Title: Red-Dog's Last Theorem Post by: kinboshi on May 02, 2009, 12:09:00 PM We all know Tom is a deep thinker, and recently a mathematical problem has been his major concern. It's caused him sleepless nights, and prompted hours of debate over the felt at the poker tables at DTD.
Basically what Tom For example, if you have 3 chips in each stack, you need to do three riffles to achieve this. The same if you have 4 in each stack. But if you have 5 in each stack, you need to do 5 riffles. 6 -takes 6 riffles. We went down the empirical evidence route, riffling larger and larger stacks:
Our problem was that we couldn't riffle beyond stacks of 10. Also Tom (being a mathematician), didn't want it based on experimental data but on a mathematical proof. So can any maths geniuses work out a formula so we can predict the number of riffles when the stacks are n chips high? ;carlocitrone; Title: Re: Red-Dog's Last Theorem Post by: tikay on May 02, 2009, 12:13:25 PM Fermat will know. I'm not entirely sure, but he may have passed away by now. Title: Re: Red-Dog's Last Theorem Post by: EvilPie on May 02, 2009, 12:14:40 PM Lol
I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Title: Re: Red-Dog's Last Theorem Post by: kinboshi on May 02, 2009, 12:16:58 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL As a scientist, I'm appalled by your lack of trust in the accuracy of my scientific method. I find it an insult that you question my findings. (I just checked and they are right). Title: Re: Red-Dog's Last Theorem Post by: tikay on May 02, 2009, 12:18:34 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Matt, don't be silly! Of COURSE there is a formula. Title: Re: Red-Dog's Last Theorem Post by: kinboshi on May 02, 2009, 12:21:53 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Matt, don't be silly! Of COURSE there is a formula. Agreed. There HAS to be a formula. I'm waiting for Jon MW to see this...he's the man for this one. He's probably out on his horse at the moment though. Title: Re: Red-Dog's Last Theorem Post by: Jon MW on May 02, 2009, 01:05:38 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Matt, don't be silly! Of COURSE there is a formula. Agreed. There HAS to be a formula. I'm waiting for Jon MW to see this...he's the man for this one. He's probably out on his horse at the moment though. tl;ds Title: Re: Red-Dog's Last Theorem Post by: Pawprint on May 02, 2009, 01:16:42 PM Looking at this sequence there will be no mathematical formulae available. UL +1 Title: Re: Red-Dog's Last Theorem Post by: EvilPie on May 02, 2009, 01:29:11 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Matt, don't be silly! Of COURSE there is a formula. Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here). I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward. What formula can possibly make both 7 and 8 = 4? I'll be thinking on this for a while. It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks. Title: Re: Red-Dog's Last Theorem Post by: tikay on May 02, 2009, 01:59:02 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Matt, don't be silly! Of COURSE there is a formula. Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here). I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward. What formula can possibly make both 7 and 8 = 4? I'll be thinking on this for a while. It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks. There will be a formula for it, 100% guaranteed. If it can happen, there's a formula. If chance permits, Matt, try & get a copy of, & read, "Fermat's Lost Theorem", upon which this Thread title is based. You'd absolutely love it. And you'd learn the mathematical formula for working out the true (actual) length of the bendiest of rivers on earth, simply by using pye. ;) Title: Re: Red-Dog's Last Theorem Post by: AndrewT on May 02, 2009, 02:04:10 PM This is more complex than you could ever possibly imagine and, to my knowledge, is unsolved.
Title: Re: Red-Dog's Last Theorem Post by: Cf on May 02, 2009, 02:05:17 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Matt, don't be silly! Of COURSE there is a formula. Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here). I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward. What formula can possibly make both 7 and 8 = 4? I'll be thinking on this for a while. It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks. Really trivial example, but: 4 + (floor (n/9)) Title: Re: Red-Dog's Last Theorem Post by: marcro on May 02, 2009, 02:13:06 PM Hmmm, does the title of this thread mean that he will never have another theorem?
Title: Re: Red-Dog's Last Theorem Post by: Grier78 on May 02, 2009, 02:20:15 PM Its not simple problem, although some of the results are easy.
2 stacks of 2 chips = 2 Riffles 2 stacks of 4 chips = 3 Riffles 2 stacks of 8 chips = 4 Riffles 2 stacks of 16 chips = 5 Riffles 2 stacks of 32 chips = 6 Riffles 2 stacks of 64 chips = 7 Riffles etc... You will also get different results if you don't always riffle in exactly the same way, i.e. when the stack cut off the top has its lowest chip at the bottom of the riffle or if the bottom stack has its lowest chip at the bottom. Title: Re: Red-Dog's Last Theorem Post by: Robert HM on May 02, 2009, 02:20:56 PM Who cares, I don't, I have a life Title: Re: Red-Dog's Last Theorem Post by: EvilPie on May 02, 2009, 02:32:08 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Matt, don't be silly! Of COURSE there is a formula. Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here). I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward. What formula can possibly make both 7 and 8 = 4? I'll be thinking on this for a while. It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks. Really trivial example, but: 4 + (floor (n/9)) (n x 0) + 4 Even trivialler. Title: Re: Red-Dog's Last Theorem Post by: tikay on May 02, 2009, 02:46:51 PM This is more complex than you could ever possibly imagine and, to my knowledge, is unsolved. Quite probably. In other words, the Formula for it has yet to be discovered. But it must exist. Title: Re: Red-Dog's Last Theorem Post by: kinboshi on May 02, 2009, 02:53:49 PM We promised Mark @ DTD that the clever bods on here would have an answer for him tonight.
[ ] He'll be devastated. Title: Re: Red-Dog's Last Theorem Post by: tikay on May 02, 2009, 02:56:16 PM We promised Mark @ DT (http://www.dusktilldawnpoker.com/)D (http://www.dusktilldawnpoker.com/) that the clever bods on here would have an answer for him tonight. [ ] He'll be devastated. We await Tighty. Andrew & JonMW are let downs. Title: Re: Red-Dog's Last Theorem Post by: EvilPie on May 02, 2009, 02:57:39 PM This is more complex than you could ever possibly imagine and, to my knowledge, is unsolved. Quite probably. In other words, the Formula for it has yet to be discovered. But it must exist. You would think so wouldn't you. It seems fairly straight forward. You have 2 equal stacks, you intersperse them equal, half them again....... Seems like quite a simple sequence. With even numbered stacks I'm sure the formula would be simple but when you throw in uneven numbered stacks it gets very complex. Add in a few prime numbered stacks and it becomes imo almost impossible. Title: Re: Red-Dog's Last Theorem Post by: 12barblues on May 02, 2009, 02:59:37 PM I have a few minutes free this afternoon, so I'll tackle Red Dog's Last Theorem just as soon as I've finished my proof of the Reimann Hypothesis.....
