Send Beginners mailing list submissions to
[email protected]
To subscribe or unsubscribe via the World Wide Web, visit
http://www.haskell.org/mailman/listinfo/beginners
or, via email, send a message with subject or body 'help' to
[email protected]
You can reach the person managing the list at
[email protected]
When replying, please edit your Subject line so it is more specific
than "Re: Contents of Beginners digest..."
Today's Topics:
1. Re: recursion problem. (Karl Voelker)
2. sum problem (Roelof Wobben)
3. Re: sum problem (yi lu)
4. Re: sum problem (Roelof Wobben)
5. any feedback on this solution (Roelof Wobben)
6. Re: any feedback on this solution (Kees Bleijenberg)
7. Re: any feedback on this solution (Roelof Wobben)
----------------------------------------------------------------------
Message: 1
Date: Fri, 06 Feb 2015 20:59:36 -0800
From: Karl Voelker <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] recursion problem.
Message-ID:
<1423285176.1254095.224241757.68c01...@webmail.messagingengine.com>
Content-Type: text/plain
On Fri, Feb 6, 2015, at 01:25 AM, Roelof Wobben wrote:
> but now when I do toDigits 0 , I see [] as output where I was expected
> [0]
In our written notation, "0" is a special case: it's the only integer
written with a leading zero. So don't be too surprised if your code
requires a special case for it.
-Karl
------------------------------
Message: 2
Date: Sat, 07 Feb 2015 09:53:34 +0100
From: Roelof Wobben <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: [Haskell-beginners] sum problem
Message-ID: <[email protected]>
Content-Type: text/plain; charset=windows-1252; format=flowed
Hello,
The exercises of CIS 194 are really hard.
Now I have to sum all the numbers in a array with a catch
If a number is greater then 10 it has to be dived in single numbers.
so [ 1 , 12, 3] will be [ 1, 1,2,3] and then sum up.
So far I have this :
-- | sum all the digits of a array
sumDigits :: [Integer] -> Integer
sumDigits n
| n == 0 = acc
| n < 10 = acc + n
| otherwise = sumDigits acc + n `mod` 10 + n `rem` 10
But now I see a message that acc is not known which is not wierd.
But when I do sumDigits n acc I see a message that sumDigits has only 1
parameter where I give it to,
When I only had to sum up i is easy , I could use map [+] xs
Anyone who can give a me tip without giving the answer otherwise I do
not learn anything.
Roelof
------------------------------
Message: 3
Date: Sat, 7 Feb 2015 17:01:48 +0800
From: yi lu <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] sum problem
Message-ID:
<CAKcmqqyMAnsv8sORoDk_JzD=dzmuorc-eypan4joywqzsnm...@mail.gmail.com>
Content-Type: text/plain; charset="iso-8859-1"
Do not try to `sum` all at first. See how you can deal with one integer.
On Sat, Feb 7, 2015 at 4:53 PM, Roelof Wobben <[email protected]> wrote:
> Hello,
>
> The exercises of CIS 194 are really hard.
>
> Now I have to sum all the numbers in a array with a catch
> If a number is greater then 10 it has to be dived in single numbers.
>
> so [ 1 , 12, 3] will be [ 1, 1,2,3] and then sum up.
>
> So far I have this :
>
> -- | sum all the digits of a array
> sumDigits :: [Integer] -> Integer
> sumDigits n
> | n == 0 = acc
> | n < 10 = acc + n
> | otherwise = sumDigits acc + n `mod` 10 + n `rem` 10
>
>
> But now I see a message that acc is not known which is not wierd.
>
> But when I do sumDigits n acc I see a message that sumDigits has only 1
> parameter where I give it to,
>
> When I only had to sum up i is easy , I could use map [+] xs
>
> Anyone who can give a me tip without giving the answer otherwise I do not
> learn anything.
>
> Roelof
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/beginners
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL:
<http://www.haskell.org/pipermail/beginners/attachments/20150207/003d8fe5/attachment-0001.html>
------------------------------
Message: 4
Date: Sat, 07 Feb 2015 10:05:16 +0100
From: Roelof Wobben <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] sum problem
Message-ID: <[email protected]>
Content-Type: text/plain; charset="us-ascii"
An HTML attachment was scrubbed...
