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PROGRAMMING IN HASKELL

PROGRAMMING IN HASKELL. Chapter 12 – Lazy evaluation and infinite lists. Slides not from Hutton. Redex. A reducible expression (redex) is a function that can be applied to its arguments.

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PROGRAMMING IN HASKELL

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

  2. Redex • 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 Run it and get a stack overflow. int :: Int inf = 1 + inf > inf where inf = 1+inf ERROR - C stack overflow But > 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.

  4. Outermost first Consider ones :: [Int] ones = 1 : ones ones = [1, 1, …] 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]

  5. > 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 > take 5 [1 ..] [1,2,3,4,5] Note: [1 .. ] = [1, 2, 3, 4, … ] primes :: [Int] primes = seive [2 .. ] sieve :: [Int]  [Int] sieve (p:xs) = p : sieve [x | x  xs, x `mod` p  0] 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]

  7. Primes The color coding lets you follow each list. [2 .. ] = 2 : [3 .. ] > take 4 primes = take 4 sieve [2 .. ] = take 4 sieve (2:[3 .. ]) = 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] sieve (p:xs) = p : sieve [x | x xs, x `mod` p 0]

  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]

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