Here is a image:
I have two vectors : os, oe
the range between them is always from os (start) to oe (end).
So in this image the range between is a angle of 270°.
Then I have two vector to check: oa, ob
As you can see the vector oa should be within the range formed by osoe and the vector ob should be outside.
I am wondering if there is a way to do the check using only vector math (such as cross product dot product).
I tried to use cross product with clockwise/counter clockwise check but it seems like when the angle in between is larger then 180°, things get complex.
Any advice will be appreciated, thanks :)
I denote vector to point p as op.
Calculate cross product
c_se = cross(os, oe)
If c_se>=0 (angle in 0..180 range), then you have to check whether
cross(os, op) >= 0 AND cross(op, oe) >= 0
If c_se < 0 (angle in 180..360 range), then you have to check whether
NOT (cross(oe, op) >= 0 AND cross(op, os) >= 0)
Related
Recently, I stumbled over the following problem in an embedded software project. I cannot yet figure out, under which conditions there is a unique solution and how it can be found.
Let us assume that we have a 16-bit MASK and a 16-bit COMPARE value and we want to set them, such that for a defined range of consecutive IDs {a, ..., b} (e.g. {0x78, 0x79, ..., 0x97}) those IDs satisfy bitwise
ID & MASK == COMPARE
while IDs outside that range do not satisfy the above equation.
As an example: If a=0x100 and b=0x1FF, the MASK is set as 0x700 and the COMPARE as 0x100.
I have the following questions:
What are the conditions for the minimal and maximal ID, so that uniquely defined MASK and COMPARE values exist?
How can they be calculated?
Looking forward to your answers!
The combination of MASK and COMPARE can be described as a string of 0 and 1 with wildcards (*) to indicate "don't care" positions. If there's a * left of a 0 or 1, the string does not describe a continuous range. Thus, the strings that represent ranges are exactly those where all the * are at the right end. The combinations of a and b that can be checked this way are then those where we can divide the 16 bits into a left part and a right part such that a and b are equal in the left part and the right part is all zero for a and all ones for b. We can test this as follows:
((((a ^ b) + 1) | a) & b) == a
As a background, I'm a computer programmer and I'm working on a software library that allows a computer to quickly search through all dates to find a set of dates that satisfies a criteria. For example:
I want a list of every possible time that has ever occurred that has occurred on a friday or a saturday that is in April or May during the first week of the month.
My library uses numerical sets to efficiently represent ranges of dates that satisfy a criteria.
I've been thinking about ways to improve the performance of some parts of the app and I think that by combining sets and some geometry, I can really improve my results. However, my geometry is a bit rusty and I was hoping you might could help.
Here's my thought:
Certain elements of time can be represented as a circular dial. For example, Minutes can be positioned on a clock with values between 0...59. We could store valid ranges as a list of arcs. For example, If we wanted all times that ended with 05..10, we could store [5,10]. If we wanted all times that end with :45-59 or :00-15, we could store [45, 15]. Notice how this last arc "loops around" the dial. Here's a mockup showing different ranges intersecting on a dial
My question is this:
Given a set of whole numbers between N...M arranged into a circle.
Given Arc1 which is representing by [A, B] and Arc2 which is represented by [C, D] where A, B, C, and D are all within in range N...M
How do I determine:
A. Whether the arcs intersect.
B. If they do, what their intersection is.
C. If they do, what their union is.
Thank you so much for your help. If you're not able to help, if you can point me in the right direction, that would be great.
Thanks!
A simple and safe approach is to split the intervals that straddle 0. Then you perform pairwise interval intersection/union (for instance if A < D and C < B then [max(A,C), min(B,D)] for the intersection), and merge them if they meet at 0.
It seems the primitive operation to implement would be something like 'is the number X contained in the arch [A,B]'. Once you have that, you could implement an [A,B]/[C,D] arch-intersection predicate by something like -
Arch intersection means exactly that at least one of the following conditions is met:
C is contained in [A,B]
D is contained in [A,B]
A is contained in [C,D]
B is contained in [C,D]
One way to implement this contained-in-arch test without any branches is with some trigonometry and vector cross product. Not sure it would be faster (the math/branches performance tradeoff is entirely empiric), but it might be worth a try.
Denote Xa = sin(X/N * 2PI), Ya = cos(X/N * 2PI) and similarly for Xb,Yb etc.
C is contained in [A,B] is equivalent to:
Xa * Yc - Ya * Xc > 0
AND
Xc * Yb - Yc * Xb > 0
You can complete the other 3 conditions in an identical manner.
Hope this turns out useful.
So in my text book there is this example of a recursive function using f#
let rec gcd = function
| (0,n) -> n
| (m,n) -> gcd(n % m,m);;
with this function my text book gives the example by executing:
gcd(36,116);;
and since the m = 36 and not 0 then it ofcourse goes for the second clause like this:
gcd(116 % 36,36)
gcd(8,36)
gcd(36 % 8,8)
gcd(4,8)
gcd(8 % 4,4)
gcd(0,4)
and now hits the first clause stating this entire thing is = 4.
