Posts for Flip


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If I have the first n natural numbers arranged in a random order, how many would you expect to be in their correct locations?
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Well I don't quite understand your initial formulae, but the information we need is where you want to end up. Regarding (3) at least, you'd have to know the coordinates of where you're going. EG if we want to go at angle Red, but the script is off by angle Blue, thus we need an angle of Purple to compensate, this will depend on how far we want to go in the intended direction. We'd need to know initial/final location.
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Just from the fact that the integers are a unique factorisation domain, we have the primes dividing a number will be the only primes dividing all powers of it; thus we cannot have one power being a prime multiple of another. IE For a given prime P P1/n= a/b P=an/bn Pbn=an Given our prime factorisation a=p1x1p2x2...plxl b=q1y1q2y2...qmym an=p1nx1p2nx2...plnxl bn=q1ny1q2ny2...qmnxm For Pbn=an, we have Pbn=q1ny1q2ny2...PnyP+1...qmnym =an=p1nx1p2nx2...PnxP...plnxl For these to be equal, due to unique factorisation of primes, we need all pi=qi, and xi=yi. But just consider the power of P, clearly we have nxP+1=/=nxP. Thus the two are not equal, and thus one power is not a prime multiple of another. Hence nth roots of primes are not rational.
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ex = 1+x+x2/2!+x3/3!+x4/4!+x5/5!+... n!/(n-m)! = n.(n-1).(n-2)....(n-m+1)=nm+{ O(n(m-1)) } -->nm as n->inf (1+1/n)n=(nC0)(1n)(n0)+(nC1)(1n-1)(n-1)+(nC2)(1n-2)(n-2)+(nC1)(1n-3)(n-3)+... =1+1+n.(n-1)/2!.(1/n2)+n.(n-1)(n-2)/3!.(1/n3)+n.(n-1)(n-2)(n-3)/4!.(1/n4)+... -->1+1+n2/2!.1/n2+n3/3!.1/n3+n4/4!.1/n4+n5/5!.1/n5+... =1+1+1/2!+1/3!+1/4!+1/5!+...=e1 =e
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Pokota wrote:
I just love this concept that it has to be perfect to be designed. Then again, I'm of the opinion that evolution is a tool used by God to develop life, so...
https://www.youtube.com/watch?v=ckJPHTo-mvM Moist air rising into cold air and condensing, is how the gods makes rain. Static electricity is how Zeus throws lightning bolts at you. Gravitational pull of the moon as it goes around the earth is how King Neptune makes waves to sink your boat. Nuclear fusion in the sun is how Ra sends out radiation to light up the world. Genetic coding, instincts, and a big wallet is how Cupid makes people fall in love. Oh, and evolution is how god makes life forms. Yes, people really are this stupid, it's embarrassing. As a society, we should be beyond religion by now.
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For those who don't know, this is a rerelease of the Amiga game Sleepwalker, just using the Eek! The Cat! license. Instead of a dog escorting your sleepwalking master around, it's now Eek and his cartoon buddies. Anyway, this run always desyncs at Level 02 Flying Saucers for me. I've tried Bizhawk 1.11.1, 1.13.1, 1.9.1, same problem. It's possible I'm using the wrong rom, but have you encountered any problems/solution to this?
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a2=b2+c2-2bcCos(A) ABC=Cos-1( (272+302-222)/(2x27x30) )=Cos-1(1145/1620)=~45.03' OBC=Cos-1( (232+302-162)/(2x23x30) )=Cos-1=Cos-1(1173/1380)=~31.79 ABO=ABC-OBC=45.03-31.79=~13.23' d2=272+232-2x27x23Cos(13.23) =~49.00000121 d=~7.0000000086
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My current data points have a mean value of 4.4, I add an additional point, 9.2 to the set, and the mean becomes 4.7. How many data points do I have?
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32, as each Knight must move/attack to a different colored square, thus place 32 on the same color.
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Warp wrote:
That got me thinking: What is the probability of getting a straight if you were to throw five different dice: A d4, a d6, a d8, a d12 and a d20?
There are 4 potential straights to make, easiest to visualize with a diagram: Here, we have 5 columns for the 5 dice, each representing their number range. The barely visible green lines represent the region where each of the 4 straights can take place, with a random example given in red. The probability for each straight occurring will be calculated separately. Each die needs to contribute some number to the straight. This needs to be in the straight's range, and has not been occupied by the previous dice. Thus they have probabilities : P(1:5)=4/4 x 4/6 x 3/8 x 2/12 x 1/20 =1/480 P(2:6)=3/4 x 4/6 x 3/8 x 2/12 x 1/20 =1/640 P(3:7)=2/4 x 3/6 x 3/8 x 2/12 x 1/20 =1/1280 P(4:8)=1/4 x 2/6 x 3/8 x 2/12 x 1/20 =1/3840 Thus total, P(Any straight)=(8+6+3+1)/3840=18/3840=3/640
Warp wrote:
And while we are at it, what's the probability of throwing 5-of-a-kind with those dice?
