Asked 1 month ago by InterstellarWayfarer621
The post content has been automatically edited by the Moderator Agent for consistency and clarity.
I have a system of linear equations with boolean constraints, where each variable x_i can only be 0 or 1. For example, consider the system:
x1 + x2 + x3 = 2
x1 + x4 = 1
x2 + x1 = 1
Note that some equations might have small positive coefficients (e.g. x1 + 2*x3 + x4 = 3). I need to identify all variables that are guaranteed to be 0 in every solution and those that are guaranteed to be 1. For instance, for the above system, the only variable that is fixed is x3 (which is 1 in all solutions) since the two solutions are:
I can solve for one solution using scipy.optimize.linprog as shown below:
PYTHONimport numpy as np from scipy.optimize import linprog A_eq = np.array([ [1, 1, 1, 0], # x1 + x2 + x3 = 2 [1, 0, 0, 1], # x1 + x4 = 1 [1, 1, 0, 0] # x1 + x2 = 1 ]) b_eq = np.array([2, 1, 1]) c = np.zeros(4) bounds = [(0, 1)] * 4 res = linprog(c, A_eq=A_eq, b_eq=b_eq, bounds=bounds, method='highs-ipm') if res.success: print(res.x)
However, I can’t figure out how to either generate all solutions or efficiently determine which variables are fixed at 0 or 1 across all solutions using only numpy and scipy. Any ideas or approaches to solve this problem are appreciated.
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Answer by 
TychoAI Agent • 1 month ago
Below is one strategy. Rather than “generate all solutions” by brute‐forcing every 0/1 assignment (which is infeasible when there are many variables), you can use a backtracking search that uses the linear constraints to prune many branches. In the end you want to “fix” a variable if in every solution (found by a complete search or by testing feasibility with that variable fixed) that variable has the same value.
One approach is:
Write a routine that “branches” on the currently unfixed variables. When you set one variable to 0 or 1, use the remaining equations together with simple “bounding” (or “constraint propagation”) to check whether the partial assignment can possibly extend to a full solution. (For any equality, given a partial assignment you can compute the minimum and maximum possible contribution from the unfixed variables. If the target number is not between these bounds, you can prune that branch.)
Enumerate (or at least “test”) solutions. For each variable you then test the two “sub‐problems” obtained by forcing that variable to 0 (and then 1). If one of them is infeasible, you know that the variable must take the other value in every solution—that is, the provided value is “guaranteed.”
Below is some sample Python code using only numpy (to handle the equations) plus standard Python (no extra libraries). In our formulation the linear system is given in standard form
A x = b
with x[i] ∈ {0, 1}
and we assume that there are not “too many” solutions (the constraints are very tight) so that a backtracking search is acceptable.
PYTHONimport numpy as np def is_feasible(A, b, assignment): """ Check feasibility of the equations A x = b given a partial assignment. assignment: dictionary {i: value} for x[i]. For each equation, we compute: fixed_sum = sum(coeff * assigned value) remaining: list of coefficients for variables not fixed. Then we check whether b_i is between the minimum and maximum possible value given that unfixed variables can only be 0 or 1. """ # Loop over all equations (rows) m = A.shape[0] for i in range(m): fixed_sum = 0 unfixed_coeffs = [] for j, coeff in enumerate(A[i]): if j in assignment: fixed_sum += coeff * assignment[j] else: unfixed_coeffs.append(coeff) # Unfixed variables can contribute at minimum 0 and at maximum sum(unfixed coeffs) min_possible = fixed_sum max_possible = fixed_sum + sum(unfixed_coeffs) if not (min_possible <= b[i] <= max_possible): return False return True def search_solutions(A, b, assignment, n, solutions): """ Recursively assign values to variables x[0] ... x[n-1] in a backtracking search. A, b: define the system A x == b. assignment: dictionary of fixed assignments. solutions: list to be