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How to solve a system of linear equations over the nonnegative integers?


How to solve a system of linear equations over the nonnegative integers?

By : Manish Ramteke
Date : November 22 2020, 03:01 PM
To fix the issue you can do You could do this using integer programming, defining a non-negative integer decision variable for every x value you have. Then you could solve the problem with constraints Ax=b and objective 0, which searches for any feasible integer solution to your system of equations.
This can easily be implemented using the pulp package in python. For instance, consider solving the following system:
code :
x+2y+z = 5
3x+y-z = 5
import pulp
A = [[1, 2, 1], [3, 1, -1]]
b = [5, 5]
mod = pulp.LpProblem('prob')
vars = pulp.LpVariable.dicts('x', range(len(A[0])), lowBound=0, cat='Integer')
for row, rhs in zip(A, b):
    mod += sum([row[i]*vars[i] for i in range(len(row))]) == rhs
mod.solve()
[vars[i].value() for i in range(len(A[0]))]
# [1.0, 2.0, 0.0]


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How to solve a system of linear equations with b=0 in R

How to solve a system of linear equations with b=0 in R


By : sourabh Gangoli
Date : March 29 2020, 07:55 AM
it fixes the issue In R I need to solve a system of linear equations (Ax=b), where b=0. By using solve() it just returns a zero vector for the answer, but I want the non-zero solutions of the system. Is there any way for it? , I think you are looking for the null space of a matrix A. Try :
code :
library(MASS)
Null(t(A))

R > (A <- matrix(c(1,2,3,2,4,7), ncol = 3, byrow = T))
     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    2    4    7
R > Null(t(A))
              [,1]
[1,] -8.944272e-01
[2,]  4.472136e-01
[3,]  7.771561e-16
R > (A <- matrix(c(1,2,3,2,4,6), ncol = 3, byrow = T))
     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    2    4    6
R > Null(t(A))
           [,1]       [,2]
[1,] -0.5345225 -0.8017837
[2,]  0.7745419 -0.3381871
[3,] -0.3381871  0.4927193
Solving Non-linear System of 3 differentail equations(Lorenz equations) using ode solve: ode23s in Matlab

Solving Non-linear System of 3 differentail equations(Lorenz equations) using ode solve: ode23s in Matlab


By : Raffy Sueno
Date : March 29 2020, 07:55 AM
around this issue I am trying to solve 3 differentail equations(Lorenz equations) using ode solver: ode23s in Matlab. Here are the 3 lorenz equations: , Your function is incorrect as far as I can see. This line:
code :
xprime=[ alpha*I*(1-c) + c*(- k_f - k_d - k_n * s - k_p*(1-q)); lambda_b * c* P_C - lambda_r *(1-q)*s;   k_p * c *(P_C / P_Q)- gamma * q];
xprime=[ alpha*I*(1-x(1)) + x(1)*(- k_f - k_d - k_n * x(2) - k_p*(1-x(3))); lambda_b * x(1)* P_C - lambda_r *(1-x(3))*x(2);   k_p * x(1) *(P_C / P_Q)- gamma * x(3)];
%initial vectors
  c=0.25; 
  s=0.02;
  q=0.02; 
Solve non-linear system of equations R

Solve non-linear system of equations R


By : Yakov Shtykov
Date : March 29 2020, 07:55 AM
I wish this help you With your addition of some data for C10 the example runs perfectly ok if one takes into account that the loop should take account of length(C10). Like this (with some changes; for why see below):
code :
library(nleqslv)

C10 <- c(1.000000e+00,9.754920e-01,9.547681e-01,9.359057e-01,9.182586e-01,9.014674e-01)
system_size <- 1024

target <- function(Coeffs){
    y <- numeric(3)
    y[1] <- Coeffs[1]*(exp(-Coeffs[2]*(C[4]-1))+exp(-Coeffs[2]*(system_size-(C[4]-1))))+Coeffs[3]^2-C[1]
    y[2] <- Coeffs[1]*(exp(-Coeffs[2]*(C[4]))+exp(-Coeffs[2]*(system_size-(C[4]))))+Coeffs[3]^2-C[2]
    y[3] <- Coeffs[1]*(exp(-Coeffs[2]*(C[4]+1))+exp(-Coeffs[2]*(system_size-(C[4]+1))))+Coeffs[3]^2-C[3]
    y
}

init <- 50*c(0.001,0.01,0)

for(i in 2:min(length(C10)-1,(system_size/2)))
{

    C <- c(C10[i-1],C10[i],C10[i+1],i-2)
  #function
    target <- function(Coeffs){
        y <- numeric(3)
        y[1] <- Coeffs[1]*(exp(-Coeffs[2]*(C[4]-1))+exp(-Coeffs[2]*(system_size-(C[4]-1))))+Coeffs[3]^2-C[1]
        y[2] <- Coeffs[1]*(exp(-Coeffs[2]*(C[4]))+exp(-Coeffs[2]*(system_size-(C[4]))))+Coeffs[3]^2-C[2]
        y[3] <- Coeffs[1]*(exp(-Coeffs[2]*(C[4]+1))+exp(-Coeffs[2]*(system_size-(C[4]+1))))+Coeffs[3]^2-C[3]
        y
    }

    cat("i=",i,"init=",init, "target(init)=",target(init),"\n")
    sol <- nleqslv(init, target,control=list(btol=.01), method="Broyden")
    print(sol)
}
Solve a system of linear equations and linear inequalities

