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@boops boops: for what it's worth, I made an effort to avoid Heron's formula since it seemed like its use just pushes the problem farther down the linewhy is Heron's formula true? On the other hand...

First, here's why the circumcenter exists. Take two sides of a triangle and consider their perpendicular bisectors. Since the sides are not parallel, neither are the perpendicular bisectors, so they...

Just to be clear, my comment about precision was directed at how precisely you wanted to draw "the maximum sized scale triangle within the bounding rectangle [subject to some given angles]", rather...

The complexity of what you want depends a lot on how precise you want it all to be and what you're willing to say is "good enough". For instance, your "acute" case in post #1 allows nonacute...

@dday9: cool beans.
Totally understandable, though in this case there was a very easy check besides looking at the original source code. The matrix form of the operation needed to have...

I see one big mistake still in #3, namely the " x" should be "+ x". That would produce extremely erratic results.
I disagree with passel in that I think the integerbased coordinates for your...

You seem to be referring to the fact that the rows of Pascal's triangle, when plotted after normalization, look very much like a normal distribution, especially as you take longer and longer rows....

Ah, my mistake. That makes the computation easier. Since 0 < N < 100, we may evidently assume every such sequence terminates, since a nonterminating sequence occurs with 0 probability. Each sequence...

Unless I'm really, really misunderstanding your question, it's not (N/100)M times. For instance, at N=100, this gives M, but as you say it should obviously give infinity. As I understand it, you...

x is a value between 0 and 1 (a float), but you're making it an int. You could also make x use percentages, I suppose.

You want what's called a "linear interpolation". If p and q are two numbers and x is a parameter between 0 and 1, you can get values in between p and q based on x using the standard formula F(x) =...

I agree with passel. The guys each have $9 at the end, the bellhop has $2, and the manager has $25. 9*3 = 27 = 2+25, which makes perfect sense. The "$1 discrepancy" is just coming from confusing +...

I had a more iterative approach in mind for the lexer. That is, you run through the code character by character adding tokens to a queue as you go. It's fundamentally similar to what you're doing,...

RPN probably isn't an option here since that would force users to enter unusual syntax.
The shuntingyard algorithm is probably what you're after here. The example and description at the link are...

I feel I should have a disclaimer at the top of this post. Namely, you probably shouldn't have to write your own parser nowadays and should instead be able to use someone else's; if you do have to...

I don't work in VB anymore, though the Python translation of that code fails on the example I gave earlier, and the VB version should be similar. Specifically, given
def dec2frac(f):
u =...

The trouble is your code is assuming the decimal representation infinitely repeats. For instance, for 6/5 = 1.2, you're assuming it's really 1.2222222... = 11/9. But it's not, it's 1.200000.... I...

I can't really understand your revised version, sorry. What I know so far: if A (the random number) is > 0.9, compute N = ceil((1+log10(1A))/log10(0.9)), which is the smallest N such that 0.1*0.9^N...

Say x is the random number. If x < 0.1, you want to know when 0.1*0.9^y = x (essentially); this occurs for y = log_(0.9) (x/0.1) = (1+log10(x))/log10(0.9). For instance, if x = 0.0324, this gives...

There's a nice StackOverflow post on this topic. Roughly, you want to pick a maximum denominator size and find the numeratordenominator pair which gets closest to your number, subject to the...

Sure, there's a lot of ways to tweak it as you wish. Divide (AB) by a magic constant, replace (AB) with sign(AB)*AB^x for some constant x (for x=3 this reduces to just (AB)^3; for x=2 you need...

Glad it worked out. For your revised problem, I would probably have gone for arctan, something like arctan(AB)*2/pi + 1. It's 1 at A=B, pi/2*2/pi + 1 = 2 at A >> B, and pi/2*2/pi + 1 = 0 at A << B.

I have trouble understanding the question, eg. I don't really know what the following means: "Then will the second pair still be derived from the first pair. I'm looking for a math function on this,...

You put the negative inside the parens when it should be outside, i.e. 2 * EXP(1*(AB)^2). For instance, if A=1, B=0, you should get 2/e = 0.735.... Without the negative the function grows very...

Your question is a bit vague, but maybe 2exp((AB)^2) is what you're after. It's 2 when A=B and is normally distributed in some sense. If you have more specific requirements, maybe those can be...

Yes, I understood your original post. I'm saying to find v or h requires solving a transcendental equation, which in general will have to be done numerically (or using a special function, though you...

Not sure why you wrote your formula in terms of v (which you didn't quite define) rather than h when you want to solve for h. Either way you have a transcendental equation, which will at best have an...

Yup, you're right. A little more formally, if pValue = min, you'll return 0, and if pValue = max, you'll return 1 * Me.Width, with values in between computed via linear interpolation. Roughly, the...

Your reply confuses me in several ways. (1) I showed you how to find the ADP corner, but you seem to have ignored this. (2) If you could do what I did in my response, why bother asking the question...

I'll assume angles that look like right angles are right angles.
Drop a perpendicular from C to the line AD, and call the point of intersection Y. This line intersects LP at one point, say Q, and...

Since the original topic is likely dead (after having been derailed twice, and having relatively little content to begin with), I'll make a few "meta" comments.
Boy, this was a painful thread to...

The timings you requested, first "GaussLCM", then naive:
In [110]: %timeit sum_quotients5([2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43], 100000)
10 loops, best of 3: 93.9 ms per loop
...

Here's a little speed comparison data using the code in my previous post on my personal machine:
: %timeit sum_quotients5([174, 192, 934, 554, 1234, 4321], 100000)
10000 loops, best of 3: 199...

Since this is "codeitbetter", I thought I'd mention the Python implementation of some of these algorithms, since many of these operations are particularly straightforward in standard Python. The...

That's wonderful, glad to be helpful in sparking some interest in math.

Let D(n) denote the number of divisors of n. Our first goal is to show D is "multiplicative". That is, suppose n and m are relatively prime. We'll show D(nm) = D(n)D(m).
If d divides n and d'...

Yup, that's true. Edit: though to be completely clear, you can also compute the other two angles from that information. The largest angle isn't privileged in any way.
Yeah, it should clarify...

As I understand it, you (at the moment) want to compute the angle with vertex in the middle of the red rectangle, with the black line and one of the pink lines forming the rays. In that case, I don't...

Yup, that's what I mean. Sorry, I totally forgot I used the factorial notation without defining it. For a more thorough example of my formula, if you have R, R, R, R, G, G, G, B, Y, the answer is
...

This is a classic problem with a conveniently straightforward solution. Using R, G, B, Y, R, first replace the second R with R'. There are 5! = 120 ways to permute those five characters, but really...

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