What i don't get is this (%)percentage sign/operator or whatever it is called in this connection. for an instance i don't get how
116 % 36 = 8
I have turned this so many times in my head now and I can't figure how this can turn into 8?
I know this is probably a silly question for those of you who knows this but I would very much appreciate your help the same.
% is a questionable version of modulo, which is the remainder of an integer division.
In the positive, you can think of % as the remainder of the division. See for example Wikipedia on Euclidean Divison. Consider 9 % 4: 4 fits into 9 twice. But two times four is only eight. Thus, there is a remainder of one.
If there are negative operands, % effectively ignores the signs to calculate the remainder and then uses the sign of the dividend as the sign of the result. This corresponds to the remainder of an integer division that rounds to zero, i.e. -2 / 3 = 0.
This is a mathematically unusual definition of division and remainder that has some bad properties. Normally, when calculating modulo n, adding or subtracting n on the input has no effect. Not so for this operator: 2 % 3 is not equal to (2 - 3) % 3.
I usually have the following defined to get useful remainders when there are negative operands:
/// Euclidean remainder, the proper modulo operation
let inline (%!) a b = (a % b + b) % b
So far, this operator was valid for all cases I have encountered where a modulo was needed, while the raw % repeatedly wasn't. For example:
When filling rows and columns from a single index, you could calculate rowNumber = index / nCols and colNumber = index % nCols. But if index and colNumber can be negative, this mapping becomes invalid, while Euclidean division and remainder remain valid.
If you want to normalize an angle to (0, 2pi), angle %! (2. * System.Math.PI) does the job, while the "normal" % might give you a headache.
Because
116 / 36 = 3
116 - (3*36) = 8
Basically, the % operator, known as the modulo operator will divide a number by other and give the rest if it can't divide any longer. Usually, the first time you would use it to understand it would be if you want to see if a number is even or odd by doing something like this in f#
let firstUsageModulo = 55 %2 =0 // false because leaves 1 not 0
When it leaves 8 the first time means that it divided you 116 with 36 and the closest integer was 8 to give.
Just to help you in future with similar problems: in IDEs such as Xamarin Studio and Visual Studio, if you hover the mouse cursor over an operator such as % you should get a tooltip, thus:
Module operator tool tip
Even if you don't understand the tool tip directly, it'll give you something to google.
I have 2 3D vectors. (objects with X, Y and Z float values)
In my diagram below, I would like to determine the length of the green line.
This is the distance along Vector 1 that Vector 2 is. Or, the distance from the origin to the end of a line on Vector 1 which is at 90' to Vector 1 and passes thorough the point at the end of Vector 2.
I am doing this in Unity3D so I have access to quite a few helper methods that enable me to get the length of a Vector3 and so on very easily.
The length is obviously
norm(v2)*cos(angle(v1,v2))
and since
cos(angle(v1,v2))=abs(dot(v1,v2))/norm(v1)/norm(v2)
the final formula is
abs(dot(v1,v2))/norm(v1)
One could also say that
e1 = v1/norm(v1)
is the unit vector in the direction of v1, and that the green vector is
dot(e1,v2)*e1
resulting in the same length formula.
This is projection of Vector2 onto Vector1 direction. The simplest way (I think) to find it - using scalar product
D = |V2| * DotProduct(V2, V1) / (|V2| * |V1|) = DotProduct(V2, V1) / |V1|
where |V1| is the length of V1 vector
Im not sure but I think this is what you wanted
Vector3 distance = Vector3.Lerp(Vector3.zero, vector_1, vector_2.sqrMagnitude / vector_1.sqrMagnitude);
http://docs.unity3d.com/ScriptReference/Vector3-sqrMagnitude.html
http://docs.unity3d.com/ScriptReference/Vector3.Lerp.html
What is the double percent (%%) used for in R?
From using it, it looks as if it divides the number in front by the number in back of it as many times as it can and returns the left over value. Is that correct?
Out of curiosity, when would this be useful?
The "Arithmetic operators" help page (which you can get to via ?"%%") says
‘%%’ indicates ‘x mod y’
which is only helpful if you've done enough programming to know that this is referring to the modulo operation, i.e. integer-divide x by y and return the remainder. This is useful in many, many, many applications. For example (from #GavinSimpson in comments), %% is useful if you are running a loop and want to print some kind of progress indicator to the screen every nth iteration (e.g. use if (i %% 10 == 0) { #do something} to do something every 10th iteration).
Since %% also works for floating-point numbers in R, I've just dug up an example where if (any(wts %% 1 != 0)) is used to test where any of the wts values are non-integer.
The result of the %% operator is the REMAINDER of a division,
Eg. 75%%4 = 3
I noticed if the dividend is lower than the divisor, then R returns the same dividend value.
Eg. 4%%75 = 4
Cheers
%% in R return remainder
for example:
s=c(1,8,10,4,6)
d=c(3,5,8,9,2)
x=s%%d
x
[1] 1 3 2 4 0