10.2 Euclidean and Manhattan distance

“Nearest” depends on how movement is allowed. A drone can travel almost straight to a point, while a robot confined to horizontal and vertical grid paths cannot. Euclidean and Manhattan distances model those two situations.

This is foundational to intermediate coordinate geometry. Learn the distance formula in OpenStax, then compare it with the simple grid-route sum below.

A basic idea: two ways to measure

From $A=(x_1,y_1)$ to $B=(x_2,y_2)$, write $\Delta x=x_2-x_1$ and $\Delta y=y_2-y_1$. Straight-line Euclidean distance uses the Pythagorean theorem. Manhattan distance counts horizontal plus vertical movement.

A quick tour of formulas

$$d_2(A,B)=\sqrt{(\Delta x)^2+(\Delta y)^2}, \qquad d_1(A,B)=|\Delta x|+|\Delta y|.$$

Both distances are nonnegative, symmetric, and zero only for identical points. For any two points, $d_2\le d_1$, with equality only when $\Delta x=0$ or $\Delta y=0$. If you only need to compare Euclidean distances, compare their squares and skip square roots; squaring preserves their order because distances are nonnegative.

Application 1: a grid route

Move from $(1,1)$ to $(4,5)$. The horizontal and vertical changes are $3$ and $4$.

1. A shortest grid route uses $3+4=7$ unit moves, so $d_1=7$.
2. A direct path has length $d_2=\sqrt{3^2+4^2}=5$.
3. Use $7$ for a robot limited to axis-aligned moves and $5$ for a straight-line route.

Application 2: which candidate is closer?

From the origin, compare $P=(3,0)$ and $Q=(2,2)$. Euclidean distances are $d_2(O,P)=3$ and $d_2(O,Q)=\sqrt8\approx2.83$, so $Q$ is nearer by straight line. Manhattan distances are $d_1(O,P)=3$ and $d_1(O,Q)=4$, so $P$ is nearer on an axis-aligned grid. State the metric before selecting a nearest point.