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updating interactive simplex method to support other fields #41

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May 24, 2024
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updating interactive simplex method to support other fields #41

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36 changes: 31 additions & 5 deletions src/sage/numerical/interactive_simplex_method.py
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Original file line number Diff line number Diff line change
Expand Up @@ -300,7 +300,11 @@ def _latex_product(coefficients, variables,
sage: var("x, y") # needs sage.symbolic
(x, y)
sage: print(_latex_product([-1, 3], [x, y])) # needs sage.symbolic
- \mspace{-6mu}&\mspace{-6mu} 1 x \mspace{-6mu}&\mspace{-6mu} + \mspace{-6mu}&\mspace{-6mu} 3 y
- \mspace{-6mu}&\mspace{-6mu} x \mspace{-6mu}&\mspace{-6mu} + \mspace{-6mu}&\mspace{-6mu} 3 y
sage: var("x, y, z")
(x, y, z)
sage: print(_latex_product([-pi, log(2), pi**0], [x, y, z]))
- \mspace{-6mu}&\mspace{-6mu} \pi x \mspace{-6mu}&\mspace{-6mu} + \mspace{-6mu}&\mspace{-6mu} \log\left(2\right) y \mspace{-6mu}&\mspace{-6mu} + \mspace{-6mu}&\mspace{-6mu} z
"""
entries = []
for c, v in zip(coefficients, variables):
Expand All @@ -312,7 +316,7 @@ def _latex_product(coefficients, variables,
if t.strip().startswith("-"):
sign = "-"
c = - c
if t.strip() == '1':
if t.strip() == '1' or t.strip() == '-1':
t = latex(v)
else:
t = latex(c)
Expand Down Expand Up @@ -673,7 +677,7 @@ def __init__(self, A, b, c, x="x",
R = PolynomialRing(base_ring, x, order="neglex")
x = vector(R, R.gens()) # All variables as a vector
self._Abcx = A, b, c, x
self._constant_term = objective_constant_term
self._constant_term = base_ring(objective_constant_term)

if constraint_type in ["<=", ">=", "=="]:
constraint_type = (constraint_type, ) * m
Expand Down Expand Up @@ -2693,6 +2697,28 @@ def run_simplex_method(self):
Entering: $x_{2}$. Leaving: $x_{3}$.
...
The optimal value: $6250.0$. An optimal solution: $\left(249.99999999999997,\,750.0\right)$.
sage: A = Matrix(([1, 1], [3, 1], [-1, -1])) * pi
sage: b = vector((1000, 1500, -400)) * pi
sage: c = vector((10, 5)) * pi
sage: P = InteractiveLPProblemStandardForm(A, b, c)
sage: P.run_simplex_method()
\begin{equation*}
...
\end{equation*}
The initial dictionary is infeasible, solving auxiliary problem.
...
Entering: $x_{0}$. Leaving: $x_{5}$.
...
Entering: $x_{1}$. Leaving: $x_{0}$.
...
Back to the original problem.
...
Entering: $x_{5}$. Leaving: $x_{4}$.
...
Entering: $x_{2}$. Leaving: $x_{3}$.
...
The optimal value: $6250 \, \pi$. An optimal solution: $\left(250,\,750\right)$.

"""
output = []
d = self.initial_dictionary()
Expand Down Expand Up @@ -4042,8 +4068,8 @@ def _latex_(self):
\renewcommand{\arraystretch}{1.5} %notruncate
\begin{array}{|rcrcrcr|}
\hline
x_{3} \mspace{-6mu}&\mspace{-6mu} = \mspace{-6mu}&\mspace{-6mu} 1000 \mspace{-6mu}&\mspace{-6mu} - \mspace{-6mu}&\mspace{-6mu} 1 x_{1} \mspace{-6mu}&\mspace{-6mu} - \mspace{-6mu}&\mspace{-6mu} 1 x_{2}\\
x_{4} \mspace{-6mu}&\mspace{-6mu} = \mspace{-6mu}&\mspace{-6mu} 1500 \mspace{-6mu}&\mspace{-6mu} - \mspace{-6mu}&\mspace{-6mu} 3 x_{1} \mspace{-6mu}&\mspace{-6mu} - \mspace{-6mu}&\mspace{-6mu} 1 x_{2}\\
x_{3} \mspace{-6mu}&\mspace{-6mu} = \mspace{-6mu}&\mspace{-6mu} 1000 \mspace{-6mu}&\mspace{-6mu} - \mspace{-6mu}&\mspace{-6mu} x_{1} \mspace{-6mu}&\mspace{-6mu} - \mspace{-6mu}&\mspace{-6mu} x_{2}\\
x_{4} \mspace{-6mu}&\mspace{-6mu} = \mspace{-6mu}&\mspace{-6mu} 1500 \mspace{-6mu}&\mspace{-6mu} - \mspace{-6mu}&\mspace{-6mu} 3 x_{1} \mspace{-6mu}&\mspace{-6mu} - \mspace{-6mu}&\mspace{-6mu} x_{2}\\
\hline
z \mspace{-6mu}&\mspace{-6mu} = \mspace{-6mu}&\mspace{-6mu} 0 \mspace{-6mu}&\mspace{-6mu} + \mspace{-6mu}&\mspace{-6mu} 10 x_{1} \mspace{-6mu}&\mspace{-6mu} + \mspace{-6mu}&\mspace{-6mu} 5 x_{2}\\
\hline
Expand Down
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