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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.
I was wondering if anyone can help with grouping the data below as I'm trying to use the subset function to filter out volumes below a certain threshold but given that the data represents groups of objects, this creates the problem of removing certain items that should be kept.
In Column F ( and I) you can see Blue, Red, and Yellow Objects. Each represent three separate colored probes on one DNA strand. Odd numbered or non-numbered Blue ,Red, and Yellow are paired with a homologous strand represented by an even numbered Blue, Red, and Yellow. Ie data in rows 2,3,and 4 are one "group" and pair with the "group" shown in rows 5,6,and 7. This then repeats, so 8,9,10 are a new group and that group pairs with the one in 11,12,13.
What I would like to do is subset the groups so that only those below a certain Distance to Midpoint (column M) are kept. The Midpoint here is the midpoint of the line that connects the blue of one group with the blue of its partner, so the subset should only apply to the Blue distance to midpoint, and that is where I'm having a problem. For instance if I ask to keep blue distances to midpoint that are less than 3, then the objects in row 3 and 4 should be kept because they are part of the group with the blue distance below 3. Right now though when I filter with the subset function I lose Red Selection and Yellow Selection. I'm confident there is a straighforward solution to this in R, but I'd also be open to some type of filtering in excel if anyone has suggestions via that route instead.
EDIT
I managed to work something out in Excel last night after posting the question. Solution isn't pretty but it works well enough. I just added a new column next to "distance to midpoint" that gives all the objects in one group the same distance so that when I filter the data I won't lose any objects that I shouldn't. If it helps anyone in the future, the formula I used in excel was:
=SQRT ( ((INDEX($B$2:$B$945,1+QUOTIENT(ROWS(B$2:B2)-1,3)*3))- (INDEX($O$2:$O$945,1+QUOTIENT(ROWS(O$2:O2)-1,3)*3)) ) ^2 +( (INDEX($C$2:$C$945,1+QUOTIENT(ROWS(C$2:C2)-1,3)*3))-(INDEX($P$2:$P$945,1+QUOTIENT(ROWS(P$2:P2)-1,3)*3)) ) ^2 +( (INDEX($D$2:$D$945,1+QUOTIENT(ROWS(D$2:D2)-1,3)*3))-(INDEX($Q$2:$Q$945,1+QUOTIENT(ROWS(Q$2:Q2)-1,3)*3)) ) ^2)
Would be easier with a reproducible example, but here's a (hacky) plyr solution:
filterframe<-function(df,threshold){
df$grouper<-rep(seq(from=1,to=6),nrow(df)/6)
dataout<-df%>%group_by(grouper)%>%summarise(keep=.[[1]]$distance_to_midpoint<threshold)
dataout[dataout$keep,]
}
filterframe(mydata)
A base R solution provided below. The idea is that once your data are in R, you (edit) keep! rows iff they meet 2 criteria. First, the Surpass column has to contain the word "blue" in it, which is done with the grepl function. Second, the distance must below a certain threshold (set arbitrarily by thresh.
fakeData=data.frame(Surpass=c('blue', 'red', 'green', 'blue'),
distance=c(1,2,5,3), num=c(90,10,9,4))
#thresh is your distance threshold
thresh = 2
fakeDataNoBlue = fakeData[which(grepl('blue', fakeData$Surpass)
& fakeData$distance < thresh),]
There's probably also a quick dplyr solution using filter, but I haven't fully explored the functionality there. Also, I may be a bit confused on if you also want to keep the other colors. If so, that's the same as saying you want to remove the blue ones exceeding a certain distance threshold, which you would just do a -which command, and turn the < operator into a > operator.
Based off of a previous question I asked, which #Andrie answered, I have a question about the usage of the cut function and labels.
I'd like get summary statistics based on the range of number of times a user logs in.
Here is my data:
# Get random numbers
NumLogin <- round(runif(100,1,50))
# Set the login range
LoginRange <- cut(NumLogin,
c(0,1,3,5,10,15,20,Inf),
labels=c('1','2','3-5','6-10','11-15','16-20','20+')
)
Now I have my LoginRange, but I'm unsure how the cut function actually works. I want to find users who have logged in 1 time, 2 times, 3-5 times, etc, while only including the user if they are in that range. Is the cut function including 3 twice (In the 2 bucket and the 3-5 bucket)? If I look in my example, I can see a user who logged in 3 times, but they are cut as '2'. I've looked at the documentation and every R book I own, but no luck. What am I doing wrong?
