At about the 0:01:50 mark in your commentary, you said, in essence, that your movement down the stairs was un-optimized; is it reasonable to go back, re-optimize that movement, and continue improving the TAS from there?
After finding this RTA run, I figured it's as good a route as any to build a TAS from:
Link to video
So far, this is the only run of the game on speedrun.com and the runner is also on Twitch.
Here are some new links for the Potential Emulators:
There are two distinct PCSX2-rr projects (Google Code link isn't dead, but it also didn't point to either of these, and Google Code projects can no longer be updated):
VICE (existing domain name was snatched by a group claiming to be an Arizona-based legal team)
Also, the link for nullDC isn't dead, but the emulator itself isn't being developed anymore; the Atari++ link timed out for me, but archive.org did say that the download link for the program itself was still live, and I could still download it, so I guess it's not dead yet.
A surd of the form a+b*sqrt(2) has the cube (a^2+6b^2)a+(3a^2+2b^2)b*sqrt(2), so if the cube roots of 7+5sqrt(2) and 7-5sqrt(2) are surds, you can find them by solving a pair of cubic equations (I'm starting with the + case):
(a^2+6b^2)a=7
(3a^2+2b^2)b=5
From the first equation, 2b^2=7/(3a)-(1/3)a^2, so substituting into the second,
(3a^2+7/(3a)-(1/3)a^2)b=5, so
(8a^3+7)b=15a, so b=15a/(8a^3+7).
Then squaring and re-substituting,
450a^2/(8a^3+7)^2=7/(3a)-(1/3)a^2, and then
1350/(8a^3+7)^2=7/a^3-1
1350a^3=(7-a^3)(8a^3+7)^2
1350a^3=(7-a^3)(64a^6+112a^3+49)
1350a^3=343+735a^3+336a^6-64a^9
64a^9-336a^6+615a^3-343=0
This is a cubic in a^3, and testing potential rational roots yields the root a^3=1, from which a=1.
Using that root, b=15/(8+7)=1; indeed, (1+sqrt(2))^3=1+3sqrt(2)+3*2+2sqrt(2)=7+5sqrt(2).
This suggests that maybe the algebraic cube root of 7-5sqrt(2) is 1-sqrt(2): (1-sqrt(2))^3=1-3sqrt(2)+3*2-2sqrt(2)=7-5sqrt(2).
Therefore, the given expression simplifies to 2.
Famously, lim(x->∞) (1 + a/x)bx = eab
This got me thinking what happens if that 1 were something else. Not a very hard problem given that even I could figure it out, but thought you might find it a mildly interesting question.
If 1>|c|, then lim((c+a/x)x,x,+∞) has the form (something that eventually is less than 1 in absolute value)(something that approaches +∞) and therefore approaches 0; the case for |c|>1 is analogous (it approaches a directed infinity with the same polar angle as c), and in the case where |c|=1 but c does not equal 1, there is no limit.
The first time I tried the multiple integration, I tried using a bunch of intermediate simplifications, and I got about 0.907; then I learned how to set up an ImplicitRegion in Mathematica to just numerically integrate in one step, and I guess I didn't figure out how to make it more accurate (that's where I got about 92% from).
2084 is the same as 1876 which cycled 40 years through 1900.
I was looking for other leap years starting on Saturday and overlooked that one; the way I figured it was just that the leap years starting on Friday (including 2072) had their previous 40-year gap the longest time ago, among those leap years in the 21st century with 40-year gaps.
The article linked from the Riddler Classic problem says the maximum board size is 6x6, but I haven't bothered to figure out a draw configuration for that size, or to check why it doesn't work for larger sizes: http://fivethirtyeight.com/features/dont-throw-out-that-calendar/#ss-1Amaraticando wrote:
A nice problem from the brazillian mathematical olympiad (2016, senior and junior).
Let ABC be a triangle. The line r is the bisector of ∠ABC and line s is the bisector of ∠ACB. D is onto r and AD||s. E is onto s and AE||r. The lines BE and CD meet at point F. I is the incenter of ABC. Prove that F, A and I are collinear if, and only if, AB = AC.
If AB=AC, then angles ACB and ABC are congruent, and then, angles ACE and ABD are congruent.
Marking the intersection between AC and BD as G, and between AB and CE as H, by ASA we have triangles CAH and BAG congruent, which means CH=BG, AH=AG, and the angles AHC and AGB are congruent.
Then angles AHE and AGD are congruent, and by alternate interior angles, angles HAD and GAE are congruent. Then because m(HAD)=m(HAG)+m(GAD) and m(GAE)=m(HAG)+m(HAE), it follows by elimination that angles GAD and HAE are congruent; then by ASA, triangles GAD and HAE are congruent, which means AE=AD.
Then by SAS, triangles ABE and ACD are congruent, which means angles ABE and ACD are congruent; by adding congruent angles, this means that angles FBC and FCB are congruent, so FB=FC.
