Red-Dog's Last Theorem

[TightEnd]

The answer, I think you will find, is 8.

[kinboshi]

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.

[kinboshi]

I found this equation for the 'stack effect'.  I'm not sure it'll help too much here unfortunately.

[tikay]

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?

[thetank]

There will be k riffles with stacks of n chips



That's about as far as I got before lack of education inhibited my progress.

[Cf]

There will be k riffles with stacks of n chips



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

[Royal Flush]

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

[david3103]

http://echochamber.me/viewtopic.php?f=17&t=33796

[TightEnd]

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

[The Baron]

Wont the formula be covered in the faro shuffle maths?

[thetank]

Chips	Riffles
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 = 2^x (where x is a positive integer) to require a relatively small number of rifles.

eg n=2,4,8,16,32

[thetank]

Would urge others to corroborate my data if possible.

[thetank]

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.

[Cf]

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

[thetank]

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.