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August 4

Even binomial coefficients

Is there an elementary way to show that all the entries except the outer two, on a row of Pascal's triangle corresponding to a power of 2, are even? 2A00:23C6:AA0D:F501:658A:3BC6:7F9D:2A4A (talk) 17:37, 4 August 2024 (UTC)

The entries in Pascal's triangle are the coefficients of aⁿb^n-i in (a+b)ⁿ. It's easy to see that (a+b)²=(a²+b²) mod 2, and consequently (a+b)⁴=(a⁴+b⁴) mod 2, (a+b)⁸=(a⁸+b⁸) mod 2, and in general (a+b)ⁿ=(aⁿ+bⁿ) mod 2 if n is a power of 2. This implies that all the coefficients of aⁿb^n-i are 0 mod 2 except when i=0 or i=n. Note, there are more complex formulas for finding aⁿb^n-i mod 2 or any other prime for any values of n and i. --RDBury (talk) 18:40, 4 August 2024 (UTC)

@RDBury: Did you mean aⁱb^n-i in the first sentence...? --CiaPan (talk) 09:44, 5 August 2024 (UTC)

Yes, good catch. RDBury (talk) 13:20, 5 August 2024 (UTC)

Another way to see it is, we divide $2^{n}$ into two equal piles of size $2^{n-1}$, so that to select k things from $2^{n}$ is to select i things from one pile and k-i things from the other. That is, we have the following special case of Vandermonde's identity:

${\binom {2^{n}}{k}}=\sum _{i=0}^{k}{\binom {2^{n-1}}{i}}{\binom {2^{n-1}}{k-i}}$

In the right-hand side, every term appears twice, except the middle term (if k is even). We thus have

${\binom {2^{n}}{2k}}\equiv {\binom {2^{n-1}}{k}}{\pmod {2}}.$

We iterate this until k is either odd (and the right hand side is even by induction), or $n=0$ in which case $k=2^{n}$ or $k=0$, which corresponds one of the two outer binomial coefficients. Tito Omburo (talk) 18:51, 4 August 2024 (UTC)

(OP) I think the first approach more my level, but thanks to both for the replies.2A00:23C6:AA0D:F501:6CF2:A683:CAFC:3D8 (talk) 07:10, 5 August 2024 (UTC)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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

August 13

Another probability question

Hello, this is not homework. It is a conundrum that arose out of something I was trying to work out for myself. If we have only one sample of a random variable, and no other information at all, then the best estimate of the mean of the distribution is the sample itself. It may be a rubbish estimate, but nothing more can be said. However, suppose we also know that the mean is >= 0. With this extra information, can any better estimate of the mean be achieved from a single data point? (If necessary, please define "better" in any sensible way that aids the question.) It seems to me that, in the event that the sample is negative, replacing it with zero would always be a better estimate. However, averaged multiple adjusted samples will no longer converge to the true mean, assuming that positive samples are left alone, which seems undesirable. It's not obvious to me anyway that positive samples have to be left alone. Can anything better be achieved? How about if we also know anything else necessary about the distribution, even down to its exact "shape", except that we do not know the mean, apart from that it is >= 0. What then? Does that extra information help at all? 2A00:23C8:7B0C:9A01:8D5A:A879:6AFC:AB6 (talk) 20:56, 13 August 2024 (UTC)

Adjusting positive samples will do you little good if you don't know how to adjust them – but if nothing else is known about the distribution than μ ≥ 0, there is no information to base an adjustment on. For an extreme case, assume all samples have sample size 1. Unknown to the sampler, the distribution is discrete with possible outcomes {−9, 1}. The sample means are them equal to the single element in each sample. Adjusting them to be nonnegative amounts to left-censoring. If μ is the true mean of the distribution, the average of the adjusted (left-censored) sample means tends to (μ + 9) / 10. For the average of the adjusted sample means to tend to the true mean, the positive sample means should be replaced by 10μ / (μ + 9). But this replacement requires already knowing both the value of μ and the outcome space – in fact, all there is to know about the distribution.  --Lambiam 07:33, 14 August 2024 (UTC)

August 14

Hypothetical US Senate reform

Suppose that the US Senate is reformed to be lightly weighted by population, with each state having either 1, 2, or 3 senators (and keeping the total number of 100 senators constant). What would be a plausible method for determining which states get 1, 2, or 3? 71.126.57.129 (talk) 03:33, 14 August 2024 (UTC)

Representatives are assigned using a greedy algorithm: First, every state is allocated one representative; then whichever state currently has the highest ratio of population to representatives is given another representative; repeat until out of representatives. One could reasonably use the same algorithm for senators, with a maximum number of senators per state allowed.--Antendren (talk) 06:31, 14 August 2024 (UTC)

