question and answer thread.

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Originally posted by Wfranc
Oh Gosh, you have forgoten what happened, Simon :) :p, I like it much,....really. ;)

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You seem to not have paid attention to the fact that I didn't trust you in the first place and therefore did not write anything in that PM you forwarded that wasn't already known by the person in question or that would go over the line, not that I ever done that in a PM. However, I don't like you much...really. :mad:

I'm not even going to pretend I understand what's going on in this thread right now.

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Originally posted by Simon666

You seem to not have paid attention to the fact that I didn't trust you in the first place and therefore did not write anything in that PM you forwarded that wasn't already known by the person in question or that would go over the line, not that I ever done that in a PM. However, I don't like you much...really. :mad:

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Fight, fight, come on, we want to see a good fight! :D:p;)

:rolleyes:

But I don't know either what this is about. :confused: :(

Let's just say that WFranc forwarded a PM about the previous issues I had with Gabriel, I didn't know for sure with what intentions but based upon "Oh Gosh, you have forgoten what happened, Simon", it were apparently no good intentions.

WFranc : Most of us who post in CodeGuru, are working adults and you should realise by now, most of them have a degree. However, when we are posting in the General Discussion forum(Chit-chat), we automatically 'shift' into our pseudo-self which seem to be very funny characters, actually most of us here are very serious people when it comes to work.

I hope you do not pm people unless it is necessary. In the very start when you pm'ed me, it is not that I'm unfriendly so I'm not replying to your PMs, it is because I'm afraid you would keep pm me more of such stuff.

[Edited some English grammer mistakes]

Next question would be "How many of us, WFranc has PMed to?"

:D

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Originally posted by CBasicNet
Next question would be "How many of us, WFranc has PMed to?"
:D

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That's a difficult one. :D

Okay let me post a real question (a killer)
So you have corridor (we can look at it as a coordinate system) with x-size X and y-size Y (X>Y>1). In the corridor there are N traps (a trap is a point) given by x and y coordinates. A fat man (or a circle) needs to pass from the one end of the corridor to the other. He mustn't touch the traps (so if the fat-man (circle) has radius R the distance between his center and a trap cannot be less than R but it can be R). Question is: If you have X,Y N and N coordinates (all numbers are integers) ot traps how to find the maximal radius that the fat man can have.

ex N is 0 than the fat man's radius can be maximally Y/2...

so I don't need an answer because there can't be such without actaually having the input numbers. I need a valid solution to the problem writen in steps. (Algorithm for a program)

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Originally posted by SeventhStar
Okay let me post a real question (a killer)
So you have corridor (we can look at it as a coordinate system) with x-size X and y-size Y (X>Y>1). In the corridor there are N traps (a trap is a point) given by x and y coordinates. A fat man (or a circle) needs to pass from the one end of the corridor to the other. He mustn't touch the traps (so if the fat-man (circle) has radius R the distance between his center and a trap cannot be less than R but it can be R). Question is: If you have X,Y N and N coordinates (all numbers are integers) ot traps how to find the maximal radius that the fat man can have.

ex N is 0 than the fat man's radius can be maximally Y/2...

so I don't need an answer because there can't be such without actaually having the input numbers. I need a valid solution to the problem writen in steps. (Algorithm for a program)

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Write an algorithm that calculates the distance between all combinations of two traps and that calculates the distance between each trap and the boundaries. The radius is equal to the shortest distance divided by two.

Simon666
Oh well, Mr Simon666, what you said here seemed to prove that you are the guy who always walks with the head down and your feet up in the air. Or I can express in a different way: Come out with feet first. YOU DID send me or did I ,a_h ?

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Most of us who post in CodeGuru, are working adults and you should realise by now, most of them have a degree. However, when we are posting in the General Discussion forum(Chit-chat), we automatically 'shift' into our pseudo-self which seem to be very funny characters, actually most of us here are very serious people when it comes to work.

