I am new to LabVIEW and I am trying to read a code written in LabVIEW. The block diagram is this:
This is the program to input x and y functions into the voltage input. It is meant to give an input voltage in different forms (sine, heartshape , etc.) into the fast-steering mirror or galvano mirror x and y axises.
x and y function controls are for inputting a formula for a function, and then we use "evaluation single value" function to input into a daq assistant.
I understand that { 2*(|-Mpi|)/N }*i + -Mpi*pi goes into the x value. However, I dont understand why we use this kind of formula. Why we need to assign a negative value and then do the absolute value of -M*pi. Also, I don`t understand why we need to divide to N and then multiply by i. And finally, why need to add -Mpi again? If you provide any hints about this I would really appreciate it.
This is just a complicated way to write the code/formula. Given what the code looks like (unnecessary wire bends, duplicate loop-input-tunnels, hidden wires, unnecessary coercion dots, failure to use appropriate built-in 'negate' function) not much care has been given in writing it. So while it probably yields the correct results you should not expect it to do so in the most readable way.
To answer you specific questions:
Why we need to assign a negative value and then do the absolute value
We don't. We can just move the negation immediately before the last addition or change that to a subtraction:
{ 2*(|Mpi|)/N }*i - Mpi*pi
And as #yair pointed out: We are not assigning a value here, we are basically flipping the sign of whatever value the user entered.
Why we need to divide to N and then multiply by i
This gives you a fraction between 0 and 1, no matter how many steps you do in your for-loop. Think of N as a sampling rate. I.e. your mirrors will always do the same movement, but a larger N just produces more steps in between.
Why need to add -Mpi again
I would strongly assume this is some kind of quick-and-dirty workaround for a bug that has not been fixed properly. Looking at the code it seems this +Mpi*pi has been added later on in the development process. And while I don't know what the expected values are I would believe that multiplying only one of the summands by Pi is probably wrong.
I am a novice at using TradingView's Pinescript and having a hard time finding an easy to understand example of a script. I am used to Java/C++ and Pinescript is very different. I am trying to build a script that will scan a stock chart and look for gaps of over 5%. Here is psuedocode for what I am trying to create:
if(difference between open of current day and previous day close > 5%) {
plot a green circle or red circle, depending on if gap was up or down
}
Thank you in advance!
You're best bet would be to go through their tutorial
There's some odds choices in this language if you have any programming background so it's probably a good idea to read it all (it's not that much). E.g.
open is the current bars open price, but open[1] is the previous bar open price (so should be read as open[current_index-1])
you can't use the plot calls inside function bodies
as for you question (not tested, but should be close enough to give the right idea):
study(title='gap detector', overlay=true)
//plotshape(<condition>, <options>) // condition must be true to plot something
is_percentage_increase = if (close-close[1])/close[1] > 0.05
true
plotshape(is_percentage_increase, style=shape.circle, color=green)
Pine scripting is easy to use; Initially it was bit hard to understand, Once started using it it becomes so useful to strategize the logic.
In your case you can use conditional operator as well to detect this.This will work in Version 2 .The version 3 is bit differrent
//version =2
study(title ="Experementing the code ",overlay =true ,shorttitle ="testing") //overlay=false to get this down of the chart as seperate layout
plotchar( (close-close[1])/close[1] >0.05 ? 1:na ,char =' ',text ="plot\nTest",textcolor=red,size.huge)
Instead of if the condition you can use ?: operator to do this job.
Please make sure plotchar(.....) coming in the same line, not in separate line.
Pine has lot of cool features to use and helped me to derive my own strategy. The tutorial is really good.
Note if you don't put char='' above it will print STAR as the default character. And in the character even if you put char='testtest' it will print the only t .
i am currenty studying BDD, but i have a different doubt, can you tell me if the following is right or not:
1) Feature = it means "the problem" isn't it?
2) Scenario = the way (the beahaviour) to resolve the feature
I find very difficult to find the "given when then" sentences.
In this problem for example:
As a student
I would like / i want to calculate the rectangle perimeters if i have 2 number
Or the circle area if i have one
So i don't make mistake with the computation
I wrote down the scenario, is that correct?
