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Asked: 2024-09-25T01:37:44+05:30

You are given a scenario where you can navigate on an infinite 2D grid starting from a specific point. The grid allows movement in eight possible directions: up, down, left, right, and the four diagonals. Your task is to determine the minimum number of moves required to reach a targeted point on the grid. You are provided with the initial coordinates (x1, y1) of your starting position and the coordinates (x2, y2) of the destination point you wish to reach. Compute the least number of moves needed to move from the starting position to the target position by utilizing the allowed movements on the grid. Consider the characteristics of the grid and how diagonal movements can help you cover more distance in fewer steps. How would you calculate the steps required to achieve this?

Imagine you’re on an infinite 2D grid, which sounds kind of cool, right? You can start at any point, and you have the freedom to move in eight possible directions: up, down, left, right, and also diagonally (up-left, up-right, down-left, down-right). Now, let’s say you start at a point (x1, y1) and you want to get to another point (x2, y2). Your mission is to figure out the least number of moves it’ll take to reach your destination.

Here’s a fun scenario to think about: You’re standing at point (2, 3) – let’s call it your “home base” – and you’ve got your eyes set on the point (5, 1) as your target. So, how do you plan your route?

To tackle this, you first want to visualize the grid. You can think of it like a chessboard where each square is a spot you can step on. The cool part? You can move diagonally, which means you’re not just restricted to moving up, down, left, or right.

Let’s break this down a bit. The horizontal distance between your starting point and the target is found by taking the absolute difference of the x-coordinates: |x2 – x1|. And the same goes for the vertical distance with the y-coordinates: |y2 – y1|. For our example, that would be |5 – 2| = 3 for the x distance and |1 – 3| = 2 for the y distance.

Now, here’s where it gets interesting: when you move diagonally, you can actually cover both the x and y distances at the same time. So, if you want to get from (2, 3) to (5, 1), you could make diagonal moves to efficiently reduce both the x and y distances simultaneously.

The minimum number of moves you’ll need to make is determined by finding the maximum of the two distances you calculated (x and y). In our situation, it’s max(3, 2), which gives you 3 as the minimum number of moves required.

So, if you were to navigate this infinite grid from (2, 3) to (5, 1), how do you think you would strategize your movements? Would you focus on those diagonal shortcuts, or do you think there are better routes? Give it a shot and let’s see how you’d tackle this grid puzzle!

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    2 Answers






    1. Grid Moves

      Moving on an Infinite Grid

      So, let’s say I’m starting at point (2, 3), like that’s my “home base.” And I wanna go to (5, 1), which sounds like a nice place, right? Here’s how I think we can figure this out.

      First, I should check the distance in the x-direction. You get that by doing |x2 - x1|. For our points, it’s |5 - 2| = 3. And then for y, it’s the same thing: |1 - 3| = 2.

      So now I have:

      • Horizontal distance (x): 3
      • Vertical distance (y): 2

      Here’s the cool trick: I can move diagonally! This means I can decrease both distances at the same time. If I go diagonal, I can cut down on both x and y with one move!

      The magic number for the least moves I need is the bigger distance of the two. So, I look at them and say, max(3, 2) = 3. That means it’ll take me 3 moves to get to my new spot.

      To think about how to move, I guess I’d try a combo of diagonal and maybe straight moves, depending on where I am on the grid. It sounds pretty straightforward, right? Just keep focusing on that big distance, and I should be good.

      So yeah, I’m thinking my best bet is to go diagonal when I can and just count those moves!


    2. To solve the grid navigation problem from point (2, 3) to point (5, 1), we start by calculating the horizontal and vertical distances. The horizontal distance is |x2 – x1| = |5 – 2| = 3, and the vertical distance is |y2 – y1| = |1 – 3| = 2. Given that we can move in eight directions, including diagonally, we can optimize our moves by reducing both x and y distances simultaneously. Diagonal movements allow us to cover one unit on both the x-axis and y-axis in a single move, making them highly efficient in reducing our overall distance to the target point.

      Having established the distances, the minimum number of moves required is determined by taking the maximum of the two computed distances: max(3, 2) = 3. This indicates that in the optimal scenario, we would need to make three moves to reach our target. For example, we could move diagonally first to (3, 2), then to (4, 1), and finally reach our destination at (5, 1). This route maximizes the efficiency of our movements on the infinite grid, demonstrating the value of utilizing diagonal shortcuts whenever possible.