# I know enough calculus to think of the question but not the answerFebruary 2, 2015 12:58 PM   Subscribe

You want to walk 10 blocks north and 10 blocks east. It would be shorter to walk along the hypotenuse, but the same distance to walk 1 block north, then 1 block east, then 1 block north, etc. it seems that if the blocks were infinitesimally short, it would approximate the hypotenuse, but alas, walking 10 blocks in one direction and then 10 blocks in the perpendicular direction is the same as dividing it up. What gives?
posted by i_am_a_fiesta to Science & Nature (20 answers total) 8 users marked this as a favorite

Even with infinitesimally short blocks, the direction of travel is still north or east. It's only shorter if your angle changes.
posted by huckit at 1:01 PM on February 2, 2015 [7 favorites]

Say the blocks start out with l=1. So the total distance is l×10+l×10=l×20=20, right?
If you shrink them, to say, l=.1, the sum is still l×20=2. However small you make them, the sum is always l×20. Even if they're infinitesimally short.
posted by signal at 1:06 PM on February 2, 2015

Best answer: Here is an explanation of the 'no shortcuts' paradox. Basically, even if you're travelling infinitesimally small horizontal then vertical legs, it's still longer than taking the diagonal.
posted by majuju at 1:08 PM on February 2, 2015 [5 favorites]

BUT IN THE REAL WORLD.... you can jaywalk intersections, so it pays to take the 1x1x1x... method!
posted by IAmBroom at 1:16 PM on February 2, 2015 [8 favorites]

it seems that if the blocks were infinitesimally short, it would approximate the hypotenuse

When the blocks are infinitesimally short, they aren't blocks, they're a line. The limit of 1/x as x approaches ∞ is 0, so at infinity the sides of your blocks have no length and the hypotenuse is the same as two sides. But this only holds at infinity, at any real number of blocks the tiny sides all still add up to 20.
posted by nicwolff at 1:55 PM on February 2, 2015

When you say things like "if the blocks were infinitesimally short", what you mean is "in the limit as the size of a block approaches 0". So let's do that.

If the size of a block is 10, you walk a distance of 20.
If the size of a block is 1, you walk a distance of 20.
If the size of a block is 0.1, you walk a distance of 20.

You see where this is going.

There's no such thing as a block being "infinitesimally short" (yes, I know about non-standard analysis, but shush); all you can do is reason about what happens in the limit as the blocks get arbitrarily short.
posted by dfan at 1:59 PM on February 2, 2015 [2 favorites]

Best answer: Another interesting discussion of the problem. This counterintuitive result reminds me of those that often arise with fractals eg how long is a coastline? It can get longer if you use a smaller 'measuring stick'
posted by firesine at 2:06 PM on February 2, 2015 [2 favorites]

To look at it in a more intuitive (albeit not mathematically rigorous) way: walking north then walking east is qualitatively different than walking northeast, regardless of how small your north and east steps are.
posted by DevilsAdvocate at 2:07 PM on February 2, 2015 [2 favorites]

The real problem here is that, even as things get arbitrarily short, the square of the distance along the edges will be more than the square of the distance along the hypotenuse — that is, (x + y)2 > x2 + y2 — by a factor of 2xy. Only if your distance is zero (in at least one direction) will you get rid of that term, and it's true that going no distance east and then going no distance west is certainly the same as going no distance diagonally.
posted by ubiquity at 2:08 PM on February 2, 2015 [1 favorite]

It might be helpful to approach the problem geometrically if the perspective of calculus isn't clicking. So you're essentially looking at isosceles right triangles. No matter how infinitely small those triangles get, the ratio of the side to the hypotenuse is constant. No matter how small, that ratio doesn't change, so the hypotenuse always remains an equally good shortcut.

As mentioned above, this constant term only disappears if the side length is actually 0.
posted by novelgazer at 3:21 PM on February 2, 2015

One reason is that the zig-zag path doesn't actually approach the shortcut: The two paths always have different slopes if you place them on a coordinate grid. The zig-zag path is pointing horizontally, vertically, or it's undefined (at the stair-steps). The shortcut is pointing diagonally always. This is true at every step, and it's true at infinity.

So, the zig-zag path can't be a straight line. A straight line is (by definition) the shortest distance between two points, so the zig-zag path (even in the limit, out to infinity) absolutely has to be longer.

The way you built it makes it exactly the Manhattan distance, but you could make it arbitrarily long! Imagine stretching out the upper left corner of the initial L-shape up and away from the original beginning and destination. Push it way out there. Now you can "fold" it in half, like you did before, making a really odd-shaped stair step that looks more like peaks. You can keep doing this, and it will behave the same way, but will be an even longer path, since you stretched it out to start.

