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Poker Forums => The Rail => Topic started by: kinboshi on May 02, 2009, 12:09:00 PM



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 wants needs to know is if there's a mathematical formula (which there will be) that can tell you how many times you have to riffle two different coloured stacks of chips so that the stacks are back to all the same colours, according to how many chips are in the stacks?

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:


ChipsRiffles
2 2
3 3
4 3
5 5
6 6
7 4
8 4
9 9
10 6

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
This is more complex than you could ever possibly imagine and, to my knowledge, is unsolved.
Who cares, I don't, I have a life


FYP


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

ChipsRiffles
2 2
3 3
4 3
5 5
6 6
7 4
8 4
9 9
10 6



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

ChipsRiffles
2 2
3 3
4 3
5 5
6 6
7 4
8 4
9 9
10 6



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

ChipsRiffles
2 2
3 3
4 3
5 5
6 6
7 4
8 4
9 9
10 6



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 boring engaging the people at his table with this problem: http://www.funtrivia.com/askft/Question23938.html That site also has the solution FWIW, so thought Red Dog might be interested.  The problem is you have 12 chips and one of the chips is of a different weight to the others (either heavier or lighter).  On a set of balancing scales, determine in 3 weighs which chip it is and whether it is heavier or lighter.


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   196
296   74
297   24
298   198
299   299
300   25
301   33
302   220
303   303
304   84
305   276
306   306
307   20
308   77
309   309
310   198
311   33
312   250
313   45
314   72
315   45
316   105
317   28
318   84
319   210
320   32
321   107
322   28
323   323
324   145
325   30
326   326
327   260
328   18
329   329
330   330
331   24
332   36
333   308
334   74
335   60
336   24
337   180
338   338
339   48
340   113
341   11
342   34
343   76
344   78
345   115
346   30
347   276
348   40
349   58
350   350
351   36
352   92
353   300
354   354
355   78
356   55
357   60
358   238
359   359
360   51
361   24
362   70
363   121
364   243
365   56
366   122
367   84
368   165
369   123
370   36
371   371
372   74
373   123
374   318
375   375
376   25
377   60
378   378
379   110
380   190
381   36
382   24
383   348
384   192
385   16
386   386
387   20
388   36
389   180
390   70
391   252
392   26
393   393
394   262
395   84
396   30
397   52
398   398
399   184
400   66
401   90
402   132
403   268
404   202
405   135
406   270
407   324
408   63
409   12
410   410
411   411
412   20
413   413
414   414
415   92
416   168
417   332
418   90
419   419
420   406
421   35
422   78
423   330
424   47
425   396
426   426
427   36
428   214
429   429
430   60
431   431
432   86
433   136
434   390
435   132
436   48
437   300
438   438
439   292
440   55
441   441
442   116
443   443
444   21
445   135
446   414
447   356
448   132
449   140
450   104
451   42
452   90
453   453
454   300
455   91
456   205
457   60
458   390
459   153
460   51
461   420
462   90
463   102
464   232
465   126
466   310
467   40
468   117
469   156
470   470
471   220
472   36
473   473
474   36
475   316
476   34
477   380
478   140
479   204
480   155
481   159
482   96
483   483
484   72
485   97
486   138
487   60
488   244
489   110
490   36
491   491
492   98
493   138
494   154
495   495
496   15
497   396
498   166
499   36
500   60
501   232
502   132
503   468
504   252
505   42
506   46
507   84
508   84
509   509
510   170
511   10
512   10
513   156
514   294
515   515
516   129
517   132
518   120
519   519
520   173
521   444
522   180
523   348
524   131
525   175
526   108
527   420
528   15
529   88
530   530
531   531
532   140
533   240
534   178
535   24
536   126
537   140
538   358
539   492
540   253
541   171
542   60
543   543
544   165
545   545
546   182
547   36
548   137
549   156
550   366
551   29
552   24
553   180
554   554
555   100
556   156
557   148
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... ::)