Archived

This topic is now archived and is closed to further replies.

Other math puzzles

This topic is 5312 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Recommended Posts

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

Share this post


Link to post
Share on other sites
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.

Share this post


Link to post
Share on other sites
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]

Share this post


Link to post
Share on other sites
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

Share this post


Link to post
Share on other sites
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.

Share this post


Link to post
Share on other sites
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

Share this post


Link to post
Share on other sites
Guest Anonymous Poster
How about this one:

What are the real value(s) of all the complex numbers n that satisfy the following:

0 = The sum from k = 1 to infinity of 1/(k^n). (Like a p-series where p = n and n is a complex number.)

Share this post


Link to post
Share on other sites
LOL :D.
Who are you kidding, AP? I will go wake up Riemann. He will figure this out! Hahahahahahahahaha!

It is 1/2. Now that I answered, why don''t you post your solution to it !

20: ...Qg2 ++

Share this post


Link to post
Share on other sites
Given a vectorial space, a scalar product for this space, and two vectors u != v from this space, prove that if you take ANY two vectors m, m'' for which ||u-m|| = ||v-m|| and ||u-m''|| = ||v-m''||, then ( u-v | m - m'' ) = 0. (IE u-v and m-m'' are orthogonal).

ToohrVyk

Share this post


Link to post
Share on other sites
quote:
Original post by GaulerTheGoat
LOL :D.
Who are you kidding, AP? I will go wake up Riemann. He will figure this out! Hahahahahahahahaha!

It is 1/2. Now that I answered, why don''t you post your solution to it !

20: ...Qg2 ++


Well techinally that hasn''t been proven yet (has it?) .

Share this post


Link to post
Share on other sites
quote:
Original post by The Heretic
quote:
Original post by GaulerTheGoat
LOL :D.
Who are you kidding, AP? I will go wake up Riemann. He will figure this out! Hahahahahahahahaha!

It is 1/2. Now that I answered, why don''t you post your solution to it !

20: ...Qg2 ++


Well techinally that hasn''t been proven yet (has it?) .

I have seen in one of my old magazines that a theorom proving program proved it with a certain probability. I didn''t understand the article but I will either find a link to it or try to copy it when I get home. It had something to do with Hyper-geometric functions which I also don''t get. I will post later, then you can decide .

20: ...Qg2 ++

Share this post


Link to post
Share on other sites
Guest Anonymous Poster
quote:
Original post by ToohrVyk
Given a vectorial space, a scalar product for this space, and two vectors u != v from this space, prove that if you take ANY two vectors m, m'' for which ||u-m|| = ||v-m|| and ||u-m''|| = ||v-m''||, then ( u-v | m - m'' ) = 0. (IE u-v and m-m'' are orthogonal).

ToohrVyk




Are the scalars real numbers, complex, or just a generic field? Is scalar product the same as a Hermitian inner product?

Share this post


Link to post
Share on other sites
I just started getting more interested in vector spaces. Tell me how I am doing :

Vector differences are independant of the origin, so choose m . So
||u -m || = ||v -m || implies that
||u || = ||v ||. Multiplying out the norms
(u -m )•(u -m ) =
(v -m )•(v -m ) or
u u -2u m +m m =
v v -2v m +m m or
u m = v m .
The other equation gives similarly,
u m'' = v m'' .
Subtracting these two and factoring out the u on the LHS and the v on the RHS, and then subtracting the RHS and factoring out the m -m'' gives
(u -v )•(m -m'' ) = 0.

20: ...Qg2 ++

Share this post


Link to post
Share on other sites
Guest Anonymous Poster
quote:
Original post by GaulerTheGoat
I just started getting more interested in vector spaces. Tell me how I am doing :

Vector differences are independant of the origin, so choose m . So
||u -m || = ||v -m || implies that
||u || = ||v ||. Multiplying out the norms
(u -m )•(u -m ) =
(v -m )•(v -m ) or
u u -2u m +m m =
v v -2v m +m m or
u m = v m .
The other equation gives similarly,
u m'' = v m'' .
Subtracting these two and factoring out the u on the LHS and the v on the RHS, and then subtracting the RHS and factoring out the m -m'' gives
(u -v )•(m -m'' ) = 0.

