mm-167 Is it true that the free group on 3 generators is isomorphic to a>subgroup>>of the free group on 2 generators?>your question with Tarski's old conjecture, on whether the elementarytheory of the free group of rank 3 is equivalent to the elementarytheory of the free group of rank 2; that was recently solved in theaf'rmative by Myasnikov and Kharlampovish (in fact, the elementarytheories of all nonabelian free groups are elementarily equivalent);their announcement was in seem to remember from years ago that it was an open question whether the>free group on 3 generators is isomorphic to some subgroup of the free group>on 2 generators. Can anyone tell me what has become of that?Nielsen, circa 1919: every subgroup of a free group is free.By looking at fundamental groups of graphs and their coverings, you caneven 'gure out the free ranks: e.g. the subgroup of generatedby x^2, xy, and y^2 is freely generated by those three.dave I was wondering of anyone has a solution to the following game:> Assume a game of pool is played on a frictionless service. If player 1> pockets a ball player 1 looses otherwise player 2 wins.> 1) Can one guarantee to hit the cue ball in such a manner that none of> the balls will go into the pockets. How would this problem be solved> mathematically?> 2) If no, Is there a special case where the number of balls on the> table are limited to a prede'ned amount such that after the break no> balls will be pocketed (obviously with only a cue and another ball on> the table one can design the system such that neither ball will be> pocketed).> cheers,> JBEHere's how to have no balls pocketed:Strike the cue ball to miss the rack, and to travel parallel to thelong rail. Since the cushions are perpendicular, the cue ball will travel until its KE is absorbed by the cushions, along the same lineup and down the table.Irrespective of the type of hit on the cue ball, it is clearthat player 2 will win.Anyone care to venture why? All the relevant information is inDale. > Assume a game of pool is played on a frictionless service. If player 1>> pockets a ball player 1 looses otherwise player 2 wins.> Irrespective of the type of hit on the cue ball, it is clear> that player 2 will win.As far as I can see, we haven't been told anything about who winsif player 1 pockets a ball. If player 1 pockets a ball he apparentlyreleases something; just what isn't speci'ed. Assume a game of pool is played on a frictionless service. If player 1>pockets a ball player 1 looses otherwise player 2 wins.>>Irrespective of the type of hit on the cue ball, it is clear>>that player 2 will win. As far as I can see, we haven't been told anything about who wins> if player 1 pockets a ball. If player 1 pockets a ball he apparently> releases something; just what isn't speci'ed.Of course we've been told that. It may not have been what the OPintended to tell us, or the way he intended to spell it, but we'vecertainly been told.Dale Assume a game of pool is played on a frictionless service. If player 1>> pockets a ball player 1 looses otherwise player 2 wins.> Irrespective of the type of hit on the cue ball, it is clear> that player 2 will win.>> As far as I can see, we haven't been told anything about who wins>> if player 1 pockets a ball. If player 1 pockets a ball he apparently>> releases something; just what isn't speci'ed.> Of course we've been told that. It may not have been what the OP> intended to tell us, or the way he intended to spell it, but we've> certainly been told.> Dale> and think of the notoriously unfair coin-tossing game called heads I win, tails you loseThen you'll probably see what I mean.Dale. =Who Oinvented' or Odevelopped' functional calculus (the one that says that if you derive a 'eld against itself you have a delta function, that say how to integrate over paths and so on). This is very important mathematical tool and natural extension of analysis of real and complex variable (kind of 'eld variable analysis) and yet there is not much recognition for it, nor popular praise or expert advocating it everywhere and constantly, and not even historical accounts (all of this being the common lot for all mathematical tools which made their proofs). I think that Euler had something to do with variational principle but it seems much too remote and in our contemporary time may be Schwartz is the name but his creation seems related partially only. So who did it? Why is it not very famous among laymen who know everything about complex analysis and yet nothing about functional analysis? Comments welcome. =**** Gambling Dice Machine probability ****A gambling machine:A possiblity table of the possiblity of each side of a dice, from 1 to 6:__________________________________________|side | 1, 2, 3, 4, 5, 6 |------------------------------------------|possiblity | 0.1, 0.2, 0.2,0.15,0.2,0.15 | (total possiblity of one)__________________________________________There are four dices on the screen. __ __ __ __ |__| |__| |__| |__|Each prize draw will roll the dices to show 4 random numbers based on the probability provided in the table.The award prize is the sum of all the numbers appeared in the 4 slots. E.g. 3351 will have a prize of 3+3+5+1=12. So the prize is in the range of 4 to 24.Q1: What is the average prize for this gambling machine?Q2: Generalize the question: n numbers, w_1, w_2, w_3 .... w_n, and their corresponding possiblities: p_1, p_2, p_3 .... p_n. With replacement, 4 boxes, each box 'lled with one number, what is the average sum of these four numbers? **** Gambling Dice Machine probability ****> A gambling machine:> Note: this is generally referred to as a probability distribution, and possibility would be probability.> A possiblity table of the possiblity of each side of a dice, from 1 to 6:> __________________________________________> |side | 1, 2, 3, 4, 5, 6 |> ------------------------------------------> |possiblity | 0.1, 0.2, 0.2,0.15,0.2,0.15 | (total possiblity of one)> __________________________________________> There are four dices on the screen.> __ __ __ __> |__| |__| |__| |__|> Each prize draw will roll the dices to show 4 random numbers based on > the probability provided in the table.> The award prize is the sum of all the numbers appeared in the 4 slots. > E.g. 3351 will have a prize of 3+3+5+1=12. So the prize is in the range > of 4 to 24.> Q1: What is the average prize for this gambling machine?The expected value.> Q2: Generalize the question: n numbers, w_1, w_2, w_3 .... w_n, and > their corresponding possiblities: p_1, p_2, p_3 .... p_n. With > replacement, 4 boxes, each box 'lled with one number, what is the > average sum of these four numbers?> The expected value.The question is: do you need more of a hint than that? If so, what have you done with this so far?-- Will Twentyman > **** Gambling Dice Machine probability ****>> A gambling machine:> Note: this is generally referred to as a probability distribution, and > possibility would be probability.>> A possiblity table of the possiblity of each side of a dice, from 1 to 6:>> __________________________________________>> |side | 1, 2, 3, 4, 5, 6 |>> ------------------------------------------>> |possiblity | 0.1, 0.2, 0.2,0.15,0.2,0.15 | (total possiblity of one)>> __________________________________________>> There are four dices on the screen.>> __ __ __ __>> |__| |__| |__| |__|>> Each prize draw will roll the dices to show 4 random numbers based on >> the probability provided in the table.>> The award prize is the sum of all the numbers appeared in the 4 >> slots. E.g. 3351 will have a prize of 3+3+5+1=12. So the prize is in >> the range of 4 to 24.>> Q1: What is the average prize for this gambling machine?> The expected value.about it again, and 'nd a easier way to express it :)I have got a formula, but I do think it is the smartest way to do ... I state as following: for a single case: like 3351, for example, it isprobability = 0.2*0.2*0.2*0.1prize = 3+3+5+1= 12weight of 3351 = probability * prize = 12 * (0.2*0.2*0.2*0.1)So the forumula of weight for a single case is: weight = (w_i+w_j+w_k+w_l)*(p_i*p_j*p_k*p_l)Enumerate all the case, from 1111 to 6666, get all the weight, and add them together, then get the average: for i=1~6 for j=1~6 for k=1~6 for l=1~6W_{avg}=sum_{l=1}^{6}sum_{k=1}^{6}sum_{j=1}^{6} sum_{i=1}^{6}(w_{i}+w_{j}+w_{k}+w_{l}) *(p_{i} * p_{j} * p_{k} * p_{l})But this algorithm will end up with a high complexity, because there is redudency: e.g. 1121 and 1112, in the enumeration, have the same sum, but were counted twice, similarly, 2111, 1211.So I guess that there will be a smarter way than mine... Any suggestion is highly appreciated! > **** Gambling Dice Machine probability ****>> A gambling machine:> Note: this is generally referred to as a probability distribution, and >> possibility would be probability.> A possiblity table of the possiblity of each side of a dice, from 1 > to 6:> __________________________________________> |side | 1, 2, 3, 4, 5, 6 |> ------------------------------------------> |possiblity | 0.1, 0.2, 0.2,0.15,0.2,0.15 | (total possiblity of one)> __________________________________________>> There are four dices on the screen.> __ __ __ __> |__| |__| |__| |__|>> Each prize draw will roll the dices to show 4 random numbers based on > the probability provided in the table.>> The award prize is the sum of all the numbers appeared in the 4 > slots. E.g. 3351 will have a prize of 3+3+5+1=12. So the prize is in > the range of 4 to 24.>> Q1: What is the average prize for this gambling machine?>> The expected value.> about it again, and 'nd a easier way to express it :)> I have got a formula, but I do think it is the smartest way to do ... I > state as following:> for a single case: like 3351, for example, it is> probability = 0.2*0.2*0.2*0.1> prize = 3+3+5+1= 12> weight of 3351 = probability * prize = 12 * (0.2*0.2*0.2*0.1) So the forumula of weight for a single case is: weight = > (w_i+w_j+w_k+w_l)*(p_i*p_j*p_k*p_l)> Enumerate all the case, from 1111 to 6666, get all the weight, and add > them together, then get the average:> for i=1~6> for j=1~6> for k=1~6> for l=1~6> W_{avg}=sum_{l=1}^{6}sum_{k=1}^{6}sum_{j=1}^{6}sum_{i= 1}^{6}> (w_{i}+w_{j}+w_{k}+w_{l}) *(p_{i} * p_{j} * p_{k} * p_{l})> But this algorithm will end up with a high complexity, because there is > redudency: e.g.> 1121 and 1112, in the enumeration, have the same sum, but were counted > twice, similarly, 2111, 1211.> So I guess that there will be a smarter way than mine... Any suggestion > is highly appreciated!Since each die rolls independently of the other dice, the expected value for each die will be identical. Given that result, you now only need to know the number of dice.For example, normally dice have probability 1/6 for each side, which gives an expected value of 3.5. For two dice, it's 7; for three dice it's 10.5; etc.-- Will Twentyman > **** Gambling Dice Machine probability ****>> A gambling machine:>> Note: this is generally referred to as a probability distribution, > and possibility would be probability.> A possiblity table of the possiblity of each side of a dice, from 1 >> to 6:>> __________________________________________>> |side | 1, 2, 3, 4, 5, 6 |>> ------------------------------------------>> |possiblity | 0.1, 0.2, 0.2,0.15,0.2,0.15 | (total possiblity of one)>> __________________________________________>> There are four dices on the screen.>> __ __ __ __>> |__| |__| |__| |__|>> Each prize draw will roll the dices to show 4 random numbers based >> on the probability provided in the table.>> The award prize is the sum of all the numbers appeared in the 4 >> slots. E.g. 3351 will have a prize of 3+3+5+1=12. So the prize is >> in the range of 4 to 24.>> Q1: What is the average prize for this gambling machine?> The expected value.>> about it again, and 'nd a easier way to express it :)>> I have got a formula, but I do think it is the smartest way to do ... >> I state as following:>> for a single case: like 3351, for example, it is>> probability = 0.2*0.2*0.2*0.1>> prize = 3+3+5+1= 12>> weight of 3351 = probability * prize = 12 * (0.2*0.2*0.2*0.1)>> So the forumula of weight for a single case is: weight = >> (w_i+w_j+w_k+w_l)*(p_i*p_j*p_k*p_l)>> Enumerate all the case, from 1111 to 6666, get all the weight, and add >> them together, then get the average:>> for i=1~6>> for j=1~6>> for k=1~6>> for l=1~6>> W_{avg}=sum_{l=1}^{6}sum_{k=1}^{6}sum_{j=1}^{6}sum_{i= 1}^{6}>> (w_{i}+w_{j}+w_{k}+w_{l}) *(p_{i} * p_{j} * p_{k} * p_{l})>> But this algorithm will end up with a high complexity, because there >> is redudency: e.g.>> 1121 and 1112, in the enumeration, have the same sum, but were >> counted twice, similarly, 2111, 1211.>> So I guess that there will be a smarter way than mine... Any >> suggestion is highly appreciated!