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

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

I am reading this book 2

Chapter 1 The following slides amplifies the content of the books 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. 3

What is Functional Algorithm Design? It is solving problems by composing functions. 4

Function Composition 5 map toUpper map toUpper +

Function Composition (cont.) map toUpper hello world HELLO WORLD 6

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. 7

The Problem We Will Solve 8

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? 9

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?

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? 11

Problem Statement Find the difference between list A and list B. List A is in ascending order; list B is in no particular order. 12

Just the First Difference We will just find the first difference, not all the differences. 13

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 14

What is the smallest number not in this list? 15 [08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06] 15

Problem Re-Statement Find the smallest natural number not in a given finite list of natural numbers. 16

We Will Solve The Problem In Two Ways 17

Solution #1 18

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

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, …) 20

21 for each x in N notElem ? x [08, 23, 09, …, 06] no discard x yes retain x

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. 22

Compose filter and notElem 23 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, …]

head head is a standard function. It selects the first value in a list. Compose it with filter and notElem to complete the solution: 24 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

Solution #1 25 head (filter (notElem xs) [0..]) xs (pronounced, exes) is the list that we are analyzing. [0..] is the natural numbers.

Solution #2 26

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. 27

Example 28 [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] no discard 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

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. 31

zip (cont.) 32 Pair this value with this value

zip the filtered set with a list of True values 33 [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)]

repeat repeat is a standard function. It repeats its argument an infinite number of times. 34 repeat True [True, True, True, …]

Create a list of pairs 35 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] [(08, True), (09, True), (00, True), (12, True), (11, True), …, (06, True)]

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. 36 [(08, True), (09, True), (00, True), (12, True), (11, True), …, (06, True)] Association List:

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: || 37 [(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 ……….

Gaps result in the creation of a pair with a value of False 38 [(0,True),(1,True),(2,True),(3,True),(4,True), …,(14,True),(17,True), (19,True)] Heres a gap in the alist. It produces a pair with a value of False.

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 39

Nearly finished! 40 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] [(0,True),(1,True),(2,True),(3,True),(4,True), …,(14,True),(15,False),(16,False),…]

checklist checklist is a (user-defined) function; it is the collection of functions shown on the previous slide (copied below). 41 accumArray (||) False (zip (filter (<=n) xs) (repeat True)) where n = length xs checklist

elems elems is a standard function. Give it an alist and it returns its values: 42 [(0,True),(1,True),(2,True),(3,True),(4,True), …,(14,True),(15,False),(16,False),…] 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.

takeWhile (cont.) 45 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 [True, True, True, True, True, …,True]

Hey, thats the answer! 47 length ___ 15 [True, True, True, True, True, …,True] 15 is the answer to the problem

From alist to answer 48 [(0,True),(1,True),(2,True),(3,True),(4,True), …,(14,True),(15,False),(16,False),…] 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

Solution #2 50 search checklist xs xs (pronounced, exes) is the list we are analyzing

Comparison of the two solutions 51

Solution #1 With a list of length n this solution takes, in the worst case, on the order of n 2 steps. This will give you an idea of idea how fast the time requirements grow: 52 ntime 11 24 39 416 525 636

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. Thats nice!

Implementation 54

Haskell Haskell is a functional programming language. The following slides show how to express the two solutions using Haskell. 55

Solution #1 56 findGap :: [Int] -> Int findGap xs = head (filter (`notElem` xs) [0..])

Solution #2 57 import Data.Array checklist :: [Int] -> Array Int Bool checklist xs = accumArray (||) False (0,n) (zip (filter (<=n) xs) (repeat True)) where n = length xs search :: Array Int Bool -> Int search = length. takeWhile id. elems findGap = search. checklist

Sort Problem: Sort a list of values 58

59 The Problem We Will Solve

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. 60

Recurring Pattern Bookshelf: On my bookshelf I have a bunch of books. I want to organize (sort) them by author. 61

62 Problem Statement Sort a list of items. There may be duplicates in the list; thats okay.

63 Abstract Representation of the Problem Represent the items in the list using the natural numbers, N = (0, 1, …)

Example of sorting a list 64 [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

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 items value is 0x<20 Assumption: Duplicates are okay. – On the previous slide there are two occurrences of 15 65

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) Thats fast! 66

Create a list of 1s using the repeat function 67 repeat 1 [1, 1, 1, …]

Create a list of pairs using the zip function 68 [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)]

Interpret each pair as (index, count) 70 [(08, 1), (15, 1), (09, 1), (00, 1), (12, 1), (11, 1), (01, 1), (10, 1)…, (06, 1)] index one occurrence

Merge pairs with the same index (add their count values) 71 [(08, 1), (15, 1), (09, 1), (00, 1), (12, 1), (11, 1), … (15, 1)…, (06, 1)] (15, 2) There are two occurrences of 15

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. 72

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 73

The result is an array 74 [(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)]

Compose accumArray, zip, and repeat 75 [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). accumArray (+) 0 (0, n) (zip xs (repeat 1)) where n = length xs countlist

The assocs Function The function takes as its argument an array and returns just the list of pairs. 77 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)]

Replicate n times the index in (index, n) 78 [(0,1),(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(7,1),…,(15,2),…,(19,1),(20,0)] (15) replicate twice (0) (1) ……………………….. replicate once

The replicate function The replicate function creates n copies of a value. It returns a list, containing n items. 79 replicate 3 "Ho" returns ["Ho","Ho","Ho"] replicate 2 15 returns [15,15]

Recall set comprehensions from your school days 80 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.

