MaplePrimes - answers and comments on Question, Phantom solution to linear inequalities using solve

Variant

Kitonum — Wed, 30 Jan 2013 00:39:59 Z

The solution will be correct, if we write

solve({ a + b = 1, a > 0, b > 1/2}, {a,b,c}) assuming c>0;

{a = -b+1, c = c, 1/2 < b, b < 1}

Well, that seems to fix that simple example.

Simon Willerton — Wed, 30 Jan 2013 01:32:17 Z

Well, that seems to fix that simple example. How does one work in general? I have a routine which generates linear equalites and inequalities and I want to see if there are any solutions. The following should give no solutions

  solve([0 < 2*f[1], 5 < f[1]+f[2], 8 < f[1]+f[3], f[1]+f[4] = 7, 0 < 2*f[2], f[2]+f[3] = 6, 
    6 < f[2]+f[4], 0 < 2*f[3], 11/2 < f[3]+f[4], 0 < 2*f[4]],{f[1],f[2],f[3],f[4]});

But gives

{f[1] = 7-f[4], f[2] = 6-f[3], f[3] = f[3], f[4] = f[4]}

Rewriting as you suggest

  solve([ 5 < f[1]+f[2], 8 < f[1]+f[3], f[1]+f[4] = 7, f[2]+f[3] = 6, 
               6 < f[2]+f[4],11/2 < f[3]+f[4]])  
      assuming And(f[1]>0,f[2]>0,f[3]>0,f[4]>0);

gives the same wrong answer.

By LinearMultivariateSystem command

Markiyan Hirnyk — Wed, 30 Jan 2013 01:39:22 Z

This works:

Download LMS.mw See ?SolveTools/Inequality for more info.

PS. The notation fk is preferable over f[k].

PPS. Relation (2) is not true.

SolveTools:-SemiAlgebraic

acer — Wed, 30 Jan 2013 02:39:57 Z

In 16.01 on 64bit Linux on an Intel i5, after about 5 minutes,

SolveTools:-SemiAlgebraic({f1+f2 = 5, f3+f4 = 11/2,
 0 <f1, 0 < f2, 0 < f3, 0 < f4,
 6 < f2+f3, 6 < f2+f4,
 7 < f1+f4, 8 < f1+f3},
 [f1,f2,f3,f4]);

 []

which, with [] as output, means no solutions.

More quickly,

SolveTools:-SemiAlgebraic({ a + b = 1, a > 0, b > 1, c > 0 }, {a,b,c});

 []

However, the computation time grows quickly with problem size, for this approach.

The `solve` command is using an older, faster, and buggier algorithm here.

acer

Mathematica

Kitonum — Wed, 30 Jan 2013 02:47:37 Z

To solve these problems, use Mathematica rather than Maple:

Reduce[{0 < 2*f[1], 5 < f[1] + f[2], 8 < f[1] + f[3], 
 f[1] + f[4] == 7, 0 < 2*f[2], f[2] + f[3] == 6, 6 < f[2] + f[4], 
 0 < 2*f[3], 11/2 < f[3] + f[4], 0 < 2*f[4]}, {f[1], f[2], f[3], 
 f[4]}]

                                         False

Splitting in eqs and ineqs

Preben Alsholm — Wed, 30 Jan 2013 03:05:56 Z

The following works on your longer example (as well as on your simple one):

restart;
sys:=[0 < 2*f[1], 5 < f[1]+f[2], 8 < f[1]+f[3], f[1]+f[4] = 7, 0 < 2*f[2], f[2]+f[3] = 6,
6 < f[2]+f[4], 0 < 2*f[3], 11/2 < f[3]+f[4], 0 < 2*f[4]];
var:={f[1],f[2],f[3],f[4]};
#Splitting:
eqs,ineqs:=selectremove(type,sys,`=`);
solve(eqs,var);
subs(%,ineqs);
solve(%,indets(%,name));
#Output NULL as is correct in this case.