Title: Re: Red-Dog's Last Theorem Post by: tikay on May 02, 2009, 03:04:14 PM I have a few minutes free this afternoon, so I'll tackle Red Dog's Last Theorem just as soon as I've finished my proof of the Reimann Hypothesis..... http://en.wikipedia.org/wiki/Riemann_hypothesis There you go, sorted. Now sort that piffling chip-riffliing thing please. Title: Re: Red-Dog's Last Theorem Post by: EvilPie on May 02, 2009, 03:06:17 PM I have a few minutes free this afternoon, so I'll tackle Red Dog's Last Theorem just as soon as I've finished my proof of the Reimann Hypothesis..... http://en.wikipedia.org/wiki/Riemann_hypothesis There you go, sorted. Now sort that piffling chip-riffliing thing please. Not until he's sorted out the second 10 trillion zeros. You can never be too sure. Title: Re: Red-Dog's Last Theorem Post by: tikay on May 02, 2009, 03:08:59 PM I have a few minutes free this afternoon, so I'll tackle Red Dog's Last Theorem just as soon as I've finished my proof of the Reimann Hypothesis..... http://en.wikipedia.org/wiki/Riemann_hypothesis There you go, sorted. Now sort that piffling chip-riffliing thing please. Not until he's sorted out the second 10 trillion zeros. You can never be too sure. Made my eyes glaze over, that! Next, the Yang-Mills existence & mass gap. Title: Re: Red-Dog's Last Theorem Post by: EvilPie on May 02, 2009, 03:12:04 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Matt, don't be silly! Of COURSE there is a formula. Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here). I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward. What formula can possibly make both 7 and 8 = 4? I'll be thinking on this for a while. It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks. There will be a formula for it, 100% guaranteed. If it can happen, there's a formula. If chance permits, Matt, try & get a copy of, & read, "Fermat's Lost Theorem", upon which this Thread title is based. You'd absolutely love it. And you'd learn the mathematical formula for working out the true (actual) length of the bendiest of rivers on earth, simply by using pye. ;) I can't find it. Title: Re: Red-Dog's Last Theorem Post by: tikay on May 02, 2009, 03:19:11 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Matt, don't be silly! Of COURSE there is a formula. Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here). I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward. What formula can possibly make both 7 and 8 = 4? I'll be thinking on this for a while. It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks. There will be a formula for it, 100% guaranteed. If it can happen, there's a formula. If chance permits, Matt, try & get a copy of, & read, "Fermat's Lost Theorem", upon which this Thread title is based. You'd absolutely love it. And you'd learn the mathematical formula for working out the true (actual) length of the bendiest of rivers on earth, simply by using pye. ;) I can't find it. I SO nearly bit.......even got as far as a Link to Amazon. I'm so easy to whoosh! Title: Re: Red-Dog's Last Theorem Post by: david3103 on May 02, 2009, 03:22:43 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Matt, don't be silly! Of COURSE there is a formula. Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here). I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward. What formula can possibly make both 7 and 8 = 4? I'll be thinking on this for a while. It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks. There will be a formula for it, 100% guaranteed. If it can happen, there's a formula. If chance permits, Matt, try & get a copy of, & read, "Fermat's Lost Theorem", upon which this Thread title is based. You'd absolutely love it. And you'd learn the mathematical formula for working out the true (actual) length of the bendiest of rivers on earth, simply by using pye. ;) I can't find it. Fermat's Last Theorem and we know where it is, it's the proof that was lost. As to Red-Dog's problem - the issue of which way you shift the stacks for the second and subsequent riffles will be crucial for odd starting stacks, (which colour chip is on the base of the left-hand stack?) you need consistency in both this, and in the form of the riffle. Did this impact on the observed results given thus far? Title: Re: Red-Dog's Last Theorem Post by: 12barblues on May 02, 2009, 03:24:57 PM I have a few minutes free this afternoon, so I'll tackle Red Dog's Last Theorem just as soon as I've finished my proof of the Reimann Hypothesis..... http://en.wikipedia.org/wiki/Riemann_hypothesis There you go, sorted. Now sort that piffling chip-riffliing thing please. Not until he's sorted out the second 10 trillion zeros. You can never be too sure. Made my eyes glaze over, that! Next, the Yang-Mills existence & mass gap. Blimey! Tough audience. Could be a busy afternoon as I have to cut the grass as well. I suppose I could fit it in after I finish my thesis entitled 'Einstein 1 Bohr 0 : Why The Copenhagen Interpretation Of Quantum Mechanics Is Ridic Com'. Title: Re: Red-Dog's Last Theorem Post by: cod meharly on May 02, 2009, 03:28:46 PM This is gna drive me crazy...im quite a nerd with numbers and to me this looks way more complex than i could possibly imagine!
Title: Re: Red-Dog's Last Theorem Post by: TightEnd on May 02, 2009, 03:36:39 PM Not too difficult really
Just input the starting numbers into the following equations α0 = wz + h + j − q = 0 α1 = (gk + 2g + k + 1)(h + j) + h − z = 0 α2 = 16(k + 1)3(k + 2)(n + 1)2 + 1 − f2 = 0 α3 = 2n + p + q + z − e = 0 α4 = e3(e + 2)(a + 1)2 + 1 − o2 = 0 α5 = (a2 − 1)y2 + 1 − x2 = 0 α6 = 16r2y4(a2 − 1) + 1 − u2 = 0 α7 = n + l + v − y = 0 α8 = (a2 − 1)l2 + 1 − m2 = 0 α9 = ai + k + 1 − l − i = 0 α10 = ((a + u2(u2 − a))2 − 1)(n + 4dy)2 + 1 − (x + cu)2 = 0 α11 = p + l(a − n − 1) + b(2an + 2a − n2 − 2n − 2) − m = 0 α12 = q + y(a − p − 1) + s(2ap + 2a − p2 − 2p − 2) − x = 0 α13 = z + pl(a − p) + t(2ap − p2 − 1) − pm = 0 The 14 equations α0, …, α13 can be used to produce a prime-generating polynomial inequality in 26 variables: (k+2)(1-\alpha_0^2-\alpha_1^2-\cdots-\alpha_{13}^2) > 0 ie: (k + 2)(1 − [wz + h + j − q]2 − [(gk + 2g + k + 1)(h + j) + h − z]2 − [16(k + 1)3(k + 2)(n + 1)2 + 1 − f2]2 − [2n + p + q + z − e]2 − [e3(e + 2)(a + 1)2 + 1 − o2]2 − [(a2 − 1)y2 + 1 − x2]2 − [16r2y4(a2 − 1) + 1 − u2]2 − [n + l + v − y]2 − [(a2 − 1)l2 + 1 − m2]2 − [ai + k + 1 − l − i]2 − [((a + u2(u2 − a))2 − 1)(n + 4dy)2 + 1 − (x + cu)2]2 − [p + l(a − n − 1) + b(2an + 2a − n2 − 2n − 2) − m]2 − [q + y(a − p − 1) + s(2ap + 2a − p2 − 2p − 2) − x]2 − [z + pl(a − p) + t(2ap − p2 − 1) − pm]2) > 0 Title: Re: Red-Dog's Last Theorem Post by: TightEnd on May 02, 2009, 03:37:01 PM The answer, I think you will find, is 8.
