Review request for my permutations implementation

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CK Kashyap

7 Jan 2010 7 Jan '10

3:37 a.m.

Hi All, I've written this piece of code to do permutations - perms :: String -> [String] perms []= [] perms (x:[])= [[x]] perms (x:xs)= concat (f [x] (perms xs)) spread :: String -> String -> [String] -- interpolate first string at various positions of second string spread str1 str2 = _spread str1 str2 (length str2) where _spread str1 str2 0= [str1 ++ str2] _spread str1 str2 n= [(take n str2) ++ str1 ++ (drop n str2)] ++ (_spread str1 str2 (n-1)) f xs = map (spread xs) The number of outcomes seem to indicate that correctness of the algo .. however, I'd be very obliged if I could get some feedback on the Haskellness etc of this ... also any performance pointers ... Regards, Kashyap

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Jochem Berndsen

7 Jan 7 Jan

4:46 a.m.

CK Kashyap wrote:

...

I've written this piece of code to do permutations -

First off, this is a recurring topic. If you search the archives, you'll find some more topics about it.

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perms :: String -> [String]

Why this type? Since a String is just a list of Char, and you don't use the fact that you're actually using a list of characters. It's better to keep this function generic, and say perms :: [a] -> [[a]]

...

perms []= []

I don't think this is what you expect or want. I would consider a permutation of X to be a bijection X -> X. The number of bijections X -> X when X is empty, is in fact 1. So I think perms [] = [[]]

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perms (x:[])= [[x]]

I think you can drop this case if you do perms [] = [[]]. (Didn't prove it, though.)

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perms (x:xs)= concat (f [x] (perms xs))

A small stylistic issue: Normally I'd write a space before the '='.

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spread :: String -> String -> [String] -- interpolate first string at various positions of second string

This function becomes easier if you define it like spread :: a -> [a] -> [[a]] since you only use it in that way.

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spread str1 str2 = _spread str1 str2 (length str2) where _spread str1 str2 0= [str1 ++ str2] _spread str1 str2 n= [(take n str2) ++ str1 ++ (drop n str2)] ++ (_spread str1 str2 (n-1))

f xs = map (spread xs)

There is a better way to write spread, something like spread str1 xs = zipWith (\x y -> x ++ str1 ++ y) (inits xs) (tails xs) with inits and tails from Data.List. HTH, regards, Jochem -- Jochem Berndsen | jochem@functor.nl | jochem@牛在田里.com

Daniel Fischer

5:46 a.m.

Am Donnerstag 07 Januar 2010 09:37:42 schrieb CK Kashyap:

...

Hi All,

I've written this piece of code to do permutations -

perms :: String -> [String]

Nothing in the algorithm needs the list elements to be Chars, there's no type class involved, so it should be perms :: [a] -> [[a]]

...

perms []= []

This should actually be perms [] = [[]]

...

perms (x:[])= [[x]]

That is then superfluous.

...

perms (x:xs)= concat (f [x] (perms xs))

'f' is a good name for a function parameter, not for a top level binding. Why not perms (x:xs) = concat (map (spread [x]) (perms xs)) whcih you can reformulate as perms (x:xs) = concatMap (spread [x]) (perms xs) or, if you like Monads, since concatMap is just the bind operator of the []-monad, perms (x:xs) = perms xs >>= spread [x] Which can be written as a simple do-block: perms (x:xs) = do prm <- perms xs spread [x] prm or a list-comprehension perms (x:xs) = [permutation | tailPerm <- perms xs, permutation <- spread [x] tailPerm]

...

spread :: String -> String -> [String] -- interpolate first string at various positions of second string spread str1 str2 = _spread str1 str2 (length str2) where _spread str1 str2 0= [str1 ++ str2] _spread str1 str2 n= [(take n str2) ++ str1 ++ (drop n str2)] ++ (_spread str1 str2 (n-1))

import Data.List spread short long = zipWith (\a b -> a ++ short ++ b) (inits long) (tails long) If you only use spread for perms, you never interpolate anything but single element lists, so you might consider spread' :: a -> [a] -> [[a]] spread' x xs = zipWith (\a b -> a ++ x:b) (inits xs) (tails xs) But if you import Data.List, you could also say perms = permutations and be done with it :) (except if you 1. need the permutations in a particular order, which is different from the one Data.List.permutations generates, or 2. you need it to be as fast as possible - Data.List.permutations was written to also cope with infinite lists, so a few things that could speed up generation of permutations for short lists couldn't be used).