Solve a system of linear equations and linear inequalities


By : jyothsna shekar
Date : March 29 2020, 07:55 AM
wish helps you Yours is a problem in linear programming, where your equality and inequalities are the limitations and you want to minimize (then later maximize) the expression y. The equality, inequalities, and expression are all linear, so that makes it linear programming. The scipy package, using the scipy.optimize.linprog function, can do this kind of linear programming.
Here is commented code to do what you want. Note that all the inequalities were slightly changed to include equality, which is necessary to have a maximum or minimum value of y. To find the maximum value of y the code instead finds the minimum value of -y then prints the additive inverse of that, since linprog minimizes the objective function. Finally, the inequality restrictions must be "less than or equal to" in linprog, so I multiplied both sides of your inequality x + y > 180 by -1 to get one, namely -x + -y <= -180. Ask if you have any questions.
code :
from scipy.optimize import linprog

# Set up values relating to both minimum and maximum values of y
coefficients_inequalities = [[-1, -1]]  # require -1*x + -1*y <= -180
constants_inequalities = [-180]
coefficients_equalities = [[3, 12]]  # require 3*x + 12*y = 1000
constants_equalities = [1000]
bounds_x = (30, 160)  # require 30 <= x <= 160
bounds_y = (10, 60)  # require 10 <= y <= 60

# Find and print the minimal value of y
coefficients_min_y = [0, 1]  # minimize 0*x + 1*y
res = linprog(coefficients_min_y,
              A_ub=coefficients_inequalities,
              b_ub=constants_inequalities,
              A_eq=coefficients_equalities,
              b_eq=constants_equalities,
              bounds=(bounds_x, bounds_y))
print('Minimum value of y =', res.fun)

# Find and print the maximal value of y = minimal value of -y
coefficients_max_y = [0, -1]  # minimize 0*x + -1*y
res = linprog(coefficients_max_y,
              A_ub=coefficients_inequalities,
              b_ub=constants_inequalities,
              A_eq=coefficients_equalities,
              b_eq=constants_equalities,
              bounds=(bounds_x, bounds_y))
print('Maximum value of y =', -res.fun)  # opposite of value of -y
Minimum value of y = 43.3333333333
Maximum value of y = 51.1111111111
How to solve non-linear system of trigonometric equations in Python (which MATLAB can solve easily)

How to solve non-linear system of trigonometric equations in Python (which MATLAB can solve easily)


By : user3497036
Date : March 29 2020, 07:55 AM
it fixes the issue I am trying to solve a system of non-linear trigonometric equations in Python. I tried the following: , In SymPy if you want numerical solutions you should use nsolve:
code :
In [97]: nsolve((eq1,eq2,eq3), (a, f, phi), [1, 1, 1])                                                                            
Out[97]: 
⎡-0.5538674055548 ⎤
⎢                 ⎥
⎢0.837453526933376⎥
⎢                 ⎥
⎣6.95538865037068 ⎦
In [101]: eq1                                                                                                                     
Out[101]: a⋅sin(11.42806⋅π⋅f + φ) - 0.347064 = 0

In [102]: (sol,) = nsolve((eq1,eq2,eq3), (a, f, phi), [1, 1, 1], dict=True)                                                       

In [103]: sol                                                                                                                     
Out[103]: {a: -0.5538674055548, f: 0.837453526933376, φ: 6.95538865037068}

In [104]: eq1.subs(sol)                                                                                                           
Out[104]: False

In [105]: eq1.lhs.subs(sol)                                                                                                       
Out[105]: -0.347064 - 0.5538674055548⋅sin(6.95538865037068 + 9.57046915300624⋅π)

In [106]: eq1.lhs.subs(sol).evalf()                                                                                               
Out[106]: -1.29025679909939e-15
In [107]: (sol,) = nsolve((eq1,eq2,eq3), (a, f, phi), [1, 1, 1], dict=True, prec=50)                                              

In [108]: sol                                                                                                                     
Out[108]: 
{a: -0.55386740555480009188439615822304411607289430639164, f: 0.83745352693337644862065403386504543698722276260565, φ: 6.9553886
503706758809942541544797040214354242211993}

In [109]: eq1.lhs.subs(sol).evalf()                                                                                               
Out[109]: -3.27785083138700e-51
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