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electronic noise?
Ciao,
luker
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Thermal noise is a common one. Interval between photons on a light detector would work well if it was low enough intensity. Radioactive decay is a classic which you can use through a web service if you're connected to the Internet.
Intervals between discrete events are generally better, as a continuous variable will always have linkage between readings taken close together in time (i.e. if it's 15.264C now, a measurement taken next time round the loop is more more likely to be 15.263, 4 or 5 than it is any other value).
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There is no such a thing as a random number in practice. I would Rather you use pseudo - random since whatever you use to generate your "random" number has a pattern unlike in theoretical mathematics. So temperature or humidity, like others have said, have certain patterns though complex but not random.
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Because Intel has incorporated a high speed random number generator into the Ivy Bridge CPU architecture? It is called Bull Mountain.
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i have 2 aspx page
1st page one drop down list 1 text box
textbox visible ="false"
drop down list 2 value , if i select 1st value it goes to 2nd page . if i select 2nd value, text box should be visible..
in my 2nd aspx page only one drop down list . if i select one value it goes to first page .. i did until this
now if i select 2nd page drop down list value , it goes first page and textbox should be visible ..
can any 1 help me pls ?
my coding :
first page:
script type="text/javascript">
function goto1(str) {
if (str =="http://www.contactus.aspx")
{
window.location = str;
}
else if (str == "Corporate") {
document.getElementById("txtdesignation").style.display = "block";
}
}
<asp:dropdownlist id=""DropDownList1"" runat=""server"" onchange=""goto1(this.value)"" height=""22px">
" <asp:listitem="" text=""Select"" value="""></asp:listitem>
" ><="" asp:listitem>
="" <="" asp:dropdownlist><="" pre="">
<asp:textbox ID="txtdesignation" style="display:none "
runat="server" Height="22px" />
2nd page only one drop down list:
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1. Format your code blocks properly.
2. Post your question in the appropriate forum.
Unrequited desire is character building. OriginalGriff
I'm sitting here giving you a standing ovation - Len Goodman
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I'm being stuck at a project on Image feature extraction. I don't have many documents about it, such as to know what 'Image feature' is, how many methods we can use to extract image features and what they are. Then I have to write some code demonstrating what I've understood about Image feature extraction (as well as show its application). I intend to code to extract color features of image which are considered as the easiest thing to do with image feature extraction. I post this question here in algorithm forum because I think I need some clear step-by-step guide of algorithm to code follow. If possible, could you please help me with the following stuffs:
1. The best documents on Image feature extraction that cover many enough the basic concepts. (please don't tell me to Google it, I've done for weeks but the obtained documents are not much useful, I hope in your searching ability or your experience).
2. Some sample code extracting image feature (I prefer color feature, but the others will be fine). I'm also in need for some code involving extracting image feature, for example before going to next stage of processing like pattern recognition or content-based image searching.
Finally, I'd like to say why I prefer color feature, just because it's the simplest of all, and I think I can code it in a page of lines without using any library, for others, I think some library like OpenCV should be used, but it's still not easy to me. (at least I have to learn about OpenCV and consume my time away, I have only more 1 week to finish all).
Your help would be highly appreciated anyway!
Thank you all, thank this forum!
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There's an article here somewhere, in which the author crafts a program that will solve any sudoku puzzle held up-to the pc's webcam.
That article definitely deals with pattern matching - it's how it works out which digits are already printed within the game board.
There's also another article here that automatically classifies images according to pre-learnt categories - Sunset, People, Animals, Flowers, Trees etc, etc.
They'd both be likely a good read for you and should help.
:eeek: "I have only more 1 week to finish all" - I won't ask how long you've had it for so far... 
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Thank you!
You were always saying "there's an article here..." , "there's also another article here..."... but I don't see where they are, I need some link to those articles, could you please fulfil your help? I remember that I searched in here with all the key words possible but there were only some 'face recognition' projects using some library like Opencv.
Anyway thank you very much!
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There is a near-infinite number of image features you could extract. Which features you choose to extract should depend on your goal(s).
E.g. if your goal is facial recognition, you'd extract features related to heads, eyes, etc... If your goal is part inspection, you'd extract features like lines with an edge-detection algorithm.
Your problem seems to stem from the lack of goals; you appear to be doing feature extraction as an end in itself.
There are a variety of edge-detection algorithms, some of which are easy to implement. One of the easiest is to shift the image one pixel then subtract it from the original image. Then set all pixels in the difference image below a threshold to zero.
