making function problem (chapter 6 of Programming in Haskell)

Roelof Wobben

8 Aug 2011 8 Aug '11

6:27 a.m.

Hello, After a short holiday I now studying chapter 6 of this book. For the first exercise I have to make the function for ^ for postitive numbers. So I did these steps. Step 1 : Defining the type ^ :: [Int] -> Int Step 2 : enumarate the cases ^ [] = ^ [n, 0] = ^ [n, ns] = Step 3 : Define the simple cases ^ [] = [] ^ [n.0] = 1 Step 4 : Define the other cases ^[n.ns] = n * ^ns Step 5 : generalize and simplify : ^[] = [] ^[-, 0] = 1 So the whole function would be : ^ :: [Int] -> Int ^[] = [] ^[-, 0] = 1 ^ [n,ns] = n * ^ns But the answer in the book only says : m ^ 0 = 1 m (n+1) = m * m ^n so no type defenition and a whole different solution to the last rule. So my question is : Is what I did wrong ? Roelof

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Ertugrul Soeylemez

8 Aug 8 Aug

6:51 a.m.

Roelof Wobben wrote:

...

After a short holiday I now studying chapter 6 of this book.

For the first exercise I have to make the function for ^ for postitive numbers.

I assume you mean the exponentiation function, and the only real indication for that is the solution you quoted later in your post. To help us help, you really should work on your problem descriptions.

...

Step 1 : Defining the type

^ :: [Int] -> Int

First of all, you have to learn proper Haskell syntax. Your type signature is invalid. But let me rewrite your type signature to correct syntax: (^) :: [Int] -> Int This is the type signature for a function (^), which expects exactly one argument, a list. Is this really what you want? Before going any further, you should come up with the right type signature for your function. Once you have that, we will continue. Greets, Ertugrul -- nightmare = unsafePerformIO (getWrongWife >>= sex) http://ertes.de/

Roelof Wobben

7:06 a.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

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To: beginners@haskell.org From: es@ertes.de Date: Mon, 8 Aug 2011 12:51:34 +0200 Subject: Re: [Haskell-beginners] making function problem (chapter 6 of Programming in Haskell)

Roelof Wobben wrote:

...
After a short holiday I now studying chapter 6 of this book.

For the first exercise I have to make the function for ^ for postitive numbers.

I assume you mean the exponentiation function, and the only real indication for that is the solution you quoted later in your post. To help us help, you really should work on your problem descriptions.

...
Step 1 : Defining the type

^ :: [Int] -> Int

First of all, you have to learn proper Haskell syntax. Your type signature is invalid. But let me rewrite your type signature to correct syntax:

(^) :: [Int] -> Int

This is the type signature for a function (^), which expects exactly one argument, a list. Is this really what you want? Before going any further, you should come up with the right type signature for your function. Once you have that, we will continue.

oke, I don't think I want that. I want to type this 2^3 and then the outcome will be 8. So next try (^) :: Int -> Int -> Int Because the first and second numbers are integers and the outcome also will be a integer. Roelof

Ertugrul Soeylemez

7:26 a.m.

Roelof Wobben wrote:

...

I don't think I want that. I want to type this 2^3 and then the outcome will be 8. So next try

(^) :: Int -> Int -> Int

Because the first and second numbers are integers and the outcome also will be a integer.

This looks more like it. Now forget about programming for a moment and think about how you can express the exponentiation operation in terms of simple equations. Let me show you how you would do that for addition of natural numbers, if you have only successor (succ) and predecessor (pred) functions available: x + 0 = x x + y = succ x + pred y Evaluation shows: 3 + 2 = succ 3 + pred 2 = 4 + 1 = succ 4 + pred 1 = 5 + 0 = 5 A similarly simple ruleset can express exponentiation, too. Greets, Ertugrul -- nightmare = unsafePerformIO (getWrongWife >>= sex) http://ertes.de/

KC

10:26 a.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

Besides the programming problem for (^) for natural numbers; there is also the algorithmic question in that can the recursive case be done better than m ^ n = m * m ^ (n -1)? -- -- Regards, KC

Roelof Wobben

10:28 a.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

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To: beginners@haskell.org From: es@ertes.de Date: Mon, 8 Aug 2011 13:26:14 +0200 Subject: Re: [Haskell-beginners] making function problem (chapter 6 of Programming in Haskell)

Roelof Wobben wrote:

...
I don't think I want that. I want to type this 2^3 and then the outcome will be 8. So next try

(^) :: Int -> Int -> Int

Because the first and second numbers are integers and the outcome also will be a integer.

