 Tags IOS SQL HTML C RUBY-ON-RAILS MYSQL ASP.NET DEVELOPMENT RUBY .NET LINUX SQL-SERVER REGEX WINDOWS ALGORITHM ECLIPSE VISUAL-STUDIO STRING SVN PERFORMANCE APACHE-FLEX UNIT-TESTING SECURITY LINQ UNIX MATH EMAIL OOP LANGUAGE-AGNOSTIC VB6 # 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)

# 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()
``````

## 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)
``````
``````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)
``````

## 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:
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):

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>>();

}

}

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. 