09 searching[1]

90-723: Data Structuresand Algorithms for

Information Processing Copyright © 1999, Carnegie Mellon. All Rights Reserved.

1Lecture 9: Searching

Data Structures and Algorithms for Information

ProcessingLecture 9: Searching




Outline

• The simplest method: serial search

• Binary search• Open-address hashing• Chained hashing




Search Algorithms

Whenever large amounts of data need to be accessed quickly, search algorithms are crucially involved.




Search Algorithms Lie at the heart of many computer

technologies. To name a few:– Databases– Information retrieval applications– Web infrastructure (file systems,

domain name servers, etc.)– String searching for patterns




Search Algorithms: Two Broad Categories

• Searching a static database– Accessing indexed Web pages– Finding a file on disk

• Evaluating a dynamically changing set of hypotheses– Computer chess (search for a move)– Speech recognition (search for text given

speech)We’ll be concerned with the first




The Simplest Search: Serial Lookup

• Items are stored in an array or list.

• To search for an item x:– Start at the beginning of the list– Compare the current item to x– If unequal, proceed to next item




Pseudocode for Serial Search// Find x in an array a of length

nint i=0;boolean found = false;while ((i < n) && !found) { if (a[i] == x) found = true; else i++;}if (found) ...




Analysis for Serial Search• Best case: Requires one array

access: Θ(1)• Worst case: Requires n array

accesses: Θ(n) • Average case: To access an item,

assuming position is random (uniform):(1+2+3+...+n)/n = n(n+1)/2n = (n+1)/2 = Θ(n)




A Useful Combinatorial Identity

1+2+3+…+n = n(n+1)/2Why?

Algebraic Proof in MainVisual Counting




Visual Counting

n*n




Visual Counting

n




Visual Counting

n*n - n




Visual Counting

(n*n - n)/2 + n = n(n+1)/2




Binary Search• Can be used whenever the data are totally

ordered -- e.g., the integers. All elements are comparable.

• Requires sorting in advance, and storing in an array

• One of the simplest to implement, often “fast enough”

• Can be tricky to handle “boundary cases”• This a classic divide-and-conquer algorithm.




Idea of Binary Search• Closely related to the natural

algorithm we use to look up a word in a dictionary – Open to the middle– If target comes before all words on

the page, search in left half of book– Otherwise, search in right half.




Interface for Binary Searchint search(int [] a, int first, int size, int target)• Parameters:

– int [] a: array to be searched over– Search over a[first,first+1,...,first+size-1]

• Precondition:– array is sorted in increasing order– first >= 0




int search (int [] a, int start, int size, int target) { if (size <= 0) return -1; else { int middle = start + size/2; if (a[middle] == target) return middle; else if (target < a[middle]) return search(a, start, size/2, target); else return search(a, middle+1, size/2, target); }}

Implementation




Implementation

Where’s the error??Suppose size is odd. Arenew sizes correct?Suppose size is even. Are new sizes correct?




int search (int [] a, int first, int size, int target) { if (size <= 0) return -1; else { int middle = first + size/2; if (a[middle] == target) return middle; else if (target < a[middle]) return search(a, first, size/2, target); else return search(a, middle+1, (size-1)/2, target); }}

Implementation




Boundary Cases• Binary search is sometimes tricky

to get right. • A common source of bugs.• Test cases are not always helpful

for checking correctness of code.• How many test cases would our

first implementation solve?




Binary Search with Other Data Structures

• Can binary search be implemented using linked lists rather than arrays?

• Are there any other data structures that could be used?




Analysis of Binary Search• Recursively dividing up array in half represents

data as a full binary tree.• Consider the simplest case -- array of size n =

2k -1, complete binary tree.• Take away one and divide by 2.• New Size = 2k-1 - 1.• We can only do that k times and k = Lg(n+1).• Thus, worst case involves Θ(log n)

operations.




Average Case• A complete binary tree with k

leaves has k-1 internal nodes.• So, about half of the n data

elements require Θ(log n) operations to find.

• Thus, assuming uniform distribution on target elements, average cost is also Θ(log n).




