Haskell list comprehension returns empty? - list

So I'm new to learning Haskell (started on Saturday) and I read a few chapters from online books and I wrote a list comprehension to find the diameter of a circle given its circumference.
ghci> let circle = [(a) | a <- [1..10], 9 / pi == a] but it returns [] I was told this is because [1..10] only counts whole integers and not all the real numbers between 1 and 10.
I tried ghci> let circle = [(a) | a <- [1..10], round (pi * a) == 9] which returns [3.0] but I was wondering if there was a way to get a more precise answer.
Thanks in advance.

Welcome to Haskell and StackOverflow!
let circle = [(a) | a <- [1..10], 9 / pi == a]
Considering all the real numbers between 1 and 10 is an uncountably infinite set (hat tip to Georg Cantor for that one) I'd be surprised, to say the least, if you managed to make a list comprehension over them. :)
Your reasoning about why this list is empty is correct -- 9/pi will never equal a whole integer.
Other than that, there's no reason to use a list comprehension for this. You can do the same thing by simply dividing by pi:
let diam c = c / pi

This is not what list comprehensions are used for. To get a precise answer to 9 / pi, just calculate 9 / pi directly.
ghci> let diam = 9/pi

diameter c = c / pi
or as you get more advanced in Haskell:
diameter = (/ pi)
And then you can call:
Prelude> diameter 9
2.864788975654116
A list comprehension may be useful if you wanted the diameters of the circles with circumferences from 1 to 10, like so:
Prelude> [diameter x | x <- [1..10]]
[0.3183098861837907,0.6366197723675814,0.954929658551372,1.2732395447351628,1.5915494309189535,1.909859317102744,2.228169203286535,2.5464790894703255,2.864788975654116,3.183098861837907]

How about:
let circle = [ 9 / pi ]

Why don't you just find the diameter of a circle with a given circumference?
diameterOfCircleWithCircumference :: Floating a => a -> a
diameterOfCircleWithCircumference = (/ pi)
Then diameterOfCircleWithCircumference 9 is 2.8947...

