Why does 2(*i.)5 evaluate to 0 2 4 6 8 ?
It's clear that 2*i.5 does, but the () creates a hook and evaluating from right to left it seems we get
(*i.)5 == 0 5 10 15 20
and 2 won't act on that list -- so where am I going wrong ?
2 (* i.) 5, is a dyadic hook, which translates to 2 * (i. 5).
(* i.) 5, is a monadic hook, which translates to 5 * (i. 5).
The dyadic hook, x (u v) y is equivalent to x u (v y), which is the same as x u v y.
The monadic hook, (u v) y is equivalent to y u (v y), which is the same as y u v y.
http://www.jsoftware.com/jwiki/Vocabulary/hook
x (u v) y <--> x u (v y) for dyadic hooks.
A use might be to reshape y into the shape x as shown at
http://www.jsoftware.com/jwiki/Vocabulary/hook
[t=.i. 3 2
0 1
2 3
4 5
2 3 ($ ,) t
0 1 2
3 4 5
Related
Why does GHCi give incorrect answer below?
GHCi
λ> ((-20.24373193905347)^12)^2 - ((-20.24373193905347)^24)
4.503599627370496e15
Python3
>>> ((-20.24373193905347)**12)**2 - ((-20.24373193905347)**24)
0.0
UPDATE
I would implement Haskell's (^) function as follows.
powerXY :: Double -> Int -> Double
powerXY x 0 = 1
powerXY x y
| y < 0 = powerXY (1/x) (-y)
| otherwise =
let z = powerXY x (y `div` 2)
in if odd y then z*z*x else z*z
main = do
let x = -20.24373193905347
print $ powerXY (powerXY x 12) 2 - powerXY x 24 -- 0
print $ ((x^12)^2) - (x ^ 24) -- 4.503599627370496e15
Although my version doesn't appear any more correct than the one provided below by #WillemVanOnsem, it strangely gives the correct answer for this particular case at least.
Python is similar.
def pw(x, y):
if y < 0:
return pw(1/x, -y)
if y == 0:
return 1
z = pw(x, y//2)
if y % 2 == 1:
return z*z*x
else:
return z*z
# prints 0.0
print(pw(pw(-20.24373193905347, 12), 2) - pw(-20.24373193905347, 24))
Short answer: there is a difference between (^) :: (Num a, Integral b) => a -> b -> a and (**) :: Floating a => a -> a -> a.
The (^) function works only on integral exponents. It will normally make use of an iterative algorithm that will each time check if the power is divisible by two, and divide the power by two (and if non-divisible multiply the result with x). This thus means that for 12, it will perform a total of six multiplications. If a multiplication has a certain rounding-off error, that error can "explode". As we can see in the source code, the (^) function is implemented as:
(^) :: (Num a, Integral b) => a -> b -> a
x0 ^ y0 | y0 < 0 = errorWithoutStackTrace "Negative exponent"
| y0 == 0 = 1
| otherwise = f x0 y0
where -- f : x0 ^ y0 = x ^ y
f x y | even y = f (x * x) (y `quot` 2)
| y == 1 = x
| otherwise = g (x * x) (y `quot` 2) x -- See Note [Half of y - 1]
-- g : x0 ^ y0 = (x ^ y) * z
g x y z | even y = g (x * x) (y `quot` 2) z
| y == 1 = x * z
| otherwise = g (x * x) (y `quot` 2) (x * z) -- See Note [Half of y - 1]
The (**) function is, at least for Floats and Doubles implemented to work on the floating point unit. Indeed, if we take a look at the implementation of (**), we see:
instance Floating Float where
-- …
(**) x y = powerFloat x y
-- …
This thus redirect to the powerFloat# :: Float# -> Float# -> Float# function, which will, normally be linked to the corresponding FPU operation(s) by the compiler.
If we use (**) instead, we obtain zero as well for a 64-bit floating point unit:
Prelude> (a**12)**2 - a**24
0.0
We can for example implement the iterative algorithm in Python:
def pw(x0, y0):
if y0 < 0:
raise Error()
if y0 == 0:
return 1
return f(x0, y0)
def f(x, y):
if (y % 2 == 0):
return f(x*x, y//2)
if y == 1:
return x
return g(x*x, y // 2, x)
def g(x, y, z):
if (y % 2 == 0):
return g(x*x, y//2, z)
if y == 1:
return x*z
return g(x*x, y//2, x*z)
If we then perform the same operation, I get locally:
>>> pw(pw(-20.24373193905347, 12), 2) - pw(-20.24373193905347, 24)
4503599627370496.0
Which is the same value as what we get for (^) in GHCi.
I am trying to do a recursive function that should give the sum of all integers between and including its two arguments. For example,
sumFromTo 5 8 is 5 + 6 + 7 + 8 = 26. If the first argument is greater than the
second, the function should return 0.
