*Main> let triangles = [ (a,b,c) | c <- [1..10], b <- [1..10], a <- [1..10] ]Now create a list of right angle triangles by
*Main> let rightAngleTriangle = [ (a,b,c)|(a,b,c) <- triangles , a^2 == b ^2 + c ^ 2]Here variable a is hypotenuse. b and c are representing other 2 sides. Now call the function.
*Main> rightAngleTriangle [(5,4,3),(5,3,4),(10,8,6),(10,6,8)]The above result contains duplicate values. For a single hypotenuse value it printed twice. 2 Methods:
Method 1:
*Main> let rightAngleTriangle = [ (a,b,c)|(a,b,c) <- triangles , a > b, b > c, a^2 == b ^2 + c ^ 2]
In the above example, I just added 2 conditions, a > b and b > c. That restricts the number of permutations.
Method 2:
*Main> let rightAngleTriangle2 = [(a,b,c) | a <- [1..10], b <- [1..a], c <- [1..b], a ^ 2 == b ^ 2 + c ^ 2]
In the second example, we are taking a,b and c values separately instead of getting from a tuple. This also produces the same result.
Why not nub?
The nub is slower when we apply it in complex programs. The following one is the nub version.
*Main> import Data.List *Main Data.List> nub rightAngleTriangle [(5,4,3),(10,8,6)]It is slow, because it works after the expression is evaluated. To confirm that, the following one produces duplicate values.
*Main Data.List> let rightAngleTriangle3 = nub [ (a,b,c)|(a,b,c) <- triangles , a^2 == b ^2 + c ^ 2]
*Main Data.List> rightAngleTriangle3 [(5,4,3),(5,3,4),(10,8,6),(10,6,8)]
So better to prevent the duplication at origin.