This looks more like it. Now forget about programming for a moment and think about how you can express the exponentiation operation in terms of simple equations. Let me show you how you would do that for addition of natural numbers, if you have only successor (succ) and predecessor (pred) functions available:

x + 0 = x x + y = succ x + pred y

Evaluation shows:

3 + 2 = succ 3 + pred 2 = 4 + 1 = succ 4 + pred 1 = 5 + 0 = 5

A similarly simple ruleset can express exponentiation, too.

Hello, I think you mean something like this : 2 ^ 3 = = 2 * 2 ^ 2 = 2 * 2 * 2 ^1 = 2 * 2 * 2 * 2 ^ 0 = 2 * 2 * 2 * 1 = 2 * 2 * 2 = 4 * 2 = 8 Roelof

Ertugrul Soeylemez

10:36 a.m.

Roelof Wobben wrote:

...

I think you mean something like this :

2 ^ 3 = = 2 * 2 ^ 2 = 2 * 2 * 2 ^1 = 2 * 2 * 2 * 2 ^ 0 = 2 * 2 * 2 * 1 = 2 * 2 * 2 = 4 * 2 = 8

Yes, but that's not a rule system. Try to come up with a rule system to express exponentiation. You need two rules for the algorithm you want to implement. Greets, Ertugrul -- nightmare = unsafePerformIO (getWrongWife >>= sex) http://ertes.de/

Roelof Wobben

10:53 a.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

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To: beginners@haskell.org From: es@ertes.de Date: Mon, 8 Aug 2011 16:36:18 +0200 Subject: Re: [Haskell-beginners] making function problem (chapter 6 of Programming in Haskell)

Roelof Wobben wrote:

...
I think you mean something like this :

2 ^ 3 = = 2 * 2 ^ 2 = 2 * 2 * 2 ^1 = 2 * 2 * 2 * 2 ^ 0 = 2 * 2 * 2 * 1 = 2 * 2 * 2 = 4 * 2 = 8

Yes, but that's not a rule system. Try to come up with a rule system to express exponentiation. You need two rules for the algorithm you want to implement.

Greets, Ertugrul

-- nightmare = unsafePerformIO (getWrongWife >>= sex) http://ertes.de/

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

Maybe this : x ^ 0 = 1 x ^ y = x * (y-1) Roelof

Ertugrul Soeylemez

11:17 a.m.

Roelof Wobben wrote:

...

Maybe this :

x ^ 0 = 1 x ^ y = x * (y-1)

No, that's wrong. Don't try guessing things, because that will bring you nowhere with Haskell. But at least you are getting closer to the idea of solving things through algebraic rules. Greets, Ertugrul -- nightmare = unsafePerformIO (getWrongWife >>= sex) http://ertes.de/

Roelof Wobben

11:24 a.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

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To: beginners@haskell.org From: es@ertes.de Date: Mon, 8 Aug 2011 17:17:18 +0200 Subject: Re: [Haskell-beginners] making function problem (chapter 6 of Programming in Haskell)

Roelof Wobben wrote:

...
Maybe this :

x ^ 0 = 1 x ^ y = x * (y-1)

No, that's wrong. Don't try guessing things, because that will bring you nowhere with Haskell.

But at least you are getting closer to the idea of solving things through algebraic rules.

Greets, Ertugrul

-- nightmare = unsafePerformIO (getWrongWife >>= sex) http://ertes.de/

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

Im not guessing. Im trying to understand what you mean by exponation rules. As far as I can imagaging it cannot be done the same way as you described the sum problem. Roelof

Ertugrul Soeylemez

11:28 a.m.

