Pearls of Functional Algorithm Design Chapter 1 1 Roger L. Costello June 2011

1

Pearls of Functional Algorithm Design

Chapter 1

Roger L. CostelloJune 2011

2

I am reading this book

3

Chapter 1

• The following slides amplifies the content of the book’s Chapter 1.

• Chapter 1 shows three ways to solve the problem of finding the smallest free number. Also, it shows a neat sorting algorithm.

• Slides 1 – 57 describes version #1 of finding the smallest free number.

• Slides 58 – 89 describes the sorting algorithm.• Slides 90 – 105 describes version #2 of finding the

smallest free number.• Slides 106 – 133 describes version #3 of finding the

smallest free number.

4

What is Functional Algorithm Design?

• It is solving problems by composing functions.

5

Function Composition

map toUpper

maptoUpper

+

6

Function Composition (cont.)

map toUpper

“hello world”

“HELLO WORLD”

7

Attention

• As you go through these slides, be alert to the functions (puzzle pieces) used.

• Observe how they are composed to solve problems (i.e., how the puzzle pieces are put together to create something new).

• Example: The previous slide composed two functions to solve a problem -- convert strings to uppercase.

8

The Problem We Will Solve

9

Recurring Problem

• Cooking: a recipe calls for this list of ingredients: eggs, flour, milk, chocolate. In my kitchen I have some ingredients. Is there a difference between what the recipe requires versus what I have in my kitchen?

10

Recurring Problem (cont.)

• Product Inventory: the inventory sheet says one thing. The actual products on the shelf says another. Is there a difference between what the inventory sheet says versus what is actually on the shelves?

11

Recurring Problem (cont.)

• Air Mission: the air mission calls for aircraft and weapons. In the military unit there are aircraft and weapons. Is there a difference between what the air mission requires versus what is in the military unit?

12

Problem Statement

• Find the difference between list A and list B.• List A is in ascending order; list B is in no

particular order.

13

Just the First Difference

• We will just find the first difference, not all the differences.

14

Abstract Representation of the Problem

• List A: represent it using the natural numbers, N = (0, 1, …)

• List B: also represent it using the natural numbers; the numbers may be in any order

15

What is the smallest number not in this list?

[08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06]

15

16

Problem Re-Statement

• Find the smallest natural number not in a given finite list of natural numbers.

17

We Will Solve The Problem In Two Ways

18

Solution #1

19

notElem

• “notElem” is a standard function.• It takes two arguments, a value and a list.• It returns True if the value is not an element of

the list, False otherwise.

notElem 23 [08, 23, 09, …, 06]

False

20

notElem (cont.)

• The notElem function can be used to help solve the problem.

• Iterate through each natural number and see if it is not an element of the list. Retain any natural number not in the list.

• See next slide. (Note: “N” denotes the natural numbers: 0, 1, 2, …)

21

for each x in N

notElem?

x

[08, 23, 09, …, 06]nodiscard x

yes

retain x

22

filter

• “filter” is a standard function.• It does all that stuff shown on the previous

slide: – it selects, one by one, the values in N– it hands the value to the notElem function, “Hey, is

this value not an element of [08, 23, 09, …, 06]?” – if notElem returns True, it retains the value.

23

Compose filter and notElem

filter (notElem ___) [0, 1, 2, ..]

[08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06]

[15, 16, 18, 20, 22, 24, 25, 26, 27, …]

24

head

• “head” is a standard function.• It selects the first value in a list.• Compose it with filter and notElem to

complete the solution:

head (filter (notElem ___) [0, 1, 2, ..])

[08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06]

15

25

Solution #1

head (filter (notElem xs) [0 ..])

xs (pronounced, ex’es) is the list that we are analyzing.[0 ..] is the natural numbers.

26

Solution #2

27

Okay to discard some values

• Recall that we are analyzing a list of values.• We are representing the values using numbers.• We represent one value as 0, another value as 2,

and so forth.• Suppose we have 20 values. Suppose one of the

values has the number 23. We can discard it.• Therefore, use the filter function to retain only

values that are less than the length of the list.

28

Example

[08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06]

length = 20

29

Example (cont.)

[08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06]

<=

x

length [08, 23, …, 06]nodiscard x

yes

retain x

30

Example (cont.)

