Fleshed out stub. Added fortify and stub for attack.

static/js/player.js

@@ -91,6 +91,55 @@ class Player {
                 this.turnPhase = PHASE_ATTACK;
             }
 
+            return false;
+        } else {
+            // End turn prematurely.
+            if (data.action === "END") {
+                return true;
+            }
+
+            if (this.turnPhase === PHASE_ATTACK && data.action === "ATTACK") {
+                if (this.attack(data)) {
+                    this.turnPhase = PHASE_FORTIFY;
+                }
+
+                return false;
+            }
+
+            this.turnPhase = PHASE_FORTIFY;
+
+            if (data.action === "FORTIFY") {
+                return this.fortify(data);
+            }
+        }
+        return false;
+    }
+
+    /**
+     * Process an action which is to attack another region.
+     *
+     * @param data Data received via socket
+     */
+    attack(data) {}
+
+    /**
+     * Process an action which is to attack another region.
+     *
+     * @param data Data received via socket
+     */
+    fortify(data) {
+        let sender = Region.getRegion(data.sender);
+        let receiver = Region.getRegion(data.receiver);
+
+        if (
+            sender.owner === this &&
+            receiver.owner === this &&
+            sender.strength > data.strength
+        ) {
+            receiver.reinforce(data.strength);
+            sender.strength -= data.strength;
+            return true;
+        } else {
             return false;
         }
     }

whitepaper/Dissertation.tex

@@ -297,7 +297,7 @@ As the prime generation routine generates primes of equal length, this property
 	We see that $(1 + n)^n \equiv 1 \mod n^2$ from binomial expansion. So $1 + n$ is invertible as required.
 \end{proof}
 
-The selection of such $g$ is ideal, as the binomial expansion property helps to optimise exponentiation. Clearly, from the same result, $g^m = 1 + mn$. This operation is far easier to perform, as it can be performed without having to take the modulus to keep the computed value within range (for $m$ of sufficiently small size).
+The selection of such $g$ is ideal, as the binomial expansion property helps to optimise exponentiation. Clearly, from the same result, $g^m = 1 + mn$. This operation is far easier to perform, as it can be performed without having to take the modulus to keep the computed value within range.
 
 \subsection{Encryption}