Re: Re: Inequalities or histogram or something.......
- To: mathgroup at smc.vnet.net
- Subject: [mg89459] Re: [mg89353] Re: [mg89338] Inequalities or histogram or something.......
- From: Curtis Osterhoudt <cfo at lanl.gov>
- Date: Tue, 10 Jun 2008 03:40:09 -0400 (EDT)
- Organization: LANL
- References: <200806061047.GAA24209@smc.vnet.net> <6AABE4A5-728F-45E8-AB45-F86C6B445438@mimuw.edu.pl> <200806070656.CAA09750@smc.vnet.net>
- Reply-to: cfo at lanl.gov
And I believe that Andrzej is correct, merely because his closed-form
solution agrees with my plot :)
Cheers,
Curtis O.
On Saturday 07 June 2008 00:56:29 Andrzej Kozlowski wrote:
> In fact, I think, I was initially too pessimistic about this problem.
> I belive that I can now prove the following:
>
> d(v,k) = (k + 1) (v - k - 2)
>
> The proof uses the same argument which I used to show that d(v,0) = v-2
> (note that this follows form the above formula). Note also that
>
> d(v,k) + d(v,v-k-3) = (k + 1) (-k + v - 2) + (k + 1) (-k + v - 2) = 2
> (k + 1) (-k + v - 2)
>
> which is just your formula. You can also check that this agrees with
> the function d[v,k] defined below.
>
> So I think your problem is now completely solved.
>
> Andrzej Kozlowski
>
> On 6 Jun 2008, at 22:39, Andrzej Kozlowski wrote:
> > I just realized that the program I wrote is usually giving wrong
> > answers (but very fast ;-)) . The answer returned last time for
> > d[10000, 2305] was absurdly low!
> > The reason for this is that I forgot (not for the first time) that
> > Reduce only returns a list of solutions of an equation with a finite
> > number of solutions if that number is less than a certain value,
> > which by default is only 10. We can change this as follows:
> >
> > SetSystemOptions[ReduceOptions -> {DiscreteSolutionBound -> 10^8}]
> >
> > d[v_, 0] := v - 2
> > d[v_, k_] /; (v - 3)/2 < k < v - 2 := d[v, k] = 2 (k + 1) (v - k -
> > 2) - d[v, v - k - 3]
> > d[v_, k_] /; 0 < k <= (v - 3)/2 := d[v, k] = Length[Union[Reduce[1
> > <= x1 < x2 < x3 <= v && k == x1 + x3 - x2 - 2, {x1, x2, x3},
> > Integers]]]
> >
> > d[200, 24] // Timing
> > {0.118747, 4350}
> >
> > This is still pretty fast and this time correct. One can try
> > somewhat larger values:
> >
> > d[1000, 500] // Timing
> > {3.47657, 249498}
> >
> > The number of solutions is already pretty large and I would
> > certainly not try much larger numbers with this method. Of course,
> > if the purpose if to find a conjecture about the value of d[v,k]
> > this ought to be quite enough.
> >
> > Andrzej Kozlowski
> >
> > On 6 Jun 2008, at 22:13, Andrzej Kozlowski wrote:
> >> I don't have a general formula, and having given it only a little
> >> thought, am not convinced an explicit formula can be found in the
> >> general case. But I may be wrong. But here are some observations.
> >>
> >> First, note that in your post you make the assumption that 1<=k but
> >> then you give an example with k==0. Presumably you mean k>=0?
> >>
> >> Also, your assumption k<v follows from 1<= x1 < x2 < x3 <= v and
> >> k = x1 + x3 - x2 - 2, since clearly
> >> k==x3 -(x2-x1)-2 must be less than x3 which is less than v. But
> >> this is not important.
> >>
> >> It also easy to see that your equation can only have solutions if
> >> v>k+2, which of course is needed for your relation d(v,k) + d(v,v-
> >> k-3) = 2(k+1)(v-k-2) to make sense. The case v=3 then k must be 0
> >> so we can assume that v>=4.
> >>
> >> Next, it is easy to prove that
> >>
> >> d[v,0]== v -2
> >>
> >> The proof goes as follows. For k= 0 your equation becomes
> >>
> >> x1 + x3 - x2 - 2==0
> >>
> >> which can be re-written as
> >>
> >> x1 = 2 -(x3-x2)
> >>
> >> Now, since x3-x2>0 and x1>=1, this only leaves the possibility that
> >> x1 = 1. So the equation becomes x3-x2=1. So d[v,0] is the number of
> >> ways of choosing two successive integers between 1 and v. You can
> >> choose any integer z such that 2<=z<=v-1 as the smaller one, and v
> >> +1 as the larger one. There are v-2 such choices.
> >> So now that we know that d[v,0]=v-2 and can write a reasonably fast
> >> program to find other values of d[v,k].
> >>
> >> d[v_, 0] := v - 2
> >>
> >> d[v_, k_] /; (v - 3)/2 < k < v - 2 := d[v, k] = 2 (k + 1) (v - k -
> >> 2) - d[v, v - k - 3]
> >>
> >> d[v_, k_] /; 0 < k <= (v - 3)/2 := d[v, k] = Length[Union[ Reduce[1
> >> <= x1 < x2 < x3 <= v && k == x1 + x3 - x2 - 2, {x1, x2, x3},
> >> Integers]]]
> >>
> >>
> >> This seems quite fast; for example
> >>
> >> d[10000, 2305] // Timing
> >> {0.283073, 4}
> >>
> >> d[10000, 7692] // Timing
> >> {0.000065, 35480112}
> >>
> >> So one could try to first explore this empirically, guess the
> >> formula and then it should not be hard to prove it inductively. But
> >> this is as much time as I can devote to this problem.
> >>
> >>
> >> Andrzej Kozlowski
> >>
> >> On 6 Jun 2008, at 19:47, Steve Gray wrote:
> >>> Can Mathematica help with this? Or can someone?
> >>>
> >>> I have positive integers x1,x2,x3,k,v.
> >>>
> >>> There are assumptions:
> >>> 1 <= k < v and
> >>> 1 <= x1 < x2 < x3 <= v.
> >>>
> >>> There is one equation:
> >>> k = x1 + x3 - x2 - 2.
> >>>
> >>> I need a symbolic solution for the number of combinations of
> >>> x1,x2,x3
> >>> that satisfy the equation under the assumptions.
> >>>
> >>> This will be a histogram of k vs. the number of solutions.
> >>> One numeric point on the histo: if v=8 and k=0, there are 6
> >>> solutions,
> >>> x1,x2,x3 = 1,2,3; 1,3,4; 1,4,5; 1,5,6; 1,6,7; 1,7,8; none with x1
> >>>
> >>> > 1.
> >>>
> >>> I need a general symbolic solution in terms of D(v,k). I need a
> >>> way to
> >>> show its derivation for a paper. I happen to know that
> >>> D(v,k) + D(v,v-k-3) = 2(k+1)(v-k-2).
> >>>
> >>> This is not overwhelmingly complicated but I don't know a decent way
> >>> to go about it. Thank you for any help, using Mathematica or not.
> >>> (I have
> >>> version 6.)
> >>>
> >>> Steve Gray
--
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Curtis Osterhoudt
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- References:
- Inequalities or histogram or something.......
- From: Steve Gray <stevebg@roadrunner.com>
- Re: Inequalities or histogram or something.......
- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
- Inequalities or histogram or something.......