Sample : Similar Monad as Functor

Sample : Similar Monad as Functor (revision 9)

similar :: (Show b, Eq b) => (a -> b) -> (a -> b) -> a -> b
 similar f g x = same $ Similar x g (f x)
 
--- samples (revision 9)
+-- samples
 ok         = map (similar twicePlus double)  [1..10]
 notOk      = map (similar twicePlus plusTwo) [1..10]
 okSpecific = map (similar twicePlus plusTwo) [2]
@@ -173,10 +197,81 @@


 
-
+
+		
+			
+				Similar Monad as Monad
+			
+			
+
+data Similar a b = Similar a (a -> b) b
+
+instance EqMonad Similar where
+  return x = Similar x id x
+  s >>= f  = mu (eqmap f s)
+
+
+
+  型変数が1つなら Monad の定義はできる
+  通常の Functor だと (a -> b) の b に constraint をかけられない
+
+
+
+
+		
+
+
+
 		
 			
-				謎
+				Similar Monad as Monad
+			
+			
+
+
+class EqFunctor f where
+  eqmap :: (Eq a, Eq b) => (a -> b) -> f a -> f b
+
+instance EqFunctor Similar where
+  eqmap f s = Similar fs id fs
+    where fs = f $ same s
+
+
+
+  (a -> b) の b が Eq 保証された Functor Class を使えばどうにか
+
+
+
+
+		
+
+
+
+		
+			
+				Sample : Similar Monad as Monad (revision 11)
+			
+			
+
+same $ return 100 >>= (\x -> Similar x twicePlus $ double x)
+200
+
+same $ return 2 >>= (\x -> Similar x plusTwo $ double x)
+4
+
+same $ return 100 >>= (\x -> Similar x plusTwo $ double x)
+*** Exception: Prelude.undefined
+
+
+
+
+		
+
+
+
+		
+			
+				Problem
 			
 			
 
diff -r 3773cf2f688d -r ce4917c50d29 slides/20140902/slide.md
--- a/slides/20140902/slide.md	Tue Sep 02 17:46:15 2014 +0900
+++ b/slides/20140902/slide.md	Tue Sep 02 18:03:20 2014 +0900
@@ -30,6 +30,8 @@
 data Similar a a = Similar a (a -> a) a
 ```
 
+* 型変数をどう扱うかで悩み中です
+
 # Usage : Similar Monad
 ```
 initObj = Similar 100 id 100
@@ -47,23 +49,70 @@
   fmap g (Similar a f b) = Similar a (g . f) $ g b
 ```
 
-* こんな感じなんだろうけれど Monad の return が定義できない
+* 型変数が2つだと Functor の定義は楽
+
+# Similar Monad as Functor
+```
+instance Monad (Similar a) where
+    return x = Similar x id x -- NG
+    s >>= f  = mu (eqmap f s)
+```
+* 型変数が2つだと Monad の return が定義できない
     * (Similar a) の a を固定していないと Monad の instance にできない
     * return :: a1 -> Similar a a1
     * を想定されているが、 id なので Similar a a でおかしい
 
-# Sample : Similar Monad as Functor
+# Sample : Similar Monad as Functor (revision 9)
 ```
 similar :: (Show b, Eq b) => (a -> b) -> (a -> b) -> a -> b
 similar f g x = same $ Similar x g (f x)
 
--- samples (revision 9)
+-- samples
 ok         = map (similar twicePlus double)  [1..10]
 notOk      = map (similar twicePlus plusTwo) [1..10]
 okSpecific = map (similar twicePlus plusTwo) [2]
 ```
 
-# 謎
+
+# Similar Monad as Monad
+```
+data Similar a b = Similar a (a -> b) b
+
+instance EqMonad Similar where
+  return x = Similar x id x
+  s >>= f  = mu (eqmap f s)
+```
+
+* 型変数が1つなら Monad の定義はできる
+* 通常の Functor だと (a -> b) の b に constraint をかけられない
+
+
+# Similar Monad as Monad
+
+```
+class EqFunctor f where
+  eqmap :: (Eq a, Eq b) => (a -> b) -> f a -> f b
+
+instance EqFunctor Similar where
+  eqmap f s = Similar fs id fs
+    where fs = f $ same s
+```
+
+* (a -> b) の b が Eq 保証された Functor Class を使えばどうにか
+
+# Sample : Similar Monad as Monad (revision 11)
+```
+same $ return 100 >>= (\x -> Similar x twicePlus $ double x)
+200
+
+same $ return 2 >>= (\x -> Similar x plusTwo $ double x)
+4
+
+same $ return 100 >>= (\x -> Similar x plusTwo $ double x)
+*** Exception: Prelude.undefined
+```
+
+# Problem
 * type variable が 1つだと
     * Functor で移す先が Eq である保証が無くて怒られる
 * type variable が 2つだと