Inb4 someone goes to fix the Wikipedia page before this movie gets judged. This would be the ultimate troll.
What a weird ending. There is a score counter and you went for "first 2048 tile" as an ending point?
Uzebox huh? Long time no see. Could you explain how this game works? All I see are numbers rapidly changing.
but tracing it backwards starting from x=N/m (going outwards) until the x value is less than or equal to 1. Note that it is not actually possible for floor(N/phi) and ceiling(N/phi) to both be maximal sequences for the same N; by alternation, one has even length and the other has odd length.
What I found very curious is that, from my calculations, most N have a unique maximal sequence, but for some N, there are actually two maximal sequences. I wasn't able to find any N that had more than two maximal sequences so I'm not sure if they exist (I checked up to N=20000 or so). No, this is not acknowledged anywhere in Riddler's solution, not even in the PDF on the page that gives a list of maximal sequences up to N=200.
There appear to be an infinite number of them but the first few with two maximal sequences as far as I can remember are N=6, 15, 32, 35, ... Considering the Fibonacci fractions 1/1, 2/1, 3/2, 5/3, 8/5, ... mark the boundaries for which values closer to phi give longer sequences, some values of N have two values m1 and m2 where N/m1 and N/m2 both fit in the same "gap". For example, for N=35:
35/23=1.5217...
35/22=1.5909...
Both conveniently fit in the gap between 3/2=1.5 (exclusive) and 8/5=1.6 (inclusive). This gives the two maximal backward sequences (35, 23, 12, 11, 1, 10) and (35, 22, 13, 9, 4, 5). You can check that 35/21=1.6666... is exactly 5/3, so it fits in the gap from 2/1=2 (exclusive) to 5/3=1.6666... (inclusive), a worse gap than 3/2 to 8/5.
I didn't think about it too much, so perhaps there is a proof that an infinite number of N have two maximal sequences, and/or a proof that there is no value of N with more than two maximal sequences, but I probably won't come back to that anymore.
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As a quick problem, I will present the newest Riddler Express (I will just copy and paste the description):
You have a case with 11 golden globes, weighing 1 kilogram, 2 kg and so on, up to 11 kg. And you know exactly which globe is which.
You have arranged to sell one of the globes to a queen. She has heard tales of these orbs, and knows for a fact that they have masses from 1 kg to 11 kg. However, she doesn’t know which is which and will not simply take your word for it. She will purchase a globe if you can demonstrate its weight.
The queen has a balancing scale with two plates, one on each side. It shows whether the total weight on either plate is equal or, if not, which side is heavier. The queen can clearly see which globes you place on which plate. However, if at any point the mass on either side exceeds 11 kg, the scale will break and the queen will refuse to buy from you.
The queen is in a hurry for her next appointment, so time is limited. What is the fewest number of weighings by which you can prove the weight of at least one globe? Which globe is it?
Answer: Only one weighing required, proves the weight of the 11kg globe. Reasoning: Put 1 2 3 4 on one side and 11 on the other. Four globes on one side weigh a minimum of 10kg. This confirms the lone globe on the other side to be 11kg, since it's the only globe that is possibly heavier than four globes.
Why does my intuitive answer (light lost to inefficiency equals light lost to polarization) closely match the exact answer (a transcendental equation based on the derivative of the function)?
(By the way, light intensity is proportional to the cosine squared, but whatever. Don't learn your physics from mathematicians.)
This problem is pretty hard to solve by hand, but there's a way to solve it using freely available software and math tables. For which natural numbers n is the sum of the first n perfect squares also a perfect square?
Of course, with an unlimited number of polarizers, you could just use an arbitrarily large number of polarizers to make a "circular turn" to orient the polarity of the light horizontally while keeping a fraction of the light that is arbitrarily close to 1.
This is where Riddler Classic throws a twist: each polarizer is only 99% effective. So, only 0.99 of the light that is supposed to pass through each polarizer actually passes through.
With this in mind, what number of polarizers should be used to orient the polarity of the light horizontally while keeping the largest possible fraction of light? And what is the largest possible fraction?
(You may wish to use WolframAlpha to help out with this.)
For fun: If S is an empty set, show that inf S = +infinity, and sup S = -infinity Is there a non-empty subset S of the reals such that inf S > sup S?
A secret society would like to poll their 60 members on some yes/no question. They are only interested in the parity of the number of yes votes. However, they will only allow a given pollster to interview 50 members, and each pollster may only report back a number between 0 and 50 inclusive. It is believed that up to one pollster may lie or make a mistake in reporting. * Find the smallest number of pollsters required (easy).
Here is a counting exercise that requires some basic musical knowledge*: How many distinct rhythms are possible in one measure of 4/4 time if you are limited to using whole, half, quarter, and eighth notes or rests and any number of ties among them? The rhythm must be contained entirely within the measure, so ties across bar lines are not allowed.
since the second term goes to 0 and is always positive. So you can just get the number of such rhythms of length n by evaluating the first term and rounding up to the next integer.
P.P.S. The sequence is OEIS A001519 (offset by 1 index), and the page mentions that this sequence can be thought of as "Number of musical compositions of Rhythm-music over a time period of n-1 units."
Edit: Fixed an error in the image.
Here is a counting exercise that requires some basic musical knowledge*: How many distinct rhythms are possible in one measure of 4/4 time if you are limited to using whole, half, quarter, and eighth notes or rests and any number of ties among them? The rhythm must be contained entirely within the measure, so ties across bar lines are not allowed. * There are numerous music theory tutorials online for those who do not know what the terms mean but want to learn.
Given a function that gives evenly distributed random real values between 0 and 1, let's call it R(), the probability distribution of sqrt(R()) is exactly the same as the probability distribution of max(R(), R()).
Based on the triangle inequality, to minimize the sum of the two distances, (x+2,0) must be collinear with (0,3) and (6,-5), giving two similar triangles. Splitting 6 in the ratio of 3:5 gives 9/4:15/4. So x+2=9/4, and so x=1/4 minimizes f(x).