gavinksong
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Posts posted by gavinksong
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On 9/16/2017 at 12:56 AM, Quantum.Mechanic said:
My thoughts exactly... except is it possible with four points?
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2 hours ago, araver said:
To clarify what I meant. Before the change / cooperation actions read like this:
Night 2:
1. Dolores - Flamebirde - Kill Plasmid
2. Teddy - Bonanova - Block Plasmid
3. Maeve - gavinksong - No action ???
5. Hector - plasmid - Kill Bonanova - blocked
6. Dr. Robert Ford - Nana - NK GavinksongNP shows Dolores removed Plasmid and Ford removed Gavinksong.
1. Gavinksong - Killed N2 by Dr. Robert Ford
2. Flamebirde
3. Nana77
4. plasmid - Killed N2 by Dolores
6. bonanova
5. Aura 2.0 (phil1882) - Lynched D1 as Man in Black
7. aura - Killed N1 by Dr. Robert FordThat would have left Flame, Nana and Bonanova alive in Day 2.
Flame couldn't verify Bona's claim just by reading the NP. Granted, he would know Dolores didn't block plasmid, but he could think plasmid lied about being Hector. After all, Aura could have been Hector. When Bona blocked Plasmid, but he was killed, he could also deduce, but not be 100% sure that plasmid was Hector (what he claimed).
I do agree goodies could have *still* won this, just saying it wasn't as clear as advertised. Because assuming goodies tell the truth and trust or cooperate fully in a Mafia game ... is a dangerous assumption to say the least
It hurts to know that Flamebirde was not on board.
Just in this case here, if he had saved me instead, nobody would have died. Even if Nana had chosen somebody else to kill, it would have been 3v1, and I would have called her out, and it would have been possible to lynch/kill both of us on the following days/nights (bona can block the remaining suspect N3 if flamebirde is dead).
Either way, plasmid's method was foolproof since the worst case ends in a tie. It would have been 100% foolproof if we had asked Flamebirde if he had used his vote manip in D1. It's weird how nobody thought to ask.
@araver random question: how does it appear on the NP when someone's been targeted for a kill by multiple roles?
(^i can't figure out how to change font color on mobile)
14 hours ago, flamebirde said:this was one of the situations I had wanted when I went for the tie. Sure, it let the baddie control the lynch, but if he used his vote manip there then in the endgame he would lose since I had both still hidden up my sleeve (of course, that all depended on whether or not I lived). The other option was that the tie went through and no one died, leading to more time for the goodies to figure out the game.
That is what you said when you originally suggested it, and I'm sure everyone agreed with the logic of it (even Nana). As planned, you must have immediately known that the baddie's ODTG vote manip was gone... which is what makes it even stranger that you didn't speak up. Deeply ingrained Mafia reflexes/paranoia, I suppose?
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5 hours ago, araver said:
(and goodies couldn't have known the baddie had already spent the ODTG vote manipulation).
Actually, reading that made me realize we definitely could have known if we had just asked Flamebirde - unless he was the baddie which still would have been fine.
So actually, a lot of plans would have worked - if we had known that. Also, slowly starting to realize that, contrary to what Nana said, plasmid was always able to clear himself if bona didn't block him since Hector's role would appear next to his kill, which means my original plan could have been a lot simpler.
Anyways, it was fun! Thank you.
Man, the end of D1 was a big twist for me. I had Flamebirde pegged as Dolores, Nana as the baddie, and bona as indy - and I couldn't help but think that Aura 2.0 was definitely not indy because of the thing. So basically, I wasn't even expecting a vote manip. I'm glad it happened though, not because it gave the goodies an advantage, but just because it was kind of cool. It wouldn't be Westworld themed without a few twists.
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4 hours ago, plasmid said:
How to mafia
Fair enough. There were a few problems with your analysis of the first plan, but as far as I can tell, baddie can't win with the new plan. In fact, since I know that I'm not the baddie, I see it as a guaranteed win.
I didn't realize that the name of the acting role appears next to every kill in the NP.
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1 hour ago, nana77 said:
Your analysis left out the scenario in which you are baddie.
It is also not clear how my being nk would clear Plas or implicate Flame. It would clear Bona though if he survived. And would be 3vs1. Every scenario that does not end tonight is 3 vs 1. 2 vs 1 requires 2 goodie deaths and that can not happen with the plan as given. I also do not see how anything blows Plas cover.
Bona can be cleared by surviving. I can be cleared by being spied. Flame can be cleared by successful save of nk. Gavin and Plas can not be cleared (other than by baddie being killed and game ending).
Odds are Bona or me die tonight and the three others will be the suspects.
Indeed, I did leave that part out. Like I said, I was operating under the assumption that I was not the baddie.