Title: Re: Red-Dog's Last Theorem Post by: kinboshi on May 02, 2009, 04:04:47 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Matt, don't be silly! Of COURSE there is a formula. Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here). I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward. What formula can possibly make both 7 and 8 = 4? I'll be thinking on this for a while. It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks. There will be a formula for it, 100% guaranteed. If it can happen, there's a formula. If chance permits, Matt, try & get a copy of, & read, "Fermat's Lost Theorem", upon which this Thread title is based. You'd absolutely love it. And you'd learn the mathematical formula for working out the true (actual) length of the bendiest of rivers on earth, simply by using pye. ;) I can't find it. Fermat's Last Theorem and we know where it is, it's the proof that was lost. As to Red-Dog's problem - the issue of which way you shift the stacks for the second and subsequent riffles will be crucial for odd starting stacks, (which colour chip is on the base of the left-hand stack?) you need consistency in both this, and in the form of the riffle. Did this impact on the observed results given thus far? Consistency has been maintained in the empirical studies. The chip that moves to the base of the riffled stack is always from the same stack. This should be assumed for any trials or repeats of the experiment. In order to find a useful pattern, it might be necessary to carry out experiments on increasingly larger stacks. Discovering this formula is going to be difficult, but as discussed last night, it will be easier than trying to understand the maths behind the size of a raise at Walsall's turbo deepstack during the opening level if you want less than 3 callers. Title: Re: Red-Dog's Last Theorem Post by: kinboshi on May 02, 2009, 04:08:59 PM I found this equation for the 'stack effect'. I'm not sure it'll help too much here unfortunately.
(http://upload.wikimedia.org/math/f/9/0/f907f6ddef9ae9b5e9e4e882953c5c93.png) Title: Re: Red-Dog's Last Theorem Post by: tikay on May 02, 2009, 04:33:12 PM Not too difficult really Just input the starting numbers into the following equations α0 = wz + h + j − q = 0 α1 = (gk + 2g + k + 1)(h + j) + h − z = 0 α2 = 16(k + 1)3(k + 2)(n + 1)2 + 1 − f2 = 0 α3 = 2n + p + q + z − e = 0 α4 = e3(e + 2)(a + 1)2 + 1 − o2 = 0 α5 = (a2 − 1)y2 + 1 − x2 = 0 α6 = 16r2y4(a2 − 1) + 1 − u2 = 0 α7 = n + l + v − y = 0 α8 = (a2 − 1)l2 + 1 − m2 = 0 α9 = ai + k + 1 − l − i = 0 α10 = ((a + u2(u2 − a))2 − 1)(n + 4dy)2 + 1 − (x + cu)2 = 0 α11 = p + l(a − n − 1) + b(2an + 2a − n2 − 2n − 2) − m = 0 α12 = q + y(a − p − 1) + s(2ap + 2a − p2 − 2p − 2) − x = 0 α13 = z + pl(a − p) + t(2ap − p2 − 1) − pm = 0 The 14 equations α0, …, α13 can be used to produce a prime-generating polynomial inequality in 26 variables: (k+2)(1-\alpha_0^2-\alpha_1^2-\cdots-\alpha_{13}^2) > 0 ie: (k + 2)(1 − [wz + h + j − q]2 − [(gk + 2g + k + 1)(h + j) + h − z]2 − [16(k + 1)3(k + 2)(n + 1)2 + 1 − f2]2 − [2n + p + q + z − e]2 − [e3(e + 2)(a + 1)2 + 1 − o2]2 − [(a2 − 1)y2 + 1 − x2]2 − [16r2y4(a2 − 1) + 1 − u2]2 − [n + l + v − y]2 − [(a2 − 1)l2 + 1 − m2]2 − [ai + k + 1 − l − i]2 − [((a + u2(u2 − a))2 − 1)(n + 4dy)2 + 1 − (x + cu)2]2 − [p + l(a − n − 1) + b(2an + 2a − n2 − 2n − 2) − m]2 − [q + y(a − p − 1) + s(2ap + 2a − p2 − 2p − 2) − x]2 − [z + pl(a − p) + t(2ap − p2 − 1) − pm]2) > 0 Are you sure you did not just make that up? No porkies, please. You made it up, right? Title: Re: Red-Dog's Last Theorem Post by: thetank on May 02, 2009, 04:58:01 PM There will be k riffles with stacks of n chips
(http://i264.photobucket.com/albums/ii177/tigmong/kisaninteger.jpg) That's about as far as I got before lack of education inhibited my progress. Title: Re: Red-Dog's Last Theorem Post by: Cf on May 02, 2009, 05:12:39 PM There will be k riffles with stacks of n chips (http://i264.photobucket.com/albums/ii177/tigmong/kisaninteger.jpg) That's about as far as I got before lack of education inhibited my progress. Love the picture, but you need to use Z+. Your set allows a negative number of riffles ;) Title: Re: Red-Dog's Last Theorem Post by: Royal Flush on May 02, 2009, 05:14:37 PM This is gna drive me crazy...im quite a nerd with numbers and to me this looks way more complex than i could possibly imagine! You sure are Title: Re: Red-Dog's Last Theorem Post by: david3103 on May 02, 2009, 05:43:13 PM http://echochamber.me/viewtopic.php?f=17&t=33796
Title: Re: Red-Dog's Last Theorem Post by: TightEnd on May 02, 2009, 05:52:05 PM Not too difficult really Just input the starting numbers into the following equations α0 = wz + h + j − q = 0 α1 = (gk + 2g + k + 1)(h + j) + h − z = 0 α2 = 16(k + 1)3(k + 2)(n + 1)2 + 1 − f2 = 0 α3 = 2n + p + q + z − e = 0 α4 = e3(e + 2)(a + 1)2 + 1 − o2 = 0 α5 = (a2 − 1)y2 + 1 − x2 = 0 α6 = 16r2y4(a2 − 1) + 1 − u2 = 0 α7 = n + l + v − y = 0 α8 = (a2 − 1)l2 + 1 − m2 = 0 α9 = ai + k + 1 − l − i = 0 α10 = ((a + u2(u2 − a))2 − 1)(n + 4dy)2 + 1 − (x + cu)2 = 0 α11 = p + l(a − n − 1) + b(2an + 2a − n2 − 2n − 2) − m = 0 α12 = q + y(a − p − 1) + s(2ap + 2a − p2 − 2p − 2) − x = 0 α13 = z + pl(a − p) + t(2ap − p2 − 1) − pm = 0 The 14 equations α0, …, α13 can be used to produce a prime-generating polynomial inequality in 26 variables: (k+2)(1-\alpha_0^2-\alpha_1^2-\cdots-\alpha_{13}^2) > 0 ie: (k + 2)(1 − [wz + h + j − q]2 − [(gk + 2g + k + 1)(h + j) + h − z]2 − [16(k + 1)3(k + 2)(n + 1)2 + 1 − f2]2 − [2n + p + q + z − e]2 − [e3(e + 2)(a + 1)2 + 1 − o2]2 − [(a2 − 1)y2 + 1 − x2]2 − [16r2y4(a2 − 1) + 1 − u2]2 − [n + l + v − y]2 − [(a2 − 1)l2 + 1 − m2]2 − [ai + k + 1 − l − i]2 − [((a + u2(u2 − a))2 − 1)(n + 4dy)2 + 1 − (x + cu)2]2 − [p + l(a − n − 1) + b(2an + 2a − n2 − 2n − 2) − m]2 − [q + y(a − p − 1) + s(2ap + 2a − p2 − 2p − 2) − x]2 − [z + pl(a − p) + t(2ap − p2 − 1) − pm]2) > 0 Are you sure you did not just make that up? No porkies, please. You made it up, right? Not at all, its a real theorem, for solving polynomial equations Title: Re: Red-Dog's Last Theorem Post by: The Baron on May 02, 2009, 05:57:06 PM Wont the formula be covered in the faro shuffle maths?