The following is speculation, but I think it works:

If you stretch out the stair steps a little bit (say, 10%) each time you fold, the overall length of the zig-zag path gets longer at each iteration. The steps still get smaller, looking closer and closer to the diagonal. Continue this, and your path is infinitely long.
posted by BungaDunga at 5:45 PM on February 2, 2015 [3 favorites]

Consider the sequence S[i] = 1/i. That is, the sequence 1, 1/2, 1/3, 1/4, 1/5...
Every single element in this sequence is greater than 0.
However, the limit of the sequence... is not greater than 0!

What? You don't look impressed. Why is that? It's not fundamentally different from the stair-stepped path. Both involve a sequence converging to something that doesn't share a property found in each term of the sequence. But we have an intuition that it should be the same in once case and not in the other.

Unfortunately, intuition is a pretty poor guide to how limits behave.
posted by baf at 6:22 PM on February 2, 2015 [1 favorite]

The harmonic series doesn't have a limit at all, it grows without bound and does not converge.
posted by BungaDunga at 6:34 PM on February 2, 2015

Oh. Sequence, not series. Sorry, yes, the limit of your sequence is 0.
posted by BungaDunga at 6:42 PM on February 2, 2015

Best answer: the "Manhattan Path" does approach the diagonal in the limit (in that the distance between them goes to zero), but that doesn't mean the length of the Manhattan path must approach the length of the diagonal (since the Manhattan path has constant length 2, where the diagonal path has constant length sqrt2 ). You could even take a path that geometrically approaches a straight line of finite length and have its (arc) length tend to infinity (imagine all sorts of fractal curlycues).
posted by youchirren at 10:38 PM on February 2, 2015 [4 favorites]

It looks like the stair-step path is approximating the straight line path, but it really isn't. Think about how you approximate a curve to measure its length: you pick a finite set of points _on_ the curve, and join those with straight segments. Use lots of points, and your straight-segment line looks a lot like the curve: the limit straight-segment length is the length of the curve.

What you aren't allowed to do is introduce additional points _off_ the curve, even if they are very close to it. Here is a simpler example, where the fact that the length doesn't converge is not counterintuitive at all: Let's say you want to measure the length of the interval from (0,0) to (0,1). Split this up into intervals of length 1/n, i.e. from (i/n,0) to ((i+1)/n,0). Don't join them directly though: between each two points, make a square-shaped detour via (i/n,1/n) and ((i+1)/n,1/n) - you can also alternate between 1/n and -1/n between points.

Now the length of each detour is 1/n+1/n+1/n=3/n, and you have n of them, so the total length is (3/n)*n=3, not the straight-line distance of 1. Even if n gets very big, and all your detours are very close to the line, you are still adding unnecessary length to the path.

The funny part is that, if the detours have height of 1/n^2 instead of 1/n, the total length is n*(1/n+1/n^2+1/n^2) = 1 + 2/n, which _does_ approach the "real" length of 1 as n goes to infinity.

The takeaway from this is that formally defining concepts such as "infinite","infinitely small" or "curve length" is hard for a reason - even if intuitively they look obvious.
posted by Dr Dracator at 12:35 AM on February 3, 2015 [1 favorite]

the "Manhattan Path" does approach the diagonal in the limit...but that doesn't mean the length of the Manhattan path must approach the length of the diagonal

This is a wonderfully consinse resolution of the seeming paradox. For this convergence, path length is a total red herring.
posted by So You're Saying These Are Pants? at 3:38 AM on February 3, 2015

I like to think of this problem as watching a very drunk guy walk down a sidewalk, making a zigzag staggering track that goes the full width of the path - staggering from foot to foot is clearly taking him on a circuitous route. if he's making 90-degree staggers, then this is an example of the "very short blocks" in your problem.

Much like with the coastline, there's a question of physical size of objects that affects the practical path length. Once the length of "very short block" becomes shorter than a step length, each step starts averaging over many 90-degree turns and coming out at a net angle, such that the path starts to approach the hypotenuse. But that's practical, not mathematical.
posted by aimedwander at 6:54 AM on February 3, 2015

youchirren: You could even take a path that geometrically approaches a straight line of finite length and have its (arc) length tend to infinity (imagine all sorts of fractal curlycues).
Math people: is the process of creating this infinite stepwise path a true fractal? The shape is determined by recursively applying an invariant change at each iteration.

Not sure of the exact definition of fractals in this sense.
posted by IAmBroom at 9:24 AM on February 3, 2015

That's the typical way of generating the Koch curve, among others. At each step, you replace every line segment with a straight bit, a pointy bit, and another straight bit. The result is a fractal; the actual path has infinite length.

It follows the simplest definition of a fractal: it is infinitely self-similar. The curve is made of four copies of itself, one for each original line segment.
posted by BungaDunga at 8:24 PM on February 3, 2015

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