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プログラミングHaskell 13章 問題7

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スタートHaskellでの発表

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プログラミングHaskell 13章 問題7

  1. 1. (定義1) map f [] = [](定義2) map f (x:xs) = f x : map f xs(定義3) (f . g) x = f (g x)map f (map g xs) = map (f . g) xs
  2. 2. map f (map g []) = map (f . g) []
  3. 3. map f (map g []) = map (f . g) []map f [] = [](定義1) map f [] = []
  4. 4. map f (map g []) = map (f . g) []map f [] = [][] = [](定義1) map f [] = []
  5. 5. (仮定1) map f (map g xs) = map (f . g) xsmap f (map g (x:xs)) = map (f . g) (x:xs)
  6. 6. map f (map g (x:xs)) = map (f . g) (x:xs)map f (g x : map g xs) = (f . g) x : map (f . g) xs(定義2) map f (x:xs) = f x : map f xs
  7. 7. map f (map g (x:xs)) = map (f . g) (x:xs)map f (g x : map g xs) = (f . g) x : map (f . g) xsf (g x) : map f (map g xs) = (f . g) x : map (f . g) xs(定義2) map f (x:xs) = f x : map f xs
  8. 8. map f (map g (x:xs)) = map (f . g) (x:xs)map f (g x : map g xs) = (f . g) x : map (f . g) xsf (g x) : map f (map g xs) = (f . g) x : map (f . g) xsf (g x) : map f (map g xs) = f (g x) : map (f . g) xs(定義3) (f . g) x = f (g x)
  9. 9. map f (map g (x:xs)) = map (f . g) (x:xs)map f (g x : map g xs) = (f . g) x : map (f . g) xsf (g x) : map f (map g xs) = (f . g) x : map (f . g) xsf (g x) : map f (map g xs) = f (g x) : map (f . g) xsf (g x) : map (f . g) xs = f (g x) : map (f . g) xs(仮定1) map f (map g xs) = map (f . g) xs
  10. 10. http://www.paraiso-lang.org/ikmsm/

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