Posts for OmnipotentEntity

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Wouldn't just shuffling the standard way, but flip one half of the deck each time, work?
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FractalFusion wrote:
What is F/A analysis? Never heard of it before.
Took a class called "Engineering Economy." It had financial formulas. That one was called F/A, which as far as I've been able to gather stands for "Future Value given Annualized Payment."
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This is just a F/A analysis right? If then you just need to solve: 500 = ((1 + i)^N - 1)/i where i = 0.01, for N. Which gives 180.07 months, or 15 years.
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https://pastebin.com/ZYbscnCu Ripped from youtube video
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A simple way to prove that a triangle's internal angles are 180 degrees to simply use the fact that the external angles sum to 360 around any closed convex polyhedra, and the internal angle + the external angle is 180, and there's three such angles, 3*180 - 360 = 180. As far as I know, this doesn't use parallel lines at all. The sum of external angles can be proven by observing that in order to walk around a polyhedra you need to end up facing the same direction you began in. And the external angle is the change in heading at each corner. Whereas the fact that an internal angle + an external angle comes from the definition of an external angle directly. Actually, upon reflection it does, because it assumes that the heading direction does not change between beginning a side and ending it, which is rather antithetical to the notion of a straight line, but proving it requires constructing parallel lines. I'm leaving it just in case someone else had this idea, to help explain why it's wrong.
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Warp wrote:
Wikipedia defines Joule as "the energy dissipated as heat when an electric current of one ampere passes through a resistance of one ohm for one second." This seems to imply that a resistor of 1 ohm will always dissipate the exact same amount of heat regardless of what material it's made of. That feels surprising and counter-intuitive. Or, perhaps, it could be restated as: It seems to imply that there's a tight unique relationship between resistance and heat dissipation, regardless of the material.
That's not strictly true though. For instance, if you have an LED, the energy is dissipated as light. If you have a speaker, it's dissipated as sound. If you have a motor, the energy becomes mechanical work. And heat makes up only the rest of the energy. However, if we're talking about a passive resistive component that does no external work, then Wikipedia is correct. This comes simply from the conservation of energy. You can think of it in reverse. There isn't any reason for two materials to have the same heat dissipation when current is ran through it. Instead the amount material and its geometry determines how much energy is lost, and from this, the value of its resistance may be determined.
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Any n-bit number that ends with a 0 is even, whether it is positive or negative. If it ends with a 1 it is odd. Let us say for the sake of argument that we're dealing with 8 bit signed integers. Our two cases are: XXXXXXX0 * XXXXXXX0 and XXXXXXX1 * XXXXXXX1 Multiplying each of these through we get: XXXXXX00 for the even side. And the 0s tend to accumulate. For the odd side it's a bit more complicated. In the 1s digit we're left with a 1, obviously (odd * odd = odd); however, in the 2s digit we have the old 2s digit + old 2s digit. Because 0 + 0 = 0 and 1 + 1 = 0b10, this means that a 0 will live in the 2s place after multiplication. XXXXXX01 On the next iteration another zero appears in the new 4s place because it is equal to the 0b10*current 4s. The twos digit does not change because it is equal to 1 * 0 + 0 * 1. So we get an additional 0 XXXXX001 And so on. Eventually the zeros will pile up until it shifts them out of the precision of the integer regardless of how many bits it is.
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A minor remark. I just found out an updated version of this game was released in Jan 2017. Among other things, it increases Iji's walk speed. Perhaps now the previously actually impossible reallyjoel's dad difficulty is possible?
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If the numerical approximation is 1.1249976996... then the formula I have isn't too terribly bad after simplification, but you're correct that it's not invertible.
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Warp wrote:
(I understand that it's probably not possible to say "proof by contradiction is impossible for this number." I suppose what I'm asking is if there's some number that can be proven to not be rational, but for which there is no known proof by contradiction, or that proof is way too complicated to be feasible and there's a much easier way of proving it.)
