Implemented the Chinese Remainder Theorem

bin/chinese.ml

@@ -0,0 +1,25 @@
+open Toy_crypto
+module M = Numbers.Make (Int) (* FIXME *)
+
+open Printf
+
+
+let () =
+  let int_arg n = int_of_string Sys.argv.(n) in
+  if Array.length Sys.argv = 5 then
+    let a, p, b, q = int_arg 1, int_arg 2, int_arg 3, int_arg 4 in
+    printf "Input problem:\n\tx = %d mod %d\n\tx = %d mod %d\n" a p b q;
+    try
+      let n = M.crt (a, p) (b, q) in
+      printf "Solution:\n\tx = %d mod %d\n" n (p * q)
+    with Failure _ ->
+      printf "No solution:\n\t%d and %d aren't coprime.\n" p q
+  else
+    let p, q = int_arg 1, int_arg 2 in
+    printf "(a, b) represents this:\n\tx = a mod %d\n\tx = b mod %d\n" p q;
+    for a=0 to p-1 do
+      for b=0 to q-1 do
+        let n = M.crt (a, p) (b, q) in
+        printf "(%d, %d) -> %d\n" a b n
+      done
+    done

bin/dune

@@ -1,5 +1,5 @@
-(executable
- (name main)
+(executables
+ (names main chinese)
  (libraries toy_crypto))
 (install
  (section bin)

lib/numbers.ml

@@ -28,6 +28,7 @@ module type S = sig
   val euclid: N.t -> N.t -> e
   val gcd: N.t -> N.t -> N.t
   val inv: N.t -> m:N.t -> N.t
+  val crt: N.t * N.t -> N.t * N.t -> N.t
   val random: bits:int -> N.t
   val random_prime: bits:int -> N.t
   val of_string : string -> N.t
@@ -102,6 +103,12 @@ module Make (N: Concrete) = struct
       (* this should not happen *)
       failwith "non-invertible value"
 
+  let crt (a, p) (b, q) =
+    let open N in
+    let p1, q1 = inv p ~m:q, inv q ~m:p in
+    let n = (a * q1 * q + b * p1 * p) mod (p * q) in
+    if n < zero then n + p * q else n
+
   let random ~bits =
     let rec f a = function
       | 0 -> a

lib/numbers.mli

@@ -30,6 +30,9 @@ module type S = sig
 
   val gcd: N.t -> N.t -> N.t
   val inv: N.t -> m:N.t -> N.t
+  val crt: N.t * N.t -> N.t * N.t -> N.t
+  (* [crt (a, p) (b, q)] solves ${x = a mod p, x = b mod q}$ *)
+
   val random: bits:int -> N.t
   val random_prime: bits:int -> N.t
   val of_string : string -> N.t