P=4x(1/4 x 1/6 x 1/8 x 1/12 x 1/20)=1/11520
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Warcraft II-Orc Expansion, being 10/12 levels in; (**not so**) easily done this year. That'd be 3/4 runs done on PSXjin, no doubt I'd prefer making Human Expansion on it too.
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Warp wrote:
I don't trivially see why this equality is true when |x| < 1: 1 + 2x + 3x2 + 4x3 + 5x4 + ... = 1/(1-x)2
Standard formula for geometic series. Sn=1+X+X2+X3+X4+...Xn S[Sub]∞[/sub]=1/(1-X) =(1-X)-1 for |X|<1 dS/dx=1 + 2x + 3x2 + 4x3 + 5x4 + ... =d/dx (1-x)-1 =1/(1-x)2
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Hmmm, who explained to them where the doors lead, was it the same person he said "Show me your door" to? Obviously asking that causes the truth teller to point to his own door, and the liar to point to the other; identifying both who is being honest and which is the safe door. Therefore, if upon asking this you find that he points away, that means you're currently speaking to the liar. And if this is the source of your original rules for the riddle, then clearly they're not true. That means the 2 doors are not Liar/Truth, but either guarded by Liar/Liar, or Truth/Truth. As we've already established the current one is lying, then that would imply both are, and that no door leads to safety. PS, does this explain why in Labyrinth, Sarah ends up in the pit? Her logic was fine, but I guess if the original set up for the riddle was performed by a liar, then that's why she still got screwed.
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Which 6 countdown numbers would give the greatest possible range of attainable answers?
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Well if "How many numbers with property X are there?" is a valid question, there's all sorts of TAS equivalents. How many games have ACE potential? How many justifiable categories can we invent for a game? Etc etc. Will new glitches ever not be found in Zelda games? Seriously, how many different routes have to be invented for OOT... What is the ideal ratio between Rerecords and movie length? Taken over all movies, does it converge on any interesting number? Imagine if it was the golden ratio. Will bots be able to TAS with greater or equal efficiency of humans? Currently we cannot brute force stuff, and they're not intelligent enough to understand what the hell they're doing, but when will they be? When will bots be able to hypothesis/test potential glitches for itself? If a NES game for example is made from finite coding, finite memory, with finite calculations going on, then is there a finite limit on the maximum amount of glitches a game can have? What is it?
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K, time for some grammar trolling. "By pressing the rewind button your rewinding current progress" "My PC is kinda slow if your looking at that FPS" "You need a little knowledge about the controllers your using" Quite a few 'you're' mistakes there, as well as 'controls' being misspelled too. Next, you frequently have black and white text over the fuzzy black and white background of Bizhawk, making it harder to read. What little information you do provide is trivial and pretty obvious I would've thought anyway; and your climactic ending TAS failed to explain what to actually accomplish in a TAS.
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Annoying trying to do that test in another language, but I got "Älykkyysosamääräsi on noin 119 , joka on korkeampi kuin 89.7 prosentilla väestöstä"="Intelligence is about 119, which is higher than 89.7 percent of the population". Pretty good. For that particular puzzle, I say Top Right. "Reason": I notice the 2 smaller squares for columns 2/3 are in the same location for a given row, yet always different from column 1. Applying that to the bottom row seems to work, as we have column 1/2 having different small square locations, so we should have column 2/3 matching their locations; which would place them in the bottom right. There is only one such picture with that, the top right option. That would only partially explain it though, so to explain the rest, we can consider the initial squares in the first column, as 'growing', with the adjacent pictures showing progress. Nothing initial in the first row, thus no growth. In second row, we have 1 left square grow, then the other. In the third row, we have the slice of the 2 smaller squares, then 2 larger ones, so we'd need a halfway point, thus a picture with 1 small 1 large, both still pointing downwards. Curiously, the top right option. Feel free to come up with a better solution, but that's what I've got.
