Comments on: About Mark Nelson http://marknelson.us Programming, mostly. Mon, 30 Jan 2012 17:56:19 +0000 hourly 1 http://wordpress.org/?v=3.2.1 By: Mark Nelson http://marknelson.us/about/comment-page-1/#comment-335418 Mark Nelson Fri, 29 Oct 2010 18:29:16 +0000 #comment-335418 Nothing too tricky here. The GIF encoder is creating nine-bit codes. Unfortunately for the encoder, the files are aligned on eight-bit boundaries. So when the bits of the first code are written out, the lower eight bits are packed into the first byte, then the most significant bit goes into the LSB position of the next output byte. That's why you see a 1 with a pipe symbol on its left hand side - that is the nine-bit boundary. The 028 is written out next. The lower seven bits are written into the upper seven bits of output byte two, then the upper two bits are written into the lower two bits of output byte 3. Repeat for each output code. - Mark Nothing too tricky here. The GIF encoder is creating nine-bit codes. Unfortunately for the encoder, the files are aligned on eight-bit boundaries. So when the bits of the first code are written out, the lower eight bits are packed into the first byte, then the most significant bit goes into the LSB position of the next output byte. That’s why you see a 1 with a pipe symbol on its left hand side – that is the nine-bit boundary.

The 028 is written out next. The lower seven bits are written into the upper seven bits of output byte two, then the upper two bits are written into the lower two bits of output byte 3.

Repeat for each output code.

- Mark

]]> By: Peter Gaczi http://marknelson.us/about/comment-page-1/#comment-335400 Peter Gaczi Fri, 29 Oct 2010 15:18:05 +0000 #comment-335400 [c] Mark Nelson, The WikiPedia article "Graphics Interchange Format" has an example of code packing as follows 9-bit binary Bytes I was able to reproduce their (hex) (hex) 11 compressed data bytes by 00000000 00 creating a GIF file that 100 matches their example. 0101000|1 51 028 I understand how the 9-bit 111111|00 FC codes on the left are obtained, 0FF but how the hell do they get 00011|011 1B from the 9-bit codes to the 103 11 bytes? 0010|1000 28 102 The diagram on the left looks 011|10000 70 there are 7 bits per code! 103 10|100000 A0 Where, for example, is the 106 100000000 binary value for the 1|1000001 C1 first code? 107 10000011 83 00000001 01 101 0000000|1 01 TIA, Peter Gaczi [/c] PLAIN TEXT
C:
  1. Mark Nelson,
  2.  
  3. The WikiPedia article "Graphics Interchange Format" has
  4. an example of code packing as follows
  5.  
  6. 9-bit   binary   Bytes         I was able to reproduce their
  7. (hex)            (hex)         11 compressed data bytes by
  8.         00000000  00           creating a GIF file that
  9. 100                            matches their example.
  10.        0101000|1  51
  11. 028                            I understand how the 9-bit
  12.        111111|00  FC           codes on the left are obtained,
  13. 0FF                            but how the hell do they get
  14.        00011|011  1B           from the 9-bit codes to the
  15. 103                            11 bytes?
  16.        0010|1000  28
  17. 102                            The diagram on the left looks
  18.        011|10000  70           there are 7 bits per code!
  19. 103
  20.        10|100000  A0           Where, for example, is the
  21. 106                            100000000 binary value for the
  22.        1|1000001  C1           first code?
  23. 107
  24.         10000011  83           
  25.         00000001  01
  26. 101
  27.        0000000|1  01           TIA, Peter Gaczi

]]> By: John http://marknelson.us/about/comment-page-1/#comment-326189 John Fri, 30 Apr 2010 19:07:04 +0000 #comment-326189 Hi mark, For the Data compression with BWT, can you explain to me the reasons why the BWT is implemented in this manner with the MTF? why is it done this way rather than in any other way? what are the advantages to it? Kind regards Hi mark,

For the Data compression with BWT, can you explain to me the reasons why the BWT is implemented in this manner with the MTF?

why is it done this way rather than in any other way?
what are the advantages to it?