Unfortunately, there is no real way to "access" irrationality directly. An irrational number is simply a number that is not rational, but a rational number is a number that can be expressed as p/q where p and q are integers. Because the definition of an irrational number is in terms of what it is not, then you have to prove that it's not a rational number, which means proof by contradiction. Other related properties of irrational numbers, such as a non-repeating decimal expansion, are consequences of this property and are themselves proven through contradiction (try it!). This means that the other standard argument to prove irrationality for numbers such as the Champernowne constant relies on proof by contradiction as well, albeit indirectly.
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I can diagonalize the matrix, but that requires fixing a value of h. I thought I could get a recurrence for h as well then somehow combine the two methods to get a single function in both f and h, but I couldn't figure it out. By continuing down this path I'm not suggesting that FractalFusion's answer is wrong or insufficient. (On the contrary, it's far and away more elegant than mine.) I was just trying to suss out why my answer wasn't working, and/or how to make it better.
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FractalFusion wrote:
Do you mean C(f+1,h) = 2 * C(f,h) + ~C(f-h, h)? This one instead would agree with your calculation C(6,2) = 2 * C(5,2) + ~C(3, 2) when f=5, h=2.
Yes, you are correct, I made a mistake copying from my notes.
FractalFusion wrote:
Some of my calculations don't seem to agree with this. For example: C(4,2)=8 (HHHH HHHT HHTH HHTT THHH THHT HTHH TTHH) C(4,1)=15 (everything but TTTT) ~C(3,1)=1 (TTT) 8≠(15-1)/2=7
That part may be entirely incorrect, as I didn't use or test out the h side in recursion, and I might have made a simple math error of some sort. Or it may simply be that the identity does not hold in certain cases. I never actually proved anything, I just saw a pattern and extrapolated. In this case: C(4, 1):
HXXX | True
THXX | True
?THX | True
??TH | True
Other | False
C(4, 2):
H(H)XX | True
TH(H)X | True
?TH(H) | True
(??TH) | False
Other | False
The assumption ? in the third lines take on different values between the lines. In C(4,1) the ? can only be T whereas in C(4,2) it can be T or H, which is where the missing combination came from. So it seems as if the relationship I gave earlier does not hold for any case where the patterns on the ? changes with the changing h, which is actually quite often. Hell, it does not even hold for the original example I gave (C(6,2) -> C(6,3) the ?s on the 4th line change from 3 to 4 possibilities.) I apologize for that and thanks for catching it.
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I approached the problem using a toy example. Let C(f, h) be the number of flip possibilities out of the 2^f that satisfy the condition that h heads appear in a row. Enumerating C(5,2) gives the following truth table (X is don't care, ? will be explained later).
HHXXX | True
THHXX | True
?THHX | True
??THH | True
Other | False
The ? marks are combinations that don't include the target sequence. So in the third line with one ? it can be either T or H. But the fourth line ?? can only be TT HT TH, not HH (otherwise it would be double counted in line 1.) Let ~C(f,h) = 2^f - C(f,h) for C(5,2) we have: 2^3 (first line) + 2^2 (second line) + 2 (third line ?) * 2 (third line X) + ~C(2,2) If we move from C(5,2) to C(6, 2) we get:
HHXXXX | True
THHXXX | True
?THHXX | True
??THHX | True
???THH | True
Other | False
Which can be examined line this:
HHXXX(X) | True
THHXX(X) | True
?THHX(X) | True
??THH(X) | True
(???THH) | True
Other | False
