# 07.31.07

## Exercise 8.7

One way to do this is to handle each constructor for a `Region` (and a `Shape` for those cases) and reflect those in the x-axis, i.e.

flipX :: Region -> Region flipX (Translate (a,b) r) = Translate (a,-b) (flipX r) flipX (Scale (a,b) r) = Scale (a,-b) (flipX r) flipX (Complement r) = Complement (flipX r) flipX (r1 `Union` r2) = (flipX r1) `Union` (flipX r2) flipX (r1 `Intersect` r2) = (flipX r1) `Intersect` (flipX r2) flipX (HalfPlane (x1,y1) (x2,y2)) = HalfPlane (x2,-y2) (x1,-y1) flipX Empty = Empty flipX (Shape s) = Shape (flipXShape s) where flipXShape (RtTriangle a b) = RtTriangle a (-b) flipXShape (Polygon vs) = Polygon (map ((x,y) -> (x,-y)) vs) flipXShape s = s |

And similarly (negating the x-coordinate) for `flipY`. Note that the order of points must change in `HalfPlane`, and should also be reversed in flipping a `Polygon` if we had the clockwise ordering constraint. Another way to solve this exercise is to use the `Scale` constructor on a `Region`:

flipX :: Region -> Region flipX r = Scale (1, -1) r flipY :: Region -> Region flipY r = Scale (-1, 1) r |

fero said,

July 22, 2008 at 11:21 am

Wow, that Scale solution is really good one.

Joe said,

June 6, 2010 at 9:38 pm

2 minor errors in the first solution:

> flipX (Scale (a,b) r) = Scale (a,-b) (flipX r)

flipX (Scale (a,b) r) = Scale (a,b) (flipX r)

> flipXShape (Polygon vs) = Polygon (map ((x,y) -> (x,-y)) vs)

flipXShape (Polygon vs) = Polygon (map (\ (x,y) -> (x,-y)) vs)