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2IL50 Data Structures. Spring 2014 Lecture 4: Sorting in linear time. Building a heap. One more time …. Building a heap. Build-Max-Heap (A) heap-size = A.length for i = A.length downto 1 do Max-Heapify(A,i). 14. 3. 8. 11. Building a heap. Build-Max-Heap2 (A) heap-size[A] = 1

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2IL50 Data Structures

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## 2IL50 Data Structures

Spring 2014Lecture 4: Sorting in linear time

One more time …

### Building a heap

Build-Max-Heap(A)

• heap-size = A.length

• for i = A.length downto 1

• do Max-Heapify(A,i)

14

3

8

11

### Building a heap

Build-Max-Heap2(A)

• heap-size[A] = 1

• for i = 2 to A.length doMax-Heap-Insert(A, A[i])

Max-Heap-Insert(A, key)

• heap-size[A] = heap-size[A] + 1

• A[heap-size[A] ] = - infinity

• Increase-Key(A, heap-size[A], key)

Lower bound worst case running time?

14

3

8

11

2

24

35

28

16

5

20

21

A[i] moves up until it reaches the correct position

### Building a heap

Build-Max-Heap2(A)

• heap-size[A] = 1

• for i = 2 to A.length doMax-Heap-Insert(A, A[i])

Running time: Θ(1) + ∑2≤i ≤ n (time for Max-Heap-Insert(A, A[i]))

• Max-Heap-Insert (A, A[i] ) takes O(log i) = O(log n) time

➨ worst case running time is O(n log n)

• If A is sorted in increasing order, then A[i] is always the largest element when Max-Heap-Insert(A, A[i])) is called and must move all the way up the tree

➨ Max-Heap-Insert (A, A[i]) takes Ω(log i) time.

Worst case running time: Θ(1) + ∑2≤i ≤ nΩ(log i) = Ω(1 + ∑2≤i ≤ n log i)

= Ω (n log n)

since ∑2≤i ≤ n log i ≥ ∑n/2≤i ≤ n log (n/2) = n/2 log (n/2)

log2 n = Θ(log n2) ?

√n = Ω(log4 n) ?

2lg n = Ω(n2) ?

2n = Ω(n2) ?

log(√n) = Θ(log n) ?

no

yes

no

yes

yes

### The sorting problem

Input: a sequence of n numbers ‹a1, a2, …, an›

Output: a permutation of the input such that ‹ai1≤ … ≤ain›

Why do we care so much about sorting?

• sorting is used by many applications

• (first) step of many algorithms

• many techniques can be illustrated by studying sorting

### Can we sort faster than Θ(n log n) ??

Worst case running time of sorting algorithms:

InsertionSort: Θ(n2)

MergeSort: Θ(n log n)

HeapSort: Θ(n log n)

Can we do this faster? Θ(n loglog n) ? Θ(n) ?

### Upper and lower bounds

Upper bound

How do you show that a problem (for example sorting) can be solved in Θ(f(n)) time?

➨ give an algorithm that solves the problem in Θ(f(n)) time.

Lower bound

How do you show that a problem (for example sorting) cannot be solved faster than in Θ(f(n)) time?

➨ prove that every possiblealgorithm that solves the problem needs Ω(f(n)) time.

### Lower bounds

Lower bound

How do you show that a problem (for example sorting) can not be solved faster than in Θ(f(n)) time?

➨ prove that every possible algorithm that solves the problem needs Ω(f(n)) time.

Model of computation: which operations is the algorithm allowed to use?

Bit-manipulations?Random-access (array indexing) vs. pointer-machines?

### Comparison-based sorting

InsertionSort(A)

• for j = 2 to A.length

• do begin

6.

7. end

Which steps precisely the algorithm executes — and hence, which element ends up where — only depends on the result of comparisons between the input elements.

key = A[ j ] ; i = j -1

while i > 0 and A[ i ] > key

do begin A[ i+1] = A[ i ]; i = i -1 end

A[ i +1] = key

or ≤, =, >, ≥

### Decision tree for comparison-based sorting

exchange of elements, assignments, etc. …

A[.] < A[.]