> Since each die rolls independently of the other dice, the expected value > for each die will be identical. Given that result, you now only need to > know the number of dice.> For example, normally dice have probability 1/6 for each side, which > gives an expected value of 3.5. For two dice, it's 7; for three dice > it's 10.5; etc.> This is really surpising! I did not expect that there could exist such a smart solution!!complexity was reduced from O(n^4) to O(n)....!then the formula will be as simple as 4*sum_{i=1}^{4}(w_{i}*p_{i}) =A similar problem, I think it might not be the most effective solution as well.... Looks like it is, but not 100% sureProblem statement:A probability table of each side of a dice, from 1 to 6:__________________________________________|side | 1, 2, 3, 4, 5, 6 |------------------------------------------|possiblity | p_1, p_2, p_3, p_4, p_5, p_6 | (total possiblity of one)__________________________________________There are four dices on the screen. __ __ __ __ |__| |__| |__| |__|Each prize draw will roll the dices to show 4 random numbers based on the probability provided in the table.The difference in that, the prize of award of each roll is the maximum number of the four dices:e.gif a roll is 3351, then the prize will be max(3,3,5,1) = 5if a roll is 1121, then the prize will be max(1,1,2,1) = 2Q: What is the average prize of award for this scheme?Solution (maybe not the best):1. caculating the probability of each possible prize, from 1 to 6. P(i),e.g. P(5), is the possiblity that a roll of four dices will have at least one 've, e.g. 1235, 1154, 5554, 1151, thus, prize is 5P(6)P(5)P(4)P(3)P(2)P(1)The caculation of P(5) is:P(5) = (p_5+p_4+p_3+p_2+p_1)^4 - (p_4+p_3+p_2+p_1)^4Where (p_5+p_4+p_3+p_2+p_1)^4 is the probability that for the 4 dices,1,2,3,4,5 could appear in each dice. (e.g the probability of all from 1111 to 5555)Minus the possiblity that there is no 've exists (p_4+p_3+p_2+p_1)^4, i.e. the probability of all from 1111 to 4444So it is the case, that at least one 've exist.Generalized formula:P(i)=(sum_(a=1)^(i)p_{a})^4 - (sum_{i=b}^{i-1}p_{b})^42 Weight each number with its probability, and sum all of them:Average = 6*P(6) + 5*P(5) + 4*P(4) + 3*P(3) + 2*P(2) + 1*P(1)Average Formula = sum_{i=1}^{6} P(i)*iI am not sure is it clear to say like appreciated!! A similar problem, I think it might not be the most effective solution > as well.... Looks like it is, but not 100% sure> Problem statement:> A probability table of each side of a dice, from 1 to 6:> __________________________________________> |side | 1, 2, 3, 4, 5, 6 |> ------------------------------------------> |possiblity | p_1, p_2, p_3, p_4, p_5, p_6 | (total possiblity of one)> __________________________________________> There are four dices on the screen.> __ __ __ __> |__| |__| |__| |__|> Each prize draw will roll the dices to show 4 random numbers based on > the probability provided in the table.> The difference in that, the prize of award of each roll is the maximum > number of the four dices:> e.g> if a roll is 3351, then the prize will be max(3,3,5,1) = 5> if a roll is 1121, then the prize will be max(1,1,2,1) = 2> Q: What is the average prize of award for this scheme?> Solution (maybe not the best):> 1. caculating the probability of each possible prize, from 1 to 6. P(i),> e.g. P(5), is the possiblity that a roll of four dices will have at > least one 've, e.g. 1235, 1154, 5554, 1151, thus, prize is 5> P(6)> P(5)> P(4)> P(3)> P(2)> P(1)> The caculation of P(5) is:> P(5) = (p_5+p_4+p_3+p_2+p_1)^4 - (p_4+p_3+p_2+p_1)^4> Where (p_5+p_4+p_3+p_2+p_1)^4 is the probability that for the 4 > dices,1,2,3,4,5 could appear in each dice. (e.g the probability of all > from 1111 to 5555)> Minus the possiblity that there is no 've exists (p_4+p_3+p_2+p_1)^4, > i.e. the probability of all from 1111 to 4444> So it is the case, that at least one 've exist.> Generalized formula:> P(i)=(sum_(a=1)^(i)p_{a})^4 - (sum_{i=b}^{i-1}p_{b})^4> 2 Weight each number with its probability, and sum all of them:> Average = 6*P(6) + 5*P(5) + 4*P(4) + 3*P(3) + 2*P(2) + 1*P(1)> Average Formula = sum_{i=1}^{6} P(i)*i> I am not sure is it clear to say appreciated!!> This is exactly the way I would approach the problem.-- Will Twentyman > A similar problem, I think it might not be the most effective solution >> as well.... Looks like it is, but not 100% sure>> Problem statement:>> A probability table of each side of a dice, from 1 to 6:>> __________________________________________>> |side | 1, 2, 3, 4, 5, 6 |>> ------------------------------------------>> |possiblity | p_1, p_2, p_3, p_4, p_5, p_6 | (total possiblity of one)>> __________________________________________>> There are four dices on the screen.>> __ __ __ __>> |__| |__| |__| |__|>> Each prize draw will roll the dices to show 4 random numbers based on >> the probability provided in the table.>> The difference in that, the prize of award of each roll is the >> maximum number of the four dices:>> e.g>> if a roll is 3351, then the prize will be max(3,3,5,1) = 5>> if a roll is 1121, then the prize will be max(1,1,2,1) = 2>> Q: What is the average prize of award for this scheme?>> Solution (maybe not the best):>> 1. caculating the probability of each possible prize, from 1 to 6. P(i),>> e.g. P(5), is the possiblity that a roll of four dices will have at >> least one 've, e.g. 1235, 1154, 5554, 1151, thus, prize is 5>> P(6)>> P(5)>> P(4)>> P(3)>> P(2)>> P(1)>> The caculation of P(5) is:>> P(5) = (p_5+p_4+p_3+p_2+p_1)^4 - (p_4+p_3+p_2+p_1)^4>> Where (p_5+p_4+p_3+p_2+p_1)^4 is the probability that for the 4 >> dices,1,2,3,4,5 could appear in each dice. (e.g the probability of all >> from 1111 to 5555)>> Minus the possiblity that there is no 've exists (p_4+p_3+p_2+p_1)^4, >> i.e. the probability of all from 1111 to 4444>> So it is the case, that at least one 've exist.>> Generalized formula:>> P(i)=(sum_(a=1)^(i)p_{a})^4 - (sum_{i=b}^{i-1}p_{b})^4>> 2 Weight each number with its probability, and sum all of them:>> Average = 6*P(6) + 5*P(5) + 4*P(4) + 3*P(3) + 2*P(2) + 1*P(1)>> Average Formula = sum_{i=1}^{6} P(i)*i>> I am not sure is it clear to say appreciated!!> This is exactly the way I would approach the problem.> Best wishes,Charlie **** Gambling Dice Machine probability ****>> A gambling machine:> Note: this is generally referred to as a probability distribution, >> and possibility would be probability.> A possiblity table of the possiblity of each side of a dice, from 1 > to 6:> __________________________________________> |side | 1, 2, 3, 4, 5, 6 |> ------------------------------------------> |possiblity | 0.1, 0.2, 0.2,0.15,0.2,0.15 | (total possiblity of one)> __________________________________________>> There are four dices on the screen.> __ __ __ __> |__| |__| |__| |__|>> Each prize draw will roll the dices to show 4 random numbers based > on the probability provided in the table.>> The award prize is the sum of all the numbers appeared in the 4 > slots. E.g. 3351 will have a prize of 3+3+5+1=12. So the prize is > in the range of 4 to 24.>> Q1: What is the average prize for this gambling machine?>> The expected value.>> about it again, and 'nd a easier way to express it :)>> I have got a formula, but I do think it is the smartest way to do ... > I state as following:> for a single case: like 3351, for example, it is>> probability = 0.2*0.2*0.2*0.1> prize = 3+3+5+1= 12>> weight of 3351 = probability * prize = 12 * (0.2*0.2*0.2*0.1)>> So the forumula of weight for a single case is: weight = > (w_i+w_j+w_k+w_l)*(p_i*p_j*p_k*p_l)> Enumerate all the case, from 1111 to 6666, get all the weight, and add > them together, then get the average:> for i=1~6> for j=1~6> for k=1~6> for l=1~6>> W_{avg}=sum_{l=1}^{6}sum_{k=1}^{6}sum_{j=1}^{6}sum_{i= 1}^{6}>> (w_{i}+w_{j}+w_{k}+w_{l}) *(p_{i} * p_{j} * p_{k} * p_{l})> But this algorithm will end up with a high complexity, because there > is redudency: e.g.> 1121 and 1112, in the enumeration, have the same sum, but were > counted twice, similarly, 2111, 1211.> So I guess that there will be a smarter way than mine... Any > suggestion is highly appreciated! Not necessarily. If the probabilities are different for the different dice (i.e. the die in slot 3 is different from the die in slot 4), then there is no simpler method.Computer time is cheap and programmer time is expensive (and analyst time even more so), so the most ef'cient solution is almost certainly to just do it this way. At least for small numbers of dice and sides (like 4 dice and 6 sides).Unless, of course, you just happen to see the answer. Or the enjoyment you get out of solving the problem exceeds the cost of solving the problem. Or there are a large number of dice. (As in computing reserves [and that's not just for insurance companies any more, since GASB 10 and the increase in self-funding of (especially) workers comp and risk management in general, both for government entities and publicly traded companies].).Jon Miller And I'm not saying> proof but a proof, which means that the mathematical argument is> true, not just that there's some yahoo claiming they have a proof,> when actually they just have a §awed argument.> James HarrisJSH has lots of experience in this area. =I'm a software engineer looking to go to grad school for biostatistics. Theschools have prerequisites of linear algebra, differential equations, andstatistics. I took all these courses 10 years ago, but hardly rememberthem. Can someone suggest a good book for catching up on all these topics? =Personally I like to use the Linear Algebra book by BernardKolman--Introductory Linear Algebra with Applications.>> I'm a software engineer looking to go to grad school for biostatistics.The> schools have prerequisites of linear algebra, differential equations, and> statistics. I took all these courses 10 years ago, but hardly remember> them. Can someone suggest a good book for catching up on all thesetopics?> =Not just trolling here... I would really like to hear your opinion:What are the ('ve to ten, say) greatest unsolved problems in mathematics?Of course, I can start the list with the Riemann Hypothesis. Not just trolling here... I would really like to hear your opinion:> What are the ('ve to ten, say) greatest unsolved problems in mathematics?> Of course, I can start the list with the Riemann Hypothesis.> How about the other Millennium Prize Problems? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ > Not just trolling here... I would really like to hear your opinion:> What are the ('ve to ten, say) greatest unsolved problems inmathematics?>> Of course, I can start the list with the Riemann Hypothesis.>> How about the other Millennium Prize Problems?> OK, A good point! Those are quite fresh problems - anyway.I prefer one of the oldest problem:We have computers and software, but we cannot proof Eqyptian fractionproblem (formulated by Erdos-Strauss):4/n=(1/x)+(1/y)+(1/z) is true for every positive integer n so that x,y,and zare also positive integers.Egyptian used unit fractions (1/integer) daily instead of the decimalexpansions, but we are still out of the general solution Erdos-Straussproblem.Tapio>> --> G. A. Edgarhttp://www.math.ohio-state.edu/~edgar/ Not just trolling here... I would really like to hear your opinion:> What are the ('ve to ten, say) greatest unsolved problems in mathematics?How to get rid of James Harris?GC> Of course, I can start the list with the Riemann Hypothesis. =I'm a big fan of the closed form solution of Kepler's Equation. That one's also been around for several centuries, and it's got a very strong practical foundation from physics.>> What are the ('ve to ten, say) greatest unsolved problems in > mathematics? I prefer one of the oldest problem:> We have computers and software, but we cannot proof Eqyptian fraction> problem (formulated by Erdos-Strauss):> 4/n=(1/x)+(1/y)+(1/z) is true for every positive integer n so that x,y,and z> are also positive integers.> Egyptian used unit fractions (1/integer) daily instead of the decimal> expansions, but we are still out of the general solution Erdos-Strauss> problem.> Tapio> Some material and links for the Egyptian fraction problem:http://www.research.att.com/projects/OEIS?Anum= A073101Hugo Pfoertner =The word greatest in mathematics has two meanings:a) Most importantb) Most dif'cultUnder the 'rst meaning, I think the seven problems submitted by theClay Institute are the most important.But under the second, in my opinion are these: (Possibly indecidables)1.- The Riemann Hypothesis. This problem is important and dif'cult.2.- The Goldbach's Conjecture3.- The twin primes conjecture.4.- The no-existence of odd perfect numbers.5.- The Firoozbakht Conjecture: nLog(p(n+1)<(n+1)Log(p(n)) ; p(n)=nthprime. This problem is important and dif'cult.6.