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Pearls of Functional Algorithm Design Chapter 1 Roger L. Costello June 2011 1 I am reading this book 2 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. 3 What is Functional Algorithm Design? • It is solving problems by composing functions. 4 Function Composition + map toUpper toUpper map 5 Function Composition (cont.) “hello world” map toUpper “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 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? 10 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? [08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06] 15 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 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 19 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 for each x in N x discard x no notElem ? [08, 23, 09, …, 06] yes retain x 21 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 [08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06] filter (notElem ___) [0, 1, 2, ..] [15, 16, 18, 20, 22, 24, 25, 26, 27, …] 23 head • “head” is a standard function. • It selects the first value in a list. • Compose it with filter and notElem to complete the solution: [08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06] head (filter (notElem ___) [0, 1, 2, ..]) 15 24 Solution #1 head (filter (notElem xs) [0 ..]) xs (pronounced, ex’es) is the list that we are analyzing. [0 ..] is the natural numbers. 25 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 [08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06] length = 20 28 Example (cont.) [08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06] x discard x no length [08, 23, …, 06] <= yes retain x 29 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 30 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.) Pair this value with this value 32 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)] 33 repeat • “repeat” is a standard function. • It repeats its argument an infinite number of times. repeat True [True, True, True, …] 34 Create a list of pairs [08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06] zip (filter (<=n) xs) (repeat True) where n = length xs [(08, True), (09, True), (00, True), (12, True), (11, True), …, (06, True)] 35 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: [(08, True), (09, True), (00, True), (12, True), (11, True), …, (06, True)] 36 “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 (||) False ………………. (||) False ………. (||) False ………. [(0,True),(1,True),(2,True),(3,True),(4,True), …,(14,True),(15,False),(16,False),…] 37 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 a value of False. 38 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! [08, 23, 09, 00, 12, 11, 01, 10, 13, 07, 41, 04, 14, 21, 05, 17, 03, 19, 02, 06] accumArray (||) False (zip (filter (<=n) xs) (repeat True)) where n = length xs [(0,True),(1,True),(2,True),(3,True),(4,True), …,(14,True),(15,False),(16,False),…] 40 checklist • “checklist” is a (user-defined) function; it is the collection of functions shown on the previous slide (copied below). checklist accumArray (||) False (zip (filter (<=n) xs) (repeat True)) where n = length xs 41 elems • “elems” is a standard function. • Give it an alist and it returns its values: [(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] 42 id • “id” is a standard function. • It is the identity function; give it a value and it returns the same value: True False id id True False 43 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. 44 takeWhile (cont.) [True, True, True, True, True, …,True, False, False, True, True] takeWhile id ___ [True, True, True, True, True, …,True] takeWhile stops when it gets to the first False. 45 length • “length” is a standard function. • It takes one argument, a list; it returns the number of values in the list: [True, True, True, True, True, …,True] length ___ 15 46 Hey, that’s the answer! [True, True, True, True, True, …,True] length ___ 15 “15” is the answer to the problem 47 From alist to answer [(0,True),(1,True),(2,True),(3,True),(4,True), …,(14,True),(15,False),(16,False),…] length takeWhile id elems ___ 15 48 search • “search” is a (user-defined) function; it is the collection of functions shown on the previous slide (copied below). search length takeWhile id elems 49 Solution #2 search checklist xs xs (pronounced, ex’es) is the list we are analyzing 50 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 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 52 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! 53 Implementation 54 Haskell • Haskell is a functional programming language. • The following slides show how to express the two solutions using Haskell. 55 Solution #1 findGap :: [Int] -> Int findGap xs = head (filter (`notElem` xs) [0..]) 56 Solution #2 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 57 Sort Problem: Sort a list of values 58 The Problem We Will Solve 59 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 Problem Statement • Sort a list of items. There may be duplicates in the list; that’s okay. 62 Abstract Representation of the Problem • Represent the items in the list using the natural numbers, N = (0, 1, …) 63 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] sort [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 15, 16, 17, 19] 64 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 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) • That’s fast! 66 Create a list of 1’s using the repeat function repeat 1 [1, 1, 1, …] 67 Create a list of pairs using 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)] 68 Compose the zip and repeat functions [08, 15, 09, 00, 12, 11, 01, 10, 13, 07, 16, 04, 14, 15, 05, 17, 03, 19, 02, 06] zip ___ (repeat 1) [(08, 1) , (15, 1), (09, 1), (00, 1), (12, 1), (11, 1), (01, 1), (10, 1)…, (06, 1)] 69 Interpret each pair as (index, count) [(08, 1) , (15, 1), (09, 1), (00, 1), (12, 1), (11, 1), (01, 1), (10, 1)…, (06, 1)] one occurrence index 70 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” 71 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” [(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)] 74 Compose accumArray, 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)] 75 countlist • “countlist” is a (user-defined) function; it is the collection of functions shown on the previous slide (copied below). countlist accumArray (+) 0 (0, n) (zip xs (repeat 1)) where n = length xs 76 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)] 77 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)] replicate once (0) replicate once ……………………….. (1) replicate twice (15) (15) 78 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] 79 Recall “set comprehensions” from your school days { 2 x | x N , x 10 } “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. 