Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
OK, we both agree on Blue=3.52, so I guess the point where the edge touches the outer circle is still in that direction? I thought it would've been perpendicular, rather than in the same direction, so ignore that final calculation. Purple should be in Blue's direction.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
By previous question, at the origin we have a smaller circle of radius r, touching the surrounding five of radius R=1. This gives the interior circle a radius of
r = R*(1/(sin(pi/n))-1=1/sin(pi/5)-1= ~0.7 (Yellow)
Distance from origin to inner circle origin D(O,Oi)=r+R=r+1 (Red)
Distance from inner to outer circle origin D(Oi,Oo)=2R=2 (Green)
Distance from outer origin to outer edge D(Oo,E)=R=1 (Green)
But these are obviously not in the same direction, so to find the overall distance we need a bit of trig. First consider the closest triangle, formed by 2 reds and a green. This is isosceles, and we know the inner angle, thus:
A(Oi,O,Oi)=2pi/5
A(Oi,Oi,O)=(pi-2pi/5)/2=(5pi/5-2pi/5)/2=3pi/10
We know the four inner/outer circles create a square, giving
A(Oi,Oi,Oo)=pi/2
A(O,Oi,Oo)=A(Oi,Oi,O)+A(Oi,Oi,Oo)=pi/2+3pi/10=5pi/10+3pi/10=8pi/10=4pi/5
Now we know this angle, we can use it for the larger triangle OoOiO (Green Red Blue). From a2=b2+c2-2bcCos(A), we have
D(O, Oo)2=22+(1+r)^2-2(2)(1+r)Cos(4Pi/5)= ~12.4
D(O, Oo)=L= ~3.52 (Blue)
Next, we have consider triangle OoOOo (blue blue green), being isosceles with lengths L, L, and 2R=2. We need the interior angle, so therefore we have
A(Oo,O,Oo)=ArcCos ((b2+c2-a2)/(2bc))=ArcCos ((L2+L2-22)/(2(L)(L)))=ArcCos ((2L2-4)/(2L2)=ArcCos (1-2/L2)
=~ArcCos(0.84) =~ 0.58 radians.
This gives the other triangles' angles as:
A(O,OO,OO)=(Pi-0.58)/2=~1.28 radians.
A(O,OO,E)=1.28+Pi/2=~2.86 radians.
Finally, consider the triangle going from OOoE (Blue Green Purple). We know 2 lengths, and one angle (2.86), giving our overall radius D (Purple) to be
a2=b2+c2-2bcCos(A)
D2=3.522+12-2(3.52)(1)Cos(2.86)=~20.1
D=~4.49
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
-Prove that a circle can be surrounded by exactly 6 others of its own radius. IE The above picture might actually be incomplete, there may be indistinguishable gaps in between each one; so prove it to make sure.
-If surrounding the central circle with smaller ones, what radius would they need to be in order to have exactly 8 surrounding it instead?
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
This is more simple than it looks. Just rearranging the power, and taking 22x-1 modulo 3, we have
22x-1=(22)x-1=(4)x-1≡(1)x-1≡1-1≡0
Hence 3 divides 22x-1. Similarly for the other numbers
23x-1=(23)x-1=(8)x-1≡(1)x-1≡0 Mod 7
25x-1=(25)x-1=(32)x-1≡(1)x-1≡0 Mod 31
27x-1=(27)x-1=(128)x-1≡(1)x-1≡0 Mod 127
213x-1=(213)x-1=(8192)x-1≡(1)x-1≡0 Mod 8191
217x-1=(217)x-1=(131072)x-1≡(1)x-1≡0 Mod 131071
219x-1=(219)x-1=(524288)x-1≡(1)x-1≡0 Mod 524287
Or more generally, just taking the numbers inside, then going modulo (2p-1) in order to reduce this value to 1, gives
2x*p-1=(2p)x-1≡(1)x-1≡0 mod (2p-1).
Hence (2p-1) divides (2x*p-1).
This works for all p, even for values not prime. EG 2x*4-1 will be divisible by 24-1=16-1=15. Now this itself is a composite number, but since composite numbers have prime factors it immediately follows that there exists a prime (3 or 5 here) which divides this expression.
Now, for the case P=11, that's not an exception, it's just that you're choosing to focus on the wrong information. We have 211-1=2048-1=2047. So the formula predicts it will be divisible by 2047, but since this is a composite number, we have 2047=23*89 then it is also true to say it is always divisible by 23 (or 89) as well.