- The Collatz problem: X(n+1)= 3X(n)+1 if X(n)is odd ; X(n+1)= X(n)/2 if even. Ends the sequence always, in the loop 1, 4, 2, 1 ....? Luis Rodriguez The word greatest in mathematics has two meanings:> a) Most important> b) Most dif'cult> Under the 'rst meaning, I think the seven problems submitted by the> Clay Institute are the most important.> But under the second, in my opinion are these: (Possibly indecidables)> 1.- The Riemann Hypothesis.> This problem is important and dif'cult.> 2.- The Goldbach's ConjectureProving, if it's true, the Goldbach's Conjecture, might be dif'cult,but is it an important? I mean, it might be true just because there areenough primes and not for any deep reason. Perhaps that can only beknown when it is proved, if it ever is.GC> 3.- The twin primes conjecture.> 4.- The no-existence of odd perfect numbers.> 5.- The Firoozbakht Conjecture: nLog(p(n+1)<(n+1)Log(p(n)) ; p(n)=nth> prime.> This problem is important and dif'cult.> 6.- The Collatz problem:> X(n+1)= 3X(n)+1 if X(n)is odd ; X(n+1)= X(n)/2 if even.> Ends the sequence always, in the loop 1, 4, 2, 1 ....? Luis Rodriguez =So would Fermat's Last Theorem be the Greatest Solved Problem?Interest in this problem continues despite it having been solved.--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm = > So would Fermat's Last Theorem be the Greatest Solved Problem?> Interest in this problem continues despite it having been solved.The same is true of the 4-colour theorem.To some extent a great solved problem (in the original thread-sense)could also be termed a Crackpot Magnet.On this basis, we would *de'nitely* have to include1) Godel's theorem;2) Cantors uncountability theorem.For some reason, Angle Trisection has dropped off the list! Dunno why.It seems to have been still going strong in the O60s, so says Dudley Underwood.--------------------------------------------------- --------------------------- Bill Taylor W.Taylor@math.canterbury.ac.nz-------------------------------- ---------------------------------------------- I'm a little crackpot short and stout. Don't ask my handle, let me spout. When I get all steamed up then I shout, Snip a poster - §ame the lout.--------------------------------------------------------- --------------------- So would Fermat's Last Theorem be the Greatest Solved Problem?> Interest in this problem continues despite it having been solved.> The same is true of the 4-colour theorem.> To some extent a great solved problem (in the original thread-sense)> could also be termed a Crackpot Magnet.> On this basis, we would *de'nitely* have to include> 1) Godel's theorem;> 2) Cantors uncountability theorem.> For some reason, Angle Trisection has dropped off the list! Dunno why.> It seems to have been still going strong in the O60s, so says Dudley Underwood.Long before I ever heard of Usenet, a fellow I workedwith brought me a manuscript from a friend of histhat he wanted my opinion on. I was allowed to lookat but not copy it.It was 20 pages of geometrical gibberish that he wantedsomeone with AutoCad to verify. Apparently, he thoughtthat a CAD program could verify a proof in geometry.It was 'lled with such nonsense as ...and if AutoCadblows up at this point, that will be very interesting.I was suspicious when, in the introduction, he said thatverifying the proof would be a feather in the cap foryour company and may even lead to an appearance onNightline. (Sound familiar?)Flipping ahead to the end, I found the give-away:...and this will prove that the enneagon is possible,from which one can trisect an angle.Wow, a genuine crackpot manuscript! It's not just folklore, mathematical cranks really do exist. I had to give the manuscript back (I wish I had made a copy)and explain to the fellow that his friend was just acommon angle trisector crank who had high-tech aspirations and that there was no point in asking anyone to waste their time on such nonsense.> ------------------------------------------------------------- -----------------> Bill Taylor W.Taylor@math.canterbury.ac.nz> ------------------------------------------------------------- -----------------> I'm a little crackpot short and stout.> Don't ask my handle, let me spout.> When I get all steamed up then I shout,> Snip a poster - §ame the lout.> ------------------------------------------------------------- ----------------- =incidentally,Dudley's book glosses-over what is apparently his favorite fermatiste,whose would-be proof amounts to only 36 pages or so. funny thing:it was clear to me that he went straight to the climactic chapter,without bothering with any of the (mostly) simple formalities, orwith any of the humorous & interesting appendices, buthe couldn't get it at exactly the same place as I couldn't,in that shrot chapter. isn't his *last* name, Dudley? > For some reason, Angle Trisection has dropped off the list! Dunno why.> It seems to have been still going strong in the O60s, so says Dudley Underwood.--Dec.2000 OWAND' Chairman Paul O'Neill, reelectedto Board. Newsish?http://www.rand.org/publications/randreview/issues/rr news@aol.com =|For some reason, Angle Trisection has dropped off the list! Dunno why.|It seems to have been still going strong in the O60s, so says Dudley|Underwood.Perhaps because Euclidean geometry is not taught the way it usedto be? I don't think I was told about these impossibilities in a class.department in the late 80s, though.Somewhat unusually, there was a postcard from someone challengingFaltings' theorem that algebraic curves of genus >1 over Q have 'nitelymany rational points on them. The writer evidently had read the formulafor the genus of a curve of degree d without realizing that it wouldn'tapply to a reducible polynomial.One outstanding unsolved problem I haven't seen mentioned here iswhether there exists an algorithm for determining whether there's apoint with rational coordinates satisfying a set of polynomials withrational coordinates. It might seem a little surprising since we knowthat the corresponding problem for points with integer coordinatesis not computable-- it's equivalent to the halting problem.I think the people who selected the Clay foundation problems did agood job. They seem to have given number theory at least its fairshare of the problems.I like a problem for which Erdos offered a $3000 prize. I don't knowwhether Graham has kept the offer open. In any sequencea10 such that for all but 'nitely many nthe number of elements less than n is >=epsilon*n) then it hasarbitrarily long arithmetic sequences in it was proven, but ittook a good solid effort to prove it.Another thing I like about that conjecture is that it's elementaryenough to be explained to just about anybody who likes math,at least with a little patience. Maybe it's not a great problem,but at least it's a pretty good problem.Keith Ramsay incidentally,> Dudley's book glosses-over what is apparently his favorite fermatiste,> whose would-be proof amounts to only 36 pages or so. funny thing:> it was clear to me that he went straight to the climactic chapter,> without bothering with any of the (mostly) simple formalities, or> with any of the humorous & interesting appendices, but> he couldn't get it at exactly the same place as I couldn't,> in that shrot chapter.> isn't his *last* name, Dudley?Yes, it's Underwood (First) Dudley (Last). A constant struggle to keep itstraight. Sometimes people call him Woody, and you'd think awareness ofthis would help, but I still have to struggle.He just retired from DePauw university this year, but I rather expect we'llsee signs of his activity for a good time to come.Give me a name like Bill Taylor, there's a name I can keep straight. =I have a quick question regarding work that has already been donetowards the ends of determining if a Turing machine halts. I haveread the material I have been able to 'nd through Google and lookingif a concept I am using is out there.The method is as follows:1)Take a binary, quintuple Turing Machine M for which the HaltingProblem is unknown2)Assume M Halts3)Create an in'nite tape which can contain 3 values = {0, 1,UnDe'ned}De'nition of Unde'ned - A postition in memory where the value is notknown and may be either 0 or 14)Starting at a chosen position on the tape, mark the conditionrequired for halting. For example, a machine may need to read a 1 instate k. Therefore, mark a 1 on the tape.5)Look for each instruction which calls state k. Take the next(or'rst) instruction that does so. If no more instructions call statek, fallback to the last value for k. If this is the 'rst value for kand all paths have returned, the machine does not halt.6)Move on the tape the direction opposite the instruction.7)If the memory at that position is Unde'ned, mark the conditionrequired for that state to run. For example, state j may call state kif it reads a 0. Go to step 5 using state k=j.8)If the memory at that position is 0 or 1, compare the value of thememory with the data which the instruction marks to the tape. Forexample, the instruction may write a 0 to the tape. If the two valuesare not equal, go to step 6 and increment to the next instructionwhich calls state k. If the values are equal, proceed as if memorywas unde'ned.I hope that makes sense, it is my attempt at translating ym recursivecode. Basically, the function runs a Turing Machine backwardsassuming that the machine halts. Id the machine halts in n stepsthere is a path backwards n steps which has the tape in the originalcon'guration. If none of the paths create that con'guration, yetthey all terminate because of a contradiction between the memory andthe instructions, the machine does not halt. If the machine runs foran arbitrary number of steps backwards in a single path with noresult, the machine's halting status is considered unde'ned.If my method is in fact valid, it is actually a very powerful methodfor determining if a Turing Machine halts. In practice it eliminatesa large percentage of 4 state Turing machine using a step ceiling of30 steps backwards. On a theoretical note, I believe that the easiestmachines to use this method on are the machines that are dif'cult todiscern a pattern in when running normally. Basically every path thatterminates when run in reverse is a sequence which cannot occur on thememory if the machine does not halt (otherwise the machine would infact halt).The longer and more numerous that paths that run backwards, the moresequences which cannot occur on memory. That means that a pettern isjust a bit more likely to occur because another sequence isimpossible. The fewer and smaller the patterns, the more optionsavailable to the machine without halting. This class of programs iswhere I think we 'nd some of the more dif'cult machines to workwith.Advice, questions, has anyone else already done this?-Chris Pitman I have a quick question regarding work that has already been done> towards the ends of determining if a Turing machine halts. I have> read the material I have been able to 'nd through Google and looking> if a concept I am using is out there.> The method is as follows:> 1)Take a binary, quintuple Turing Machine M for which the Halting> Problem is unknown> 2)Assume M Halts> 3)Create an in'nite tape which can contain 3 values = {0, 1,> UnDe'ned}> De'nition of Unde'ned - A postition in memory where the value is not> known and may be either 0 or 1> 4)Starting at a chosen position on the tape, mark the condition> required for halting. For example, a machine may need to read a 1 in> state k. Therefore, mark a 1 on the tape.> 5)Look for each instruction which calls state k. Take the next(or> 'rst) instruction that does so. If no more instructions call state> k, fallback to the last value for k. If this is the 'rst value for k> and all paths have returned, the machine does not halt.> 6)Move on the tape the direction opposite the instruction.> 7)If the memory at that position is Unde'ned, mark the condition> required for that state to run. For example, state j may call state k> if it reads a 0. Go to step 5 using state k=j.> 8)If the memory at that position is 0 or 1, compare the value of the> memory with the data which the instruction marks to the tape. For> example, the instruction may write a 0 to the tape. If the two values> are not equal, go to step 6 and increment to the next instruction> which calls state k. If the values are equal, proceed as if memory> was unde'ned.> I hope that makes sense, it is my attempt at translating ym recursive> code. Basically, the function runs a Turing Machine backwards> assuming that the machine halts. Id the machine halts in n steps> there is a path backwards n steps which has the tape in the original> con'guration. If none of the paths create that con'guration, yet> they all terminate because of a contradiction between the memory and> the instructions, the machine does not halt. If the machine runs for> an arbitrary number of steps backwards in a single path with no> result, the machine's halting status is considered unde'ned.