A[.] < A[.]

A[.] < A[.]

A[.] < A[.]

A[.] < A[.]

A[.] < A[.]

A[.] < A[.]

### Comparison-based sorting

• every permutation of the input follows a different path in the decision tree

➨ the decision tree has at least n! leaves

• the height of a binary tree with n!leaves is at least log(n!)

• worst case running time

≥ longest path from root to leaf

≥ log(n!) = Ω(n log n)

### Lower bound for comparison-based sorting

TheoremAny comparison-based sorting algorithm requires Ω(n log n) comparisons in the worst case.

➨ The worst case running time of MergeSort and HeapSort is optimal.

### Sorting in linear time …

Three algorithms which are faster:

• CountingSort

• BucketSort

(not comparison-based, make assumptions on the input)

### CountingSort

Input: array A[1..n] of numbers

Assumption: the input elements are integers in the range0 to k, for some k

Main idea:count for every A[i] the number of elements less than A[i]➨ position of A[i] in the output array

Beware of elements that have the same value!

position(i) = number of elements less than A[i] in A[1..n]

+ number of elements equal to A[i] in A[1..i]

### CountingSort

position(i) = number of elements less than A[i] in A[1..n]

+ number of elements equal to A[i] in A[1..i]

5

3

10

5

4

5

7

7

9

3

10

8

5

3

3

8

3

3

3

3

4

5

5

5

5

7

7

8

8

9

10

10

numbers < 5

third 5 from left position: (# less than 5) + 3

### CountingSort

position(i) = number of elements less than A[i] in A[1..n]

+ number of elements equal to A[i] in A[1..i]

LemmaIf every element A[i] is placed on position(i), then the array is sorted and the sorted order is stable.

Numbers with the same value appear in the same order in the output array as they do in the input array.

### CountingSort

C[i] will contain the number of elements ≤ i

CountingSort(A,k)

►Input: array A[1..n] of integers in the range 0..k

►Output: array B[1..n] which contains the elements of A, sorted

• for i = 0 to k do C[i] = 0

• for j = 1 to A.length do C[A[j]] = C[A[j]] + 1

• ►C[i] now contains the number of elements equal to i

• for i = 1 to k do C[i] = C[i ] + C[i-1]

• ►C[i] now contains the number of elements less than or equal to i

• for j = A.length downto 1

• do B[C[A[ j ] ] ] = A[j]; C[A[ j ]] = C[A[ j ]] – 1

### CountingSort

CountingSort(A,k)

►Input: array A[1..n] of integers in the range 0..k

►Output: array B[1..n] which contains the elements of A, sorted

• for i = 0 to k do C[i] = 0

• for j = 1 to A.length do C[A[j]] = C[A[j]] + 1

• ►C[i] now contains the number of elements equal to i

• for i = 1 to k do C[i] = C[i ] + C[i-1]

• ►C[i] now contains the number of elements less than or equal to i

• for j = A.length downto 1

• do B[C[A[ j ] ] ] = A[j]; C[A[ j ]] = C[A[ j ]] – 1

Correctness lines 6/7: Invariant

Inv(m): for m ≤ i ≤ n: B[position(i)] contains A[i]

for 0 ≤ i ≤ k: C[i] = ( # numbers smaller than i )

+ ( # numbers equal to i in A[1..m-1])

Inv(m+1) holds before loop is executed with j =m, Inv(m) holds afterwards

### CountingSort: running time

CountingSort(A,k)

►Input: array A[1..n] of integers in the range 0..k

►Output: array B[1..n] which contains the elements of A, sorted

• for i = 0 to k do C[i] = 0

• for j = 1 to A.length do C[A[j]] = C[A[j]] + 1

• ►C[i] now contains the number of elements equal to i

• for i = 1 to k do C[i] = C[i ] + C[i-1]

• ►C[i] now contains the number of elements less than or equal to i

• for j = A.length downto 1

• do B[C[A[ j ] ] ] = A[j]; C[A[ j ]] = C[A[ j ]] – 1

line 1: ∑0≤i≤kΘ(1) = Θ(k)

line 2: ∑1≤i≤nΘ(1) = Θ(n)

line 4: ∑0≤i≤kΘ(1) = Θ(k)

lines 6/7: ∑1≤i≤nΘ(1) = Θ(n)

Total: Θ(n+k) ➨ Θ(n) if k = O(n)

### CountingSort

TheoremCountingSort is a stable sorting algorithm that sorts an array of n integers in the range 0..k in Θ(n+k) time.

