I have to write a program to solve quadratics, returning a complex number result.
I've gotten so far, with defining a complex number, declaring it to be part of num, so +,- and * - ing can take place.
I've also defined a data type for a quadratic equation, but im now stuck with the actual solving of the quadratic. My math is quite poor, so any help would be greatly appreciated...
data Complex = C {
re :: Float,
im :: Float
} deriving Eq
-- Display complex numbers in the normal way
instance Show Complex where
show (C r i)
| i == 0 = show r
| r == 0 = show i++"i"
| r < 0 && i < 0 = show r ++ " - "++ show (C 0 (i*(-1)))
| r < 0 && i > 0 = show r ++ " + "++ show (C 0 i)
| r > 0 && i < 0 = show r ++ " - "++ show (C 0 (i*(-1)))
| r > 0 && i > 0 = show r ++ " + "++ show (C 0 i)
-- Define algebraic operations on complex numbers
instance Num Complex where
fromInteger n = C (fromInteger n) 0 -- tech reasons
(C a b) + (C x y) = C (a+x) (b+y)
(C a b) * (C x y) = C (a*x - b*y) (b*x + b*y)
negate (C a b) = C (-a) (-b)
instance Fractional Complex where
fromRational r = C (fromRational r) 0 -- tech reasons
recip (C a b) = C (a/((a^2)+(b^2))) (b/((a^2)+(b^2)))
root :: Complex -> Complex
root (C x y)
| y == 0 && x == 0 = C 0 0
| y == 0 && x > 0 = C (sqrt ( ( x + sqrt ( (x^2) + 0 ) ) / 2 ) ) 0
| otherwise = C (sqrt ( ( x + sqrt ( (x^2) + (y^2) ) ) / 2 ) ) ((y/(2*(sqrt ( ( x + sqrt ( (x^2) + (y^2) ) ) / 2 ) ) ) ) )
-- quadratic polynomial : a.x^2 + b.x + c
data Quad = Q {
aCoeff, bCoeff, cCoeff :: Complex
} deriving Eq
instance Show Quad where
show (Q a b c) = show a ++ "x^2 + " ++ show b ++ "x + " ++ show c
solve :: Quad -> (Complex, Complex)
solve (Q a b c) = STUCK!
EDIT: I seem to have missed out the whole point of using my own complex number datatype is to learn about custom datatypes. I'm well aware that i could use complex.data. Any help that could be given using my solution so far would be greatly appreciated.\
EDIT 2: It seems that my initial question was worded horribly. I'm aware that the quadratic formula will return both (or just the one) root to me. Where I am having trouble is returning these roots as a (complex, complex) tuple with the code above.
I'm well aware that I could use the built in quadratic functions as have been displayed below, but this is not the exercise. The idea behind the exercise, and creating ones own complex number data type, is to learn about custom data types.
Like newacct said, it's just the quadratic equation:
(-b +- sqrt(b^2 - 4ac)) / 2a
module QuadraticSolver where
import Data.Complex
data Quadratic a = Quadratic a a a deriving (Show, Eq)
roots :: (RealFloat a) => Quadratic a -> [ Complex a ]
roots (Quadratic a b c) =
if discriminant == 0
then [ numer / denom ]
else [ (numer + root_discriminant) / denom,
(numer - root_discriminant) / denom ]
where discriminant = (b*b - 4*a*c)
root_discriminant = if (discriminant < 0)
then 0 :+ (sqrt $ -discriminant)
else (sqrt discriminant) :+ 0
denom = 2*a :+ 0
numer = (negate b) :+ 0
in practice:
ghci> :l QuadraticSolver
Ok, modules loaded: QuadraticSolver.
ghci> roots (Quadratic 1 2 1)
[(-1.0) :+ 0.0]
ghci> roots (Quadratic 1 0 1)
[0.0 :+ 1.0,(-0.0) :+ (-1.0)]
And adapting to use your terms:
solve :: Quad -> (Complex, Complex)
solve (Q a b c) = ( sol (+), sol (-) )
where sol op = (op (negate b) $ root $ b*b - 4*a*c) / (2 * a)
Although I haven't tested that code
Since Haskell's sqrt can also handle complex numbers, rampion's solution can even be further simplified:
import Data.Complex
-- roots for quadratic equations with complex coefficients
croots :: (RealFloat a) =>
(Complex a) -> (Complex a) -> (Complex a) -> [Complex a]
croots a b c
| disc == 0 = [solution (+)]
| otherwise = [solution (+), solution (-)]
where disc = b*b - 4*a*c
solution plmi = plmi (-b) (sqrt disc) / (2*a)
-- roots for quadratic equations with real coefficients
roots :: (RealFloat a) => a -> a -> a -> [Complex a]
roots a b c = croots (a :+ 0) (b :+ 0) (c :+ 0)
You can also use this croots function with your own datatype, if you change the types to fit your implementation (and call your root function instead of sqrt).