...

f xs = map (spread xs)

The number of outcomes seem to indicate that correctness of the algo ..

Apart from the case of empty input, it is correct.

...

however, I'd be very obliged if I could get some feedback on the Haskellness etc of this ... also any performance pointers ...

Re performance: I think the repeated (take k) and (drop k) in your spread are likely to be slower than using inits and tails, but it would need measuring the performance to be sure. I don't see anything that would automatically give bad performance. But there's the question of repeated elements. perms "aaaaabbbbb" spills out 3628800 permutations, but there are only 252 distinct permutations, each of them appearing 120^2 = 14400 times. If your input may contain repeated elements and you're 1. only interested in the distinct permutations (and 2.) or 2. don't care about the order in which the permutations are generated, distinctPerms :: Ord a => [a] -> [[a]] distinctPerms = foldr inserts [[]] . group . sort inserts :: [a] -> [[a]] -> [[a]] inserts xs yss = yss >>= (mingle xs) mingle :: [a] -> [a] -> [[a]] mingle xs [] = [xs] mingle [] ys = [ys] mingle xxs@(x:xs) yys@(y:ys) = [x:zs | zs <- mingle xs yys] ++ [y:zs | zs <- mingle xxs ys] generates the distinct permutations much faster if there are many repeated elements; if you want each distinct permutation repeated the appropriate number of times, the modification is easy.

...

Regards, Kashyap

Rafael Gustavo da Cunha Pereira Pinto

6:39 a.m.

Hi, Is there an entry in the haskell wiki for permutations? Since this is a recurring topic, as primes, shouldn't we create a topic for that in the wiki? Regards, Rafael On Thu, Jan 7, 2010 at 08:46, Daniel Fischer wrote:

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Am Donnerstag 07 Januar 2010 09:37:42 schrieb CK Kashyap:

...
Hi All,

...
...
I've written this piece of code to do permutations -

...
...
perms :: String -> [String]

Nothing in the algorithm needs the list elements to be Chars, there's no type class involved, so it should be

perms :: [a] -> [[a]]

...
perms []= []

This should actually be

perms [] = [[]]

...
perms (x:[])= [[x]]

That is then superfluous.

...
perms (x:xs)= concat (f [x] (perms xs))

...
'f' is a good name for a function parameter, not for a top level binding.

Why not

perms (x:xs) = concat (map (spread [x]) (perms xs))

whcih you can reformulate as

perms (x:xs) = concatMap (spread [x]) (perms xs)

or, if you like Monads, since concatMap is just the bind operator of the []-monad,

perms (x:xs) = perms xs >>= spread [x]

Which can be written as a simple do-block:

perms (x:xs) = do

prm <- perms xs

spread [x] prm

or a list-comprehension

perms (x:xs) = [permutation | tailPerm <- perms xs, permutation <- spread [x] tailPerm]

...
spread :: String -> String -> [String] -- interpolate first string at

...
various positions of second string spread str1 str2 = _spread str1 str2

...
(length str2)

...
where

...
_spread str1 str2 0= [str1 ++ str2]

...
_spread str1 str2 n= [(take n str2) ++ str1 ++ (drop n str2)] ++ (_spread

...
str1 str2 (n-1))

...
import Data.List

spread short long = zipWith (\a b -> a ++ short ++ b) (inits long) (tails long)

If you only use spread for perms, you never interpolate anything but single element lists, so you might consider

spread' :: a -> [a] -> [[a]]

spread' x xs = zipWith (\a b -> a ++ x:b) (inits xs) (tails xs)

But if you import Data.List, you could also say

perms = permutations

and be done with it :) (except if you 1. need the permutations in a particular order, which is different from the one Data.List.permutations generates, or 2. you need it to be as fast as possible - Data.List.permutations was written to also cope with infinite lists, so a few things that could speed up generation of permutations for short lists couldn't be used).

...
f xs = map (spread xs)

...
...
...
The number of outcomes seem to indicate that correctness of the algo ..

Apart from the case of empty input, it is correct.

...
however, I'd be very obliged if I could get some feedback on the

...
Haskellness etc of this ... also any performance pointers ...