So there always exists a prime which divides every term in such a sequence.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
This will be assuming the game consists of cards just getting blindly dealt to the players, then they show them. Without being too familiar with the rules, whether or not they could swap cards, exchange etc, which just adds too much complexity, here's a rough model.
Easy answer: Given those 5 shared cards first, then deal 2 cards each, we have only one set of specific cards completing the 4A/RF hands. Thus P(4A n RF)=47C2 * 45C2, multiplied by 2 if you also count Person B beating Person A as a desired outcome.
Difficult answer: If you also want to factor the probability of having community cards which allow for such a setup in the first place, then that complicates things. One way to work this out is by using the probability of having the cards needed for this setup being the 9 cards in play, and then the probability of them being in locations between players/middle which allows for a climax.
P(Climax)=P(Cards being dealt)*P(Dealt to right players|given these cards)
Clearly there are 52C9 cards which could be in play at any time, and those could be split between the player/table/player in 9C2*7C5*2C2=9!/2!5!2!=756 combinations, only a few of which allow for the climax. We need to work out how many ways this is.
For the significant cards required for each hand, the royal flush requires 5, the 4xAce requires 4, but they are not independent. It requires an intersection of 1 card, thus 5+4-1=8 unique cards needed for this setup, leaving 1 random dummy card (9s in video). Also remember though that there is only 1 unique set of 4 significant cards which produce the 4xAces, yet the significant cards forming a royal flush have 4 different variations; one for each suit. Thankfully though, each of these still allows for 4A combo, as they would not require additional cards in play.
Total combinations for cards being in play=52C9
Amount of 4A combos=1
Amount of RF combos=4
Amount of Dummy Cards=44
P(Cards being dealt)=(1*4*44)/52C9
This is the chance that the interesting cards are dealt onto the table in some fashion, but we still need to work out the probability of them actually being distributed within the correct slots. For example the assignment AA...TJQKA...A9 would not give the climax we saw, despite having the correct cards in play.
A Royal Flush uses 5 cards, the player holding at most 2 of these means we need at least 3 in the middle. 4 Aces uses 4 cards, the player holding at most 2 means we need at least 2 in the middle. One card needed in both sets, IE must be in middle, thus 3+2-1=4 significant cards in the middle minimum. The 5th card could be part of the player's pattern, or the dummy. So given 9 cards in play, we know 1 dummy card might be in the middle, or in player A or B's hand. Therefore to work out the total probability, we can consider just one of the three locations, and consider the probabilities of creating a good setup with or without the dummy card in that particular pile. (The 'without' option would be the sum of the other 2, as a small shortcut). Here we will consider the effect of having the potential dummy in Player A's hand, thus having only 1 RF card with it.
P(Dealt to right players)=P(Setup with Dummy in A)+P(Setup without Dummy in A)
1 Dummy card to be in first hand
5 potential RF cards to be paired with it =5C1=5
4 remaining RF cards plus 1 random ace in middle =3C1=3
Other 2 aces in final hand =2C2=1
P(Setup with Dummy in A)=5/9C2 * 3/7C5 * 1/1 =~ 0.02
Next, we consider climactic hands with the dummy not in Player A's hand, which could be broken down into specific locations within Middle/B, but quicker to consider it as a single calculation. This is because having 2xRFs in Hand A means there are no Aces, and thus they all must lie somewhere within Mid/B. Thus the dummy's exact location in either Mid or B makes no difference to player B's ability to make 4A using the Mid/B selection; and so won't affect if it's a climax or not.
We have potentially 9C2 combos for first hand, 7C5 for middle, 2C2=1 for second hand.
2 RF cards in hand =5C2 ways
Amount of ways to pair 3 remaining RF cards with 2 others=4C2 ways
Therefore we have
P(1st person getting 2xRF cards)=5C2/9C2
P(3xRF cards in middle, plus 2 random)=4C2/7C5
P(2 others going to 2nd player)=1
P(Setup without Dummy in A)=5C2/9C2 * 4C2/7C5 * 1 =~ 0.08
So finally this should give us
P(Climax)=P(Cards being dealt)*P(Dealt to right players|given these cards)
=176/52C9 * ( P(Setup with Dummy in A) + P(Setup without Dummy in A) )
=176/52C9 * ( ( 5/9C2 * 3/7C5 ) + ( 5C2/9C2 * 4C2/7C5 ) )
=~4.7x10^-9
That finally gives us the probability that Player A makes a Royal Flush, and B gets 4 Aces using the provided community cards to help. Again, multiply this by 2 if you want to consider the other dude winning instead.