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In the Articles section you can find general information on Memory Search and RAM Search. To find particular addresses, define your search criteria on the lower half of the Ram Search window, then press Search. This will filter out all addresses not matching your criteria, then repeat the process until you find the one you want. This may require a few assumptions, which if incorrect will not yield the result you want, so it may require multiple tries. If you don't quite understand, then start with super easy values to find. EG the lives value, since you have access to this value in games at all time anyway, it's not like it's hidden data you need to extract, but it's a good start. IE -Start game, try "Equal to", "Specific Value", "3". Search. Obviously plenty of address will have this number, so need to filter it down even more. Do a bunch of random stuff in the game (without dieing), then press Search again to use the same criteria. -This time die, and now change the criteria to say "Equal to", "Specific Value", "2". Obviously the lives value has changed, where as not every one of the current results will have, and so it filters out the incorrect values. That is of course, assuming lives value gets stored like that; but without being an expert in cryptography, I'm pretty sure that's how it's done. -Also trying "Equal to", "Number of Changes", "1" would be stronger criteria, and would filter out more results. By coincidence they'll be other results, eg 0006BA4E| 3, 5, 7, 2, 6, 6, 2, 16, 80 0008AEE4| 3, 3, 3, 3, 2, 2, 2, 2, 2 Pressing 'search' on bolded frames would clearly yield both as valid results, yet we can obviously conclude the above sequence does not keep track of lives. Searching for values which have only changed once would filter out the top one, but still give the bottom one. -If you still haven't found it, then repeat the process. Die again, search for Equal to 1, plus Equal to Number of Changes 2. By which time you should find it, you can then click WATCH to save this value to your Watch list, to keep track of later. That was super easy, since we knew what we were looking for. For XYZ coords, things are more difficult, because you don't know "Ah, clearly I'm at x=59041 right now, so I know what to search for", which means your search criteria needs to be more vague. IE you can search for "equal to previous value" when stationary, or "greater than previous value" when moving, maybe clicking specific addresses and 'eliminate' if you can see already that they're not the address you want. However, be warned of the aforementioned assumptions. For example it's common to have the Y Axis on 2D games pointing downwards. So if you jumped, and search for 'greater than previous value', clearly you would not find the value you want.
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Small problem with the description, it refers to him as "Dr. Robotnik", where as his current name should be "Dr.Eggman".
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I've since discontinued my work on Orc Expansion to pursue other projects. I've got to level 6, but got demotivated by the lack of strategy emerging. The entire place seems too fortified, guard towers everywhere which cannot be dodged, so it appears to be the first level where an early attack is not possible. I'll get back to it eventually.
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Voting yes, and agreeing with this being an Any% run, to be published alongside the current 100% run.
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Since it's arbitrary code, you could just rewrite every flag which keeps track of what gets done, and thus give yourself 100% completion without the need for a modified playthrough. It could be done for fun, but certainly not the fastest means of getting 100%.
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First note that on the right we have only a linear combination of f, but on the right we have a function of a function. This means that for a polynomial, the left side increases much faster than the right side does. EG a quadratic of a quadratic turns it into a quartic, degree 3 to degree 6, etc. This means to control this growth and equalize both sides we cannot have our function above degree 2, hence we have f(x)=ax+b. 2f(x)=2ax+2b 2f(x)+xy=2ax+2b+xy=(2a+y)x+2b f(y)=ay+b. xf(y)=axy+bx. xf(y)+f(x)=axy+bx+ax+b=(ay+b+a)x+b f(xf(y)+f(x))=a((ay+b+a)x+b)+b=(ay+b+a)ax+ab+b=(a2y+ab+a2)x+(ab+b) These need to be true for all x,y, eg x=0. 2f(x)+xy=f(xf(y)+f(x)) 2b=ab+b 0=ab-b=b(a-1) b=0 or a=1 For b=0 2f(x)+xy=f(xf(y)+f(x)) (2a+y)x=(a2y+a2)x 2a+y=a2y+a2=a2(y+1) True for all values, eg y=0 2a+y=a2(y+1) 2a=a2 0=a(a-2) a=0, or 2. {a,b}={0,0},{2,0} For a=1 2f(x)+xy=f(xf(y)+f(x)) (2+y)x+2b=(y+b+1)x+(b+b) (2+y)x+2b=(y+b+1)x+2b (2+y)x=(y+b+1)x 2+y=y+b+1 2=b+1 1=b {a,b}={1,1} TESTING {a,b}={0,0} f(x)=0x+0=0 2f(x)+xy=0+xy=xy f(y)=0 2f(y)+f(x)=0+0=0 f(2f(y)+f(x))=f(0)=0 =/=0+xy BAD {a,b}={2,0} f(x)=2x+0=2x 2f(x)=4x 2f(x)+xy=4x+xy xf(y)+f(x)=x(2y)+2x=2xy+2x f(2f(y)+f(x))=f(2xy+2x)=4xy+4x =/=2f(x)+xy BAD {a,b}={1,1} f(x)=1x+1=x+1 2f(x)=2x+2 2f(x)+xy=2x+2+xy xf(y)+f(x)=x(y+1)+(x+1)=xy+x+x+1=xy+2x+1 f(xf(y)+f(x))=f(xy+2x+1)=(xy+2x+1)+1=xy+2x+2 =2f(x)+xy GOOD Hence the only function is f(x)=x+1. The other 2 generated values failed, because they only satisfied the equation at the local value x=0, (or y=0, IE along main axes).
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Find the area of an intersection, both are the same.
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33=27=26+1=1 mod 13 31974=3(3x658)=(33)658=1658=1 mod 13 52=25=26-1=-1 mod 13 51974=5(2x987)=(52)987=(-1)975=-1 mod 13 31974+51974=1+(-1)=0 mod 13
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