Related

Backtraking with list Monad in Haskell

I'm trying to solve a decomposition problem with backtracking and list Monad in Haskell. Here is the problem statement: given a positive integer n, find all lists of consecutive integers (in range i..j) whose sum is equal to n.
I came out with the following solution which seems to work fine. Could someone suggest a better/more efficient implementation using list Monad and backtracking?
Any suggestions are welcome. Thanks in advance.
import Control.Monad
decompose :: Int -> [[Int]]
decompose n = concatMap (run n) [1 .. n - 1]
where
run target n = do
x <- [n]
guard $ x <= target
if x == target
then return [x]
else do
next <- run (target - n) (n + 1)
return $ x : next
test1 = decompose 10 == [[1,2,3,4]]
test2 = decompose 9 == [[2,3,4],[4,5]]
The sum of a range of numbers k .. l with k≤l is equal to (l×(l+1)-k×(k-1))/2. For example: 1 .. 4 is equal to (4×5-1×0)/2=(20-0)/2=10; and the sum of 4 .. 5 is (5×6-4×3)/2=(30-12)/2=9.
If we have a sum S and an offset k, we can thus find out if there is an l for which the sum holds with:
2×S = l×(l+1)-k×(k-1)
0=l2+l-2×S-k×(k-1)
we can thus solve this equation with:
l=(-1 + √(1+8×S+4×k×(k-1)))/2
If this is an integral number, then the sequence exists. For example for S=9 and k=4, we get:
l = (-1 + √(1+72+48))/2 = (-1 + 11)/2 = 10/2 = 5.
We can make use of some function, like the Babylonian method [wiki] to calculate integer square roots fast:
squareRoot :: Integral t => t -> t
squareRoot n
| n > 0 = babylon n
| n == 0 = 0
| n < 0 = error "Negative input"
where
babylon a | a > b = babylon b
| otherwise = a
where b = quot (a + quot n a) 2
We can check if the found root is indeed the exact square root with by squaring the root and see if we obtain back the original input.
So now that we have that, we can iterate over the lowerbound of the sequence, and look for the upperbound. If that exists, we return the sequence, otherwise, we try the next one:
decompose :: Int -> [[Int]]
decompose s = [ [k .. div (sq-1) 2 ]
| k <- [1 .. s]
, let r = 1+8*s+4*k*(k-1)
, let sq = squareRoot r
, r == sq*sq
]
We can thus for example obtain the items with:
Prelude> decompose 1
[[1]]
Prelude> decompose 2
[[2]]
Prelude> decompose 3
[[1,2],[3]]
Prelude> decompose 3
[[1,2],[3]]
Prelude> decompose 1
[[1]]
Prelude> decompose 2
[[2]]
Prelude> decompose 3
[[1,2],[3]]
Prelude> decompose 4
[[4]]
Prelude> decompose 5
[[2,3],[5]]
Prelude> decompose 6
[[1,2,3],[6]]
Prelude> decompose 7
[[3,4],[7]]
Prelude> decompose 8
[[8]]
Prelude> decompose 9
[[2,3,4],[4,5],[9]]
Prelude> decompose 10
[[1,2,3,4],[10]]
Prelude> decompose 11
[[5,6],[11]]
We can further constrain the ranges, for example specify that k<l, with:
decompose :: Int -> [[Int]]
decompose s = [ [k .. l ]
| k <- [1 .. div s 2 ]
, let r = 1+8*s+4*k*(k-1)
, let sq = squareRoot r
, r == sq*sq
, let l = div (sq-1) 2
, k < l
]
This then gives us:
Prelude> decompose 1
[]
Prelude> decompose 2
[]
Prelude> decompose 3
[[1,2]]
Prelude> decompose 4
[]
Prelude> decompose 5
[[2,3]]
Prelude> decompose 6
[[1,2,3]]
Prelude> decompose 7
[[3,4]]
Prelude> decompose 8
[]
Prelude> decompose 9
[[2,3,4],[4,5]]
Prelude> decompose 10
[[1,2,3,4]]
Prelude> decompose 11
[[5,6]]
NB This answer is slightly tangential since the question specifically calls for a direct backtracking solution in Haskell. Posting it in case there is some interest in other approaches to this problem, in particular using off-the-shelf SMT solvers.
These sorts of problems can be easily handled by off-the-shelf constraint solvers, and there are several libraries in Haskell to access them. Without going into too much detail, here's how one can code this using the SBV library (https://hackage.haskell.org/package/sbv):
import Data.SBV
decompose :: Integer -> IO AllSatResult
decompose n = allSat $ do
i <- sInteger "i"
j <- sInteger "j"
constrain $ 1 .<= i
constrain $ i .<= j
constrain $ j .< literal n
constrain $ literal n .== ((j * (j+1)) - ((i-1) * i)) `sDiv` 2
We simply express the constraints on i and j for the given n, using the summation formula. The rest is simply handled by the SMT solver, giving us all possible solutions. Here're a few tests:
*Main> decompose 9
Solution #1:
i = 4 :: Integer
j = 5 :: Integer
Solution #2:
i = 2 :: Integer
j = 4 :: Integer
Found 2 different solutions.
and
*Main> decompose 10
Solution #1:
i = 1 :: Integer
j = 4 :: Integer
This is the only solution.
While this may not provide much insight into how to solve the problem, it sure leverages existing technologies. Again, while this answer doesn't use the list-monad as asked, but hopefully it is of some interest when considering applications of SMT solvers in regular programming.