Title: Re: Red-Dog's Last Theorem Post by: thetank on May 02, 2009, 06:27:49 PM
I'd like to question this data. I simulated the effect of a regimented riffling system on two stacks of chips under laboratory conditions and oserved the following 2 chips = 2 riffles 3 chips = 4 riffles 4 chips = 3 riffles 5 chips = 6 riffles 6 chips = 10 riffles 7 chips = 12 riffles 8 chips = 4 riffles As a working hypothesis, I would expect any n where n = 2x (where x is a positive integer) to require a relatively small number of rifles. eg n=2,4,8,16,32 Title: Re: Red-Dog's Last Theorem Post by: thetank on May 02, 2009, 06:30:06 PM Would urge others to corroborate my data if possible.
Title: Re: Red-Dog's Last Theorem Post by: thetank on May 02, 2009, 06:32:41 PM It wouldn't surprise me if k=5 where n =16, k=6 where n=32 and k=7 where n = 64
Have to go out though, no time to check that. Title: Re: Red-Dog's Last Theorem Post by: Cf on May 02, 2009, 06:35:34 PM Lol I thought I was the only saddo that wanted to work this out. Are you sure about you 7 and 8? They look a bit low to me. Looking at this sequence there will be no mathematical formulae available. UL Matt, don't be silly! Of COURSE there is a formula. Well if there is it's going to have lots of log's, ln's, <'s, >'s and upside down i's (haven't got a factorial button on here). I honestly don't see how this sequence can have a mathematical formula to it. I agree it should have but the sequence looks too awkward. What formula can possibly make both 7 and 8 = 4? I'll be thinking on this for a while. It's interesting that 3, 6 and 9 lead to 3, 6 and 9. Could be a starting point me thinks. Really trivial example, but: 4 + (floor (n/9)) (n x 0) + 4 Even trivialler. Not bad. I can beat that tho: 4 :) Title: Re: Red-Dog's Last Theorem Post by: thetank on May 02, 2009, 06:49:23 PM If someone could write a quick computer program to simulate riffles and give us an observed k for a given n then spotting patterns and deriving forumla (then proving said forumlas) would be a bit easier.
Title: Re: Red-Dog's Last Theorem Post by: thetank on May 02, 2009, 07:19:28 PM There will be k riffles with stacks of n chips (http://i264.photobucket.com/albums/ii177/tigmong/kisaninteger.jpg) That's about as far as I got before lack of education inhibited my progress. Love the picture, but you need to use Z+. Your set allows a negative number of riffles ;) I was just gonna use Z, but go on to specify that k must fall into the range 1<k<(2n)! (With < denoting less than or equal to but cba going to paint again to use the proper symbol :) ) Title: Re: Red-Dog's Last Theorem Post by: Grier78 on May 02, 2009, 07:45:46 PM
I'd like to question this data. I simulated the effect of a regimented riffling system on two stacks of chips under laboratory conditions and oserved the following 2 chips = 2 riffles 3 chips = 4 riffles 4 chips = 3 riffles 5 chips = 6 riffles 6 chips = 10 riffles 7 chips = 12 riffles 8 chips = 4 riffles I can corroborate your data Title: Re: Red-Dog's Last Theorem Post by: UpTheMariners on May 02, 2009, 08:48:39 PM the answer is 3.5 bags of sugar
Title: Re: Red-Dog's Last Theorem Post by: ACE2M on May 02, 2009, 10:47:00 PM this battered my brain a few years a go, way beyond us.
Title: Re: Red-Dog's Last Theorem Post by: Jon MW on May 02, 2009, 11:04:18 PM There will be k riffles with stacks of n chips (http://i264.photobucket.com/albums/ii177/tigmong/kisaninteger.jpg) That's about as far as I got before lack of education inhibited my progress. Love the picture, but you need to use Z+. Your set allows a negative number of riffles ;) I was just gonna use Z, but go on to specify that k must fall into the range 1<k<(2n)! (With < denoting less than or equal to but cba going to paint again to use the proper symbol :) ) ffs I was trying to keep out of it, wtf is Z+ ? Do you mean (http://upload.wikimedia.org/math/6/2/4/624e4cf68723f677d53e8cf2272f348a.png) ? Title: Re: Red-Dog's Last Theorem Post by: thetank on May 03, 2009, 01:16:44 AM Oooh, is that for Natural numbers.
Wil use that then for kay is less than or equal to (two times enn) factorial :P Title: Re: Red-Dog's Last Theorem Post by: StuartHopkin on May 03, 2009, 06:17:36 AM
I'd like to question this data. I simulated the effect of a regimented riffling system on two stacks of chips under laboratory conditions and oserved the following 2 chips = 2 riffles 3 chips = 4 riffles 4 chips = 3 riffles 5 chips = 6 riffles 6 chips = 10 riffles 7 chips = 12 riffles 8 chips = 4 riffles I can corroborate your data 3 chips = 3 riffles :dontask: Title: Re: Red-Dog's Last Theorem Post by: thetank on May 03, 2009, 07:54:00 AM 3 chips = 3 riffles :dontask: after 0... b y b y b y after 1... b y y b b y after 2... y y y b b b after 3... (not quite there yet) y y b y b b after 4... (now we're home) b y b y b y Title: Re: Red-Dog's Last Theorem Post by: thetank on May 03, 2009, 08:01:40 AM Only way to do it in 3 is if you break from the regimented system and suddenly change the direction on the 3rd riffle.
