In principle, zero is probably the correct guess if everyone plays perfectly - this would also yield a tie between everyone because everyone will be correct.
In reallity though, I'd probably venture to guess slightly more then zero, perhaps even one. As eluded to, all it takes is one person...
My bid is:
\omega^{\omega^\omega}
Though my grasp of transfinite numbers is shakey at best, I think it's bigger then "infinity" (i.e. plain omega) and 3^infinity :-).
I've pondered the same "overanalytical" thing before when faced with similar questions. It certainly doesn't take much to realize where the problem means to aim with it, but it is a little dubious. Even with one wife, {6,6,1} could be a posibillity, in fact with the way my sisters and I were...
I assume it would be solvable by some combination of iffs (if and only if) or "if I asked you X, what would you say" type questions. Not having heard the question before, I don't understand the rules. "at" in 1. and 2. could mean "living at" or "currently at", 3. I'm not even sure what it might...
<joins in the applause> Very nice. I still feel that given continous pieces of one color each, five still ought to be impossible as per the four color problem. Yes, I agree, the original problem is stated a bit haphazardly and doesn't prevent that and possibly other things. I figured that this...
Sorry about not seeing the sticky at the time. I picked e9e9e9 by simply checking what the background rendered to, though with color profiles, websafeness and what not it may well be something else when it starts out (I just hit printscreen, tabbed over to photoshop, pasted in a screenshot and...
I was kind of thinking the same thing, i.e. if a single point is considered ok then the original 4 slice pie would have been fine. However, it might well be that non-single point contact can be made but reshaping the slices a bit and it cetainly ruins my previous argument.
None of the pre-set colors match the new skin, but if you use e9e9e9 it'll be invisible. I.e. <bracket>color="#e9e9e9"<bracket> Text <bracket>/color<bracket> (not sure how to make a bracket actually show up instead of being interpreted). Demo: Invisible text
Hmm, interesting. On an intuitive level, it certainly feels like four would be impossible, any attempt to "cross connect" the remaining pairs of the "three each" solution would encircle one of the others, something which obviously cannot happen with identical tiles. I'd suspect a proof would...
Same with the previous "70 cents... 3 quarters... 75 cents... 4 nickles... 20 cents... 4 pennies...", the three quaters, two dimes and a nickle would work as would numerious other combinations.
There are indeed half-dollar coins, though quater, dime, nickle and penny (25, 10, 5 and 1, respectively) are *way* the most common. The half-dollar is used pretty rarely, though they are definitly around. There have also been rounds of dollar coins for ages (gold/silver dollars, etc) though...
I can't figure out a way to get above $1.19 (such as 50+25+4*10+4*1). The only "split" that would take me above a dollar is the 25 which keeps it from happening since there's no 25 in 4*10 and there isn't enough pennies to make another 5. I may be wrong, but that'd be my guess.
Answer:
15 112 113
42 40 58
70 24 74
(leg leg hypotenuse)
All three have area 840.
Gotten by:
I essentially used the old formula for generating primitive (i.e. non-reducable by division)pythagorian triangles:
(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2
...where m>n and m and n are relativily prime...
This may possibly be the wrong place for this, but I figure it's the closest area around since it's mostly software related I think. It may be a lame question, I'm not entirely at home in this area to say the least - this is a horizon streaching game more then anything on my part.
For reasons...
If you're ok with .\bar{4} = \frac{4}{9} then you have:
51 = \frac{4!}{4}-\frac{4}{.\bar{4}}
53 = 44+\frac{4}{.\bar{4}}
82 = \frac{4!}{.\bar{4}}+4!+4
87 = 4!*4-\frac{4}{.\bar{4}}
Possibly others, I haven't had time to seach around that much.
Been fiddling around a bit, here's one
1000000
*7
-7
/7
/777
-777
+7
+7
+7
+7
*7
+7
/7
/77
20 7s, 33 keystrokes not counting the million. It could probably be polished down a bit I'm sure. The essential trick here is that one can use *7 (several +-7, 77, 777) /7 to accomplish...
Wow, I exahausive searched 7k, 77k, 777k.. with each other over +-10 and not a single number is divisible by 7 (or 77, 777 etc, though that would have been kind of a bounus). Unless I did it wrong (and it's entirely possible that I did) I can't find anything. There are several +- solutions that...
Very interesting question.
1000000 - 777777 - 77777 (repeat twice) - 7777 (repeat 8 times) - 777 (five times) - 77 (seven times) - 7 (four times) *7
would be one rather convoluted solution. I'm sure there's something better though, some number up there ought to be an even multiple of 7...
He is essentially saying that 6k will be divisible by 2 and 3 (6 is, after all, a factor). Hence, 6k+2 will also be divisible by 2 (adding two doesn't change divisibility by 2), 6k+3 by 3, 6k+4 by 2 (adding +2+2). Thus, only 6k+1 and 6k+5 have the option of being prime. Thus, the number between...
Well, what I'm essentially saying is that your original function is only true when M is distributed. It's not at all true for a fixed M, it's a function to calculate the gain for *all* M when they are distributed over p. If you fix M to anything (lincluding m) then it's no longer distributed...
<nods> You're right, it was a very bad way to put it.
I still feel, despite the P(x) changing thing being bogus, that m=M violates the equation. You will, long term, win
\int _m ^{\infty} xP(x)\, dx - \int _0 ^m mP(x)\, dx
*but* that assumes that M is randomly distributed per P(x)...
I've had probably a half dozen false starts on this. I see somthing wrong and then, after some thought, realize that it's right.
The only problems I see are ones that I can't quite formalize. In the OP there is really no solid definition for what is supposed to happen at m=M, i.e. a tie. One...
Very interesting question. Though I can't think of an obvious solution nor, in fact, one that I can work all the way through I've decided after mature consideration to go over what I think so far.
I belive the answer lies in the assumption that the odds of winning is 50% regardless of the...
I belive (although I'm at this time not at all in a position to prove it) that the optimal solutions will all wind up with f(N)*log(B) for N>=2, i.e. every static N>=2 has a k*log(B) solution (on average, best and worst may be further away and closer to it depending). My two "hint 2" solutions...
This is a problem that was posed by our professor when taking algorithms years and years ago. Unfortunately I can't recall terribly much about his solutions and (rather lengthy) musings on it, so I don't have a "correct" or "optimal" answer as such. Since I don't have a more normally worded...
If X is random, the answer is random but this is delt with later. You're right in that the two first sections only deal with T and F, however the random case is delt with by using a god to choose between the other two thus choosing one of T and F at random if the questioned god is R. This has...
Glad I'm not (at lease initially) ridiculed for saying this :-). Low-voltage (such as electronics) is my game, I rarely mess with the other stuff even though I hold it in the highest regard. It could well be that 40kV is rather out of range for normal buildup, I wouldn't know. I've been told...
I (too?) feel that the definition is inadequit or at the very least non-optimal. "15 deg increments" isn't well specified in 3d. It makes more sense to use a polyhedra confined to a sphere with angles being normals to the sphere.