Comments on: Math on the Mississippi http://bit-player.org/2007/math-on-the-mississippi An amateur's outlook on computation and mathematics. Fri, 29 Aug 2008 05:35:09 +0000 http://wordpress.org/?v=1.5.2 Comment on Math on the Mississippi by: Jasper Crowne http://bit-player.org/2007/math-on-the-mississippi#comment-1440 Fri, 09 Mar 2007 22:15:32 +0000 http://bit-player.org/2007/math-on-the-mississippi#comment-1440 I just realized that my comment was cut off (most likely due to use of the less than sign), and that I misspelled my name... It follows from the triangle inequality that ¦(b’-Y’)/y’/¦+¦(b-Y)/y/¦+¦V¦ is less than or equal to ¦U¦; since /y/ and /y'/ are arbitrary in the general case, we have equality only when b=Y, b'=Y', and hence V=U. QED I just realized that my comment was cut off (most likely due to use of the less than sign), and that I misspelled my name…

It follows from the triangle inequality that

|(b’-Y’)/y’/|+|(b-Y)/y/|+|V| is less than or equal to |U|; since /y/ and /y’/ are arbitrary in the general case, we have equality only when b=Y, b’=Y’, and hence V=U. QED

]]> Comment on Math on the Mississippi by: Jasper Crowe http://bit-player.org/2007/math-on-the-mississippi#comment-1409 Sat, 13 Jan 2007 08:29:27 +0000 http://bit-player.org/2007/math-on-the-mississippi#comment-1409 I think the following proves Merten's algorithm is the best possible: Let the point in the (x',y') lattice be at (X',Y') and let the point in the (x,y) lattice be (X,Y). Any path from (X',Y') to (X,Y) has three segments: 1) we travel from (X',Y') through the (x',y') lattice until we reach x'=0. Call this point (0,b'). 2) we walk through the park to the line x=0. Call this point (0,b). 3) we travel through the (x,y) lattice from (0,b) to (X,Y). In 1), note that any direct path you take from (X',Y') to (0,b') (by this I mean any path such that the Manhattan distance to (0,b') is strictly decreasing) has the same distance as just walking from (X',Y') to (0,Y') and then walking from (0,Y') to (0,b'). And similarly for 2). This reduces the problem to finding b, b' so that the sum of the following distances is minimized: 1) from (0,Y') to (0,b') (along the y' axis) 2) from (0,b') to (0,b) (through the park) 3) from (0,b) to (0,Y) (along the y axis) The distance we are trying to minimize is (b'-Y')+(b-Y)+¦V¦ where V is the vector in 2). We'd like to show that this is less than ¦U¦ where U is the vector from (0,Y') to (0,Y) (i.e. where b'=Y' and b=Y). Let /y'/ be the unit vector in the +y' direction and /y/ be the unit vector in the +y direction. Then we have the following vector sum: (b'-Y')/y'/+(b-Y)/y/+V=U It follows from the triangle inequality that ¦(b'-Y')/y'/¦+¦(b-Y)/y/¦+¦V¦ I think the following proves Merten’s algorithm is the best possible:
Let the point in the (x’,y’) lattice be at (X’,Y’) and let the point in the (x,y) lattice be (X,Y). Any path from (X’,Y’) to (X,Y) has three segments:
1) we travel from (X’,Y’) through the (x’,y’) lattice until we reach x’=0. Call this point (0,b’).
2) we walk through the park to the line x=0. Call this point (0,b).
3) we travel through the (x,y) lattice from (0,b) to (X,Y).