(ec) You can think of the states as each being one party in a multi-party system, and of the population of each state as being voters for their own party (voting for the party, not for a specific candidate). Several multi-winner voting systems can be used, specifically party-list proportional representation, tweaked to keep the number of seats for each party in a given 1 to n range – in the question as posed n = 3. One possible way is the D'Hondt method, which will assign the seats one by one to the state that is proportionally the least represented and still has fewer than n seats. As you can see at Party-list proportional representation § Apportionment of party seats, there are many other methods.  --Lambiam 06:36, 14 August 2024 (UTC)

One plausible way is to start moving senators from the smallest states to the largest. So pick the smallest state first, take one of their senators and assign it to the largest. Then repeat with the next smallest and largest. Keep going until you run out of states that are small or large enough for it to make sense. This works as if you require the total to be unchanged, with 50 states and 100 senators, the number of states with 1 and 3 must be equal. --2A04:4A43:90FF:FB2D:F067:D97A:B158:E36B (talk) 12:44, 14 August 2024 (UTC)

One approach is to define what is meant by a fair distribution of Senate seats. A reasonable ansatz is to minimize the sum of seats/population subject to the constraints that there are 100 seats total and each state gets between 1 and 3 seats. This is a variant of the knapsack problem. Tito Omburo (talk) 13:15, 14 August 2024 (UTC)

Another theory-based approach is to minimize total dissatisfaction, assuming that the members of a group will feel less happy if the group is underrepresented. Let, for group $i,$ the quantity $p_{i}$ stand for the fraction the group takes up in the total population, and $s_{i}$ for the fraction of the seats allocated to the group (so $\textstyle {\sum _{i}p_{i}=\sum _{i}s_{i}=1}.$) Ideally, $p_{i}=s_{i}=1$ for all groups, but this is rarely posible, given the constraints. For a model of total dissatisfaction, choose some function $f(p,s)$ that represents the dissatisfaction of a group with population weight $p$ and representation weight $s;$ then $\textstyle {\sum _{i}p_{i}f(p_{i},s_{i})}$ stands for the proportionally weighted combined dissatisfaction. Plausibly, $s\geq p,$ meaning that a group is not underrepresented, should imply that $f(p,s)=0;$ furthermore, for fixed $p,$ the dissatisfaction should weakly monotonically decrease with increasing $s;$ conversely, for fixed $s,$ the dissatisfaction should weakly monotonically increase with increasing $p.$ A possible formula is given by $f(p,s)=(p{\dot {-}}s)^{2},$ using truncated subtraction.  --Lambiam 16:55, 14 August 2024 (UTC)

August 15

Can two polygons have the same interior angles (in order) if they are noncongruent and have no parallel sides?

Was thinking about the question of whether the interior angles, in order, determine a polygon up to congruence, when I realized that the answer is obviously no if there are parallel sides which you can contract and lengthen at will without changing any of the angles. Barring these polygons with parallel sides and their lengthenings/contractions, then, are there any polygons that share the same interior angles in order without being congruent? GalacticShoe (talk) 17:30, 15 August 2024 (UTC)

I'm not exactly sure what you're asking, but it seems to me that an example can be obtained by starting with a regular pentagon, and lengthening two adjacent sides, and the third opposite side, and shrinking the remaining two sides, while maintaining all angles. Tito Omburo (talk) 17:37, 15 August 2024 (UTC)

I'm realizing now that this question is embarrassingly simple; if we consider extending polygon sides to lines, with the sides then being line segments between the intersections of adjacent sides, then moving lines always preserves angles. For example, in the pentagon example, the lengthening/shrinking you mentioned is akin to moving two lines further away from the center of the pentagon. GalacticShoe (talk) 17:48, 15 August 2024 (UTC)

Upon further searching online, the answer is very much no yes; see this Math StackExchange post. In particular, there are many polygons where you can simply intersect the polygon with a planar half-space such that 1. the line L demarcating the planar half-space is parallel to one of the sides S, and 2. all sides of the polygon either fall on the interior of the half-space, share a vertex with S and intersect L, or are S. GalacticShoe (talk) 17:39, 15 August 2024 (UTC)

I think you meant to say that the answer is, Yes – they can have the same interior angles and yet be noncongruent (from tetragons on).  --Lambiam 22:25, 15 August 2024 (UTC)

Yup, silly mistake on my part; amended GalacticShoe (talk) 03:51, 16 August 2024 (UTC)