If plasmid claimed that he had been blocked and NK'd you, it would clear bona and obviously you since you'd be dead. That just leaves me and Flamebirde as the other two suspects, and again, I was operating under the assumption that I was not the baddie. If plasmid NK'd Flamebirde instead, he would have to claim he was blocked, clearing bona, while I would clear you, leaving only me and plasmid as suspects. And again, if you operate under the assumption that I am not the baddie, then plasmid's cover is blown.
And yes, if bona blocks plasmid, then every time I wrote 2 v 1 for D2 in my original analysis should instead be read as 3 v 1. Also, the case of "Flamebirde baddie / Nana NK" would result in an identical scenario to the one above where "plasmid baddie / Nana NK", which is why I was saying it would implicate Flamebirde (as one of two suspects). Anyways, these are just minor updates.
It might seem strange that I formulated and shared an entire strategy based on the assumption that I wasn't the baddie. However, as it is literally the only assumption I've made in an otherwise foolproof strategy, to reject it is equivalent to saying that it is more likely for that assumption to be false than it is for an alternative strategy to fail. Intuitively, based on the fact that I am a role spy who can't be cleared without being killed, I don't think there would be another strategy that hinges on fewer assumptions (although you're welcome to try to come up with one). But even if there were, given that I claimed first without any counter-claims, I think I'm currently the least likely to be the baddie. And if I am, then I at least deserve to win for being ballsy.
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1 hour ago, plasmid said:
@bonanovaYour block can come in handy after all... you get to prove your claim by blocking me tonight or else end up full of lead.
I guess that's true. I initially thought this would allow a baddie plasmid to claim that he was blocked, NK Nana, and then blame Flamebirde. But I guess in that case we can just lynch Flamebirde and see if he saves himself.
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Just to be clear, here's my analysis of how my strategy might play out.
However, this is all assuming everyone believes that I am who I say I am.
SpoilerIf Nana is the baddie:
- I will know, and we can win 2 v 1 standoff.If bonanova is the baddie:
- He gets killed by Hector.If plasmid is the baddie:
- bonanova NK: Nana is cleared. Gavin and Nana vs Flamebirde vs plasmid (see below).
- other NK: plasmid cover blown. We win 3 v 1 standoff.If Flamebirde is the baddie:
- gavin NK: Flamebirde cover blown. We win 2 v 1 standoff.
- bonanova NK: Gavin and Nana vs Flamebirde vs plasmid (see below).
- Nana NK: plasmid is cleared. We win 2 v 1 standoff.
- plasmid NK: Nana is cleared. We win 2 v 1 standoff.me and Nana vs Flamebirde vs plasmid:
- Unclear in this situation whether baddie is Flamebirde or plasmid.
- Lynch Flamebirde. If Flamebirde is baddie, we win. If Flamebirde is Dolores, she can save herself. We can win 2 v 1 standoff against plasmid the next day. -
6 hours ago, bonanova said:
So ... if Dolores or Hector is still around ... then ... grab the gun and get it done?
Gav and I have claimed goodie roles. I don't know what (1 of 2 roles) Nana has claimed.
Flame and plas: just need one of you to claim and it's done. Wow.
Actually, I'm pretty sure Nana's claim defaults to Clementine since I claimed Maeve.
Since you've claimed Teddy, Plasmid must claim Hector.
I think Flamebirde is in the clear, both because of his behavior and since Aura 1.0 probably wasn't Dolores (she forgot about Dolores' block in her jab at plasmid).
Also, hopefully, I'm mostly in the clear for claiming first.
plasmid will be cleared if Hector acts tonight, although it could be a bit confusing if Hector and the baddie choose the same target.
Therefore, I'm going to role spy Nana and ask Flamebirde to save me. I suggest Hector target bonanova since Teddy is a non-factor. Also, if Nana is Clementine she should hold off on using her save since we can't afford to potentially lose three goodie votes for tomorrow.
As long as Flamebirde and I are telling the truth, that should net us a goodie win.
Actually, we would win even if Flamebirde was the baddie as long as Nana holds off on her save. I guess you guys mainly just have to trust me.
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That was really unexpected.
Anyways, I suppose the game is pretty much set if we all reveal our roles?
I claim Maeve.
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WHAT
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5 hours ago, Donald Cartmill said:
Someone else spoke up making the statement, that in fact these were the planes that made it back meaning these areas were less damaging to the plane ability to fly ,and the armor plating could best be used else where
It was a man by the name of Abraham Wald. This story happened to be in the preface of a book that my dad bought me that I never read.
Edit: also, he specifically suggested that the engine should be armored since it was statistically the least damaged area for returning planes.
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On the other hand, I would really like to hear from bonanova.