Related
I'm teaching myself OCaml, and the main resources I'm using for practice are some problem sets Cornell has made available from their 3110 class. One of the problems is to write a function to reverse an int (i.e: 1234 -> 4321, -1234 -> -4321, 2 -> 2, -10 -> -1 etc).
I have a working solution, but I'm concerned that it isn't exactly idiomatic OCaml:
let rev_int (i : int) : int =
let rec power cnt value =
if value / 10 = 0 then cnt
else power (10 * cnt) (value/10) in
let rec aux pow temp value =
if value <> 0 then aux (pow/10) (temp + (value mod 10 * pow)) (value / 10)
else temp in
aux (power 1 i) 0 i
It works properly in all cases as far as I can tell, but it just seems seriously "un-OCaml" to me, particularly because I'm running through the length of the int twice with two inner-functions. So I'm just wondering whether there's a more "OCaml" way to do this.
I would say, that the following is idiomatic enough.
(* [rev x] returns such value [y] that its decimal representation
is a reverse of decimal representation of [x], e.g.,
[rev 12345 = 54321] *)
let rev n =
let rec loop acc n =
if n = 0 then acc
else loop (acc * 10 + n mod 10) (n / 10) in
loop 0 n
But as Jeffrey said in a comment, your solution is quite idiomatic, although not the nicest one.
Btw, my own style, would be to write like this:
let rev n =
let rec loop acc = function
| 0 -> acc
| n -> loop (acc * 10 + n mod 10) (n / 10) in
loop 0 n
As I prefer pattern matching to if/then/else. But this is a matter of mine personal taste.
I can propose you some way of doing it:
let decompose_int i =
let r = i / 10 in
i - (r * 10) , r
This function allows me to decompose the integer as if I had a list.
For instance 1234 is decomposed into 4 and 123.
Then we reverse it.
let rec rev_int i = match decompose_int i with
| x , 0 -> 10 , x
| h , t ->
let (m,r) = rev_int t in
(10 * m, h * m + r)
The idea here is to return 10, 100, 1000... and so on to know where to place the last digit.
What I wanted to do here is to treat them as I would treat lists, decompose_int being a List.hd and List.tl equivalent.
Is it possible to have nested if without else statements. I wrote the following useless program to demonstrate nested ifs. How do I fix this so it's correct in terms of syntax. lines 5 and 6 gives errors.
let rec move_helper b sz r = match b with
[] -> r
|(h :: t) ->
if h = 0 then
if h - 1 = sz then h - 1 ::r
if h + 1 = sz then h + 1 ::r
else move_helper t sz r
;;
let move_pos b =
move_helper b 3 r
;;
let g = move_pos [0;8;7;6;5;4;3;2;1]
You can't have if without else unless the result of the expression is of type unit. This isn't the case for your code, so it's not possible.
Here's an example where the result is unit:
let f x =
if x land 1 <> 0 then print_string "1";
if x land 2 <> 0 then print_string "2";
if x land 4 <> 0 then print_string "4"
You must understand that if ... then is an expression like any other. If no else is present, it must be understood as if ... then ... else () and thus has type unit. To emphasize the fact that it is an expression, suppose you have two functions f and g of type, say, int → int. You can write
(if test then f else g) 1
You must also understand that x :: r does not change r at all, it constructs a new list putting x in front of r (the tail of this list is shared with the list r). In your case, the logic is not clear: what is the result when h=0 but the two if fail?
let rec move_helper b sz r = match b with
| [] -> r
| h :: t ->
if h = 0 then
if h - 1 = sz then (h - 1) :: r
else if h + 1 = sz then (h + 1) :: r
else (* What do you want to return here? *)
else move_helper t sz r
When you have a if, always put an else. Because when you don't put an else, Java will not know if the case is true or false.
I found an algorithm to do trilateration in 2-D this link. But the formulas are too complicated. What is happening here? Can you please break it down to terms like dot product, cross product, distance etc?