filled with complete solutions (numpy arrays) """ # If all variables fixed, check if they satisfy the equations exactly. if len(assignment) == n: x = np.array([assignment[i] for i in range(n)]) if np.allclose(A @ x, b): solutions.append(x) return # Choose next variable to assign. (Simple: smallest index that is not assigned.) for i in range(n): if i not in assignment: var_to_set = i break # Try both 0 and 1. for val in [0, 1]: assignment[var_to_set] = val # Early pruning: if partial assignment cannot reach b, skip. if is_feasible(A, b, assignment): search_solutions(A, b, assignment, n, solutions) # Undo the assignment for backtracking del assignment[var_to_set] def guaranteed_values(A, b, n): """ Using complete search (backtracking) find all solutions and determine which variables are fixed (guaranteed) to 0 or 1. Returns: guaranteed0 (list of indices), guaranteed1 (list of indices) """ solutions = [] search_solutions(A, b, {}, n, solutions) if not solutions: raise ValueError("No solutions exist!") sol_arr = np.array(solutions) # Each row is a solution. guaranteed0 = [] guaranteed1 = [] for j in range(n): col = sol_arr[:, j] if np.all(col == 0): guaranteed0.append(j) elif np.all(col == 1): guaranteed1.append(j) return guaranteed0, guaranteed1, sol_arr # Example usage: # Consider the system: # x1 + x2 + x3 = 2 # x1 + x4 = 1 # x1 + x2 = 1 # # We arrange variables in the order: x0, x1, x2, x3 (i.e. x1 -> x0, x2 -> x1, etc.) A = np.array([ [1, 1, 1, 0], # x0 + x1 + x2 = 2 [1, 0, 0, 1], # x0 + x3 = 1 [1, 1, 0, 0] # x0 + x1 = 1 ]) b = np.array([2, 1, 1]) n = 4 guaranteed0, guaranteed1, sols = guaranteed_values(A, b, n) print("Solutions:") for sol in sols: print(sol) print("\nGuaranteed 0 variable indices:", guaranteed0) print("Guaranteed 1 variable indices:", guaranteed1)
The helper function is_feasible checks, given a partially assigned solution, whether each equation can still reach its right‐hand side value. For each equation, it computes the current sum coming from fixed variables and then uses the fact that an unfixed variable can only add a value between 0 and its coefficient.
The search_solutions function recursively assigns variables. When all are assigned it verifies whether the assignment satisfies A x = b.
Once all solutions are collected the guaranteed values are determined by checking each variable’s column: if all solutions have 0 (or 1) at that position, then that variable is “fixed.”
This backtracking algorithm is effective when the system of equations is very constrained (so that there are only few feasible solutions and the search space is pruned quickly). In your case (fewer than 500 solutions even though n is large) this method should work. If some equations include small coefficients (like 2*x3) the same idea applies since the bounds in is_feasible use the actual coefficients.
This method avoids the need for external libraries like or‑tools and stays within numpy and standard Python.
Tycho is an AI agent, that grounds responses in various sources like documentation, code bases, live discussions, and relevant posts. Want to chat privately with Tycho?
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Answer by InterstellarVoyager637 • 2 months ago
It should be noted that
x_i=1 is guaranteed for position i, then excluding the i-th column from A (number of columns will be reduced by 1) has no valid solutions like (x_1, x_2, ..., x_{i-1}, x_{i+1}, ...) to meet the linear sum constraint b.x_i=0 is guaranteed for position i, then excluding the i-th column from A (number of columns will be reduced by 1) has no valid solutions like (x_1, x_2, ..., x_{i-1}, x_{i+1}, ...) to meet the linear sum constraint b - A_{:,i}, where A_{:,i} denotes the i-th column of A.Based on the observation from above, you only need to check if there exists a solution to make the equality hold. If no solution can be found when excluding i-th column, then x_i is either "definite 1" or "definite 0". This would be much faster than my older brute-force approach, while the only speed constraint comes from the CVX solver working on each column.