[08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06]

filter (<=n) xs where n = length xs

[08, 09, 00, 12, 11, 01, 10, 13, 07, 04, 14, 05, 17, 03, 19, 02, 06]

These values are discarded: 23, 21

31

zip

• “zip” is a standard function.• It takes two lists and zips them up (just like a

zipper zips up two pieces of clothing). That is, it pairs: – the first value in the first list with the first value in

the second list– the second value in the first list with the second

value in the second list – etc.

32

zip (cont.)

Pair this value with this value

33

zip the filtered set with a list of True values

[08, 09, 00, 12, 11, 01, 10, 13, 07, 04, 14, 05, 17, 03, 19, 02, 06] [True, True, …]

zip

[(08, True), (09, True), (00, True), (12, True), (11, True), …, (06, True)]

34

repeat

• “repeat” is a standard function.• It repeats its argument an infinite number of

times.

repeat True

[True, True, True, …]

35

Create a list of pairs

zip (filter (<=n) xs) (repeat True)where n = length xs

[08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06]


36

Association List

• A list of pairs is called an “association list” (or alist for short)

• The first value in a pair is the index. The second value is the value.

• A value can be quickly obtained given its index.


Association List:

37

“OR” each value in the alist with False

• For all indexes from 0 to the length of the list, OR the value with False. OR is represented by this symbol: ||

[(0,True),(1,True),(2,True),(3,True),(4,True), …,(14,True),(17,True), (19,True)]

(||) False

[(0,True),(1,True),(2,True),(3,True),(4,True), …,(14,True),(15,False),(16,False),…]

(||) False (||) False………………. (||) False………. ……….

38

Gaps result in the creation of a pair with a value of False

[(0,True),(1,True),(2,True),(3,True),(4,True), …,(14,True),(17,True), (19,True)]

Here’s a gap in the alist.It produces a pair with avalue of False.

39

accumArray

• “accumArray” is a standard function.• It does all the stuff shown on the previous two

slides: For each index from 0 to the length of the list do

Apply a function (e.g., ||) to the value

40

Nearly finished!

accumArray (||) False (zip (filter (<=n) xs) (repeat True)) where n = length xs

[08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06]


41

checklist

• “checklist” is a (user-defined) function; it is the collection of functions shown on the previous slide (copied below).

accumArray (||) False (zip (filter (<=n) xs) (repeat True)) where n = length xs

checklist

42

elems

• “elems” is a standard function.• Give it an alist and it returns its values:


elems

[True, True, True, True, True, …,True, False, False, True, True]

43

id

• “id” is a standard function.• It is the identity function; give it a value and it

returns the same value:

True

id

True

False

id

False

44

takeWhile

• “takeWhile” is a standard function.• It takes two arguments, a function and a list; it

starts at the beginning of the list and retains each value until it arrives at a value for which the function returns False.

45

takeWhile (cont.)

takeWhile id ___

[True, True, True, True, True, …,True]

[True, True, True, True, True, …,True, False, False, True, True]

takeWhile stops when it gets to the first False.

46

length

• “length” is a standard function.• It takes one argument, a list; it returns the

number of values in the list:

length ___

15


47

Hey, that’s the answer!

length ___

15


“15” is theanswer to

the problem

48

From alist to answer


length takeWhile id elems ___

15

49

search

• “search” is a (user-defined) function; it is the collection of functions shown on the previous slide (copied below).

length takeWhile id elems

search

50

Solution #2

search checklist xs

xs (pronounced, ex’es) is the list we are analyzing

51

Comparison of the two solutions

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Solution #1

• With a list of length “n” this solution takes, in the worst case, on the order of n2 steps.

• This will give you an idea of idea how fast the time requirements grow:

n time

1 1

2 4

3 9

4 16

5 25

6 36

53

Solution #2

• With a list of length “n” this solution takes on the order of n steps.

• That is, it is a linear-time solution for the problem. That’s nice!

54

Implementation

55

Haskell

• Haskell is a functional programming language.• The following slides show how to express the

two solutions using Haskell.

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Solution #1

findGap :: [Int] -> IntfindGap xs = head (filter (`notElem` xs) [0..])