Terminology 81 output function variable input set predicate

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: 82 [2*x|x <- [1..10]] [2,4,6,8,10,12,14,16,18,20]

Explanation 83 [2*x|x <- [1..10]] x is drawn from [1.. 10] and for every value drawn, that value is doubled.

Create a list of the indexes 84 [(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.

The concat function The concat function creates a single list out of a list of lists 86 [[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]

array -> sorted list 87 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]

sort sort is a (user-defined) function; it is the collection of functions shown on the previous slide (copied below). 88 concat [replicate k x | (x, k) <- assocs (countlist xs)] sort

Heres the Solution 89 import Data.Array countlist :: [Int] -> Array Int Int countlist 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]

Find the smallest free number Version #2 90

91 The Problem We Will Solve

92 Problem Statement Find the smallest natural number not in a given finite list of natural numbers.

93 What is the smallest number not in this list? 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 items value is 0x<19 Assumption: Duplicates are okay. – On the previous slide there are three occurrences of 14

Recall the countlist function 95 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 accumArray (+) 0 (0, n) (zip xs (repeat 1)) where n = length xs countlist xs is the list of Natural numbers.

Recall the checklist function The first version (see slides 27-41) to the problem used the checklist function: 97 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)]

We will use countlist to solve the problem 99

A number that was not in the input will have the form (_, 0) 100 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]

Select all pairs with (_,0) 101 [(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]

Explanation of this list comprehension 102 [(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).

Select just the first value in each pair 103 [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]

Select the first value 104 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 Heres the Solution import Data.Array countlist :: [Int] -> Array Int Int countlist 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]

Find the smallest free number Version 3 106

107 The Problem We Will Solve

108 Problem Statement Find the smallest natural number not in given finite list of natural numbers.

109 What is the smallest number not in this list? 15 [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 24, 34, 5, 17, 3, 19, 2, 6]

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

Assumption The list we are processing has no duplicates If the list has duplicates, the algorithm may enter into an infinite recursion 111

Partition using this function: (<b) where b = half the length of the list 112 [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

Does the left list have gaps? 113 [8, 9, 0, 1, 7, 4, 5, 3, 2, 6] How do we tell if the list is any missing numbers? Heres an easy way to tell: 1.Get the length of the list 2.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?

Variable names we will use 114 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 xs b = a + 1 + (n `div` 2) m = length us partition

Does the left list have gaps? 115 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 xs b = 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

Lets trace an example The following slides traces the processing of a list using the divide-and-conquer algorithm. 118

119 xs = [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 24, 34, 5, 17, 3, 19, 2, 6] a = 0 n = length xs = 19 b = 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] a = 0 n = length xs = 19 b = a + 1 + (n `div` 2) = 0 + 1 + (19 `div` 2) = 10 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] a = 0 n = length xs = 19 b = a + 1 + (n `div` 2) = 0 + 1 + (19 `div` 2) = 10 partition (< b) xs m = length us = 10 m == b – a == 10 – 0 == True Therefore, this list has no missing numbers and we should recurse on the 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 = 10 n = 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 = 10 n = 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 = 10 n = n – m = 19 – 10 = 9 b = a + 1 + (n `div` 2) = 10 + 1 + (9 `div` 2) = 15 partition (< b) xs m = length us = 5 m == b – a == 15 – 10 == True Therefore, 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] set a to the value of b (15) set n to the value of n - m (4)

128 us = [17]vs = [24,34,19 ] partition (< b) xs m = length us = 1 m == b – a == 18 – 15 == False Therefore, 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] set a to the value of b (15) set n to the value of n - m (4)

129 xs = [17] set a to the old value of a (15) set n to the value of m (1)

130 xs = [17] set a to the old value of a (15) set n to the value of m (1) b = a + 1 + (n `div` 2) = 15 + 1 + (1 `div` 2) = 16

131 xs = [17] set a to the old value of a (15) set n to the value of m (1) b = a + 1 + (n `div` 2) = 15 + 1 + (1 `div` 2) = 16 us = [ ]vs = [17] partition (< b) xs m = length us = 0 m == b – a == 16 – 15 == False Therefore, this list has a missing number and we should recurse on it.

132 If n == 0 then return a set a to the old value of a (15) set n to the value of m (0) Done! The answer is: 15

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. 133

Heres the Solution 134 import List minfree :: [Int] -> Int minfree xs = minfrom 0 (length xs, xs) minfrom :: Int -> (Int, [Int]) -> Int minfrom 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

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