> If my method is in fact valid, it is actually a very powerful method> for determining if a Turing Machine halts. In practice it eliminates> a large percentage of 4 state Turing machine using a step ceiling of> 30 steps backwards. On a theoretical note, I believe that the easiest> machines to use this method on are the machines that are dif'cult to> discern a pattern in when running normally. Basically every path that> terminates when run in reverse is a sequence which cannot occur on the> memory if the machine does not halt (otherwise the machine would in> fact halt).> The longer and more numerous that paths that run backwards, the more> sequences which cannot occur on memory. That means that a pettern is> just a bit more likely to occur because another sequence is> impossible. The fewer and smaller the patterns, the more options> available to the machine without halting. This class of programs is> where I think we 'nd some of the more dif'cult machines to work> with.yeah, your method in not invalid, it just doesn't solve the haltingproblem. Your method solves the problem Does there exist any tapethat causes the TM to halt?. The real halting problem is Does runningthe TM on this speci'c tape halt? which is a different problem alltogether.-- mark@biggar.orgmark.a.biggar@comcast.net [ Chris Pitman presented an algorithm ]> yeah, your method in not invalid, it just doesn't solve the halting> problem. Your method solves the problem Does there exist any tape> that causes the TM to halt?. No it doesn't. How do I know, not having read Pitman's post? Take any instance of the halting problem, e.g. does the TM phi halt for input p. Take any TM which never halts for any input (there are plenty of these) and modify it so that for inputs other than p it still does not halt, but if the input is p, it emulates phi. Call this new machhine psi. Clearly now the question whether psi halts for any tape is equivalent to whether phi halts for input phi.-- Aatu Koskensilta (aatu.koskensilta@xortec.')Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus No it doesn't. How do I know, not having read Pitman's post? Take any > instance of the halting problem, e.g. does the TM phi halt for input p. > Take any TM which never halts for any input (there are plenty of these) > and modify it so that for inputs other than p it still does not halt, > but if the input is p, it emulates phi. Call this new machhine psi. > Clearly now the question whether psi halts for any tape is equivalent > to whether phi halts for input phi.This should read: ... to whether phi halts for input p.-- Aatu Koskensilta (aatu.koskensilta@xortec.')Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus 5)Look for each instruction which calls state k. Take the next(or>'rst) instruction that does so. If no more instructions call state>k, fallback to the last value for k. If this is the 'rst value for k>and all paths have returned, the machine does not halt.Exhaustive search.>6)Move on the tape the direction opposite the instruction.You have not resolved the in'nite search with many possible states j callingstate k. So the instruction seems to be unde'ned at this point.>7)If the memory at that position is Unde'ned, mark the condition>required for that state to run. What state? Which instruction?>Basically, the function runs a Turing Machine backwards>assuming that the machine halts. Sorry, the process is multiple branch and the tape is linear. Only works forthose Turing machines which call one state from every other state, I think. Notmuch of a universal halting problem there, is it?>If my method is in fact valid, it is actually a very powerful method>for determining if a Turing Machine halts. I don't think it is valid.>In practice it eliminates>a large percentage of 4 state Turing machine using a step ceiling of>30 steps backwards. Maybe it does. That may be signi'cant.What you are essentially saying is that there is a transition matrix that saysif the tape reads x and the machine state is y then go to state z. That is partof the de'nition of the Turing machine and is 'ne. But to make this approachwork, there would have to be another matrix that says if the state is z thenconsider all states y and al inputs x and branch and branch....Now, Mathcad will let me create a 213x217 matrix M with each element equal to a3-column row vector representing RGB or anything I want. But, and I think forthe same reason as here, they will NOT allow me to say Let element (3, 2) of matrix M now be equal to a 4-column row vector.In fact, they prohibit ALL such multiple index assignments. You can't even sayLet element (4, 5) of matrix M now be a different 3-column row vector.And I think they must have a very wise reason for this. Hey, I'm not sayingit's not frustrating. But it is real.Basically, you're saying that state x can now have multiple simultaneous valuesand so can input y. This is NOT allowed in TM. There is no parallel TM that Iknow of.A quantum TM? Hmmmmm.....I guess if you reserve x times y for each entry in z, then you have areversible TM. But then z is no longer a state, is it?Yours,Doug Goncz, Replikon Research, Seven Corners, VA Fair use and Usenet distribution without restriction or feeCivil and criminal penalties for circumvention of any embedded encryption =I knew I must have no explained something well. While running themachine backwards you check for the starting tape in question. Forexample, you may be looking for a tape where all de'ned positions are0 and the current state is state 1. (For my application I am focusingon blank tapes, but there should be no difference as long as you lookfor the correct starting con'guration)I can easily construct a machine where there is a tape that it haltson but it does not halt on the input tape, and have this method work. Its simple, but it at least shows that the method is also applicableto machines that do have inputs that they halt on. 0 1S1 |1 S2 R|1 S3 RS2 |0 S1 R|0 S3 RS3 |1 H R|1 H RThe above machine will halt when started on any mark, yet does nothalt on a blank tape. Like I said, the machine is overtly simple, butit shows the method of reverse running is more powerful then thequestion Does there exist any tape that causes the TM to halt?.-Chris PitmanPS - My replys may be a bit on the slower side, because I must wait I have a quick question regarding work that has already been done> towards the ends of determining if a Turing machine halts. I have> read the material I have been able to 'nd through Google and looking> if a concept I am using is out there.> The method is as follows:> 1)Take a binary, quintuple Turing Machine M for which the Halting> Problem is unknown> 2)Assume M Halts> 3)Create an in'nite tape which can contain 3 values = {0, 1,> UnDe'ned}> De'nition of Unde'ned - A postition in memory where the value is not> known and may be either 0 or 1> 4)Starting at a chosen position on the tape, mark the condition> required for halting. For example, a machine may need to read a 1 in> state k. Therefore, mark a 1 on the tape.> 5)Look for each instruction which calls state k. Take the next(or> 'rst) instruction that does so. If no more instructions call state k, fallback to the last value for k. If this is the 'rst value for k> and all paths have returned, the machine does not halt.> 6)Move on the tape the direction opposite the instruction.> 7)If the memory at that position is Unde'ned, mark the condition> required for that state to run. For example, state j may call state k> if it reads a 0. Go to step 5 using state k=j.> 8)If the memory at that position is 0 or 1, compare the value of the> memory with the data which the instruction marks to the tape. For> example, the instruction may write a 0 to the tape. If the two values> are not equal, go to step 6 and increment to the next instruction> which calls state k. If the values are equal, proceed as if memory> was unde'ned.> I hope that makes sense, it is my attempt at translating ym recursive> code. Basically, the function runs a Turing Machine backwards assuming that the machine halts. Id the machine halts in n steps> there is a path backwards n steps which has the tape in the original> con'guration. If none of the paths create that con'guration, yet> they all terminate because of a contradiction between the memory and> the instructions, the machine does not halt. If the machine runs for> an arbitrary number of steps backwards in a single path with no> result, the machine's halting status is considered unde'ned.> If my method is in fact valid, it is actually a very powerful method> for determining if a Turing Machine halts. In practice it eliminates> a large percentage of 4 state Turing machine using a step ceiling of> 30 steps backwards. On a theoretical note, I believe that the easiest> machines to use this method on are the machines that are dif'cult to> discern a pattern in when running normally. Basically every path that> terminates when run in reverse is a sequence which cannot occur on the> memory if the machine does not halt (otherwise the machine would in> fact halt).> The longer and more numerous that paths that run backwards, the more> sequences which cannot occur on memory. That means that a pettern is> just a bit more likely to occur because another sequence is> impossible. The fewer and smaller the patterns, the more options> available to the machine without halting. This class of programs is> where I think we 'nd some of the more dif'cult machines to work> with.> yeah, your method in not invalid, it just doesn't solve the halting> problem. Your method solves the problem Does there exist any tape> that causes the TM to halt?. The real halting problem is Does running> the TM on this speci'c tape halt? which is a different problem all> together. =Again, I aplogize for the lag in my responses, as I am responding toyour post before Google has even put my last response on the server.code up. I will try to explain in English again too, but this may beeasier for a programmer to understand.This method does in fact trace a tree type structure backwards throughthe machine, which is why I have implemented it as a recursivefunction. For each state that calls the state in question, we trace itback and continue until a terminating condition.As for the data structure I use to keep track of the tree, I traverseit in-order, use the call stack to keep track of what was done on eachmove, and a linear array to store the status of the memory. I havenot found a need to use any matrices so far.[code]enum TERM_STATUS{TERM, NONTERM, RUNNING};int revmem[reverse_max*2];TERM_STATUS ReverseRun(TRANS* info){int i;for (i=0; ireverse_max || pos<1 || pos>=reverse_max*2-1) return RUNNING; if (state == 1 && BlankTape()) return TERM; int i, temp; for (i=0; i'rst) instruction that does so. If no more instructions call state>k, fallback to the last value for k. If this is the 'rst value for k>and all paths have returned, the machine does not halt.> Exhaustive search.>6)Move on the tape the direction opposite the instruction.> You have not resolved the in'nite search with many possible states j calling> state k. So the instruction seems to be unde'ned at this point.>7)If the memory at that position is Unde'ned, mark the condition>required for that state to run. > What state? Which instruction?>Basically, the function runs a Turing Machine backwards>assuming that the machine halts. > Sorry, the process is multiple branch and the tape is linear. Only works for> those Turing machines which call one state from every other state, I think. Not> much of a universal halting problem there, is it?>If my method is in fact valid, it is actually a very powerful method>for determining if a Turing Machine halts. > I don't think it is valid.>In practice it eliminates>a large percentage of 4 state Turing machine using a step ceiling of>30 steps backwards. > Maybe it does. That may be signi'cant.> What you are essentially saying is that there is a transition matrix that says> if the tape reads x and the machine state is y then go to state z. That is part> of the de'nition of the Turing machine and is 'ne. But to make this approach> work, there would have to be another matrix that says if the state is z then> consider all states y and al inputs x and branch and branch....> Now, Mathcad will let me create a 213x217 matrix M with each element equal to a> 3-column row vector representing RGB or anything I want. But, and I think for> the same reason as here, they will NOT allow me to say Let element (3, 2) of matrix M now be equal to a 4-column row vector.