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+(*
+Jose Gabriel Perez
+ex 1: could save gcd on reduce to avoid double calc on big nums
+Also, finding common denom instead to reduce computing could be better for large numbers
+*)
+
+//ex1
+let rec gcd = function
+| (a,0) -> a
+| (a,b) -> gcd (b, a % b)
+
+let reduce = function
+| (a, b) -> (a/gcd(a,b), b/gcd(a,b))
+
+let (.+) (a,b) (c,d) = reduce (a*d + b*c, b*d)
+let (.*) (a,b) (c,d) = reduce (a*c, b*d)
+
+//ex2
+let revlists l = List.map List.rev l
+
+//ex3 Really banking on l1 and l2 having equal length, otherwise it will just do it up to the smaller size
+let rec interleave = function
+| ([], l) -> []
+| (l, []) -> []
+| (x::xs, y::ys) -> x:: y :: interleave(xs, ys)   
+
+//ex4
+let gencut(n, l1) = 
+    let rec helpcut (n, a, b) = 
+        if List.isEmpty b then (n, a, b)
+        elif List.length a < n then helpcut (n, a @ (List.head b :: []), List.tail b)
+        else (n, a, b)
+    let (_, f1, f2) = helpcut (n, [], l1)
+    (f1, f2)
+
+let cut l1 = gencut((List.length l1)/2, l1)
+
+//ex5
+let shuffle l = interleave (cut l)
+
+//ex6
+let countshuffles n = 
+    let rec countaux(deck, target) =
+        if deck = target then 0
+        else 1 + countaux(shuffle deck, target)
+    countaux(shuffle [1..n],[1..n]) + 1
+
+//samples
+interleave ([1;2;3],[4;5;6]);;
+cut [1;2;3;4;5;6];;
+shuffle [1;2;3;4];;
+countshuffles 8;;
+countshuffles 52;;
+
+shuffle [1; 2; 3; 4; 5; 6; 7; 8];;
+shuffle [1; 5; 2; 6; 3; 7; 4; 8];;
+shuffle [1; 3; 5; 7; 2; 4; 6; 8];;
diff --git a/HW2.fsx b/HW2.fsx
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+(* Homework 2
+
+*)
+
+/// 1. Returns a list of pairs that represents the cartesian product
+let rec cartesian(xs, ys) =
+    match xs, ys with
+    | (xs, []) -> []
+    | ([], ys) -> []
+    | (x::xs, ys) -> (List.map(fun y -> x,y) ys) @ (cartesian (xs,ys));;
+
+/// 2. Powerset set returns the set of all subsets of set
+let rec powerset = function
+    | [] -> [[]]
+    | x::xs -> List.collect (fun subset -> [subset; x::subset]) (powerset xs)
+
+/// 3. Transpose an m-by-n matrix
+let rec transpose = function
+    | (_::_)::_ as M -> List.map List.head M :: transpose (List.map List.tail M)
+    | _ -> []
+    
+//// 4. Correctness of sort recursive function with respect to the Checklist for Programming with Recursion
+(*
+Step One: It is okay, because empty and single item lists are automatically sorted
+Step Two: wrong, because even assuming that sort (x2::xs) returns the correct answer,  there is no guarantee
+  that x1 is the minimun since it was only compared to x2 and  is not checked again. 
+  For a counter example sort [3;2;1];; will return [2;1;3;]. 
+  All that this sort function really guarantees is that the largest element goes to the   end of the list. 
+  (Note: works like the insertion step of insertion sort, if it gets called n-1 times the list would be sorted)
+Step Three: Each recursive call gets an input that is smaller than the original input
+*)
+let rec sort = function
+  | []         -> []
+  | [x]        -> [x]
+  | x1::x2::xs -> if x1 <= x2 then x1 :: sort (x2::xs)
+                              else x2 :: sort (x1::xs)
+
+/// 5. Analyze mergesort with respect to the Checking for Programming with Recursion (merge and split work correctly)
+(*
+Step One: It is okay, because empty lists are automatically sorted
+Step Two: If mergesort of M and N return the correct answer, then since split and merge are correct the non-base case
+   is correct too. There is a missing base case that leads to a bug, but step 2 is still satisfied.