Re performance:

I think the repeated (take k) and (drop k) in your spread are likely to be slower than using inits and tails, but it would need measuring the performance to be sure.

I don't see anything that would automatically give bad performance.

But there's the question of repeated elements.

perms "aaaaabbbbb"

spills out 3628800 permutations, but there are only 252 distinct permutations, each of them appearing 120^2 = 14400 times.

If your input may contain repeated elements and you're

1. only interested in the distinct permutations (and 2.) or

2. don't care about the order in which the permutations are generated,

distinctPerms :: Ord a => [a] -> [[a]]

distinctPerms = foldr inserts [[]] . group . sort

inserts :: [a] -> [[a]] -> [[a]]

inserts xs yss = yss >>= (mingle xs)

mingle :: [a] -> [a] -> [[a]]

mingle xs [] = [xs]

mingle [] ys = [ys]

mingle xxs@(x:xs) yys@(y:ys)

= [x:zs | zs <- mingle xs yys] ++ [y:zs | zs <- mingle xxs ys]

generates the distinct permutations much faster if there are many repeated elements;

if you want each distinct permutation repeated the appropriate number of times, the modification is easy.

...
...
...
Regards,

...
Kashyap

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

-- Rafael Gustavo da Cunha Pereira Pinto

CK Kashyap

11:48 p.m.

Thanks everyone, Thanks Daniel for this really detailed explanation - thank you very much. Regards, Kashyap

...

From: Daniel Fischer To: haskell-cafe@haskell.org Cc: CK Kashyap Sent: Thu, January 7, 2010 4:16:33 PM Subject: Re: [Haskell-cafe] Review request for my permutations implementation

...
Hi All,

I've written this piece of code to do permutations -

perms :: String -> [String] Nothing in the algorithm needs the list elements to be Chars, there's no type class involved, so it should be

...
perms []= [] This should actually be

Am Donnerstag 07 Januar 2010 09:37:42 schrieb CK Kashyap: perms :: [a] -> [[a]] perms [] = [[]]

...
perms (x:[])= [[x]] That is then superfluous. perms (x:xs)= concat (f [x] (perms xs))

'f' is a good name for a function parameter, not for a top level binding. Why not perms (x:xs) = concat (map (spread [x]) (perms xs)) whcih you can reformulate as perms (x:xs) = concatMap (spread [x]) (perms xs) or, if you like Monads, since concatMap is just the bind operator of the []-monad, perms (x:xs) = perms xs >>= spread [x] Which can be written as a simple do-block: perms (x:xs) = do prm <- perms xs spread [x] prm or a list-comprehension perms (x:xs) = [permutation | tailPerm <- perms xs, permutation <- spread [x] tailPerm]

...
spread :: String -> String -> [String] -- interpolate first string at various positions of second string spread str1 str2 = _spread str1 str2 (length str2) where _spread str1 str2 0= [str1 ++ str2] _spread str1 str2 n= [(take n str2) ++ str1 ++ (drop n str2)] ++ (_spread str1 str2 (n-1))

...
f xs = map (spread xs)

The number of outcomes seem to indicate that correctness of the algo .. Apart from the case of empty input, it is correct. however, I'd be very obliged if I could get some feedback on the Haskellness etc of this ... also any performance pointers ... Re performance: I think the repeated (take k) and (drop k) in your spread are likely to be slower than using inits and tails, but it would need measuring the performance to be sure. I don't see anything that would automatically give bad performance. But there's the question of repeated elements.

import Data.List spread short long = zipWith (\a b -> a ++ short ++ b) (inits long) (tails long) If you only use spread for perms, you never interpolate anything but single element lists, so you might consider spread' :: a -> [a] -> [[a]] spread' x xs = zipWith (\a b -> a ++ x:b) (inits xs) (tails xs) But if you import Data.List, you could also say perms = permutations and be done with it :) (except if you 1. need the permutations in a particular order, which is different from the one Data.List.permutations generates, or 2. you need it to be as fast as possible - Data.List.permutations was written to also cope with infinite lists, so a few things that could speed up generation of permutations for short lists couldn't be used). perms "aaaaabbbbb" spills out 3628800 permutations, but there are only 252 distinct permutations, each of them appearing 120^2 = 14400 times. If your input may contain repeated elements and you're 1. only interested in the distinct permutations (and 2.) or 2. don't care about the order in which the permutations are generated, distinctPerms :: Ord a => [a] -> [[a]] distinctPerms = foldr inserts [[]] . group . sort inserts :: [a] -> [[a]] -> [[a]] inserts xs yss = yss >>= (mingle xs) mingle :: [a] -> [a] -> [[a]] mingle xs [] = [xs] mingle [] ys = [ys] mingle xxs@(x:xs) yys@(y:ys) = [x:zs | zs <- mingle xs yys] ++ [y:zs | zs <- mingle xxs ys] generates the distinct permutations much faster if there are many repeated elements; if you want each distinct permutation repeated the appropriate number of times, the modification is easy.