Also - As a usage question - should I attach the LoginRange to my data frame? If so, what's the best way to do so?
DF <- data.frame(NumLogin, LoginRange)
?
Thanks
The intervals defined by the cut() function are (by default) closed on the right. To see what that means, try this:
cut(1:2, breaks=c(0,1,2))
# [1] (0,1] (1,2]
As you can see, the integer 1 gets included in the range (0,1], not in the range (1,2]. It doesn't get double-counted, and for any input value falling outside of the bins you define, cut() will return a value of NA.
When dealing with integer-valued data, I tend to set break points between the integers, just to avoid tripping myself up. In fact, doing this with your data (as shown below), reveals that the 2nd and 3rd bins were actually incorrectly named, which illustrates the point quite nicely!
LoginRange <- cut(NumLogin,
c(0.5, 1.5, 3.5, 5.5, 10.5, 15.5, 20.5, Inf),
# c(0,1,3,5,10,15,20,Inf) + 0.5,
labels=c('1','2-3','4-5','6-10','11-15','16-20','20+')
)
Say we have a set of 3D (integer) coordinates from (0,0,0) to (100,100,100)
We want to visit each possible coordinate (100^3 possible coordinates to visit) without visiting each coordinate more than once.
The sum of the differences between each coordinate in adjacent steps cannot be more than 2 (I don't know if this is possible. If not, then minimized)
for example, the step from (0,2,1) to (2,0,0) has a total difference of 5 because |x1-x2|+|y1-y2|+|z1-z2| = 5
How do we generate such a sequence of coordinates?
for example, to start:
(0,0,0)
(0,0,1)
(0,1,0)
(1,0,0)
(1,0,1)
(0,0,2)
(0,1,1)
(0,2,0)
(1,1,0)
(2,0,0)
(3,0,0)
(2,0,1)
(1,0,2)
(0,0,3)
etc...
Anyone know an algorithm that will generate such a sequence to an arbitrary coordinate (x,y,z) where x=y=z or can prove that it is impossible for such and algorithm to exist? Thanks
Extra credit: Show how to generate such a sequence with x!=y!=z :D
One of the tricks (there are other approaches) is to do it one line [segment] at a time, one plane [square] at a time. Addressing the last part of the question, this approach works, even if the size of the volume visited is not the same in each dimension (ex: a 100 x 6 x 33 block).
In other words:
Start at (0,0,0),
move only on the Z axis till the end of the segment, i.e.
(0,0,1), (0,0,2), (0,0,3), ... (0,0,100),
Then move to the next line, i.e.
(0,1,100)
and come backward on the line, i.e.
(0,1,99), (0,1,98), (0,1,97), ... (0,1,0),
Next to the next line, going "forward"
And repeat till the whole "panel is painted", i.e ending at
... (0,100,99), (0,100,100),
Then move, finally, by 1, on the X axis, i.e.
(1,100,100)
and repeat on the other panel,but on this panel going "upward"
etc.
Essentially, each move is done on a single dimension, by exactly one. It is a bit as if you were "walking" from room to room in a 101 x 101 x 101 building where each room can lead to any room directly next to it on a given axis (i.e. not going joining diagonally).
Implementing this kind of of logic in a programming language is trivial! The only mildly challenging part is to deal with the "back-and-forth", i.e. the fact that sometimes, some of the changes in a given dimension are positive, and sometimes negative).
Edit: (Sid's question about doing the same diagonally):
Yes! that would be quite possible, since the problem states that we can have a [Manhattan] distance of two, which is what is required to go diagonally.
The path would be similar to the one listed above, i.e. doing lines, back-and-forth (only here lines of variable length), then moving to the next "panel", in the third dimension, and repeating, only going "upward" etc.
(0,0,0) (0,0,1)
(0,1,0) first diagonal, only 1 in lengh.