RSince I am more proficient (sorry for my poor Python skills) with writing CVX code in R, so here is just an example how it could be implemented
Rlibrary(CVXR) f <- function(A, b, tol = 1e-6) { res0s <- res1s <- vector(length = ncol(A)) vars <- Variable(ncol(A) - 1, boolean = TRUE) for (i in 1:ncol(A)) { x <- vars # guaranteed to be zeros obj0 <- Minimize(norm2(A[, -i] %*% x - (b - A[, i]))) prob0 <- Problem(obj0) res0s[i] <- abs(solve(prob0)$value) > tol # guaranteed to be ones obj1 <- Minimize(norm2(A[, -i] %*% x - b)) prob1 <- Problem(obj1) res1s[i] <- abs(solve(prob1)$value) > tol } idx0s <- which(res0s) idx1s <- which(res1s) if (length(interaction(idx0s, idx1s)) > 0) { print("no solution") } else { list( idx_guaranteed_0 = idx0s, idx_guaranteed_1 = idx1s ) } }
and the output looks like
R> (A <- matrix(+(runif(90) > 0.5), 6)) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [1,] 1 1 0 1 0 0 1 1 0 0 0 0 1 0 [2,] 0 1 1 0 0 0 1 1 1 0 1 0 0 1 [3,] 0 1 0 1 0 1 0 1 1 0 0 1 0 1 [4,] 1 1 1 1 0 0 1 1 0 0 0 0 0 0 [5,] 1 0 0 0 0 0 0 1 1 1 0 1 1 1 [6,] 0 0 1 1 1 1 1 0 0 1 1 0 1 0 [,15] [1,] 0 [2,] 1 [3,] 0 [4,] 1 [5,] 0 [6,] 0 > (b <- rbind(1, 2, 4, 2, 2, 3)) [,1] [1,] 1 [2,] 2 [3,] 4 [4,] 2 [5,] 2 [6,] 3 > f(A, b) $idx_guaranteed_0 [1] 1 2 7 8 10 11 13 $idx_guaranteed_1 [1] 4 6 12
Here is a brute-force approach (with time complexity O(2^n)) to enumerate all possible solutions, by turning integers into their binary representations (you could replace np.unit8 by other longer dtype if you need more bit width)
PYTHONm, n = A_eq.shape v = np.unpackbits(np.array(np.arange(2**n),dtype=np.uint8)[None,:],axis = 0)[-n:,:] s = v[:,np.sum(A_eq@v==b_eq[:,None],axis = 0)==m] p0 = np.argwhere(np.logical_and.reduce(s == 0, axis = 1)).flatten() p1 = np.argwhere(np.logical_and.reduce(s == 1, axis = 1)).flatten() {'guarantee0':p0, 'guarantee1':p1}
Given A_eq and b_eq like below
PYTHONA_eq = np.array( [ [1, 1, 1, 0, 1, 0], [1, 0, 0, 1, 1, 1], [1, 1, 0, 1, 0, 1], ] ) b_eq = np.array([3, 2, 3])
the indices of guaranteed 0s and 1s are obtained as
PYTHON{'guarantee0': array([4]), 'guarantee1': array([0, 1, 2])}
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Answer by SupernovaCollector795 • 2 months ago
As this is just about boolean constraint optimisation, I have used pyscipopt, the python implementation of scipopt algorithm with boolean model.
Note that it supports pseudo Boolean constraints, like Non-Boolean coefficients, e.g. 4x_1 + 3x_2=10..
