General Math Stuff

[Ryan Thunder]

I figure we don't have a thread for this and we should, so I'll leave it up to the mods to sticky it or let it die.

Anyway, this isn't meant for people to ask others to do their homework, but more like a study group where we can just talk about what we know and reinforce the right ideas, think of applications for them, and so on.

For instance, I was just considering whether you could use the vectors [0, 1] and [1, 1] as a basis for a hexagonal grid. It seems to work.

[Jaepheth]

It wouldn't be a regular hexagon. The lengths of those two vectors are different.

Or are you asking if the vectors form a basis for a vector space on which a hexagonal grid could be plotted? I think they do (it's been a long time since I've taken linear algebra), but it's not nearly as nice as just using [0,1] [1,0].


I believe several general math threads have popped up from time to time, but they don't tend to last given the shear vastness of the topic as well as the wide variation in subject level expertise. I don't know if this thread is meant to be at a high school math level, undergraduate level, or graduate math level (which would be over my head).

[Jaepheth]

EDIT:

Also, if you're talking about making an irregular hexagonal grid you'd need a third vector similar to [-1, 1]

[Feil]

To position a point on a regular hexagonal grid by way of a whole numbered sum of basis vectors, the most convenient means would be to use two sides of the triangle which, if arrayed six in a circle, points-inward, would construct the hexagon.

A = [1, 0]
B = [1, sqrt(3)]/2

Now, these two basis vectors can move us between points, but aren't very good for counting position. The following composite vectors can be used for counting hexes on a hex grid as it is usually set up (such that the vector A describes one side of the hexagon).

a movement of Y = -A+2B moves us one hex up
a movement of X1 = A+B moves us one hex up-right
a movement of X2 = 2A-B moves us one hex down-right

-Y will move us down, -X1 will move us down-left, -X2 will move us up-left, as one would expect.

To move around the hexagon from point to point, note that |A-B| = |A| = |B| = 1. Starting at the bottom left of the hex and going ccw from vertex to vertex, the path around the hex is: B-A, A, B, A-B, -B, -A.

[Ryan Thunder]

Ah, thought I fucked up somewhere. It'd work for mimicing six-directions movement of a hexagonal grid using squares, though, right? So the end result would've been something like this, I think;

Code: Select all

    O O O O O O
  O O O O O O
O O O O O O

Where yours actually produces something more accurate like this;

Code: Select all

  O O O O O O
 O O O O O O
O O O O O O

Except, well, with uniform distances that is.

EDIT: and this thread is meant for anything, really. Whatever math you feel like bringing up, be it addition or advanced calculus, geometry, etc.

[Surlethe]

Hey, am I the only person on this forum who knows anything about hyperbolic Dehn surgery?

[Purple]

Could be. Here is an interesting question that has been bugging me for a while now. And since it's math, and this thread is for math...

Say I have an array of decimal numbers. Call it X. And say it has N elements. Now, say I take these decimal numbers and round each of them off separately to the closest. So 0.5 becomes 0 and 0.6 becomes 1. That way, I now have two arrays. One made up of decimals and one made up of integers. Let's call this new one Y. Now, say I add all the elements of each of those arrays together to form two sums. SumX = X1+X2+...+XN and SumY = Y1+Y2+...+YN. My question is. Will SumX equal SumY? And if yes, is there any known law or hypothesis or how ever these things are called that defines this as being so?

[Shadow6]

Will SumX equal SumY?

Not in general, no. Obvious counter example:

X = [0.5] => Y = [0]
SumX = 0.5 =/= SumY = 0

[Purple]

I was assuming an array of more than one number, hence array. The logic being to see if over a large enough number of numbers the value lost and value gained through rounding would have a tendency to even out. Converging in a way I guess.

[Eternal_Freedom]

Also, you round 0.5 to 1 not 0. Unless that was a typo on your part or a deliberate assumption, in which case I'll shut up.

[Purple]

Well, the reason I am asking is because it's something that I have ran into several times recently while running various numbers for some work I have been having in excel. Basically, each of the decimals is in my case a fraction of 1 whole. Like 0.25 or 0.31 etc. And the rounded numbers seemed to always sum up to 1 just like the decimals. But it's probably a fluke due to having large arrays and always working with data that comes from a similar source.

[Jaepheth]

Also, you round 0.5 to 1 not 0. Unless that was a typo on your part or a deliberate assumption, in which case I'll shut up.

Not in all cases.
If you always round .5 up you end up skewing your results. This can effect accounting, statistics, or anything else where compounding errors could accrue; so there are different methods of rounding .5 up or down depending on another variable.

[Feil]

The sums of the sets will only be equal if we impose the condition on the initial set that the numbers must be chosen so that the sums will be equal, which is tautological.

Counterexamples:
Adding and subtracting a constant: Consider the set of all integers from 1 to n (representing our 'rounded' values). Take the sum. Add c, such that -0.5<c<0.5 to each integer (the new integers would all round to their original values, since they are within 0.5 of them). Take the new sum. The two sums clearly differ by c*n.