> In fact, they prohibit ALL such multiple index assignments. You can't even say> Let element (4, 5) of matrix M now be a different 3-column row vector.> And I think they must have a very wise reason for this. Hey, I'm not saying> it's not frustrating. But it is real.> Basically, you're saying that state x can now have multiple simultaneous values> and so can input y. This is NOT allowed in TM. There is no parallel TM that I> know of.> A quantum TM? Hmmmmm.....> I guess if you reserve x times y for each entry in z, then you have a> reversible TM. But then z is no longer a state, is it?> Yours,> Doug Goncz, Replikon Research, Seven Corners, VA > Fair use and Usenet distribution without restriction or fee> Civil and criminal penalties for circumvention of any embedded encryption I have a quick question regarding work that has already been done> towards the ends of determining if a Turing machine halts. I have> read the material I have been able to 'nd through Google and looking> if a concept I am using is out there.> The method is as follows:> 1)Take a binary, quintuple Turing Machine M for which the Halting> Problem is unknown> 2)Assume M Halts> 3)Create an in'nite tape which can contain 3 values = {0, 1,> UnDe'ned}> De'nition of Unde'ned - A postition in memory where the value is not> known and may be either 0 or 1That can be seen as a notion for an (in'nite) set of tape contents.> 4)Starting at a chosen position on the tape, mark the condition> required for halting. For example, a machine may need to read a 1 in> state k. Therefore, mark a 1 on the tape.> 5)Look for each instruction which calls state k. Take the next(or> 'rst) instruction that does so. If no more instructions call state> k, fallback to the last value for k. If this is the 'rst value for k> and all paths have returned, the machine does not halt.> 6)Move on the tape the direction opposite the instruction.> 7)If the memory at that position is Unde'ned, mark the condition> required for that state to run. For example, state j may call state k> if it reads a 0. Go to step 5 using state k=j.> 8)If the memory at that position is 0 or 1, compare the value of the> memory with the data which the instruction marks to the tape. For> example, the instruction may write a 0 to the tape. If the two values> are not equal, go to step 6 and increment to the next instruction> which calls state k. If the values are equal, proceed as if memory> was unde'ned.> I hope that makes sense, it is my attempt at translating ym recursive> code. Basically, the function runs a Turing Machine backwards> assuming that the machine halts. Id the machine halts in n steps> there is a path backwards n steps which has the tape in the original> con'guration. If none of the paths create that con'guration, yet> they all terminate because of a contradiction between the memory and> the instructions, the machine does not halt. If the machine runs for> an arbitrary number of steps backwards in a single path with no> result, the machine's halting status is considered unde'ned.> If my method is in fact valid,It appears to be valid (to me).Running a TM backwards can be thought of as equivalent to run anondeterministic TM (suitably de'ned). We build and extend a treeof possible predecessor con'gurations (really: set of them),and that tree may turn out to be of 'nite size due to contradictions.> it is actually a very powerful method> for determining if a Turing Machine halts. In practice it eliminates> a large percentage of 4 state Turing machine using a step ceiling ofA large percentage? Really? How large?> 30 steps backwards. On a theoretical note, I believe that the easiest> machines to use this method on are the machines that are dif'cult to> discern a pattern in when running normally. Basically every path that> terminates when run in reverse is a sequence which cannot occur on the> memory if the machine does not halt (otherwise the machine would in> fact halt).> The longer and more numerous that paths that run backwards, the more> sequences which cannot occur on memory. That means that a pettern is> just a bit more likely to occur because another sequence is> impossible. The fewer and smaller the patterns, the more options> available to the machine without halting. This class of programs is> where I think we 'nd some of the more dif'cult machines to work> with.> Advice, questions, has anyone else already done this?Yes, others have done that kind of backward reasoning.See e.g. http://www.drb.insel.de/~heiner/BB/mabu90.html# Nonterminationcase (b).-- Heiner Marxen I have a quick question regarding work that has already been done> towards the ends of determining if a Turing machine halts. I have> read the material I have been able to 'nd through Google and looking> if a concept I am using is out there. The method is as follows:> 1)Take a binary, quintuple Turing Machine M for which the Halting> Problem is unknown> 2)Assume M Halts> 3)Create an in'nite tape which can contain 3 values = {0, 1,> UnDe'ned}> De'nition of Unde'ned - A postition in memory where the value is not> known and may be either 0 or 1> That can be seen as a notion for an (in'nite) set of tape contents.> 4)Starting at a chosen position on the tape, mark the condition> required for halting. For example, a machine may need to read a 1 in> state k. Therefore, mark a 1 on the tape.> 5)Look for each instruction which calls state k. Take the next(or> 'rst) instruction that does so. If no more instructions call state k, fallback to the last value for k. If this is the 'rst value for k> and all paths have returned, the machine does not halt.> 6)Move on the tape the direction opposite the instruction.> 7)If the memory at that position is Unde'ned, mark the condition> required for that state to run. For example, state j may call state k> if it reads a 0. Go to step 5 using state k=j.> 8)If the memory at that position is 0 or 1, compare the value of the> memory with the data which the instruction marks to the tape. For> example, the instruction may write a 0 to the tape. If the two values> are not equal, go to step 6 and increment to the next instruction> which calls state k. If the values are equal, proceed as if memory> was unde'ned.> I hope that makes sense, it is my attempt at translating ym recursive> code. Basically, the function runs a Turing Machine backwards assuming that the machine halts. Id the machine halts in n steps> there is a path backwards n steps which has the tape in the original> con'guration. If none of the paths create that con'guration, yet> they all terminate because of a contradiction between the memory and> the instructions, the machine does not halt. If the machine runs for> an arbitrary number of steps backwards in a single path with no> result, the machine's halting status is considered unde'ned.> If my method is in fact valid,> It appears to be valid (to me).> Running a TM backwards can be thought of as equivalent to run a> nondeterministic TM (suitably de'ned). We build and extend a tree> of possible predecessor con'gurations (really: set of them),> and that tree may turn out to be of 'nite size due to contradictions.> it is actually a very powerful method for determining if a Turing Machine halts. In practice it eliminates> a large percentage of 4 state Turing machine using a step ceiling of> A large percentage? Really? How large?I didn't keep track of the number, by I am currently working with 5state machines, where I use the method of backwards reasoning onmachines that have not been resolved one way or another in 30 steps. At that point I call my backwards reasoning function for a max of 30steps. This conveniently makes it unnecesary to see if the begginingof the program has been reached, since we already know the machinedoes not halt in 30 swteps from running it forward.Of all machines I have run so far, 10% are caught so far. This isless than initially, because of running it 30 steps into the program. It is possible to run it earlier, but it is more useful to determinenon-halting through greater con'guration than reverse reasoning, soI wanted to give a chance for that to happen. The combination isworking so far, as I have scanned about 9.64% of 5 state machines withno holdouts.All 4 state machines produced 3 holdouts, which I am looking at. Ibelieve the three would be eliminated if I gave my reverse reasoningfunction the ability to determine when a pattern in reverse isactually capable of producing a halting machine or when the pattern isin'nite and therefore impossible when running forward. After 15hours of running the program on a standard computer it had proven BusyBeaver 4 to be 21 marks productive. I am currently working to improvethat time.> 30 steps backwards. On a theoretical note, I believe that the easiest> machines to use this method on are the machines that are dif'cult to> discern a pattern in when running normally. Basically every path that> terminates when run in reverse is a sequence which cannot occur on the> memory if the machine does not halt (otherwise the machine would in> fact halt).> The longer and more numerous that paths that run backwards, the more> sequences which cannot occur on memory. That means that a pettern is> just a bit more likely to occur because another sequence is> impossible. The fewer and smaller the patterns, the more options> available to the machine without halting. This class of programs is> where I think we 'nd some of the more dif'cult machines to work> with.> Advice, questions, has anyone else already done this?> Yes, others have done that kind of backward reasoning.> See e.g. http://www.drb.insel.de/~heiner/BB/mabu90.html#Nontermination> case (b).I can't believe I forgot that part of your paper, I have read itmultiple times but for some reason your backward reasoning method didnot come to mind when I thought of my method. They do differ, butthat is only because of our different methods of machine enumeration. You enumerate durring the process of running the machine, and I fullyenumerate each machine than isolate the elements that create earlyhalting and in'nite machines. I think that currently your method ofenumeration is superior, but I am hoping that a shortcut enumerationroutine I have developed will speed the process up when dealing withdisconnected machines.I am glad to see you are still around Usenet, Mr. Marxen. I havealready read your other posts on the Busy Beaver problem clear back tothe late 80's, when I assume you started? Material is scarce, andyours has been great help. I was just wondering, what do you think ismost preventing you from proving your 4098 champion is the mostproductive machine?-Chris Pitman =[...]> I am glad to see you are still around Usenet, Mr. Marxen. I have> already read your other posts on the Busy Beaver problem clear back to> the late 80's, when I assume you started? Material is scarce, and> yours has been great help. I was just wondering, what do you think is> most preventing you from proving your 4098 champion is the most> productive machine?The most critical resource is my time. I do BB reasearch in my spare time,and I'm not completely dedicated to it (e.g. I also work on a chessproblem solver).Also, there are lots of intersting sub-topics. E.g. I've spent considerabletime to write a program for the simulation of single machines (in awk).I've used it to produce my simulation results pages.Currently I'm busy developing a signi'cantly different tape representation,and how to ef'ciently run a machine on top of it.But I have also started a complete redesign of the old software,with the goal to settle sigma(5) in the end. Interestingly, serveralothers currently appear to aim at the same goal (including you :-).> -Chris PitmanGood luck!-- Heiner Marxen =I am in need of maths help!My problem involves 'nding the centre of the smallest circle which can bedrawn through one point and touching two other circles. We know thecoordinates of the point and the coordinates and radii of the two circles.It seems to me that there is only one solution for each set of data so Iassume it must be possible to 'nd it. The answer has to be expressed in assimple an equation as possible in order to translate the solution to a lowlevel programming language (assembly lanuage).I've drawn the problem and it is on line atwww.telco.demon.co.uk/targ/target.jpgAny help greatly appreciated,Paul I am in need of maths help!> My problem involves 'nding the centre of the smallest circle which can be> drawn through one point and touching two other circles. We know the> coordinates of the point and the coordinates and radii of the two circles.> It seems to me that there is only one solution for each set of data so I> assume it must be possible to 'nd it. The answer has to be expressed in as> simple an equation as possible in order to translate the solution to a low> level programming language (assembly lanuage).