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Today's Topics:
1. State Monad stack example (Animesh Saxena)
2. Re: State Monad stack example
(Sumit Sahrawat, Maths & Computing, IIT (BHU))
3. Re: State Monad stack example (Benjamin Edwards)
----------------------------------------------------------------------
Message: 1
Date: Sat, 07 Mar 2015 04:42:21 +0000 (GMT)
From: Animesh Saxena <[email protected]>
To: [email protected]
Subject: [Haskell-beginners] State Monad stack example
Message-ID: <[email protected]>
Content-Type: text/plain; charset="iso-8859-1"; Format="flowed"
I am trying to relate the state monad to a stack example and somehow found it
easy to get recursively confused!
instance Monad (State s) where
? ? return x = State $ \s -> (x,s)
? ? (State h) >>= f = State $ \s -> let (a, newState) = h s
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (State g) = f a
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?in g newState
Considering the stack computation
stackManip stack = let?
? ? ? ? ? ? ? ((), newStack1) = push 3 stack
? ? ? ? ? ? ? (a, newStack2) = pop newStack1
? ? ? ? ? ? ? ?in pop newStack2
in do notation this would become?
do ?
? ?push 3 ?
? ?a <- pop ?
? ?pop
If I consider the first computation push 3 >>= pop and try to translate it to
the definition there are problems....
Copy paste again, I have?
? ? (State h) >>= f = State $ \s -> let (a, newState) = h s
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (State g) = f a
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?in g newState
f is the push function to which we are shoving the old state. I can't exactly
get around to what exactly is the state computation h? Ok assuming it's
something which gives me a new state, but then thats push which is function f.?
Then push is applied to a which is assuming 3 in this case. This gives me a new
state, which I would say newStack1 from the stockManip above.
Then somehow I apply g to newState?? All the more confusion. Back to the
question what exactly is state computation and how is it different from f? It
seems to be the same function?
-Animesh
? ? ? ? ? ? ? ? ? ? ? ? ? ?
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Message: 2
Date: Sat, 7 Mar 2015 10:39:00 +0530
From: "Sumit Sahrawat, Maths & Computing, IIT (BHU)"
<[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] State Monad stack example
Message-ID:
<cajbew8mcyux_41bpq+bmhqoo7khlk5av126jxwmvumebhvu...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
I won't comment on what state exactly is, but you can read up on that and
gain some intuition here:
https://en.wikibooks.org/wiki/Haskell/Understanding_monads/State
It's helpful to implement it using a pen and paper, and consider how the
state flows and gets transformed.
According to the below example,
stackManip stack =
let ((), newStack1) = push 3 stack
(a, newStack2) = pop newStack1
in pop newStack2
We get,
push :: a -> Stack a -> ((), Stack a) -- Assuming 'Stack a' is a
defined datatype
pop :: Stack a -> (a, Stack a) -- Representing a stack with
elements of type 'a'
Thus,
push 3 >>= pop
~~ (Stack a -> ((), Stack a)) >>= (Stack a -> (a, Stack a))
{ Replacing by types }
Bind (>>=) has the type, (for "State s")
(>>=) :: State s a -> (a -> State s b) -> State s b
This is a type mismatch. The conversion to do syntax is at fault here.
First, you must write the computation using bind (>>=), and then convert to
do-notation.
On 7 March 2015 at 10:12, Animesh Saxena <[email protected]> wrote:
> I am trying to relate the state monad to a stack example and somehow found
> it easy to get recursively confused!
>
> instance Monad (State s) where
> return x = State $ \s -> (x,s)
> (State h) >>= f = State $ \s -> let (a, newState) = h s
> (State g) = f a
> in g newState
>
> Considering the stack computation
>
> stackManip stack = let
> ((), newStack1) = push 3 stack
> (a, newStack2) = pop newStack1
> in pop newStack2
>
> in do notation this would become
> do
> push 3
> a <- pop
> pop
>
> If I consider the first computation push 3 >>= pop and try to translate it
> to the definition there are problems....
