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Everything posted by harey

  1. 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.
  2. 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?
  3. 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!!!
  4. 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
  5. What does it mean? I have never been to an orgy
  6. 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?
  7. 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.
  8. Please post, I think that you cannot get more than 3:
  9. 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.
  10. We already had this problem, something like drawing two different fruits from a handbag. I just cannot locate it.
  11. If it worked, it would be great for fractal pictures. The bad news is that multiplications are in the processor heavily optimized, a fair guess 4-5 additions. Googling: The big question is how the 3 temporary variables were treated by the compiler. If we have to access the RAM...
  12. @bonanova - sorry, but I believe you have a sign error with the factor bd: I = (a+b)(c+d) - R =ac+bc+ad+bd - (ac - bd) ... = ... +2*bd ....
  13. @bmad As nobody found the solution, can you post it?
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