Problems with FindInstance in v. 11

Question

I have been trying to solve a system of linear equations and linear inequalities with 18 unknowns. I would think that for Mathematica giving an answer to such question will be a matter of milliseconds. Instead, the kernel runs for ages, uses up the entire CPU and RAM and then stops without providing any message. Just as if I did not let run the program at all.

Does anyone know what could be the problem? (and maybe how can I get ao some answer to my exercise?). Or is it a software bug? Thank you!!

Here is my code:

FindInstance[{1/5 t12 + 1/5 t22 + 3/5 t32 == 0, 
  1/5 t1 + 1/5 t2 + 3/5 t3 == 0,
  Min[1/5 t12 + 1/5 t22 + 3/5 t32, 1/5 t1 + 1/5 t2 + 3/5 t3] + 
    Min[1/5 t1 + 1/5 t4 + 3/5 t7, 1/5 t12 + 1/5 t42 + 3/5 t72] >= 
   Min[1/5 t42 + 1/5 t52 + 3/5 t62, 1/5 t4 + 1/5 t5 + 3/5 t6] + 
    Min[ 1/5 t2 + 1/5 t5 + 3/5 t8, 1/5 t22 + 1/5 t52 + 3/5 t82],
  Min[1/5 t12 + 1/5 t22 + 3/5 t32, 1/5 t1 + 1/5 t2 + 3/5 t3] + 
    Min[1/5 t1 + 1/5 t4 + 3/5 t7, 1/5 t12 + 1/5 t42 + 3/5 t72] >= 
   Min[1/5 t72 + 1/5 t82 + 3/5 t92, 1/5 t7 + 1/5 t8 + 3/5 t9] + 
    Min[ 1/5 t3 + 1/5 t6 + 3/5 t9, 1/5 t32 + 1/5 t62 + 3/5 t92],
  Min[1/5 t42 + 1/5 t52 + 3/5 t62, 1/5 t4 + 1/5 t5 + 3/5 t6] < 0,
  Min[1/5 t72 + 1/5 t82 + 3/5 t92, 1/5 t7 + 1/5 t8 + 3/5 t9] < 0,
  Min[1/2 (1/5 + 1/9) t12 + 1/2 (1/5 + 1/9) t22 + 1/2 (3/5 + 7/9) t32,
     1/2 (1/5 + 1/9) t1 + 1/2 (1/5 + 1/9) t2 + 1/2 (3/5 + 7/9) t3] < 0,
  1/2 (1/5 + 1/9) t42 + 1/2 (1/5 + 1/9) t52 + 1/2 (3/5 + 7/9) t62 == 0,
  1/2 (1/5 + 1/9) t4 + 1/2 (1/5 + 1/9) t5 + 1/2 (3/5 + 7/9) t6 == 0,
  Min[1/2 (1/5 + 1/9) t7 + 1/2 (1/5 + 1/9) t8 + 1/2 (3/5 + 7/9) t9, 
    1/2 (1/5 + 1/9) t72 + 1/2 (1/5 + 1/9) t82 + 1/2 (3/5 + 7/9) t92] <0,
  Min[1/9 t12 + 1/9 t22 + 7/9 t32, 1/9 t1 + 1/9 t2 + 7/9 t3] < 0,
  Min[1/9 t42 + 1/9 t52 + 7/9 t62, 1/9 t4 + 1/9 t5 + 7/9 t6] < 0,
  1/9 t72 + 1/9 t82 + 7/9 t92 == 0, 1/9 t7 + 1/9 t8 + 7/9 t9 == 0},
 {t12, t22, t32, t42, t52, t62, t72, t82, t92, t1, t2, t3, t4, t5, t6,t7, t8, t9}, Reals]