> Copy paste again, I have
> (State h) >>= f = State $ \s -> let (a, newState) = h s
> (State g) = f a
> in g newState
>
> f is the push function to which we are shoving the old state. I can't
> exactly get around to what exactly is the state computation h? Ok assuming
> it's something which gives me a new state, but then thats push which is
> function f.
> Then push is applied to a which is assuming 3 in this case. This gives me
> a new state, which I would say newStack1 from the stockManip above.
>
> Then somehow I apply g to newState?? All the more confusion. Back to the
> question what exactly is state computation and how is it different from f?
> It seems to be the same function?
>
>
> -Animesh
>
>
>
>
>
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
>
--
Regards
Sumit Sahrawat
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Message: 3
Date: Sat, 07 Mar 2015 09:45:00 +0000
From: Benjamin Edwards <[email protected]>
To: [email protected], The Haskell-Beginners Mailing
List - Discussion of primarily beginner-level topics related to
Haskell <[email protected]>
Subject: Re: [Haskell-beginners] State Monad stack example
Message-ID:
<CAN6k4nhfPpzRE0g-X_dN7mFBqpmyi_Zg9xO=7=l5al1-ccu...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
I gave an expansion of the state monad for a different computation on
Reddit some time ago. Perhaps it will be useful to you:
http://www.reddit.com/r/haskell/comments/25fnrj/nicta_course_help_with_state_exercise/
Best,
Ben
On Sat, 7 Mar 2015 5:09 am Sumit Sahrawat, Maths & Computing, IIT (BHU) <
[email protected]> wrote:
> I won't comment on what state exactly is, but you can read up on that and
> gain some intuition here:
> https://en.wikibooks.org/wiki/Haskell/Understanding_monads/State
> It's helpful to implement it using a pen and paper, and consider how the
> state flows and gets transformed.
>
> According to the below example,
>
> stackManip stack =
> let ((), newStack1) = push 3 stack
> (a, newStack2) = pop newStack1
> in pop newStack2
>
> We get,
>
> push :: a -> Stack a -> ((), Stack a) -- Assuming 'Stack a' is a
> defined datatype
> pop :: Stack a -> (a, Stack a) -- Representing a stack with
> elements of type 'a'
>
> Thus,
>
> push 3 >>= pop
> ~~ (Stack a -> ((), Stack a)) >>= (Stack a -> (a, Stack a))
> { Replacing by types }
>
> Bind (>>=) has the type, (for "State s")
>
> (>>=) :: State s a -> (a -> State s b) -> State s b
>
> This is a type mismatch. The conversion to do syntax is at fault here.
> First, you must write the computation using bind (>>=), and then convert
> to do-notation.
>
> On 7 March 2015 at 10:12, Animesh Saxena <[email protected]> wrote:
>
>> I am trying to relate the state monad to a stack example and somehow
>> found it easy to get recursively confused!
>>
>> instance Monad (State s) where
>> return x = State $ \s -> (x,s)
>> (State h) >>= f = State $ \s -> let (a, newState) = h s
>> (State g) = f a
>> in g newState
>>
>> Considering the stack computation
>>
>> stackManip stack = let
>> ((), newStack1) = push 3 stack
>> (a, newStack2) = pop newStack1
>> in pop newStack2
>>
>> in do notation this would become
>> do
>> push 3
>> a <- pop
>> pop
>>
>> If I consider the first computation push 3 >>= pop and try to translate
>> it to the definition there are problems....
>> Copy paste again, I have
>> (State h) >>= f = State $ \s -> let (a, newState) = h s
>> (State g) = f a
>> in g newState
>>
>> f is the push function to which we are shoving the old state. I can't
>> exactly get around to what exactly is the state computation h? Ok assuming
>> it's something which gives me a new state, but then thats push which is
>> function f.
>> Then push is applied to a which is assuming 3 in this case. This gives me
>> a new state, which I would say newStack1 from the stockManip above.
>>
>> Then somehow I apply g to newState?? All the more confusion. Back to the
>> question what exactly is state computation and how is it different from f?
>> It seems to be the same function?
>>
>>
>> -Animesh
>>
>>
>>
>>
>>
>>
>> _______________________________________________
>> Beginners mailing list
>> [email protected]
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
>>
>
>
> --
> Regards
>
> Sumit Sahrawat
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
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