I am trying to create a WHERE statement, where excludes a square of coordinates.
I have saved on a table some records with its X and Y coordinates.
I want to exclude the records that are between x1 and x2 and y1 and y1
I was using
select * from dbo.records where (x not between x1 and x2) and (y not between y1 and y2)
but the first not between is deleting me a lot of coordinates
enter image description here
Do you know a way to get the ordinates inside the square?
Thanks
I think that you want ored conditions:
where (x not between x1 and x2) or (y not between y1 and y2)
I find that it is clearer to phrase this as:
where not (x between x1 and x2 and y between y1 and y2)
Conditions x between x1 and x2 and y between y1 and y2 define coordinates that belong to the square; you want coordinates that do not satisfy these conditions.
Related
In particle physics, we have to compute the invariant mass a lot, which is for a two-body decay
When the momenta (p1, p2) are sometimes very large (up to a factor 1000 or more) compared to the masses (m1, m2). In that case, there is large cancellation happening between the last two terms when the calculation is carried out with floating point numbers on a computer.
What kind of numerical tricks can be used to compute this accurately for any inputs?
The question is about suitable numerical tricks to improve the accuracy of the calculation with floating point numbers, so the solution should be language-agnostic. For demonstration purposes, implementations in Python are preferred. Solutions which reformulate the problem and increase the amount of elementary operations are acceptable, but solutions which suggest to use other number types like decimal or multi-precision floating point numbers are not.
Note: The original question presented a simplified 1D dimensional problem in form of a Python expression, but the question is for the general case where the momenta are given in 3D dimensions. The question was reformulated in this way.
With a few tricks listed on Stackoverflow and the transformation described by Jakob Stark in his answer, it is possible to rewrite the equation into a form that does not suffer anymore from catastrophic cancellation.
The original question asked for a solution in 1D, which has a simple solution, but in practice, we need the formula in 3D and then the solution is more complicated. See this notebook for a full derivation.
Example implementation of numerically stable calculation in 3D in Python:
import numpy as np
# numerically stable implementation
#np.vectorize
def msq2(px1, py1, pz1, px2, py2, pz2, m1, m2):
p1_sq = px1 ** 2 + py1 ** 2 + pz1 ** 2
p2_sq = px2 ** 2 + py2 ** 2 + pz2 ** 2
m1_sq = m1 ** 2
m2_sq = m2 ** 2
x1 = m1_sq / p1_sq
x2 = m2_sq / p2_sq
x = x1 + x2 + x1 * x2
a = angle(px1, py1, pz1, px2, py2, pz2)
cos_a = np.cos(a)
if cos_a >= 0:
y1 = (x + np.sin(a) ** 2) / (np.sqrt(x + 1) + cos_a)
else:
y1 = -cos_a + np.sqrt(x + 1)
y2 = 2 * np.sqrt(p1_sq * p2_sq)
return m1_sq + m2_sq + y1 * y2
# numerically stable calculation of angle
def angle(x1, y1, z1, x2, y2, z2):
# cross product
cx = y1 * z2 - y2 * z1
cy = x1 * z2 - x2 * z1
cz = x1 * y2 - x2 * y1
# norm of cross product
c = np.sqrt(cx * cx + cy * cy + cz * cz)
# dot product
d = x1 * x2 + y1 * y2 + z1 * z2
return np.arctan2(c, d)
The numerically stable implementation can never produce a negative result, which is a commonly occurring problem with naive implementations, even in double precision.
Let's compare the numerically stable function with a naive implementation.
# naive implementation
def msq1(px1, py1, pz1, px2, py2, pz2, m1, m2):
p1_sq = px1 ** 2 + py1 ** 2 + pz1 ** 2
p2_sq = px2 ** 2 + py2 ** 2 + pz2 ** 2
m1_sq = m1 ** 2
m2_sq = m2 ** 2
# energies of particles 1 and 2
e1 = np.sqrt(p1_sq + m1_sq)
e2 = np.sqrt(p2_sq + m2_sq)
# dangerous cancelation in third term
return m1_sq + m2_sq + 2 * (e1 * e2 - (px1 * px2 + py1 * py2 + pz1 * pz2))
For the following image, the momenta p1 and p2 are randomly picked from 1 to 1e5, the values m1 and m2 are randomly picked from 1e-5 to 1e5. All implementations get the input values in single precision. The reference in both cases is calculated with mpmath using the naive formula with 100 decimal places.
The naive implementation loses all accuracy for some inputs, while the numerically stable implementation does not.
If you put e.g. m1 = 1e-4, m2 = 1e-4, p1 = 1 and p2 = 1 in the expression, you get about 4e-8 with double precision but 0.0 with single precision calculation. I assume, that your question is about how one can get the 4e-8 as well with single precision calculation.
What you can do is a taylor expansion (around m1 = 0 and m2 = 0) of the expression above.
e ~ e|(m1=0,m2=0) + de/dm1|(m1=0,m2=0) * m1 + de/dm2|(m1=0,m2=0) * m2 + ...
If I calculated correctly, the zeroth and first order terms are 0 and the second order expansion would be
e ~ (p1+p2)/p1 * m1**2 + (p1+p2)/p2 * m2**2
This yields exactly 4e-8 even with single precision calculation. You can of course do more terms in the expansion if you need, until you hit the precision limit of a single float.