We won't be able to come up with a formula if people are changing the direction of the riffle at arbitary times. Title: Re: Red-Dog's Last Theorem Post by: kinboshi on May 03, 2009, 09:56:37 AM I still get 3 riffles for stacks of 3 chips.
The top three chips always going off to the left and being the first chip at the bottom: £ $ £ $ £ $ First riffle gives: $ £ £ $ $ £ 2nd: £ £ $ £ $ $ 3rd: £ $ £ $ £ $ Title: Re: Red-Dog's Last Theorem Post by: Cf on May 03, 2009, 10:04:45 AM There will be k riffles with stacks of n chips (http://i264.photobucket.com/albums/ii177/tigmong/kisaninteger.jpg) That's about as far as I got before lack of education inhibited my progress. Love the picture, but you need to use Z+. Your set allows a negative number of riffles ;) I was just gonna use Z, but go on to specify that k must fall into the range 1<k<(2n)! (With < denoting less than or equal to but cba going to paint again to use the proper symbol :) ) ffs I was trying to keep out of it, wtf is Z+ ? Do you mean (http://upload.wikimedia.org/math/6/2/4/624e4cf68723f677d53e8cf2272f348a.png) ? Z+ is the set of positive integers, {1,2,3...} N is the set of natural numbers, {1,2,3...} or {0,1,2,3...}. Using Z+ saves this ambiguity. I believe 0 is commonly said to be a natural number these days. So if you want to allow 0 riffles in your calculations use N, if not use Z+. Title: Re: Red-Dog's Last Theorem Post by: Rupert on May 03, 2009, 10:31:52 AM Hmm pretty sure someone from poker society at Warwick did a paper on this. Will see if I can get in touch with them.
At Walsall GUKPT this year, Red Dog was Title: Re: Red-Dog's Last Theorem Post by: RichEO on May 03, 2009, 10:33:43 AM If I have 3 chips and cut them to the left every time it takes 4 riffles. If I cut them to the right everytime it takes 3 riffles. It is to do with having the same colour chip on the bottom (for the 2nd riffle) as you started with or a different colour one.
Title: Re: Red-Dog's Last Theorem Post by: sharky_uk on May 03, 2009, 12:05:54 PM I only have a 2000 piece chipset so had to stop at n=1000
2 2 3 3 4 3 5 5 6 6 7 4 8 4 9 9 10 6 11 11 12 10 13 9 14 14 15 5 16 5 17 12 18 18 19 12 20 10 21 7 22 12 23 23 24 21 25 8 26 26 27 20 28 9 29 29 30 30 31 6 32 6 33 33 34 22 35 35 36 9 37 20 38 30 39 39 40 27 41 41 42 8 43 28 44 11 45 12 46 10 47 36 48 24 49 15 50 50 51 51 52 12 53 53 54 18 55 36 56 14 57 44 58 12 59 24 60 55 61 20 62 50 63 7 64 7 65 65 66 18 67 36 68 34 69 69 70 46 71 60 72 14 73 42 74 74 75 15 76 24 77 20 78 26 79 52 80 33 81 81 82 20 83 83 84 78 85 9 86 86 87 60 88 29 89 89 90 90 91 60 92 18 93 40 94 18 95 95 96 48 97 12 98 98 99 99 100 33 101 84 102 10 103 66 104 45 105 105 106 70 107 28 108 15 109 18 110 24 111 37 112 60 113 113 114 38 115 30 116 29 117 92 118 78 119 119 120 12 121 81 122 84 123 36 124 41 125 25 126 110 127 8 128 8 129 36 130 84 131 131 132 26 133 22 134 134 135 135 136 12 137 20 138 46 139 30 140 35 141 47 142 36 143 60 144 68 145 48 146 146 147 116 148 45 149 132 150 42 151 100 152 30 153 51 154 102 155 155 156 78 157 12 158 158 159 140 160 53 161 72 162 30 163 36 164 69 165 15 166 36 167 132 168 21 169 28 170 10 171 147 172 44 173 173 174 174 175 36 176 44 177 140 178 24 179 179 180 171 181 55 182 36 183 183 184 60 185 156 186 186 187 100 188 42 189 189 190 14 191 191 192 60 193 21 194 194 195 88 196 65 197 156 198 22 199 18 200 100 201 60 202 108 203 180 204 102 205 68 206 174 207 164 208 69 209 209 210 210 211 138 212 40 213 60 214 60 215 43 216 36 217 28 218 198 219 73 220 42 221 221 222 44 223 148 224 112 225 20 226 30 227 12 228 38 229 72 230 230 231 231 232 20 233 233 234 66 235 52 236 35 237 180 238 156 239 239 240 18 241 66 242 48 243 243 244 81 245 245 246 56 247 60 248 105 249 83 250 166 251 251 252 50 253 156 254 254 255 9 256 9 257 204 258 230 259 172 260 130 261 261 262 60 263 40 264 253 265 87 266 30 267 212 268 89 269 210 270 270 271 180 272 18 273 273 274 60 275 252 276 39 277 36 278 278 279 84 280 40 281 281 282 14 283 54 284 142 285 57 286 190 287 220 288 72 289 96 290 246 291 260 292 12 293 293 294 90 295 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558 558 559 372 560 261 561 561 562 300 563 231 564 282 565 84 566 510 567 452 568 189 569 264 570 162 571 42 572 38 573 180 574 382 575 575 576 144 577 60 578 132 579 180 580 63 581 83 582 116 583 388 584 249 585 585 586 88 587 460 588 265 589 195 590 118 591 156 592 156 593 593 594 70 595 44 596 149 597 476 598 18 599 180 600 150 601 200 602 24 603 280 604 60 605 516 606 606 607 324 608 76 609 572 610 180 611 611 612 420 613 204 614 614 615 615 616 204 617 36 618 618 619 174 620 72 621 140 622 164 623 28 624 78 625 69 626 534 627 100 628 209 629 629 630 48 631 420 632 220 633 180 634 414 635 20 636 99 637 40 638 638 639 639 640 60 641 641 642 16 643 60 644 161 645 645 646 86 647 36 648 324 649 72 650 650 651 651 652 84 653 653 654 120 655 198 656 150 657 524 658 146 659 659 660 30 661 126 662 130 663 221 664 221 665 605 666 70 667 44 668 285 669 204 670 444 671 312 672 134 673 224 674 630 675 96 