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

*To*: mathgroup at smc.vnet.net*Subject*: [mg89350] Re: [mg89338] Inequalities or histogram or something.......*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sat, 7 Jun 2008 02:55:57 -0400 (EDT)*References*: <200806061047.GAA24209@smc.vnet.net> <C131242B-F293-41C3-86B9-233078332582@mimuw.edu.pl>

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 >> >

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

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

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