> I've drawn the problem and it is on line at> www.telco.demon.co.uk/targ/target.jpg> Any help greatly appreciated,> Paula general suggestion:let A, B be the two circles given. parametrize A in s, B in t. thenthe distance from the point in question, x, to A is given by somefunction d1(s). similarly, the distance from x to B is given by somed2(t). consider the function d(s,t) = d1(s) + d2(t), whose minimum canbe found via standard calculus techniques. i am sure other folks onhere will have more sophisticatedsolutions, but if you try this, please let me know if it works. M.T. =Well at least someone else thinks it's possible!Will I de'nately need to use calculus functions to solve this? I wouldrather avoid them because of the problems involved in programming themicroprocessor. I expect it would be possible but it's outside my ability todo at the moment.Paul > My problem involves 'nding the centre of the smallest circle which can be>> drawn through one point and touching two other circles. We know the>> coordinates of the point and the coordinates and radii of the two circles. >a general suggestion:>let A, B be the two circles given. parametrize A in s, B in t. then>the distance from the point in question, x, to A is given by some>function d1(s). similarly, the distance from x to B is given by some>d2(t). consider the function d(s,t) = d1(s) + d2(t), whose minimum can>be found via standard calculus techniques. i am sure other folks on>here will have more sophisticated>solutions, but if you try this, please let me know if it works.No, there's no minimization involved. Suppose the 'rst point is p_0, the centres of the given circles are p_1 and p_2, and the radiiof those circles are r_1 and r_2. Suppose the circle to be constructedwill not contain those given circles, and has radius r. Then the distances from the centre c of the new circle to p_0, p_1 and p_2 arer, r-r_1 and r-r_2 respectively (if any of the given circles is contained in the new circle, the - will become a +). In general the points whose distances from two given points differ by a constant form a hyperbola, and to get c you must 'nd the intersection of two hyperbolas.However, I think things will be simpli'ed considerably if you 'rst apply a fractional-linear transformation which will make the radii of the twogiven circles equal, as then the hyperbola becomes a straight line, the perpendicular bisector of the line joining the centres.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 > My problem involves 'nding the centre of the smallest circle which can be>> drawn through one point and touching two other circles. We know the>> coordinates of the point and the coordinates and radii of the two circles.>a general suggestion:>let A, B be the two circles given. parametrize A in s, B in t. then>the distance from the point in question, x, to A is given by some>function d1(s). similarly, the distance from x to B is given by some>d2(t). consider the function d(s,t) = d1(s) + d2(t), whose minimum can>be found via standard calculus techniques. i am sure other folks onhere will have more sophisticated>solutions, but if you try this, please let me know if it works.> No, there's no minimization involved. Suppose the 'rst point is > p_0, the centres of the given circles are p_1 and p_2, and the radii> of those circles are r_1 and r_2. Suppose the circle to be constructed> will not contain those given circles, and has radius r. Then the > distances from the centre c of the new circle to p_0, p_1 and p_2 are> r, r-r_1 and r-r_2 respectively (if any of the given circles is > contained in the new circle, the - will become a +). In general the > points whose distances from two given points differ by a constant form a > hyperbola, and to get c you must 'nd the intersection of two hyperbolas.> However, I think things will be simpli'ed considerably if you 'rst apply > a fractional-linear transformation which will make the radii of the two> given circles equal, as then the hyperbola becomes a straight line, the > perpendicular bisector of the line joining the centres.> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2the professor is right. i stand corrected. M.T. =Ok,This sounds like progress but the maths is far beyond me so please forgivemy questions for being simple.Can you suggest a source of further reading (or perhaps give examples) offractional-linear transformation? As I'm not familiar with www.telco.demon.co.uk/targ/target.jpg> No, there's no minimization involved. Suppose the 'rst point is> p_0, the centres of the given circles are p_1 and p_2, and the radii> of those circles are r_1 and r_2. Suppose the circle to be constructed> will not contain those given circles, and has radius r. Then the> distances from the centre c of the new circle to p_0, p_1 and p_2 are> r, r-r_1 and r-r_2 respectively (if any of the given circles is> contained in the new circle, the - will become a +). In general the> points whose distances from two given points differ by a constant form a> hyperbola, and to get c you must 'nd the intersection of two hyperbolas.> However, I think things will be simpli'ed considerably if you 'rst apply> a fractional-linear transformation which will make the radii of the two> given circles equal, as then the hyperbola becomes a straight line, the> perpendicular bisector of the line joining the centres.> Ok,> This sounds like progress but the maths is far beyond me so please forgive> my questions for being simple.> Can you suggest a source of further reading (or perhaps give examples) of> fractional-linear transformation? As I'm not familiar with the concept.For complex numbers z, a fractional linear transformation has the form:f(z) = (az+b)/(cz+d) for complex constants a,b,c,d. Discussion can befound in texts on complex analysis. The point here is that a LFT mapscircles to circles. > Ok,>> This sounds like progress but the maths is far beyond me so pleaseforgive> my questions for being simple.>> Can you suggest a source of further reading (or perhaps give examples)of> fractional-linear transformation? As I'm not familiar with the concept.>> For complex numbers z, a fractional linear transformation has the form:> f(z) = (az+b)/(cz+d) for complex constants a,b,c,d. Discussion can be> found in texts on complex analysis. The point here is that a LFT maps> circles to circles.I would strongly recommend a wonderful small inexpensive book,Elements of the Theory of Functionsby Konrad Knopp, Frederick Bagemihl (Translator) (Paperback - December 1952)Learn everything in this book and you will not regret it.At Amazon:List Price: $7.95Buy new: $7.95 Used & new from $3.50 =At www.telco.demon.co.uk/targ/target.jpg is a description of a problem Ihave for a personal project.Are there any kind mathematicians out there who would like to solve thisproblem for me?Several people have suggested methods but unfortunately it is beyond mymaths to use them!Ideally the solution will avoid calculus as I have to program amicrocontroller to do this.Am I asking the impossible or is it a reasonable challenge for the sci.mathgroup?Many great =Fiddling around on a computer appears to indicate the truth of thebelow, but I'd like to know how to prove it analytically. I wonder ifanyone can help?if a,b both are real with a > 1 and b > 0, then (b-1)*a^b - b*a^(b-1) + 1 < 0 < b < 1 if a,b both are real with a > 1 and b > 0, then > (b-1)*a^b - b*a^(b-1) + 1 < 0 < b < 1 Fix a > 1. For any x in R, consider (x-1)*a^x - x*a^(x-1) + 1. Divide by a^x to see this is < 0 iff f(x) = (x-1) - x/a + a^(-x) < 0. We have f(0) = 0 = f(1). Also f''(x) = (ln(a)^2)*a^(-x) > 0. So you have a strictly convex function f on R that is zero at 0 and 1. That implies f < 0 on (0,1) and f > 0 on (-oo,0) U (1,oo). > if a,b both are real with a > 1 and b > 0, then> (b-1)*a^b - b*a^(b-1) + 1 < 0 < b < 1>> Fix a > 1. For any x in R, consider (x-1)*a^x - x*a^(x-1) + 1. Divide by> a^x to see this is < 0 iff f(x) = (x-1) - x/a + a^(-x) < 0. We have f(0) 0 = f(1). Also f''(x) = (ln(a)^2)*a^(-x) > 0. So you have a strictly convex> function f on R that is zero at 0 and 1. That implies f < 0 on (0,1) and f> 0 on (-oo,0) U (1,oo).Very nice :-)Dirk Vdm > if a,b both are real with a > 1 and b > 0, then> (b-1)*a^b - b*a^(b-1) + 1 < 0 < b < 1>> Fix a > 1. For any x in R, consider (x-1)*a^x - x*a^(x-1) + 1. Divide by> a^x to see this is < 0 iff f(x) = (x-1) - x/a + a^(-x) < 0. We have f(0) 0 = f(1). Also f''(x) = (ln(a)^2)*a^(-x) > 0. So you have a strictly convex> function f on R that is zero at 0 and 1. That implies f < 0 on (0,1) and f> 0 on (-oo,0) U much! =I have been trying to 'nd a short proof for the two distributivity of GCDand LCM:[a,(b,c)]=([a,b],[a,c])(a,[b,c])=[(a,b),(a,c)]Anyone?p.s. I proved this in a rather troublesome way and was very long...I want tosee what other people come up with... I have been trying to 'nd a short proof for the two distributivity> of GCD and LCM:> [a,(b,c)]=([a,b],[a,c])> (a,[b,c])=[(a,b),(a,c)]>> Anyone?>> p.s. I proved this in a rather troublesome way and was very long...I> want to see what other people come up with...There is a nice isomorphism with union and intersection of (multi)sets : GCD(p_1^a_1p_2^a_2..., p_1^b_1p_2^b_2...)= p_1^min (a_1,b_1) p_2^min(a_2,b_2),so the distrbutivity is in fact Morgan's Laws =I am reading the book Element of Algebra which introduces that allintegers have unique prime factorization only later...I wonder if there is aprove without know this....suppose you only know the following:x|a < There is k <- N, a = x*kGCD(a,b)=max{x <- N | x|a, x|b}LCM(a,b)=min{x <- N | a|x, b|x}and one more:GCD(a,b)=u a + v b for some u, v <- ZI am trying to avoid writing a long prove with prime factorization (which isthe proof that I mention I came up with)...because like I said...it camelater...> I have been trying to 'nd a short proof for the two distributivity> of GCD and LCM:> [a,(b,c)]=([a,b],[a,c])> (a,[b,c])=[(a,b),(a,c)]>> Anyone?>> p.s. I proved this in a rather troublesome way and was very long...I> want to see what other people come up with...>> There is a nice isomorphism with union and intersection of (multi)sets :GCD> (p_1^a_1p_2^a_2..., p_1^b_1p_2^b_2...)= p_1^min (a_1,b_1)p_2^min(a_2,b_2),> so the distrbutivity is in fact Morgan's Laws> = >[a,(b,c)] = ([a,b],[a,c]) >(a,[b,c]) = [(a,b),(a,c)]Use ab = (a,b)[a,b][a,(b,c)] = a(b,c)/(a,(b,c)) = a(b,c)/(a,b,c)([a,b],[a,c]) = (ab/(a,b), ac/(a,c)) = a(b/(a,b), c/(a,c))Is it any easier to prove (b/(a,b), c/(a,c)) = (b/(a,b,c), c/(a,b,c)) ?Assume x | lhs x(a,b) | b; x(a,c) | c x(a,b,c) | b; x(a,b,c) | cx | rhsAssume x | rhs x(a,b,c) | b, cis it possible to prove x divides left hand side using: >suppose you only know the following: >x|a < There is k <- N, a = x*k >GCD(a,b)=max{x <- N | x|a, x|b} >LCM(a,b)=min{x <- N | a|x, b|x} >GCD(a,b)=u a + v b for some u, v <- Z---- =I need to write an algorithm in Java that to generate coef'cients ofexponential regression function f(x) = A*exp(B*x) +C.This is an interesting problem, but it must have been solved many timesbefore, and I don't want to reinvent the wheel, so if you can suggest anygood resources, I'll be very grateful... > I need to write an algorithm in Java that to generate coef'cients of> exponential regression function f(x) = A*exp(B*x) +C.>> This is an interesting problem, but it must have been solved many times> before, and I don't want to reinvent the wheel, so if you can suggest any> good resources, I'll be very grateful...sci.math.num-analysis would be a more receptive forum. What Iwould do:With a 'xed guess at B, the least squares problem leads to alinear 2-by-2 system with straightforward solution. That createsa single-variable (namely B) optimization problem which can besolved approximately by many methods, from Golden Section (slowbut easy to perform) or, with some more effort, a modi'cationof Newton's Method.Or: With a 'xed guess at C, take logarithms to linearize: log(f(x)-C) = log(A) + B*xso it becomes an optimization problem (variable C), but this isshakier - it re-scales the variables in a non-linear way. Ifnothing else, it can give a reasonable initial guess at B fromthe previous problem. > I need to write an algorithm in Java that to generate coef'cients of> exponential regression function f(x) = A*exp(B*x) +C.>> This is an interesting problem, but it must have been solved many times> before, and I don't want to reinvent the wheel, so if you can suggestany> good resources, I'll be very grateful...>> sci.math.num-analysis would be a more receptive forum. What I> would do:>> With a 'xed guess at B, the least squares problem leads to a> linear 2-by-2 system with straightforward solution. That creates> a single-variable (namely B) optimization problem which can be> solved approximately by many methods, from Golden Section (slow> but easy to perform) or, with some more effort, a modi'cation> of Newton's Method.