Given 1 number
Or 2 number
When i have 1 positive number
Or 2 positive number
Then calculate the area
Or the Perimeters
About the terminology:
1) feature is not a "problem". It would rather be a solution.
In software programming, a feature is a something that your program does to solve a problem.
A feature could be the ability to compute the area of a rectangle.
2) a scenario is a description of the usage of your feature. Like an example.
Like a test case, but usually in a more human-readable form.
3) a story (in Agile terminology, in which BDD stands) is a way to describe the a need/problem.
Your problem ("as a student...") is presented as a story.
This story will lead to a new feature in your soft.
This new feature will be tested by scenarios.
About your scenarios.
Yours are not correct.
There is no way to know that if you have 1 nb you should compute an area.
You should have several scenarios, like
Given I send the number 2
When I launch the computation
Then I get the result 12,56
Given I send the number 2 and 3
When I launch the computation
Then I get the result 10
Given I send the number -4
When I launch the computation
Then I get the result error
Given I send the number 1 3 7
When I launch the computation
Then I get the result error
I have created a game, (4 in a row), in prolog. My heuristic function requires me to know how many Player's and Opponent's chips are in each possible 4-row combination on the board. The method I am using is as follows (in psuedocodish):
I have 1 list of all possible fours of the board (ComboList) =of the form==> [[A,B,C,D]|Rest].
I have 1 list of all the moves of the 1st player (List1) =of the form==> [[1],[7],[14]]
And 1 for opponent's moves (List2).
Step 1: obtain the first combo from ComboList, 2:
Check all of List1 to see how many are in this combo, 3:
Check all of List2 to see how many are in this combo,
Move onto the next combo from ComboList and start over...
This PROCESS takes waay too much runtime for what is required.
Please can someone suggest something better and more efficient! Much thanks in advance!
The following code uses member/3, which has also to become
known as nth1/3. See here:
http://storage.developerzen.com/fourrow.pro.txt
The predicate is nowadays found in library(lists) and has
possibly native support or a fast implementation:
http://www.swi-prolog.org/pldoc/man?predicate=nth1/3
But I guess asserting some facts and relying on argument
indexing might get you an even better result. See for
example here:
http://www.mxro.de/applications/four-in-a-row
Hope this helps.
Bye
I'm attempting to gauge the percentage difference between two images.
Having done a lot of reading I seem to have a number of options but I'm not sure what the best method to follow for:
Ease of coding
Performance.
The methods I've seen are:
Non language specific - academic Image comparison - fast algorithm and Mac specific direct pixel access http://www.markj.net/iphone-uiimage-pixel-color/
Does anyone have any advice about what solutions make most sense for the above two cases and have code samples to show how to apply them?
I've had success calculating the difference between two images using the histogram technique mentioned here. redmoskito's answer in the SO question you linked to was actually my inspiration!
The following is an overview of the algorithm I used:
Convert the images to grayscale—compare one channel instead of three.
Divide each image into an n * n grid of "subimages". Then, for subimage pair:
Calculate their colour composition histograms.
Calculate the absolute difference between the two histograms.
The maximum difference found between two subimages is a measure of the two images' difference. Other metrics could also be used (e.g. the average difference betwen subimages).
As tskuzzy noted in his answer, if your ultimate goal is a binary "yes, these two images are (roughly) the same" or "no, they're not", you need some meaningful threshold value. You could produce such a value by passing images into the algorithm and tweaking the threshold based on its output and how similar you think the images are. A form of machine learning, I suppose.
I recently wrote a blog post on this very topic, albeit as part of a larger goal. I also created a simple iPhone app to demonstrate the algorithm. You can find the source on GitHub; perhaps it will help?
It is really difficult to suggest something when you don't tell us more about the images or the variations. Are they shapes? Are they the different objects and you want to know what class of objects? Are they the same object and you want to distinguish the object instance? Are they faces? Are they fingerprints? Are the objects in the same pose? Under the same illumination?