676 20 677 540 678 638 679 30 680 340 681 644 682 12 683 683 684 666 685 76 686 686 687 100 688 216 689 588 690 690 691 460 692 46 693 18 694 462 695 636 696 99 697 60 698 70 699 233 700 233 701 660 702 140 703 66 704 352 705 328 706 156 707 188 708 18 709 35 710 84 711 237 712 180 713 713 714 42 715 468 716 179 717 60 718 478 719 719 720 65 721 36 722 136 723 723 724 66 725 725 726 726 727 48 728 115 729 243 730 486 731 90 732 146 733 81 734 42 735 245 736 245 737 580 738 210 739 56 740 185 741 741 742 180 743 743 744 372 745 210 746 746 747 132 748 83 749 749 750 234 751 498 752 84 753 340 754 502 755 755 756 88 757 100 758 90 759 105 760 156 761 761 762 30 763 508 764 345 765 765 766 18 767 204 768 182 769 27 770 66 771 771 772 204 773 24 774 774 775 230 776 97 777 620 778 516 779 779 780 111 781 260 782 78 783 783 784 261 785 785 786 660 787 60 788 369 789 263 790 40 791 791 792 158 793 506 794 678 795 252 796 261 797 140 798 266 799 60 800 200 801 228 802 212 803 803 804 201 805 267 806 26 807 72 808 210 809 809 810 810 811 540 812 150 813 271 814 180 815 87 816 385 817 36 818 818 819 740 820 273 821 260 822 276 823 180 824 48 825 84 826 252 827 60 828 46 829 78 830 30 831 831 832 36 833 833 834 834 835 556 836 357 837 660 838 84 839 99 840 410 841 120 842 84 843 24 844 281 845 198 846 846 847 28 848 424 849 283 850 162 851 780 852 20 853 284 854 122 855 812 856 57 857 588 858 200 859 570 860 215 861 287 862 220 863 260 864 36 865 144 866 866 867 692 868 96 869 828 870 870 871 246 872 174 873 873 874 260 875 408 876 73 877 36 878 150 879 879 880 293 881 140 882 88 883 90 884 210 885 330 886 588 887 140 888 37 889 148 890 102 891 891 892 24 893 893 894 298 895 198 896 405 897 716 898 598 899 48 900 25 901 50 902 684 903 276 904 99 905 181 906 252 907 220 908 429 909 424 910 606 911 911 912 180 913 84 914 290 915 305 916 276 917 732 918 830 919 612 920 393 921 144 922 60 923 923 924 301 925 77 926 72 927 156 928 309 929 780 930 930 931 594 932 186 933 933 934 66 935 935 936 468 937 500 938 938 939 939 940 45 941 804 942 42 943 72 944 236 945 60 946 90 947 756 948 135 949 105 950 950 951 860 952 28 953 953 954 902 955 84 956 239 957 764 958 630 959 900 960 56 961 64 962 60 963 460 964 107 965 965 966 322 967 84 968 222 969 276 970 646 971 924 972 194 973 145 974 974 975 975 976 30 977 88 978 306 979 652 980 234 981 60 982 260 983 210 984 445 985 18 986 986 987 780 988 329 989 989 990 282 991 660 992 22 993 993 994 24 995 180 996 498 997 36 998 998 999 333 1000 308 Title: Re: Red-Dog's Last Theorem Post by: david3103 on May 03, 2009, 12:29:21 PM Some interesting stuff here... http://echochamber.me/viewtopic.php?f=17&t=33796
Title: Re: Red-Dog's Last Theorem Post by: tikay on May 03, 2009, 02:55:33 PM Impressive stiuff Sharky. Two points..... 1) So we know "how many riffles", but we still need a formula, 2) You must have HUGE hands. Photo of your hands, please. Title: Re: Red-Dog's Last Theorem Post by: sharky_uk on May 03, 2009, 03:01:24 PM Found a few more chips....
1001 143 1002 200 1003 222 1004 420 1005 201 1006 60 1007 60 1008 168 1009 48 1010 322 1011 408 1012 540 1013 1013 1014 1014 1015 676 1016 477 1017 180 1018 48 1019 1019 1020 78 1021 339 1022 102 1023 11 1024 11 1025 876 1026 1026 1027 68 1028 440 1029 140 1030 228 1031 1031 1032 348 1033 156 1034 1034 1035 36 1036 115 1037 820 1038 330 1039 90 1040 520 1041 1041 1042 276 1043 1043 1044 29 1045 40 1046 132 1047 836 1048 174 1049 1049 1050 190 1051 700 1052 210 1053 42 1054 36 1055 1055 1056 22 1057 276 1058 252 1059 324 1060 300 1061 480 1062 200 1063 708 1064 266 1065 1065 1066 234 1067 60 1068 534 1069 110 1070 1070 1071 51 1072 60 1073 252 1074 102 1075 714 1076 538 1077 172 1078 718 1079 56 1080 540 1081 102 1082 72 1083 980 1084 24 1085 996 1086 130 1087 140 1088 465 1089 363 1090 242 1091 1044 1092 396 1093 729 1094 990 1095 156 1096 56 1097 292 1098 1014 1099 244 1100 35 1101 367 1102 84 1103 1103 1104 1081 1105 165 1106 1106 1107 884 1108 123 1109 948 1110 1110 1111 36 1112 220 1113 520 1114 742 1115 528 1116 420 1117 148 1118 1118 1119 1119 1120 369 1121 1121 1122 224 1123 318 1124 258 1125 375 1126 750 1127 20 1128 90 1129 75 1130 72 1131 45 1132 60 1133 1133 1134 1134 1135 756 1136 284 1137 60 1138 330 1139 364 1140 95 1141 380 1142 38 1143 381 1144 36 1145 1092 1146 1146 1147 72 1148 574 1149 495 1150 348 1151 483 1152 230 1153 384 1154 1154 1155 1155 1156 48 1157 924 1158 30 1159 772 1160 105 1161 1100 1162 20 1163 1068 1164 136 1165 36 1166 1166 1167 932 1168 180 1169 1169 1170 390 1171 70 1172 132 1173 391 1174 756 1175 47 1176 90 1177 52 1178 1178 1179 21 1180 393 1181 552 1182 140 1183 786 1184 561 1185 1185 1186 84 1187 900 1188 594 1189 60 1190 238 1191 397 1192 156 1193 30 1194 1194 1195 796 1196 299 1197 956 1198 184 1199 1199 1200 1029 1201 198 1202 18 1203 1148 1204 90 1205 241 1206 126 1207 132 1208 604 1209 580 1210 804 1211 1211 1212 240 1213 404 1214 1038 1215 120 1216 135 1217 972 1218 1218 1219 270 1220 305 1221 348 1222 324 1223 1223 1224 195 1225 63 1226 370 