>> Or: With a 'xed guess at C, take logarithms to linearize:> log(f(x)-C) = log(A) + B*x>> so it becomes an optimization problem (variable C), but this is> shakier - it re-scales the variables in a non-linear way. If> nothing else, it can give a reasonable initial guess at B from> the previous problem.>I'm cross-posting this reply to sci.math.num-analysis just in case...The second part is pretty clear. As a matter of fact, that was my quick anddirty implementation.Let me ask you some questions about the 'rst part, and forgive myignorance.Is it something like this?1. Let n = 02. Guess B by doing linear regression of ln(f(x))3. Find best A and C by using the guess value of B, let n = n + 1, let S[n]= {error of f(x) = A*exp(B*x) +C with respect to actual data points}4. If S diverges, abort5. If S neither diverges nor converges or is suf'ciently small, return A,B, C6. go to step 2Or:6. guess B again by some other method and go to step 3If this is generally the correct approach, how do we decide in steps 4-5whether the sequence S converges or diverges or neither? Is it somethingsimple like comparing |S[n] - S[n-1]| and |S[n-1] - S[n-2]| ?How should we guess B in step 6?Right now speed of computation doesn't matter, so let's consider the GoldenSection method. Is it a special way of doing successive approximations? I doknow what the Golden Section means, but not the method... =I have a polynomial that has an a radical in it and I was wondering if therewas a way to solve it without squaring out the radical...For example when trying to solve for x:a*x^2 - b*x + c - (x^2 - d^2)^(1/2) = 0What I've been doing is solving via:a*x^2 - b*x + c = (x^2 - d^2)^(1/2)Squaring both sides(a*x^2 - b*x + c)^2 = (x^2 - d^2)and then solving the equation... problem is that you can't solve foranything higher than a quadratic this way since you're doubling theexponents to remove the radical. Unfortunately I'm trying to solve anequation that's formatted like this but is 4th order and when I do thesquaring it's 8th order... which I don't know how to solve.Btw, all the coef'cients are rational and I'm only interested in therational solutions that can be expressed algebraically.-Tom =whoops.. I meant to say the coef'cients and the solutions are real numbersnot that they had to be rational :)> I have a polynomial that has an a radical in it and I was wondering ifthere> was a way to solve it without squaring out the radical...>> For example when trying to solve for x:>> a*x^2 - b*x + c - (x^2 - d^2)^(1/2) = 0>> What I've been doing is solving via:>> a*x^2 - b*x + c = (x^2 - d^2)^(1/2)>> Squaring both sides>> (a*x^2 - b*x + c)^2 = (x^2 - d^2)>> and then solving the equation... problem is that you can't solve for> anything higher than a quadratic this way since you're doubling the> exponents to remove the radical. Unfortunately I'm trying to solve an> equation that's formatted like this but is 4th order and when I do the> squaring it's 8th order... which I don't know how to solve.>> Btw, all the coef'cients are rational and I'm only interested in the> rational solutions that can be expressed algebraically.> -Tom> equation that's formatted like this but is 4th order and when I do the> squaring it's 8th order... which I don't know how to solve.>> Btw, all the coef'cients are rational and I'm only interested in the> rational solutions that can be expressed algebraically.In your case (all the coef'cients are rationals) you have to square etc..(then you'll obtain a 8th degree polynomial equation with integercoef'cients) then you can apply the Rational Root Test, which should giveyou all the rational solutions in no time (if the coef'cients are nottoo... composed...).> rational solutions that can be expressed algebraically.Do you know any rationals that can't be expressed algebraically ?-- Julien Santini,CMI Technop.99le de Ch.89teau-Gombert,France mean to say rational, meant to say real :)Since the solutions might not be rational is there another solution to solvehigher order polynomials as long as you know the solutions are notimaginary?> equation that's formatted like this but is 4th order and when I do the> squaring it's 8th order... which I don't know how to solve.>> Btw, all the coef'cients are rational and I'm only interested in the> rational solutions that can be expressed algebraically.>> In your case (all the coef'cients are rationals) you have to squareetc..> (then you'll obtain a 8th degree polynomial equation with integer> coef'cients) then you can apply the Rational Root Test, which should give> you all the rational solutions in no time (if the coef'cients are not> too... composed...).>> rational solutions that can be expressed algebraically.>> Do you know any rationals that can't be expressed algebraically ?>> -- > Julien Santini,> CMI Technop.99le de Ch.89teau-Gombert,> France>> Is it possible to strectch this logic to Spherical Trigonometry ?>> [ sides of the spherical triangle a,b,c now are a-> a/R, b->a/R, c->c/R>> [ where>> R is radius of sphere ]>> Is the area of a sphereical triangle with sides a, b and c equal to>> sqrt(p*(p-a)*(p-b)*(p-c)) where p = (a+b+c)/2 ?>>No, as any fule kno, the area is A + B + C - pi>(A, B and C angles).Ya, bud as sum doan, tan(E/4) = sqrt( tan(s/2) tan((s-a)/2) tan((s-b)/2) tan((s-b)/2) )where E = A + B + C - pi, an dat look a lot like Heron.Rob Johnsontake out the trash before replying i'd like to know something about homogeneous coordinates and parametric > representation of a line.> if i have two points p1(x, y, z, w) and p2(x, y, z, w) with w != 1; i want to avoid dividing coordinates of this two points by w before > calculating the parametric equation of the line (p1p2), and i'd like to > know if this is a possible result : x = p1x + k * (p2x - p1x)> y = .....> ......> w = p1w + k * (p2w - p1w) ?There is no linear representation of a line with one parameter in the3-dimensional projective space. Every linear representation of theform P1 + k * (P2-P1) will miss exactly one point.i.e the point with coordinates P2-P1For instance P2 + k*P1would be a representation of all points of the line P1P2 but P1.A full linear parametric representation of the line P1P2 would needtwo linear (homogeneous) parameters k1*P1 + k2*P2 ( k1 and k2 not both 0 )where (k1,k2) produce the same point as (c*k1,c*k2) for any nonzerofactor c.Or you may split the representation P1P2 : {P2+kP1} u {P1}You may visualize P1 as lim (P2 + k*P1) ~ lim (1/k*P2 + P1) k->oo k->oo -- Horst =Rate my picture at www.hotornot.comI'm in the big oval mirror...One of the alexa movers this week, you just click 1 to 10 then itgives you the average rating and you see the next guy or gal, andthey can see your pic if you both click yes or something then somechemical reaction occurs and 1000 people getting introduced every hourthere at the moment.... whooaHerc-- __ / / / / / / / / / / / /__/__ /________ ___________/ www.A1Sites.com =In cyrilic, HOT reads NOT.> Rate my picture at www.hotornot.com> I'm in the big oval mirror...>> One of the alexa movers this week, you just click 1 to 10 then it> gives you the average rating and you see the next guy or gal, and> they can see your pic if you both click yes or something then some> chemical reaction occurs and 1000 people getting introduced every hour> there at the moment.... whooa>> Herc>> --> __> / /> / / > / / / > / / / > / /__/__ > /________ > ___________/> www.A1Sites.com>> Rate my picture at www.hotornot.com> I'm in the big oval mirror...>Unless you are a very attractive black teenage girl, I doubt that simplyclicking on www.hotornot.com shows your picture ... Please report problems or inappropriate use to the =Jenn from alt.2600 is a pedophile and a troll !jenn =So I am getting my crossposts to alt.2600, my newslist is corrupted somedon't show up, hardly worth downloading again !!Why don't you add your URLS to www.a1sites.com ? Takes a minute andits just free traf'c, nothing at this stage but I need some sites listed to enticepeople to link back so it takes off. I'll be putting it 1st in every free for all linkpage there is next month A1! so even without people linking there to start it offyou'd get a tonne of free visitors. And if you send traf'c back it raises yourposition in the list, most topsites send back more than they get.Herc-- __ / / / / / / / / / / / /__/__ /________ ___________/ www.A1Sites.com Rate my picture at www.hotornot.com> I'm in the big oval mirror...> Unless you are a very attractive black teenage girl, I doubt that simply> clicking on www.hotornot.com shows your picture ...>You were 90% of the way there then missed out, try again and rate the'rst person that comes up!Herc Rate my picture at www.hotornot.com> I'm in the big oval mirror...> One of the alexa movers this week, you just click 1 to 10 then it> gives you the average rating and you see the next guy or gal, and> they can see your pic if you both click yes or something then some> chemical reaction occurs and 1000 people getting introduced every hour> there at the moment.... whooa> Herc> --> __> / /> / / > / / / > / / / > / /__/__ > /________ > ___________/> www.A1Sites.comI'm not seeing you... male or female? =If we have a regular m-gon, how many ways can we pick n vertexes sothat n straight lines from these vertexes, each drawn to a singleinterior point P, divides the m-gon into n equal-area pieces?For example, for m = 6 and n = 3:(This enumeration might be a little controversial.)Let the hexagon's vertexes be labeled, in order, from 1 to 6.3 cases:1) If we let the vertexes be 1, 3, 5, and P be the hexagon's center,this works.Rotating by one vertex gets the same solution rotated.(2 sets of n vertexes)2) If we let vertexes be 1, 2, 4, and P be the midpoint of theedge(4,5), we get a division which corresponds to my puzzle I posted alittle while ago:http://mathforum.org/discuss/sci.math/a/t/501734With rotations and re§ections we get 12 vertex-sets.This depends, of course, on exactly what we refer to as an interiorpoint.If we literally mean a non-exterior point, then this should be anacceptable set of vertexes to include in our count.3) The only other (after rotations and re§ections) set of vertexes isif we have vertexes 1, 2, 3.Now, here we have an interesting situation. The only place to put Pthat I can think of is JUST INSIDE the vertex directly opposite thecenter vertex of this set. Placing point P ON the opposing vertex getsthe hexagon divided into 3 equal pieces, BUT one piece is notsimply-connected. And placing P anywhere just inside the opposingvertex, as to connect the 3rd piece with itself, gets 3 pieces whichare not all equal to 1/3 the hexagon in area.So, I would count this as not adding anything to the count of totalvertex-sets which divide the hexagon. But some might still. If theydo, 6 vertex-sets are to be added to our total.So, we have at least 2 vertex-sets, more likely we have 14vertex-sets, and perhaps 20 vertex-sets, all depending on how wede'ne the problemLeroy Quet =What is a direct proof or a counterexample to say that l^1 ( space ofabsolutely converging successions ) is not re§exive? And about thenot-re§exivity of c_0 ( space of succession converging to zero ) ? What is a direct proof or a counterexample to say that l^1 ( space of> absolutely converging successions ) is not re§exive? And about the> not-re§exivity of c_0 ( space of succession converging to zero ) ?Just compute the second dual. For example, the second dual of c_0 is l_in'nity, which is not c_0.-- Stephen Montgomery-Smithstephen@math.missouri.eduhttp:// www.math.missouri.edu/~stephen > What is a direct proof or a counterexample to say that l^1 ( space of>> absolutely converging successions ) is not re§exive? And about the>> not-re§exivity of c_0 ( space of succession converging to zero ) ?>>Just compute the second dual. For example, the second dual of c_0 is >l_in'nity, which is not c_0.Hmm, can one really compute the second dual of l^1? (Of courseone can show that it's not l^1 fairly easily, say by applying theHahn-Banach theorem to a certain functional on the subspaceof l^in'nity consisting of sequences with limits...)************************David C. Ullrich >What is a direct proof or a counterexample to say that l^1 ( space of>absolutely converging successions ) is not re§exive? And about the>not-re§exivity of c_0 ( space of succession converging to zero ) ?>>Just compute the second dual. For example, the second dual of c_0 is >>l_in'nity, which is not c_0.> Hmm, can one really compute the second dual of l^1? (Of course> one can show that it's not l^1 fairly easily, say by applying the> Hahn-Banach theorem to a certain functional on the subspace> of l^in'nity consisting of sequences with limits...)