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Is it possible to find all integer solutions?


By : José Maciel
Date : September 16 2020, 06:00 AM
wish help you to fix your issue This is a linear, integer constraint satisfaction problem, which can be solved efficiently by OR Tools' CP-SAT. I've modified their example to solve your problem in Python:
code :
from ortools.sat.python import cp_model

class VarArraySolutionPrinter(cp_model.CpSolverSolutionCallback):
    """Print intermediate solutions."""

    def __init__(self, variables):
        cp_model.CpSolverSolutionCallback.__init__(self)
        self.__variables = variables
        self.__solution_count = 0

    def on_solution_callback(self):
        self.__solution_count += 1
        for v in self.__variables:
            print('%s=%i' % (v, self.Value(v)), end=' ')
        print()

    def solution_count(self):
        return self.__solution_count

def SearchForAllSolutionsSampleSat():
    """Showcases calling the solver to search for all solutions."""
    # Creates the model.
    model = cp_model.CpModel()

    p = [1, 2, 3, 4]
    ceq = 30
    cgeq = 2

    N = len(p)

    # Creates the variables
    x = [model.NewIntVar(0, 100, f'x{i}') for i in range(N)]

    # Create the constraints.
    model.Add(sum([xi*pi for xi, pi in zip(x, p)]) == ceq)
    model.Add(sum(x) >= cgeq)

    # Create a solver and solve.
    solver = cp_model.CpSolver()
    solution_printer = VarArraySolutionPrinter(x)
    status = solver.SearchForAllSolutions(model, solution_printer)

    print('Status = %s' % solver.StatusName(status))
    print('Number of solutions found: %i' % solution_printer.solution_count())


SearchForAllSolutionsSampleSat()


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integer to string conversion / integer-string concatenation in C++ - more compact solutions?


By : smack081
Date : March 29 2020, 07:55 AM
This might help you How to do integer -> string conversion has been answered many times on the internet... however, I'm looking for the most compact "C++-way" to do this.
code :
#include <string>

Using Python to Find Integer Solutions to System of Linear Equations


By : Štefan Lacko
Date : March 29 2020, 07:55 AM
will be helpful for those in need You can use Numpy Linear Algebra to solve a system of equations, the least-squares solution to a linear matrix equation. In your case, you can consider the following vectors:
code :
import numpy as np
# range(T): coefficients of the first equation
# np.ones(T): only 'ones' as the coefficients of the second equation 
A = np.array([range(T), np.ones(T)) # Coefficient matrix
B = np.array([a, b]) # Ordinate or “dependent variable” values
x = np.linalg.lstsq(A, B)[0]
import numpy as np
def solve(T, a, b):
    A = np.array([range(T), np.ones(T)])
    B = np.array([a, b])
    return np.linalg.lstsq(A, B)[0]

how to find all integer solutions to sum(xi) =b, with linear constraints


By : Rpitre
Date : March 29 2020, 07:55 AM
will be helpful for those in need Recursion is best. Here is the natural Python solution with generators:
code :
def solutions(variables, sum_left, max_value):
    if 0 == variables:
        if 0 == sum_left:
            yield []
    else:
        for i in range(0, max_value + 1):
            if sum_left < i:
                break
            else:
                for partial_solution in solutions(variables - 1, sum_left - i,
                                                  max_value):
                    yield [i] + partial_solution

for x in solutions(10, 10, 2):
    print(x)
def do_something_for_solutions(variables, sum_left, max_value, known=None):
    if known is None:
        known = []
    if 0 == variables:
        if 0 == sum_left:
            do_something(known)
    else:
        for i in range(0, max_value + 1):
            if sum_left < i:
                break
            else:
                do_something_for_solutions(variables - 1, sum_left - i,
                                           max_value, known + [i])

def do_something(solution):
    print(solution)

do_something_for_solutions(10, 10, 2)
def solutions(variables, sum_left, max_value):
    if 0 == variables:
        if 0 == sum_left:
            return [[]]
        else:
            return []
    else:
        answer = []
        for i in range(0, max_value + 1):
            if sum_left < i:
                break
            else:
                for partial_solution in solutions(variables - 1, sum_left - i,
                                                  max_value):
                    answer.append([i] + partial_solution)
        return answer

for x in solutions(10, 10, 2):
    print(x)

How to find all positive integer solutions to an cubic equation?


By : anchordown
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further You can group by all the possible sums and print out groups which contain more than one item. This is O(N**2) algorithm:
code :
  // key is sum and value is a list of representations
  Dictionary<int, List<Tuple<int, int>>> solutions = 
    new Dictionary<int, List<Tuple<int, int>>>();

  for (int a = 1; a <= 1000; ++a)
    for (int b = a; b <= 1000; ++b) {
      int sum = a * a * a + b * b * b;

      List<Tuple<int, int>> list = null;

      if (!solutions.TryGetValue(sum, out list)) {
        list = new List<Tuple<int, int>>();

        solutions.Add(sum, list);
      }

      list.Add(new Tuple<int, int>(a, b));
    }

  String report = String.Join(Environment.NewLine,
    solutions.Values
    .Where(list => list.Count > 1) // more than one item
    .Select(list => String.Join(", ", 
      list.Select(item => String.Format("({0}, {1})", item.Item1, item.Item2)))));

  Console.Write(report);
(1, 12), (9, 10)
(1, 103), (64, 94)
(1, 150), (73, 144)
(1, 249), (135, 235)
(1, 495), (334, 438)
...
(11, 493), (90, 492), (346, 428) // <- notice three representations of the same sum
...
(663, 858), (719, 820)
(669, 978), (821, 880)
(692, 942), (720, 926)
(718, 920), (816, 846)
(792, 901), (829, 870)
   1**3 + 12**3 == 9**3 + 10**3
   ...
   11**3 + 493**3 == 90**3 + 492**3 == 346**3 + 428**3   
   ...
   792**3 + 901**3 == 829**3 + 870**3 
(11, 493), (90, 492), (346, 428)
(15, 930), (198, 927), (295, 920)
(22, 986), (180, 984), (692, 856)
(70, 560), (198, 552), (315, 525)
(111, 522), (359, 460), (408, 423)
(167, 436), (228, 423), (255, 414)
(300, 670), (339, 661), (510, 580)
(334, 872), (456, 846), (510, 828)

Find integer solutions to a set of equations with more unknowns than equations


By : Abhishek Kundu
Date : March 29 2020, 07:55 AM
I hope this helps . Use the Lenstra–Lenstra–Lovász lattice basis reduction algorithm to find the Hermite normal form of your system.
It's straaightforward in matlab http://fr.mathworks.com/help/symbolic/mupad_ref/linalg-hermiteform.html
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