Haskell assigning value to a variable inside a comprehension

I'm having some issues with the following exercise:
I'm supposed to write a function that thats 3 numbers, x y n, being x and y the bottom and upper bounds of a list comprehension (respectively) and n being the number of partitions that comprehension will have.
E.g:
λ> partition 10 20 4
[10.0, 12.5, 15.0, 17.5, 20.0]
What I have done is the following:
partition :: Double -> Double -> Double -> [Double]
partition x y n = [a+b | b = (y-x) / n , a -> [x,b..y]]
I don't understand why i can't define the value of the b variable inside the comprehension, since when i try to run it i get the following error message:
parse error on input `='
NOTE: This is supposed to be a beginners exercise and this should have a simple resolution
You just need to use the let keyword:
partition x y n = [a+b | let b = (y-x)/n , a <- [x,x+b..y]]
Then you can use b as desired:
λ partition 10 20 4
[12.5,15.0,17.5,20.0,22.5]
What variables? Everything is immutable :)
What you want is a let expression:
partition x y n = [let b = (y-x)/n in a+b | a <- [x,b..y]]
Also note the direction of the arrow: it comes from the list expression to a, not the other way around.
It looks more natural to move the common sub-expression away from the comprehension:
partition x y n = let b = (y-x)/n in [a+b | a <- [x,b..y]]

How to calculate the sine function in Haskell?