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MR CBASICNET,
Yes, I have realized that already since the time I joined the forums not NOW.
Let me ask you some questions, Do YOU or anybody here have any self_introductions before posting something ? Why dont YOU use your real name instead of CBASICNET ?Did I say something to hurt or to insult anyone just like some of the guys here did to me in some of my previous posts ?
MR CBASICNET, What do you mean by ' most of the people here are working adults' ? Did I ever say you or anyone were a kid in my post till now ???? you did automatically 'shift' to prove that YOU ARE A WORKING ADULT with CHILDISH BEHAVIORS, didn't you ? Oh, let me say one thing which I have learnt from someone in here:If a clown has 5 apples, I'll take 2. But in this case, I took 3, one to eat, one to laugh, one to send you an PM to share with the Guy the fun I had when I was still in CA. I did not care about whether you sent me a PM back or not but I knew you lied to me about it. Thatz it.
You donot have to say hope that I willnot do this or that, I send PM only to those I like and I think those will be not like some of the liars or bast@rds I met.
By the way, I would like to tell some of those guys here again that if YOU disagreed with something I wrote in my posts, let me know directly and right away. Insulting just proves you to have a disgusting rude behavior and that YOU are nothing but a LOSER, lose for ME. YOU could not argue for your opinions to convince me and people and YOU turned to insult, is it right ? WHAT AN ADULT YOU ARE !!! If I were you I should have felt so much ashamed of myself. YOU ! so proud of WHO YOU ARE and what you are possessing, YOU turn to be SELFISH or maybe this is YOUR NATURE !!! I absolutely know and admit that YOU are much well informed about life and perhaps have very good educations but unfortunately what you did shew you werenot deserving for any respects from me about your success indeed. This is what I would like to tell some of the guys who did insult me before. Hope you will take these words as reminds. I am not teaching YOU anything here as YOU KNOW SO MUCH ABOUT THEORY about life, BUT it is now true that YOU COULDNOT BRING IT INTO PRACTICE, and because I don't want to be thought 'student outgrows teacher'.

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Next question would be "How many of us, WFranc has PMed to?"

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That's a difficult one.

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The next question will be 'Do you guys have any comments on what I wrote here ? I repeat if any, let me know, teach me right from wrong, or correct the mistakes I made. Anything ? I appreciate your help !!!'
It is not difficult really, it is up to your tongue, MR SIMON666 !!! :D

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Originally posted by Wfranc
YOU DID send me or did I ,a_h ?

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Yep. But as far as I know, a private message is a private message, nomatter what its content is. Sending it around to other people shows little respect and can be proof of bad intentions.

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Originally posted by Wfranc
Why dont YOU use your real name instead of CBASICNET ?

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I don't know about CBasicNet but if you want my real name, look at my profile and go to my website. My name is just too long to use as nick. Why people choose to have a nick other than their real name is supposed to be nobody's business, and as I've been told too, using some other person's avatar than your own picture is apparently also.

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Originally posted by Wfranc
Did I say something to hurt or to insult anyone just like some of the guys here did to me in some of my previous posts ?

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How about using "I HATE YOU GABRIEL FLESH" as location? I have had issues with Gabriel but I don't hate him or spread hate.

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Originally posted by Wfranc
MR CBASICNET, What do you mean by ' most of the people here are working adults' ? Did I ever say you or anyone were a kid in my post till now ????

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Depending on your use of language and other indicators, I also assumed you are a kid too. And what is the stuff aboutthe apples? You're not clear in most of what you write. I think solarflare intended to make fun of the fact that asking trivial questions is pointless, and I thought it was a good joke. Why do you see it as an insult?

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Originally posted by Wfranc
[Senseless rambling nobody understands what it is about, where it came from and where it is going to.]

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Simon666:
Yes, PM is PM :D. But 'no matter what the content' is an unaceptable phrase. Why ???... You have to answer yourself !?!
What I wrote in my post didnot directly aim at anyone but YOU and CBasicNet. And moreover, I didnot say Solarflare insulted me in her/his posts. Are you trying to make the 'war' flame burn more between me and SolarFlare ? Or are you crying out for help from others when you seemed to lose in this nice argument ? what a shame !!!
About the fact I used that statement as my location some time ago: Do you need any explanation about the difference between 'hate' and 'insult' ? Can you spell the words ? That was the private problem between me and MR Fleseriu which there was also some guy secretly trying to poke his nose in.
What you are trying to list up here is just to provoke other people hatre towards me, I know that, Sim. But you must know for sure that people here are not kids to be that easily provoked. If you want a fair play and you think you can win, I am always here to wait for you:) !!!
If you want peace, stop it from now on.

I think you have to think twice before you post the reply in order that I could call you an adult.
One more thing, I donot need YOU to understand the whole things I wrote in my post, what I wrote about YOU could be understood by you was my intention. And that is enough for your part. Thanks !!!

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Originally posted by WFranc
Why dont YOU use your real name instead of CBASICNET ?

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My real name is Bob Lim as stated in the readme file inside my screensaver.zip

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Originally posted by WFranc
Did I say something to hurt or to insult anyone just like some of the guys here did to me in some of my previous posts ?
MR CBASICNET, What do you mean by ' most of the people here are working adults' ? Did I ever say you or anyone were a kid in my post till now ????