Input: array A[1..n] of numbers

Assumption: the input elements are integers with ddigits

example(d = 4): 3288, 1193, 9999, 0654, 7243, 4321

• for i = 1 to d

• do use a stable sort to sort array A on digit i

dth digit

1st digit

720

355

436

457

657

329

839

720

329

436

839

355

457

657

329

355

436

457

657

720

839

Correctness: Assignment 3

329

457

657

839

436

720

355

sort on 1st digit

sort on 2nd digit

sort on 3rd digit

Running time: If we use CountingSort as stable sorting algorithm

➨ Θ(n + k) per digit

TheoremGiven nd-digit numbers in which each digit can take up to k possible values, RadixSort correctly sorts these numbers in Θ(d (n + k)) time.

each digit is an integer in the range 0..k

### BucketSort

Input: array A[1..n] of numbers

Assumption: the input elements lie in the interval [0..1) (no integers!)

BucketSort is fast if the elements are uniformly distributed in [0..1)

1

0

0.792

2

1

0.1

0.1

0.15

0.13

0.287

0.287

0.256

0.15

0.346

0.346

0.734

0.5

0.53

0.5

0.13

0.792

0.734

0.256

n-1

n

0.53

### BucketSort

• Throw input elements in “buckets”, sort buckets, concatenate …

input array A[1..n]; numbers in [0..1)

auxiliary array B[0..n-1]

bucket B[i] contains numbers in [i/n … (i+1)/n]

1

0.792

2

0.1

0.1

0.13

0.15

0.287

0.256

0.287

0.15

0.346

0.346

0.346

0.734

0.5

0.5

0.53

0.53

0.5

0.13

0.734

0.792

0.256

n

0.53

### BucketSort

• Throw input elements in “buckets”, sort buckets, concatenate …

input array A[1..n]; numbers in [0..1)

auxiliary array B[0..n-1]

bucket B[i] contains numbers in [i/n … (i+1)/n]

0

1

0.1

0.15

0.13

0.287

0.256

0.792

0.734

n-1

### BucketSort

BucketSort(A)

► Input: array A[1..n] of numbers with 0 ≤ A[i ] < 1

► Output: sorted list, which contains the elements of A

• n = A.length

• initialize auxiliary array B[0..n-1]; each B[i] is a linked list of numbers

• for i = 1 to n

• do insert A[i] into list B[ n∙A[i ] ]

• for i = 0 to n-1

• do sort list B[i], for example with InsertionSort

• concatenate the lists B[0], B[1], …, B[n-1] together in order

### BucketSort

Running time?

Define ni = number of elements in bucket B[i]

➨ running time = Θ(n) + ∑0≤i≤n-1Θ(ni2)

• worst case:

• best case:

• expected running time if the numbers are randomly distributed ?

all numbers fall into the same bucket ➨ Θ(n2)

all numbers fall into different buckets ➨Θ(n)

### BucketSort: expected running time

Define ni = number of elements in bucket B[i]

➨ running time = Θ(n) + ∑0≤i≤n-1Θ(ni2)

Assumption: Pr { A[j] falls in bucket B[i] } = 1/n for each i

E [ running time ] = E [Θ(n) + ∑0≤i≤n-1Θ(ni2) ]

= Θ ( n + ∑0≤i≤n-1 E [ni2 ] )

What is E [ni2 ] ?

(some math with indicator random variables – see book for details)

➨ E [ni2 ] = 2 - 1/n = Θ(1)

➨ expected running time = Θ(n)

but E [ni2 ] ≠ E [ni ]2

We have E [ni] = 1 …