When you say performance, what exactly do you mean? How large are the images? All in all it really depends. With what you've said if it is only ease of coding and performance I would suggest to just find the absolute value of the difference of pixels. That is super easy to code and about as fast as it gets, but really unlikely to work for anything other than the most synthetic examples.
That being said I would like to point you to: DHOG, GLOH, SURF and SIFT.
You can use fairly basic subtraction technique that the lads above suggested. #carlosdc has hit the nail on the head with regard to the type of image this basic technique can be used for. I have attached an example so you can see the results for yourself.
The first shows a image from a simulation at some time t. A second image was subtracted away from the first which was taken some (simulation) time later t + dt. The subtracted image (in black and white for clarity) then shows how the simulation has changed in that time. This was done as described above and is very powerful and easy to code.
Hope this aids you in some way
This is some old nasty FORTRAN, but should give you the basic approach. It is not that difficult at all. Due to the fact that I am doing it on a two colour pallette you would do this operation for R, G and B. That is compute the intensities or values in each cell/pixal, store them in some array. Do the same for the other image, and subtract one array from the other, this will leave you with some coulorfull subtraction image. My advice would be to do as the lads suggest above, compute the magnitude of the sum of the R, G and B componants so you just get one value. Write that to array, do the same for the other image, then subtract. Then create a new range for either R, G or B and map the resulting subtracted array to this, the will enable a much clearer picture as a result.
* =============================================================
SUBROUTINE SUBTRACT(FNAME1,FNAME2,IOS)
* This routine writes a model to files
* =============================================================
* Common :
INCLUDE 'CONST.CMN'
INCLUDE 'IO.CMN'
INCLUDE 'SYNCH.CMN'
INCLUDE 'PGP.CMN'
* Input :
CHARACTER fname1*(sznam),fname2*(sznam)
* Output :
integer IOS
* Variables:
logical glue
character fullname*(szlin)
character dir*(szlin),ftype*(3)
integer i,j,nxy1,nxy2
real si1(2*maxc,2*maxc),si2(2*maxc,2*maxc)
* =================================================================
IOS = 1
nomap=.true.
ftype='map'
dir='./pictures'
! reading first image
if(.not.glue(dir,fname2,ftype,fullname))then
write(*,31) fullname
return
endif
OPEN(unit2,status='old',name=fullname,form='unformatted',err=10,iostat=ios)
read(unit2,err=11)nxy2
read(unit2,err=11)rad,dxy
do i=1,nxy2
do j=1,nxy2
read(unit2,err=11)si2(i,j)
enddo
enddo
CLOSE(unit2)
! reading second image
if(.not.glue(dir,fname1,ftype,fullname))then
write(*,31) fullname
return
endif
OPEN(unit2,status='old',name=fullname,form='unformatted',err=10,iostat=ios)
read(unit2,err=11)nxy1
read(unit2,err=11)rad,dxy
do i=1,nxy1
do j=1,nxy1
read(unit2,err=11)si1(i,j)
enddo
enddo
CLOSE(unit2)
! substracting images
if(nxy1.eq.nxy2)then
nxy=nxy1
do i=1,nxy1
do j=1,nxy1
si(i,j)=si2(i,j)-si1(i,j)
enddo
enddo
else
print *,'SUBSTRACT: Different sizes of image arrays'
IOS=0
return
endif
* normal finishing
IOS=0
nomap=.false.
return
* exceptional finishing
10 write (*,30) fullname
return
11 write (*,32) fullname
return
30 format('Cannot open file ',72A)
31 format('Improper filename ',72A)
32 format('Error reading from file ',72A)
end
! =============================================================
Hope this is of some use. All the best.
Out of the methods described in your first link, the histogram comparison method is by far the simplest to code and the fastest. However key point matching will provide far more accurate results since you want to know a precise number describing the difference between two images.
To implement the histogram method, I would do the following:
Compute the red, green, and blue histograms of each image
Add up the differences between each bucket
If the difference is above a certain threshold, then the percentage is 0%
Otherwise the colors found in the images are similar. So then do a pixel by pixel comparison and convert the difference into a percentage.
I don't know any precise algorithms for finding the key points of an image. However once you find them for each image you can do a pixel by pixel comparison for each of the key points.