Can someone explain this formula for calculating Manhattan distance?

Question

This is from a Kickstart problem:

Note: The Manhattan distance between two squares (r1,c1) and (r2,c2) is defined as |r1 - r2| + |c1 - c2|, where |*| operator denotes the absolute value.

Then in the analysis:

Note that the manhattan distance has an equivalent formula:

dist((x1, y1), (x2, y2)) = max(abs(x1 + y1 - (x2 + y2)), abs(x1 - y1 - (x2 - y2)))

This formula is based on the fact that for any point, the set of points within a manhattan distance of K form a square rotated by 45 degrees. The benefit of this formula is that if we fix (x2, y2), the distance will be maximized when x1 + y1 and x1 - y1 are either maximized or minimized.

Could someone explain in more details how this formula can be derived?

Answer 1

Lemma. $|a|+|b|=\max(|a+b|, |a-b|)$ for any real number $a$ and $b$.

Proof 1. $|x|=\max(x, -x)$ for all real number $x$. So $$\begin{aligned} |a|+|b| &=\max(a, -a) + \max(b, -b)\\ &=\max(a+b, a-b, -a+b, -a-b)\\ &=\max(\max(a+b, -a-b), \max(a-b, -(a-b))\\ &=\max(|a+b|, |a-b|) \end{aligned}$$

Proof 2. There are $2 \times 2 = 4$ cases.

• $a\ge 0$
  • $b\gt 0$. LHS is $a+b$, RHS is $a+b$.
  • $b\le 0$. LHS is $a-b$, RHS is $a-b$.
• $a\lt 0$
  • $b\gt 0$. LHS is $-a+b$, RHS is $-(a-b)$.
  • $b\le 0$. LHS is $-a-b$, RHS is $-(a+b)$.

One dimensionality of Manhattan-distance.

The Manhattan-distance of two points $(x_1, y_1)$ and $(x_2, y_2)$ is either $|(x_1+y_1)-(x_2+y_2)|$ or $|(x_1-y_1)-(x_2-y_2)|$, whichever is larger. That is, $ d((x_1, y_1),(x_2, y_2))= \max(|(x_1+y_1)-(x_2+y_2)|, |(x_1-y_1)-(x_2-y_2)|)$$

Proof: By definition,
$$d((x_1, y_1),(x_2, y_2))=|x_1-x_2| + |y_1-y_2|.$$ Now apply the lemma above. QED.

_{This answer also serves as a complement to another answer of mine.}