harey

What about:

I do not follow here. Worst case: the cards are sorted. Bonanova wrote: What is the expected number of cards in the left pile after all N cards have been drawn? OK, that answer is not difficult to determine, since the definitions of the piles are symmetrical.

100% loss

Even here, it increases the travel time. A non-zero component of the plane's speed is used to compensate the wind.

This assumes that a point is (at least partially) illuminated if a straight line can be drawn uninterrupted between it and a light source.

These thoughts suggest what I think is an interesting question: for each N, find an example of a convex solid that can be illuminated with the fewest lights. That's the start to the original question What bothers me with points and lines: We do not illuminate an object as such, we illuminate it's surfaces. (Already a problem with a sphere and alike: in this case, we may not consider the sphere has one surface, but infinity.) If a surface is illuminated, does it imply that it's edges and vertexes are illuminated, too? One answer would be that this question does not have a sense, but it seems a little week as argument.

Not so quickly... If we need (in 3D) four lights for a sphere, a cylinder can be illuminated by three lights only. (I spent a long time to find a way to show that the edges can be smoothened and still illuminated until I consulted http://en.wikipedia.org/wiki/Convex_body and realized that a cylinder is a convex solid). For a cone, two lights are enough. What bothers me at most: Consider a unit cube in the xyz coordinates and a line (0;0;0) - (1;1;1). If we need 0 lights to illuminate a point or a line, the whole cube would be illuminated by just two lights placed on this line just slightly outside the cube.

c - a = -1 You mistakenly have this as positive one in your post. This is just a typo, sorry for that: d) ac=2; (c-a)=1; I would find very tasteful a=2; c=1; (1-2)=-1 I'm looking for you to show the steps in solving the system of simultaneous equations above. Sorry, I fear I will disappoint you (as I already said, there are few possibilities, so a brute force approach will solve the system in seconds while I would spend hours due to algebra errors): for a in [-2,-1,1,2]: c=int(2/a) for b in [-2,-1,1,2]: d=int(-2/b) if(d-b==3): if(a*d+b*c==3): print("a=",a,"c=",c,"b=",b,"d=",d,"ad+bc=",a*d+b*c,"d-b=",d-b) print(a*c,"x2+",a*d+b*c,"xy+",b*d,"y2+",c-a,"x+",d-b,"y-1") The problem does not have to be solved by a system of equations, As no one else posts and I have no other idea how to solve it, can you post the solution?

Can you prove it? I hesitate between 1 and 2...

Well, I am just not very sure concerning the assumption in b). I do not consider whether there is a solution of another kind - when solving a problem, there is a little bit of intuition on the beginning. This does not make me doubt about the solution if I find one, except in the case I would claim it is unique. As for the remainder, there are so few possible solutions (we look for integers) that you can try them all. Otherwise, you could solve it as a system of multiple equations with multiple variables, but I think it is too much hassle here. I am awaiting the official solution!!!

I get the same result as bonanova and Yoruichi-san. Wanting to avoid algebraic errors, I asked Wolfram Alpha: http://www.wolframalpha.com/input/?i=minimize+%28%28x*x%29%2B4%29**%281%2F2%29%2B%28%28x*x%29%2B9%29**%281%2F2%29%2B5-x&lk=4&num=5

What does it mean? I have never been to an orgy

I do not really fancy this kind of problem. The solution usually is "What would answer the god B if the god A asked him what would answer the god C if..." Did you think about something like that?

I am assuming we are looking for the smallest CONSTANT speed. If we want to maximize the travel time (at varying speed), he can travel almost forever: he follows the shadow line moving imperceptibly to the south.

Please post, I think that you cannot get more than 3:

N-1 (except for N=3 or 4 where you can do better). Let's put M=(N-1)/2. If we ask all i to test all i+1, the most unlucky case occurs when all answers are GOOD. Consider that all bad computers are followed by all good ones. We only can be sure the last one is good - there is no bad computer after a good one. If the answer is BAD: - discard both (you never discard two good computers) - substract 1 from i (if i=0, take anyone) - renumber... Now, each computer said the next one is GOOD, so we are in the case (B)(B)(B)..GGGG. In the worst case: a) we always discarded a good computer and a bad one b) both were tested c) there were M bad computers and therefore M discardments d) b+c imply 2*M tests. If a bad computer says BAD, the above still applies. I just hope I did not forget something... @Rainman Suppose: a) BGBXX b) BBGXX c) GGBXX The results of the tests can be GOOD and BAD in each case. 3 and 4 are special cases, so you cannot use them for induction.