URL:
<http://www.haskell.org/pipermail/beginners/attachments/20150207/8db36552/attachment-0001.html>
------------------------------
Message: 5
Date: Sat, 07 Feb 2015 10:37:08 +0100
From: Roelof Wobben <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: [Haskell-beginners] any feedback on this solution
Message-ID: <[email protected]>
Content-Type: text/plain; charset=windows-1252; format=flowed
Hello,
I finally solved exercise 1 where I had to write a programm which checks
if a creditcard number is valid.
I solved it this way :
toDigits :: Integer -> [Integer]
toDigits n
| n < 0 = []
| n < 10 = [n]
| otherwise = toDigits (n `div` 10) ++ [n `mod` 10]
-- | convert a number to a reversed array where a negative number will
be a empty array
toDigitsRev :: Integer -> [Integer]
toDigitsRev 0 = [0]
toDigitsRev n
| n < 0 = []
| n < 10 = [n]
| otherwise = n `mod` 10 : toDigitsRev (n `div` 10)
-- | Doubles every second number from the right.
doubleEveryOther :: [Integer] -> [Integer]
doubleEveryOther [] = []
doubleEveryOther (x:[]) = [x]
doubleEveryOther (x:(y:zs))
| length (x:(y:zs)) `mod` 2 /= 0 = [x] ++ (y * 2) : doubleEveryOther zs
| otherwise = [x *2] ++ y : doubleEveryOther zs
-- | sum all the digits of a array
sumDigits :: [Integer] -> Integer
sumDigits [] = 0
sumDigits (x:zs)
| x < 10 = x + sumDigits zs
| otherwise = x `mod` 10 + x `div` 10 + sumDigits zs
-- | validate a number by looking if a number can be divided by 10
validate :: Integer -> Bool
validate n = sumDigits(doubleEveryOther(toDigits(n))) `mod` 10 == 0
-- | The main entry point.
main :: IO ()
main = do
print $ validate 4012888888881881
Any remarks about this solution.
I know there are higher function solutions but that part is not
descrived in chapter 1 .
Roelof
------------------------------
Message: 6
Date: Sat, 7 Feb 2015 10:56:11 +0100
From: "Kees Bleijenberg" <[email protected]>
To: "'The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell'" <[email protected]>
Subject: Re: [Haskell-beginners] any feedback on this solution
Message-ID: <[email protected]>
Content-Type: text/plain; charset="us-ascii"
Roelof,
Maybe it's better to reuse toDigits in sumDigits.
Kees
-----Oorspronkelijk bericht-----
Van: Beginners [mailto:[email protected]] Namens Roelof Wobben
Verzonden: zaterdag 7 februari 2015 10:37
Aan: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell
Onderwerp: [Haskell-beginners] any feedback on this solution
Hello,
I finally solved exercise 1 where I had to write a programm which checks
if a creditcard number is valid.
I solved it this way :
toDigits :: Integer -> [Integer]
toDigits n
| n < 0 = []
| n < 10 = [n]
| otherwise = toDigits (n `div` 10) ++ [n `mod` 10]
-- | convert a number to a reversed array where a negative number will
be a empty array
toDigitsRev :: Integer -> [Integer]
toDigitsRev 0 = [0]
toDigitsRev n
| n < 0 = []
| n < 10 = [n]
| otherwise = n `mod` 10 : toDigitsRev (n `div` 10)
-- | Doubles every second number from the right.
doubleEveryOther :: [Integer] -> [Integer]
doubleEveryOther [] = []
doubleEveryOther (x:[]) = [x]
doubleEveryOther (x:(y:zs))
| length (x:(y:zs)) `mod` 2 /= 0 = [x] ++ (y * 2) : doubleEveryOther zs
| otherwise = [x *2] ++ y : doubleEveryOther zs
-- | sum all the digits of a array
sumDigits :: [Integer] -> Integer
sumDigits [] = 0
sumDigits (x:zs)
| x < 10 = x + sumDigits zs
| otherwise = x `mod` 10 + x `div` 10 + sumDigits zs
-- | validate a number by looking if a number can be divided by 10
validate :: Integer -> Bool
validate n = sumDigits(doubleEveryOther(toDigits(n))) `mod` 10 == 0
-- | The main entry point.
main :: IO ()
main = do
print $ validate 4012888888881881
Any remarks about this solution.
