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-rw-r--r--calculette_aoo/lib/build.ml2
-rw-r--r--calculette_aoo/lib/roll.ml85
2 files changed, 59 insertions, 28 deletions
diff --git a/calculette_aoo/lib/build.ml b/calculette_aoo/lib/build.ml
index 5c2ac0b..2119fb6 100644
--- a/calculette_aoo/lib/build.ml
+++ b/calculette_aoo/lib/build.ml
@@ -80,7 +80,7 @@ let repr : env -> Format.formatter -> build -> unit =
let score : env -> build -> float =
fun env build ->
- let d, v = eval build env in
+ let d, _ = eval build env in
d
(* Upgrade each caracteristic and keep only the values in range *)
diff --git a/calculette_aoo/lib/roll.ml b/calculette_aoo/lib/roll.ml
index 00d8465..14488e6 100644
--- a/calculette_aoo/lib/roll.ml
+++ b/calculette_aoo/lib/roll.ml
@@ -1,24 +1,52 @@
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
+(** Build the frequencies table for the given number of dices to roll. The
+ function return a table with the number of occurences for the each index.
+
+ Example :
+ table[2] = 1
+ table[3] = 4
+
+ Tell us than there the probabily to get the value 3 is 4× higher to get the
+ value 2.
+ *)
+let build_frequencies n =
+ let length = n * 3 in
+
+ let arr_source = Array.init length ~f:(fun i -> if i < 3 then 1 else 0) in
+
+ (* Recursive function, evaluate the odd by adding a new dice to the previous
+ distribution.
+
+ The probabily to get the value V with N dice is equal to :
+ - The probabilyt to get the value V - 1 with N - 1 dices and having 1 with
+ the new dice
+ - The probabilyt to get the value V - 2 with N - 1 dices and having 2 with
+ the new dice
+ - The probabilyt to get the value V - 3 with N - 1 dices and having 3 with
+ the new dice
-(** 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
+ As the dice is fair, the probability to get the new value is equal in each
+ case, and we can just ignore this part.
+
+ This give us this formula :
+
+ P(V)_N = P(V - 1)_(N-1) + P(V - 2)_(N-1) + P(V - 3)_(N-1)
+ *)
+ let rec apply level arr_source =
+ match level with
+ | 0 -> arr_source
+ | _ ->
+ let arr_target = Array.init length ~f:(fun _ -> 0) in
+ let depth = n - level + 1 in
+ for index = max 1 (depth - 1) to (3 * depth) - 1 do
+ for j = max 0 (index - 3) to index - 1 do
+ arr_target.(index) <- arr_target.(index) + arr_source.(j)
+ done
+ done;
+ apply (level - 1) arr_target
+ in
+ apply (n - 1) arr_source
(** Evaluate the odd to win agains a give difficulty *)
let against : int -> int array -> Q.t array =
@@ -26,18 +54,19 @@ let against : int -> int array -> Q.t array =
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
+ (* Create the polynomial with the odd ratio for the given
+ caracteristic *)
+ let frequencies = build_frequencies 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. *)
+ (* 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
@@ -51,7 +80,6 @@ let against : int -> int array -> Q.t array =
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)
@@ -63,8 +91,11 @@ let compare : int -> int -> Q.t =
(* 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
+ let frequencies_1 =
+ build_frequencies carac1 |> Array.map ~f:(fun v -> Z.(of_int v))
+ and frequencies_2 =
+ build_frequencies 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