def solve(A, B, C, firstBase, invert=[], auxiliar=False, iniVal=0):
n, m = A.shape
A = tableau(A, B, C, invert, iniVal)
baseCols = []
for row, col in firstBase: # Compensating VERO
baseCols.append([row + 1, col + n])
# Canonize for base
for row, col in baseCols:
if not np.isclose(A[0][col], 0):
pivot(A, row, col)
while (1):
# Se cN < 0, X é ótima
if (not less(np.min(A[0][n:-1]), 0)):
value = A[0][-1]
cert = A[0][:n]
base = recoverBase(A, np.array(baseCols), n, m)
return [value, base, cert, A, baseCols]
# Find an entry smaller than 0
for j in range(n, A.shape[1]):
if less(A[0][j], 0):
col = j
break
# Se Ak < 0, A PL é ilimitada.
if (less(np.max(A[:, col][1:]), 0)):
base = recoverBase(A, np.array(baseCols), n, m)
cert = recoverIli(A, np.array(baseCols), col, n, m)
return [base, cert, A, baseCols]
# Iterate over column and find Index
minRatio = float('inf')
row = -1
start = (2 if auxiliar else 1)
for i in range(start, A.shape[0]):
if (np.isclose(A[i][col], 0) or less(A[i][col], 0)): continue
if (less(A[i][-1] / A[i][col], minRatio)):
row = i
minRatio = A[i][-1] / A[i][col]
# Sanity Check
assert (row > 0)
# Finds Col to be replaced
for i in range(len(baseCols)):
if baseCols[i][0] == row:
baseCols.pop(i)
break
# Adds new Col
baseCols.append([row, col])
# Pivot
pivot(A, row, col)
def recoverBase(A, baseCols, n, m, compensate=True):
base = np.zeros(m)
for row, col in baseCols:
if compensate: colP = col - n # Compensating for VERO columns
else: colP = col
# Sanity check
assert (not less(1, A[:, col].sum()))
base[colP] = A[row][-1]
return base
def auxiliar(A, B, C, n, m):
invert = []
# Making B positive
for i in range(B.shape[0]):
if less(B[i], 0):
B[i] = -B[i]
A[i] = -A[i]
invert.append(i + 1)
AAux = np.append(A, np.identity(A.shape[0]), axis=1)
BAux = np.append(0, B)
CAux = np.append(np.zeros(A.shape[1]), np.full(A.shape[0], -1))
CAlong = np.append(-C, np.zeros(A.shape[0]))
AAux = np.vstack([CAlong, AAux])
baseAux = [[i + 1, i + n + m] for i in range(n)]
ans = solve(AAux, BAux, CAux, baseAux, invert=invert, auxiliar=True)
return ans
def tableau(A, B, C, invert=[], iniVal=0):
# Base for VEROtm
VERO = np.identity(A.shape[0])
# Inverting
for i in invert:
VERO[i] = -VERO[i]
VERO = np.vstack([np.zeros(A.shape[0]), VERO])
# Add restriction values
A = np.append(A, np.array([B]).T, axis=1)
# Add cost in A
C = -C
C = np.append(C, np.zeros(A.shape[1] - C.shape[0]))
A = np.vstack([C, A])
# Append VERO
A = np.append(VERO, A, axis=1)
if less(0, iniVal):
A[0][-1] = iniVal
return A
import numpy as np
from utils import less
from utils import matprint
import simplex
from simplex import solve
n, m, A, B, C = simplex.read()
A, B, C = simplex.fpi(A, B, C)
if (less(B.min(), 0)):
ans = simplex.auxiliar(A, B, C, n, m)
# Unviable
if (less(ans[0], 0)):
print("inviavel")
for i in ans[2][1:n + 1]:
print(i, end=' ')
print()
exit()
# Transforms Auxiliar Back
finalA = ans[3].copy()
finalA = np.delete(finalA, [0], axis=1)
finalA = np.delete(finalA, [0], axis=0)
finalA = np.delete(finalA, slice(A.shape[0] + A.shape[1], -1), axis=1)
finalBase = [[x - 1, y - 1] for x, y in ans[-1]]
ans = simplex.solveTableau(finalA, n, m, finalBase)
if (len(ans) == 5):
print("otima")
print(ans[0])