Yes, Manhattan distance exists, it has a name. But I don’t understand how having a name makes it faster “a straight line distance” also has a name, euclidean distance. And always euclidean ≤ Manhattan.
So if on both routes you go at the same speed, it is faster to take the one of the euclidean distance.
Well, we were literally walking in Manhattan when it came up, and couldn’t take the euclidean straight path. We could only walk on the grid of streets.
(This is setting aside factors like waiting to cross, or a busier street)
Yes, in a grid where the Manhattan distance is the minimum one, taking a single 90° turn is the fastest, since that path will have the length of the Manhattan distance.
However, it’s not the only path. The “ladder” one you said will be the same length.
While we are at it, if you wanna search for more. The same flawed assumption of “a ladder approximates a straight line” can also lead to π=2. Since you can enclose a circle in a square (Wich has perimeter 4R), then fold the corners recursively so there is a “stair” along the circle’s perimeter. That “stair” would have a length of 4R, but the circle’s perimeter is 2πR.
I don’t understand your conclusion.
Yes, Manhattan distance exists, it has a name. But I don’t understand how having a name makes it faster “a straight line distance” also has a name, euclidean distance. And always euclidean ≤ Manhattan.
So if on both routes you go at the same speed, it is faster to take the one of the euclidean distance.
Well, we were literally walking in Manhattan when it came up, and couldn’t take the euclidean straight path. We could only walk on the grid of streets.
(This is setting aside factors like waiting to cross, or a busier street)
Ah. I originally missread your original comment.
Yes, in a grid where the Manhattan distance is the minimum one, taking a single 90° turn is the fastest, since that path will have the length of the Manhattan distance.
However, it’s not the only path. The “ladder” one you said will be the same length.
While we are at it, if you wanna search for more. The same flawed assumption of “a ladder approximates a straight line” can also lead to π=2. Since you can enclose a circle in a square (Wich has perimeter 4R), then fold the corners recursively so there is a “stair” along the circle’s perimeter. That “stair” would have a length of 4R, but the circle’s perimeter is 2πR.