> ************************> I do recall seeing somewhere a theorem that if a Banach space was not re§exive than neither is its dual. I think it was the book by Koether - Topological Vector Spaces. There was some theorem to the effect that if the 2nth and 2mth duals were the same, or the (2m+1)th and (2n+1)th duals were the same, for any distinct non-negative integers m and n, then the space was re§exive. (Here same means that the natural embedding is an isomorphism.) What is a direct proof or a counterexample to say that l^1 ( space of>absolutely converging successions ) is not re§exive? And about the>not-re§exivity of c_0 ( space of succession converging to zero ) ?If X is re§exive, its unit ball is weakly compact.You can show the unit ball of c_0 is not weakly compact, as follows:Consider the weakly open sets U_n = {x in c_0: |x_n| < 1/2}. Thesecover the unit ball of c_0 since for any x in c_0, x_n -> 0. Butthere is no 'nite subcover: given any 'nite family F ={U_{n_1}, U_{n_2}, ..., U_{n_k}}, there is x in the unit ball ofc_0 which is not in any member of F. For example, take x_{n_i} = 1 fori=1...k and x_n = 0 otherwise. Similarly, the unit ball of l_1 is not weakly compact. Considerthe weakly open sets U_n = {x in l_1: sum_{m >= n} x_m < 1/2}. These cover the unit ball of l_1. But there is no 'nite subcover:given any 'nite family {U_{n_1}, ..., U_{n_k}}, take N > max{n_1,...,n_k}, and x such that x_N = 1 and x_i = 0 otherwise.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 >What is a direct proof or a counterexample to say that l^1 ( space of>>absolutely converging successions ) is not re§exive? And about the>>not-re§exivity of c_0 ( space of succession converging to zero ) ?>>Just compute the second dual. For example, the second dual of c_0 is >l_in'nity, which is not c_0.>> Hmm, can one really compute the second dual of l^1? (Of course>> one can show that it's not l^1 fairly easily, say by applying the>> Hahn-Banach theorem to a certain functional on the subspace>> of l^in'nity consisting of sequences with limits...)> ************************>I do recall seeing somewhere a theorem that if a Banach space was not re§exive >than neither is its dual. Sure. (But I don't think that _that_ is what we had in mind as adirect proof...)>I think it was the book by Koether - Topological >Vector Spaces. There was some theorem to the effect that if the 2nth and 2mth >duals were the same, or the (2m+1)th and (2n+1)th duals were the same, for any >distinct non-negative integers m and n, then the space was re§exive. (Here >same means that the natural embedding is an isomorphism.)>************************David C. Ullrich >What is a direct proof or a counterexample to say that l^1 ( space of>>absolutely converging successions ) is not re§exive? And about the>>not-re§exivity of c_0 ( space of succession converging to zero ) ?>>If X is re§exive, its unit ball is weakly compact.I don't know if I'd call this a direct proof either. (Doing nothingbut complain today, sorry...)>You can show the unit ball of c_0 is not weakly compact, as follows:>>Consider the weakly open sets U_n = {x in c_0: |x_n| < 1/2}. These>cover the unit ball of c_0 since for any x in c_0, x_n -> 0. But>there is no 'nite subcover: given any 'nite family F {U_{n_1}, U_{n_2}, ..., U_{n_k}}, there is x in the unit ball of>c_0 which is not in any member of F. For example, take x_{n_i} = 1 for>i=1...k and x_n = 0 otherwise. >>Similarly, the unit ball of l_1 is not weakly compact. Consider>the weakly open sets U_n = {x in l_1: sum_{m >= n} x_m < 1/2}. >These cover the unit ball of l_1. But there is no 'nite subcover:>given any 'nite family {U_{n_1}, ..., U_{n_k}}, take N >max{n_1,...,n_k}, and x such that x_N = 1 and x_i = 0 otherwise.>>Robert Israel israel@math.ubc.ca>Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia >Vancouver, BC, Canada V6T 1Z2> ************************David C. Ullrich =I found this hint to proof l^1 is not re§exive:Consider the linear form de'ned from h : (l^1)' - > R, where h is inthe bi-dual of l^1 and is so de'ned: h(x) := lim_( i -> in'nity )x(i) , where x is an element of l^(in'nity). The fact that l^1 is notre§exive is that there's not any element x in (l^in'nity) so thatf(x) = h(f), for every f in (l^1)'. So the problem is how to proofthat there isn't any element in l^in'nity that satisfy the condition.I thought to use the isomorphism: l^in'nity -> (l^1)', so de'ned: x |-> Sum_(1 <= i <= in'nity) x(i)*y(i), where x is in l^in'nity and yis in l^1. I found this hint to proof l^1 is not re§exive:>Consider the linear form de'ned from h : (l^1)' - > R, where h is in>the bi-dual of l^1 and is so de'ned: h(x) := lim_( i -> in'nity )>x(i) , where x is an element of l^(in'nity).This is not well-de'ned: an element of l^in'nity need not have a limit.But there are elements h of (l^in'nity)' such that lim_i x(i) = h(x) for every x such that the limit exists (use Hahn-Banach).> The fact that l^1 is not>re§exive is that there's not any element x in (l^in'nity) so that>f(x) = h(f), for every f in (l^1)'.No, the fact that l^1 is not re§exive is implied by the fact thatthere's no v in l^1 such that h(x) = sum_i x_i v_i for all x inl^in'nity. And that's easy.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =I was wondering how I wouldprove this : if there is a 'xed number ofpoints in 2D, then the numberof triangles will be the sameno matter how you triangulatethe points. So for anytriangulation, the number oftriangles will be the same.Is induction the right way toprove this? Any help isJohn I was wondering how I would> prove this : > if there is a 'xed number of> points in 2D, then the number> of triangles will be the same> no matter how you triangulate> the points. So for any> triangulation, the number of> triangles will be the same.> Is induction the right way to> prove this? Any help is> JohnIt's easy.Think with combinatorial numbers ;-)-- Saludos,Alejandro Torras. =As long as no 3 points lie on a straight line ...Yes, induction will easily solve this.>> I was wondering how I would> prove this :>> if there is a 'xed number of> points in 2D, then the number> of triangles will be the same> no matter how you triangulate> the points. So for any> triangulation, the number of> triangles will be the same.>> Is induction the right way to> prove this? Any help is>> John I was wondering how I would> prove this : > if there is a 'xed number of> points in 2D, then the number> of triangles will be the same> no matter how you triangulate> the points.Depends a bit on what exactly you mean by triangulate. Suppose you have 4 points forming the vertices of a square. To triangulate the points, do you consider it suf'cient to draw in one diagonal? If so, then you get 2 triangles that way, 4 if one of the four points is inside the triangle formed by the other three, so what you're trying to prove is false. But if your understanding of triangulate makes it insuf'cient to just draw the one diagonal, then, yes, induction should do it.-- =In a particular retail store the average price of a typical two-seater sofahas been 200 and average sales were 1000 per year. This year the store hasrun a half price sale on such sofas and found that sales increased by 50%.Question: Find a hyperbolic model for demand in terms of price. In a particular retail store the average price of a typical two-seater sofa>has been 200 and average sales were 1000 per year. This year the store has>run a half price sale on such sofas and found that sales increased by 50%.>>Question: Find a hyperbolic model for demand in terms of price.What have you done to try to solve this problem on your own, aside from posting it in duplicate to a whole bunch of newsgroups? It sure looks like homework. Where precisely do you get stuck? textbook?)-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/You 'nd yourself amusing, Blackadder.I try not to §y in the face of public opinion. =A question---it is my layman's guess---for an algebraic geometrician (orjust a geometrician with basic knowledge of graph theory, which most arelikely to have):When describing reasons for replacing a particular solid-geometry metaphorwith a network metaphor (that is, a labeled-graph metaphor) recently, Ichose as an example for my argument the graphical representation of what isknown as a four-dimensional hypercube. I found myself suggesting that:[S->]Whereas it is still possible to draw a (single) two-dimensional projectionof a three-dimensional cube that shows all its vertices and edges (be itdashed when hidden) determining its faces in a consistent way from which,given that it is a cube, all its geometrical properties can bereconstructed, the same is no longer possible for cubes of dimensionalityhigher than three.[<-S]This, I went on, is what forces one to resort to representing a hypercube aswhat amounts to a labeled graph that is the diagram of a relation betweenthe members of a set of 2^n vertices, in which, having labeled the verticeswith bitstrings of length n, no two vertices are connected that differ inmore than one position, but all that differ in precisely one are.My point is that, to be honest, I don't really know whether my suggestion[S] is correct, because I would not know how to go about proving it. Emburg,Leiden, The Netherlands. Whereas it is still possible to draw a (single) two-dimensional projection> of a three-dimensional cube that shows all its vertices and edges (be it> dashed when hidden) determining its faces in a consistent way from which,> given that it is a cube, all its geometrical properties can be> reconstructed, the same is no longer possible for cubes of dimensionality> higher than three.>Take a polar projection of a cube from a point above the center of ahorizontal face of a cube onto a horizontal plane below the. You get asquare within a square with corner points respectively connected.Do the same with a hypercube and you get a cube within a cube which, withcare can be squished onto a plane so that each line distictly showswithout overlapping anyother except possibly at a point while preservingconnectivity. So depending on what Oreconstruction' means, I think I'veshow some pause for thought about your claim about hypercubes.> This, I went on, is what forces one to resort to representing a hypercube as> what amounts to a labeled graph that is the diagram of a relation between> the members of a set of 2^n vertices, in which, having labeled the vertices> with bitstrings of length n, no two vertices are connected that differ in> more than one position, but all that differ in precisely one are.>Looks alright as a de'nition of a n-cube as viewed as only consisting ofedges and vertices and likely can be extended to in'nite dimensions, bothcountable and uncountable. > Whereas it is still possible to draw a (single) two-dimensionalprojection> of a three-dimensional cube that shows all its vertices and edges (be it> dashed when hidden) determining its faces in a consistent way fromwhich,> given that it is a cube, all its geometrical properties can be> reconstructed, the same is no longer possible for cubes ofdimensionality> higher than three.>> Take a polar projection of a cube from a point above the center of a> horizontal face of a cube onto a horizontal plane below the. You get a> square within a square with corner points respectively connected.>> Do the same with a hypercube and you get a cube within a cube which, with> care can be squished onto a plane so that each line distictly shows> without overlapping anyother except possibly at a point while preserving> connectivity. So depending on what Oreconstruction' means, I think I've> show some pause for thought about your claim about hypercubes.>> This, I went on, is what forces one to resort to representing ahypercube as> what amounts to a labeled graph that is the diagram of a relationbetween> the members of a set of 2^n vertices, in which, having labeled thevertices> with bitstrings of length n, no two vertices are connected that differin> more than one position, but all that differ in precisely one are.>> Looks alright as a de'nition of a n-cube as viewed as only consisting of> edges and vertices and likely can be extended to in'nite dimensions, both> countable and uncountable.Nothing like a swift reply to make one reconsider one's question, I suppose.William Elliot puts his 'nger on the sore spot: what do I mean by'reconstruction'. Indeed, particularly in combination with my phrase Ogiventhat it is a cube'. Because if a cube is given so are its geometricalproperties, implying that they do not need to be reconstructed. Elliot'sremark about the polar projection makes me suspect that all 'nite sets ofn-dimensional points can be projected in a k