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If I had wanted to insult you, I would have said "most of the people here are not kids" insteads.:D Please note I did not say you are a kid. I have also never say you said anyone was a kid.

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Originally posted by WFranc
I did not care about whether you sent me a PM back or not but I knew you lied to me about it.

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No I did not lie to you; I did not really go to the URL you pm'ed me. Since you are so adamant, well I can say "I did visit the site you sent me." to you, just to humor you. :eek:


I have a feeling that arguing with you, brings me down to your level.

I have to remind you that this thread belongs to the SeventhStar. If this thread dis-integrate into flaming, this thread may get locked, which many do not like to see this happen, especially SeventhStar.

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Please note I did not say you are a kid

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, did I say you said I was a kid ?
But now I think you have to explain what you meant by this: 'I have a feeling that arguing with you, brings me down to your level.'

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Since you are so adamant, well I can say "I did visit the site you sent me." to you, just to humor you.

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Oh well, now it turns out to be truer about who you really are.

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If this thread dis-integrate into flaming, this thread may get locked, which many do not like to see this happen, especially SeventhStar.

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Yes, I agree with you that this thread belongs to SeventhStar and I also would like to remind you that this thread is about 'questions and answers'. I did make some questions for you to answer did't I ?
Let me ask you :who is the first to start for this argument ? YOU, SIMON666 OR ME ?
You should remember that if you think you are old enough, you must say and do something much more right for anyone younger like me to pay some respects to.
If you said something wrong about not only me but anyone or anything, I think I 'd have the right to say it WRONG, no matter who you are, this is who I am. I very much hope that what I wrote here could be thoroughly understood. Thanks. :)

From myself

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Please note I did not say you are a kid

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From WFranc

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did I say you said I was a kid ?

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That was not what I meant. Don't play around with semantics.

From WFranc

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But now I think you have to explain what you meant by this: 'I have a feeling that arguing with you, brings me down to your level.'

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Your statement before this statement explains it all.

From WFranc

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Let me ask you :who is the first to start for this argument ?

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The one who typed in bold, started the flaming first.

From WFranc

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If you said something wrong about not only me but anyone or anything, I think I 'd have the right to say it WRONG, no matter who you are, this is who I am.

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In that case, then I must say you are rather uprighteous for standing up for anyone or anything. :D

Guys,

this flamewar ends right here. If you have some issues to clarify, do it by mail or PM. If you have a problem with a post, report that post.

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Originally posted by SeventhStar
Okay let me post a real question (a killer)
So you have corridor (we can look at it as a coordinate system) with x-size X and y-size Y (X>Y>1). In the corridor there are N traps (a trap is a point) given by x and y coordinates. A fat man (or a circle) needs to pass from the one end of the corridor to the other. He mustn't touch the traps (so if the fat-man (circle) has radius R the distance between his center and a trap cannot be less than R but it can be R). Question is: If you have X,Y N and N coordinates (all numbers are integers) ot traps how to find the maximal radius that the fat man can have.

ex N is 0 than the fat man's radius can be maximally Y/2...

so I don't need an answer because there can't be such without actaually having the input numbers. I need a valid solution to the problem writen in steps. (Algorithm for a program)

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I do not think Simon666's algorithm will actually discover the truest maximal size for the fat man, though it will definitely generate a fat man size that would work. Take for example a corridor of size 1 wide, z long (which any general corridor can be scaled to) with one trap at location (.2, z/2). Simon666's algorihm would give .1 as the maximal radius, when actually a man of radius .4 could squeeze through on the opposite side.

Therefore, I think the problem requires a little more work for an algorithm. I gave a similar algorithm in this thread, but I do not think this exactly applies to this problem. I think a better approach is that one calculates for each trap the maximum fat man radius needed to completely pass it (to get the last fold past the trap down the corridor, to graphically speak). This is done by calculating the distance from each trap to every other obstacle (trap or wall) and separating results into travelling pass-to-left and pass-to-right results. Minimize only to each side and pick the maximal side result. Then one calculates a path of traps around which one will touch (not all traps may be used in a path) using a search algorithm on the tree of possible path passingss to maximize the decision of whether to pass to the left or right of a trap in a given path. The most difficult part of this process will be to determine the tree of possible path passings, but I believe the problem can be decomposed into three-obstacle decisions. It seems a hard problem to me, but I guess you could use the fact that three points uniquely determine a circle that touches them (if not colinear -- which is a degenerate circle) and similarly, two points and a line (for a circle tangent to the line). This combined with the above minimax should give the connection criteria for the possible path tree tree.

