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dimension reduction - Linear map of unit cube



Let c be a vector of n constants and X = Table[Symbol["x" <> ToString[i]], {i, n}] be a vector of n variables.


Assuming that 0 <= X <= 1, I would like to compute the linear map c.X. This can be expressed with Reduce:


formula = Exists[Evaluate[X], y == c.X && Fold[#1 && 0 <= #2 <= 1 &, True, X]];
Reduce[formula, {y}, Reals]

With increasing n, the performance of this solution drops rapidly. Is there a better way to compute the solution y?



Answer



This is a typical linear programming problem. For a given vector c you can quickly find the minimum and maximum values of y that allow a solution X.


Example for n=10:


n = 10;

c = RandomVariate[NormalDistribution[], n] (* random, to have an example *)
(* {0.105518, -0.439788, -0.33041, -1.93738, -0.810956, 1.43839, -0.529671, 0.668688, 0.879621, 0.458327} *)
X = Array[x, n]
(* {x[1], x[2], x[3], x[4], x[5], x[6], x[7], x[8], x[9], x[10]} *)
Minimize[{c.X, And @@ Thread[0 <= X <= 1]}, X]
(* {-4.0482, {x[1] -> 0., x[2] -> 1., x[3] -> 1., x[4] -> 1., x[5] -> 1.,
x[6] -> 0., x[7] -> 1., x[8] -> 0., x[9] -> 0., x[10] -> 0.}} *)
Maximize[{c.X, And @@ Thread[0 <= X <= 1]}, X]
(* {3.55054, {x[1] -> 1., x[2] -> 0., x[3] -> 0., x[4] -> 0., x[5] -> 0.,
x[6] -> 1., x[7] -> 0., x[8] -> 1., x[9] -> 1., x[10] -> 1.}} *)


So for this particular choice of the vector c there exists solutions X if −4.0482≤y≤3.55054. The convexity of the problem assures that all values of y in this range give rise to solutions X.


The above code is fast and executes in about one second for n=104. For ultimate speed, call the LinearProgramming function directly: you get the same results, but much faster:


X1 = LinearProgramming[c,
-IdentityMatrix[n, SparseArray],
ConstantArray[-1, n]];
X1.c
(* -4.0482 *)
X2 = LinearProgramming[-c,
-IdentityMatrix[n, SparseArray],

ConstantArray[-1, n]];
X2.c
(* 3.55054 *)

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