80 Terminology variable input set { 2 x | x N , x 10 } output function predicate 81 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] 82 Explanation [2*x|x <- [1..10]] “x is drawn from [1 .. 10] and for every value drawn, that value is doubled.” 83 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. 84 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. 85 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] 86 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] 87 sort • “sort” is a (user-defined) function; it is the collection of functions shown on the previous slide (copied below). sort concat [replicate k x | (x, k) <- assocs (countlist xs)] 88 Here’s the Solution 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] 89 Find the smallest free number Version #2 90 The Problem We Will Solve 91 Problem Statement • Find the smallest natural number not in a given finite list of natural numbers. 92 What is the smallest number not in this list? [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6] 15 93 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 94 Recall the countlist function [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6] countlist ___ 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)] See slides 67-76 for an explanation of the countlist function. 95 The countlist function countlist accumArray (+) 0 (0, n) (zip xs (repeat 1)) where n = length xs xs is the list of Natural numbers. 96 Recall the checklist function • The first version (see slides 27-41) to the problem used the checklist function: [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)] 97 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) [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6] countlist ___ 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)] These numbers were not in the input list, as indicated by 0 in the second value of their pairs 100 Select all pairs with (_,0) [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6] [(x, k) | (x, k) <- assocs (countlist ___), k == 0] [(15,0), (16,0), (18,0)] 101 Explanation of this list comprehension input set [(x, k) | (x, k) <- assocs (countlist xs), k == 0] 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).” 102 Select just the first value in each pair [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6] [x | (x, k) <- assocs (countlist ___), k == 0] [15, 16, 18] For each pair, output only the first value. 103 Select the first value [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 14, 14, 5, 17, 3, 19, 2, 6] head [x | (x, k) <- assocs (countlist ___), k == 0] 15 head returns the first item in the list. 104 Here’s 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] 105 Find the smallest free number Version 3 106 The Problem We Will Solve 107 Problem Statement • Find the smallest natural number not in given finite list of natural numbers. 108 What is the smallest number not in this list? [8, 9, 0, 12, 11, 1, 10, 13, 7, 4, 14, 24, 34, 5, 17, 3, 19, 2, 6] 15 109 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]) 110 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 [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 b==10 [8,9,0,1,7,4,5,3,2,6] [12,11,10,13,14,24,34,17,19] 112 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 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? 113 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] partition 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 114 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] partition 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 If m == b – a then the left list has no gaps n = length xs b = a + 1 + (n `div` 2) m = length us 115 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] partition 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 116 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] partition 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 117 Let’s trace an example • The following slides traces the processing of a list using the divide-and-conquer algorithm. 118 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) = 10 119 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 partition (< b) xs us = [8,9,0,1,7,4,5,3,2,6] vs = [12,11,10,13,14,24,34,17,19] 120 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 partition (< b) xs us = [8,9,0,1,7,4,5,3,2,6] vs = [12,11,10,13,14,24,34,17,19] 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. 121 set a to the value of b (10) set n to this value: n – m (19 – 10 = 9) xs = [12,11,10,13,14,24,34,17,19] 122 a = b = 10 n = n – m = 19 – 10 = 9 xs = [12,11,10,13,14,24,34,17,19] b = a + 1 + (n `div` 2) = 10 + 1 + (9 `div` 2) = 15 123 a = b = 10 n = n – m = 19 – 10 = 9 xs = [12,11,10,13,14,24,34,17,19] b = a + 1 + (n `div` 2) = 10 + 1 + (9 `div` 2) = 15 partition (< b) xs us = [12,11,10,13,14] vs = [24,34,17,19] 124 a = b = 10 n = n – m = 19 – 10 = 9 xs = [12,11,10,13,14,24,34,17,19] b = a + 1 + (n `div` 2) = 10 + 1 + (9 `div` 2) = 15 partition (< b) xs us = [12,11,10,13,14] vs = [24,34,17,19] m = length us = 5 m == b – a == 15 – 10 == True Therefore, this list has no missing numbers and we should recurse on the other. 125 set a to the value of b (15) set n to the value of n - m (4) xs = [24,34,17,19] 126 set a to the value of b (15) set n to the value of n - m (4) xs = [24,34,17,19] b = a + 1 + (n `div` 2) = 15 + 1 + (4 `div` 2) = 18 127 set a to the value of b (15) set n to the value of n - m (4) xs = [24,34,17,19] b = a + 1 + (n `div` 2) = 15 + 1 + (4 `div` 2) = 18 partition (< b) xs us = [17] vs = [24,34,19 ] m = length us = 1 m == b – a == 18 – 15 == False Therefore, this list has a missing number and we should recurse on it. 128 set a to the old value of a (15) set n to the value of m (1) xs = [17] 129 set a to the old value of a (15) set n to the value of m (1) xs = [17] b = a + 1 + (n `div` 2) = 15 + 1 + (1 `div` 2) = 16 130 set a to the old value of a (15) set n to the value of m (1) xs = [17] b = a + 1 + (n `div` 2) = 15 + 1 + (1 `div` 2) = 16 partition (< b) xs us = [ ] vs = [17] m = length us = 0 m == b – a == 16 – 15 == False Therefore, this list has a missing number and we should recurse on it. 131 set a to the old value of a (15) set n to the value of m (0) If n == 0 then return a Done! The answer is: 15 132 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 Here’s the Solution import List minfree :: [Int] -> Int minfree xs = minfrom 0 (length xs, xs) minfrom :: Int -> (Int, [Int]) minfrom a (n, xs) | n == 0 | m == b - a | otherwise where (us, b m -> Int = a = minfrom b (n - m, vs) = minfrom a (m, us) vs) = partition (<b) xs = a + 1 + (n `div` 2) = length us 134