Roelof Wobben wrote:

...

Im not guessing.

Im trying to understand what you mean by exponation rules.

As far as I can imagaging it cannot be done the same way as you described the sum problem.

Try with multiplication first. If you can do multiplication, then exponentiation should be clear. Greets, Ertugrul -- nightmare = unsafePerformIO (getWrongWife >>= sex) http://ertes.de/

Roelof Wobben

11:35 a.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

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To: beginners@haskell.org From: es@ertes.de Date: Mon, 8 Aug 2011 17:28:08 +0200 Subject: Re: [Haskell-beginners] making function problem (chapter 6 of Programming in Haskell)

Roelof Wobben wrote:

...
Im not guessing.

Im trying to understand what you mean by exponation rules.

As far as I can imagaging it cannot be done the same way as you described the sum problem.

Try with multiplication first. If you can do multiplication, then exponentiation should be clear.

Greets, Ertugrul

-- nightmare = unsafePerformIO (getWrongWife >>= sex) http://ertes.de/

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

Oke, Just to be sure. I only have to use suc and pred ? I think you want something like this : x + 0 = x x + y = succ x + pred y Roelof

Roelof Wobben

2:12 p.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

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From: rwobben@hotmail.com To: beginners@haskell.org Date: Mon, 8 Aug 2011 15:35:55 +0000 Subject: Re: [Haskell-beginners] making function problem (chapter 6 of Programming in Haskell)

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To: beginners@haskell.org From: es@ertes.de Date: Mon, 8 Aug 2011 17:28:08 +0200 Subject: Re: [Haskell-beginners] making function problem (chapter 6 of Programming in Haskell)

Roelof Wobben wrote:

...
Im not guessing.

Im trying to understand what you mean by exponation rules.

As far as I can imagaging it cannot be done the same way as you described the sum problem.

Try with multiplication first. If you can do multiplication, then exponentiation should be clear.

Greets, Ertugrul

-- nightmare = unsafePerformIO (getWrongWife >>= sex) http://ertes.de/

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

Oke,

Just to be sure.

I only have to use suc and pred ?

I think you want something like this :

x + 0 = x x + y = succ x + pred y

Roelof

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

oke, Maybe you are looking something like this : x * 0 = 0 0 * y = 0 x * y = z Roelof

David Beacham

2:18 p.m.

On 08/08/11 16:17, Ertugrul Soeylemez wrote:

...

...
...
Maybe this :

x ^ 0 = 1 x ^ y = x * (y-1) Hi Roelof,

I don't think you're far away - it could just be a typo/lack of care in writing it down. Have you tried putting some actual values into these rules? - that should help you understand what is wrong with what you've written down/get you closer to the correct solution. For example: (x=2, y=4) 2^4 = 16 != 6 = 2 * 3 David.

Roelof Wobben

2:58 p.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

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Date: Mon, 8 Aug 2011 19:18:55 +0100 From: dbeacham@dbeacham.co.uk To: beginners@haskell.org Subject: Re: [Haskell-beginners] making function problem (chapter 6 of Programming in Haskell)

On 08/08/11 16:17, Ertugrul Soeylemez wrote:

...
...
...
Maybe this :

x ^ 0 = 1 x ^ y = x * (y-1) Hi Roelof,

I don't think you're far away - it could just be a typo/lack of care in writing it down. Have you tried putting some actual values into these rules? - that should help you understand what is wrong with what you've written down/get you closer to the correct solution.

For example: (x=2, y=4) 2^4 = 16 != 6 = 2 * 3

David.