(PS, **** me this seems overly complicated, there must be an easier way)
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
On Penn and Teller's: Fool us, there was a neat card trick which predicts the location of 2 chosen cards in a deck. But, as shown by explanation videos, the solution is that the positions are worked out in advance. Then, through manipulation of each section; without the need to know where the 2 cards originally were, they can be forced into the arranged locations and thus the trick always works.
However, that vid only explains how to replicate it, not the logic behind why it works. How could the joker positions force cards of unknown locations into a known one? Prove it however you can.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
kusoman wrote:
ive had this game on my gamecube ever since the
game came out if you have any RTA tricks questions
ask me
oh yeah I also want to see a TAS that gets the
early ice beam to
(no sarcasm its actually possible on console)
should I explain this?
I have a lot to say :)
Wow, you can
get early icebeam
too? I highly doubt
that's possible. After all,
nobody in this huge thread
has ever seen a Metroid Prime
Speed run before.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
Again, since there was no direct answer, am I screwed on my current game? The game cannot be completed with the Compatibility core, which the TAS was made with up to this point. There appears to be no way to change midway, and changing to the only other core available (=Performance) requires the game to start over. Even worse is the fact it complains that movie recording cannot even be done using this core; so am I forced to just give up now?
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
Meh, changing the SNES core from Compatibility to Performance does indeed fix it. The problem now is the fact that the Compatibility core was incorporated into my TAS originally, so it auto runs under that. I don't know how to change that, I really hope I don't have to start the TAS from scratch now just to fix that error.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
I don't have any specific info as to the cause, but incorrect emulation on SNES game Eek! The Cat is causing the game to be incompletable sadly. The characters fall through the right hand side of lift platforms, and cannot walk across a moving barrel, rending the game impossible.
https://www.youtube.com/watch?v=EFXoOGhoSMk
Playing the game in the SNES9x emulator works fine in this section, but not Bizhawk 1.7.4 currently.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
You can continue creating a TAS by playing the movie on an emulator in Read+Write mode (IE not Read-only), then loading a savestate. That will resume recording from that point, and you may continue.
If you don't have a savestate during this movie which you'd like to continue from, then of course you first need to create one. Play the movie in Read only mode, so all you're doing is viewing it. Loading a state in this mode will simply jump the movie and start viewing at that point, rather than editing from that point. Use the existing savestates, or Pause+Frame Advance tools to get to the exact section you want, then create a savestate. Now load the movie in Read+write, and loading from this new savestate will enable you to continue working from that exact point as requested.
Do that any time you have to stop working on your TAS, and continue it another time. The final product of a TAS needs to be just 1 long movie file, rather than independent sections.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
Very nice trick, using your own corpse as a projectile to finish off the boss. That's definitely counts as a game completion, as we see the boss's health go to 0, the ending cutscene at least starts to play, as well as the highscore chart at the end showing 'Ed'=end as your progress.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
Using Bizhawk 1.7.3, on Ram Search, you can resize the window, but you can shrink it beyond where the buttons are, and thus can longer click the [/O Search ] [Z /O] buttons. Fixable by making the window larger again of course, but it'd help if there were proper minimum sizes defined.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
p4wn3r wrote:
(a) Prove that the topology is well defined. In other words, prove that it satisfies axioms (1)-(3)
For (1), already defined φ and S=S(1,0) are open. For (2), already defined opens sets if they're unions of the form S(a,b). For (3), we want to prove that the intersection of two sets S(a,b) and S(c,d) is still another set, and thus by induction any finite intersection is. To do that, consider these sequences as congruences modulo a or c respectively. We have each element of the set of the form Si=b mod a, Sj=d mod c. We want to know about their intersection, IE when they satisfy both, and thus by the Chinese remainder theorem there exists a unique e<ac such that e mod ac satisfies both; with e= b mod a = d mod c . This describes S(ac, e), and thus is a still a linear progression, IE member of our set; thus any finite intersection is. In the event no intersection exists, then we have the intersection as φ, which was still defined to be open.
p4wn3r wrote:
(b) Can a finite non-empty set be open?
Given how each set s is infinite, their unions are infinite, as are their (non empty) intersections, so no.
p4wn3r wrote:
(c) Are the sets S(a,b) also closed?
