Mathematics desk
< August 5	<< Jul \| August \| Sep >>	Current desk >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

August 6

Bishop and generalised leaper checkmate

We know from the 1983 work of J. Telesin that king, bishop, and knight can force mate against the lone king on any square n×n board in O(n²) moves (a clear presentation), provided that the board has a corner of the colour the bishop travels on.

So, what if we replace the knight with a general (p,q)-leaper with gcd(p,q)=1, so that it can travel to any square on the board? (I guess I'm using the convention gcd(p,0)=|p|.) Is it still the case that mate is forceable in O(n²) moves on an n×n board, for sufficiently large n to stop the leaper from getting stuck by board edges?

(An almost identical question was asked on Kirill Kryukov's forum in 2015 by "byakuugan", but no general answer came.) Double sharp (talk) 11:52, 6 August 2024 (UTC)[reply]

Note that the king and bishop may be used to pen the black king into a corner, unless the black king "waits" (just moves back and forth without attempting to escape the enclosure). This is the only time the knight (or leaper) needs to take action (and it doesn't matter how many moves it takes to check the king). If you can devise conditions such that half of the board above a main diagonal is connected under knight-moves, I think basically the solution given in that arxiv link works, although I am confused about the second pane of Figure 1 since this does not appear to be a chess position that can be arrived at legally. Presumably checkmate can only happen in the corner? Tito Omburo (talk) 16:08, 6 August 2024 (UTC)[reply]

	a	b	c	d	e	f	g	h
8									8
7									7
6									6
5									5
4									4
3									3
2									2
1									1
	a	b	c	d	e	f	g	h

Black to move

@Tito Omburo: Thanks, that sounds logical to me.

The second pane of Fig. 1 could be reached from the position I give to the right: 1...Kh1 2.Bg2+ Kh2 3.Nf3#. Double sharp (talk) 08:46, 19 August 2024 (UTC)[reply]

Identical partitions of cells of the 24-cell?

I'm wondering whether for each divisor d of 24, whether the cells of a 24-cell can be split into identical pieces (not necessarily connected) with d cells (octahedra). Note since the 24-cells is self dual, an identical partition of d cells is equivalent to a partition of 24/d vertices which can be viewed as the partition of 24/d cells from the vertex in the dual corresponding to each cell. (mostly using the +/- 1,+/- 1,0,0 arrangements here) 1 and 24 area trivial. For 2 cells, pairing a cell with the flip of signs works, For 12 cells, grabbing one of each of the pairs in the 2 cell split will give the partition. this leaves the 3/8 cell pieces in partitions and the 4/6 cell pieces in partitions. For the 4, the vertices can be split based on which two dimensions are zero. so the faces in the dual can be split that way. But I don't think the same flip of 2/12 works. I think for the 8 cell pieces, that a consistent coloring of the octahedra can be done by using three different colors on a single edge, and then continuing the coloring so that each edge had three colors or by pairing 4 cell, but I'm not quite sure how to do 3 or 6.Naraht (talk) 14:44, 6 August 2024 (UTC)[reply]

I wasn't quite able to follow everything, so I'll settle for restating and expanding slightly on what you have so far. For 12 sets of 2, as you pointed out, you can pair each vertex with it's negative. For 2 sets of 12 an explicit partition can be given as {vertices of type (±1, ±1, 0, 0) with first sign +} and {vertices of type (±1, ±1, 0, 0) with first sign -}. For 6 sets of 4, as you also pointed out, you can divide up the vertices of (±1, ±1, 0, 0) according to the positions of the 0's. For 3 sets of 8 you can use an equivalent formulation of the 24-cell using 8 vertices of type (±2, 0, 0, 0) and 16 of type (±1, ±1, ±1, ±1). Divide these into the (±2, 0, 0, 0) type vertices, the (±1, ±1, ±1, ±1) type with even +'s and (±1, ±1, ±1, ±1) type with odd +'s. Geometrically this corresponds to a compound of 3 hypercubes whose intersection is the 24-cell. I assume you can turn this into the octahedra coloring you were talking about. That leaves 8 sets of 3 and 4 sets of 6. Anyway, this doesn't really add much new information, but mainly I wanted to give some sort of response to show your question isn't being ignored. My feeling is that there are no such partitions, but I don't see a way of proving this without a lot of computation. --RDBury (talk) 19:51, 9 August 2024 (UTC)[reply]

