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Knights:

[countably infinitely many are needed. Place them as in the following board (king is placed on only legal square) and move any knight to check the king. knight solution](#sp)

I think you missed the fact that you have to pay a dollar if you guess wrong (so your expected profit is definitely lower than 26, though still greater than 0).

You can compute the expected profit using [dynamic programming. The base case is f(i,0) = 1 and f(0,j) = 1 for all i,j <=N because you know that if one type of card is depleted, you can always guess right. The recurrence relation is f(i,j) = (|i-j| + i f(i-1,j) + j f(i,j-1))/(i+j). The |i-j|/(i+j) term is simply equal to max(i,j)/(i+j) - min(i,j)/(i+j) since the strategy makes you win with probability max(i,j)/(i+j) and lose with probability min(i,j)/(i+j).](#sp) I implemented this in C++ (spoilers obviously). You can expect to profit around [$7.1554](#sp).

I wrote a quick python script to test different ranges and changing the number of players. The results are pretty interesting. Unfortunately it brute forces the solution without any pruning so it can't really explore very large ranges or more than like 4 or 5 players.

http://ideone.com/fZ1RST

I was wondering if player C could come out with a better result by changing the range of values and 0.25 was actually the worst result for C. The best result was 0.333 when the range was just 1-3.

http://ideone.com/YgIjvH

In the output it shows only the first bid scenario encountered which is counted in the average payout for the optimal choice.

There are only six times throughout the day the minute hands would line up, including Midnight/Ten and Noon/Five. None of them have the same hour hand placement. You can work out the details easier with this graph.

No proof here, but a bit of messing around on Desmos suggests no. It seems to start very well-matched, by then there a horizontal stretch towards the right which increases as you move along. At least, that's what I see.

https://www.desmos.com/calculator/asqunbiaft

I don't entirely follow this, especially "[There are always exactly N wins of the first type](#sp)" - can you elaborate?

Your approximation gives [11.78...](#sp), however simulating the game a bunch (to cheat out an approximation of what the answer ought to look like) gives something more like [8.08](#sp), the same as the dynamic programming solution posted elsewhere.

I hope your proof just has a minor error in it, because it's cool!

That is a lot easier than I was expecting, haha. What you have is even simpler then Euler-Maclaurin summation, your just bounding a sum by an integral.

My solution is about the same length if you are generous enough with skipping steps: https://www.overleaf.com/read/dsgytbxvrczd

You appear to be off by a full year. You actually want [31,112,739,000](#sp) seconds. Divided by 240 is [129636412.5](#sp) seconds. You would therefore finish on [Saturday, March 9, 2019 at 6:06:52 AM](#sp).

I would phrase want you want as a parameterization with the same parameter t of a square A(t) and shape B(t) such that the distance between A(t) and B(t) is a constant r for all t. You can do this by taking a parameterization of a square A(t) = (x(t), y(t)) and adding the vectors (r*cos( f(t) ), r*sin( f(t) )) for some function f(t) to get B(t) = ( x(t) + r*cos( f(t) ), y(t) + r*sin( f(t) ) ). If t comes from [a,b] then we need f(a) = f(b). If you want the shape to be a simple curve contained by the square you'll have to make appropriate choices for r and f(t).

I made this on desmos so you can visualize what a family of such shapes looks like. r is the constant distance and k changes the parameter. The way I set things up, you can also change the function F(x) = x a different function, which would work (I think) as long as F(0) = 0, F(1) = 1, and F is increasing between 0 and 1. For example you could change F(x) to x^(2), sqrt(x), sin(pi*x/2), or x+0.2sin(2pix) and it would work.

So I think this paper might have the answer, if I only I could access it.

In a perfectly transitive tournament there would be a first place winner who beat everyone else, then a second place who beat everyone except the person who came first and so on and so forth. Every 'upset' would be an occasion which disrupted this perfect transitivity. So your question is basically 'how intransitive can a round robin competition be?'

This paper is all about transitivity of round robin tournaments. However I can't access it myself to actually check. In particular, they might (and probably do) have a different definition for what counts as intransitive.

If anyone can access the paper, it might help with the question - or at least give you some ideas on how to think about this sort of question.

It is obviously impossible to do this on Euclidean space.

But if we do it on a cylinder, draw a circle with diameter d which is bigger than the cylinder's circumference, C. Then draw a smaller circle with diameter 2C - d that it just touches the bigger circle. There you go -- two concentric circles that intersect at exactly 2 points.