A very difficult problem indeed! I can even see situations where it could be advantageous for a path to pass a trap on one side and backtrack slightly on the opposite side to pass obstacles ahead (not to the point of center going back in front of a trap, though). But I do not see a solution without a search on the possible path tree...

Its amazing how one second I will think a problem is particularly difficult, and shortly after posting so, I will figure it out. Its like going to ask the teacher for help, and after the question is asked, getting the epiphany and having to blurt out "oh never mind, I got it after all..." Happened all throughout my schooling.

This problem is actually a (very thinly) disguised problem that is used as the canonical use of complementary graphs and networks on them, and the answer is basically isomorphic. There are four basic steps to solving it.

• Start with the given problem. Draw it out as specified. Label the traps (a simple rule like T1, T2, ... works fine) and the left and right walls (WL, WR).
• Now for each trap, draw lines segments connecting it to each of the other traps and perpendicularly connecting it to each wall. If any of these line segments pass through another trap, only draw the line segment to that trap and not through the trap to the original target. At this point you have a graph with the traps and wall intersections as the vertices and the drawn segments and wall seegments as the edges. Label the regions formed on the plane (I use 1, 2, ... for simplicity). The special regions at the start and end of the corridor can be labelled "start" and "end".
• Now you draw the complementary graph to the one formed in the previous step. Each region becomes a vertex, and regions that share a bounding edge (not a point) are connected by edges. We turn this into a network by associating a weight to each edge, that weight being half the distance of the full drawn line segment of which the boundary is a subsegment (not just half the subsegment length).
• Finally, we apply a path traversal search algorithm to maximize the cost for the problem on paths that start at "start" and end at "end". The cost for this problem is cost(path) = min(edge(E)) where the minimization is over the edges of the given path. Finding the path that maximizes this cost gives the widest path a fat man could take, and the cost of that path is his radius.

I've attached a little jpeg that illustrates this for a sample with four traps. Any graph traversal enumerating algorithm can be used for the final step, or a more optimum algorithm like Dijkstra's can be used to perform the maximization (the standard form of Dijkstra is a minimization, so one can merely take the reciprocal weights). Anyways, it appears I should study more on network problems, because once the idea comes to draw the graph, it is amazing how closely this follows the canonical network cost-minimizing problem...

Well, my original picture was a bit too big, so I've decreased the jpeg quality and the dpi of the scan. I hope it is still possible to figure it out.

All right galathaea you got it!! :D ;)
There are 2 or 3 other solves (including mine which is different) but I like yours. You managed to 'standartize' (or something like that) the problem.
This problem was given at the Bulgarian National Programing Olympics and was solved only by eleven (out of 60) people (including me ;) ). You desereve the point :D
So will you ask a question or shall i proceed?

As for Simon: you didn't think the problem over very good this 'obvious' solve gives a possible but not optimal answer.

I gotta tell you though! my solve was more optimized for speed and made less verifications ;) :p :D

Okay, I have one problem question and one general question.

Problem
We make strings with only the characters '0' and '1'. How many strings of length N can we make if we additionally require that we can never have any adjacent '0's.

ie. 11011110 is a legitimate string of length 8
and 0001 is not allowed.

General
I'm curious what type of problems everyone is interested in. Some problems have been the classic word problems that require some logical and arithmetic insight, while a few others have been more calculational. In particular, I have a bunch of problems that others might find interesting, but looking through them I see that many of them are proofs or problems in specialized domains (combinatorics, group theory, number theory, topology, series manipulation, etc.). Proofs seem like they may cause a problem, since there is likely differing opinions on what constitutes a proof, but they can be quite interesting to work out as well. The specialized domains could probably be handled by given specific definitions that are required for the solution, but this may take away some of the insight of trying to work it out. So I am a little stumped. In particular, I'd probably like to post a few series manipulation problems if people were interested, since I particularly love manipulating the little things, but then the questions would probably be something like "prove" or "what are the series of steps" to show a particular identity (which has the problem of proofs in a way, unless a collection of atomic manipulations were agreed on) or they would be something like "calculate" a particular sum (which could leave open the use of Mathematica and its ilk).

So basically, my second question is: what type of problems do all of you enjoy most?

well

let a(n) = the number of such sequences of length n.

a(1) = 2
a(2) = 3

a(n) = a(n-1) + a(n-2)

The Fibonacci Sequence

Yep! Your turn.