20: ...Qg2 ++


I''m not sure where you''re getting

||u || = ||v ||

but you don''t need it anyway. If you subtract the two equations at the end those two terms drop out because they are in both equations:

This:
u u -2u m +m m =
v v -2v m +m m

And this:
u u -2u m'' +m'' m'' =
v v -2v m'' +m'' m''

Share this post


Link to post
Share on other sites
Heya, folks. I just realized that what I did above was wrong .
Anyway, I tried to find that article on the web but couldn't. Try googling for something like "Zeilberger+WZ+proof+Riemann+Hypothesis". I turned up a ps doc at "www.math.rutgers.edu/~zeilberg/mamarim/mamarimPS/rh.ps" and had to download GhostView to read it. Or you can try "www.math.rutgers.edu/~zeilberger/papers1.html", which also has alot of other cool stuff on it BTW. Sorry, 'bout the let down. That's all I could find without looking too hard. Like I said before, I don't understand what Zeilberger is talking about, so I can't say whether the Riemann Hypothesis is really proven or not.

EDIT: I didn't know you couldn't link outside of GameDev!?

20: ...Qg2 ++

[edited by - GaulerTheGoat on May 31, 2003 10:40:20 PM]

Share this post


Link to post
Share on other sites
quote:
Original post by Anonymous Poster
quote:
Original post by GaulerTheGoat
I just started getting more interested in vector spaces. Tell me how I am doing :

Vector differences are independant of the origin, so choose m . So
||u -m || = ||v -m || implies that
||u || = ||v ||. Multiplying out the norms
(u -m )•(u -m ) =
(v -m )•(v -m ) or
u u -2u m +m m =
v v -2v m +m m or
u m = v m .
The other equation gives similarly,
u m'' = v m'' .
Subtracting these two and factoring out the u on the LHS and the v on the RHS, and then subtracting the RHS and factoring out the m -m'' gives
(u -v )•(m -m'' ) = 0.

20: ...Qg2 ++


I''m not sure where you''re getting

||u || = ||v ||

but you don''t need it anyway. If you subtract the two equations at the end those two terms drop out because they are in both equations:

This:
u u -2u m +m m =
v v -2v m +m m

And this:
u u -2u m'' +m'' m'' =
v v -2v m'' +m'' m''


Yeah, that made its way into my brain while I was typing up this other post. For some reason ||u || = ||v || just popped into my head and I went with it. I am researching affine spaces right now and I kept thinking, "origins don''t matter." They were feeling negleted . Of course, like you said, it isn''t even neccessary . Can I give myself partial credit on this one?

20: ...Qg2 ++

Share this post


Link to post
Share on other sites
AP: scalars are elements from any fully ordered body, because you need the positivity result to define the scalar product. Also, IIRC, Hermitte inner product is defining using a cosinus, and so is the scalar product (in fact, you can define a scalar product using a formula and from that formula define the cosinus in a particular space. For example, cosinus in C: [0,1] -> IR).

GaulerTheGoat: To link outside of gamedev, don't forget the "http://"... Besides, you can indeed call in an affine space (you can usually build one from an euclidean space anyway). But for the "origin doesn't matter" thing, you applied it wrong.

v = v + 0 if origin is 0, v = (v - m) + m if origin is m
m = m + 0 if origin is 0, m = 0 + m if origin is m

Therefore, ||v - m|| = ||v - m + 0|| : indeed vector differences DO NOT change. You almost got it right, but credits go to AP...

EDIT : additional question : let IM be the set of all vectors m defined the previous way, what is the dimension of IM?

ToohrVyk



[edited by - ToohrVyk on June 1, 2003 4:55:55 AM]

Share this post


Link to post
Share on other sites