Why do you think Mathematica should be able to do this instantly? What is your reasoning? Not all problems can be easily computed. — ktm, Commented Mar 6, 2018 at 20:47
Mathematica does well with a portion of the equations, but as soon as I include too many of them memory spikes. For example, assuming eqns is your list of equations, MemoryConstrained[FindInstance[eqns[[3 ;;]], {t1, t2, t3, t12, t22, t32, t42, t52, t62, t4, t5, t6, t7, t8, t9, t72, t82, t92}, Reals], 4000000000] // AbsoluteTiming quits after ~15 seconds on my machine. So it uses 4GB of RAM in 15 seconds, and from my tests would only keep climbing. This is with only 16 of your 18 equations. — ktm, Commented Mar 6, 2018 at 21:03
I would do what you can to reduce the number of variables/equations. — ktm, Commented Mar 6, 2018 at 21:10
Take a half of the system and find instance for it. Then verify all the system for these values. Then add to the half of the system the false equations/inequalities and find instance again and so on. — user64494, Commented Mar 6, 2018 at 21:19
Thank you. @user6014 I found that 4th inequality (with Min) is the game stopper, without it the solution comes within milliseconds. I was thinking that it would be easy because just yesterday I was running a highly non-linear polynomial approximation with 24 unknowns and CPU and memory use was about 30% and at the end of 4 hours all was solved. So it kills all intuition for why a system of linear equations-inequalities takes that much of memory/time. — Kass, Commented Mar 6, 2018 at 22:52

Bob Hanlon · Accepted Answer · 2018-05-06 18:39:25Z

eqns = {1/5 t12 + 1/5 t22 + 3/5 t32 == 0, 1/5 t1 + 1/5 t2 + 3/5 t3 == 0, 
    Min[1/5 t12 + 1/5 t22 + 3/5 t32, 1/5 t1 + 1/5 t2 + 3/5 t3] + 
      Min[1/5 t1 + 1/5 t4 + 3/5 t7, 1/5 t12 + 1/5 t42 + 3/5 t72] >= 
     Min[1/5 t42 + 1/5 t52 + 3/5 t62, 1/5 t4 + 1/5 t5 + 3/5 t6] + 
      Min[1/5 t2 + 1/5 t5 + 3/5 t8, 1/5 t22 + 1/5 t52 + 3/5 t82], 
    Min[1/5 t12 + 1/5 t22 + 3/5 t32, 1/5 t1 + 1/5 t2 + 3/5 t3] + 
      Min[1/5 t1 + 1/5 t4 + 3/5 t7, 1/5 t12 + 1/5 t42 + 3/5 t72] >= 
     Min[1/5 t72 + 1/5 t82 + 3/5 t92, 1/5 t7 + 1/5 t8 + 3/5 t9] + 
      Min[1/5 t3 + 1/5 t6 + 3/5 t9, 1/5 t32 + 1/5 t62 + 3/5 t92], 
    Min[1/5 t42 + 1/5 t52 + 3/5 t62, 1/5 t4 + 1/5 t5 + 3/5 t6] < 0, 
    Min[1/5 t72 + 1/5 t82 + 3/5 t92, 1/5 t7 + 1/5 t8 + 3/5 t9] < 0, 
    Min[1/2 (1/5 + 1/9) t12 + 1/2 (1/5 + 1/9) t22 + 1/2 (3/5 + 7/9) t32, 
      1/2 (1/5 + 1/9) t1 + 1/2 (1/5 + 1/9) t2 + 1/2 (3/5 + 7/9) t3] < 0, 
    1/2 (1/5 + 1/9) t42 + 1/2 (1/5 + 1/9) t52 + 1/2 (3/5 + 7/9) t62 == 0, 
    1/2 (1/5 + 1/9) t4 + 1/2 (1/5 + 1/9) t5 + 1/2 (3/5 + 7/9) t6 == 0, 
    Min[1/2 (1/5 + 1/9) t7 + 1/2 (1/5 + 1/9) t8 + 1/2 (3/5 + 7/9) t9, 
      1/2 (1/5 + 1/9) t72 + 1/2 (1/5 + 1/9) t82 + 1/2 (3/5 + 7/9) t92] < 0, 
    Min[1/9 t12 + 1/9 t22 + 7/9 t32, 1/9 t1 + 1/9 t2 + 7/9 t3] < 0, 
    Min[1/9 t42 + 1/9 t52 + 7/9 t62, 1/9 t4 + 1/9 t5 + 7/9 t6] < 0, 
    1/9 t72 + 1/9 t82 + 7/9 t92 == 0, 1/9 t7 + 1/9 t8 + 7/9 t9 == 0} // 
   Simplify;