(though I'd like to know your reasoning if you cared to share... not necessary, just curious which direction you took) :)

Hi Galathaea,

I have seen this question as homework in many text books, but have never bothered to solve it. Like any honest person I was trying to guess at a general formula and prove it by induction. Needless to say I could not come up with a general formula, that would be quite an accomplishment (Is that a solution which is worth quite a bit of money?). The question I ended up asking myself was: "How do I build sequences of length n with no adjacent zeros from sequences of length n-1 with no adjacent zeros". The answer was I can add a 1 to every sequence of length n-1 and I can add a 0 to all the sequences of length n-1 which end in 1. It is easy to see by this same reasonong that the number of sequences of length n-1 which end in 1 is the number of sequences of length n-2. I then recognized the Fibonacci sequence.

I am having troublr thinking of a good problem. Here is one
Problem:

You have 9 brass rings, three of which are gold. can you find the three gold rings using a balance scale three times?

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Originally posted by souldog
You have 9 brass rings, three of which are gold. can you find the three gold rings using a balance scale three times?

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Number of possible ways to have 3 gold and 6 brass rings:
9!/(3!6!)=84
Number of possible data to aquire from scale:
3^3=27

84>27

So I say it is impossible.

[edit: "3^3=81" changed to "3^3=27," thank you souldog for pointing that out. But my answer still holds ;)]

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Originally posted by souldog
You have 9 brass rings, three of which are gold. can you find the three gold rings using a balance scale three times?

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Also, it is impossible to have brass rings which are gold ;).

Yes, it is not possible.
Note: 3^3=27 not 81.

There is not a one-to-one correspondance between the possible outcomes of the weighings and the number of possibilities for the three gold rings. So more than one configuration for the three gold rings will lead to the same result in three weighings.

YOur turn

I say brass rings can be gold. Just my opinion.

Maybe you can look at the rings and separate out the gold-colored rings from the brassy-colored rings, and then use the scale to weigh three pouches of diamonds that a black-market dealer is exchanging for the rings. But you have to make sure that you leave real quickly, before the dealer scratches off the gold paint and discovers that the ring is really brass...

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Originally posted by souldog
Needless to say I could not come up with a general formula, that would be quite an accomplishment (Is that a solution which is worth quite a bit of money?).

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I'm not sure how general you mean here. There is a simple function for the Fibonacci sequence that gives the function explicitly (its 1/sqrt(5)*(((sqrt(5)+1)/2)^n - ((sqrt(5)-1)/2)^n) or something like that where you differ powers of the roots of the characteristic) instead of just defining it recursively. But if you meant sequence problems in general, then I was actually thinking of a solution method that does generalize this somewhat. I was thinking of the use of generating functions and the 4 foundational theorems of combinatorial enumeration. In particular, this problem is a simple use of the composition theorem and the fact that there exists an isomorphism from the composition of the collection of sequences I specified and the collection of sequences of zeroes to the collection of all sequences of ones and zeroes. You get the generating function right away, and similar methods work for all kinds of sequence problems (as well as enumerating all kinds of cool other combinatorial structures).

I take it you did not like my problem. I did not like it either.

I had fogotton that there was a general formula for the Fibonnaci sequence. It has been a while. Although I have had to teach basic finite combinatorics a few times, I am no expert and so did not use any theorems. In fact I work in servohydraulic control now and so am kind of out of the loop as far as theoretical mathematics goes. It seemed more expediant to just solve the problem at hand. Please show the solution using the theory.

souldog: pm me the solution and I will come up with a combination of rings for which it will not work.

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Originally posted by souldog
I take it you did not like my problem. I did not like it either.

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I didn't mind it. I was trying to figure out if there was some trick way to solve the problem, because it wasn't solvable through the standard way as Solarflare has pointed out. But I hadn't noticed the gold / brass thing that Solarflare posted about, or that the question was not to solve but whether one could solve. So I just posted my $m@rt@$$ response on color as a joke... not unhappy with it...

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Originally posted by souldog
Please show the solution using the theory.

