So I am working on DSA processor that must multiply large signed values to check if they fit onto a curve. The function I am using works in python given the same values, but when passed in Verilog I am getting incorrect results. I believe it has to deal with my signed values for y & x. During testbenching multiplying x or y by itself gave incorrect values.
module on_curve (
input wire signed [511:0] point
);
reg signed [255:0] x, y;
always#*
begin
x = point[511:256];
y = point[255:0];
if ((y * y - x * x * x - 0 * x - 7) % 256'hFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F == 0) begin
$write("This value is on the curve");
end
if ((y * y - x * x * x - 0 * x - 7) % 256'hFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F !== 0) begin
$write("This value is not on the curve");
end
end
endmodule
The value i'm using for "point" is
512'h79BE667EF9DCBBAC55A06295CE870B07029BFCDB2DCE28D959F2815B16F81798483ADA7726A3C4655DA4FBFC0E1108A8FD17B448A68554199C47D08FFB10D4B8;
which is the generator point for the DSA so it should fall onto the curve.
Related
I have a nonconvex optimization problem for which I am calculating a lower bound using the McCormick envelope. Each bilinear term is replaced with an auxiliary variable which has the following constraints defined:
w_{ij} >= x_i^L * x_j + x_i * x_j^L - x_i^L * x_j^L
w_{ij} >= x_i^U * x_j + x_i * x_j^U - x_i^U * x_j^U
w_{ij} <= x_i^U * x_j + x_i * x_j^L - x_i^U * x_j^L
w_{ij} <= x_i^L * x_j + x_i * x_j^U - x_i^L * x_j^U
where
x_U <= x <= x_L
I am given a function taking in several arguments:
def convex_bounds(n,m,c,H,Q,A,b,lb,ub):
# n is the number of optimization variables
# m is the number of eq constraints
# H = positive, semidefinite matrix from objetcive function (n x n)
# Q is (mxn) x n
# A is m x n
# b is RHS of non linear eq constraints (m x 1)
# c,lb,ub are vectors size (n x 1)
......................................
# Create matrix B & b_ineq for inequality constraints
# where B*x <= b_ineq
B = np.eye(3)
b_ineq = np.array((10,10,10))
## these values would work in a scenario with no bilinear terms
My problem is that I don't know how to specify the inequality constraints matrix B and vector b_ineq. For this particular exercise my variables are x1, x2 and x3 with bounds 0 (x_L) and 10 (x_U). My bilinear terms are x_12 and x_23 (which will lead to auxiliary variables w_12 and w_23). How can I specify the known bounds (0 and 10) for x1,x2 and x3 and the calculated ones (as in the theory pasted above) in B and b_ineq?
I don't actually know how to proceed with this.
These are the conditions:
if(x > 0)
{
y >= a;
z <= b;
}
It is quite easy to convert the conditions into Linear Programming constraints if x were binary variable. But I am not finding a way to do this.
You can do this in 2 steps
Step 1: Introduce a binary dummy variable
Since x is continuous, we can introduce a binary 0/1 dummy variable. Let's call it x_positive
if x>0 then we want x_positive =1. We can achieve that via the following constraint, where M is a very large number.
x < x_positive * M
Note that this forces x_positive to become 1, if x is itself positive. If x is negative, x_positive can be anything. (We can force it to be zero by adding it to the objective function with a tiny penalty of the appropriate sign.)
Step 2: Use the dummy variable to implement the next 2 constraints
In English: if x_positive = 1, then y >= a
However, if x_positive = 0, y can be anything (y > -inf)
y > a - M (1 - x_positive)
Similarly,
if x_positive = 1, then z <= b
z <= b + M * (1 - x_positive)
Both the linear constraints above will kick in if x>0 and will be trivially satisfied if x <=0.
I need to solve the following Least Squares Problem where A and B and X are all matrices:
cvx_begin quiet;
variable X(len_x) nonnegative;
minimize ( norm(X * A - B , 2));
subject to
X >= 0;
for i=1: size(X,2)
for j= i + 1: size(X,2)
transpose(X(:,i)) * X(:,j) <= epsilon
end
end
cvx_end
I choose CVX, but it doesn't require me to transform the problem into standard form. But with CVX, I get the following error:
Error using cvx/quad_form (line 230)
The second argument must be positive or negative semidefinite.
Error in * (line 261)
[ z2, success ] = quad_form( xx, P, Q, R );
Error in sanaz_opt (line 28)
transpose(X(:,i)) * X(:,j) <= 0.1
I'm wondering how I can solve this problem? I'm trying to use Gurobi or least squares function in Matlab, but it seems they can't handle the transpose(X(:,i)) * X(:,j) constraint.
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 7 years ago.
As I was debugging my VBA code, I came across this weird phenomenon:
This loop
Dim x,y as Double
x = 0.7
y = 0.1
For x = x - y To x + y Step y
Next x
runs only twice!
I tried many variations of this code to nail down the problem, and here is what I came up with:
Replacing the loop boundaries with simple numbers (0.6 to 0.8) - helped.
Replacing variables with numbers (all the combinations) - didn't help.
Replacing the for-loop with do while/until loops - helped.
Replacing the values of x and y (y=0.01, 0.3, 0.4, 0.5, 0.7, 0.8, 0.9 - helped. 0.2, 0.6 -didn't help. x=1, 2 ,3 helped. x=4, 5, 6, 7, 8, 9 - didn't help.
Converting the Double to Decimal with CDec() - helped.
Using the Currency data type instead of Double - helped.
So what we have here is a floating-point rounding-error that happens on mysterious conditions.
What I'm trying to find out is what are those conditions, so we can avoid them.
Who will unveil this mystery?
(Pardon my English, it's not my mother tongue).
GD Falcon,
Generally in solving a For...Next loop it would not be advisable to use 'double' or 'decimal' or 'currency' variables as they provide a level of uncertainty in their accuracy, it's this level of inaccuracy that is wrecking havoc on your code as the actual stop parameter (when x-y, plus (n x y) = x+y) is, in terms of absolutes, an insolvable equation unless you limit the number of decimals it uses.
It is generally considered better practice to use integers (or long) variables in a For...Next loop as their outcome is more certain.
See also below post:
How to make For loop work with non integers
If you want it to run succesfully and iterate 3 times (as I expect you want)
Try like below:
Dim x, y As Double
x = 0.7
y = 0.1
For x = Round(x - y, 1) To Round(x + y, 1) Step Round(y, 1)
Debug.Print x
Next x
Again, it is better not to use Doubles in this particular way to begin with but if you must you would have to limit the number of decimals they calculate with or set a more vague end point (i.e. x > y, rather than x = y)
The coding you use implies that you wish to test some value x against a tolerance level of y.
Assuming this is correct it would imply testing 3 times where;
test_1: x = x - y
test_2: x = x
test_3: x = x + y
The below code would do the same but it would have a better defined scope.
Dim i As Integer
Dim x, y, w As Double
x = 0.7
y = 0.1
For i = -1 To 1 Step 1
w = x + (i * y)
Debug.Print w
Next i
Good luck !
How can I divide two NSIntegers, for instance, 13 / 4 and round the result to the next integer = 3?
I have seen some samples converting the integers to float and back to integer.
But what is the recommended way with the least amount of code to do it?
Assuming x >= 0 and y > 0:
If you want to round down: x / y
If you want to round up: (x + y - 1) / y
If you want to round to nearest: (x + y / 2) / y