diff options
| -rw-r--r-- | calculette_aoo/bin/dune | 4 | ||||
| -rw-r--r-- | calculette_aoo/bin/main.ml | 60 | ||||
| -rw-r--r-- | calculette_aoo/lib/build.ml | 2 | ||||
| -rw-r--r-- | calculette_aoo/lib/roll.ml | 85 |
4 files changed, 97 insertions, 54 deletions
diff --git a/calculette_aoo/bin/dune b/calculette_aoo/bin/dune index 1177b56..3fee2aa 100644 --- a/calculette_aoo/bin/dune +++ b/calculette_aoo/bin/dune @@ -1,3 +1,5 @@ (executable (name main) - (libraries aoo)) + (libraries + zarith + aoo)) diff --git a/calculette_aoo/bin/main.ml b/calculette_aoo/bin/main.ml index c738cd7..37d015b 100644 --- a/calculette_aoo/bin/main.ml +++ b/calculette_aoo/bin/main.ml @@ -3,34 +3,33 @@ let cout_m = (110, 55, 35) let cout_rm = (50, 40, 20) let cout_pm = (5, 3, 1) let cout_fm = (100, 50, 30) -let fm_oponent = 10 +let fm_oponent = 1 +let n = 3 +let ratio = Float.(to_int (pow 3.0 (of_int n))) + +let print_arr arr = + Format.printf "@[<v>%a@]@." (Format.pp_print_array Format.pp_print_int) arr + +let () = print_arr (Aoo.Roll.build_frequencies n) + +(* Calcule les probabilités de toucher contre les 10 prochains niveaux *) +let compare_result = Aoo.Roll.compare 9 1 + +let () = + Format.printf "%.4f - %s\n" + (compare_result |> Q.to_float) + (compare_result |> Q.to_string) let env : Aoo.Build.env = { cost_max = 1800 - ; max_tours = 5.0 + ; max_tours = 5 ; cout_sort = 8 ; degat_sort = 6 ; fm_oponent ; frequencies = Aoo.Build.buil_freq_table 9 fm_oponent } -(* -let () = Random.self_init () - -let roll_and_accumulate dices = - Seq.repeat () |> Seq.take dices - |> Seq.fold_left (fun res _ -> res + (1 + Random.int 6)) 0 - -let delta_carac level = - Seq.repeat () |> Seq.take 1000000 - |> Seq.fold_left - (fun acc _ -> - if roll_and_accumulate level > roll_and_accumulate (level - 2) then acc - else acc + 1) - 0 -*) - (* Définition des caractéristiques *) let a = Aoo.Carac.create 2 cout_a let m = Aoo.Carac.create 5 cout_m @@ -38,12 +37,23 @@ let rm = Aoo.Carac.create 7 cout_rm let pm = Aoo.Carac.create 40 cout_pm let fm = Aoo.Carac.create 9 cout_fm let default = Aoo.Build.{ a; m; rm; pm; fm } +let score1 = Aoo.Build.score env default +let cost1 = Aoo.Build.cost default +let () = print_endline @@ Printf.sprintf "Score du build : %f" score1 +let build3 = { default with fm = Aoo.Carac.incr ~step:1 fm } +let score3 = Aoo.Build.score env build3 +let cost3 = Aoo.Build.cost build3 +let () = print_endline @@ Printf.sprintf "Score du build : %f" score3 +let () = print_endline @@ Printf.sprintf "%b" (score1 = score3) +let () = print_endline @@ Printf.sprintf "%b" (cost1 < cost3) -let () = - let build = - Aoo.Build.traverse env - (Aoo.Build.cost default, Aoo.Build.score env default) - (Aoo.Build.upgrade env default) - in +(* + let () = + let build = + Aoo.Build.traverse env + (Aoo.Build.cost default, Aoo.Build.score env default) + (Aoo.Build.upgrade env default) + in - Aoo.Build.repr env Format.std_formatter build + Aoo.Build.repr env Format.std_formatter build +*) 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 |