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Well, just to add to my previous post, here is the basic outline:

• First, you start with two-dimensional generating functions, so you can keep track of the number of 0's and 1's separately. For ease of putting them on the boards, I will call them F[{0,1}*](x, y), F[S](x, y), and F[0{0}*](x, y) for the generating functions of all sequences of 1's and 0's, the sequences I am enumerating, and the sequences of 0's only (at least one 0) -- with x being the zeroes and y the ones.
• Combinatorial composition is the replacement of certain combinatorial subobjects in one collection with elements of another collection, and the composition theorem (commonly just referred to as a lemma, since its pretty straightforward to prove) is that the combinatorial composition generating function is the functional composition.
• So for this problem, it is easy to see that replacing any zero in our collection with a member of the collection of zero sequences will give any member of the collection of all sequences of 0s and 1s, and conversely that this is uniquely done. So we get the generating function equation: F[{0,1}*](x, y) = F[S](F[0{0}*](x, y), y).
• Now the generating function for the two sequences we are not solving for can be taken as known or solved for immediately using the combinatorial sum and product theorems. Without going into the details, I will just say that F[{0,1}*](x, y) = (1 - x - y)^(-1) and F[0{0}*](x, y) = x (1 - x)^(-1).
• This gives (1 - x - y)^(-1) = F[S](x (1 - x)^(-1), y). You then use the substitution u = x (1 - x)^(-1) to get the nice form F[S](u, y) = {1 - u (1 + u)^(-1) - y}^(-1).
• And finally, to get the generating function for sequence of length n regardless of the number of 0s and 1s, you just set the two arguments equal and the convolution sums the various terms nicely for you. In other words, you get F[S](x, x) = (1 + x) (1 - x - x^2)^(-1), which is the generating function for the Fibonacci sequence, starting at Fibonacci(2) instead of (1). You can see this by multiplying the denominator over and solving it as a series problem, from which you get the recurrence relationships.

However, it is obvious that your solution is much more straightforward for the problem. This is just the way I was thinking of because I was trying to come up with a combinatorial problem that might present some insight to the general way of solving these things and actually gave a problem with several different ways of attack specific to the question. Another method I thought of after I posted the question was to divide the sequence into two subsequences, both of which must also obey the problem. Counting out the places where you can divide up the sequence and using the fact that the split point must not combine two different sequences the first of which ends on zero and the other that begins, you get a relationship that also solves to Fibonacci fairly quickly. But I think I like your solution best...

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Originally posted by solarflare
souldog: pm me the solution and I will come up with a combination of rings for which it will not work.

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There is no solution. You were right. Why don't you ask another question so this keeps going.

I guess SolarFlare is not going to ask a question so I will.

Be warned. I do not know the answer to this question. I do not know if there is an answer, i.e. if the solution can be found with standard algebraic techniques.

Problem: (working in 3 dimensions) Six vectors are are attached to six distinct points in the XY-plane (if you like put them on the vertices of a hexagon centered at the origin). They all have a positive (>0) z-coordinate. The other end of the six vectors (call it the free end) are coplanar. If all you know is the lengths of the six vectors, find the equation of the plane determined by the free ends. Restrict the lengths of the vectors to be in some range, say between 8 and 10. The distance between the free ends of any two of the vectors is constant.

The distance between the free ends of any two of the vectors is constant

This problem is a bit unclear. First, I do not understand what is meant by the distances between endpoints being constant, since I do not see this as a dynamic problem. And it cannot mean that all the distances are equal, since at most four points can have this property in 3-space. So I will assume that is superfluous information. Also, I do not understand the restriction on lengths of the vectors, since according to the problem, one of the informations given is the length of the vectors, so these are not variable or dynamic either... Again, I will assume this is superfluous information. So basically, I have transformed this problem into:

1) You are given six points in the x-y plane. In fact, they lie on the regular hexagon x=cos(n p / 3), y=sin(n p / 3) for n = [0, 5] e Z.

2) You are given six distances that represent the lengths of vectors from these points. Additionally, all points must lie in the positive-z subspace.

3) You know that the points of these vectors are coplanar.

Q) Determine the equation that these vectors must obey to satisfy these conditions.

Now, I still am confused as to whether you are given in 2) distances that are assigned to a particular collection of vertices (distance 1 to vertex 1, etc.) or just a collection of distances in which assignment must be made. First it is important to realize that there may be none, one, many, or infinite solutions to a particular set of distances, so the equation one arrives at may not be constructive. But a general brute force attack would be to enumerate all 15 groups of four vertices. The coplanarity condition can be seen as
| x1 x2 x3 x4 |
| y1 y2 y3 y4 |
| z1 z2 z3 z4 | = 0
| 1 1 1 1 |
since the determinant is proportional to the volume of the parallelipiped described by the four points (which would be zero if they are coplanar). This will give 15 equations on the points. Additionally you get the six distance equations (of the form (xp - xv)^2 + (yp - yv)^2 + (zp - zv)^2 = d^2 where the p's refer to point coordinates and the v's refer to vertex coordinates on the hexagon). This gives 21 equations for 18 coordinates, and thus is over determined (thus allowing all possibilities mentioned above).