Extract equations, i.e., remove inequalities

eqns2 = Cases[eqns, Equal[_, 0]]

(* {t12 + t22 + 3 t32 == 0, t1 + t2 + 3 t3 == 0, 7 t42 + 7 t52 + 31 t62 == 0, 
 7 t4 + 7 t5 + 31 t6 == 0, t72 + t82 + 7 t92 == 0, t7 + t8 + 7 t9 == 0} *)

n = Length[eqns2];

vars = Variables[Level[eqns, {-1}]];

Find a subset of vars that solves eqns2

varsn = Subsets[vars, {n}];

lvn = Length[varsn];

ptr = 0;
finished = False;
While[! (finished), ptr++; sol1 = Solve[eqns2, varsn[[ptr]]]; 
   finished = sol1 =!= {} || ptr == lvn] p;
sol1

(* {{t1 -> -t2 - 3 t3, t12 -> -t22 - 3 t32, t4 -> 1/7 (-7 t5 - 31 t6), 
  t42 -> 1/7 (-7 t52 - 31 t62), t7 -> -t8 - 7 t9, t72 -> -t82 - 7 t92}} *)

eqns2 = DeleteCases[eqns /. sol1[[1]], True];

vars2 = Variables[Level[eqns2, {-1}]];

sol2 = FindInstance[eqns2, vars2][[1]]

{t2 -> 0, t22 -> 0, t3 -> -1, t32 -> 0, t5 -> 0, t52 -> 0, t6 -> 1, t62 -> -1,
  t8 -> -(50/7), t82 -> 0, t9 -> 1, t92 -> 38/147}

sol = Join[sol2, sol1[[1]] /. sol2]

(* {t2 -> 0, t22 -> 0, t3 -> -1, t32 -> 0, t5 -> 0, t52 -> 0, t6 -> 1, t62 -> -1,
  t8 -> -(50/7), t82 -> 0, t9 -> 1, t92 -> 38/147, t1 -> 3, t12 -> 0, 
 t4 -> -(31/7), t42 -> 31/7, t7 -> 1/7, t72 -> -(38/21)} *)

Verifying that sol satisfies all equations and inequalities

And @@ (eqns /. sol)

(* True *)

Daniel Lichtblau · Accepted Answer · 2018-05-06 20:34:16Z

It can be handled by recasting the Min expressions as a convex combination of their constituents, with the added inequalities that the convex combination be less-equal to both constituents. The function below will assist in this transformation.

ineq[Min[a_, b_]] := 
 With[{nv = Unique[m]}, {nv[1]*a + nv[2]*b, {nv[1] >= 0, nv[2] >= 0, 
    nv[1] + nv[2] == 1, nv[1]*a + nv[2]*b <= a, 
    nv[1]*a + nv[2]*b <= b}}]

