Re: Inequalities or histogram or something.......

*To*: mathgroup at smc.vnet.net*Subject*: [mg89353] Re: [mg89338] Inequalities or histogram or something.......*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sat, 7 Jun 2008 02:56:29 -0400 (EDT)*References*: <200806061047.GAA24209@smc.vnet.net> <C131242B-F293-41C3-86B9-233078332582@mimuw.edu.pl> <6AABE4A5-728F-45E8-AB45-F86C6B445438@mimuw.edu.pl>

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 >>> >> >

**Follow-Ups**:**Re: Re: Inequalities or histogram or something.......***From:*Curtis Osterhoudt <cfo@lanl.gov>

**References**:**Inequalities or histogram or something.......***From:*Steve Gray <stevebg@roadrunner.com>

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**Re: Inequalities or histogram or something.......**

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