From here I would try to simplify somewhat, but I find it difficult to proceed without knowing whether the distances are tied to a particular vertex or not...

I would consider that a sufficient answer.

But even if it's not what souldog was looking for, it would be tedious to describe exactly what it was, since he doesn't have a solution anyway.

I am pretty bad at asking these question I realize. Let me not try and phrase this without some technical jargon.

What I had in mind here was the forward static analysis of the nearly general stuart platform. This is a problem from robotics.

First some background.
Let the X-Y plane represent the ground. Then you have six actuators (A device which can generate motion in One dimension)
connected to the Ground on the vertices of a Hexagon. These actuators are the vectors in the problem. The other end of the actuators are connected to another plane (Call this the platform). The connections are fixed, only the angles the actuators (vectors) make with the ground and the platform can change as the length of the actuators change, the connections are made with universal joints, The connections of the actuators on the platform again lie on the vertices of a hexagon, but one of a much smaller size than the ground (In reality the Heaxgons are not regular, but it does not matter) These devices are used to move an object connected to the platform to simualte motion. As the length of the six actuators changes the platform can acheive any orientation and position within the range of motion. Think of this as a solid body problem. There are six degrees of freedom for a solid body moving in three space. The position of its center of mass and the orientation of a coordinate system connected to the center of mass relative to a fixed reference frame. You need six linear actuators to generate all six degrees of freedom. Devices like this are used for flight simulators and earthquake simulators (though often a different configuration is used for earthquake simulation).

The device I have described here is often called a Stuart platform (Gough actually used it first).

As in any such problem you can perform a dynamic or static analysis. I was mixing in dynamic features which do not matter.

In the real world you usually measure the lengths of the actuators and send them to a computer which generates command signals based on these lengths (The control is actually through a servo mechanism, but that is another story). There are two problems related to this for a static analysis.

1. Inverse Static analysis : Given that you know where the center of the platform is (this is the center of the hexagon determined by the connection points for the actuators) what are the length of the actuators. Here I assume you have a fixed reference frame and you know the coordinates of the center of the platform as well as the Eular angles for the plane.
This is easy.

2. Forward Static Analysis: Given the length of the six actuators, where is the center of the platform and what is its orientation?

This is the question I wanted to ask.

Let me now start by saying that I do know the solution to this. It runs along the methods Galathaea was outlining and results in solving a 21st order polynomial by numerical techniques. This method uses no restriction on the lengths of the actuators and results in many solutions which are not meaningful. Notice Galathaea you had not used in any way that the vectors were in the positive Z subspace.

So is it possible to make the above problem determined (One solution) if you further introduce that the lengths you are given satisfy 8<=Length(i)<=10 where i = 1,2,3,4,5,6.


I am sorry you find this tedious SolarFlare, but you were not asking a question.

Quote:

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Originally posted by souldog
I am sorry you find this tedious SolarFlare, but you were not asking a question.

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I'm terribly sorry. I didn't mean that at all. If I actually thought the questions were tedious, I wouldn't post in this thread, would I? I guess I was just tired when I made that post, and couldn't express myself well. What I basically meant is that I expected there might be a long argument as to what the problem specifically meant, when it didn't really matter that much about the specifics. But you actually cleared it up quite nicely.

Sorry i didn't respond earlier, but I've been under the weather the past few days (I hope none of you catch anything from me!). Anyways, looking at this I have a clearer idea how to proceed, but just wanted one more clarification. I think the important part here is found in the statement

Quote:

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Originally posted by souldog
The distance between the free ends of any two of the vectors is constant

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From the description of the problem, it seems like what is really meant here is that the distance between the free ends of any two vectors whose origins are adjacent vertices of the base hexagon are all equal. But before I proceed on something like this, I'd like to know if this is the correct interpretation.

PS: souldog, I actually enjoy working out problems like these, so don't get discouraged. It is sometimes just as difficult to ask a question correctly as it is to answer it, as the numerous errors that have popped up in the popular mathematics competitions show.