Kind regards

]]> By: Otto http://marknelson.us/about/comment-page-1/#comment-321270 Otto Tue, 12 Jan 2010 09:15:24 +0000 #comment-321270 Carlos Eduardo Olivieri: I wrote some code for your problem. While this may still be regarded as a brute force solution, it is somewhat more efficient than the most naive solutions as it does early pruning. Maybe you can get some ideas from it. A further idea could be to try to find a formula that can be used to determine if a valid string can be built from given frequencies. [cpp] #include #include #include using namespace std; vector start; vector curPerm; map frequencies; const int MAX_REPEAT = 2; void process(){ //Do whatever, the current permutation is in curPerm vector::iterator it; for(it = curPerm.begin(); it!= curPerm.end(); it++) cout ::iterator it; for(it = frequencies.begin(); it != frequencies.end(); it++){ if(it->second == 0) continue; //If no items of this type are free int curItem = it->first; it->second--; //Decrease the available frequency of this item curPerm.push_back(curItem); if(curItem == lastItem){ if(runLength + 1 second++; //Revert the frequency as the item is not in use anymore } } int main(int argc, char* argv[]){ //Any starting permutation, e.g. 6667778 (even exceeding //the limit of allowed repeats) start.push_back(6); start.push_back(6); start.push_back(6); start.push_back(7); start.push_back(7); start.push_back(7); start.push_back(8); for(vector::iterator it = start.begin(); it != start.end(); it++) frequencies[*it]++; generator(-1, 0); } [/cpp] Carlos Eduardo Olivieri:

I wrote some code for your problem. While this may still be regarded as a brute force solution, it is somewhat more efficient than the most naive solutions as it does early pruning. Maybe you can get some ideas from it. A further idea could be to try to find a formula that can be used to determine if a valid string can be built from given frequencies.

PLAIN TEXT
C++:
  1. #include
  2. #include
  3. #include
  4. using namespace std;
  5.  
  6. vector start;
  7. vector curPerm;
  8. map frequencies;
  9. const int MAX_REPEAT = 2;
  10.  
  11. void process(){ //Do whatever, the current permutation is in curPerm
  12.     vector::iterator it;
  13.     for(it = curPerm.begin(); it!= curPerm.end(); it++)
  14.         cout ::iterator it;
  15.     for(it = frequencies.begin(); it != frequencies.end(); it++){
  16.         if(it->second == 0) continue; //If no items of this type are free
  17.         int curItem = it->first;
  18.        
  19.         it->second--; //Decrease the available frequency of this item
  20.         curPerm.push_back(curItem);
  21.         if(curItem == lastItem){
  22.             if(runLength + 1 second++; //Revert the frequency as the item is not in use anymore
  23.     }   
  24. }
  25.  
  26. int main(int argc, char* argv[]){
  27.    
  28.     //Any starting permutation, e.g. 6667778 (even exceeding
  29.     //the limit of allowed repeats)
  30.     start.push_back(6); start.push_back(6); start.push_back(6);
  31.     start.push_back(7); start.push_back(7); start.push_back(7);
  32.     start.push_back(8);
  33.    
  34.     for(vector::iterator it = start.begin(); it != start.end(); it++)
  35.         frequencies[*it]++;
  36.    
  37.     generator(-1, 0);
  38. }

]]> By: Ronald Garlit http://marknelson.us/about/comment-page-1/#comment-255923 Ronald Garlit Wed, 24 Dec 2008 06:44:34 +0000 #comment-255923 Just a note to say Hi. I don'tknow if you remember me from the 80's. Did a bunch of beta stuff for Don Killen in the old days on the C desktop side of things. I was in medical electronics then. Happy Holidays Ron You can delete this reply since it doesn't go with the article. I didn't find your email on the site. My info is: ----------- Personal Email: ron.garlit@gmail.com Business contact info: Ronald Garlit Lead Technical Architect | Distributed Computing .NET TEC Technologies, Strategy and Operations | American Express Company Ph: 856.848.7908 | Email: Ronald.A.Garlit@aexp.com Just a note to say Hi. I don'tknow if you remember me from the 80's.

Did a bunch of beta stuff for Don Killen in the old days on the C desktop side of things. I was in medical electronics then.

Happy Holidays
Ron

You can delete this reply since it doesn't go with the article. I didn't find your email on the site.

