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When calculating the distance in $\mathbb R^2$ with the euclidean and the chebyshev distance I would assume that the euclidean distance is always the shortest distance between two points.

Considering the following example

$P_{1}=(1,2)$, $P_{2}=(7,6)$

$Euclidean_{distance} = \sqrt{(1-7)^2+(2-6)^2} = \sqrt{52} \approx 7.21$

$Chebyshev_{distance} = max(|1-7|, |2-6|) = max(6,4)=6$

the chebyshev distance seems to be the shortest distance. Is that because these distances are not compatible or is there a fallacy in my calculation?

xvzwx
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1 Answers1

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Er... the phrase "the shortest distance" doesn't make a lot of sense. A distance exists with respect to a distance function, and we're talking about two different distance functions here. It's not as if there is a single distance function that is the distance function. I don't know what you mean by "distances are not compatible."

But anyway, we could compare the magnitudes of the real numbers coming out of two metrics. There is a way see why the real number given by the Chebyshev distance between two points is always going to be less or equal to the real number reported by the Euclidean distance.

Both distances are translation invariant, so without loss of generality, translate one of the points to the origin. Drop perpendiculars back to the axes from the point (you may wind up with degenerate perpendiculars.)

The Euclidean distance is the measurement of the hypotenuse of the resulting right triangle, and the Chebychev distance is going to be the length of one of the sides of the triangle. Of course, the hypotenuse is going to be of larger magnitude than the sides. (Or equal, if you have a degenerate triangle.)

rschwieb
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