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Free Video Lectures for Free Video Lectures for MBAMBA
By:video.edhole.com
Keys into Buckets: Keys into Buckets: Lower bounds, Linear-time sort, Lower bounds, Linear-time sort,
& Hashing& Hashing
Comp 122, Spring 2004video.edhole.com
Comparison-based SortingComparison-based Sorting
Comparison sort◦Only comparison of pairs of elements may be
used to gain order information about a sequence.◦Hence, a lower bound on the number of
comparisons will be a lower bound on the complexity of any comparison-based sorting algorithm.
All our sorts have been comparison sortsThe best worst-case complexity so far is (n lg n) (merge sort and heapsort).
We prove a lower bound of (n lg n) for any comparison sort: merge sort and heapsort are optimal.
The idea is simple: there are n! outcomes, so we need a tree with n! leaves, and therefore lg(n!) =
Comp 122video.edhole.com
Decision TreeDecision TreeComp 122
For insertion sort operating on three elements.
1:2
2:3 1:3
1:3 2:31,2,3
1,3,2 3,1,2
2,1,3
2,3,1 3,2,1
>
>
>>
Contains 3! = 6 leaves.
Simply unroll all loops for all possible inputs.
Node i:j means compare A[i] to A[j].
Leaves show outputs;
No two paths go to same leaf!
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Decision Tree (Contd.)Decision Tree (Contd.)
Execution of sorting algorithm corresponds to tracing a path from root to leaf.
The tree models all possible execution traces.At each internal node, a comparison ai aj is
made.◦If ai aj, follow left subtree, else follow right subtree.◦View the tree as if the algorithm splits in two at each
node, based on information it has determined up to that point.
When we come to a leaf, ordering a(1) a (2) … a (n) is established.
A correct sorting algorithm must be able to produce any permutation of its input.◦Hence, each of the n! permutations must appear at one
or more of the leaves of the decision tree.
Comp 122video.edhole.com
A Lower Bound for Worst CaseA Lower Bound for Worst Case
Worst case no. of comparisons for a sorting algorithm is◦Length of the longest path from root to any of the
leaves in the decision tree for the algorithm. Which is the height of its decision tree.
A lower bound on the running time of any comparison sort is given by◦A lower bound on the heights of all decision trees
in which each permutation appears as a reachable leaf.
Comp 122video.edhole.com
Optimal sorting for three Optimal sorting for three elementselements
Comp 122
Any sort of six elements has 5 internal nodes.
1:2
2:3 1:3
1:3 2:31,2,3
1,3,2 3,1,2
2,1,3
2,3,1 3,2,1
>
>
>>
There must be a worst-case path of length ≥ 3.
video.edhole.com
A Lower Bound for Worst CaseA Lower Bound for Worst Case
Proof:Suffices to determine the height of a decision
tree.The number of leaves is at least n! (#
outputs)The number of internal nodes ≥ n!–1The height is at least lg (n!–1) = (n lg n)
QEDComp 122
Theorem 8.1:Any comparison sort algorithm requires (n lg n) comparisons in the worst case.
Theorem 8.1:Any comparison sort algorithm requires (n lg n) comparisons in the worst case.
video.edhole.com
Beating the lower boundBeating the lower bound
We can beat the lower bound if we don’t base our sort on comparisons:◦Counting sort for keys in [0..k], k=O(n)◦Radix sort for keys with a fixed number of
“digits”◦Bucket sort for random keys (uniformly
distributed)
Comp 122video.edhole.com
Counting SortCounting Sort
Assumption: we sort integers in {0, 1, 2, …, k}.
Input: A[1..n] {0, 1, 2, …, k}n. Array A and values n and k are given.
Output: B[1..n] sorted. Assume B is already allocated and given as a parameter.
Auxiliary Storage: C[0..k] countsRuns in linear time if k = O(n).