Let P be the unknown point. (Bold for 2D vectors.)
Write the implicit equations of circles 1 and 2:
(P - P1)² = d1²
(P - P2)² = d2²
Substract memberwise and rearrange:
2.(P2 - P1).P = d1² - d2² + P2² - P1²
Similarly with circles 1 and 3:
2.(P3 - P1).P = d1² - d3² + P3² - P1²
Looking closely, you will notice that this forms a system of two linear equations in two unknowns:
2.(X2 - X1).X + 2.(Y2 - Y1).Y = d1² - d2² + P2² - P1²
2.(X3 - X1).X + 2.(Y3 - Y1).Y = d1² - d3² + P3² - P1²
Use Cramer's rule, or if you insist on using vector calculus, work it out as follows.
Rewrite the system as:
A.P = a
B.P = b
Compute vectors perpendicular to A and B in the xy plane, using cross products A' = A /\ 1z and B' = B /\ 1z, and express P as a linear combination of these:
P = u . A' + v . B'
Performing a dot product with A and B gives, after simplification:
A.P = a = v . A.B'
B.P = b = u . B.A'
Note that A.B' = A.(B /\ 1z) = 1z.(A /\ B) = -1z.(B /\ A) = -B.(A /\ 1z) = -B.A' (mixed product).
All in all:
P = [ (- b.A + a.B) /\ 1z ] / [ 1z.(A /\ B) ]
(which is a rewrite of Cramer's result.)
So I have a bit of code*, that when taking three points, is supposed to return a direction. I've written this solution, but every time I try to run it, it causes the GHCi to freeze, so I'm wondering what I'm doing wrong. Here's the code:
--chapter 3 question 9
data Point x y = Point x y deriving (Eq, Show)
data Vector x y = Vector x y deriving (Eq, Show)
sub (Point x y) (Point a b) = (Vector (x-a) (y-b))
dot (Vector x y) (Vector a b) = (x*a)+(y*b)
perp (Vector x y) = (Vector (-y) x)
mag (Vector x y) = sqrt (dot v v) where v = (Vector x y)
data Direction = LeftTurn | RightTurn | Straight | Reverse | Stop | Undefined
deriving (Eq, Show)
getDirection (Point a b) (Point c d) (Point e f)
| a/=c && b/=d && c==e && d==f = Stop
| a==c && b==d || c==e && d==f || e==a && f==b = Undefined
| d > 0 = LeftTurn
| d < 0 = RightTurn
| otherwise = Straight
where d = dot (sub p1 p0) (perp (sub p2 p1))
where p0 = (Point a b)
p1 = (Point c d)
p2 = (Point e f)
There's no recursion that I can see, so I don't understand why it's behaving this way. So far the Haskell compiler has been very vocal about telling me when I'm doing something dumb, but this compiles just fine.
*This is Question 9 from Chapter 3 of "Real World Haskell" in case you're wondering.
You're binding the name twice. First in the pattern Point c d than in the where clause.
Thus if you're trying to access the d bound by the pattern, you're actually referring to the d from the where clause recursively.
So I have finished creating my own complex number data type in haskell.
I've also, thanks to another question on here, got a function that will solve a quadratic equation.
The only problem now is that the code generates a parsing error in hugs, when trying to solve a quadratic with complex roots.
i.e. In hugs...
Main> solve (Q 1 2 1)
(-1.0,-1.0)
Main> solve (Q 1 2 0)
(0.0,-2.0)
Main> solve (Q 1 2 2)
(
Program error: pattern match failure: v1618_v1655 (C -1.#IND -1.#IND)
It looks to my like its a problem after the square-root has been applied, but I'm really not sure. Any help trying to pick up what is going wrong or any indications as to what this error means would be brilliant.