57

Solution #2

import Data.Array

checklist :: [Int] -> Array Int Boolchecklist xs = accumArray (||) False (0,n) (zip (filter (<=n) xs) (repeat True)) where n = length xs

search :: Array Int Bool -> Intsearch = length . takeWhile id . elems

findGap = search . checklist

58

Sort

Problem: Sort a list of values

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60

Recurring Pattern

• Kitchen: In my kitchen I have a 2 quart sauce pan, a 1 quart sauce pan, a 5 quart sauce pan, and another 2 quart sauce pan. I want to organize (sort) them by increasing size.

61

Recurring Pattern

• Bookshelf: On my bookshelf I have a bunch of books. I want to organize (sort) them by author.

62

Problem Statement

• Sort a list of items. There may be duplicates in the list; that’s okay.

63

Abstract Representation of the Problem

• Represent the items in the list using the natural numbers, N = (0, 1, …)

64

Example of sorting a list

[08, 15, 09, 00, 12, 11, 01, 10, 13, 07, 16, 04, 14, 15, 05, 17, 03, 19, 02, 06]

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 15, 16, 17, 19]

sort

65

Important Assumptions

• Assumption: Each item in the list has a value that is less than the length of the list – On the previous slide the list contains 20 elements.

Thus, each item’s value is 0≤x<20

• Assumption: Duplicates are okay. – On the previous slide there are two occurrences

of 15

66

Time Required

• Given the assumptions on the previous slide, the algorithm shown on the following slides performs a sort in a time proportional to the length of the list.

• That is, the time to sort is linear, i.e., O(n)• That’s fast!

67

Create a list of 1’susing the repeat function

repeat 1

[1, 1, 1, …]

68

Create a list of pairsusing the zip function

[08, 15, 09, 00, 12, 11, 01, 10, 13, 07, 16, 04, …, 19, 02, 06] [1, 1, …]

zip

[(08, 1) , (15, 1), (09, 1), (00, 1), (12, 1), (11, 1), (01, 1), (10, 1)…, (06, 1)]

69

Compose the zip and repeat functions

zip ___ (repeat 1)

[08, 15, 09, 00, 12, 11, 01, 10, 13, 07, 16, 04, 14, 15, 05, 17, 03, 19, 02, 06]

[(08, 1) , (15, 1), (09, 1), (00, 1), (12, 1), (11, 1), (01, 1), (10, 1)…, (06, 1)]

70

Interpret each pair as(index, count)

[(08, 1) , (15, 1), (09, 1), (00, 1), (12, 1), (11, 1), (01, 1), (10, 1)…, (06, 1)]

index

one occurrence

71

Merge pairs with the same index(add their count values)

[(08, 1) , (15, 1), (09, 1), (00, 1), (12, 1), (11, 1), … (15, 1)…, (06, 1)]

(15, 2)

“There are two occurrences of 15”

72

The accumArray Function

• The accumArray function goes through a list of pairs and merges the pairs that have a duplicate index.

• accumArray is flexible in how it merges – you supply it a function and it will use that function to merge the pairs’ values.

• In our problem we supply it the plus (+) function because we want the values added.

73

The accumArray Function (cont.)

• The accumArray function has four arguments. I describe them in reverse order:– A list of pairs, such as that shown two slides back– A pair, (0, n), where n is the length of the list– An initial value for the function (see next)– A function to be applied on the values of pairs with

duplicate indexes

74

The result is an “array”

[(08, 1) , (15, 1), (09, 1), (00, 1), (12, 1), (11, 1), … (15, 1)…, (06, 1)]

accumArray (+) 0 (0, n) (___)where n = length xs

array (0,20) [(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(7,1),…,(19,1),(20,0)]

75

ComposeaccumArray, zip, and repeat

[08, 15, 09, 00, 12, 11, 01, 10, 13, 07, 16, 04, 14, 15, 05, 17, 03, 19, 02, 06]

accumArray (+) 0 (0, n) (zip xs (repeat 1))where n = length xs

array (0,20) [(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(7,1),…,(15,2),…,(19,1),(20,0)]

76

countlist

• “countlist” is a (user-defined) function; it is the collection of functions shown on the previous slide (copied below).


countlist

77

The assocs Function

• The function takes as its argument an array and returns just the list of pairs.

