# The "No Shortcuts" Paradox

## The Ancient Fallacy

The world has long been fooled by a certain Mr.

Pythagoras
whose

Pythagorean theorem
(

c^{2}=a^{2}+b^{2}, where

c is the length of the hypotenuse in a

right
triangle and

a &

b
are the lengths of the two legs) can be used to infer that the

*shortest distance*
s = √((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2})
(also called the

Euclidean
distance) between two points

P_{1}=(x_{1},y_{1})
&

P_{2}=(x_{2},y_{2})
is shorter than the

*Manhattan
distance* m = h + v, where

h
= |x_{2}-x_{1}| and

v = |y_{2}-y_{1}|,
between the same two points.

## The Truth

The above suggests that the shortest distance

s represents
a

*shortcut* in comparison to the path constructed by following

|P_{1}P|
(

h = |x_{2}-x_{1}|) and then

|PP_{2}|
(

v = |y_{2}-y_{1}|). However, nothing
could be further from the truth. Apart from empirical results that show that shortcuts
even result in prolonged traveling, it can be mathematically proven that the alleged
shortcut

|P_{1}P_{2}| with length

s is

*exactly* as long as the Manhattan distance

h + v.

## The Proof

Consider the above figure. The horizontal distance between

P_{1}
&

P_{2} is

h
and the vertical distance

v. The total distance between

P_{1} and

P_{2}
via

P hence is

h + v.
Let us start approximating the straight line between

P_{1}
and

P_{2} by replacing the large triangle

P_{1},

P,

P_{2}
with two triangles where the legs have length

h/2
and

v/2 correspondingly (see above). A better approximation
is obtained by using four triangles with leg lengths

h/4
and

v/4. The more triangles we use, the better the
triangles will approximate the straight line between

P_{1}
and

P_{2}. In the limit, the leg lengths
will be zero and the infinite number of triangles will match the straight line
between

P_{1} and

P_{2}.
However, note that the sum of the horizontal distances — independently of
the granularity of the division — is always

h
and the sum of the vertical distances is always

v.
As a result, the length of

s is

h
+ v as opposed to

√(h^{2}+v^{2})
as commonly believed. Consequently, it does not matter whether one proceeds from

P_{1} to

P_{2}
following a straight line or via

P, the length of
the path is always

h + v. In other words, there are
no shortcuts.

## The Fun Part

*Hint*: As is typical for a falsidical paradox, the above rationalization suggests
a result that is not only counterintuitive but also wrong. The fun part is in
figuring out where the flaw in the apparently logical reasoning is.

*Spoiler: *Think about in what way the breaking up into small horizontal
and vertical stretches constrains a particle moving along them.