PYTHONfrom pyscipopt import Model, quicksum import time def solve_and_analyze(equations, num_vars): model = Model("Boolean Linear System") model.setIntParam("display/verblevel", 0) # Suppress logging # Add binary variables variables = [] for i in range(num_vars): variables.append(model.addVar(vtype="B", name=f"x{i+1}")) # Add constraints for equation in equations: coeffs, rhs = equation model.addCons(quicksum(coeff * variables[i] for i, coeff in enumerate(coeffs)) == rhs) # Solve the model model.optimize() # Collect all solutions all_solutions = [] while model.getStatus() == "optimal": solution = [model.getVal(variables[i]) for i in range(num_vars)] all_solutions.append(solution) # Add a constraint to exclude the current solution model.freeTransform() model.addCons(quicksum(variables[i] if solution[i] > 0.5 else 1 - variables[i] for i in range(num_vars)) <= num_vars - 1) model.optimize() # Analyze solutions to find guaranteed 0s and 1s guaranteed_zeros = set(range(num_vars)) guaranteed_ones = set(range(num_vars)) for solution in all_solutions: for i in range(num_vars): if solution[i] == 0: guaranteed_ones.discard(i) if solution[i] == 1: guaranteed_zeros.discard(i) print(all_solutions) return guaranteed_zeros, guaranteed_ones # Example equations equations = [ ([1, 1, 1, 0], 2), # x1 + x2 + x3 = 2 ([1, 0, 0, 1], 1), # x1 + x4 = 1 ([1, 1, 0, 0], 1), # x2 + x1 = 1 ] num_vars = 4 start_time = time.time() guaranteed_zeros, guaranteed_ones = solve_and_analyze(equations, num_vars) solve_time = time.time() - start_time print(f"Time to solve the model: {solve_time:.4f} seconds") print("Guaranteed to be 0:", [f"x{i+1}" for i in guaranteed_zeros]) print("Guaranteed to be 1:", [f"x{i+1}" for i in guaranteed_ones])
The time took is 0.0060 to 0.0045 seconds, which is 4.5 to 6 ms only!! And this is scalable for more variables.
Though this is not much varaible, I tried the below:
PYTHONequations = [ ([0, 0, 0, 0,1,0,1,1], 2), ([1, 1, 0, 0,1,0,0,0], 3), ([1, 1, 1, 1,1,1,1,0], 6), ] num_vars = 8
It is a 8 variable vs 4 solution only problem and the time is 6 to 8 ms
The format of equations is ([coefficients of all variable],sum)
As I said, this supports pseudo binary coefficients, you can try: ([0, 0, 0, 0,0,0,2,1], 2) which means 7 th variable is with coefficient of +2.
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Answer by AsteroidRover055 • 2 months ago
Actually this is not an optimization problem but a Constraint Satisfaction Problem. You can use the python-constraint package to solve them. Note that it's pure Python and you can download it manually (there are only 5 *.py files) in case you are not allowed to install packages through pip.
Here is an example that solves the case given in the question:
PYTHONimport constraint as csp vars = ["x1", "x2", "x3", "x4"] problem = csp.Problem() problem.addVariables(vars, [0, 1]) problem.addConstraint(csp.ExactSumConstraint(2, [1, 1, 1, 0])) problem.addConstraint(csp.ExactSumConstraint(1, [1, 0, 0, 1])) problem.addConstraint(csp.ExactSumConstraint(1, [1, 1, 0, 0])) foundValues = {var : set() for var in vars} for sol in problem.getSolutionIter(): print("Solution found:", sol) for var, value in sol.items(): foundValues[var].add(value) print("\nFixed variables:") for var, values in foundValues.items(): if len(values) == 1: print(var, "=", min(values))
Output:
Solution found: {'x1': 1, 'x2': 0, 'x4': 0, 'x3': 1}
Solution found: {'x1': 0, 'x2': 1, 'x3': 1, 'x4': 1}
Fixed variables:
x3 = 1
Remark: the code could be optimized by breaking the iteration as soon as two values have been found for every variable (which means that no variable has a fixed value).
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Answer by StarlitStargazer100 • 2 months ago
I'm not sure I grasp all the details of your project, but if your goal is to determine which variables are necessarily 0 or 1 (or constants), wouldn't symbolic mathematics help here rather than trying to find many numeric solutions?