Binary Search is Limited When we have a large number

of items that will be accessed in part of the program, where efficiency is crucial, binary search may be too slow.



Improving Binary SearchTry to guess more precisely where the

key is located in the interval.Generalize middle = first + size/2

(key – a[first])

middle = ------------------------------ * size (a[first+size-1] – a[first])




Interpolation Search• This modifies method is called

interpolation search.• Uses fewer than log(log(N))

comparisons in the average caes.• But uses Θ(N) in the worst case.• For analysis, see Perl, Ital, Avni

“Interpolation Search – A Log Log N search” CACM 21 (1978) Pages 550 – 553• Is log (log (N)) better that log (N)?




Comparing Log N to Log(Log N)

• Suppose N = 2^100 Log N = 100 Log (Log N) = Log (100) = 6.65• Suppose N = 2^(2^100) Log N = 2^100 Log (Log N) = Log 2^100 = 100




Comparing Log(N) to Log(Log N)

• Or, by taking limits…• Lim Log(Log(n)) / Log(n) n->∞

is of the form inf. / inf.• Apply L’Hopital and take derivatives.• Lim 1/(Log N) * 1/n n->∞ -------------------- = 0 1/n





Hashing• Fortunately, we can often do better • Hashing is a technique that where

the access time can be O(1) rather than O(log n)




Open Address HashingThe basic technique:• Items are stored in an array of size N• The preferred position in the array is

computed using a hash function of the item’s key

• When adding an item, if the preferred position is occupied, the next open position in the array is used instead.




Open Address HashingMain ’s presentation for Chapter 11




A Basic Hash Table• We keep arrays for the keys and

data, and a bit indicating whether a given position has been occupiedprivate class Table { private int numItems; private Object[] keys; private Object[] data; private boolean[] hasBeenUsed; ....}




The Hash Function• We can use the built in hashCode()

method that Java provides private int hash (Object key) { return Math.abs(key.hashCode()) %

data.length; }




Calculating the Index// If found return value is index of key private int findIndex(Object key) { int count=0; int i=hash(key); while ((count < data.length) && (hasBeenUsed[i]))

{ if (key.equals(keys[i])) return i; i = nextIndex(i); count++; } return -1;}




Inserting an Itempublic Object put (Object key, Object element) { int index = findIndex(key); if (index != -1) { Object answer = data[index]; data[index] = element; return answer; } else if (numItems < data.length) { ....




Inserting an Itempublic Object put (Object key, Object element) { ... else if (numItems < data.length) { index = hash(key); while (keys[index] != null) index = nextIndex(index); keys[index] = key; data[index] = element; hasBeenUsed[index] = true; numItems++; return null; } else throw new IllegalStateException(“Table full”) ....




Two Hashes are Better than One

• Collisions can result in long stretches of positions with keys not in their “preferred” position

• This is called clustering• To address this problem, when a

collision results we jump a “random” number of positions, using a second hash function




Double Hashing• Find the first position using hash1(key)• If there’s a collision, step through the

array in steps of size hash2(key):i = (i + hash2(key)) % data.length

• To avoid cycles, hash2(key) and the length of the array must be relatively prime (no common factors)




Double Hashing• Knuth’s technique to avoid cycles:• Choose the length of the array so that

both data.length and data.length-2 are prime

hash1(key) = Math.abs(key.hashCode()) % lengthhash2(key) = 1 + (Math.abs(key.hashCode()) %

(length-1)




Issues with O-A Hashing• Each array cell holds only one

element• Collisions and clustering can

degrade performance• Once the array is full, no more

elements can be added, unless we:– create a new array with the right size

and hash functions– re-hash the original elements




Chained Hashing• Each array cell can hold more than

one element of the hash table• Hash the key of each element to

obtain the array index• When a collision happens, the

element is still placed at the original hash index

• How is this handled?




Answer• Each array location must be

implemented with a data structure that can hold a group of elements with the same hash index

• Most common approach– each array location stores the head of

a linked list– items in the list all have the same has

index




Chained Hashingtable …

[0] [1] [2] [3]

elementkeylink

elementkeylink

elementkeylink

elementkeylink

Any number of elements can beadded to the table without a need to rehash

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