This is what I got currently but I am a beginner and I don't think I did it right
sumFromTo :: Int -> Int -> Int
sumFromTo x y
| x == 1 = 1
| y == 1 = 1
| x > 1 && y > 1 = x + y sumFromTo (x - 1)
| otherwise = 0
Any help please?
You can use a list comprehension to do this very simply:
sumFromTo :: Int -> Int -> Int
sumFromTo x y = sum [x..y]
However, I'm not sure what you'd want to do if x == y.
In the code you've given, your recursive definition isn't correct syntax. You've only given sumFromTo one argument, when it needs two, and you appear to have missed a + between the y and the function.
λ> sumFromTo 8 5
0
λ> sumFromTo 5 8
26
λ> sumFromTo 8 8
8
λ>
[a..b] means a list with a step of 1 between a and b, so [1..4] is [1,2,3,4]
Thanks to Phylogenesis, here is a recursive definition:
sumFromTo :: Int -> Int -> Int
sumFromTo x y
| x > y = 0
| x == y = x
| otherwise = x + sumFromTo (x + 1) y
I think you make the problem too complex if you want to implement this with recursion. Basically there are two cases:
one where the lower bound is greater than the upper bound, in which case the sum is zero; and
one where the lower bound is less than or equal to the upper bound.
The first case (1) can be expressed by writing:
sumFromTo x y | x > y = 0
In the second case, the result is the lower bound plus the sum of the lowerbound plus one to the upper bound, so:
| otherwise = x + sumFromTo (x+1) y
and putting these together:
sumFromTo :: (Num a, Ord a) => a -> a -> a
sumFromTo x y | x > y = 0
| otherwise = x + sumFromTo (x+1) y
For example, how to write the function g=(x-y)/(x-z)?
I know how to write the function with 2 parameters.
One way is to use variable matching:
f =: 3 : 0
'x y z' =. y
(x-y)%(x-z)
)
f 1; 2; 3
0.5
f 1 2 3
0.5
f 1.5; 2; 0.5
_0.5
Another way is to treat your variables as an array v -> x y z and define your function as a series of array operations. For example:
multiply and add +/ 1 _1 0 * x y z,
multiply and add +/ 1 0 _1 * x y z,
divide %/
This can be written as:
g =: 3 :'%/ F (+/ . *) y'
where F is
1 _1 0
1 0 _1
:
g 1 2 3
0.5
g 1.5 2 0.5
_0.5
You can take this too far and write:
h =: 3 : '((0{y) - (1{y)) % ((0{y) - (2{y))'
but you probably shouldn't.
I was reading the Haskell Prelude and finding it pretty understandable, then I stumbled upon the exponention definition:
(^) :: (Num a, Integral b) => a -> b -> a
x ^ 0 = 1
x ^ n | n > 0 = f x (n-1) x
where f _ 0 y = y
f x n y = g x n where
g x n | even n = g (x*x) (n `quot` 2)
| otherwise = f x (n-1) (x*y)
_ ^ _ = error "Prelude.^: negative exponent"
I do not understand the need for two nested wheres.
What I understood so far:
(^) :: (Num a, Integral b) => a -> b -> a
The base must be a number and the exponent intege, ok.
x ^ 0 = 1
Base case, easy.
g x n | even n = g (x*x) (n `quot` 2)
| otherwise = f x (n-1) (x*y)
Exponention by squaring... kind of ... Why is the f helper needed? Why are f and g given single letter names? Is it just optimization, am I missing something obvious?
_ ^ _ = error "Prelude.^: negative exponent"
N > 0 was checked before, N is negative if we arrived here, so error.
My implementation would be a direct translation to code of:
Function exp-by-squaring(x, n )
if n < 0 then return exp-by-squaring(1 / x, - n );
else if n = 0 then return 1; else if n = 1 then return x ;
else if n is even then return exp-by-squaring(x * x, n / 2);
else if n is odd then return x * exp-by-squaring(x * x, (n - 1) / 2).
Pseudocode from wikipedia.
To illustrate what #dfeuer is saying, note that the way f is written it either:
f returns a value
or, f calls itself with new arguments
Hence f is tail recursive and therefore can easily be transformed into a loop.
On the other hand, consider this alternate implementation of exponentiation by squaring:
-- assume n >= 0
exp x 0 = 1
exp x n | even n = exp (x*x) (n `quot` 2)
| otherwise = x * exp x (n-1)
The problem here is that in the otherwise clause the last operation performed is a multiplication. So exp either:
returns 1
calls itself with new arguments
calls itself with some new arguments and multiplies the result by x.
exp is not tail recursive and therefore cannot by transformed into a loop.
f is indeed an optimization. The naive approach would be "top down", calculating x^(n `div` 2) and then squaring the result. The downside of this approach is that it builds a stack of intermediate computations. What f lets this implementation do is to first square x (a single multiplication) and then raise the result to the reduced exponent, tail recursively. The end result is that the function will likely operate entirely in machine registers. g seems to help avoid checking for the end of the loop when the exponent is even, but I'm not really sure if it's a good idea.
