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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
>>
>



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