array (0,20) [(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(7,1),…,(15,2),…,(19,1),(20,0)]

assocs ____

[(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(7,1),…,(15,2),…,(19,1),(20,0)]

78

Replicate n times the index in (index, n)

[(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(7,1),…,(15,2),…,(19,1),(20,0)]

(15) (15)

replicate twice

(0) (1)………………………..

replicate oncereplicate once

79

The replicate function

• The replicate function creates n copies of a value. It returns a list, containing n items.

replicate 3 "Ho" returns ["Ho","Ho","Ho"]

replicate 2 15 returns [15,15]

80

Recall “set comprehensions”from your school days

}10,|2{ xNxx

“The set of the first ten even numbers”

A set comprehension builds a more specific set out of a general set. In this example, the more general set is N, the set of natural numbers.

81

Terminology

}10,|2{ xNxx

outputfunction

variableinputset

predicate

82

List Comprehensions

• List comprehensions are similar to set comprehensions.

• The set comprehension on the previous slide is equivalently expressed in Haskell using this list comprehension:

[2*x|x <- [1..10]]

[2,4,6,8,10,12,14,16,18,20]

83

Explanation

[2*x|x <- [1..10]]

“x is drawn from [1 .. 10] and for every value drawn, that value is doubled.”

84

Create a list of the indexes

[(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),…(13,1),(14,1),(15,2),(16,1),(17,1),(18,0),(19,1),(20,0)]

[x | (x,y) <- ____ ]

[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]

Oops! There should be two of these. Need to replicate.

85

Replicate the indexes the proper number of times

[(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),…(13,1),(14,1),(15,2),(16,1),(17,1),(18,0),(19,1),(20,0)]

[replicate y x | (x,y) <- ____ ]

[[0],[1],[2],[3],[4],[5],[6],[7],[8],[9],[10],[11],[12],[13],[14],[15,15],[16],[17],[],[19],[]]

Now we need to merge (concat) the list of lists.

86

The concat function

• The concat function creates a single list out of a list of lists

[[0],[1],[2],[3],[4],[5],[6],[7],[8],[9],[10],[11],[12],[13],[14],[15,15],[16],[17],[],[19],[]]

concat ____

[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,15,16,17,19]

87

array -> sorted list

array (0,20) [(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(7,1),…,(15,2),…,(19,1),(20,0)]

concat [replicate k x | (x, k) <- assocs (countlist xs)]

[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,15,16,17,19]

88

sort

• “sort” is a (user-defined) function; it is the collection of functions shown on the previous slide (copied below).

concat [replicate k x | (x, k) <- assocs (countlist xs)]

sort

89

Here’s the Solution

import Data.Array

countlist :: [Int] -> Array Int Intcountlist xs = accumArray (+) 0 (0, n) (zip xs (repeat 1)) where n = length xs

sort :: [Int] -> [Int] sort xs = concat [replicate k x | (x, k) <- assocs $ countlist xs]

90

Find the smallest free number

Version #2

91


92

Problem Statement

• Find the smallest natural number not in a given finite list of natural numbers.

93


15

[8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6]

94

Important Assumptions

• Assumption: Each item in the list has a value that is less than the length of the list – On the previous slide the list contains 19 elements.

Thus, each item’s value is 0≤x<19

• Assumption: Duplicates are okay. – On the previous slide there are three occurrences

of 14

95

Recall the countlist function

array (0,19) [(0,1),(1,1),(2,1),(3,1),(4,1), … , (14,3),(15,0),(16,0),(17,1),(18,0),(19,1)]

countlist ___

[8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6]

See slides 67-76 for an explanation of the countlist function.

96

The countlist function


countlist

xs is the list of Natural numbers.