As far as I understand it exponentiation is solved by squaring as long as the exponent is even.
This leads to the answer why f is needed in case of an odd number - we use f to return the result in the case of g x 1, in every other odd case we use f to get back in the g-routine.
You can see it best I think if you look at an example:
x ^ n | n > 0 = f x (n-1) x
where f _ 0 y = y
f x n y = g x n
where g x n | even n = g (x*x) (n `quot` 2)
| otherwise = f x (n-1) (x*y)
2^6 = -- x = 2, n = 6, 6 > 0 thus we can use the definition
f 2 (6-1) 2 = f 2 5 2 -- (*)
= g 2 5 -- 5 is odd we are in the "otherwise" branch
= f 2 4 (2*2) -- note that the second '2' is still in scope from (*)
= f 2 4 (4) -- (**) for reasons of better readability evaluate the expressions, be aware that haskell is lazy and wouldn't do that
= g 2 4
= g (2*2) (4 `quot` 2) = g 4 2
= g (4*4) (2 `quot` 2) = g 16 1
= f 16 0 (16*4) -- note that the 4 comes from the line marked with (**)
= f 16 0 64 -- which is the base case for f
= 64
Now to your question of using single letter function names - that's the kind of thing you have to get used to it is a way most people in the community write. It has no effect on the compiler how you name your functions - as long as they start with a lower case letter.
As others noted, the function is written using tail-recursion for efficiency.
However, note that one could remove the innermost where while preserving tail-recursion as follows: instead of
x ^ n | n > 0 = f x (n-1) x
where f _ 0 y = y
f x n y = g x n
where g x n | even n = g (x*x) (n `quot` 2)
| otherwise = f x (n-1) (x*y)
we can use
x ^ n | n > 0 = f x (n-1) x
where f _ 0 y = y
f x n y | even n = f (x*x) (n `quot` 2) y
| otherwise = f x (n-1) (x*y)
which is also arguably more readable.
I have however no idea why the authors of the Prelude chose their variant.
My homework was to provide a function that computes 'x^y mod n' -for any n < (sqrt maxint32)
So I started by writing doing this:
modPow :: Int -> Int -> Int -> Int
modPow x y n = (x `mod` n) ^ (y `mod` n) `mod` n
Which seemed to work fine, for any number of n, although my next homework question involved using x^n mod n = x (Camichael numbers) and I could never get modPow to work.
So I made another modPow using pseudocode for mod exponentiation, -from wikipedia:
modPow2 :: Int -> Int -> Int -> Int
modPow2 x y n
= loopmod 1 1
where
loopmod count total = if count > y
then total
else loopmod (count+1) ((total*x) `mod` n)
Which now correctly produces the right answer for my next question, (x^n mod n = x) -for checking for Camichael numbers.
ALTHOUGH, modPow2 does not work for big numbers of 'y' (STACK-OVERFLOW!!)
How could I adjust modPow2 so it no longer gets a stackoverflow in the cases where y > 10,000 (but still less than sqrt of maxint 32 -which is around 46,000)
Or is there a fix on my original modPow so it works with x^n mod n = x? (I always do 560 561 561 as inputs and it gives me back 1 not 560 (561 is a carmichael number so should give 560 back)
Thanks alot.
Your formula for modPow is wrong, you can't just use y mod n as the exponent, it will lead to wrong results. For example:
Prelude> 2^10
1024
Prelude> 2^10 `mod` 10
4
Prelude> 2^(10 `mod` 10) `mod` 10
1
For a better modPow function you could use that x2n+1 = x2n ⋅ x and x2n = xn ⋅ xn and that for multiplication you actually can simply use the mod of the factors.
Where did you get your formula for modPow from?
(x ^ y) `mod` n = ((x `mod` n) ^ (y `mod` φ n)) `mod` n where φ is Euler's totient function.
This is probably because the argument total is computed lazily.
If you use GHC, you can make loopmod strict in total by placing a ! in frontof the argument, i.e.
loopmod count !total = ...
Another way would be to force evaluation of total like so: Replace the last line with
else if total == 0 then 0 else loopmod (count+1) ((total*x) `mod` n)
This does not change semantics (because 0*xis 0 anyway, so the reminder must be 0 also) and it forces hugs to evaluate total in every recursion.
If you are looking for implementation ( a^d mod n ) then
powM::Integer->Integer->Integer->Integer
powM a d n
| d == 0 = 1
| d == 1 = mod a n
| otherwise = mod q n where
p = powM ( mod ( a^2 ) n ) ( shiftR d 1 ) n
q = if (.&.) d 1 == 1 then mod ( a * p ) n else p