Comp 122video.edhole.com
Counting-Sort (Counting-Sort (A, B, kA, B, k))
CountingSort(A, B, k)1. for i 1 to k2. do C[i] 03. for j 1 to length[A]4. do C[A[j]] C[A[j]] + 15. for i 2 to k6. do C[i] C[i] + C[i –1] 7. for j length[A] downto 18. do B[C[A[ j ]]] A[j]9. C[A[j]] C[A[j]]–1
CountingSort(A, B, k)1. for i 1 to k2. do C[i] 03. for j 1 to length[A]4. do C[A[j]] C[A[j]] + 15. for i 2 to k6. do C[i] C[i] + C[i –1] 7. for j length[A] downto 18. do B[C[A[ j ]]] A[j]9. C[A[j]] C[A[j]]–1
Comp 122
O(k) init counts
O(k) prefix sum
O(n) count
O(n) reorder
video.edhole.com
Radix SortRadix Sort
Used to sort on card-sorters:Do a stable sort on each column,
one column at a time.The human operator is
part of the algorithm!
Key idea: sort on the “least significant digit” first and on the remaining digits in sequential order. The sorting method used to sort each digit must be “stable”.◦If we start with the “most significant digit”, we’ll
need extra storage.
Comp 122video.edhole.com
An ExampleAn Example
392 631 928 356356 392 631 392446 532 532 446928 495 446 495631 356 356 532532 446 392 631495 928 495 928
Comp 122
Input After sortingon LSD
After sortingon middle digit
After sortingon MSD
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Radix-Sort(Radix-Sort(A, dA, d))
Correctness of Radix SortBy induction on the number of digits sorted.Assume that radix sort works for d – 1 digits. Show that it works for d digits. Radix sort of d digits radix sort of the low-
order d – 1 digits followed by a sort on digit d .
Comp 122
RadixSort(A, d)
1. for i 1 to d
2. do use a stable sort to sort array A on digit i
RadixSort(A, d)
1. for i 1 to d
2. do use a stable sort to sort array A on digit i
video.edhole.com
Algorithm AnalysisAlgorithm Analysis
Each pass over n d-digit numbers then takes time (n+k). (Assuming counting sort is used for each pass.)
There are d passes, so the total time for radix sort is (d (n+k)).
When d is a constant and k = O(n), radix sort runs in linear time.
Radix sort, if uses counting sort as the intermediate stable sort, does not sort in place. ◦ If primary memory storage is an issue, quicksort or other
sorting methods may be preferable.
Comp 122video.edhole.com
Bucket SortBucket Sort
Assumes input is generated by a random process that distributes the elements uniformly over [0, 1).
Idea:◦Divide [0, 1) into n equal-sized buckets.◦Distribute the n input values into the buckets.◦Sort each bucket.◦Then go through the buckets in order, listing
elements in each one.
Comp 122video.edhole.com
An ExampleAn ExampleComp 122video.edhole.com
Bucket-Sort (Bucket-Sort (AA))
BucketSort(A)1. n length[A]2. for i 1 to n3. do insert A[i] into list B[ nA[i] ]4. for i 0 to n – 1 5. do sort list B[i] with insertion sort6. concatenate the lists B[i]s together in
order7. return the concatenated lists
BucketSort(A)1. n length[A]2. for i 1 to n3. do insert A[i] into list B[ nA[i] ]4. for i 0 to n – 1 5. do sort list B[i] with insertion sort6. concatenate the lists B[i]s together in
order7. return the concatenated lists
Comp 122
Input: A[1..n], where 0 A[i] < 1 for all i.Auxiliary array: B[0..n – 1] of linked lists, each list initially empty.
video.edhole.com
AnalysisAnalysis
Relies on no bucket getting too many values.All lines except insertion sorting in line 5 take
O(n) altogether.Intuitively, if each bucket gets a constant
number of elements, it takes O(1) time to sort each bucket O(n) sort time for all buckets.
We “expect” each bucket to have few elements, since the average is 1 element per bucket.
But we need to do a careful analysis.
Comp 122video.edhole.com
Analysis – Contd.Analysis – Contd.
RV ni = no. of elements placed in bucket B[i].
Insertion sort runs in quadratic time. Hence, time for bucket sort is:
Comp 122
1
0
2
1
0
2
1
0
2
1
0
2
)][][( ])[()(
n)expectatio oflinearity (by )]([)(
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have wen,expectatio
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)()()(
n
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n
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n
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XaEaXEnEOn
nOEn
nOnEnTE
nOnnT
(8.1)
video.edhole.com
Analysis – Contd.Analysis – Contd.