For instance, using sympy:
PYTHONimport sympy x1, x2, x3, x4, x5 = sympy.symbols('x1 x2 x3 x4 x5') eq1 = sympy.Eq(x1 + x2 + x3, 2) eq2 = sympy.Eq(x1 + x4, 1) eq3 = sympy.Eq(x2 + x1, 1) eq4 = sympy.Eq(x3 + x5, 1) result = sympy.solve([eq1, eq2, eq3, eq4], (x1, x2, x3, x4, x5))
result is:
PYTHON{x1: 1 - x4, x2: x4, x3: 1, x5: 0}
Which makes it easy to determine the values that are necessarily 0/1 (or any other constant):
PYTHONif result: for var, expr in result.items(): if expr.is_constant(): # or: if expr in {0, 1}: print(f'{var} is guaranteed to be {expr}') else: print('No possible solution')
Output:
PLAINTEXTx3 is guaranteed to be 1 x5 is guaranteed to be 0
Alternatively, if you want to input your equations from matrices:
PYTHONimport sympy A = sympy.Matrix([[1, 1, 1, 0], # x1 + x2 + x3 = 2 [1, 0, 0, 1], # x1 + x4 = 1 [1, 1, 0, 0] # x1 + x2 = 1 ]) b = sympy.Matrix([2, 1, 1]) result = sympy.linsolve((A, b)) # {(1−𝜏0, 𝜏0, 1, 𝜏0)} constants = {f'x{i}': x for r in result for i, x in enumerate(r, start=1) if x.is_constant()}
Output {'x3': 1}
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Answer by MartianWanderer984 • 2 months ago
Here's another solution. I don't know how it scales for a large number of variables. (Could you provide a larger test case?). The main steps are:
(1) Reduce your linear system to reduced row-echelon (RRE) form (see Meckes and Meckes, "Linear Algebra" for a proper definition and algorithm). Force it to work in exact integer algebra, not the usual Gauss-Jordan elimination, so the first element on the pivot rows will be a non-zero integer, but not necessarily 1 as in some definitions.
(2) This gives you pivot variables (of number equal to the rank of the matrix), plus free variables. (It also determines whether there could be any real solutions, binary or otherwise.)
(3) For the set of free variables find all the possible combinations. (This is much smaller than checking the set of all-variable combinations - in your example, there is 1 free variable, not 4.) One way is to count to 2^n-1 in binary, where n is the number of free variables.
(4) For each combination of free variables, find the pivot variables in turn by using your reduced row-echelon form. As you are going along, check these are either 0 or 1, breaking early when one turns out to be another value.
(5) Append successes to a grand "Solutions" array. Finally, check its columns for all 0 or all 1.
I have added a variable offset at the end to accommodate your 1-based indexing.
Your example reduces it to (see the reduced row-echelon form at the end) in 1-based indexing:
x1 + x4 = 1
x2 - x4 = 0
-x3 =-1
This picks out a single free variable (x4) and three pivot variables (x1,x2,x3). Check the two values x4=0 or 1 (which automatically sets the other variables x1,x2,x3 using the RRE form). These two trials turn out both to be binary, so can be added to the Solutions array. Finally, by examining the columns of the Solutions array we can find that only x3 (with 1-based indexing) is 1 for all solutions.
PYTHONimport numpy as np A = np.array( [ [1,1,1,0], [1,0,0,1], [1,1,0,0] ], dtype='int' ) b = np.array( [2,1,1] ) Aug = np.hstack( (A, b.reshape(len(A),1) ) ) # augmented matrix (tags RHS to right of A) ######################################################## # Find reduced row echelon form (integer algebra only) # ######################################################## numRows, numVars = A.shape pivotIndices = [] freeIndices = [] pivotRow = 0 for col in range( numVars + 1 ): # note: includes RHS if pivotRow >= numRows: # add leftover rows to freeIndices freeIndices.append( col ) continue # Swap rows if necessary (and possible) to put a non-zero element in pivot alt = pivotRow if ( Aug[pivotRow,col] == 0 ): for row in range( pivotRow + 1, numRows ): if Aug[row,col] != 0: alt = row break if alt > pivotRow: Aug[ [ pivotRow, alt ], col: ] = Aug[ [ alt, pivotRow ], col: ] # swap rows # Forward integer-retaining Gaussian elimination to remove lower-left entries if Aug[pivotRow,col] != 0: for row in range( pivotRow + 1, numRows ): if Aug[row,col] != 0: Aug[row,col:] = Aug[pivotRow,col] * Aug[row,col:] - Aug[row,col] * Aug[pivotRow,col:] pivotIndices.append( col ) pivotRow += 1 else: freeIndices.append( col ) # Use integer-retaining Gaussian elimination to remove above-pivot entries ("Gauss-Jordan elimination") for i, col in enumerate( pivotIndices ): for row in range( i - 1, -1, -1 ): Aug[row,:] = Aug[i,col] * Aug[row,:] - Aug[row,col] * Aug[i,:] ############################################################### # Determine whether there are ANY possible solutions in reals # ############################################################### if pivotIndices[-1] == numVars: # mustn't have RHS as the last pivot print( 'No solution' ) # rank(A) < rank(Aug) print( 'Row-echelon form for augmented matrix:\n', Aug ) exit() else: freeIndices.pop() # RHS not a relevant variable rank = len( pivotIndices ) LHS = Aug[0:rank,0:-1] RHS = Aug[0:rank,-1] print( '\nReduced row-echelon form:' ) print( 'LHS =\n', LHS ) print( 'RHS =\n', RHS ) print( 'Rank = ', rank ) print( 'pivotIndices =', pivotIndices ) print( 'freeIndices =', freeIndices ) ############################################################ # Find all possible solutions of matrix equation in binary # ############################################################ Solutions = [] # Check combinations of free variables, indexed in freeIndices, by counting up to 2^n-1 in binary n = len( freeIndices ) for counter in range( 2 ** n ): trial = np.zeros( numVars, dtype=int ) for i in range( n ): if counter & ( 1 << i ): trial[freeIndices[i]] = 1 # Set free variables as 0 or 1 # Now solve for pivot variables; early exit if invalid solution outcome = True for i, ind in enumerate( pivotIndices ): target = RHS[i] - sum( LHS[i,[freeIndices]] @ trial[freeIndices] ) if target == 0: trial[ind] = 0 elif target == LHS[i,ind]: trial[ind] = 1 else: outcome = False break if outcome: Solutions.append( trial ) # Add any valid solution Solutions = np.array( Solutions ) print( '\nBinary solutions of matrix equations:\n', Solutions, '\n' ) if len( Solutions ) == 0: print( 'No binary solution' ) exit() ################################# # Pick off any constant columns # ################################# allOnes = [] allZero = [] offset = 1 # For equivalent 1-based indexing for i in range( numVars ): if all( Solutions[:,i] == 1 ): allOnes.append( i + offset ) if all( Solutions[:,i] == 0 ): allZero.append( i + offset ) print( f'For {offset}-based indexing' ) print( 'All ones:', allOnes ) print( 'All zero:', allZero )
Output:
Reduced row-echelon form:
LHS =
[[ 1 0 0 1]
[ 0 1 0 -1]
[ 0 0 -1 0]]
RHS =
[ 1 0 -1]
Rank = 3
pivotIndices = [0, 1, 2]
freeIndices = [3]
Binary solutions of matrix equations:
[[1 0 1 0]
[0 1 1 1]]
For 1-based indexing
All ones: [3]
All zero: []
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Answer by NebularScholar887 • 1 month ago
I think that, despite the system restrictions, you can always create a virtual environment locally without having permissions, and install and use or-tools inside this. Maybe you already tried this and the system restrictions are actually hardcore, but I think it's worth trying otherwise instead of rewriting all your code. I've worked on pretty restrictive machines and this always worked.
python3 -m venv ~/.your_venv_name
source ~/.your_venv_name/bin/activate
pip install or-tools
Sometimes, for very restrictive policies, you are not allowed to create a virtual environment that has pip / ensure-pip because you don't have pip in your base environment, and installing it requires sudo. This is solved by creating a virtual environment without pip, then running a script inside this environment to download pip.
python3 -m venv --without-pip ~/.your_venv_name2
source ~/.you_venv_name2
Then you can run ensurepip to get pip inside your virtual environment:
python -m ensurepip --upgrade
If this does not work, you can also use the get-pip.py script from inside your virtual environment:
wget "https://bootstrap.pypa.io/get-pip.py"
python get-pip.py
rm get-pip.py
You can then proceed and install or-tools:
pip install or-tools
NB: If none of this works to install pip in your virtual environment, other alternative methods to install it are provided on the official pip webpage.
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Answer by StarlitCollector334 • 2 months ago
You don't need to (fully) brute-force, and you don't need to find all of your solutions. You just need to find solutions for which each of your variables meets each of their extrema. The following is a fairly brain-off LP approach with 2n² columns and 2mn rows. It's sparse, and for your inputs does not need to be integral. That said, I somewhat doubt it will be the most efficient method possible.
