
■ This is called a direct-address table
○ Operations take O(1) time!](https://image.slidesharecdn.com/hashtablesh3-170210091749/85/Hash-table-8-320.jpg)
Hash table
- 1. David Luebke 1 02/10/17 CS 332: Algorithms Introduction to Hashing
- 2. David Luebke 2 02/10/17 Review: Hashing Tables ● Motivation: symbol tables ■ A compiler uses a symbol table to relate symbols to associated data ○ Symbols: variable names, procedure names, etc. ○ Associated data: memory location, call graph, etc. ■ For a symbol table (also called a dictionary), we care about search, insertion, and deletion ■ We typically don’t care about sorted order
- 3. David Luebke 3 02/10/17 Review: Hash Tables ● Hash table: ■ Given a table T and a record x, with key (= symbol) and satellite data, we need to support: ○ Insert (T, x) ○ Delete (T, x) ○ Search(T, x) ■ We want these to be fast, but don’t care about sorting the records ■ In this discussion we consider all keys to be (possibly large) natural numbers
- 4. David Luebke 4 02/10/17 Review: Hash Tables ● More formally: ■ Given a table T and a record x, with key (= symbol) and satellite data, we need to support: ○ Insert (T, x) ○ Delete (T, x) ○ Search(T, x) ■ We want these to be fast, but don’t care about sorting the records ● The structure we will use is a hash table ■ Supports all the above in O(1) expected time!
- 5. David Luebke 5 02/10/17 Hashing: Keys ● In the following discussions we will consider all keys to be (possibly large) natural numbers ● How can we convert floats to natural numbers for hashing purposes? ● How can we convert ASCII strings to natural numbers for hashing purposes?
- 6. David Luebke 6 02/10/17 Direct Addressing ● Suppose: ■ The range of keys is 0..m-1 ■ Keys are distinct ● The idea: ■ Set up an array T[0..m-1] in which ○ T[i] = x if x∈ T and key[x] = i ○ T[i] = NULL otherwise ■ This is called a direct-address table ○ Operations take O(1) time! ○ So what’s the problem?
- 7. David Luebke 7 02/10/17 The Problem With Direct Addressing ● Direct addressing works well when the range m of keys is relatively small ● But what if the keys are 32-bit integers? ■ Problem 1: direct-address table will have 232 entries, more than 4 billion ■ Problem 2: even if memory is not an issue, the time to initialize the elements to NULL may be ● Solution: map keys to smaller range 0..m-1 ● This mapping is called a hash function
- 8. David Luebke 8 02/10/17 Review: Direct Addressing ● Suppose: ■ The range of keys is 0..m-1 ■ Keys are distinct ● The idea: ■ Set up an array T[0..m-1] in which ○ T[i] = x if x∈ T and key[x] = i ○ T[i] = NULL otherwise ■ This is called a direct-address table ○ Operations take O(1) time!
- 9. David Luebke 9 02/10/17 Review: The Problem With Direct Addressing ● Direct addressing works well when the range m of keys is relatively small ● But what if the keys are 32-bit integers? ■ Problem 1: direct-address table will have 232 entries, more than 4 billion ■ Problem 2: even if memory is not an issue, the time to initialize the elements to NULL may be ● Solution: map keys to smaller range 0..m-1 ● This mapping is called a hash function
- 10. David Luebke 10 02/10/17 Hash Functions ● Next problem: collision T 0 m - 1 h(k1) h(k4) h(k2) = h(k5) h(k3) k4 k2 k3 k1 k5 U (universe of keys) K (actual keys)
- 11. David Luebke 11 02/10/17 Resolving Collisions ● How can we solve the problem of collisions? ● Solution 1: chaining ● Solution 2: open addressing
- 12. David Luebke 12 02/10/17 Open Addressing ● Basic idea (details in Section 12.4): ■ To insert: if slot is full, try another slot, …, until an open slot is found (probing) ■ To search, follow same sequence of probes as would be used when inserting the element ○ If reach element with correct key, return it ○ If reach a NULL pointer, element is not in table ● Good for fixed sets (adding but no deletion) ■ Example: spell checking ● Table needn’t be much bigger than n
- 13. David Luebke 13 02/10/17 Chaining ● Chaining puts elements that hash to the same slot in a linked list: —— —— —— —— —— —— T k4 k2 k3 k1 k5 U (universe of keys) K (actual keys) k6 k8 k7 k1 k4 —— k5 k2 k3 k8 k6 —— —— k7 ——
- 14. David Luebke 14 02/10/17 Chaining ● How do we insert an element? —— —— —— —— —— —— T k4 k2 k3 k1 k5 U (universe of keys) K (actual keys) k6 k8 k7 k1 k4 —— k5 k2 k3 k8 k6 —— —— k7 ——
- 15. David Luebke 15 02/10/17 Chaining —— —— —— —— —— —— T k4 k2 k3 k1 k5 U (universe of keys) K (actual keys) k6 k8 k7 k1 k4 —— k5 k2 k3 k8 k6 —— —— k7 —— ● How do we delete an element? ■ Do we need a doubly-linked list for efficient delete?