system = {1/5 t12 + 1/5 t22 + 3/5 t32 == 0, 
   1/5 t1 + 1/5 t2 + 3/5 t3 == 0, 
   Min[1/5 t12 + 1/5 t22 + 3/5 t32, 1/5 t1 + 1/5 t2 + 3/5 t3] + 
     Min[1/5 t1 + 1/5 t4 + 3/5 t7, 1/5 t12 + 1/5 t42 + 3/5 t72] >= 
    Min[1/5 t42 + 1/5 t52 + 3/5 t62, 1/5 t4 + 1/5 t5 + 3/5 t6] + 
     Min[1/5 t2 + 1/5 t5 + 3/5 t8, 1/5 t22 + 1/5 t52 + 3/5 t82], 
   Min[1/5 t12 + 1/5 t22 + 3/5 t32, 1/5 t1 + 1/5 t2 + 3/5 t3] + 
     Min[1/5 t1 + 1/5 t4 + 3/5 t7, 1/5 t12 + 1/5 t42 + 3/5 t72] >= 
    Min[1/5 t72 + 1/5 t82 + 3/5 t92, 1/5 t7 + 1/5 t8 + 3/5 t9] + 
     Min[1/5 t3 + 1/5 t6 + 3/5 t9, 1/5 t32 + 1/5 t62 + 3/5 t92], 
   Min[1/5 t42 + 1/5 t52 + 3/5 t62, 1/5 t4 + 1/5 t5 + 3/5 t6] < 0, 
   Min[1/5 t72 + 1/5 t82 + 3/5 t92, 1/5 t7 + 1/5 t8 + 3/5 t9] < 0, 
   Min[1/2 (1/5 + 1/9) t12 + 1/2 (1/5 + 1/9) t22 + 
      1/2 (3/5 + 7/9) t32, 
     1/2 (1/5 + 1/9) t1 + 1/2 (1/5 + 1/9) t2 + 1/2 (3/5 + 7/9) t3] < 
    0, 1/2 (1/5 + 1/9) t42 + 1/2 (1/5 + 1/9) t52 + 
     1/2 (3/5 + 7/9) t62 == 0, 
   1/2 (1/5 + 1/9) t4 + 1/2 (1/5 + 1/9) t5 + 1/2 (3/5 + 7/9) t6 == 0, 
   Min[1/2 (1/5 + 1/9) t7 + 1/2 (1/5 + 1/9) t8 + 1/2 (3/5 + 7/9) t9, 
     1/2 (1/5 + 1/9) t72 + 1/2 (1/5 + 1/9) t82 + 
      1/2 (3/5 + 7/9) t92] < 0, 
   Min[1/9 t12 + 1/9 t22 + 7/9 t32, 1/9 t1 + 1/9 t2 + 7/9 t3] < 0, 
   Min[1/9 t42 + 1/9 t52 + 7/9 t62, 1/9 t4 + 1/9 t5 + 7/9 t6] < 0, 
   1/9 t72 + 1/9 t82 + 7/9 t92 == 0, 1/9 t7 + 1/9 t8 + 7/9 t9 == 0};

Now rewrite the system by extracting and replacing all Min terms and augmenting with the new inequalities. Also extract the full set of variables,

mins = Cases[system, Verbatim[Min][__], Infinity];
newEqsAndIneqsMin = Map[ineq, mins];
{newEqsMin, newIneqsMin} = Transpose[newEqsAndIneqsMin ];
newsys = system /. Thread[mins -> newEqsMin];
fullsys = Flatten[{newsys, newIneqsMin}];
allVars = Variables[Apply[Subtract, fullsys, {1}]];

FindInstance will handle this just fine.

AbsoluteTiming[vals = FindInstance[fullsys, allVars]]

(* Out[372]= {0.493288, {{t1 -> 103/8, t12 -> 1, t2 -> -(55/8), 
   t22 -> -1, t3 -> -2, t32 -> 0, t4 -> 31/7, t42 -> -(31/7), t5 -> 0,
    t52 -> 0, t6 -> -1, t62 -> 1, t7 -> -7, t72 -> 0, t8 -> 0, 
   t82 -> 0, t9 -> 1, t92 -> 0, m$96625[1] -> 1, m$96625[2] -> 0, 
   m$96626[1] -> 1, m$96626[2] -> 0, m$96627[1] -> 0, m$96627[2] -> 1,
    m$96628[1] -> 1, m$96628[2] -> 0, m$96629[1] -> 1, 
   m$96629[2] -> 0, m$96630[1] -> 1, m$96630[2] -> 0, m$96631[1] -> 1,
    m$96631[2] -> 0, m$96632[1] -> 1, m$96632[2] -> 0, 
   m$96633[1] -> 0, m$96633[2] -> 1, m$96634[1] -> 1, m$96634[2] -> 0,
    m$96635[1] -> 1, m$96635[2] -> 0, m$96636[1] -> 1, 
   m$96636[2] -> 0, m$96637[1] -> 1, m$96637[2] -> 0, m$96638[1] -> 1,
    m$96638[2] -> 0}}} *)