Claim: E[ni2] = 2 – 1/n.
Proof:Define indicator random variables.
◦Xij = I{A[j] falls in bucket i}◦Pr{A[j] falls in bucket i} = 1/n.
◦ni =
Comp 122
n
jijX
1
(8.2)
video.edhole.com
Analysis – Contd.Analysis – Contd.Comp 122
njkj
nkikij
n
jij
n
j njkj
nkikijij
ik
n
j
n
kij
n
jiji
XXEXE
XXX
XXE
XEnE
1 11
2
1 1 1
2
1 1
2
1
2
n.expectatio oflinearity by , ][][
E
][
(8.3)
video.edhole.com
Analysis – Contd.Analysis – Contd.2
2
22
1
11
][][][
variables.
randomt independen are and , Since
:for ][
1
11
110
}bucket in falls ][Pr{1
}bucket in fallt doesn' ][Pr{0][
nnn
XEXEXXE
XXkj
kjXXEn
nn
ijA
ijAXE
ikijikij
ikij
ikij
ij
Comp 122video.edhole.com
Analysis – Contd.Analysis – Contd.)(
)()(
)/12()()]([
.1
2
11
1)1(
1
11][
1
0
2
1 1 12
2
n
nOn
nOnnTE
n
nn
nnn
nn
nnnE
n
i
n
j njjk
nki
Comp 122
Substituting (8.2) in (8.1), we have,
(8.3) is hence,
video.edhole.com
Hash Tables – 1Hash Tables – 1
Comp 122, Spring 2004video.edhole.com
Dictionary Dictionary
Dictionary:◦Dynamic-set data structure for storing items
indexed using keys.◦Supports operations Insert, Search, and Delete.◦Applications: Symbol table of a compiler. Memory-management tables in operating systems. Large-scale distributed systems.
Hash Tables:◦Effective way of implementing dictionaries.◦Generalization of ordinary arrays.
Comp 122video.edhole.com
Direct-address Tables Direct-address Tables
Direct-address Tables are ordinary arrays.Facilitate direct addressing.
◦Element whose key is k is obtained by indexing into the kth position of the array.
Applicable when we can afford to allocate an array with one position for every possible key.◦i.e. when the universe of keys U is small.
Dictionary operations can be implemented to take O(1) time.◦Details in Sec. 11.1.
Comp 122video.edhole.com
Hash TablesHash Tables
Notation:◦U – Universe of all possible keys.◦K – Set of keys actually stored in the dictionary.◦ |K| = n.
When U is very large,◦Arrays are not practical.◦|K| << |U|.
Use a table of size proportional to |K| – The hash tables.◦However, we lose the direct-addressing ability.◦Define functions that map keys to slots of the hash
table.
Comp 122video.edhole.com
HashingHashing
Hash function h: Mapping from U to the slots of a hash table T[0..m–1]. h : U {0,1,…, m–1}
With arrays, key k maps to slot A[k].With hash tables, key k maps or “hashes” to
slot T[h[k]].h[k] is the hash value of key k.
Comp 122video.edhole.com
HashingHashingComp 122
0
m–1
h(k1)
h(k4)
h(k2)=h(k5)
h(k3)
U(universe of keys)
K(actualkeys)
k1
k2
k3
k5
k4
collision
video.edhole.com
Issues with HashingIssues with Hashing
Multiple keys can hash to the same slot – collisions are possible.◦Design hash functions such that collisions are
minimized.◦But avoiding collisions is impossible. Design collision-resolution techniques.
Search will cost Ө(n) time in the worst case.◦However, all operations can be made to have an
expected complexity of Ө(1).
Comp 122video.edhole.com
Methods of ResolutionMethods of Resolution
Chaining: ◦Store all elements that hash to the
same slot in a linked list.◦Store a pointer to the head of the
linked list in the hash table slot.Open Addressing:
◦All elements stored in hash table itself.
◦When collisions occur, use a systematic (consistent) procedure to store elements in free slots of the table.