Here is my problem: I need a Haskell function that computes an approximation of the sine of some number, using the associated Taylor serie ...
In C++ I wrote this:
double msin(double number, int counter = 0, double sum = 0)
{
// sin(x) = x - (x'3 / 3!) + (x'5 / 5!) - (x'7 / 7!) + (x'9 / 9!)
if (counter <= 20)
{
if (counter % 2 == 0)
sum += mpow(number, counter * 2 + 1) / mfak(counter * 2 + 1) ;
else
sum -= mpow(number, counter * 2 + 1) / mfak(counter * 2 + 1) ;
counter++;
sum = msin(number, counter, sum);
return sum;
}
return (sum* 180.0 / _PI);
}
Now I am trying to do it in Haskell and I have no idea how... For now I was trying something like this (it doesn't really work, but it is work in progress ;) ):
This works:
mfak number = if number < 2
then 1
else number *( mfak (number -1 ))
mpow number potenca = if potenca == 0
then 0
else if potenca == 1
then 1
else (number * (mpow number (potenca-1)))
This doesn't work:
msin :: Double -> Int -> Double -> Double
msin number counter sum = if counter <= 20
then if counter `mod` 2==0
then let sum = sum + (msin 1 (let counter = counter+1 in counter) sum) in sum
else let sum = sum + (msin 1 (let counter = counter+1 in counter) sum) in sum
else sum* 180.0 / 3.14
Updated....doesn't compile :/ "Couldn't match expected type Double' with actual type Int'"
msin :: Double -> Int -> Double -> Double
msin number counter sum = if counter <= 20
then if counter `mod` 2==0
then let sum' = sum + ((mpow number (counter*2+1))/(mfak counter*2+1)) in msin number (counter+1) sum'
else let sum' = sum - ((mpow number (counter*2+1))/(mfak counter*2+1)) in msin number (counter+1) sum'
else sum* 180.0 / 3.14
As you can see, the biggest problem is how to add something to "sum", increase "counter" and go in recursion again with these new values...
P.S. I am new to Haskell so try to explain your solution as much as you can please. I was reading some tutorials and that, but I can't find how to save the result of some expression into a value and then continue with other code after it... It just returns my value each time I try to do that, and I don't want that....
So thanks in advance for any help!
I would rework the algorithm a bit. First we can define the list of factorial inverses:
factorialInv :: [Double]
factorialInv = scanl (/) 1 [1..] -- 1/0! , 1/1! , 1/2! , 1/3! , ...
Then, we follow with the sine coefficients:
sineCoefficients :: [Double]
sineCoefficients = 0 : 1 : 0 : -1 : sineCoefficients
Then, given x, we multiply both the above lists with the powers of x, pointwise:
powerSeries :: [Double] -- ^ Coefficients
-> Double -- ^ Point x on which to compute the series
-> [Double] -- ^ Series terms
powerSeries cs x = zipWith3 (\a b c -> a * b * c) cs powers factorialInv
where powers = iterate (*x) 1 -- 1 , x , x^2 , x^3 , ...
Finally, we take the first 20 terms and sum them up.
sine :: Double -> Double
sine = sum . take 20 . powerSeries sineCoefficients
-- i.e., sine x = sum (take 20 (powerSeries sineCoefficients x))
The problem is expressions like let stevec = stevec+1 in stevec. Haskell is not an imperative language. This does not add one to stevec. Instead it defines stevec to be a number that is one more than itself. No such number exists, thus you will get an infinite loop or, if you are lucky, a crash.
Instead of
stevec++;
vsota = msin(stevilo, stevec, vsota);
You should use something like
let stevec' = stevec + 1
in msin stevilo stevec' vsota
or just
msin stevilo (stevec + 1) vsota
(There's also something here that I don't understand. You are going to need mpow and mfak. Where are they?)
As you can see the biggest problem is how to add something to "vsota",
In a functional language you would use recursion here - the variable vstota is implemented as a function parameter which is passed from call to call as a list is processed.
For example, to sum a list of numbers, we would write something like:
sum xs = go 0 xs
where go total [] = total
go total (x:xs) = go (total+x) xs
