
When Math Gets Impossibly Hard - bryanrasmussen
https://www.quantamagazine.org/when-math-gets-impossibly-hard-20200914/
======
3pt14159
I know a military contractor working on some pretty cool stuff, especially
back in the day. On one of their projects they had to solve some really gnarly
partial differential equation and weren't really getting anywhere. Since what
they actually wanted were values (not the algebraic solution) what they ended
up doing is building the equivalent circuit and measured what the voltage and
current was in the circuit to get the values that they needed for some
completely different domain.

I always thought that that way of thinking was pretty neat. Can the same
abstraction have a different physical manifestation that's cheap to create? If
you kinda squint a bit, it's what we do with computers.

~~~
curiousgal
Banks do that all the time with monte carlo simulation. Can't get a closed
form solution? Simulate every possible outcome and aggregate them.

~~~
filmor
This is neither how a Monte-Carlo process works (it's sampling the possible
outcomes randomly, not simulating all outcomes), nor what is described here
(building a physical system where an observable value is described by the
differential equation one is trying to solve).

~~~
curiousgal
In practice the number of simulated scenarii is quite large which is what I
meant by "every". Not rigorous though, I concur.

Monte Carlo methods are used to solve partial differential equations, I found
both approaches similar in the sense that they look at how the system
actually/could behave, instead of looking at the equations solely.

------
bentona
This might be incredibly pedantic, but since this is a mathematical philosophy
article, I'm going to go ahead.

These aren't "hard" problems, they're simply questions that don't make sense.
Is finding an odd number that is divisible by 2 a "hard problem"? I don't
think so, it's just that it turns out that it doesn't make sense to ask the
question.

Some of these "hard problems" actually have solutions that demonstrate they
aren't solvable, and these solutions are fairly elegant, in which case I'd
also say these are not "hard".

~~~
LolWolf
I think that's the joke of the title, though; perhaps rephrased slightly: "no
matter how hard you try, this mathematical problem cannot actually be solved
(and we can prove it, too)."

------
eat_veggies
One interesting way of detecting gerrymandering that has been used in court
[0] is to do a random walk on the graph of districting plans, where two plans
have an edge between them if you can flip a precinct (the basic building block
of a districting plan) in one plan to get the other plan.

By randomly walking this graph, starting from the current plan and randomly
flipping precincts, you can build a distribution of how compact the plans you
saw were (there are some measures of compactness defined in TFA and the
document linked below), and identify outliers, which are likely to be
gerrymandered.

[0] [https://sites.tufts.edu/vrdi/files/2018/06/md-
report.pdf](https://sites.tufts.edu/vrdi/files/2018/06/md-report.pdf)

------
roywiggins
The solution to gerrymandering is probably social, not mathematical, because
the goals we want to pursue in district-drawing aren't strictly mathematical.

~~~
thaumasiotes
It bothers me that two arguments are commonly advanced for the idea that
gerrymandering is a menace:

1\. It lets politicians create districts that will vote for them no matter
what.

2\. It lets politicians waste opposition votes by putting them into districts
that will, overall, just barely vote the other way.

The problem is that these two visions of "the problem" directly oppose each
other. #1 says that the problem is districts that vote an 80/20 split. #2 says
the problem is districts that vote a 51/49 split.

~~~
andrewem
The core problem is the ability to leverage approximately 50% of votes into
winning far more than 50% of seats.

~~~
dane-pgp
Which suggests that a simplistic "solution" to gerrymandering is that, after
the total vote share of the best-performing party is calculated (call it V%),
and their share of seats is also calculated (S%), the second-best-performing
party should be able to propose a change in district winners that reduces the
gap between V% and S%. (And so on for the other parties competing).

Of course that would be outrageously unfair, with politicians choosing which
districts they win rather than voters choosing the politicians to represent
them, and yet that is somehow an improvement over the current system of
gerrymandering.

A slightly less extreme option would be for the districts that get flipped to
be decided by where a given party came closest to winning (the "best near-
winner" system used in Baden-Württemberg), but really the aim of this system
is to remove all incentive to gerrymander in the first place, so this result
flipping rule would never have to be applied.

