-- ASSIGNMENT #2 SOLUTIONS

-- Problem #1
-- Keep all two-tuples from the input list whose elements sum to zero
    
tuplesSummingToZero lst =
    [ (n,m) | (n,m) <- lst, (n+m) == 0 ]


-- Problem #2
-- Find the average distance from the origin to all points in
-- the first quadrant (x,y positive).

averageDist lst = 
    (sum distances) / fromIntegral(length distances)
    where
      distances = [ sqrt (x*x + y*y) | (x,y)<-lst, x>0, y>0 ]
      

-- Problem #3, parts 1 and 2

-- Here are two different ways to compute the area of a "slice".
-- The areaRect appraoch approximates the area of the slice with 
-- a trapezoid, while areaRect uses a simple rectangle with 
-- height of y1.
 
areaTrap y1 y2 delta = (y1 + y2) / 2 * delta

areaRect y1 y2 delta = y1 * delta


-- Problem #4
  
-- The integrate function takes an approximation method, the 
-- width of the "slices" to use, the lower and upper bounds for
-- the X-coordinate, and a function of one argument to be
-- integrated.  My solution goes until low and high cross, on the
-- assumption that delta is small enough that the extra fraction
-- of a slice isn't a big deal.
  
integrate method delta low high fn
  | low > high       = error "Illegal range specification"
  | high-low < delta = method (fn low) (fn high) (high-low)
  | otherwise        = first + rest
      where
        first = method (fn low) (fn (low+delta)) delta
        rest = integrate method delta (low+delta) high fn
         
         
-- Problem #5
         
-- Partial application of integrate has specialized it to use the
-- trapezoid rule, and a "slice" width of 0.1.  All that's left
-- are the bounds and the function
         
trapInt = integrate (areaTrap) 0.1


-- Problem #6:

-- Here's one version that breaks it down into lots of smaller
-- steps, giving each one a name as part of the where expression.

mostFrequent [] = error "Not defined on empty lists"
mostFrequent lst = head winners
  where
    freqs = map (\item->count item lst) lst
    largest = foldr1 max freqs
    winners = filter (\item->(count item lst)==largest) lst
    count item lst = length (filter (==item) lst)


-- Here's a much more compact solution, though it ends up re-
-- calculating the frequency with which each item occurs.

mostFreq lst =
  foldr1 (\n m->if (count n lst) > (count m lst) then n else m) lst
  where
    count item lst = length (filter (==item) lst)