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

Oke, what Im trying to say it this : 2 ^ 4 = 2 * 2 ^ 3 = 2 * 2 * 2 ^ 2 = 2 * 2 * 2 * 2 ^ 1 = 2 * 2 * 2 * 2 * 2 ^ 0 So the y is on each number one less on each run. So maybe it's x ^ y = x * x ^ y Roelof

KC

5:59 p.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

Hint positivePower :: Int -> Int -> Int positivePower x 1 = x positivePower x y = dododoo positivePower yadayada -- -- Regards, KC

Roelof Wobben

9 Aug 9 Aug

8:16 a.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

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Date: Mon, 8 Aug 2011 14:59:39 -0700 From: kc1956@gmail.com To: beginners@haskell.org Subject: Re: [Haskell-beginners] making function problem (chapter 6 of Programming in Haskell)

Hint

positivePower :: Int -> Int -> Int positivePower x 1 = x positivePower x y = dododoo positivePower yadayada

-- -- Regards, KC

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

oke. I see it I'm going for this : postivePower :: Int -> Int -> Int PostivePower x 1 = 1 because when you multiply something by one is remains the same value. PostivePower x y = x + postivePower x y for example x = 4 y = 3 4 * 3 = 4 + 4 * 2 4 + 4 + 4 * 1 4 + 4 + 4 4 + 8 12 Roelof

Daniel Seidel

8:41 a.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

Hi Roelof, On Tue, 2011-08-09 at 12:16 +0000, Roelof Wobben wrote:

...

oke.

I see it

unfortunately its still not true. Your example with the concrete numbers is right, and it captures the scheme well. But try to check your code with it. I'd like to say its typos, but its actually the same typos as you had for the actual main example, calculating the lists. Try to answer the questions I ask below and then use the answers for a correct definition (that is very similar to the recent one).

...

I'm going for this :

postivePower :: Int -> Int -> Int

PostivePower x 1 = 1 because when you multiply something by one is remains the same value.

Does this really return the value you want? Is 5 * 1 = 1 ?

...

PostivePower x y = x + postivePower x y

Do you really want to call positivePower with the same y again, or more concretely: Is 4 * 3 = 4 + (4 * 3)?, or should there be a "2" somewhere?

...

for example

x = 4

y = 3

4 * 3 =

4 + 4 * 2

4 + 4 + 4 * 1

4 + 4 + 4

4 + 8

12

Roelof Wobben

8:44 a.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

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Subject: Re: [Haskell-beginners] making function problem (chapter 6 of Programming in Haskell) From: ds@iai.uni-bonn.de To: rwobben@hotmail.com CC: beginners@haskell.org Date: Tue, 9 Aug 2011 14:41:13 +0200

Hi Roelof,

On Tue, 2011-08-09 at 12:16 +0000, Roelof Wobben wrote:

...
oke.

I see it

unfortunately its still not true. Your example with the concrete numbers is right, and it captures the scheme well. But try to check your code with it. I'd like to say its typos, but its actually the same typos as you had for the actual main example, calculating the lists. Try to answer the questions I ask below and then use the answers for a correct definition (that is very similar to the recent one).

...
I'm going for this :

postivePower :: Int -> Int -> Int

PostivePower x 1 = 1 because when you multiply something by one is remains the same value.

Does this really return the value you want? Is 5 * 1 = 1 ?

...
PostivePower x y = x + postivePower x y

Do you really want to call positivePower with the same y again, or more concretely: Is 4 * 3 = 4 + (4 * 3)?, or should there be a "2" somewhere?

It should be 2 so it would then be postivepower x y = x + postivepower x (y-1) Roelof

Daniel Seidel

9:12 a.m.

New subject: making function problem (chapter 6 of Programming in Haskell)

On Tue, 2011-08-09 at 12:44 +0000, Roelof Wobben wrote:

...

...
...
PostivePower x y = x + postivePower x y

Do you really want to call positivePower with the same y again, or more concretely: Is 4 * 3 = 4 + (4 * 3)?, or should there be a "2" somewhere?

It should be 2 so it would then be postivepower x y = x + postivepower x (y-1)

Right! That's it. Now it's hopefully not a problem to write the (^) function with the same scheme. Cheers, Daniel.

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Daniel Seidel
David Beacham
Ertugrul Soeylemez
KC
Roelof Wobben