In an imperative language total would be a variable which gets updated. Here is is a function parameter which gets passed to the next recursive call to go.
In your case, I would first write a function which generates the terms of the power series:
sinusTerms n x = ... -- the first n terms of x - (x'3 / 3!) + (x'5 / 5!) - (x'7 / 7!) ...
and then use the sum function above:
sinus n x = sum (sinusTerms n x)
You may also use recursive lists definitions to get [x, x^3, x^5 ...] and [1, 1/3!, 1/5! ...] infinite sequences. When they are done, the rest is to multiply their items each by other and take the sum.
sinus count x = sum (take count $ zipWith (*) ifactorials xpowers)
where xpowers = x : map ((x*x)*) xpowers
ifactorials = 1 : zipWith (/) ifactorials [i*(i+1) | i <- [2, 4 .. ]]
Also, it would be better to define xpowers = iterate ((x*x)*) x, as it seems to be much more readable.
I’ve tried to follow your conventions as much as I could. For mfak and mpow, you should avoid using if as it is clearer to write them using pattern matching :
mfak :: Int -> Int
mfak 0 = 1
mfak 1 = 1
mfak n = n * mfak (n - 1)
mpow :: Double -> Int -> Double
mpow _ 0 = 1
mpow x 1 = x
mpow x p = x * mpow x (p - 1)
Before calculating the sinus, we create a list of coefficients [(sign, power, factorial)] :
x - (x^3 / 3!) + (x^5 / 5!) - (x^7 / 7!) + (x^9 / 9!)
→ [(1,1,1), (-1,3,6), (1,5,120), (-1,7,5040), (1,9,362880)]
The list is created infinite by a list comprehension. First we zip the lists [1,-1,1,-1,1,-1...] and [1,3,5,7,9,11...]. This gives us the list [(1,1), (-1,3), (1,5), (-1,7)...]. From this list, we create the final list [(1,1,1), (-1,3,6), (1,5,120), (-1,7,5040)...]:
sinCoeff :: [(Double, Int, Double)]
sinCoeff = [ (fromIntegral s, i, fromIntegral $ mfak i)
| (s, i) <- zip (cycle [1, -1]) [1,3..]]
(cycle repeats a list indefinitely, [1,3..] creates an infinite list which starts at 1 with a step of 2)
Finally, the msin function is near the definition. It also uses a list comprehension to achieve its goeal (note that I kept the * 180 / pi though I’m not sure it should be there. Haskell knows pi).
msin :: Int -> Double -> Double
msin n x = 180 * sum [ s * mpow x p / f | (s, p, f) <- take n sinCoeff] / pi
(take n sinCoeff returns the first n elements of a list)
You may try the previous code with the following :
main = do
print $ take 10 sinCoeff
print $ msin 5 0.5
print $ msin 10 0.5
The expression is of the form x*P(x2).
For maximal efficiency, the polynomial in x2 must be evaluated using the Horner rule rather than computing the powers of x2 separately.
The coefficient serie with the factorial values can be expressed recursively in Haskell, just like is commonly done for the Fibonacci series. Using the ghci interpreter as our testbed, we have:
$ ghci
GHCi, version 8.8.4: https://www.haskell.org/ghc/ :? for help
λ>
λ>
λ> nextCoeffs d c = c : (nextCoeffs (d+1) ((-c)/(fromIntegral $ (2*d+2)*(2*d+3))))
λ>
λ> allCoeffs = nextCoeffs 0 1.0
λ>
where d is the depth inside the serie and c the current coefficient.
Sanity check: the coefficient at depth 3 must be the inverse of 7!
λ>
λ> 1.0 /(allCoeffs !! 3)
-5040.0
λ>
The Horner rule can be rendered in Haskell thru the foldr1 :: (a -> a -> a) -> [a] -> a library function.
As is customary in Haskell, I take the liberty to put the term count as the leftmost argument because it is the one most likely to be held constant. This is for currying (partial evaluation) purposes.
So we have:
λ> :{
|λ> msin count x = let { s = x*x ; cs = take count allCoeffs ;
|λ> stepFn c acc = acc*s + c ; }
|λ> in x * (foldr1 stepFn cs)
|λ> :}
Sanity checks, taking 20 terms:
λ>
λ> pi
3.141592653589793
λ>
λ> msin 20 (pi/6)
0.49999999999999994
λ>
λ> msin 20 (pi/2)
1.0
λ>
Side note 1: final multiplication by 180 / π is only of interest for inverse trigonometric functions.
Side note 2: in practice, to get a reasonably fast convergence, one should reduce the input variable x into the [-π,+π] interval using the periodicity of the sine function.

