1. 0 PROGRAMMING IN HASKELL Chapter 12 – Lazy evaluation and infinite lists Slides not from Hutton

2. Redex 1 A reducible expression (redex) is a function that can be applied to its arguments. The expression (1+2)*(3+4) is not reducible since both arguments must be evaluated before it can be applied. The redexs are (1+2) and (3+4) but not (1+2)*(3+4). But once (1+2) and (3+4) have been reduced to 3 and 7 respectively, 3*7 is a redex.

3. Outermost first Consider int :: Int inf = 1 + inf But Run it and get a stack overflow. > inf where inf = 1+inf ERROR - C stack overflow > fst (0, inf) where inf = 1 + inf 0 > snd (inf, 0) where inf = 1 + inf 0 fst (0, _) and snd (_, 0) are both reducible. No need to reduce the other tuple element. 2

4. 3 Outermost first ones :: [Int] ones = 1 : ones Consider Run it and get an infinite list. (Use Actions -> stop to stop it.) > take 5 ones where ones = 1:ones [1,1,1,1,1] ones = [1, 1, …]

5. 4 > take 5 ones = take 5 1: ones = take 5 1:1:ones = take 5 1:1:1:ones = take 5 1:1:1:1:ones = take 5 1:1:1:1:1:ones = [1, 1, 1, 1, 1] At this point take 5 xs can be evaluated. So it is. Inner expressions are evaluated only until outer expressions can be.

6. primes :: [Int] primes = seive [2.. ] sieve :: [Int]  [Int] sieve (p:xs) = p : sieve [x | x  xs, x `mod` p  0] 5 Note: [1.. ] = [1, 2, 3, 4, … ] > take 5 [1..] [1,2,3,4,5] Sieve of Eratosthenes. Generate the primes one by one, eliminating all multiples of generated primes. > take 5 primes where primes = sieve [2.. ]; sieve (p:xs) = p : sieve [x | x <- xs, x `mod` p /= 0] [2,3,5,7,11] Primes

7. sieve (p:xs) = p : sieve [x | x  xs, x `mod` p  0] 6 Primes > take 4 primes = take 4 sieve [2.. ] = take 4 (2 : sieve [x | x  [3.. ], x `mod` 2  0]) = take 4 (2 : sieve (3: [x | x  [4.. ], x `mod` 2  0])) = take 4 (2 : 3 : sieve [y | y  [x | x  [4.. ], x `mod` 2  0], y `mod` 3  0] = take 4 (2 : 3 : sieve [y | y  [x | x  [5.. ], x `mod` 2  0], y `mod` 3  0] = take 4 (2 : 3 : sieve [y | y  5:[x | x  [6.. ], x `mod` 2  0], y `mod` 3  0] = take 4 (2 : 3 : sieve (5: [y | y  [x | x  [6.. ], x `mod` 2  0], y `mod` 3  0])) = take 4 (2 : 3 : 5 : sieve [z | z  [y | y  [x | x  [6.. ], x `mod` 2  0], y `mod` 3  0], z `mod` 5  0]) = take 4 (2 : 3 : 5 : sieve [z | z  [y | y  [x | x  [7.. ], x `mod` 2  0], y `mod` 3  0], z `mod` 5  0]) = take 4 (2 : 3 : 5 : sieve [z | z  [y | y  (7:[x | x  [8.. ], x `mod` 2  0]), y `mod` 3  0], z `mod` 5  0]) = take 4 (2 : 3 : 5 : sieve [z | z  7: [y | y  [x | x  [8.. ], x `mod` 2  0], y `mod` 3  0], z `mod` 5  0]) = take 4 (2 : 3 : 5 : sieve (7: [z | z  [y | y  [x | x  [8.. ], x `mod` 2  0], y `mod` 3  0], z `mod` 5  0])) = take 4 (2 : 3 : 5 : 7 : sieve [w | w  [z | z  [y | y  [x | x  [8.. ], x `mod` 2  0], y `mod` 3  0], z `mod` 5  0], w `mod` 7  0] ) = [2, 3, 5, 7]

8. Fibonacci numbers Exercise 4. > take 9 fibs where fibs = 0 : 1 : [x + y | (x, y) <- zip fibs (tail fibs)] [0,1,1,2,3,5,8,13,21] 7