Check:

In[373]:= system /. vals

(* Out[373]= {{True, True, True, True, True, True, True, True, True, 
  True, True, True, True, True}} *)

Akku14 · Accepted Answer · 2018-05-07 17:27:35Z

You get a full solution set and a lot of instance solutions, if you substitute for all Min[a,b], Min[c,d],... by all possible combinations of a,b,c,d,...

(eqs = {1/5 t12 + 1/5 t22 + 3/5 t32 == 0, 
1/5 t1 + 1/5 t2 + 3/5 t3 == 0, 
Min[1/5 t12 + 1/5 t22 + 3/5 t32, 1/5 t1 + 1/5 t2 + 3/5 t3] + 
  Min[1/5 t1 + 1/5 t4 + 3/5 t7, 1/5 t12 + 1/5 t42 + 3/5 t72] >= 
 Min[1/5 t42 + 1/5 t52 + 3/5 t62, 1/5 t4 + 1/5 t5 + 3/5 t6] + 
  Min[1/5 t2 + 1/5 t5 + 3/5 t8, 1/5 t22 + 1/5 t52 + 3/5 t82], 
Min[1/5 t12 + 1/5 t22 + 3/5 t32, 1/5 t1 + 1/5 t2 + 3/5 t3] + 
  Min[1/5 t1 + 1/5 t4 + 3/5 t7, 1/5 t12 + 1/5 t42 + 3/5 t72] >= 
 Min[1/5 t72 + 1/5 t82 + 3/5 t92, 1/5 t7 + 1/5 t8 + 3/5 t9] + 
  Min[1/5 t3 + 1/5 t6 + 3/5 t9, 1/5 t32 + 1/5 t62 + 3/5 t92], 
Min[1/5 t42 + 1/5 t52 + 3/5 t62, 1/5 t4 + 1/5 t5 + 3/5 t6] < 0, 
Min[1/5 t72 + 1/5 t82 + 3/5 t92, 1/5 t7 + 1/5 t8 + 3/5 t9] < 0, 
Min[1/2 (1/5 + 1/9) t12 + 1/2 (1/5 + 1/9) t22 + 
   1/2 (3/5 + 7/9) t32, 
  1/2 (1/5 + 1/9) t1 + 1/2 (1/5 + 1/9) t2 + 1/2 (3/5 + 7/9) t3] < 
 0, 1/2 (1/5 + 1/9) t42 + 1/2 (1/5 + 1/9) t52 + 
  1/2 (3/5 + 7/9) t62 == 0, 
1/2 (1/5 + 1/9) t4 + 1/2 (1/5 + 1/9) t5 + 1/2 (3/5 + 7/9) t6 == 0,
 Min[1/2 (1/5 + 1/9) t7 + 1/2 (1/5 + 1/9) t8 + 1/2 (3/5 + 7/9) t9,
   1/2 (1/5 + 1/9) t72 + 1/2 (1/5 + 1/9) t82 + 
   1/2 (3/5 + 7/9) t92] < 0, 
Min[1/9 t12 + 1/9 t22 + 7/9 t32, 1/9 t1 + 1/9 t2 + 7/9 t3] < 0, 
Min[1/9 t42 + 1/9 t52 + 7/9 t62, 1/9 t4 + 1/9 t5 + 7/9 t6] < 0, 
1/9 t72 + 1/9 t82 + 7/9 t92 == 0, 
1/9 t7 + 1/9 t8 + 7/9 t9 == 0}) // TableForm