PS. I was able to find a partition of 8 sets of 3 as follows:

{(1, 1, 0, 0), (1, 0, 1, 0), (1, 0, 0, 1)}

{(1, 0, 0, -1), (1, 0, -1, 0), (1, -1, 0, 0)}

{(0, 1, 1, 0), (0, 1, 0, 1), (0, 0, 1, 1)}

{(0, 1, 0, -1), (0, 1, -1, 0), (0, 0, -1, -1)}

{(0, 0, 1, -1), (0, -1, 1, 0), (0, -1, 0, -1)}

{(0, 0, -1, 1), (0, -1, 0, 1), (0, -1, -1, 0)}

{(-1, 1, 0, 0), (-1, 0, 1, 0), (-1, 0, 0, 1)}

{(-1, 0, 0, -1), (-1, 0, -1, 0), (-1, -1, 0, 0)}

Each triple forms an equilateral triangle of adjacent vertices, which shows they are identical, but there are many more that 8 such triangles and I used a certain amount of trial and error to find this. That leaves 4 sets of 6 still unsolved --RDBury (talk) 06:05, 13 August 2024 (UTC)[reply]

Natural next step would be to look for sets of 6 that mark out octahedra, but that won't include both the (all w=1) and (all w=-1) since the remaining w=0 can't be split that way.Naraht (talk) 18:17, 13 August 2024 (UTC)[reply]

I don't think regular octahedra will work, but I found the following set of four irregular octahedra:

{(±1, ±1, 0, 0), (1, 0, 1, 0), (-1, 0, -1, 0)}

{(0, ±1, ±1, 0), (0, 1, 0, 1), (0, -1, 0, -1)}

{(0, 0, ±1, ±1), (1, 0, -1, 0), (-1, 0, 1, 0)}

{(±1, 0, 0, ±1), (0, 1, 0, -1), (0, -1, 0, 1)}

Each set can be transformed to any other set by a combination of coordinate permutations and sign changes so they are congruent. Again, finding this was more a matter of trial and error than an actual method. --RDBury (talk) 21:55, 13 August 2024 (UTC)[reply]

I'll want to sketch those out (yeah sketching out in 4-D. :) ), but at first glance I agree. instead of the points being above and below the center, they are above and below opposite edges of the square.Naraht (talk) 17:38, 20 August 2024 (UTC)[reply]

Calculate percent below bell curve for a given standard deviation

Assume a normal bell curve. I know that 34.1% of data is 1 standard deviation above mean (and another 34.1% is 1 standard deviation below mean). What is the method for calculating the percent above/below mean for any given standard deviation, such as 0.6 standard deviations above (or below) mean? 12.116.29.106 (talk) 16:43, 6 August 2024 (UTC)[reply]

There is an analytical expression that makes use of the error function Erf(x). Specifically, the function

Erf(z/{\sqrt {2}})

gives the fraction of data within z standard deviations (above+below) of the mean of a normal distribution. Get the amount above (or below) by dividing by two. Tito Omburo (talk) 17:11, 6 August 2024 (UTC)[reply]

The "method for calculating" it is not answered there. Of course, the easy way to is to invoke an implementation of the error function that you find in all kinds of standard libraries, spreadsheets, etc. But if you really want to know how to calculate it, then you need to do the integral, numerically, and maybe built a table of polynomial approximants from doing the integral at a bunch of places. Or look it up in a book from a trusted source. Dicklyon (talk) 04:39, 8 August 2024 (UTC)[reply]

The section Error function § Numerical approximations gives formulas for calculating approximations with lower and lower maximum errors, down to a maximum relative error less than

2^{-53}.

--Lambiam 07:12, 8 August 2024 (UTC)[reply]

Nice! I should have known WP would have the comprehensive answer in the article Omburo linked! Dicklyon (talk) 14:13, 8 August 2024 (UTC)[reply]