I know there are higher function solutions but that part is not
descrived in chapter 1 .
Roelof
_______________________________________________
Beginners mailing list
[email protected]
http://www.haskell.org/mailman/listinfo/beginners
-----
No virus found in this message.
Checked by AVG - www.avg.com
Version: 2015.0.5646 / Virus Database: 4281/9071 - Release Date: 02/07/15
------------------------------
Message: 7
Date: Sat, 07 Feb 2015 11:01:55 +0100
From: Roelof Wobben <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] any feedback on this solution
Message-ID: <[email protected]>
Content-Type: text/plain; charset=windows-1252; format=flowed
Why,
They have opposite terms.
SumDigits takes a array and has a integer as output.
ToDigits takes a integer and puts out a array.
Roelof
Kees Bleijenberg schreef op 7-2-2015 om 10:56:
> Roelof,
>
> Maybe it's better to reuse toDigits in sumDigits.
>
> Kees
>
> -----Oorspronkelijk bericht-----
> Van: Beginners [mailto:[email protected]] Namens Roelof Wobben
> Verzonden: zaterdag 7 februari 2015 10:37
> Aan: The Haskell-Beginners Mailing List - Discussion of primarily
> beginner-level topics related to Haskell
> Onderwerp: [Haskell-beginners] any feedback on this solution
>
> Hello,
>
> I finally solved exercise 1 where I had to write a programm which checks
> if a creditcard number is valid.
>
> I solved it this way :
>
> toDigits :: Integer -> [Integer]
> toDigits n
> | n < 0 = []
> | n < 10 = [n]
> | otherwise = toDigits (n `div` 10) ++ [n `mod` 10]
>
> -- | convert a number to a reversed array where a negative number will
> be a empty array
> toDigitsRev :: Integer -> [Integer]
> toDigitsRev 0 = [0]
> toDigitsRev n
> | n < 0 = []
> | n < 10 = [n]
> | otherwise = n `mod` 10 : toDigitsRev (n `div` 10)
>
> -- | Doubles every second number from the right.
> doubleEveryOther :: [Integer] -> [Integer]
> doubleEveryOther [] = []
> doubleEveryOther (x:[]) = [x]
> doubleEveryOther (x:(y:zs))
> | length (x:(y:zs)) `mod` 2 /= 0 = [x] ++ (y * 2) : doubleEveryOther zs
> | otherwise = [x *2] ++ y : doubleEveryOther zs
>
>
> -- | sum all the digits of a array
> sumDigits :: [Integer] -> Integer
> sumDigits [] = 0
> sumDigits (x:zs)
> | x < 10 = x + sumDigits zs
> | otherwise = x `mod` 10 + x `div` 10 + sumDigits zs
>
>
> -- | validate a number by looking if a number can be divided by 10
> validate :: Integer -> Bool
> validate n = sumDigits(doubleEveryOther(toDigits(n))) `mod` 10 == 0
>
>
> -- | The main entry point.
> main :: IO ()
> main = do
> print $ validate 4012888888881881
>
> Any remarks about this solution.
>
> I know there are higher function solutions but that part is not
> descrived in chapter 1 .
>
> Roelof
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/beginners
>
>
> -----
> No virus found in this message.
> Checked by AVG - www.avg.com
> Version: 2015.0.5646 / Virus Database: 4281/9071 - Release Date: 02/07/15
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://www.haskell.org/mailman/listinfo/beginners
>
------------------------------
Subject: Digest Footer
_______________________________________________
Beginners mailing list
[email protected]
http://www.haskell.org/mailman/listinfo/beginners
------------------------------
End of Beginners Digest, Vol 80, Issue 12
*****************************************