97

Recall the checklist function

• The first version (see slides 27-41) to the problem used the checklist function:

checklist xs = accumArray (||) False (0,n) (zip (filter (<=n) xs) (repeat True)) where n = length xs

[8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6]

array (0,19) [(0,True),(1,True),(2,True),(3,True),(4,True),…,(14,True),(15,False),(16,False),(17,True),(18,False),(19,True)]

Compare countlist and checklist

[8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6]

countlist xs = accumArray (+) 0 (0, n) (zip xs (repeat 1)) where n = length xs

array (0,19) [(0,1),(1,1),(2,1),(3,1),(4,1), … , (14,3),(15,0),(16,0),(17,1),(18,0),(19,1)]

[8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6]

checklist xs = accumArray (||) False (0,n) (zip (filter (<=n) xs) (repeat True)) where n = length xs

array (0,19) [(0,True),(1,True),(2,True),(3,True),(4,True),…,(14,True),(15,False),(16,False),(17,True),(18,False),(19,True)]

99

We will use countlistto solve the problem

100

A number that was not in the inputwill have the form (_, 0)

array (0,19) [(0,1),(1,1),(2,1),(3,1),(4,1), … , (14,3),(15,0),(16,0),(17,1),(18,0),(19,1)]

countlist ___

These numbers were not in the input list, as indicated by 0 in the second value of their pairs

[8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6]

101

Select all pairs with (_,0)

[(x, k) | (x, k) <- assocs (countlist ___), k == 0]

[(15,0), (16,0), (18,0)]

[8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6]

102

Explanation of this list comprehension

[(x, k) | (x, k) <- assocs (countlist xs), k == 0]

input set

predicate

“The (x, k) pairs are drawn from the list of pairs returned by the assocs function (after applying the condition that the second value in each pair equal zero).”

103

Select just the first value in each pair

[x | (x, k) <- assocs (countlist ___), k == 0]

[15, 16, 18]

For each pair, output only the first value.

[8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6]

104

Select the first value

head [x | (x, k) <- assocs (countlist ___), k == 0]

[8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6]

15

head returns the first item in the list.

105


import Data.Array

countlist :: [Int] -> Array Int Intcountlist xs = accumArray (+) 0 (0, n) (zip xs (repeat 1)) where n = length xs

findGap :: [Int] -> Int findGap xs = head [x | (x, k) <- assocs $ countlist xs, k == 0]

106

Find the smallest free number

Version 3

107


108

Problem Statement

• Find the smallest natural number not in given finite list of natural numbers.

109


15

[8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 24, 34, 5, 17, 3, 19, 2, 6]

110

Divide and Conquer

• We will split the list in half, determine if the left half contains a missing number and if it does we will recurse on that, otherwise we recurse on the right half.

• We will use the Haskell partition function. It divides a list into a pair consisting of two lists. It has two arguments:– A Boolean function– A list

For each element in the list, if the Boolean function evaluates it to True then it goes in the first list, otherwise it goes in the second list.

partition (<10) [2, 45, 5, 18, 12] returns ([2,5],[45,18,12])

111

Assumption

• The list we are processing has no duplicates• If the list has duplicates, the algorithm may

enter into an infinite recursion

112

Partition using this function: (<b)where b = half the length of the list

[8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 24, 34, 5, 17, 3, 19, 2, 6]

partition (<b) ___ where b = 1 + n `div` 2 n = length xs

[8,9,0,1,7,4,5,3,2,6] [12,11,10,13,14,24,34,17,19]

b==10

113

Does the left list have gaps?

[8, 9, 0, 1, 7, 4, 5, 3, 2, 6]

How do we tell if the list is any missing numbers?Here’s an easy way to tell: 1. Get the length of the list2. If there are any missing numbers then the length is less than b.

Example: on the previous slide b = length xs `div` 2, which is 10. The length of the above list is 10. Thus, it must not have any gaps. Pretty neat, aye?

114

Variable names we will use

xs = [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 24, 34, 5, 17, 3, 19, 2, 6]

us = [8,9,0,1,7,4,5,3,2,6] vs = [12,11,10,13,14,24,34,17,19]

a = index of the first item

m

n = length xsb = a + 1 + (n `div` 2)m = length us

partition

115

Does the left list have gaps?

xs = [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 24, 34, 5, 17, 3, 19, 2, 6]

us = [8,9,0,1,7,4,5,3,2,6] vs = [12,11,10,13,14,24,34,17,19]

a = index of the first item

n = length xsb = a + 1 + (n `div` 2)m = length us

partition

If m == b – a then the left list has no gaps

116

If we recurse on the left list …

xs = [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 24, 34, 5, 17, 3, 19, 2, 6]

xs = [8,9,0,1,7,4,5,3,2,6] vs = [12,11,10,13,14,24,34,17,19]

a = the previous value of a

n = the previous value of m

partition

117

If we recurse on the right list …

xs = [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 24, 34, 5, 17, 3, 19, 2, 6]

us = [8,9,0,1,7,4,5,3,2,6] xs = [12,11,10,13,14,24,34,17,19]

a = the previous value of b

n = the previous value of n minus the previous value of m

partition

118

Let’s trace an example

• The following slides traces the processing of a list using the divide-and-conquer algorithm.