This means that angles BAC and BFC are vertex angles of the respective isosceles triangles BAC and BFC, so their angle bisectors are also the perpendicular bisector of the base, BC, which means that F, A, and I are collinear.
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Conversely, if F, A, and I are collinear, then angles CAF and BAF are congruent (being supplementary to the congruent angles CAI and BAI). Angles DAE and DIE, being opposite angles in a parallelogram, are congruent; by alternate interior angles, angles IAD and EIA are congruent, as are angles AID and IAE.
Again marking the intersection of AC and BD as G, and AB and CE as H, it is clear that by SAS, triangles HAI and AIG are congruent, so angles AHI and AGI are congruent, so their supplementary angles AHE and AGD are congruent. Then by alternate interior angles, angles HAD and GAE are congruent, and by subtracting a common angle, angles HAE and GAD are congruent; then by adding congruent angles, angles EAI and DAI are congruent; their supplementary angles, EAF and DAF, are also congruent.
Then AI is the base of two congruent isosceles triangles EAI and DAI, so EA=DA and EI=DI. By ASA, triangles EAF and DAF are congruent, so EF=DF and angles EFA and DFA are congruent, which means AI, which F is also on, is actually the bisector of angle BFC.
By ASA, triangles HAE and GAD are congruent, so AH=AG and EH=DG. Also by ASA, triangles AHI and AGI are congruent, so HI=GI. By alternate interior angles, angles EIB and DIC are congruent. By adding congruent angles, angles CEF and BDF are congruent, and then by ASA, triangles CEF and BDF are congruent, which means CE=BD, and by subtracting congruent line segments, CH=BG.
Then by ASA, triangles HIB and GIC are congruent, so HB=GC; adding congruent segments, AB=AC.
FractalFusion, the way I understood it, you had to make the seven bets ahead of time, and if you label them from A to G (where positive numbers are for bets that the Cubs win), you get this system for the first five bets, looking at a series won in four or five games:
(This system is for the Cubs winning, but the equations for the Cubs losing are obtained by multiplying both sides by -1, like "-A-B-C-D=-100", so they are the same equations.)
Add the last four equations to get 2(A+B+C+D)+4E=400, and substitute in A+B+C+D=100 to get 200+4E=400, from which E=50; then back-substitute and add the last four equations in pairs and simplify to get A+B=50, A+D=50, C+D=50, and B+C=50, from which B=D, A=C, and C=B. Then substitute into the first equation to get A=B=C=D=25.
Two of the equations for a series won in six games look like this:
A+B+C-D-E+F=100
A+B-C-D+E+F=100
Back-substitution (from the four-or-five-game case) yields
F=100
50+F=100
and the second of these equations has solution F=50, which is inconsistent with the first one.
This all assumes that if you placed a bet for a particular game, and it isn't played because one team or the other won the series, you just get your money back.
This Riddler question has an obvious and straightforward extension, but the Extra Credit section uses a much more difficult extension: http://fivethirtyeight.com/features/how-big-a-table-can-the-carpenter-build/
I figured out approximately the maximal radius for N=3, and the N=4 case was easy, but I didn't find a real pattern, just a sense that the radius increases with N.
I tried summing an expression involving binomial coefficients (in terms of the net number of cycles around the circle, the number of lefts, and the number of statisticians) over the number of lefts (considering only the case where there are at least as many rights as lefts, and treating the other case as symmetric), and I got an expression in Wolfram|Alpha involving a hypergeometric function, which it was able to express as an exponential when N=1, and when it could presume the remaining parameter was an integer.
Specifically, for a sequence in which the bill goes right around the circle k times, where k is a whole number, it takes m lefts and m+N*k rights, where m is a whole number; the probability of the bill taking this kind of sequence is nCr(2m+N*k,m)/3^(2m+N*k+1).
Wolfram|Alpha gave the sum over m from 0 to infinity (using "Binomial" instead of nCr) as F((N*k+1)/2,N*k/2+1;N*k+1;4/9)/3^(N*k+1), where F is the standard (2,1) hypergeometric function.
When I tried evaluating that directly, for the case N=1, I got ((3-sqrt(5))/2)^k/sqrt(5). Also, regardless of the value of N,
when k=0, both former and latter expressions are 1/sqrt(5).
There probably are some elegant identities with binomial coefficients that would allow you to calculate this directly, without involving a hypergeometric function first.
Anyway, it's easy to see that because N*k is also a whole number, that expression involving the hypergeometric function simplifies to ((3-sqrt(5))/2)^(N*k)/sqrt(5), and the probability is two times the sum of that expression over k from 0 to infinity, minus the k=0 case, to avoid double-counting.
It's a geometric sum, and it's then easy to see that the probability goes to 1/sqrt(5) as N approaches infinity; what looks trickier is that it seems like all of the probabilities are rational for finite values of N, but it looks like a slog to prove it.