1. gavinksong - voting for bonanova
2. Flamebirde - voting for Aura 2.0
3. Nana77 - voting for plasmid
4. plasmid - voting for Aura 2.0
5. Aura 2.0
6. bonanova - voting for Aura 2.0 -
2 hours ago, araver said:
IMHO, it shouldn't. I agree that it looks as if one player has a bit of extra info, but outing the N1 kill would disrupt much more in balance.
Maybe it's bad to make inferences from this, but that makes it sound like Aura isn't indy.
Edit: it also makes it sound like Aura 1.0 probably wasn't Hector.
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16 minutes ago, flamebirde said:
one more thing: could you do me a favor and change the colors of the votes so that each person has a different color? I would do it, but mobile makes it rather difficult.
Sure thing! I didn't know that was the convention.
1. Gavinksong - voting for Flamebirde
2. Flamebirde - voting for bonanova
3. Nana77 - voting for plasmid
4. plasmid - voting for Nana
5. phil1882
6. bonanova -
After some more thought,
SpoilerThree points. With each point, we get the equation:
R = |p| + |p - f|
where R and f are unknown values, and p is the new point in vector form.
There are three unknowns (since f is 2-dimensional), which we need at least three equations to solve.
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Minor grammar correction: the singular form of foci is focus.
Without really thinking about it, I'm going to guess:
SpoilerTwo points. The intuition is that one point would contain less information than the minimal definition of an ellipse (minus a focus), whereas two points contains more.
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This is classic.
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Just at a glance,
SpoilerThe gambler goes bankrupt if he loses four times in a row. That occurs with some non-zero probability, so if he keeps going on forever, he'll eventually go bankrupt.
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7 minutes ago, flamebirde said:
Whoever it was had to be active
Is it a rule that you need to post publicly before N1 to be active?
I either didn't know that or was late.
7 minutes ago, flamebirde said:Of course, we could have gotten ridiculously lucky and Aura was the Indy. But that's obviously pretty unlikely.
No, it's impossible; indy is invulnerable N1. Can't tell if feigning ignorance is a ploy to trick us into thinking you're not indy.
AND you asked before if indy's trap prevents voting when the rules state that the trap doesn't silence. What to think?
6 hours ago, nana77 said:Aura was nk. Maybe it was random, but if not, Plas is most likely culprit.
I vote Plasmid.
Not sure how this works, but shouldn't we be going after indy first? Or is a simultaneous goodie and indy win possible?
Tentatively, my vote:
1. Gavinksong - voting for Flamebirde
2. Flamebirde - voting for bonanova
3. Nana77 - voting for plasmid
4. plasmid - voting for Nana
5. phil1882
6. bonanova -
The classic board game Connect Four has a winning strategy for the first player.
The traditional dimensions of the board is 7 wide by 6 tall.
For which dimensions does the first player have a winning strategy?
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The classic board game Connect Four has a winning strategy for the first player.
The traditional dimensions of the board is 7 wide by 6 tall.
For which dimensions does the first player have a winning strategy?
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15 hours ago, CaptainEd said:
First, I define Restricted and Unrestricted coins, then describe and prove the algorithm for Restricted coins. Next, I request a slight tweak to the OP conditions--one additional coin known to be good. Then comes the definition of the procedure for Unconstrained coins, followed by the edge case (how many coins with only one weighing), and the formula for the Unconstrained case, since the OP requests the Unconstrained formula.
Then come the first few values.a Restricted coin is one that has been in a pan that failed to balance. A coin that was in the heavy pan is H (good or heavy), a coin that was in a light pan is an L (good or light). An Unconstrained coin is one that has not yet been weighed.
Induction hypothesis: One misweighted coin can be found from 3^p Restricted coins in p weighings.
Proof by induction on P
P=1: given 3 Restricted coins, pick 2 that match (i.e. Are both H or both L). Weigh the two against each other. If they balance, the remaining coin is the bad one. If they don't balance, the bad coin is the one that matches the behavior. That is, if both coins are H, then the one in the heavy pan; or if both coins are L, then the one in the light pan. The coin was isolated in 1 weighing.P>1: given 3^p Restricted coins, set aside 3^(p-1) coins, and divide the others into two pans, in any pattern as long as number of H coins in each pan are equal, and number of L coins in each pan are equal (either number could be zero, it's ok)
Now weigh the two pans. If they balance, the bad coin is in one of the 3^(p -1) restricted coins that were set aside. According to induction hypothesis, it will take an additional p-1 weighings to find the bad coin.
If they don't balance, one side is heavier than the other. There are exactly 3^(p -1) coins that could cause that side to be heavy: the H coins in that pan, plus the L coins in the opposite pan. Thus, the bad coin is one of those 3^(p -1) coins, and once again it will take p-1 weighings to find the bad coin.So in both cases, we have performed one weighing, and it will take p-1 more weighings. Therefore it will take p weighings to locate the bad coin out of 3^p Restricted coins. Induction step is proved.