They're closed if their compliment S' is open. S'(a,b)=S(a,1) U S(a,2) U ... U S(a,b-1) U S(a,b+1) ... U S(a,(a-1). This is the finite union of all other values of 'b', and thus by axiom (2) this is open; thus the initial sets S are both closed and open. Thus since S is open, we have S' closed, so both S and S' are clopen sets.
p4wn3r wrote:
(d) Write the union of all S(p,0), with p prime, in a more explicit form. Try to reason whether the resulting set is open or closed. What does that say about the number of primes?
This set X=U S(p,0)=Z - {-1, 1} . This set would be all integers expressible in the form nP for some integer n, prime p. This would generate all numbers apart from +-1, as they have no prime factors and thus cannot be expressed in the form nP. The compliment X'={-1,1} is a finite set, thus closed by question (b). This means our set X is open.
By (2) we know that the union of open sets S(a,b) and S(c,d) will be open, and similar to (3) can be shown to generate a cyclic pattern modulo (ac). This means their compliment is also a cyclic pattern modulo (ac), and thus can be written as a union of S(ac, k) for each occurring element k in this pattern. Thus since we have another finite union of sets S, by (2) this is still open. So if this compliment set is open, our initial set S, the union of finite open sets, must be closed. But since by (C) these open sets are also closed, that proves that these finite intersections of closed sets are still closed.
But if X was a finite union of S(p,0)s, it would still be closed, implying X' open, which cannot be true if X' finite. This reaches a contradiction, which can only occur due to applying the axiom with FINITE unions, and thus X must be an infinite union, IE infinite primes exist.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
Warp wrote:
AFAIK CDA is not enough to show that real numbers are uncountable. The only thing it does is demonstrate that there exist sets that are uncountable.
But these sets are contained in |R, and if you can't even count a subset of |R, then clearly you cannot count the whole thing.
Warp wrote:
You need an additional step to demonstrate that there exists a one-to-one correspondence between the set of infinite digit strings and a subset of R.
Digit representations are by definition just Cauchy sequences, of the form 3x10^0 + 1x10^-1 + 4x10^-2 + 1x10^-3 + 5x10^-4 + 9x10^-5+ etc. Real numbers are by definition the completion of |Q, IE the invented limits of these sequences. These digit representation are what allow real numbers to exist in the first place. If we have a given sequence of any digit strings, it is a Cauchy sequence which is automatically convergent in |R, and thus represents a real finite number. Every digit sequences is a real number, and every real number has a decimal expansion.
Warp wrote:
But how about this: Let's take a list of all irrational numbers...
...which cannot be done, thus any upcoming conclusion is not necessarily correct. Countable unions of countable sets are countable. We have |Q countable, so if irrational numbers were countable, their union |Q n (|R / |Q ) = |R would be countable too, which it isn't, thus irrational numbers are an uncountable set.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
The rational numbers are known to be countable, proven by other means.
Cantor's argument is what showed that real numbers are uncountable. If you try to list the decimal expansions of all the real numbers between 0 and 1, you can still generate via the diagonal algorithm, an additional real number not in the initial set. This contradiction means that the original list is therefore not complete, and so |R is not countable.
Doing that on rational numbers |Q, does not lead to a contradiction, considering we know that these are in fact countable, so the list itself is not a problem. By using the diagonal algorithm you have in fact constructed a not in the set, but this would merely be an irrational numbers, which are known to exist. If trying to find an extra number not in |R, that is not possible considering |R is not countable, thus there's no list to perform the diagonal argument on in the first place.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
Vault exists, so there's hope yet for this game. However, in doing some levels, mysteriously I'm getting slower times than videos I've watched, despite taking the exact same route. Is there something obvious that I'm missing here, such as a Run button, or speed settings which I haven't found?
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
I believe I said that, loading a savestate in Read+Write mode is how you resume recording from that point. Naturally, if you don't have a suitable point to record from, you need to create one first. This would be done by viewing the movie in Read-Only mode, and creating one at the appropriate moment; as you apparently did. Strangely enough you don't have to view the entire movie just to get to very the end section. As explained before, when in Read-Only mode you should be able to load any nearby savestate to basically fast forward to that point. EG load a savestate which was at the start of the level/area (which you should have if using them wisely), and you only need to watch from that point on. When near the bit you want, then use Pause+Frame Advance to get to the exact bit, then make a savestate. Then just un-check Read Only mode, and load said savestate to resume recording.