And so we have C(6,2) = 2 * C(5,2) (X on each line from 1-4) + ~C(3, 2) (last line) From here the recurrence is simply: C(f+1,h) = 2 * C(f,h) + ~C(f-(h+1), h) And then we can use the initial conditions: C(f=h, h) = 1 C(f<h, h) = 0 To find any given C(f, h) using a recurrence relationship. Now that we have a recurrence relationship and we can use eigendecomposition to find an explicit formula. First we have to get rid of the annoying constant: C(f+1, h) = 2 * C(f, h) - 2^(f-(h+1)) + C(f-(h+1), h) C(f, h) = 2 * C(f-1, h) - 2^(f-(h+2)) + C(f-(h+2), h) C(f+1, h) - 2 * C(f, h) = 2 * C(f, h) - 2^(f-(h+1)) + C(f-(h+1), h) - 2 * (2 * C(f-1, h) - 2^(f-(h+2)) + C(f-(h+2), h)) C(f+1, h) = 4 * C(f, h) - 4 * C(f-1, h) + C(f-(h+1), h) - 2 * C(f-(h+2), h) So for any given h we can construct a matrix A that when you multiply {C(f, h), C(f-1, h), ... C(f-(h+2), h)} against it you receive {C(f+1, h), C(f, h), ... C(f-(h+1), h)}, or the next iteration. Given that matrix, we can perform an eigendecomposition, let S be a matrix of the eigenvectors of A, and Λ be a diagonal matrix of the corresponding eigenvalues of A. Then SΛS^-1 = A (by some matrix math) Then because Λ is diagonal, it's very simple to take the nth power of and A^n = S Λ^n S^-1 (Because S and S^-1 cancel when multiplied.) I allowed Mathematica to chew on the condition where h=2, and it spit out: -(1/(4 Sqrt[11]))i (2 i Sqrt[11] (1+2^(3+f))+Root[-1-#1-#1^2+#1^3&,2]^(f-2) Root[704-10128 #1^2+42340 #1^4+#1^6&,2]+Root[-1-#1-#1^2+#1^3&,1]^(f-2) Root[704-10128 #1^2+42340 #1^4+#1^6&,3]+Root[-1-#1-#1^2+#1^3&,3]^(f-2) Root[704-10128 #1^2+42340 #1^4+#1^6&,6]) It's interesting that FractalFusion's answer does contain the x^3-x^2-x-1 pattern, but I'm curious where the strange x^6 formula comes from. You can construct a similar recurrence on h using the following: C(6,2)
HHXXXX | True
THHXXX | True
?THHXX | True
??THHX | True
???THH | True
Other | False
C(6,3)
HH(H)XXX | True
THH(H)XX | True
?THH(H)X | True
??THH(H) | True
(???THH) | False
Other | False
So we can say: C(f, h+1) = (C(f, h) - ~C(f-h, h))/2 With the initial conditions: C(f, h=0) = 2^f C(f, h=1) = 2^f - 1 So: 2 * C(f, h+1) = C(f, h) - 2^(f-h) + C(f-h, h) 2 * C(f, h) = C(f, h-1) - 2^(f-(h+1)) + C(f-(h+1), h) 2 * C(f, h+1) - 4 * C(f, h) = C(f, h) - 2^(f-h) + C(f-h, h) - 2 * (C(f, h-1) - 2^(f-(h+1)) + C(f-(h+1), h)) 2 * C(f, h+1) = 5 * C(f, h) - 2 * C(f, h-1) + C(f-h, h) - 2 * C(f-(h+1), h) C(f, h+1) = 5/2 * C(f, h) - C(f, h-1) + 1/2 * C(f-h, h) - C(f-(h+1), h) And the matrix can be set up in the same way, let's call this one B with eigenvector matrix T, and eigenvalue matrix M. Now I'm stuck. I have two matrices, and two degrees of freedom, but the matrix size of the f formula depends on the size of h, and vice versa. How do I put these together to find an explicit formula in two variables of C(f, h)?
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This is one I ran into on the Reddit a few days ago, and it's burrowed into my brain. The original question is: What is the probability that if I flip 15000 coins, at least 15 consecutive heads appears at least once. The answer is a very large irreducible fraction that's about 20.45%. I won't explain in case someone wants to attack this problem. My question is: If C(f, h) is the number of combinations for "f" flips and "h" consecutive heads. (So that P(f, h) = C(f, h) / 2^f, and in particular P(15000, 15) = 20.45%), is there an explicit formula (ie, not recursive) for C(f, h) in terms of f and h. I have a method of finding an explicit formula in f for a given h. But I don't know how to extend it to an arbitrary h.
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I apologize, unless it allows you to attack more often, it does the same damage as the mega buster. I thought I remembered there being a slight difference, but I was mistaken.
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Would switching to the Air Cutter speed up both Quickman and the Sniper Joe section?
