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If you can't find the equivalent of int__64 (ULONGLONG for me) in C# you can break your sum into a set of essentially 'multi-digit' digits.
That is: an accumulator for 0-999, another for 1,000-999,000, another for 1,000,000-999,000,000, etc. Now, every time your 0-999 accumulator goes greater than 1000, subtract the 1,000's off that accum and add 1 to the 1,000-999,000 accum.
Do you see it?
Tadeusz Westawic
An ounce of Clever is worth a pound of Experience.
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Hi everyone,
I have a small question, for my school project i need to make a program for counting brick in image.
So I have a picture with nothing but bricks on it, and i need to count them,
the idea i have is to use Sobel algorithm, blur, binaryzation and region search, something like that.
Is is the right way to go, or is there any better one?
Thank you all for your help!
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I would say you have it spot on... filter the image and then apply a region counting algorithm.
The filtering is relatively simple... but the region finding could be harder (I've never used/made one).
If you get the original image down to just the outlines of the bricks (via the filters), then you could use a Monte-Carlo based search to look for borders surrounding each point.
[I wish I had a school project as interesting as yours!]
Matthew Butler
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Hmm ya all i need is some guidance where i can get this algorithms implemented in C#.
Thnx for your reply.
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Member 3897207 wrote: all i need is some guidance where i can get this algorithms implemented in C#.
First, you ask for some advice on if your approach to the problem, which is fine. But now you respond to Mathew's advice saying you just want help on how to cheat to do your homework. I was going add a litte to his advice but now I won't and you'll probably not get any further help.
2 75 22 6
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Well what can i say...basically this came out wrong, because i found some algorithms, but they are written in java, so i wondered if there is any...ok...not just for this problem, but if there is any site with algorithms (general) where code is in C#.
Well if i don't get any help any more...so be it.
Thnx anyway at least for confirming that the approach is at least a bit correct.
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C# is very similar to Java (C# partly originated from it). You should be able to translate the code with little hassle. If you get stuck, post your code, and so long as your question is reasonable, we'd be happy to help.
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Yes, to echo what Mark said. The way to approach us on this is to present the problem, describe what you've done to solve it, and ask why you're not getting the results you expect. Then you'll probably get some help. But to post like you're simply looking for a place to cut and paste your homework won't fly. We get plenty of queries like that and they're generally all told to take a hike, some less politely than others.
One potential problem with the method you sketched is that the lines you get after you run the Sobel and threshold is they may contain gaps for one reason or another. You may need a way to look at the partial lines and fill in the blanks. Since the problem is fairly artificial, you have a lot of prior knowledge about what you're looking for.
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Tim Craig wrote: to post like you're simply looking for a place to cut and paste your homework won't fly
I ditto that Tim. I was going to give some input, but not now
"The clue train passed his station without stopping." - John Simmons / outlaw programmer
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
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Well project done...basically I managed to get all algorithems done, except region search, but well in the end I used AForge .NET Framework and got the job done in the fraction of time.
Thank you all for your help and assistance.
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Hi to all,
I want to send a video file on network (UDP) in C++. But when I do that, it doesn't receive properly.
So can anyone help me or give me an algorithm/code to send data on IP continously at a particular bitrate in C++?
Thanks & Regards,
Aniket A. Salunkhe
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Hi
I work with some very large arrays (millions of items) in an algorithm, however, since most fields contains the value 0 I decided to use a Dictionary<int, int=""> to represent only the array's fields where the value is not 0.
Dictionary<int, int=""> means that the first int is the index of the array in which the value would appear, and the second int is the value of this index.
Now, I have two of these "arrays" that I need to compare. For each _existing_ index I need to find the difference between the arrays.
E.g. index i exists in dict1 (first dictionary) but not in dict2, thus the difference should be the value of dict1[i] alone. index i exists in both dictionaries thus the difference should be dict1[i] - dict2[i].
Now, in theory this should be possible to achieve with iterating each unique index in BOTH dictionaries only once (e.g. if an index occurs in both dictionaries it should only be iterated once)
My suggestion was that I start by getting the first item from each and compare the index of these.if they are equal I would increase both iterators.if one is higher than the other I would increase the iterator until they are equal or the other one is lower. This would be repeated until both are iterated through. Any ideas how I can do this with C# and Dictionary<int, int="">?