...
Regards, Kashyap

Maciej Piechotka

8:04 a.m.

On Thu, 2010-01-07 at 00:37 -0800, CK Kashyap wrote:

...

Hi All,

I've written this piece of code to do permutations -

I assume that it's training piece not real code as in such I'd recommend:

...

import Data.List perms = permutations

...

perms :: String -> [String] perms []= []

As pointed out perms [] = [[]]. You can note that: length . perms == factorial

...

perms (x:[])= [[x]] perms (x:xs)= concat (f [x] (perms xs))

Don't call function f. I'd look into parameters - i.e. map f xs is = ... ok as f is some user function but f xs does not say what f is (except it is some function ;) ). Also pointed out concatMap:

...

perms (x:xs) = concatMap spread

...

spread :: String -> String -> [String] -- interpolate first string at various positions of second string spread str1 str2 = _spread str1 str2 (length str2)

I'd always be careful with (!!) and length. Both have O(n) complexity and I've seen code in which it shifted from O(n^3) to O(n^6) by uncareful usage (usually something like O(n) to O(n^2) per function).

...

where _spread str1 str2 0= [str1 ++ str2] _spread str1 str2 n= [(take n str2) ++ str1 ++ (drop n str2)] ++ (_spread str1 str2 (n-1))

Please note that the first list (here list of chars) have always length 1. It would be nice to indicate it by rewriting into:

...

spread :: a -> [a] -> [[a]] spread a b = _spread a b 0 where _spread a b 0 = [a:b] _spread a b n = ((take n b) ++ a:(drop n b)):(_spread a b (n-1)) perms (x:xs) = concatMap (spread x) (perms xs)

I took the liberty of rewriting [a] ++ b into a:b. In fact (:) is base constructor as the list [1,2,3,4] is syntax sugar for 1:2:3:4:[]. Hence it should be marginally more effective w/out optimalizations Further clarification is

...

spread a b = map (\n -> (take n b) ++ a:(drop n b)) Or: spread a b = zipWith (\i e -> i ++ a:e) (inits b) (tails b)

(\n -> something) is lambda function and is shorthand of ... f ... where f n = something

...

f xs = map (spread xs)

Why make separate function (not in where)

...

The number of outcomes seem to indicate that correctness of the algo .. however, I'd be very obliged if I could get some feedback on the Haskellness etc of this ... also any performance pointers ...

Minor difficulty with algorithm - it diverges for:

...

head $ head $ perms [1..]

...

Regards, Kashyap

Regards

Daniel Fischer

8:12 a.m.

New subject: Review request for my permutations implementation

Am Donnerstag 07 Januar 2010 14:04:20 schrieb Maciej Piechotka:

...

As pointed out perms [] = [[]]. You can note that: length . perms == factorial

Surely you meant genericLength . perms == factorial . (genericLength :: [a] -> Integer)

Maciej Piechotka

8:21 a.m.

New subject: Review request for my permutations implementation

On Thu, 2010-01-07 at 14:12 +0100, Daniel Fischer wrote:

...

Am Donnerstag 07 Januar 2010 14:04:20 schrieb Maciej Piechotka:

...
...
As pointed out perms [] = [[]]. You can note that:

...
length . perms == factorial

Surely you meant

genericLength . perms == factorial . (genericLength :: [a] -> Integer)

Ups. Sorry. About genericLength - I can use length as it is postcondition not in haskell (please note that ¬Eq ([a] -> Integer) ;)) Regards

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CK Kashyap
Daniel Fischer
Jochem Berndsen
Maciej Piechotka
Rafael Gustavo da Cunha Pereira Pinto