Going up by decimals: Consider the set {0.1, 0.2, 0.3, ..., 1}. The sum is 5.5. Round each value in the set to the nearest integer value to obtain {0, 0, 0, 0, 1, 1, 1, 1, 1, 1}. The sum is 6. The cause of the discrepancy is found in the existence of a middle term - 0.5 - which we have to put in one camp or the other. Any sum that involves rounding will necessarily differ by the value of the number being biased in one direction or the other. If we used hundredths instead of tenths, our sum would be much greater - 50.5 - but our rounded value would still differ by the same amount: it would be 51, differing from the original value by 0.5. We could increase towards a continuum of numbers, in which case our sum would increase without bound, but for as long as we used a finite number of steps (and therefore had a finite sum), the rounded sum and the unrounded sum would differ by 0.5.

Adding more terms outside the endpoints of the initial set: Consider the set {0.1, 0.2, 0.3, ..., n}. We know that for 0.1 to 1, the sums differ by 0.5, and the distributive property tells us that the sum of the next larger set of ten numbers will just be the sum of the lesser set plus ten, and so on. Therefore, the sum of the rounded set where n = 2 will be 6+6+10, where n = 3 it will be 6+6+6+10+20, where n = m it will be 6m + 10*(m-1)(m)/2. The sum of the unrounded set will be 5.5m + 10*(m-1)(m)/2. Not only do they not converge - the sums clearly diverge.

[Feil]

To move around the hexagon from point to point, note that |A-B| = |A| = |B| = 1. Starting at the bottom left of the hex and going ccw from vertex to vertex, the path around the hex is: B-A, A, B, A-B, -B, -A.

I meant clockwise, not counterclockwise, here.

[Feil]

I've a question of my own, if anybody cares to answer it -

Suppose we have a map composed of a number of countries with a bunch of noncontinuous territories, which must be the same color. (For instance, the little section of Russia between Poland and Lithuania, or, for a better example, "water," which is usually painted all blue). What conditions determine whether we need to add an extra color to our map-painting palette, beyond the 4 necessitated by the 4-color theorem?

[Grog]

Then there is no upper bound. Consider a map with n countries, each have n territories, one spherical and one tiny part just on each of the other countries boundary spherical boundary. Then all the n countries share borders, so you have to make more conditions to say something.

In general there is no known description of what the graphs that can be colored with k colors look like.

If you start with something 4 colorable and then just allow some local moves like changing the border of only one country for example then something more might be possible to say, would you be interested in something like this?

[Surlethe]

Are you thinking something precise, or just an asymptotic analysis as the number of vertices grows?

[Feil]

Grog:
"If you start with something 4 colorable and then just allow some local moves like changing the border of only one country for example then something more might be possible to say, would you be interested in something like this?"

Sure. Let's take the Russia example, and presume that Russia is the only territorially noncontiguous state in Europe. If Europe is to be 4-colored in red, green, pink, and yellow, is there a way that we can know, without resorting to trial and error, whether adding the bit of Russia between Lithuania and Poland means that Russia has to be colored purple, or whether it is still 4-colorable?

Surlethe: From Grog's post it looks like there probably isn't a precise description - which in itself is something of a description An asymptotic analysis could be interesting, although you'd have to clarify what you mean by the number of vertices. Are you referring to the representation sometimes used for the 4-colorable map where one color is at the center of a triangle and three other colors are at the vertices?

[Surlethe]

You convert a map into a graph by defining each country to be a vertex and each border to be an edge. Then the question of colorability is, how many colors do we need to color the vertices of the graph so that no two colors are adjacent? For graphs where each country is simply connected (no holes like South Africa, no multiple regions like Russia), the graph is planar and the answer is famously "4." You're asking for an analysis of non-planar graphs.

So I guess one answer for your description is, "If the resulting graph is planar, then the answer is yes." For example, if you can isotope the map to glue the disconnected regions together, then you can color it with four colors.

[Feil]

I had to read those two little paragraphs about eight times before I started to make sense of them, but I think I see what you're saying now, and it does make sense, so thanks

Does the opposite hold? If you can't isotope the map to glue the disconnected regions together, do you need a fifth color?

[Feil]

Does the opposite hold? If you can't isotope the map to glue the disconnected regions together, do you need a fifth color?

No, obviously not. Yellow circle within blue circle within green circle within yellow circle. Never-mind that question.

[Surlethe]

Yeah, the opposite almost certainly doesn't hold. The procedure "isotope the map to glue disconnected regions together" is a procedure to take an existing immersed graph and explicitly produce an embedding of the graph into $R^2$, like in this game, but that's probably not the only way of showing a graph is planar. And for all I know there are $4$-colorable graphs that aren't planar (maybe not, but I haven't really thought much about these questions).

[Kuroneko]

Could do better: there exists a non-planar 2-colorable graph. Spoiler

Take the complete 5-vertex graph, color every vertex red, and for each edge, put an extra vertex, thus subdividing each into two edges. Color each new vertex blue. The graph is non-planar by Kuratowski's theorem.

[Surlethe]

Very nice.

While we're talking about graph colorings, I'm reminded of a fact used in Ian Agol's proof of Virtually Haken. Take a uniformly locally finite graph $\Gamma$ (that is, uniformly bounded valence, so it is $k$-colorable), and a nice (locally finite, maybe cocompact?) group action by a group $G$. Observe that $G$ acts on the space of $(k+1)$-colorings of $\Gamma$. Theorem: There is a $G$-invariant measure on the space of $(k+1)$-colorings of $\Gamma$.

More can be found here in paragraphs 7 and 8, with proof following.

[Ryan Thunder]

I'm a little confused by the dollar signs. What does that mean?