(0,2,0) "turn around"
(0,1,1) (0,0,2) second diagonal: 2 in length
(0,0,3) "turn around"
(0,1,2) (0,2,1) (0,3,0) third diagonal: 3 in length
(0,4,0) turn around
etc.
It is indeed possible to mix-and-match these approaches, both at the level of complete "panel", for example doing one diagonally and the next one horizontally, as well as within a given panel, for example starting diagonally, but when on the top line, proceeding with the horizontal pattern, simply stopping a bit earlier when looping on the "left" side, since part of that side has been handled with the diagonals.
Effectively this allows a rather big (but obviously finite) number of ways to "paint" the whole space. The key thing is to avoid leaving (too many) non painted adjacent area behind, for getting back to them may either bring us to a dead-end or may require a "jump" of more than 2.
Maybe you can generalize Gray Codes, which seem to solve a special case of the problem.
Seems trivial at first but once started, it is tricky! Especially the steps can be 1 or 2.
This is not an answer but more of a demostration of the first 10+ steps for a particular sequence which hopefully can help others to visualise. Sid, please let me know if the following is wrong:
s = No. of steps from the prev corrdinates
c1 = Condition 1 (x = y = z)
c2 = Condition 2 (x!= y!= z)
(x,y,z) s c1 c2
---------------
(0,0,0) * (start)
(0,0,1) 1
(0,1,0) 2
(1,0,0) 2
(1,0,1) 1
(1,1,0) 2
(1,1,1) 1 *
(2,1,1) 1
(2,0,1) 1 *
(2,0,0) 1
(2,1,0) 1 *
(2,2,0) 1
(2,2,1) 1
(2,2,2) 1 *
(2,3,2) 1
(2,3,3) 1
(3,3,3) 1 *
(3,3,1) 2
(3,2,1) 1 *
(3,2,0) 1 *
.
.
.
Ok here's what I'm trying to accomplish. Say I have 100 items. I want to create a "grid"(each Item consisting of an x, y point). I want the grid to be as close to a square as possible.
Is there any kind of math to determine the grid width, and grid height i'd need by just a single number?(By grid width and height I mean the number of x items, and the number of Y items)
Now that I think about it would it be efficient to take the square root of the number, say varI=sqrt(45), remove the decimal place from varI...X=varI...then Y would be varI+1?
The square root is precisely what you need.
N
x=floor(sqrt(N))
y=raise(N/x)
This is the minimum rectangle that has more than N places and is closest to a square.
Now... if you want to find a rectangle that has exactly N places and is closest to a square...that's a different problem.
You need to find a factor of N, x, that's closest
You have to run through the factors of N and find the closest to sqrt(N). Then the rectangle is x by N/x, both integers.
There are several issues to consider here. If you want your grid to be as square as possible, for many Ns it will have empty cells in it. A simple example is N=10. You can create a 3x4 grid for it, but it will have two empty cells. A 2x5 grid, on the other hand, will have no empty cells. Some Ns (prime numbers) will always have empty cells in the grid.
But if you just want the square and don't care about empty fields then generally yes, you should take the square root. Say your number is N. Then, take R = int(sqrt(N)). Next, do an integer division N/R, take the quotient and add 1 to it. This is C. The grid is RxC. Note that when N is a square (like 100), this is a special case so don't add 1 to the quotient.
Example:
N = 40
R = int(sqrt(N)) = 6
C = int(40 / 6) + 1 = 7
grid is 6x7
I was looking to solve this problem too for a grid in html/css that had fixed dimensions and where N items would fit. I ended up creating my own script for that in javascript.
If you're interested in the method and maths I used, you can read http://machinesaredigging.com/2013/05/21/jgridder-how-to-fit-elements-in-a-sized-grid/, it's all documented there. I used recursion and it works really well, you can use the same method for your own language. Hope this helps.
I explored Eli's answer and found something I'd like to point out. For the sake of generality, one must add 1 to C only if R x C (C = int(N/R)) is not exactly N. So, the exception includes both numbers with square root and numbers which are exactly the product of two integers.
For instance:
N = 12
R = 3
C = 4 (int(N/R))
Hope it helps.