Edit
If the mi are not always much smaller than the pi you could further massage the equation to get
The complicated part is now the one in the square brackets. It essentially is sqrt(x+1)-1 for a wide range of x values. If x is very small, we can use the taylor expansion of the square root (e.g. like here). If the x value is larger, the formula works just fine, because the addition and subtraction of 1 are no longer discarding the value of x due to floating point precision. So one threshold for x must be choosen below one switches to the taylor expansion.
I have a temp table with two columns that I would directly copy to another table without explicitly stating the column names (and ultimately rename these columns).
Temp table:
X Y
x1 y1
x2 y2
x3 y3
Desired output:
A B C D
x1 y1 X Y
x2 y2 X Y
x3 y3 X Y
I assume this has to be done via Dynamic SQL since I'm trying to use the ordinal position to populate my new table. I already have #Column_Name1 and #Column_Name2 that points to the column names of the temp table (but not the actual data) for columns C and D in the desired output table.
Would greatly appreciate if you could provide some of the actual coding as I have very little knowledge of dynamic sql. Thank you for your help in advance!
Lets say I have pre-defined 3 variables, x1, x2, and x3, each of which is a different co-ordinate on the screen. I have a whole chunk of code to decide whether another variable, a will equal 1, 2, or 3. Now, I want to include the value of a in a variable name, allowing me to 'dynamically' change between x1, x2, and x3.
E.g. a is set to 2. Now I want to move the mouse to xa, so if a=2, xa is x2, which is a predefined variable.
Its probably clear I am very new to Lua, I have tried googling the issue, but I'm not really sure what I am looking for, terminology wise and such.
Anyhow, is anyone able to help me out?
If you can change the code where x1, x2 and x3 are defined, a cleaner approach is to use arrays (i.e. array-like tables). This is the general approach when you need a sequence of variables indexed by a number.
Therefore, instead of x1, x2 and x3 you could define:
local x = {}
x[1] = 10 -- instead of x1
x[2] = 20 -- instead of x2
x[3] = 30 -- instead of x3
Now instead of using xa you simply use x[a].
If xa are global variables, you can use the table _G like this:
x1 = 42
x2 = 43
x3 = 44
local a = 2
print(_G['x' .. a])
Output:
43
The following inequality system is solved for x1 and x2 over the integers.
x1 + x2 = l
x1 >= y1
x2 >= y2
x1 <= z1
x2 <= z2
l - z1 <= x2
l - z2 <= x1
l,y1,y2,z1,z2 are arbitrary but fixed and >= 0.
With the example values
l = 8
y1 = 1
y2 = 2
z1 = z2 = 6
I solve the system and get the following equations:
2 <= x1 <= 6
x2 = 8 - x1
When I tell WolframAlpha that it should solve it over the integers, it only outputs all possible values.
My question is whether I can derive equations/ranges for x1 and x2 for any given l,y1,y2,z1,z2 programmatically. This problem is related to constraint programming and I found an old paper about this problem: "Compiling Constraint Solving using Projection" by Harvey et al.
Is this approach used in any modern constraint solving libraries?
The reason I ask this is that I need to solve systems like the above several thousand times with different parameters and this takes a long time if the whole system is read/optimized/solved over and over again. Therefore, if I could compile my parameterized systems once and then just use the compiled versions I expect a massive speed gain.
I have a cubic bezier curve where the first and last points are given (namely P0(0,0) and P3(1,1)).
The other two points are defined like this: cubic-bezier(0.25, 0.1, 0.25, 1.0) (X1, Y1, X2, Y2, also those values must not be smaller or larger than 0 or 1, respectively)
Now what would I have to do to get the Y coordinate for a given X, assuming there's only one? (I know that under certain circumstances there can be multiple values, but let's just put them aside. I'm not doing rocket science over here, I just want to be able to get Y multiple times per second to do transitions)
I managed to dig up this: y coordinate for a given x cubic bezier, but I don't understand what xTarget stands for.
Oh, also this is no homework whatsoever, I'm just a bit annoyed at the fact that there's no comprehensible stuff about cubic bezier curves on the internet.
If you have
P0 = (X0,Y0)
P1 = (X1,Y1)
P2 = (X2,Y2)
P3 = (X3,Y3)
Then for any t in [0,1] you get a point on the curve given by the coordinates
X(t) = (1-t)^3 * X0 + 3*(1-t)^2 * t * X1 + 3*(1-t) * t^2 * X2 + t^3 * X3
Y(t) = (1-t)^3 * Y0 + 3*(1-t)^2 * t * Y1 + 3*(1-t) * t^2 * Y2 + t^3 * Y3
If you are given an x value, then you need to find which t values in [0,1] correspond to that point on the curve, then use those t values to find the y coordinate.
In the X(t) equation above, set the left side to your x value and plug in X0, X1, X2, X3. This leaves you with a cubic polynomial with variable t. You solve this for t, then plug that t value into the Y(t) equation to get the y coordinate.
Solving the cubic polynomial is tricky but can be done by carefully using one of the methods to solve a cubic polynomial.
P0 is your first point in the curve where t=0
P3 is your last point in the curve where t=1
P1 and P2 are your control points.