How do you make a list in Haskell using two numbers?

Here is my problem: Declare type and define a function that takes two numbers m and n as input and returns a list containing the doubled values of all odd integers between m and n. For instance, fun 2 11 would return [6, 10, 14, 18, 22].
I don't know how I can take the two number 2 and 11 and make it into a list [2..11]. Does anyone know how to do this?
Use sequence generation (range syntax):
Prelude> [2 .. 11]
[2,3,4,5,6,7,8,9,10,11]
Works for symbolic values, too:
Prelude> let [m,n] = [2,11]
Prelude> [m .. n]
[2,3,4,5,6,7,8,9,10,11]
Didn't work with Haskell for almost two years, so correct me if I'm wrong and it doesn't work:
getDoubledOdd :: Int -> Int -> [Int]
getDoubledOdd m n = map (2*) $ filter odd [m..n]
A combination of list comprehension and range would be the most standard way to do it.
[ 2*x | x <- [2..11], odd x ]
The code basically says "let x loop from 2 to 11 (x <- [2..11]), and if x is odd (odd x), put 2*x into the list that will be returned".
Hope that explains.

Recursion using lists - Haskell

I am trying to write a recursive function that will take a list containing a list of integers as an input and return a tuple of type ([Int],Int).
([Int],Int)
This is for a "board game" where you are supplied with a board:
[[5,4,3,8,6],
[0,2,1,0,7],
[0,1,9,4,3],
[2,3,4,0,9]]
This would be a board with 4 rows and 5 columns. The numbers inside the list are "coin values".
The objective of this board game would be to go from the top of the list to the bottom collecting the coins. You are able to start in any position from the top row and to move down, you can go straight down, or diagonally to left or right. You would want the path that will give you the largest total coin values.
I've created a first function where you input a list of paths [([Int],Int)] and it returns the path ([Int],Int) with maximum coin value.
Now I need to create a function to actually generate the list of paths that I will input into the first function.
I know that I will have to use recursion.
I will input the board (like one above) and a starting column.
I will have to take the column number and then create a list of all possible paths.
If I start with a column number, my next possible steps are positions (in the next row)- same column number, column num -1 and column num +1. I would need to recursively call this until I reach the bottom.
How would I be able to store these path steps as I go and then store the final - list of all possible paths?
([Int],Int) - [Int] is the position in list / column numbers or the rows and the Int is the coin value.
I'm new to haskell and while I understand what I have to do, it's really difficult to write the code.
You don't "store" intermediate values in some variable in idiomatic functional code. Rather, you keep them as an accumulating parameter which you pass along using a function such as foldr.
http://hackage.haskell.org/packages/archive/base/latest/doc/html/Prelude.html#v:foldr
I guess I am now in a position to (easily) adapt my answer for another question to this one. I listed the allowed index combinations and mapped the board to them. (pat's comment helped me improve index_combinations)
*Main> :load "new1.hs"
[1 of 1] Compiling Main ( new1.hs, interpreted )
Ok, modules loaded: Main.
*Main> result
([8,7,4,9],28)
*Main> path
[3,4,3,4]
import Data.List
import Data.Ord
import Data.Maybe
r = [[5,4,3,8,6],
[0,2,1,0,7],
[0,1,9,4,3],
[2,3,4,0,9]]
r1 = r !! 0
r2 = r !! 1
r3 = r !! 2
r4 = r !! 3
index_combinations =
[[a,b,c,d] | a <- [0..4], b <- [max 0 (a-1)..min 4 (a+1)],
c <- [max 0 (b-1)..min 4 (b+1)], d <- [max 0 (c-1)..min 4 (c+1)]]
mapR xs = [r1 !! (xs !! 0), r2 !! (xs !! 1),
r3 !! (xs !! 2), r4 !! (xs !! 3)]
r_combinations = map mapR index_combinations
r_combinations_summed = zip r_combinations $ map (foldr (+) 0) r_combinations
result = maximumBy (comparing snd) r_combinations_summed
path = index_combinations !! fromJust (elemIndex result r_combinations_summed)