But since the bigger circle self-intersects, and if you think these count, draw the smaller circle with diameter smaller than 2C - d. That makes no intersection between them, plus two self-intersections. That depends on your definition of 'intersection'.

And I did it on a 2d surface since you said draw instead of construct.

https://scratch.mit.edu/projects/235477532/

I did a writeup which analyzes a lot of ins and outs of this problems, seems liek this is the appropriate place to share it: https://www.overleaf.com/read/rfzfckmxswfw

Here is an article on the subject. The name "linear" is in the sense of linear algebra. A sequence is defined in terms of a linear operator applied to itself (built up as a linear combination of shift operations). If this is unfamiliar to you, don't worry, the main point is that it satisfies the following property: if f(n) and g(n) are sequences satisfying the condition, and a and b are constants, then af(n)+bg(n) is also a solution.

As for why to look for solution of that form? They work. r is a constant here. There are more systematic approaches, e.g., generating functions, but for a set up like this, it is easier to just use what I know will work.

Roughly 1/3 of the tiles can be left empty by placing a domino horizontally, then one empty space, then another tile. The same with the next row, but shifted, so the empty spaces do not align. That is a construction, I am not sure if the best one. But generally you would find some construction, and then try to figure a counting argument that there can not be more empty space without two of them being adjacent. In school math competitions there is this type of problems, they are called coloring problems. You can see some of nice examples here - https://brilliant.org/wiki/coloring-definition/ . You can try to use this technique to prove the optimality of the coloring. Will try to look at it more later. Another approach is to use some computer scripts to find the best placement for some concrete n and try to deduce some rules.

Yes, you must be able to make it. I made it with GeoGebra. I will write the definitions of the parametric curves and the points.
> > Curve((-2 + 4t, sin(2π t) / (2π), (1 - cos(2π t)) / (2π)), t, 0, 1)
> > Curve((-2 + 3t, (2sqrt(2) sin(2π t)) / (2π), 2sqrt(2) (1 - cos(2π t)) / (2π)), t, 0, 1)
> > Curve((-1 + 3t, (2sqrt(2) sin(2π t)) / (2π), 2sqrt(2) (1 - cos(2π t)) / (2π)), t, 0, 1)
> > Curve((-2 + 2t, (sqrt(13) sin(2π t)) / (2π), sqrt(13) (1 - cos(2π t)) / (2π)), t, 0, 2)
> > Curve((-1 + 2t, (sqrt(13) sin(2π t)) / (2π), sqrt(13) (1 - cos(2π t)) / (2π)), t, 0, 1)
> > Curve((-2 + t, (4sin(2π t)) / (2π), 4(1 - cos(2π t)) / (2π)), t, 0, 4)
> > A = (-2, 0, 0)
> > B = (-1, 0, 0)
> > C = (0, 0, 0)
> > D = (1, 0, 0)
> > E = (2, 0, 0)

Ooh, I think I found an elementary-school-level proof, but I don't know how to go from "list of 9, set of 8" to "list of 100, set of 99" (or "list of N, set of N - 1"), so there might be a pitfall I'm missing.

First, note that it's trivial to go from a finite group of gamers to an infinite one if at least one game on at least one list is undefined by making an arbitrary unique game for each new gamer, but that any group/list setup that is fully defined cannot be both infinite and unique.

Next, >!buy a Spot It deck and have the cards represent gamers and the symbols represent games. Each of the 57 cards has eight of a set of 57 symbols, and each card shares exactly one symbol with each other card. You now have a finite set of 57 gamers, each with 8 known games (and 1 undefined game). Your set of games is the set one card/gamer has right now (excluding the undefined one).!<

After that, extend it to infinite gamers by >!adding new ones. Note that they must have one of the 57 defined patterns of eight ~~symbols~~ games: If the card on the left of the cover image didn't have a puppy, that card wouldn't share an item with the card on the right, and would therefore break the "any two gamers have at least one favorite game in common" rule. The ninth game/symbol can be anything, either one of the 57 defined symbols or an arbitrary one.!<

One difficulty of doing this in practice is that >!they don't have to make it easy to figure out which 57 symbols/games are the relevant ones so that you can choose the right set of 8. I'm guessing there would be thousands of relevant symbols with a set of 99.!<

Also note that >!this is a worst-case scenario. It's entirely possible to have two (or more) undefined games, in which case you would need a set of 7 to capture every gamer.!<