- 16. David Luebke 16 02/10/17 Chaining ● How do we search for a element with a given key? —— —— —— —— —— —— T k4 k2 k3 k1 k5 U (universe of keys) K (actual keys) k6 k8 k7 k1 k4 —— k5 k2 k3 k8 k6 —— —— k7 ——
- 17. David Luebke 17 02/10/17 Analysis of Chaining ● Assume simple uniform hashing: each key in table is equally likely to be hashed to any slot ● Given n keys and m slots in the table: the load factor α = n/m = average # keys per slot ● What will be the average cost of an unsuccessful search for a key?
- 18. David Luebke 18 02/10/17 Analysis of Chaining ● Assume simple uniform hashing: each key in table is equally likely to be hashed to any slot ● Given n keys and m slots in the table, the load factor α = n/m = average # keys per slot ● What will be the average cost of an unsuccessful search for a key? A: O(1+α)
- 19. David Luebke 19 02/10/17 Analysis of Chaining ● Assume simple uniform hashing: each key in table is equally likely to be hashed to any slot ● Given n keys and m slots in the table, the load factor α = n/m = average # keys per slot ● What will be the average cost of an unsuccessful search for a key? A: O(1+α) ● What will be the average cost of a successful search?
- 20. David Luebke 20 02/10/17 Analysis of Chaining ● Assume simple uniform hashing: each key in table is equally likely to be hashed to any slot ● Given n keys and m slots in the table, the load factor α = n/m = average # keys per slot ● What will be the average cost of an unsuccessful search for a key? A: O(1+α) ● What will be the average cost of a successful search? A: O(1 + α/2) = O(1 + α)
- 21. David Luebke 21 02/10/17 Analysis of Chaining Continued ● So the cost of searching = O(1 + α) ● If the number of keys n is proportional to the number of slots in the table, what is α? ● A: α = O(1) ■ In other words, we can make the expected cost of searching constant if we make α constant
- 22. David Luebke 22 02/10/17 Choosing A Hash Function ● Clearly choosing the hash function well is crucial ■ What will a worst-case hash function do? ■ What will be the time to search in this case? ● What are desirable features of the hash function? ■ Should distribute keys uniformly into slots ■ Should not depend on patterns in the data
- 23. David Luebke 23 02/10/17 Hash Functions: The Division Method ● h(k) = k mod m ■ In words: hash k into a table with m slots using the slot given by the remainder of k divided by m ● What happens to elements with adjacent values of k? ● What happens if m is a power of 2 (say 2P )? ● What if m is a power of 10? ● Upshot: pick table size m = prime number not too close to a power of 2 (or 10)
- 24. David Luebke 24 02/10/17 Hash Functions: The Multiplication Method ● For a constant A, 0 < A < 1: ● h(k) = m (kA - kA) What does this term represent?
- 25. David Luebke 25 02/10/17 Hash Functions: The Multiplication Method ● For a constant A, 0 < A < 1: ● h(k) = m (kA - kA) ● Choose m = 2P ● Choose A not too close to 0 or 1 ● Knuth: Good choice for A = (√5 - 1)/2 Fractional part of kA
- 26. David Luebke 26 02/10/17 Hash Functions: Worst Case Scenario ● Scenario: ■ You are given an assignment to implement hashing ■ You will self-grade in pairs, testing and grading your partner’s implementation ■ In a blatant violation of the honor code, your partner: ○ Analyzes your hash function ○ Picks a sequence of “worst-case” keys, causing your implementation to take O(n) time to search ● What’s an honest CS student to do?
- 27. David Luebke 27 02/10/17 Hash Functions: Universal Hashing ● As before, when attempting to foil an malicious adversary: randomize the algorithm ● Universal hashing: pick a hash function randomly in a way that is independent of the keys that are actually going to be stored ■ Guarantees good performance on average, no matter what keys adversary chooses
- 28. David Luebke 28 02/10/17 The End