If you're interested in using my package grid (userguide)
here as an example to get you started.
(And if you don't want to use it, you may find some of the
source code helpful.)
Create a grid with 4 rows and 5 columns.
λ> :m + Math.Geometry.Grid
λ> let g = rectSquareGrid 4 5
λ> indices g
[(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3),(2,0),(2,1),(2,2),(2,3),(3,0),(3,1),(3,2),(3,3),(4,0),(4,1),(4,2),(4,3)]
We want to be able to map "coin values" to grid positions, so we'll
create a GridMap.
λ> :m + Math.Geometry.GridMap
λ> let m = lazyGridMap g [5,4,3,8,6,0,2,1,0,7,0,1,9,4,3,2,3,4,0,9]
λ> m
lazyGridMap (rectSquareGrid 4 5) [5,4,3,8,6,0,2,1,0,7,0,1,9,4,3,2,3,4,0,9]
λ> toList m
[((0,0),5),((0,1),4),((0,2),3),((0,3),8),((1,0),6),((1,1),0),((1,2),2),((1,3),1),((2,0),0),((2,1),7),((2,2),0),((2,3),1),((3,0),9),((3,1),4),((3,2),3),((3,3),2),((4,0),3),((4,1),4),((4,2),0),((4,3),9)]
We can find out the neighbours of any cell in the grid,
but for your application, we run into a bit of a problem: my
RectSquareGrid type doesn't allow diagonal moves.
λ> neighbours (1,2) m
[(0,2),(1,3),(2,2),(1,1)]
Now, I'd be happy to create a new type of Grid that would meet your
needs. Alternatively, you could write your own function
which would include diagonal neighbours:
λ> let neighbours2 (x, y) g = filter (`inGrid` g) [(x-1,y-1), (x-1,y), (x-1,y+1), (x,y-1), (x,y+1), (x+1,y-1), (x+1,y), (x+1,y+1)]
λ> neighbours2 (1,2) m
[(0,1),(0,2),(0,3),(1,1),(1,3),(2,1),(2,2),(2,3)]
But you're only interested in allowing downward moves, either straight down or diagonal, so here's a more useful function:
λ> let allowedMoves (x, y) g = filter (`inGrid` g) [(x+1,y-1), (x+1,y), (x+1,y+1)]
λ> allowedMoves (1,2) m
[(2,1),(2,2),(2,3)]
So now we can write a function that gives you all possible paths from a given index to the bottom row of the grid.
allPathsFrom a g | fst a == fst (size g) = [[a]]
| otherwise = Prelude.map (a:) xs
where xs = concatMap (\x -> allPathsFrom x g) ys
ys = allowedMoves a g
For example:
λ> allPathsFrom (0,1) m
[[(0,1),(1,0),(2,0),(3,0),(4,0)],[(0,1),(1,0),(2,0),(3,0),(4,1)],[(0,1),(1,0),(2,0),(3,1),(4,0)],[(0,1),(1,0),(2,0),(3,1),(4,1)],[(0,1),(1,0),(2,0),(3,1),(4,2)],[(0,1),(1,0),(2,1),(3,0),(4,0)],[(0,1),(1,0),(2,1),(3,0),(4,1)],[(0,1),(1,0),(2,1),(3,1),(4,0)],[(0,1),(1,0),(2,1),(3,1),(4,1)],[(0,1),(1,0),(2,1),(3,1),(4,2)],[(0,1),(1,0),(2,1),(3,2),(4,1)],[(0,1),(1,0),(2,1),(3,2),(4,2)],[(0,1),(1,0),(2,1),(3,2),(4,3)],[(0,1),(1,1),(2,0),(3,0),(4,0)],[(0,1),(1,1),(2,0),(3,0),(4,1)],[(0,1),(1,1),(2,0),(3,1),(4,0)],[(0,1),(1,1),(2,0),(3,1),(4,1)],[(0,1),(1,1),(2,0),(3,1),(4,2)],[(0,1),(1,1),(2,1),(3,0),(4,0)],[(0,1),(1,1),(2,1),(3,0),(4,1)],[(0,1),(1,1),(2,1),(3,1),(4,0)],[(0,1),(1,1),(2,1),(3,1),(4,1)],[(0,1),(1,1),(2,1),(3,1),(4,2)],[(0,1),(1,1),(2,1),(3,2),(4,1)],[(0,1),(1,1),(2,1),(3,2),(4,2)],[(0,1),(1,1),(2,1),(3,2),(4,3)],[(0,1),(1,1),(2,2),(3,1),(4,0)],[(0,1),(1,1),(2,2),(3,1),(4,1)],[(0,1),(1,1),(2,2),(3,1),(4,2)],[(0,1),(1,1),(2,2),(3,2),(4,1)],[(0,1),(1,1),(2,2),(3,2),(4,2)],[(0,1),(1,1),(2,2),(3,2),(4,3)],[(0,1),(1,1),(2,2),(3,3),(4,2)],[(0,1),(1,1),(2,2),(3,3),(4,3)],[(0,1),(1,2),(2,1),(3,0),(4,0)],[(0,1),(1,2),(2,1),(3,0),(4,1)],[(0,1),(1,2),(2,1),(3,1),(4,0)],[(0,1),(1,2),(2,1),(3,1),(4,1)],[(0,1),(1,2),(2,1),(3,1),(4,2)],[(0,1),(1,2),(2,1),(3,2),(4,1)],[(0,1),(1,2),(2,1),(3,2),(4,2)],[(0,1),(1,2),(2,1),(3,2),(4,3)],[(0,1),(1,2),(2,2),(3,1),(4,0)],[(0,1),(1,2),(2,2),(3,1),(4,1)],[(0,1),(1,2),(2,2),(3,1),(4,2)],[(0,1),(1,2),(2,2),(3,2),(4,1)],[(0,1),(1,2),(2,2),(3,2),(4,2)],[(0,1),(1,2),(2,2),(3,2),(4,3)],[(0,1),(1,2),(2,2),(3,3),(4,2)],[(0,1),(1,2),(2,2),(3,3),(4,3)],[(0,1),(1,2),(2,3),(3,2),(4,1)],[(0,1),(1,2),(2,3),(3,2),(4,2)],[(0,1),(1,2),(2,3),(3,2),(4,3)],[(0,1),(1,2),(2,3),(3,3),(4,2)],[(0,1),(1,2),(2,3),(3,3),(4,3)]]
Note that since GridMaps are also Grids, we can invoke all of the above functions on m or g.
λ> allPathsFrom (0,1) m
Let me know (amy at nualeargais dot ie) if you would like me to add
a grid allowing diagonal moves to my grid package.