Erlang Mini Project #1
Distributed 4/9/2015     Due 4/14/2015

Implement the following in Erlang

(1) count(X, L)
return the number of times X appears in the list L (at the top level of L)
count (2, [2, 3, 2, 1, 2, 5])  ----> 3
count (a, [ a, b, c, d]) ----> 1
count (a, [ [a, a], b, {a, b, c}]) ----> 0      %% only consider top level of L


(2) substring ( L1, L2 )

return the index (index starts at 0), where the first list
appears in the second list

substring ([2, 3, 4], [0, 1, 2, 3, 4, 3, 4 ]) ----> 2
substring ([1, 7], [0, 1, 2, 3, 1, 7, 6])     ----> 4
substring ([1], [3, 2, a, {1, 1}, 1] ---->  4


(3) get_last (L) 

return the final element in the list

get_last ([1, 2, 3, 5]) ----> 5



(4) zip ( L1, L2)
%% This is simplified from the Haskell problem

Given two lists, return a new list of 2-ary tuples where the
first value of the resulting tuple is from the first list
and the last value of  tuple is from the second list.

zip ( [ 1, 3, a ], [ 100, b, c ] ) ----> 
   [ {1, 100}, {3, b}, {a, c} ]
 
zip ([ {"aa", 0}, {"bb", 1}, {"cc", 2} ] [ 1, [10,11] ] ) ---->
   [ { {"aa", 0}, 1}, { {"bb", 1}, [10, 11] } ]
   
zip ( [ a, b ] [ c, d, 1, 2] )  -> [ {a, c}, {b, d} ]


(5) permute ( L )

Given a list of values, return a list of all permutations

permute ([1, 2, 3] ) ----> 
    [ [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1] ]
    

(6) ackermann 
Implement Ackermann's function.  (Everyone should implement Ackermann's at
some point)



Give commented source code, using the function names given above.
You can define multiple function to simplify the implementation of 1 - 6
above.  Functions must use recursion.

Show output for (1-6) that has demonstrates the correct functioning of the code.
That will include demonstrations of base cases and recursive calls.