+Step Three: Each recursive call gets an input that is smaller than the original input
+
+Clue something is wrong: The mergesort function is missing a case, which causes mergesort to be seen by 
+the compiler as 'a list -> 'b list instead of 'a list -> 'a list
+*)
+let rec merge = function
+  | ([], ys)       -> ys
+  | (xs, [])       -> xs
+  | (x::xs, y::ys) -> if x < y then x :: merge (xs, y::ys)
+                               else y :: merge (x::xs, ys)
+
+let rec split = function
+  | []       -> ([], [])
+  | [a]      -> ([a], [])
+  | a::b::cs -> let (M,N) = split cs
+                (a::M, b::N)
+
+let rec mergesort = function
+  | []  -> []
+  | [x] -> [x]  //Bug corrected
+  | L   -> let (M, N) = split L
+           merge (mergesort M, mergesort N)
+
+/// 6. Define F# function curry f that converts an uncurried function to a curried function, and 
+(* an F# uncurry f that does the opposite conversion *)
+let curry f a b = f (a, b)
+let uncurry f (a,b) = f a b
+
+(*
+val curry : ('a * 'b -> 'c) -> 'a -> 'b -> 'c
+val uncurry : ('a -> 'b -> 'c) -> 'a * 'b -> 'c
+*)
diff --git a/HW3.fsx b/HW3.fsx
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+// 1
+let rec inner xs ys = 
+    match (xs, ys) with
+    | ([x],[y]) -> x*y
+    | (x::xs, y::ys) -> x*y + inner xs ys
+    | (_, _)-> failwith "inner defined for equal sized lists only"
+
+
+
+// 2 using transpose from last HW
+let rec transpose = function
+    | (_::_)::_ as M -> List.map List.head M :: transpose (List.map List.tail M)
+    | _ -> []
+
+// Posible to improve this by transposing beforehand
+let rec multiply = function
+|([], _) -> []
+|(x::xs, ys) -> List.map (inner x) (transpose ys) :: multiply (xs, ys)
+    
+// 3
+let flatten1 xs = List.fold (@) [] xs
+let flatten2 xs = List.foldBack (@) xs []
+let makelistlist n = List.map (fun x -> [x]) [1..n]
+#time
+(*In practice flatten1 behaves as O(n^2) and flatten2 is (O(n))*)
+
+// 4
+(*
+    twice successor 0 would be 2 = 2^1
+    twice twice successor 0 would be 4 = 2^2
+    twice twice twice successor 0 would be 16 = 2^4
+    And so on..
+
+    The current value becomes equals to 2 to the power of the previous value
+
+    The function will be defined recursively, as follows:
+        f(n) = 
+            if n = 0 then 1
+            else 2 ^ (f(n-1))
+
+    We can write this function as:
+    let rec f = function
+        | 0 -> 1.0
+        | n -> 2.0 ** (f (n-1));;
+*)
+
+// 5
+type 'a stream = Cons of 'a * (unit -> 'a stream);;
+
+let rec map f (Cons(x, xsf)) = Cons (f x, fun() -> map f (xsf()));;
+
+// 6
+type Exp = Num of int
+         | Neg of Exp
+         | Sum of Exp * Exp
+         | Diff of Exp * Exp
+         | Prod of Exp * Exp
+         | Quot of Exp * Exp
+
+let rec evaluate = function
+| Num n -> Some n
+| Neg e -> match evaluate e with
+           | None -> None
+           | Some n -> Some (-1 * n)
+|Sum (e1, e2) -> match (evaluate e1, evaluate e2) with
+                 |(None, _) -> None
+                 |(_, None) -> None
+                 |(Some n1, Some n2) -> Some (n1 + n2)
+|Diff (e1, e2) -> match (evaluate e1, evaluate e2) with
+                  |(None, _) -> None
+                  |(_, None) -> None
+                  |(Some n1, Some n2) -> Some (n1 - n2)
+|Prod (e1, e2) -> match (evaluate e1, evaluate e2) with
+                  |(None, _) -> None
+                  |(_, None) -> None
+                  |(Some n1, Some n2) -> Some (n1 * n2)
+|Quot (e1, e2) -> match (evaluate e1, evaluate e2) with
+                  |(None, _) -> None
+                  |(_, None) -> None
+                  |(_, Some 0) -> None
+                  |(Some n1, Some n2) -> Some (n1 - n2)