harey
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Posts posted by harey
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100% loss
if the bin contains even numbers starting at 4
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by ignoring travel perpendicular to the direction of the wind, where it has no effect on the plane's speed
Even here, it increases the travel time.
A non-zero component of the plane's speed is used to compensate the wind.
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This assumes that a point is (at least partially) illuminated if a straight line can be drawn uninterrupted between it and a light source.
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Make the sides as small as possible and you approach a circle.
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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.
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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.
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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)=-1I'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?
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Point - 1 light
Can you prove it?
I hesitate between 1 and 2...
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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!!!
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a) -1 on end means (...+1)(....-1)
b) as there is no term with power 3, I made the bet there only are 2 termsc) multiplied (ax+by+1)(cx+dy-1)
d) ac=2; (c-a)=1; I would find very tasteful a=2; c=1;
e) bd=-2; (d-b)=3; b=-1 and d=2 ???
f) (ad+bc)=(2*2-1)=3: It works!!!
FInal form:(2x-y+1)(x+2y-1)
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Bob can refine this strategy. Suppose:
a) the number of apples is limited to N (a very large one) and
b) Alice places places a random number of apples in each room
- if the number of apples in the first room is "small", Bob should refuse it
- if the number of apples in the first room "->N" and the number of apples in the second room is "just a little bit smaller than the number of apples in the first room", Bob should accept it
If we lift a), it will still apply.
Alice can counter this strategy by placing respectively K, K+1, K+2 apples. This brings us back to the strategy of k-man.
The mathematical formulation is left to the reader as exercise
However, if Bob always accepts the first room and decides to accept or reject a second room based on whether the number of apples in the second room is larger or smaller compared to the first room then his chances of winning become 50%.
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Call the ball with the lowest mass L.
Call the ball with the highest mass H.
D=mass(H)-mass(L)
D=sum of masses of greens going from L to H clockwise
D=sum of masses of greens going from L to H counterclockwise
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I get the same result as bonanova and Yoruichi-san.
Wanting to avoid algebraic errors, I asked Wolfram Alpha:
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Let's call Wilhelm the kid
- in the same column as Pete (Pete is older than Wilhelm)
AND
- in the same row as Jane (Wilhelm is older than Jane)-
1
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Draw a circle. Inscribe a square ABCD, at best AC being horizontal. Move the points A and C some degrees towards D, delete the arc AC. Move D some 4 units from A. You get a kind of hot air baloon or an ice cream cone.
#define RAC = Remaining Arc of the Circle
- 3 points on the RAC: 4th point must be on the circle, at least 90 degrees missing;
- 3 points on the cone: 4th point cannot be on the cone (lines not parallel) nor on the RAC;
- 2 points on the cone and 2 points on the RAC: draw two parallel lines, resulting segments can never have the same length
It is hard to inscribe a square into a regular pentagon. I conjecture that many polygons will fit.
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1) number the computers 1 to N
2) test by the computer i=max(1,last_tested) the computer (i+1)
3) if the answer is bad, discard both (and renumber)
4) go to 2 unless one of these cases occurs:
A) You discarded 2*k computers => as you always discarded a good one and a bad one, the remaining [one is /are] good.
B) You got k answers "GOOD", There cannot be a bad computer after a good one; there might well be a series a bad ones (even 0) followed by good ones (even 0). As there are at most k bad ones, (k+1)-th is good.
C) A mix of A and B... (Easy to deduct, hard to describe w/o loosing simplicity. Do not forget to adapt k.))
In any case, k tests are enough.
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To be clear the sexual activity will occur in a round robin sense
What does it mean?
I have never been to an orgy
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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?
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Assuming that Santa must always get progressively closer to the south pole. What is the slowest (longest duration of travel) he can travel and remain forever at night? So far both you (bonanova and TSLF) claim it is 1 year. I am asking if this is correct.
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.
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I did not really get why the 12 hours...
- by symmetry, we can calculate only the trajectory to the Equator
- if Santa moves faster then the speed of the shadow line, he should move West as much as possible (I can develop if necessary)
- his move to the West is limited by the shadow line
(- when his speed equals the speed of the shadow line, he should make some turns around the Earth until the South pole gets into the shadow - solves Bonanova's remark, but we can omit this in the calculation)
- when his speed is smaller than the speed of the shadow line, he should move plain South
When he moves to the West, he trades somehow time against distance, but I cannot find the equivalence.
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The fastest 5 runners can be determined in 7 heats.
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I'll wait for discussion before posting the 7-heat solution
Please post, I think that you cannot get more than 3:
Make them run 5 by 5, attribute a letter to each race and number them in the order of arrival. You get:
A1 A2 A3 A4 A5
B1 B2 B3 B4 B5
...
E1 E2 E3 E4 E5
6th heat: Make run A1, B1, C1, D1, E1. If they do not arrive in that order, re-letter them so A1 is faster then B1 who is faster than C1...
Gold medal: A1
Other medals: make run (7th heat): A2, A3, B1, B2, C1.
If they arrive in this order, the 4th fastest can be A4, D1...
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No.
Behind the first bullet, there always is a combination of bullets which contains one that will reach the first one.
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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) BBGXXc) 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.
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near N*2.
Test A by B and B by A. One of the following happens:
a) bad/bad: at least one computer is bad, put them both aside;
b) good/bad (same as bad/good): idem;
c) good/good: both are good or both are bad; put one aside and keep one.
You never put aside 2 good computers, so you still have more good computers that bad computers. As the worst case is c), we put aside N/2 computers with N tests. (If the number is odd, keep one it for the next round.)
Proceed the same way with the computers you kept.
When you have only 3 or 4 computers, one (reciprocal) test is enough.
Highest of the lows
in New Logic/Math Puzzles
Posted · Edited by harey
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.