Maximizing number of positive elements in formatted output

Question

The following code

$Assumptions = {0 < x < 1, 0 < y < 1, z > 1};
myexpr = Expand[(1 - x) (1 - y) (z - 1)];
TraditionalForm[Simplify[myexpr]]

outputs

(x-1)(y-1)(z-1)

but I want to output

(1-x)(1-y)(z-1)

i.e., I want every block in my expression to be a positive number. If the expression itself is negative, that should be shown by a minus in front of it. (The $Assumptions guarantee that Mathematica knows what it needs to know to make sure every block is positive).

Is it possible?

I found this answer but it doesn't quite cut it because (a) it's way too cumbersome and (b) in my case the expression to be reformatted comes out of a solver and I have no control over it.

Answer 1

Here is one idea. Create a function that converts products into the desired form. This function will test each multiplicand to see if it is definitely negative, and if so, negate it, keeping track of how many terms were negated. I also added support for integer powers:

Options[positiveMultiplicands] = {Assumptions->$Assumptions};

positiveMultiplicands[a_Times, OptionsPattern[]] := Block[
    {$Assumptions = OptionValue[Assumptions]},

    Times @@ Replace[
        a,
        {
            Power[t_, e_Integer] :> If[Simplify[t>0], 
                {1, Power[t, e]},
                {(-1)^e, Power[-t, e]}, 
                {1, Power[t, e]}
            ],
            t_ :> If[Simplify[t>0], {1, t}, {-1, -t}, {1, t}]
        },
        {1}
    ]
]

Your example:

positiveMultiplicands[
    (x - 1) (y - 1) (z - 1),
    Assumptions -> x < 1 && y < 1 && z > 1
]

(1 - x) (1 - y) (-1 + z)

The example from the linked question:

positiveMultiplicands[(-2+a) y/(-1+a), Assumptions->a<1]

((2 - a) y)/(1 - a)

And finally, an example with powers:

positiveMultiplicands[(2-a)^3/(1-a)^2, Assumptions->a>2]

-((-2 + a)^3/(-1 + a)^2)

Answer 2

$Assumptions = {0 < x < 1, 0 < y < 1, z > 1};

Define a helper function to test whether an expression is negative assuming $Assumptions

negQ[expr_] := Simplify[expr < 0];

f[expr_] := Module[{expr2 = expr // Simplify},
  (-1)^Count[expr2, _?negQ] *(If[negQ[#], -#, #, #] & /@ expr2)]

EDIT: changed If statement in definition of f to handle undecided condition

myexpr = Expand[(1 - x) (1 - y) (z - 1)];

f[myexpr] // TraditionalForm

(* (1-x) (1-y) (z-1) *)

EDIT: With assumption z > 1 changed to z > 0

$Assumptions = {0 < x < 1, 0 < y < 1, z > 0};

f[myexpr] // TraditionalForm

(* (1-x) (1-y) (z-1) *)