In 1), note that any direct path you take from (X’,Y’) to (0,b’) (by this I mean any path such that the Manhattan distance to (0,b’) is strictly decreasing) has the same distance as just walking from (X’,Y’) to (0,Y’) and then walking from (0,Y’) to (0,b’). And similarly for 2). This reduces the problem to finding b, b’ so that the sum of the following distances is minimized:
1) from (0,Y’) to (0,b’) (along the y’ axis)
2) from (0,b’) to (0,b) (through the park)
3) from (0,b) to (0,Y) (along the y axis)

The distance we are trying to minimize is (b’-Y’)+(b-Y)+|V| where V is the vector in 2). We’d like to show that this is less than |U| where U is the vector from (0,Y’) to (0,Y) (i.e. where b’=Y’ and b=Y). Let /y’/ be the unit vector in the +y’ direction and /y/ be the unit vector in the +y direction. Then we have the following vector sum:

(b’-Y’)/y’/+(b-Y)/y/+V=U
It follows from the triangle inequality that

|(b’-Y’)/y’/|+|(b-Y)/y/|+|V|

]]> Comment on Math on the Mississippi by: brian http://bit-player.org/2007/math-on-the-mississippi#comment-1408 Thu, 11 Jan 2007 15:50:00 +0000 http://bit-player.org/2007/math-on-the-mississippi#comment-1408 "Incommensurable" was sloppy wording on my part. I simply meant that you cannot make *all* the intersections line up neatly (unless the angle is a multiple of 90 degrees). Incidentally, although the blocks in New Orleans are roughly square, other cities favor elongated rectangles; Manhattan, for example, has short blocks along the avenues and longer blocks along the cross streets. If you build a city by knitting together adjacent patches of grid rotated by 45 degrees, there's a very nice solution with blocks whose aspect ratio is 1:√2. “Incommensurable” was sloppy wording on my part. I simply meant that you cannot make *all* the intersections line up neatly (unless the angle is a multiple of 90 degrees).

Incidentally, although the blocks in New Orleans are roughly square, other cities favor elongated rectangles; Manhattan, for example, has short blocks along the avenues and longer blocks along the cross streets. If you build a city by knitting together adjacent patches of grid rotated by 45 degrees, there’s a very nice solution with blocks whose aspect ratio is 1:√2.

]]> Comment on Math on the Mississippi by: Barry Cipra http://bit-player.org/2007/math-on-the-mississippi#comment-1407 Wed, 10 Jan 2007 18:08:43 +0000 http://bit-player.org/2007/math-on-the-mississippi#comment-1407 "If the two lattices have the same spacing but different orientations, then they are incommensurable..." This isn't quite correct. Integer solutions of the Pythagorean formula a^2 + b^2 = c^2 allow for commensurable orientations -- e.g., a slope of 3/4 permits a 5-way intersection at every fifth corner for the tilted grid. (One would presumably extend each untilted street till it meets the boundary of the tilted grid. Otherwise there could be some fairly long detours.) “If the two lattices have the same spacing but different orientations, then they are incommensurable…” This isn’t quite correct. Integer solutions of the Pythagorean formula a^2 + b^2 = c^2 allow for commensurable orientations — e.g., a slope of 3/4 permits a 5-way intersection at every fifth corner for the tilted grid. (One would presumably extend each untilted street till it meets the boundary of the tilted grid. Otherwise there could be some fairly long detours.)

]]> Comment on Math on the Mississippi by: Stephan Mertens http://bit-player.org/2007/math-on-the-mississippi#comment-1406 Tue, 09 Jan 2007 15:29:26 +0000 http://bit-player.org/2007/math-on-the-mississippi#comment-1406 Let us assume equal lattice spacings in both parts of the grid (like in your counter example). Can you think of an example where the straight line heuristics is off by *more* than one block? Let us assume equal lattice spacings in both parts of the grid (like in your counter example).
Can you think of an example where the straight line heuristics is off by *more* than one block?

]]> Comment on Math on the Mississippi by: Stephan Mertens http://bit-player.org/2007/math-on-the-mississippi#comment-1405 Tue, 09 Jan 2007 11:14:29 +0000 http://bit-player.org/2007/math-on-the-mississippi#comment-1405 Draw a straight line on the map between your hotel and the destination in the tilted grid. Take the transition point that is closest to the point where this line intersects the discontinuity. This should give the shortest distance, and apparently Google knows that. Stephan Draw a straight line on the map between your hotel and the destination in the tilted grid. Take the transition point that is closest to the point where this line intersects the discontinuity. This should give the shortest distance, and apparently Google knows that.

Stephan

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