First solve all equations an insert solutions

e1 = Cases[eqs, aa_ == 0]

sol1 = First@Solve[e1]

(*   {t1 -> -t2 - 3 t3, t12 -> -t22 - 3 t32, t4 -> -t5 - (31 t6)/7, 
      t42 -> -t52 - (31 t62)/7, t7 -> -t8 - 7 t9, t72 -> -t82 - 7 t92}   *)

param = {t12, t22, t32, t42, t52, t62, t72, t82, t92, t1, t2, t3, t4, 
         t5, t6, t7, t8, t9} // Sort

paramred = {t22, t32, t52, t62, t82, t92, t2, t3, t5, t6, t8, t9}

The remanining inequations

(eqs2 = DeleteCases[eqs /. sol1 // Simplify, True]) // TableForm

Prepare combinations for the first two inequations and get it

(perm = {{a < b, c < d, e < f, a >= c + e}, {b < a, c < d, e < f, 
 b >= c + e}, {a < b, d < c, e < f, a >= d + e}, {b < a, d < c, 
 e < f, b >= d + e}, {a < b, c < d, f < e, a >= c + f}, {b < a, 
 c < d, f < e, b >= c + f}, {a < b, d < c, f < e, 
 a >= d + f}, {b < a, d < c, f < e, b >= d + f}}) // TableForm

cas12 = Cases[eqs2[[1 ;; 2]], 
      Min[a_, b_] >= Min[c_, d_] + Min[e_, f_] -> (And @@ # & /@ perm), 
        3] // Simplify

cas12 // Dimensions

(*   {2, 8}   *)

ta = Table[
     cas12[[1, i]] && cas12[[2, i]], {i, 1, Length[cas12[[1]]]}] // 
       Simplify

Do the same for the remaning more simple structured six inequations

cas38 = Cases[eqs2[[3 ;; 8]], 
        Min[a_, b_] -> {{a < 0, a < b}, {b < 0, b < a}}, 3] // Simplify

(*   {{{t6 > 0, t62 < t6}, {t62 > 0, t6 < t62}}, {{t9 > 0, 
   t92 < t9}, {t92 > 0, t9 < t92}}, {{t3 < 0, t3 < t32}, {t32 < 0, 
   t32 < t3}}, {{t9 > 0, t92 < t9}, {t92 > 0, t9 < t92}}, {{t3 < 0, 
   t3 < t32}, {t32 < 0, t32 < t3}}, {{t6 < 0, t6 < t62}, {t62 < 0, 
   t62 < t6}}}   *)

seq = Sequence @@ Map[And @@ # &, cas38, {2}]

The 12 possible inequations reduce to 8

dc = DeleteCases[Flatten[Outer[And, seq], 5] // Simplify, False]

Now build all combinations all inequations, the invalid ones later evaluate to False. You get 16 combinations of inequations, that are all valid to give solutions.

out = Outer[And, ta, dc]

dout = DeleteCases[out // Flatten // Simplify, False]

Now FindInstance a lot of soutions. Use randomsample of paramed to not always get the same solution (here 80 different ones).

(ff = Flatten[Table[FindInstance[#, 
    RandomSample[paramred, Length[paramred]]] & /@ dout, {10}], 
      2]);

Dimensions /@ {ff, Union[ff]}

(*   {{80, 12}, {80, 12}}   *)

param /. sol1 /. ff // MatrixForm

Proove all of them to satisfy all eqs.

And @@ (And @@ eqs /. sol1 /. ff)

(*   True   *)

If you have a few minutes time, you get the full solution of this system set by

redList = 
   DeleteCases[Reduce[dout[[#]]] & /@ Range[Length[dout]] // Simplify, 
     False]

(*   A very large output was generated...   *)

user64494 · Accepted Answer · 2018-03-07 11:02:23Z

0

An answer is as follows.