1227 980 1228 36 1229 1229 1230 1166 1231 820 1232 56 1233 1233 1234 822 1235 264 1236 309 1237 60 1238 1238 1239 396 1240 413 1241 1140 1242 420 1243 828 1244 585 1245 1196 1246 276 1247 332 1248 565 1249 168 1250 30 1251 1251 1252 332 1253 396 1254 96 1255 270 1256 537 1257 1004 1258 838 1259 380 1260 630 1261 812 1262 50 1263 342 1264 105 1265 1265 1266 296 1267 156 1268 203 1269 1269 1270 330 1271 1271 1272 254 1273 141 1274 1274 1275 1275 1276 396 1277 36 1278 1278 1279 852 1280 294 1281 290 1282 36 1283 120 1284 183 1285 428 1286 410 1287 1020 1288 429 1289 1289 1290 308 1291 60 1292 460 1293 396 1294 862 1295 1295 1296 81 1297 172 1298 1092 1299 308 1300 408 1301 612 1302 130 1303 390 1304 652 1305 372 1306 132 1307 1044 1308 654 1309 144 1310 1310 1311 420 1312 300 1313 1260 1314 1190 1315 876 1316 658 1317 40 1318 876 1319 84 1320 207 1321 110 1322 1012 1323 1323 1324 441 1325 120 1326 378 1327 348 1328 83 1329 1329 1330 886 1331 1331 1332 30 1333 42 1334 104 1335 445 1336 405 1337 1060 1338 1338 1339 414 1340 573 1341 1341 1342 356 1343 79 1344 112 1345 132 1346 1346 1347 420 1348 140 1349 1349 1350 36 1351 104 1352 270 1353 1353 1354 42 1355 1355 1356 678 1357 180 1358 180 1359 1359 1360 453 1361 1164 1362 90 1363 900 1364 682 1365 13 1366 182 1367 1092 1368 264 1369 205 1370 1370 1371 420 1372 60 1373 660 1374 458 1375 390 1376 688 1377 252 1378 306 1379 55 1380 25 1381 51 1382 156 1383 461 1384 420 1385 648 1386 1334 1387 180 1388 694 1389 132 1390 306 1391 110 1392 278 1393 464 1394 1394 1395 465 1396 126 1397 84 1398 1398 1399 930 1400 700 1401 1401 1402 40 1403 600 1404 1378 1405 234 1406 336 1407 1124 1408 156 1409 1409 1410 60 1411 940 1412 70 1413 80 1414 220 1415 1332 1416 59 1417 108 1418 1418 1419 664 1420 473 1421 1421 1422 142 1423 36 1424 180 1425 1425 1426 948 1427 228 1428 51 1429 68 1430 1430 1431 204 1432 380 1433 1380 1434 90 1435 420 1436 312 1437 1100 1438 204 1439 1439 1440 231 1441 310 1442 144 1443 1443 1444 477 1445 1218 1446 1310 1447 96 1448 724 1449 444 1450 966 1451 1451 1452 492 1453 72 1454 1454 1455 140 1456 97 1457 260 1458 486 1459 138 1460 77 1461 468 1462 60 1463 1463 1464 350 1465 488 1466 1254 1467 1172 1468 110 1469 1469 1470 344 1471 36 1472 180 1473 420 1474 982 1475 1356 1476 246 1477 196 1478 1478 1479 1340 1480 138 1481 1481 1482 74 1483 154 1484 371 1485 55 1486 990 1487 120 1488 114 1489 15 1490 270 1491 468 1492 396 1493 1428 1494 420 1495 332 1496 180 1497 1196 1498 108 1499 1499 1500 750 1501 60 1502 100 1503 240 1504 232 1505 1505 1506 1430 1507 132 1508 129 1509 1509 1510 468 1511 1511 1512 220 1513 504 1514 348 1515 72 1516 42 1517 1212 1518 1518 1519 92 1520 760 1521 712 1522 84 1523 460 1524 381 1525 252 1526 70 1527 276 1528 509 1529 198 1530 102 1531 340 1532 306 1533 1533 1534 30 1535 1476 1536 219 1537 20 1538 360 1539 1539 1540 156 1541 1541 1542 308 1543 294 1544 386 1545 35 1546 1030 1547 1236 1548 81 1549 129 1550 1326 1551 1484 1552 396 1553 1428 1554 222 1555 120 1556 235 1557 132 1558 1038 1559 1559 1560 78 1561 519 1562 1250 1563 1508 1564 444 1565 100 1566 24 1567 180 1568 392 1569 126 1570 348 1571 672 1572 72 1573 131 1574 1518 1575 748 1576 175 1577 180 1578 60 1579 324 1580 126 1581 527 1582 420 1583 1583 1584 792 1585 30 1586 1494 1587 140 1588 264 1589 680 1590 530 1591 1060 1592 84 1593 1593 1594 1062 1595 55 1596 255 1597 420 1598 1518 1599 228 1600 240 1601 1601 1602 64 1603 356 1604 802 1605 468 1606 72 1607 428 1608 402 1609 252 1610 322 1611 1460 1612 140 1613 1380 1614 538 1615 1074 1616 390 1617 1292 1618 492 1619 780 1620 231 1621 506 1622 580 1623 760 1624 171 1625 325 1626 1626 1627 60 1628 407 1629 543 1630 1086 1631 300 1632 326 1633 495 1634 1398 1635 545 1636 545 1637 260 1638 14 1639 364 1640 96 1641 462 1642 36 1643 1548 1644 660 1645 137 1646 396 1647 1316 1648 156 1649 1649 1650 330 1651 366 1652 330 1653 1653 1654 58 1655 210 1656 414 1657 24 1658 530 1659 1659 1660 540 1661 1661 1662 180 1663 1108 1664 832 1665 111 1666 100 1667 308 1668 805 1669 156 1670 48 1671 557 1672 148 1673 1673 1674 392 1675 1116 1676 717 1677 60 1678 372 1679 1679 1680 84 1681 261 1682 48 1683 36 1684 561 1685 1685 1686 562 1687 900 1688 255 1689 180 1690 462 1691 792 1692 338 1693 564 1694 242 1695 113 1696 84 1697 48 1698 546 1699 510 1700 801 1701 820 1702 452 1703 1703 1704 243 1705 189 1706 1706 1707 44 1708 264 1709 1572 1710 310 1711 162 1712 170 1713 1628 1714 126 1715 207 1716 858 1717 76 1718 1470 1719 180 1720 180 1721 780 1722 78 1723 1146 1724 431 1725 168 1726 1150 1727 460 1728 288 1729 288 1730 1730 1731 577 1732 60 1733 1733 1734 1734 1735 132 1736 165 1737 1380 1738 180 1739 105 1740 1711 1741 189 1742 40 1743 1580 1744 83 1745 1745 1746 498 1747 116 1748 402 1749 1749 1750 1164 1751 140 1752 350 1753 498 1754 1540 1755 1755 1756 585 1757 36 1758 1758 1759 264 1760 753 1761 540 1762 460 1763 1763 1764 441 1765 265 1766 1766 1767 300 1768 585 1769 1769 1770 118 1771 236 1772 354 1773 1773 1774 156 1775 1716 1776 360 1777 156 1778 1778 1779 1779 1780 593 