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p4wn3r wrote:
By the way, for the problem I proposed, the answer is: When you run a secure RNG from a secure cipher, it's guaranteed that all sequential outputs will be different from each other. That's because if you had two values x and y with f(x)=f(y), it would be impossible to decrypt the cipher, even with the key. The result is that, if you read all 2^n numbers generated, you'd see that they are a permutation of all values in the range [0, 2^n-1]. That's very improbable to arise from random chance, you'd expect some values to repeat! It's counter intuitive, but it's more probable that you see a value repeating even if you read a square root of the total number. See here for an explanation. So, if you simply read the numbers, at around the square root of total possibilities you should start seeing repetitions. If that does not happen, it's likely that the source of these numbers is not really random, but generated by a block cipher in counter mode!
This is clever, but this isn't really unique to secure block cipher PRNGs, this is true of any PRNG that doesn't keep a significant amount of hidden state. So for instance, the MT is safe from this attack because its internal state is much larger than the output state, but a basic LCG would be vulnerable because its output is its state. This threw me because I was looking for something specific and unique to it being a cipher, rather than just having a state that is fully utilized in the output phase.
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PikachuMan wrote:
I am going for the good ending, possibly the fastest route, but I may need assistance
Then you're way off route. You should immediately eject the first 11 missions, play 1, eject 1, play 1, eject 1, play 2, eject the last 4. Here's the RTA of this route: https://www.youtube.com/watch?v=RHGDJW_0LVA EDIT: it seems this route plays Rostov 1 instead of ejecting, double check if it's possible to eject from Rostov 1. I'm pretty sure it is.
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PikachuMan, If the goal is just "Good Ending," why aren't you taking the fly only 4/21 missions route that goes through Hubble's Star and Rostov? Do you have some other goal in mind? I'm just trying to help you submit a run that will be accepted. And it seems like without a solid goal like "Max Medals" it'll be rejected like the "Bad Ending" run.
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What about a max medals route? Only certain missions can give a medal, and you have to actually play the missions to get them. (And ejecting only gives you a medal the first time.) This might force an interesting route. This is (as far as I can tell) what medals are available. This means that the best route is to win every system except Kurasawa.
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It might seem a bit unsatisfying though. Because you only need to complete 4/21 missions in order to get the good ending. This diagram is how the game decides how the game proceeds and whether or not you get the good ending (Enyo has 2 missions, Hell's Kitchen and Venice both have 4, all others have 3): So the fastest route for *just* the good ending is eject every mission until you reach Hubble's Star. Then play Hubble's Star missions 1 and 3 (eject from 2). And play Rostov missions 2 and 3 (eject from 1). Finally eject from all 4 missions in Venice. (This is probably faster than an alternate route where you win Port Hedland, because Port Hedland requires that you achieve every objective in order to win, so you have to play all three missions.) Considering the TAS has already gone to McAliffe, PikachuMan is probably intending on either an all metals or max kills route or just play through the game "as well as possible." Completing all objectives in the shortest amount of time on the route that naturally follows from completing all objectives.
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Looks wonderful! I'm excited about this run. Shame about the escort mission being so slow.
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Ah, this is true. Forgive me. If we look at Amaraticando's solution, (all numbers in the form of m^n), then: 4 = 2^2, 8 = 2^3, 9 = 3^2, 16 = 2^4 = 4^2, 25 = 5^2, 27 =3^3 https://oeis.org/A001597 So the squares of these numbers can be represented, so the complement can not. https://oeis.org/A007916 And then simply square those: https://oeis.org/A153158
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@Warp, essentially the criteria for a is that it must be squarefree. A list of these numbers may be found in the OEIS. https://oeis.org/A005117
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Warp wrote:
Isn't even the notion that the universe might be infinite in contradiction with the hypothesis that at one point the universe, in its entirety, was a singularity that expanded? I don't think you can have both. That would be impossible.
It's not a contradiction. Space and time did not exist before the universe existed. So it's entirely possible that at the instant the big bang occurs and all space comes into existence, it comes into existence as infinite.
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