~~~
aaronblohowiak
This is how the German system works. Not unfair at all. You cast two votes:
one for a person, one for a party. At the end of the day the party percentages
have to roughly line up (with a minimum percent for any representative at
all.) This system has some very nice balancing properties in that you can vote
for the person you want to represent you in the present race as well as the
party you most strongly identify with.

~~~
dane-pgp
My understanding is that the German system is Mixed-Member Proportional
Representation (MMP) [0] which usually involves having a list of top-up
candidates who don't belong to any specific district. The Baden-Württemberg
system is a clever way around this, although it does lead to some districts
being represented by multiple representatives.

Having more representatives than districts seems like it would be an overly-
ambitious cultural shift in the US, whereas my proposal keeps the familiar 1:1
correspondence, even if there is gerrymandering. It also doesn't change the
ballot papers or counting process; and, if gerrymandering is successfully
disincentivised, the process works exactly the same as under the existing
system.

[0] [https://en.wikipedia.org/wiki/Mixed-
member_proportional_repr...](https://en.wikipedia.org/wiki/Mixed-
member_proportional_representation)

~~~
aaronblohowiak
yes, i dont believe i contradicted anything there, but you are right i didnt
tell the whole story and thank you for spelling out some of the details. more
representatives than districts would be a shift and i agree this may be a
harder pill to swallow.

------
ouid
>Many states require that districts be “compact,” a term with no fixed
mathematical definition.

Every open cover has a finite subcover.

~~~
reallyeli
Logged in just to upvote this. I shudder to think what happens in those other
states!

------
f154hfds
I've always found the perimeter over area close to the mark for jerrymandering
but not quite good enough (because of the coastline paradox - forces picking
an arbitrary segment size [1]).

My idea - I welcome counter arguments - is to calculate the district's center
of mass and find the furthest point the district still contains from that
center of mass. Then create a circle with that distance as the radius and find
the percentage of _land_ (maybe even privately owned land?) within that circle
that is in the district. My view is that the percentage should be something
like 50% (maybe higher? Hard to say without actually designing districts and
checking the numbers).

50% would mean that half of the habitable land remaining in the circle could
be a different district but no more.

[1]
[https://en.wikipedia.org/wiki/Coastline_paradox](https://en.wikipedia.org/wiki/Coastline_paradox)

~~~
bijection
What about perimeter vs area but let coastlines be counted as straight lines?

~~~
f154hfds
The coastline paradox extends to much more than coastlines - pretty much all
natural boundaries I would think. It's certainly doable, it just seems to add
extra variables to what really needs to be as foolproof as possible.

------
lordnacho
Nitpick: the bridge problem is impossible because there aren't 2 or 0 nodes of
odd degree.

If you have 2 islands with three bridges connecting them, you could just walk
them trivially, yet there are an uneven number of connections on each node.

~~~
dpbriggs
I may be misremembering my graph theory class, but isn't the necessary and
sufficient conditions for an eulerian path just that every vertex needs to
have even degree?

I'm not sure how it could work if any nodes had odd degree.

~~~
ogogmad
For an Eulerian circuit, you need the graph to be connected and for every node
to have even degree. For an Eulerian path (which doesn't have to start and end
at the same point) you may also allow two vertices to have odd degree.

~~~
dpbriggs
Got it, thanks!

------
m3kw9
Math is hard for me because once a step of a proof is lost on me, the whole
thing is lost unless I branch out into a rabbit hole, and learn what just
happened

~~~
ouid
ok, but all of those rabbit holes have bottoms. It just may be that the space
in between represents a year or more of learning.

~~~
iso8859-1
They don't have bottoms, because at the bottom you have logical assumptions
like e.g. classical logic. These only exist because of history, only because
it was the easiest way for someone to say something "insightful". Going down
the rabbit hole, you have to choose some arbitrary point to stop at; it would
be a point at which you think that you can apply the knowledge you have gained
so far, in a way that other people will "appreciate".

Or you can just not stop, and try to go even deeper. But then you stop being a
mathematician, and you turn into a madman, a philosopher or a theologist.

~~~
drdeca
The "rabbit hole" being discussed is just "understanding a step in an
argument", not, providing a foundation for logic.

It's not impossible to gain understanding of a step in someone else's
argument.

