Other math puzzles
Author
1°/ You choose at random 0 <= x <= 1 and 0 <= y <= 1. What is the probability that x² + y² < 1?
2°/ Given a cube ABCDEFG, you must start from A and jump from a vertice to another through all of the 12 edges (AB, BC, CD, DA...) once (and only once). You are allowed to include any other jumps in your path (for instance, AC). How many such jumps do you need to accomplish this? (You do not need to come back to A).
3°/ Given the same cube and the same way to move, how can you move through ALL edges formed by vertices of the cube once and only once?
ToohrVyk
1. Pi/4
2. Well, you''re missing a vertex, but how about:
A-B-C-D-A-E/B-F/C-G/D-H-E-F-G-H
3. That''s all edges formed by two vertices? All 28 of them? 12 adjacent edges, 12 diagonals on the faces, and 4 opposite-corner diagonals? You can''t.
2. Well, you''re missing a vertex, but how about:
A-B-C-D-A-E/B-F/C-G/D-H-E-F-G-H
3. That''s all edges formed by two vertices? All 28 of them? 12 adjacent edges, 12 diagonals on the faces, and 4 opposite-corner diagonals? You can''t.
Author
1°/ Indeed, pi/4. Could you explain why?
2°/ Yes, I did miss a vertex, sorry... of course it was ABCDEFG + H. However, I was expecting the minimum number of jumps, and you posted a way to do it (15 jumps). Are you sure you can't do it with fewer jumps.
3°/ Indeed, all 28 of them. Why can't you?
EDIT : 20001th post on this forum!!
ToohrVyk
[edited by - ToohrVyk on May 28, 2003 3:37:43 PM]
2°/ Yes, I did miss a vertex, sorry... of course it was ABCDEFG + H. However, I was expecting the minimum number of jumps, and you posted a way to do it (15 jumps). Are you sure you can't do it with fewer jumps.
3°/ Indeed, all 28 of them. Why can't you?
EDIT : 20001th post on this forum!!
ToohrVyk
[edited by - ToohrVyk on May 28, 2003 3:37:43 PM]
Hmm, wouldn''t the answer for Q1 be arbitrary close to, but not equal to pi/4 (ans->pi/4), since x*x+y*y<1, or am I mistaken?
/John
/John
In reguards to 1: If you draw the region on an xy plane, you get a circle. We are looking at the region inside the circle. The answer is exactly Pi/4 because the perimiter of the circle has no area. (Even though there are an infinite number of points on it). The other two problems are graph theory problems (or at least they can be thought of in that way).
Brendan
Brendan
1. I simply rephrased the question. You''re randomly picking a point inside a unit square, and checking to see if it is inside a unit circle with the center at 0,0. I simply divided the area of the part of the circle that is inside the square by the area of the square.
FunkyTune, the odds that a random number will fall exactly on the edge of a circle is infinitesmal, essentially zero. In a computer simulation though, The odds of landing on (1,0) or (0,1) may be much greater. Nothing else would hit exactly on the edge.
2. Yes, I''m sure. There are the twelve edges that must be used. Then each point except for the beginning and end must have an even number of edges connected to it. That means that 6 points must use another edge. Two points per edge is three edges. 12 + 3 = 15
3. Each vertex has 7 edges connected to it. In order to use each edge, each vertex, except for the beginning and end points, must have an even number of edges connected to it.
FunkyTune, the odds that a random number will fall exactly on the edge of a circle is infinitesmal, essentially zero. In a computer simulation though, The odds of landing on (1,0) or (0,1) may be much greater. Nothing else would hit exactly on the edge.
2. Yes, I''m sure. There are the twelve edges that must be used. Then each point except for the beginning and end must have an even number of edges connected to it. That means that 6 points must use another edge. Two points per edge is three edges. 12 + 3 = 15
3. Each vertex has 7 edges connected to it. In order to use each edge, each vertex, except for the beginning and end points, must have an even number of edges connected to it.
Author
Jedi got it right... Listen to him guys !!
Since you like it, another problem :
let u be a series, where u(n) = #{ (a,b) in Z² | a² + b² < n } ( #E is the number of elements in E). What does u(n) / n tend towards when n tends towards infinity?
ToohrVyk
Since you like it, another problem :
let u be a series, where u(n) = #{ (a,b) in Z² | a² + b² < n } ( #E is the number of elements in E). What does u(n) / n tend towards when n tends towards infinity?
ToohrVyk
u(n) is an approximation of the area of a circle with radius sqrt(n), which gets better as n approaches infinity. u(n) tends toward Pi*n, so u(n)/n tends toward Pi
This topic is closed to new replies.
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