Any ideas? need to do this iteration as fast as possible. In C I could do some pointer magic to achieve this, but I am not familiar with C# dictionary iterators/enumerators so I have no idea whatsoever how to solve this.
modified on Friday, May 9, 2008 2:50 PM
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I was going to suggest using pointers until I saw your last paragraph. It seems to me this is a language-specific question with regard to C# rather than an algorithm question, per se. Thus, you could try the C# forum[^]. (I have zero experience with C#.)
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If you could store both arrays in the same dictionary it would solve your problem. So for each index, the dictionary item would be a PAIR of ints (one from each array).
Then you go through each element, and get the difference between the pair of values at that index. This structure would also use less memory than two dictionaries.
Using a struct for the pair of ints would avoid the implicit use of a pointer that would result from a class/reference.
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SortedDictionary might help here. Theres also SortedList which is very similar. Check the API notes. SortedDictionary IIRC is quicker but uses more memory. Main speedup is in adhoc query by key, which isn't what you are doing if you iterate the keys collection.
Probably SortedList will help you out. The actual algorithm could be as you said, quite simple.
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Mark Churchill wrote: Main speedup is in adhoc query by key, which isn't what you are doing if you iterate the keys collection.
There are two approaches I can see. If what you want is to produce a new dictionary containing the differences, then your best bet is probably something like:
if dict1reader.eof then
output dict2reader
advance dict2reader
else if dict2reader.eof then
output dict1reader
advance dict1reader
elseif dict1reader.key = dict2reader.key then
output(dict1reader.key, dict1reader.value-dict2reader.value)
advance dict1reader
advance dict2reader
elseif dict1reader.key < dict2reader.key then
output(dict1reader.index, dict1reader.value)
advance dict1reader
else
output(dict2reader.index, dict2reader.value)
advance dict2reader
endif
Iterate that until both readers report eof.
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I have a slight maths problem - Maybe you guys can help me out
I have 94 unique placeholders, with a possible digit on the end.
I have 16581375 total possibilities.
I'm wondering for a way to calculate which placeholder goes with which possibility.
Eg:
Possibility 1 - PlaceHolder 1 (With a 0)
Possibility 95 - PlaceHolder 1 (With 1)
Possibility 189 - PlaceHolder 1 (With a 2)
- Reelix
-= Reelix =-
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Reelix wrote: I have 94 unique placeholders, with a possible digit on the end.
With a single digit ? Then it means that you have 10 choices for each unique placeholder. So, a total of 940 combinations.
Reelix wrote: I have 16581375 total possibilities.
It's a bit more than 940, don't you think ? So, you cannot fit all your possibilities in your 940 combinations.
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Sorry, my mistake
Multiple Digits
-= Reelix =-
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It's not clear what your problem is. You need to define it better if you want some help with it.
For example, what's the relationship between "placeholders" and "possibilities"? What are your constraints? How is the digit related to the other items?
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Please elaborate (maybe you too will gain some insight).
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
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Reelix wrote: I'm wondering for a way to calculate which placeholder goes with which possibility.
Then you will need a relationship between placeholders and possibilities. Do you have such a relationship?
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Hi all. I've been working on a model in Excel that I'm considering porting into C/C++. I've been using the "Goal Seek" function in Excel to find a value in one cell that causes the results in another cell to equal some target value.
Is there an existing numerical recipe/algorithm to perform a similar function in C without reinventing the wheel?
Thanks for any help!
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Not in C/C++, but this is a custom implementation in VB:
http://www.bmsltd.ie/DLCount/DLCount.asp?file=GoalSeek.zip[^]
Goal seeking is based on linear interpolation, perhaps you can adapt the VB code to C++?
I'm not too familiar with COM, etc. so I don't know the details but I do know that you can make calls to Excel from C++, maybe you can call the Excel Goal Seek function using COM (by loading the Excel .dll)?
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"...but I do know that you can make calls to Excel from C++..."
Well, THAT would solve my problem! I've been doing a lot of processing in Excel (nicer, because I can see the numbers all laid out), then saving data to a text file, reading it into my C program and continuing my model that way. If I could just make calls right into a spreadsheet I could save myself a lot of dumb grunt labor. I'll check into that. Thanks!
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