Working 2nd part of algo

Author: romellem

2019/19/tractor-beam.js

@@ -21,26 +21,13 @@ class TractorBeam {
 		return this.grid.sum();
 	}
 
-	partTwo() {
+	partTwo(square_size = 100) {
 		/**
 		 * First, we need to find when the tractor beam begins after the origin.
 		 * We do this by diagonalizing from 1,0 (or 0,1) until we hit 1 (being pulled).
 		 * Rather than just following a simple `y = x` line from the origin, we diagonalize
 		 * because the "slope" may be thin enough to not add pulled markers near the origin
 		 * (as I see with my puzzle input).
-		 *
-		 * Once we have a start, we then mark two points, TOP and BOTTOM. Think of TOP
-		 * as the top-right edge and BOTTOM as the bottom-left edge. If we can determine
-		 * those values, we don't need to run the program for the points in between: we
-		 * know they'll be filled in.
-		 *
-		 * TOP and BOTTOM will both start out at the same point. BOTTOM will then go down
-		 * until we hit a `0` (more than likely will be after the first down move). When
-		 * that happens, then move inward towards the right until we hit a `1`. That is
-		 * the new BOTTOM edge.
-		 *
-		 * Similarly, for TOP, move down 1, then move right until we hit a `0`. The cell
-		 * immediately to the left of that is the right edge.
 		 */
 		this.grid.set(0, 0, 1);
 		let start;
@@ -54,7 +41,7 @@ class TractorBeam {
 
 			if (output) {
 				this.grid.set(x, y, 1);
-				start = [x, y];
+				start = {x, y};
 				break;
 			}
 
@@ -66,7 +53,57 @@ class TractorBeam {
 				y++;
 			}
 		}
-		console.log(start);
+
+		/**
+		 * Once we have a start, we then mark two points, TOP and BOTTOM. Think of TOP
+		 * as the top-right edge and BOTTOM as the bottom-left edge. If we can determine
+		 * those values, we don't need to run the program for the points in between: we
+		 * know they'll be filled in.
+		 *
+		 * TOP and BOTTOM will both start out at the same point. BOTTOM will then go down
+		 * until we hit a `0` (more than likely will be after the first down move). When
+		 * that happens, then move inward towards the right until we hit a `1`. That is
+		 * the new BOTTOM edge.
+		 *
+		 * Similarly, for TOP, move down 1, then move right until we hit a `0`. The cell
+		 * immediately to the left of that is the right edge.
+		 */
+		let top_edge = {...start};
+		let bottom_edge = {...start};
+		while (top_edge.x - bottom_edge.x <= square_size) {
+			// First, move down until we hit 0
+			let output;
+			do {
+				bottom_edge.y++;
+				let computer = new Computer({ memory: this.memory, inputs: [bottom_edge.x, bottom_edge.y] });
+				[output] = computer.run();
+				this.grid.set(bottom_edge.x, bottom_edge.y, output);
+			} while (output === 1);
+			
+			// Then, move inward until we hit a `1`
+			do {
+				bottom_edge.x++;
+				let computer = new Computer({ memory: this.memory, inputs:  [bottom_edge.x, bottom_edge.y] });
+				[output] = computer.run();
+				this.grid.set(bottom_edge.x, bottom_edge.y, output);
+			} while (output === 0);
+
+			// Now do top edge
+			do {
+				// Double loops in case bottom calc mo
+				top_edge.y++;
+				do {
+					// Move right until we hit a `0`
+					top_edge.x++;
+					let computer = new Computer({ memory: this.memory, inputs: [top_edge.x, top_edge.y] });
+					[output] = computer.run();
+					this.grid.set(top_edge.x, top_edge.y, output);
+				} while (output === 1);
+			} while (top_edge.y !== bottom_edge.y);
+
+			// The cell immediately to the left of that is the right edge.
+			top_edge.x--;
+		}
 	}
 }