Taxicab geometry
Tonight's idea
In a city of square blocks a taxi cannot cut across. To get from \((x_1, y_1)\) to \((x_2, y_2)\) it drives the horizontal gap and then the vertical gap, in some order. So the taxicab distance is
\[d = |x_1 - x_2| + |y_1 - y_2|,\]
where the bars \(|\;|\) mean "drop the sign". The usual straight-line distance, from Pythagoras, is \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\). The taxi distance is never smaller, and is equal only when the two points share a row or a column.
Change the ruler and geometry changes. A circle is the set of points at a fixed distance \(r\) from a centre. Measured the taxi way, the points at distance \(r\) from the origin satisfy \(|x| + |y| = r\): a square standing on its corner, with corners at \((\pm r, 0)\) and \((0, \pm r)\). Shortest routes are no longer unique either: any path of \(a\) blocks east and \(b\) blocks north, in any order, has length \(a + b\).
Everything built from distance changes with it: midpoints, perpendicular bisectors, even \(\pi\).
Watch one: from \((1, 2)\) to \((4, 6)\)
Horizontal gap \(|1 - 4| = 3\), vertical gap \(|2 - 6| = 4\). Taxicab distance \(3 + 4 = 7\). Straight-line distance \(\sqrt{3^2 + 4^2} = 5\). The taxi drives 7 blocks for a journey a crow makes in 5.
Which 7-block route? Any one: three blocks east and four north in any order. East, east, east, north, north, north, north is one. North, east, north, east, north, east, north is another. All are shortest paths, so "the" shortest path is the wrong phrase in this city.
Slide the radius and watch the diamond grow: mathdetours.com/demos/taxicab.html.
Your turn
- Find the taxicab distance and the straight-line distance (a) from \((-2, 1)\) to \((3, 5)\) and (b) from \((0, 0)\) to \((4, 4)\). For which pairs of points are the two distances equal?
- (a) List every grid point at taxicab distance 2 from the origin. (b) Explain why a taxicab circle of radius \(r\) (a whole number) passes through exactly \(4r\) grid points. Check the figure: \(r = 3\) gives 12.
- A midpoint of \(A\) and \(B\) is a point whose distance to each is half the distance \(AB\). Let \(A = (0, 0)\) and \(B = (4, 2)\), so \(d(A, B) = 6\). Find every grid point at taxicab distance 3 from both.
- Let \(A = (0, 0)\) and \(B = (2, 2)\). (a) Check that \((0, 2)\), \((1, 1)\) and \((2, 0)\) are all taxicab midpoints. (b) Now take any point \((x, y)\) with \(x \le 0\) and \(y \ge 2\). Show it is equidistant from \(A\) and \(B\), so the "perpendicular bisector" of \(AB\) contains an entire quarter of the plane.
Stretch: taxicab \(\pi\)
For an ordinary circle, circumference divided by diameter is \(\pi\). Do the same for the taxicab circle of radius \(r\): measure its perimeter the way a taxi would, going along the grid, and divide by the diameter \(2r\). What number do you get?
Answers
- (a) Taxicab \(5 + 4 = 9\); straight-line \(\sqrt{25 + 16} = \sqrt{41} \approx 6.40\). (b) Taxicab 8; straight-line \(\sqrt{32} = 4\sqrt2 \approx 5.66\). Equal exactly when one of the gaps is zero: the points lie on the same horizontal or the same vertical line.
- (a) \((2,0), (1,1), (0,2), (-1,1), (-2,0), (-1,-1), (0,-2), (1,-1)\): eight points. (b) The circle has four sides. Along the side from \((r, 0)\) to \((0, r)\) the grid points are \((r,0), (r-1,1), \dots, (1, r-1), (0, r)\): \(r + 1\) points. Four sides give \(4(r+1)\), but each corner is counted twice, so \(4(r+1) - 4 = 4r\). For \(r = 3\): 12.
- We need \(|x| + |y| = 3\) and \(|x - 4| + |y - 2| = 3\). The grid points that work are \((3, 0)\), \((2, 1)\) and \((1, 2)\): three midpoints, all on the line from \((3,0)\) to \((1,2)\). (In fact every point of that segment is a midpoint.)
- (a) Each is at distance 2 from \((0,0)\) and distance 2 from \((2,2)\): for \((0,2)\), \(0 + 2 = 2\) and \(2 + 0 = 2\); for \((1,1)\), \(1 + 1 = 2\) both ways; for \((2,0)\), symmetric to the first. (b) With \(x \le 0\) and \(y \ge 2\): \(d(A, P) = |x| + |y| = -x + y\). And \(d(B, P) = |x - 2| + |y - 2| = (2 - x) + (y - 2) = -x + y\). The same. For example \((-5, 10)\) is at distance 15 from both.
Stretch. Each side of the diamond runs from \((r, 0)\) to \((0, r)\) or the like: \(r\) blocks across and \(r\) blocks up, taxicab length \(2r\). Four sides make a perimeter of \(8r\). Divide by the diameter \(2r\) and you get 4. In taxicab geometry, \(\pi = 4\), exactly, with no decimals.