calculette_aoo: now evaluate the exact frequency, some corrections

1 files changed, 81 insertions, 0 deletions

diff --git a/calculette_aoo/lib/roll.ml b/calculette_aoo/lib/roll.ml
new file mode 100644
index 0000000..00d8465
--- /dev/null
+++ b/calculette_aoo/lib/roll.ml
@@ -0,0 +1,81 @@
+open StdLabels
+
+(** Evaluate the coefficient for the product of two polynamials *)
+let mul p1 p2 =
+  let n1 = Array.length p1 and n2 = Array.length p2 in
+  let n = n1 + n2 in
+  let res = Array.make n 0 in
+  for i = 0 to n1 - 1 do
+    for j = 0 to n2 - 1 do
+      res.(i + j + 1) <- res.(i + j + 1) + (p1.(i) * p2.(j))
+    done
+  done;
+  res
+
+(** Apply a power function over a polynomial *)
+let pow p1 n =
+  let result = ref p1 in
+  for _ = 1 to n - 1 do
+    result := mul !result p1
+  done;
+  !result
+
+(** Evaluate the odd to win agains a give difficulty *)
+let against : int -> int array -> Q.t array =
+ fun n difficulties ->
+  match n with
+  | 0 -> Array.map ~f:(fun _ -> Q.zero) difficulties
+  | _ ->
+      (* Create the polynomial with the odd ratio for the given caracteristic *)
+      let arr = Array.make 3 1 in
+      let frequencies = pow arr n in
+      let ratio = Z.(of_int 3 ** n) in
+
+      let get_chances difficulty =
+        (* Evaluate the ratio to win the roll *)
+        let chances = ref Z.zero in
+        for i = difficulty - 1 to Array.length frequencies - 1 do
+          (* The index in the table is the odd to get exactly this value in the roll.
+             Here, we just add every value in the array from the index strarting from
+             the given difficulty. *)
+          chances := Z.(!chances + of_int frequencies.(i))
+        done;
+        Q.make !chances ratio
+      in
+      Array.map ~f:get_chances difficulties
+
+(** Compare two caracteristics. 
+    [compare v1 v2] will evaluate the odds to win if the player has the
+    caracteristics [v1] against the openent [v2]. 
+
+    In case of equality, the win is given to [v1] *)
+let compare : int -> int -> Q.t =
+ fun carac1 carac2 ->
+  let arr = Array.make 3 1 in
+  let z3 = Z.of_int 3 in
+
+  let ordinal1 = Z.(z3 ** carac1)
+  and ordinal2 = Z.(z3 ** carac2)
+  (*Number of possibles values for each caracteristc *)
+  and elements1 = 3 * carac1
+  and elements2 = 3 * carac2 in
+
+  (* The number of iterations to do is elements1 × elements2. For every iteration of elements2, we have ordinal1 points to get.  *)
+  let cases = Z.(ordinal1 * ordinal2) in
+
+  let frequencies_1 = pow arr carac1 |> Array.map ~f:(fun v -> Z.(of_int v))
+  and frequencies_2 = pow arr carac2 |> Array.map ~f:(fun v -> Z.(of_int v)) in
+
+  (* Now, compare for each values the probabily to win *)
+  let res = ref Z.zero in
+  let () =
+    for i = 0 to elements1 - 1 do
+      for j = 0 to elements2 - 1 do
+        if i >= j then
+          (* Number of times this combinaison is occuring *)
+          let freq = Z.(frequencies_1.(i) * frequencies_2.(j)) in
+          Z.(res := !res + freq)
+      done
+    done
+  in
+  Q.make !res cases