{t12 -> 0, t22 -> 26/15, t32 -> -(26/45), t42 -> 0, t52 -> 217/18, t62 -> -(49/18), t72 -> 229/135, t82 -> -(1403/270), t92 -> 1/2, t1 -> 0, t2 -> 369/20, t3 -> -(123/20), t4 -> 0, t5 -> -(93/10), t6 -> 21/10, t7 -> -(8/5), t8 -> -(73/20), t9 -> 3/4}

I solved

{Min[1/5 t12 + 1/5 t22 + 3/5 t32, 1/5 t1 + 1/5 t2 + 3/5 t3] + 
Min[1/5 t1 + 1/5 t4 + 3/5 t7, 1/5 t12 + 1/5 t42 + 3/5 t72] >= 
Min[1/5 t42 + 1/5 t52 + 3/5 t62, 1/5 t4 + 1/5 t5 + 3/5 t6] + 
Min[ 1/5 t2 + 1/5 t5 + 3/5 t8, 1/5 t22 + 1/5 t52 + 3/5 t82],
Min[1/5 t12 + 1/5 t22 + 3/5 t32, 1/5 t1 + 1/5 t2 + 3/5 t3] + 
Min[1/5 t1 + 1/5 t4 + 3/5 t7, 1/5 t12 + 1/5 t42 + 3/5 t72] >= 
Min[1/5 t72 + 1/5 t82 + 3/5 t92, 1/5 t7 + 1/5 t8 + 3/5 t9] + 
Min[ 1/5 t3 + 1/5 t6 + 3/5 t9, 1/5 t32 + 1/5 t62 + 3/5 t92]}

with Maple, obtaining, in particular,

{ t8 <= 2*t1*(1/3)+t3+t7-2*t5*(1/3)-t6, t9 <= (1/3)*t1+(1/6)*t2+(1/3)*t3+(1/6)*t4+(1/3)*t7-(1/6)*t8-(1/6)*t6, -(1/3)*t12+(1/3)*t1-(1/3)*t22+(1/3)*t2+t3 <= t32, -(1/3)*t22-(1/3)*t52+(1/3)*t2+(1/3)*t5+t8 <= t82, -(1/3)*t32-(1/3)*t62+(1/3)*t3+(1/3)*t6+t9 <= t92, -(1/3)*t42-(1/3)*t52+(1/3)*t4+(1/3)*t5+t6 <= t62, t6 < -(1/3)*t4-(1/3)*t5, (1/3)*t22+(1/3)*t52-(1/3)*t2-(1/3)*t5+t32+t62-t3-t6+t7 < t72}

The above expression was subsituted in the original system instead of the two complicated inequalities and FindInstance did the rest of the job.

edited Mar 7, 2018 at 11:02

answered Mar 7, 2018 at 10:02

user64494

29.1k4 gold badges29 silver badges56 bronze badges

$\begingroup$ Seems that Mathematica will not manage to Reduce the inequalities? It does run forever... $\endgroup$
– gwr
Commented Mar 7, 2018 at 11:38
$\begingroup$ @gwr: I think Min commands cause problems. $\endgroup$
– user64494
Commented Mar 7, 2018 at 11:56
$\begingroup$ I would argue that "use maple to do half of my solution" is not an answer in a Mathematica forum. $\endgroup$
– ktm
Commented Mar 7, 2018 at 14:21
$\begingroup$ @user6014: I could not solve it in Mathematica because the kernel connection was lost. $\endgroup$
– user64494
Commented Mar 7, 2018 at 17:46
$\begingroup$ @user64494 Interesting, thank you! It clarifies at least what Mathematica has to work out for their future versions. I will wait a bit if there is no other suggestion as how to deal with the issue for those who have no Maple (me) and then I will accept your solution. $\endgroup$
– Kass
Commented Mar 7, 2018 at 20:26

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