Comp 122
k2
0
m–1
k1 k4
k5 k6
k7 k3
k8
video.edhole.com
Collision Resolution by ChainingCollision Resolution by ChainingComp 122
0
m–1
h(k1)=h(k4)
h(k2)=h(k5)=h(k6)
h(k3)=h(k7)
U(universe of keys)
K(actualkeys)
k1
k2
k3
k5
k4
k6
k7k8
h(k8)
X
X
X
video.edhole.com
Collision Resolution by ChainingCollision Resolution by ChainingComp 122
k2
0
m–1
U(universe of keys)
K(actualkeys)
k1
k2
k3
k5
k4
k6
k7k8
k1 k4
k5 k6
k7 k3
k8
video.edhole.com
Hashing with ChainingHashing with Chaining
Dictionary Operations:Chained-Hash-Insert (T, x)
◦Insert x at the head of list T[h(key[x])].◦Worst-case complexity – O(1).
Chained-Hash-Delete (T, x)◦Delete x from the list T[h(key[x])].◦Worst-case complexity – proportional to length of
list with singly-linked lists. O(1) with doubly-linked lists.
Chained-Hash-Search (T, k)◦Search an element with key k in list T[h(k)].◦Worst-case complexity – proportional to length of
list.
Comp 122video.edhole.com
Analysis on Chained-Hash-SearchAnalysis on Chained-Hash-SearchLoad factor =n/m = average keys per slot.
◦m – number of slots.◦ n – number of elements stored in the hash table.
Worst-case complexity: (n) + time to compute h(k).
Average depends on how h distributes keys among m slots.
Assume ◦Simple uniform hashing. Any key is equally likely to hash into any of the m
slots, independent of where any other key hashes to.◦O(1) time to compute h(k).
Time to search for an element with key k is (|T[h(k)]|).
Expected length of a linked list = load factor = = n/m.
Comp 122video.edhole.com
Expected Cost of an Unsuccessful Expected Cost of an Unsuccessful SearchSearch
Proof:Any key not already in the table is equally
likely to hash to any of the m slots.To search unsuccessfully for any key k, need
to search to the end of the list T[h(k)], whose expected length is α.
Adding the time to compute the hash function, the total time required is Θ(1+α).
Comp 122
Theorem:An unsuccessful search takes expected time Θ(1+α).
video.edhole.com
Expected Cost of a Successful SearchExpected Cost of a Successful Search
Proof:The probability that a list is searched is proportional to the
number of elements it contains.Assume that the element being searched for is equally
likely to be any of the n elements in the table.The number of elements examined during a successful
search for an element x is 1 more than the number of elements that appear before x in x’s list.◦ These are the elements inserted after x was inserted.
Goal:◦ Find the average, over the n elements x in the table, of how
many elements were inserted into x’s list after x was inserted.
Comp 122
Theorem:A successful search takes expected time Θ(1+α).
video.edhole.com
Expected Cost of a Successful SearchExpected Cost of a Successful Search
Proof (contd):Let xi be the ith element inserted into the table, and
let ki = key[xi].Define indicator random variables Xij = I{h(ki) =
h(kj)}, for all i, j.Simple uniform hashing Pr{h(ki) = h(kj)} = 1/m E[Xij] = 1/m.Expected number of elements examined in a
successful search is:
n
i
n
ijijX
nE
1 1
11
Comp 122
Theorem:A successful search takes expected time Θ(1+α).
No. of elements inserted after xi into the same slot as xi.video.edhole.com
Proof – Contd.Proof – Contd.n
mn
nnn
nm
innm
innm
mn
XEn
Xn
E
n
i
n
i
n
i
n
i
n
ij
n
i
n
ijij
n
i
n
ijij
221
21
1
2)1(1
1
11
)(1
1
11
1
][11
11
2
1 1
1
1 1
1 1
1 1
Comp 122
(linearity of expectation)
Expected total time for a successful search = Time to compute hash function + Time to search= O(2+/2 – /2n) = O(1+ ).
video.edhole.com
Expected Cost – InterpretationExpected Cost – Interpretation
If n = O(m), then =n/m = O(m)/m = O(1). Searching takes constant time on average.Insertion is O(1) in the worst case.Deletion takes O(1) worst-case time when
lists are doubly linked.Hence, all dictionary operations take O(1)
time on average with hash tables with chaining.
Comp 122video.edhole.com