This zeroes out solid regions, leaving you with just the edges. After edge detection, you can take the edges and use them to identify higher-level features, based on their positions.
If I were you, I wouldn't start learning a new technology one week before a deadline; that's asking for trouble. Just focus on what you can do with straightforward programming.
"Microsoft -- Adding unnecessary complexity to your work since 1987!"
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I will assume color processing is what you want... i can advise that you start with color opponent process as occurs in the retinal ganglion cells see http://en.wikipedia.org/wiki/Opponent_process[^]
opponent process results in Red vs Green, Blue vs Yellow and Black vs White channels, if you are wondering why go through this process?
Well color sensor response overlaps (even the photo receptors in animals & humans) , and it is biologically proven to work.
but only Red vs Green and Blue vs Yellow are what is needed for color processing. you can obtain a two dimensional vector at any position (one sample per channel) normalize the vector to unity magnitude and store it with the associated color (in an indexing structure like the Kd tree see http://en.wikipedia.org/wiki/Kd_tree[^]) as a prototype. So get such descriptors for all colors. then at run time repeat the whole process except that you run the indexing structure and label each pixel with a given color based on it's closest prototype in the indexing structure. Then you can do some statistics such as create a histogram to count the frequency of occurrence of each color and normalize the histogram, and that will be a descriptor vector for an image hence the resulting color descriptor can be used to retrieve images or compute their similarity score.
or and 1 week, where have you been? 
And as for feature extraction, there are low,intermediate and high level features, most of computer vision uses either low or intermediate level features there are so many algorithms for that such as SIFT (Scale Invariant Feature Transform) see http://en.wikipedia.org/wiki/Scale-invariant_feature_transform[^] and http://www.cs.ubc.ca/~lowe/papers/ijcv04.pdf[^]
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The problem is:
given a set of numbers:
xi > d and
xi+1 > xi and
xi+1 - xi > d, i=1..N and
"distance" d
find such the greatest number v which satisfies the condition for each xi-d/2 and xi+d/2 to be entirely included between kv and (k+1)v where k is integer > 0.
I found an empiric algorithm first to be evaluated manually in spreadsheet and then I coded it but I am not sure whether it is reliable without some theory.
modified 16-Mar-12 15:41pm.
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The problem is not completely described, is it?
I guess "d" is greater than 0. Any further conditions for d?
The minimum value for xi-d/2 is Min(xi)-d/2. When d is greater than 2 times the smallest x value, that expression becomes negativ, and there is no v fulfilling the conditions.
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Right. I should write: xi > d.
I know that it is not solvable in any case of d and set of xi. It would be interesting to find limitations when it is.
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Are x, d, and v also integers?
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That doesn't make much sense to me. As the set xi is an ordered set, all that matters are the values of the smallest and largest of the xi numbers.
So it boils down to:
given natural numbers d, min, and max, where
d < min
min + d < max
find values k and v such that
k*v < min - d/2
(k+1)*v > max + d/2
which can't be hard to solve, by picking any v that satisfies
v > max - min + d
then finding k through division.
So I'd guess you want something entirely different.
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Let me show this on the example:
You're given
x1=22
x2=50
x3=100
and d=10
We can find v=15 which satisfies this:
we can divide the range <0,100(or at least)> into equal intervals of length 15 because our "10" fits in:
1*15 and 2*15 with the middle in 22 i.e. <17,27>
3*15 and 4*15 with the middle in 50 i.e. <45,55>
6*15 and 7*15 with the middle in 100 i.e. <95,105>
I hope it's clearer now and excuse me if the latter was not.
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it now sounds like you are accepting (or even requiring) a different k value for each new x sample...
if so, I don't know an algorithm that does it right away, however it reminds me of hash function optimisation, maybe a little Google could help you.
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That means d <= v <= x<small>1</small> - d/2, v must not be a divider of any xi.
There is no solution when you set x2=49 or x3=99. Are there some more constraints on the series?
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So you know the following:
1. d ≤ v ≤ x1 - d/2
2. ∀xn, d/2 ≤ xn % v ≤ v - d/2
My dumb algorithm in C# would be (untested):
int FindV(List<int> x, int d) {
int halfD = (d + 1) / 2;
d = 2 * halfD;
for (int v = x[0] - halfD; v >= d; --v) {
int upper = v - halfD;
bool solved = true;
foreach (int xn in x) {
int mod = xn % v;
if (mod < halfD || upper < mod) {
solved = false;
break;
}
}
if (solved) return v;
}
return -1;
}
Sounds like somebody's got a case of the Mondays
-Jeff
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can any developer help me
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