Well let me say this in a different way. The vectors connect to the platform on a hexagon and you know the coordinates of the vertices of that hexagon relative to a reference fixed to the center of the platform (note: in a dynamics problem this would be the moving reference frame, but again this is not dynamics). So you can take the distance between any two ends of the vectors on the plane determined by the platform as known constants. (dynamically the ends of the vectors on the platform form a hexagon which moves around as the lengths of the vectors change, but does not change shape. Rigid body motion)

Note: The solution of this problem is not necessary for the control of a stuart platform. Numeric algorithms can be found which work fine. Why this problem interests me is: Why is it hard? All solutions that I have seen have not taken constraints on the lengths of the actuators into account. These constraints, by the way, can also be interpreted as a limitation on the angles the vectors form with the ground and platform (i.e. the scalor product) Physically if you tell the platformto go some where (i.e. specify the six lengths of the actuators), then it goes to one place.
With the constriants the problem is determined.

To me the question is how to introduce the constraints into the problem tomake it determined. If this turns out not to be easy then to determine why.

A solution to this problem would almost certainly be publishable, by the way.

Galathaea, why don't you post another problem. Maybe something a little more reasonable

Although I am still messing around with the vector problem, I will post a new question that I though might be enjoyable for some to work on. It is kind of a turn of pace.

Some here may know that the torus may be topologically represented as a square, where opposite sides are specified as having been glued. As well, th Klein bottle may be represented as a square with one pair of opposite edges glued directly, and one pair glued with "opposite orientation", ie. if we say the square's left and right edges have this distinction, the left edge's points going up correspond with right edge's points going down.

Now, let us look at a similar 3-dimensional space problem and the topology induced by a gluing. Take a cube. Each face viewed from above is glued to the opposite face turned one-quarter clockwise (in the view). A picture is attached to illustrate, which I hope is sufficient. Call the strange manifold built up from this gluing a Q-manifold.

In normal 3-space, normal cubes will fill space and have corresponding faces and edges when 8 cubes meet at a corner. The question I have is this: how many of the Q-manifold's cubes meet at a corner in the gluing? If you wish, you can also say what type of geometry this imposes on the Q-manifold, as it follows fairly quickly, but it is not necessary. One hint on the last optional part is that of the eight homogeneous geometries in 3-space, the 5 "lesser" ones are immediately ruled out in this case, leaving the decision betwen the 3 more popular geometries: euclidian, elliptic, or hyperbolic.

2

(since nobody else seems to know - or hasnt read the question, more likely - i'll take my best unfounded guess ;))

Incorrect ;)

infinite

Nope. That's not it either. :)

It looks like there might be some guessing go on here...:p
Until now, I guess the answer 1 would have been one technically correct answer, since only one box comprises this space, but that's not what I meant. Maybe a better way to look at this problem is: corners become identified by the gluing. Start with one corner and find out how many other corners coincide with it because of the gluing. Add them up (including the original).

If a correct answer is not given by tonight, I'll post the solution and let the next question be asked.

1?

I guess I misunderstood this question. I assumed you were asking: Well the manifold is Euclidean and R^3 is the covering space and we are acting on the boundaries of the fundamental polyhedron with one of the crystalligraphic groups (translation by one and rotation by 90 degress clockwise about each axis). If I take any corner on any cube in a tessellation of the cube over R^3 then it is identified with at least one corner on every other cube. Therefore, all the cubes meet at a corner, ie an infinite number.

To the question: It seems to me that four corners are identified as a single point. If I take any corner then the the other three corners not adjacent to it are identified.

Quote:

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Originally posted by souldog
To the question: It seems to me that four corners are identified as a single point. If I take any corner then the the other three corners not adjacent to it are identified.

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Yes! That's the one! The three corners diagonally across from a chosen corner are identified with it, making the total 4 corners which meet. Similarly, three edges coincide in the gluing. Because each corner has a solid angle of p / 2, the sum is 2p < 4p and so the geometry is elliptic. It couldn't have been hyperbolic at all since the cube has only 8 corners...

Quote:

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Originally posted by souldog
I guess I misunderstood this question. I assumed you were asking: Well the manifold is Euclidean and R^3 is the covering space and we are acting on the boundaries of the fundamental polyhedron with one of the crystalligraphic groups (translation by one and rotation by 90 degress clockwise about each axis). If I take any corner on any cube in a tessellation of the cube over R^3 then it is identified with at least one corner on every other cube. Therefore, all the cubes meet at a corner, ie an infinite number.

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Aha! I see where you were coming from here, but the other corners in the tesselation aren't actually coincident in the covering space. In this approach, there are then 8 cubes adjoining a particular vertex and they are identified in pairs by their particular symmetry orientation, so again the answer is four.

By the way, that view of the manifold is why it is commonly called the quaternionic manifold (my "Q-manifold"), because its symmetry group is the quaternion group.

Okay! Your up, souldog!