Thanks,
Thomas
The Code:
-- A complex number z = (re +im.i) is represented as a pair of Floats
data Complex = C {
re :: Float,
im :: Float
} deriving Eq
-- Display complex numbers in the normal way
instance Show Complex where
show (C r i)
| i == 0 = show r
| r == 0 = show i++"i"
| r < 0 && i < 0 = show r ++ " - "++ show (C 0 (i*(-1)))
| r < 0 && i > 0 = show r ++ " + "++ show (C 0 i)
| r > 0 && i < 0 = show r ++ " - "++ show (C 0 (i*(-1)))
| r > 0 && i > 0 = show r ++ " + "++ show (C 0 i)
-- Define algebraic operations on complex numbers
instance Num Complex where
fromInteger n = C (fromInteger n) 0 -- tech reasons
(C a b) + (C x y) = C (a+x) (b+y)
(C a b) * (C x y) = C (a*x - b*y) (b*x + b*y)
negate (C a b) = C (-a) (-b)
instance Fractional Complex where
fromRational r = C (fromRational r) 0 -- tech reasons
recip (C a b) = C (a/((a^2)+(b^2))) (b/((a^2)+(b^2)))
root :: Complex -> Complex
root (C x y)
| y == 0 && x == 0 = C 0 0
| y == 0 && x > 0 = C (sqrt ( ( x + sqrt ( (x^2) + 0 ) ) / 2 ) ) 0
| otherwise = C (sqrt ( ( x + sqrt ( (x^2) + (y^2) ) ) / 2 ) ) ((y/(2*(sqrt ( ( x + sqrt ( (x^2) + (y^2) ) ) / 2 ) ) ) ) )
-- quadratic polynomial : a.x^2 + b.x + c
data Quad = Q {
aCoeff, bCoeff, cCoeff :: Complex
} deriving Eq
instance Show Quad where
show (Q a b c) = show a ++ "x^2 + " ++ show b ++ "x + " ++ show c
solve :: Quad -> (Complex, Complex)
solve (Q a b c) = ( sol (+), sol (-) )
where sol op = (op (negate b) $ root $ b*b - 4*a*c) / (2 * a)
Your numbers seem denormalized in your error :
(C -1.#IND -1.#IND)
In this case, you can't assume that any comparison on float are valid anymore. This is in the definition of floating point numbers. Then your definition of show
show (C r i)
| i == 0 = show r
| r == 0 = show i++"i"
| r < 0 && i < 0 = show r ++ " - "++ show (C 0 (i*(-1)))
| r < 0 && i > 0 = show r ++ " + "++ show (C 0 i)
| r > 0 && i < 0 = show r ++ " - "++ show (C 0 (i*(-1)))
| r > 0 && i > 0 = show r ++ " + "++ show (C 0 i)
leave opportunity for a pattern failure, because of denormalized numbers. You can add the following condition
| otherwise = show r ++ "i" ++ show i"
Now for the why is it like that, when you evaluate
b * b - 4 * a * c
with Q 1 2 2, you obtain -4, and then in root, you fall in your last case, and in the second equation :
y
-----------------------------
________________
/ _______
/ / 2 2
/ x + \/ x + y
2 * \ / ----------------
\/ 2
-4 + sqrt( (-4) ^2) == 0, from there, you're doomed, division by 0, followed by a "NaN" (not a number), screwing everything else
Dave hit the nail on the head.
With the original code in GHCi, I get:
*Main> solve (Q 1 2 2)
(*** Exception: c.hs:(11,4)-(17,63): Non-exhaustive patterns in function show
If we update the show block:
instance Show Complex where
show (C r i)
| i == 0 = show r
| r == 0 = show i++"i"
| r < 0 && i < 0 = show r ++ " - "++ show (C 0 (i*(-1)))
| r < 0 && i > 0 = show r ++ " + "++ show (C 0 i)
| r > 0 && i < 0 = show r ++ " - "++ show (C 0 (i*(-1)))
| r > 0 && i > 0 = show r ++ " + "++ show (C 0 i)
| otherwise = "???(" ++ show r ++ " " ++ show i ++ ")"
then we get this information in GHCi:
*Main> :l c.hs
[1 of 1] Compiling Main ( c.hs, interpreted )
c.hs:22:0:
Warning: No explicit method nor default method for `abs'
In the instance declaration for `Num Complex'
c.hs:22:0:
Warning: No explicit method nor default method for `signum'
In the instance declaration for `Num Complex'
Ok, modules loaded: Main.
*Main> solve (Q 1 2 2)
(???(NaN NaN),???(NaN NaN))
I was "born and raised" on GHCi, so I don't know exactly how Hugs compares in verbosity of warnings and errors; but it looks like GHCi is a clear winner in telling you what went wrong.
Off the top of my head: It could be a problem with your definition of show for Complex.
I notice you don't have default case like this:
| otherwise = ...
Therefore if your conditions with r and i are non exhaustive you'll get a pattern match failure.