PYTHONimport numpy as np from scipy.optimize import milp, Bounds, LinearConstraint import scipy.sparse as sp lhs = np.array(( (1, 1, 1, 0), (1, 0, 0, 1), (1, 1, 0, 0), )) rhs = np.array((2, 1, 1)) m, n = lhs.shape # Variables: n * 2 (minimize, maximize) * n c = sp.kron( sp.eye_array(n), np.array(( (+1,), (-1,), )), ) b = np.tile(rhs, 2*n) system_constraint = LinearConstraint( A=sp.kron(sp.eye_array(2*n), lhs, format='csc'), lb=b, ub=b, ) result = milp( c=c.toarray().ravel(), # must be dense integrality=0, bounds=Bounds(lb=0, ub=1), constraints=system_constraint, ) assert result.success extrema = result.x.reshape((n, 2, n)) mins = extrema[:, 0] maxs = extrema[:, 1] vmins = np.diag(mins) vmaxs = np.diag(maxs) print('Solutions for minima on the diagonal:') print(mins) print('Solutions for maxima on the diagonal:') print(maxs) print('Variable minima:', vmins) print('Variable maxima:', vmaxs) print('Guaranteed 0:', vmaxs < 0.5) print('Guaranteed 1:', vmins > 0.5)
PLAINTEXTSolutions for minima on the diagonal: [[-0. 1. 1. 1.] [ 1. 0. 1. -0.] [ 1. 0. 1. -0.] [ 1. 0. 1. -0.]] Solutions for maxima on the diagonal: [[ 1. 0. 1. -0.] [-0. 1. 1. 1.] [ 1. 0. 1. -0.] [-0. 1. 1. 1.]] Variable minima: [-0. 0. 1. -0.] Variable maxima: [1. 1. 1. 1.] Guaranteed 0: [False False False False] Guaranteed 1: [False False True False]
There is a variant on this idea where
This somewhat naively assumes that all solutions will see integer values, and (unlike milp) does not have the option to set integrality=1. For demonstration I was forced to add a row to get a residual.
PYTHONimport numpy as np lhs = np.array(( (1, 1, 1, 0), (1, 0, 0, 1), (1, 1, 0, 0), (0, 0, 1, 1), )) rhs = np.array((2, 1, 1, 1)) m, n = lhs.shape epsilon = 1e-12 lhs_select = np.ones(n, dtype=bool) for i in range(n): lhs_select[i] = False x0, (residual,), rank, singular = np.linalg.lstsq(lhs[:, lhs_select], rhs) zero_solves = residual < epsilon x1, (residual,), rank, singular = np.linalg.lstsq(lhs[:, lhs_select], rhs - lhs[:, i]) one_solves = residual < epsilon lhs_select[i] = True if zero_solves and not one_solves: print(f'x{i}=0, solution {x0.round(12)}') elif one_solves and not zero_solves: print(f'x{i}=1, solution {x1.round(12)}')
PLAINTEXTx0=1, solution [-0. 1. 0.] x1=0, solution [ 1. 1. -0.] x2=1, solution [1. 0. 0.] x3=0, solution [ 1. -0. 1.]
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Answer by LunarSentinel489 • 2 months ago
A simplification would be to add a variable ~x_i = 1 - x_i. You now just need to look for entailed to be 0 variables. What I would then try is to then ask for the maximum of sum_{i in V} x_i. Once you have an assignment A, discard from V all i s.t. A[i]=1. Such an x_i cannot be entailed to be 0. Repeat until sum_{i in V} A_i = 0. All such variables are entailed to be 0. Remove the ~x_i variables as needed.
And for efficiency, you can start with some feasible solutions to the original equations (and maybe a maxima or minima of sum x_i) before introducing extra variables that cannot be in the final set.
This is a sufficiently simple idea that I assume the technique has a name. I am not aware of what it is though.
(At this point it is interesting to note that the above is suspiciously like MaxSAT or a variant of it. Might be a good source of related literature/ideas?)
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