Re: [Help-glpk] [Fwd: I: Modelling binary variable]

[Nigel Galloway]

Michael,

Does this not imply that we could just say sum = 2a + z where a is an
integer >=0 and z is binary?

-- 
  Nigel Galloway
  address@hidden

On Wed, Oct 9, 2013, at 06:00 AM, Meketon, Marc wrote:
> Michael.  This is also very clever.
> 
> Another explanation of your code is the following:
> 
> Variable  a  is a non-negative integer (and always <= N2hi),  b  is
> binary, z  is binary.  There are 4 constraints:
> (1) sum=2a+b
> (2) z>=b-a
> (3) z<=b
> (4) z<=(1-N2hi/N2lo)b - a/N2lo + N2hi/N2lo
> 
> When sum=1 we must have a=0, b=1.  Constraints (2) becomes z >= 1, (3)
> becomes z <= 1, and (4) becomes z <=1.  Hence z = 1.
> 
> For any sum that is even,  a = sum/2 and  b=0.  Constraint (2) is then
> non-binding since b-a <=0 and we know z >=0.  Constraint (3) is z <=0. 
> Constraint (4) (since b=0) is z <= (N2hi - a)/N2lo.  Since  0 <= a <=
> N2hi, this is a non-binding since constraint (3) is tighter.  Hence z=0.
> 
> For any sum that is odd, with sum >= 3, we know that 1 <= a <= N2lo and b
> = 1.  Constraint (2) is non-binding because b-a <= 0.  Constraint (3)
> with z <= 1 is non-binding.  Constraint (4) becomes z <= 1-a/N2lo.  Since
> 1 <= a <= N2lo, we know 0 <= 1-a/N2lo < 1, implying z < 1 (strict
> inequality), and then the binary constraint forces z=0.
> 
> 
> -----Original Message-----
> From: Michael Hennebry [mailto:address@hidden
> Sent: Tuesday, October 08, 2013 1:43 PM
> To: Meketon, Marc
> Cc: Nigel Galloway; Andrew Makhorin; address@hidden;
> address@hidden
> Subject: Re: [Help-glpk] [Fwd: I: Modelling binary variable]
> 
> On Tue, 8 Oct 2013, Meketon, Marc wrote:
> 
> > Are you sure that Z = Q[2]-Q[1]?  For the case where x[1]=1, x[2]=0, 
> > x[3]=0, we have Q[1]=0, Q[2]=0, Q[3]=1, and then Q[3]-Q[2] = 1 which is the 
> > correct answer.
> 
> Z = Q[1]-Q[2]
> he has Q sorted in non-ascending order.
> 00 no ones  0-0=0
> 10 one one  1-0=1
> 11 two or more ones 1-1=0
> exactly what you want
> The size of N does not matter.
> 
> Meketon's code has the advantage of ease of coding and understanding, but
> it doubles the dimensionality.
> 
> Assume one has an integer expression sum:
> 0<=sum<=N
> One wants z==1 iff sum==1 else 0
> Define N2lo=floor(N/2), N2hi=ceil(N/2)
> Note N=N2lo+N2hi, H2hi-N2lo in {0, 1}
> Add two (not N) more integer variables:
> 0<=a<=N2hi
> b binary
> 
> require
> sum=2a+b
> z>=b-a
> z<=b
> z<=(1-N2hi/N2lo)b - a/N2lo + N2hi/N2lo
> 
> The last constraint on z should be multiplied by N2lo to ensure
> integrality of the coefficients.
> 
> Done.
> 
> The first two constraints on z are fairly obvious.
> The last needs more explanation.
> 
> The diagram below is for N==7.
> 
>     3  0 -
>     2  0 0
>   a 1  0 0
>     0  0 1
> 
>        0 1
>        b
> 
> The rectangle gives the values of z for all valid combinations of a and
> b.
> The given constraints on z are all facets of the polyhedron.
> The first is for facet (0, 0, 0)(0, 1, 1)(1, 0, 0).
> The second for facet (0, 0, 0)(0, 1, 1)(N2hi, 0, 0).
> The third for facet (N2lo, 1, 0)(0, 1, 1)(N2hi, 0, 0).
> Substitution will verify.
> 
> 
> Note that exhaustive testing is possible:
> The number of combinations that need testing is at most 2*(N+1)**2 .
> 
> --
> Michael   address@hidden
> "On Monday, I'm gonna have to tell my kindergarten class, whom I teach
> not to run with scissors, that my fiance ran me through with a
> broadsword."  --  Lily
> 
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