So imagine the function R(n), the number of Restricted coins from which the bad coin can be found in n weighings.
R(n) = 3^nBut we had to perform one weighing in order to make all these Restricted coins. Can't we learn something additional from that weighing? Yes! If they balance, then we could pursue additional coins.
I need to request an OP boon ("a boon, Sire!") :
please allow me to have one additional coin that is known to be good. My justification is uniformity: at every level, we need a known Good coin, and we've already identified some. But the initial weighing could also benefit from 1 known good coin, and we don't have one.
Now consider U(n), the number of Unrestricted coins from which the bad coin can be found in n weighings.
Follow this strategy. Create 3^n Restricted coins by weighing 1 Good coin and 3^n Unrestricted coins in the two pans. If they don't balance, then you have identified 3^n Restricted coins, and it will take n more weighings to find it.
If they do balance, the remaining coins are Unrestricted coins, and we can perform the same algorithm, with one fewer weighings.
So U(n) = R(n-1) + U(n-1)
Recursively, this works out to
U(n) = R(n-1) + R(n-2) + ... + R(1) + U(1).U(1) is interesting (and perhaps open to debate). Notice that U(0) = 1. That is, if you're given 1 coin and you're told one coin is bad, you can identify it with zero weighings.
How many can you distinguish with 1 weighing? If The Boon is granted, you can distinguish 2 Unrestricted coins in 1 weighing, as follows.
Place one coin in one pan , place the Known Good coin in the other pan, and set another coin aside.
If they balance, the bad coin is the one that was set aside. If they don't, the bad coin is the one opposite the Good coin.
So, with this OP-granted boon, U(1) = 2.The first few values of U(n):
U(2) = R(1) + U(1) = 3 + 2 = 5
U(3) = R(2) + R(1) + U(1) = 9 + 3 + 2 = 14
U(4) = 27 + 9 + 3 + 2 = 41
U (5) = 81 + 27 + 9 + 3 + 2 = 122
Congratulations! Unfortunately, the answer changes if I grant you this boon, so request denied.
But no worries! You've still solved it. The only thing that changes...
Spoiler... is that you must weigh one less gold coin in the first weighing instead. This means that
U(n) = R(n-1) + R(n-2) + ... + R(1) + U(1) - 1
But since U(1) = 2 = R(0) + 1, this works out very cleanly.
U(n) = R(n-1) + R(n-2) + ... + R(0)
And so the answer becomes the one you initially gave.
I'd also like to give my own explanation of the recursive step for "constrained" coins. Some may find it interesting and/or more intuitive.
SpoilerImagine that you've weighed some coins against each other and that the scale has tipped to one side.
Divide these coins into three groups. Then,
- Leave each coin in the first group on its original plate.
- Move each coin in the second group to the opposite plate.
- Take each coin in the third group off the scale.
Assuming you have a few authentic coins on reserve, you can add coins to either plate to make the numbers equal.
Then, activate the scale.
- If the fake was in the first group, the scale should tip in the same direction as before.
- If the fake was in the second group, the scale should tip in the opposite direction.
- If the fake was in the third group, the scale should balance.
If these three groups are roughly of equal size, it should cut down the number of suspects by a factor of three.
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On August 14, 2017 at 2:50 AM, CaptainEd said:Spoiler
in N weighings, can distinguish bad coin out of M=Sum(3^i), i=0...N-1
Can't always tell whether heavy or light
This is actually correct! Another way to write it:
SpoilerWe can always pick out the fake coin out of a bag of (3^n - 1)/2 gold coins.
However, your rationale has a few loose ends. I am interested in seeing your full solution.
Spoiler- You claimed that "We can distinguish among 3^N restricted coins in N weighings." Can you prove that this is always possible?
- In the first step, how many coins do you leave unweighed?
Equivalently, what is an explicit algorithm?
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Wow, I feel pretty stupid.
You were literally asking for the three lowest solutions. I don't know what I was thinking.
Determining an ellipse given a foci
in New Logic/Math Puzzles
Posted · Edited by gavinksong
I stand by my answer.
Although I gave an algebraic answer, my original thought process was exactly yours. I'm a spatial thinker. However, I think it's a dubious claim that when "you’re given three points... [the remaining focus] would probably at best be narrowed down to two points".
Given any distinct pair of points on the ellipse, you can map out all the possible locations of the second focus on a hyperbolic curve. Given three points, we can select three distinct pairs and produce three curves. For your claim to be true, all three curves to would have to meet at two locations. With the foci of the hyperbolic curves forming a triangle, my intuition tells me this is probably pretty unlikely.