Finally, games are not supposed to randomly accelerate, perhaps you are pressing the either + or the TAB button? That fast-forwards the emulator, even though the sound emulation doesn't quite keep up when doing so, hence the inconsistency you reported.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
OK, so you've created the first part of a movie. If you load that movie, it'll show you everything that you've recorded. If during this playback, you load one of your savestates, it'll function as a DVD's title select, and simply jump to that point in the movie and resume playing from there. If instead of viewing you want to resume recording from a particular point, then first you'll need to deselect 'read only' (either before loading, or from the menu during playback), and then load your desired savestate. Loading a savestate when in Read+Write mode is what causes you to resume recording from that point, rather than needing to wait for your whole movie play through.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
5 Characters, 4 maps, so 20 combos to potentially cover, so approx 3-4 hours worth. I for one wouldn't enjoy that as a single movie, nor would I enjoy having 20 different publications on the GC list. I suggest just creating an RE4 resources/records page, and just link individual movies on that, similar to BTT records on SSBM pages.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
Without knowledge of specific GC emulator, but generally rerecord counts usually show on the window where you select a movie to play. When loaded a file, it should show you author ID, rerecords, plus Rom used etc.
To resume recording, when you load a video, make sure you do not load it as 'read only', because this is for viewing purposes only. When loading savestates in Read Only move, it'll act as a DVDs scene selection and will jump to that state, and resume watching from there. But if you load the movie without Read-Only, loading a savestate will then just continue recording from that point, and enable you to continue TASing your game.
Experienced Forum User, Published Author, Skilled player
(1535)
Joined: 7/25/2007
Posts: 299
Location: UK
Well that doesn't discourage a TAS surely, considering that clearly proves that these movies are in fact being made. Notice how he didn't say "This glitch would take 11 hours to set up, so no way in hell am I gonna TAS that".
If you look at his channel, you see we have a lot of body parts. So all we need now is a Frankenstein to sew them all together. Granted it would be long, but hopefully any encode would simply skip all the boring sections such that that video has done, giving us a vid which is only 2-3 hours long.
By previous question, at the origin we have a smaller circle of radius r, touching the surrounding five of radius R=1. This gives the interior circle a radius of
r = R*(1/(sin(pi/n))-1=1/sin(pi/5)-1= ~0.7 (Yellow)
Distance from origin to inner circle origin D(O,Oi)=r+R=r+1 (Red)
Distance from inner to outer circle origin D(Oi,Oo)=2R=2 (Green)
Distance from outer origin to outer edge D(Oo,E)=R=1 (Green)
But these are obviously not in the same direction, so to find the overall distance we need a bit of trig. First consider the closest triangle, formed by 2 reds and a green. This is isosceles, and we know the inner angle, thus:
A(Oi,O,Oi)=2pi/5
A(Oi,Oi,O)=(pi-2pi/5)/2=(5pi/5-2pi/5)/2=3pi/10
We know the four inner/outer circles create a square, giving
A(Oi,Oi,Oo)=pi/2
A(O,Oi,Oo)=A(Oi,Oi,O)+A(Oi,Oi,Oo)=pi/2+3pi/10=5pi/10+3pi/10=8pi/10=4pi/5
Now we know this angle, we can use it for the larger triangle OoOiO (Green Red Blue). From a2=b2+c2-2bcCos(A), we have
D(O, Oo)2=22+(1+r)^2-2(2)(1+r)Cos(4Pi/5)= ~12.4
D(O, Oo)=L= ~3.52 (Blue)
Next, we have consider triangle OoOOo (blue blue green), being isosceles with lengths L, L, and 2R=2. We need the interior angle, so therefore we have
A(Oo,O,Oo)=ArcCos ((b2+c2-a2)/(2bc))=ArcCos ((L2+L2-22)/(2(L)(L)))=ArcCos ((2L2-4)/(2L2)=ArcCos (1-2/L2)
=~ArcCos(0.84) =~ 0.58 radians.
This gives the other triangles' angles as:
A(O,OO,OO)=(Pi-0.58)/2=~1.28 radians.
A(O,OO,E)=1.28+Pi/2=~2.86 radians.
Finally, consider the triangle going from OOoE (Blue Green Purple). We know 2 lengths, and one angle (2.86), giving our overall radius D (Purple) to be
a2=b2+c2-2bcCos(A)
D2=3.522+12-2(3.52)(1)Cos(2.86)=~20.1
D=~4.49
-Prove that a circle can be surrounded by exactly 6 others of its own radius. IE The above picture might actually be incomplete, there may be indistinguishable gaps in between each one; so prove it to make sure.
-If surrounding the central circle with smaller ones, what radius would they need to be in order to have exactly 8 surrounding it instead?