My info is:
-----------
Personal Email: ron.garlit@gmail.com

Business contact info:
Ronald Garlit
Lead Technical Architect | Distributed Computing .NET TEC
Technologies, Strategy and Operations | American Express Company
Ph: 856.848.7908 | Email: Ronald.A.Garlit@aexp.com

]]> By: Carlos Eduardo Olivieri http://marknelson.us/about/comment-page-1/#comment-253893 Carlos Eduardo Olivieri Thu, 18 Dec 2008 18:49:52 +0000 #comment-253893 Anyway, thank you very much Mark. In fact, brute force I also already think :), but I'm trying to find another way, bacause the processing time has been a big problem to me. Hours, days running just one program over very small input dataset. Regards. Anyway, thank you very much Mark.

In fact, brute force I also already think :), but I'm trying to find another way, bacause the processing time has been a big problem to me. Hours, days running just one program over very small input dataset.

Regards.

]]> By: Mark Nelson http://marknelson.us/about/comment-page-1/#comment-253861 Mark Nelson Thu, 18 Dec 2008 14:46:57 +0000 #comment-253861 @Carlos: I don't have an algorithm for your "at most 2 repeats" requirement, I don't think I've ever run across it. Brute force is my only suggestion :-) @Carlos:

I don't have an algorithm for your "at most 2 repeats" requirement, I don't think I've ever run across it. Brute force is my only suggestion :-)

]]> By: Carlos Eduardo Olivieri http://marknelson.us/about/comment-page-1/#comment-253812 Carlos Eduardo Olivieri Thu, 18 Dec 2008 10:57:54 +0000 #comment-253812 Hi, Mark. Nice to meet you. I'm Carlos Eduardo Olivieri. A brazilian student programmer since 12 years old (currently, I have 24). I started with BASIC (QBASIC), have passed through PASCAL (Turbo Pascal), C/C++, Assembly and some other things :). I have always been interested in algorithms, sort, cripto, binaries trees, data compression, memory operations, improve instruction set use, etc. I read your publication about permutations in C++ with function of STL next_perm(), very interesting and useful, after that I wrote one class method to get the next permutation in Delphi that is the tool most used for me in present days. This function works on lexographic order, I got the algo idea in another site, but now I have with big problem. I'm working with permutations with repetead elements in vector and there are lot of permutations that I don't need. For example, I have this first permutation for 7 elements in lexographic order: 6667778 (6 = 3 times consecutively, 7 = 3 times consecutively) For my work I consider valid perm only those with at most 2 elements repeated consecutively, like that: 6676778 (6 = 2 times consecutively at most, 7 = 2 times consecutively at most) Short, I need a function that return to me only permutations that have at most N consecutive repetitions, according parameter received. Do you know if there is some kind of algorithm to do that already done? I intend to write to you more times. Wait for your contact and sorry for any mistakes in text, I still don't speak English very well. Thank you so much, Carlos Hi, Mark.

Nice to meet you.

I'm Carlos Eduardo Olivieri. A brazilian student programmer since 12 years old (currently, I have 24). I started with BASIC (QBASIC), have passed through PASCAL (Turbo Pascal), C/C++, Assembly and some other things :). I have always been interested in algorithms, sort, cripto, binaries trees, data compression, memory operations, improve instruction set use, etc.

I read your publication about permutations in C++ with function of STL next_perm(), very interesting and useful, after that I wrote one class method to get the next permutation in Delphi that is the tool most used for me in present days. This function works on lexographic order, I got the algo idea in another site, but now I have with big problem. I'm working with permutations with repetead elements in vector and there are lot of permutations that I don't need. For example, I have this first permutation for 7 elements in lexographic order:

6667778 (6 = 3 times consecutively, 7 = 3 times consecutively)

For my work I consider valid perm only those with at most 2 elements repeated consecutively, like that:

6676778 (6 = 2 times consecutively at most, 7 = 2 times consecutively at most)

Short, I need a function that return to me only permutations that have at most N consecutive repetitions, according parameter received.

Do you know if there is some kind of algorithm to do that already done?

I intend to write to you more times.

Wait for your contact and sorry for any mistakes in text, I still don't speak English very well.

Thank you so much,
Carlos

]]>