Fix bug in random generator. Correct a proof

static/js/paillier.js

@@ -3,13 +3,14 @@ let p, q, pubKey, privKey;
 class PubKey {
     constructor(p, q) {
         this.n = p * q;
-        this.g = this.n + 1n;
+        // this.g = this.n + 1n;
     }
 
     encrypt(m) {
         // Compute g^m r^n mod n^2
         let r = random2048();
 
+        // Resample to avoid modulo bias.
         while (r >= this.n) {
             r = random2048();
         }

static/js/random_primes.js

@@ -21,8 +21,8 @@ function generate_bigint() {
     // Drop the MSB to force into range from above
     intRepr >>= 1n;
 
-    // Add 2^127 to force into range from below
-    intRepr += 2n ** 127n;
+    // Add 2^2047 to force into range from below
+    intRepr += 2n ** 2047n;
 
     return intRepr;
 }

whitepaper/Dissertation.tex

@@ -37,8 +37,6 @@
 
 Risk is a strategy game developed by Albert Lamorisse in 1957. It is a highly competitive game, in which players battle for control over regions of a world map by stationing units within their territories in order to launch attacks on neighbouring territories that are not in their control.
 
-
-
 \section{Existing solutions}
 
 For playing games over an internet connection, multiple solutions already exist. These can roughly be broken down into those that are centralised and those that are decentralised, although many decentralised systems rely on federated or centralised communications for peer discovery.
@@ -286,7 +284,7 @@ The Paillier cryptosystem is otherwise generic over the choice of primes $p, q$.
 \end{proposition}
 
 \begin{proof}
-	WLOG, assume $p > q$. Suppose $\gcd(pq, (p - 1)(q - 1)) \neq 1$. Then, $q - 1 \mid p$. However, the bit-lengths of $p, q$ are identical. So $\lfloor \frac{1}{2}p \rfloor < q$. This is a contradiction to $q - 1 \mid p$, and so $\gcd(pq, (p - 1)(q - 1)) = 1$ as required.
+	WLOG, assume $p > q$. Suppose $\gcd(pq, (p - 1)(q - 1)) \neq 1$. Then, $q \mid p - 1$. However, the bit-lengths of $p, q$ are identical. So $\frac{1}{2}(p - 1) < q$. This is a contradiction to $q \mid p - 1$ (as 2 is the smallest possible divisor), and so we must have $\gcd(pq, (p - 1)(q - 1)) = 1$ as required.
 \end{proof}
 
 As the prime generation routine generates primes of equal length, this property is therefore guaranteed. The next optimisation is to select $g = 1 + n$.