1781 1524 1782 220 1783 140 1784 287 1785 1785 1786 132 1787 60 1788 63 1789 149 1790 1790 1791 1791 1792 476 1793 840 1794 144 1795 18 1796 898 1797 1436 1798 180 1799 1740 1800 138 1801 300 1802 204 1803 601 1804 600 1805 572 1806 1806 1807 24 1808 904 1809 690 1810 280 1811 1811 1812 350 1813 60 1814 1710 1815 605 1816 516 1817 484 1818 1818 1819 1212 1820 15 1821 1821 1822 972 1823 780 1824 220 1825 152 1826 420 1827 56 1828 572 1829 1829 1830 522 1831 180 1832 122 1833 288 1834 1222 1835 1835 1836 459 1837 420 1838 1838 1839 564 1840 204 1841 28 1842 660 1843 1228 1844 120 1845 1845 1846 1230 1847 492 1848 924 1849 612 1850 1850 1851 759 1852 36 1853 210 1854 1854 1855 1236 1856 897 1857 1484 1858 174 1859 1859 1860 1830 1861 72 1862 370 1863 1863 1864 140 1865 60 1866 1866 1867 492 1868 450 1869 267 1870 28 1871 1764 1872 636 1873 156 1874 1782 1875 110 1876 207 1877 1500 1878 408 1879 534 1880 94 1881 1820 1882 100 1883 1883 1884 942 1885 627 1886 1470 1887 60 1888 629 1889 1889 1890 198 1891 48 1892 378 1893 540 1894 420 1895 296 1896 948 1897 220 1898 1898 1899 1820 1900 180 1901 1901 1902 190 1903 1242 1904 438 1905 612 1906 20 1907 36 1908 865 1909 99 1910 382 1911 637 1912 120 1913 154 1914 546 1915 1276 1916 479 1917 348 1918 1278 1919 1740 1920 913 1921 60 1922 384 1923 1923 1924 641 1925 1925 1926 1926 1927 16 1928 252 1929 904 1930 180 1931 1931 1932 386 1933 322 1934 468 1935 273 1936 645 1937 100 1938 1938 1939 258 1940 194 1941 440 1942 36 1943 1716 1944 324 1945 648 1946 152 1947 180 1948 72 1949 1668 1950 1886 1951 1300 1952 140 1953 1953 1954 1302 1955 1955 1956 84 1957 252 1958 1958 1959 1959 1960 653 1961 1961 1962 130 1963 120 1964 982 1965 1965 1966 198 1967 1572 1968 35 1969 300 1970 1686 1971 219 1972 524 1973 1973 1974 1790 1975 438 1976 957 1977 84 1978 1318 1979 1908 1980 232 1981 60 1982 30 1983 1983 1984 378 1985 855 1986 238 1987 260 1988 240 1989 1892 1990 442 1991 852 1992 398 1993 663 1994 1994 1995 204 1996 605 1997 184 1998 114 1999 70 2000 500 2001 2001 2002 132 2003 2003 2004 315 2005 570 2006 2006 2007 180 2008 204 2009 2009 2010 2010 2011 1332 2012 660 2013 671 2014 312 2015 1932 2016 18 2017 268 2018 1830 2019 144 2020 672 2021 1860 2022 202 2023 630 2024 253 2025 25 2026 96 2027 540 2028 169 2029 60 2030 130 2031 952 2032 540 2033 1722 2034 78 2035 638 2036 1018 2037 1620 2038 90 2039 2039 2040 780 2041 680 2042 252 2043 660 2044 644 2045 2045 2046 2046 2047 12 2048 12 Title: Re: Red-Dog's Last Theorem Post by: thetank on May 03, 2009, 11:15:50 PM Cool, so my prediction for n=1,024 was correct.
I feel warm and fuzzy now Title: Re: Red-Dog's Last Theorem Post by: Robert HM on May 04, 2009, 12:57:19 AM what happens if you can't riffle?
Title: Re: Red-Dog's Last Theorem Post by: mondatoo on May 04, 2009, 10:56:56 AM Where am i and what happened to blonde ?
Title: Re: Red-Dog's Last Theorem Post by: bhoywonder on May 04, 2009, 10:12:07 PM ;popcorn;
nothing to add.....but its damn interesting Title: Re: Red-Dog's Last Theorem Post by: cod meharly on May 04, 2009, 10:22:12 PM ;popcorn; nothing to add.....but its damn interesting Title: Re: Red-Dog's Last Theorem Post by: The Camel on May 05, 2009, 02:45:52 AM This is all bollocks and one huge level.
The whole theorem falls down because Red Dog never has enough chips to riffle in the first place. Title: Re: Red-Dog's Last Theorem Post by: dik9 on May 05, 2009, 03:54:41 AM Is there not a floor in these calculations as the chips are not returning to their original positions
If you start with yellow and blue stacks with the yellow on the left, and always splitting to the right, are you taking the calculations for them to return to two stacks of same colour, or till the yellow ones are back on the left. Or even for the purists even then the chips would not be in their original positions, only way to do this is to number each chip and put them in order so even numbers would be in one pile and odd in the other. i.e. 1 2 3 4 5 6 7 8 Always split to the same direction and always have the same stack grounded first [ ] can be arsed to do it :) Title: Re: Red-Dog's Last Theorem Post by: cod meharly on May 05, 2009, 04:45:00 AM and so the plot thickens.....
Title: Re: Red-Dog's Last Theorem Post by: kinboshi on May 05, 2009, 09:27:56 AM Is there not a floor in these calculations as the chips are not returning to their original positions If you start with yellow and blue stacks with the yellow on the left, and always splitting to the right, are you taking the calculations for them to return to two stacks of same colour, or till the yellow ones are back on the left. Or even for the purists even then the chips would not be in their original positions, only way to do this is to number each chip and put them in order so even numbers would be in one pile and odd in the other. i.e. 1 2 3 4 5 6 7 8 Always split to the same direction and always have the same stack grounded first [ ] can be arsed to do it :) No, that's Red-Dog's Second to Last Theorem you're talking about there... ::) |