119

xs = [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 24, 34, 5, 17, 3, 19, 2, 6]

a = 0n = length xs = 19b = a + 1 + (n `div` 2) = 0 + 1 + (19 `div` 2) = 10

120

xs = [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 24, 34, 5, 17, 3, 19, 2, 6]

us = [8,9,0,1,7,4,5,3,2,6] vs = [12,11,10,13,14,24,34,17,19]


partition (< b) xs

121

xs = [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 24, 34, 5, 17, 3, 19, 2, 6]

us = [8,9,0,1,7,4,5,3,2,6] vs = [12,11,10,13,14,24,34,17,19]


partition (< b) xs

m = length us = 10m == b – a == 10 – 0 == TrueTherefore, this list has no missingnumbers and we should recurse onthe other list, vs.

122

xs = [12,11,10,13,14,24,34,17,19]

set a to the value of b (10)set n to this value: n – m (19 – 10 = 9)

123

xs = [12,11,10,13,14,24,34,17,19]

a = b = 10n = n – m = 19 – 10 = 9

b = a + 1 + (n `div` 2) = 10 + 1 + (9 `div` 2) = 15

124

xs = [12,11,10,13,14,24,34,17,19]

a = b = 10n = n – m = 19 – 10 = 9

b = a + 1 + (n `div` 2) = 10 + 1 + (9 `div` 2) = 15

us = [12,11,10,13,14] vs = [24,34,17,19]

partition (< b) xs

125

xs = [12,11,10,13,14,24,34,17,19]

a = b = 10n = n – m = 19 – 10 = 9

b = a + 1 + (n `div` 2) = 10 + 1 + (9 `div` 2) = 15

partition (< b) xs

m = length us = 5m == b – a == 15 – 10 == TrueTherefore, this list has no missing numbers and we should recurse on the other.

us = [12,11,10,13,14] vs = [24,34,17,19]

126

xs = [24,34,17,19]

set a to the value of b (15)set n to the value of n - m (4)

127

xs = [24,34,17,19]

b = a + 1 + (n `div` 2) = 15 + 1 + (4 `div` 2) = 18

xs = [24,34,17,19]


128

us = [17] vs = [24,34,19 ]

partition (< b) xs

m = length us = 1m == b – a == 18 – 15 == FalseTherefore, this list has a missing number and we should recurse on it.

xs = [24,34,17,19]

b = a + 1 + (n `div` 2) = 15 + 1 + (4 `div` 2) = 18

xs = [24,34,17,19]


129

xs = [17]

set a to the old value of a (15)set n to the value of m (1)

130

xs = [17]


b = a + 1 + (n `div` 2) = 15 + 1 + (1 `div` 2) = 16

131

xs = [17]


b = a + 1 + (n `div` 2) = 15 + 1 + (1 `div` 2) = 16

us = [ ] vs = [17]

partition (< b) xs

m = length us = 0m == b – a == 16 – 15 == FalseTherefore, this list has a missing number and we should recurse on it.

132

If n == 0 then return a


Done! The answer is: 15

133

Time Requirements

• With a list of length “n” this algorithm takes on the order of n steps.

• That is, it is a linear-time solution for the problem.

134


import List

minfree :: [Int] -> Intminfree xs = minfrom 0 (length xs, xs)

minfrom :: Int -> (Int, [Int]) -> Intminfrom a (n, xs) | n == 0 = a | m == b - a = minfrom b (n - m, vs) | otherwise = minfrom a (m, us) where (us, vs) = partition (<b) xs b = a + 1 + (n `div` 2) m = length us

Documents

Pearls of Functional Algorithm Design Chapter 1 1 Roger L. Costello June 2011