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• Creating A Lossless SMC Encoder

  26 avril 2011, par Multimedia Mike — General

  Look, I can’t explain how or why I come up with this stuff. For some reason, I thought it would be interesting to write a new encoder for the Apple SMC video codec. I can’t even remember why. I just sat down the other day, started writing, and now I have a lossless SMC encoder that I’m not sure what to do with. Maybe this is to be my new thing— writing encoders for marginal multimedia formats.

  Introduction
  SMC is a vector quantizer (a lossy method) but I decided to attack it from the angle of lossless encoding. A.k.a. Apple Graphics Codec, SMC operates on 4x4 blocks in an 8-bit paletted colorspace. Each 4x4 block can be encoded with 1, 2, 4, 8, or 16 colors. Blocks can also be skipped (copied from previous frame) or copied from blocks rendered immediately prior within the same frame.

  Step 1 : Validating Infrastructure
  The goal of this step is to encode the most braindead SMC frame possible and see if FFmpeg/libav’s QuickTime muxer can create a valid file. I think the simplest frame would be one in which each vector is encoded with the single-color mode, starting with color 0 and incrementing through the palette.

  Status : Successful. The only ’trick’ was to set avctx->bits_per_coded_sample to 8. (For fun, this can also be set to 40 (8 | 0x20) to specify a grayscale palette.)
  

  
  
  

  Step 2 : Preprocessing
  The video frames will arrive at the encoder as 32-bit RGB. These will need to be converted to a paletted colorspace before encoding. I don’t want to use FFmpeg’s default dithering approach as this will result in a substantial loss of quality as described in this post. I would rather maintain a palette built from observed colors throughout successive frames. If the total number of unique observed colors ever exceeds 256, error out.

  That’s what I would like to do. However, I noticed that FFmpeg/libav’s QuickTime muxer has never taken into account the possibility of encoding palettes. The path of least resistance in this case is to dither the input to match QuickTime’s default 8-bit palette (if a paletted QuickTime file does not specify a palette, a default 1-, 2-, 4-, or 8-bit palette is selected).

  Status : Successful, if slow. I definitely need to optimize this step later.

  Step 3 : Most Naive Encoding
  The most basic encoding is to "encode" each block as a 16-color block. This will actually result in a slightly larger frame size than a raw encoding since each 4x4 block will be prepended by a byte opcode (0xE0 in this case) to indicate encoding mode. This should demonstrate that the encoder is functioning at the most basic level.

  Status : Successful. Try not to laugh too hard at the Big Buck Bunny dithered to an 8-bit palette :

  
  
  

  Step 4 : Better Representation
  It seems to me that encoding this format (losslessly) will entail performing vector operations on lots of 16-element (4x4-pixel) vectors. These could be done on the frame as-is, but it strikes me as more efficient and perhaps less error prone to rearrange the input images into a vector of vectors (or array of arrays if you prefer) :

    0  1  2  3  w ...
    4  5  6  7  x ...
    8  9  A  B  y ...
    C  D  E  F  z ...
  

    0 : [0 1 2 3 4 5 6 7 8 9 A B C D E F]
    1 : [...]
  

  Status : Successful.

  Step 5 : Add Interframe Skip Codes
  Time to add a bit of brainpower to the proceedings : On non-keyframes, compare the current vector to the vector at the same position from the previous frame.

  Test this by encoding a pair of identical frames. Ideally, all codes should be skip codes.

  Status : Successful, though my vector matching function could probably be improved.

  Step 6 : Analyze Blocks For Optimal Color Coding
  This is where things get potentially interesting, algorithmically. At least, I need to figure out (or look up) an algorithm to count the unique elements in a vector.

  Naive algorithm (i.e., first thing I can think of) :

  • initialize a count variable to 0
  • initialize an array of 256 flags to false
  • for each 8-bit element in vector :
    • if flag array[element] is 0, set array[element] to true and increment count

  Status : Successful. Here is the distribution for the 640x360 Big Buck Bunny title :

  1194 4636 4113 2140 1138 568 325 154 80 36 9 5 2 0 0 0

  Or, in pretty graph form, demonstrating that vectors with few distinct elements dominate :
  

  
  
  

  Step 7 : Encode Monochrome Blocks
  At this point, the structure is starting to come together pretty well. This phase involves encoding a 0x60 opcode and a palette index when the count_distinct() function returns 1.

  Status : Absolutely no problem.

  Step 8 : Encode 2-, 4-, and 8-color Modes
  This step is a little more involved. This is where SMC’s 2-, 4-, and 8-color circular palette caches come into play. E.g., when the first 2-color block is encoded, the pair of colors it uses will be inserted into entry 0 of the 2-color cache. During the next 2-color block encoding, if the block uses a pair of colors that already occurs in the cache, the encoding can reference that cache entry. Otherwise, it adds the pair to the next available cache entry, looping back around to 0 as necessary.

  I think I should modify the count_distinct() function to also return a 16-byte array that contains a sorted list of the palette indicies used in the vector. The color pair cache will contain 256 16-bit, 32-bit ints for the quads and 64-bit ints for the octets. This will allow a slightly faster linear cache search.

  Status : The 2-color encoding wasn’t too much trouble and I was able to adapt it to the 4-color mode pretty quickly afterward. I’m still having trouble with the insane 8-color coding mode, though. So that’s commented out for the time being.

  Step 9 : Run Encoding and Putting It All Together
  For each frame, convert the input pixels to a paletted format via one method or another (match to default QuickTime palette for first pass). Then, preprocess each vector to determine the minimum number of elements that can be used to represent it, storing the sorted list of distinct colors in a separate array. The number of elements can either be 0 (only for interframes and indicates a skip block), 1, 2, 4, 8, or 16. Also during this phase, for each vector after the first, test if the vector is the same as the previous vector. If it is, denote this fact in the preprocessed encoding (set the high bit of the element count number).

  Finally, pack it into the bytestream. Iterate through the element count array and search for the longest runs of elements that are encoded with the same mode (up to 256 for skip modes, up to 16 for other modes). If the high bit of an element count is set, that indicates that a copy mode can be encoded. Look for the longest run of element counts with the high bit set and encode a copy mode.

  Status : In-process. Will finish this as motivation strikes.

• The 11th Hour RoQ Variation

  12 avril 2012, par Multimedia Mike — Game Hacking, dreamroq, Reverse Engineering, roq, Vector Quantization

  I have been looking at the RoQ file format almost as long as I have been doing practical multimedia hacking. However, I have never figured out how the RoQ format works on The 11th Hour, which was the game for which the RoQ format was initially developed. When I procured the game years ago, I remember finding what appeared to be RoQ files and shoving them through the open source decoders but not getting the right images out.

  I decided to dust off that old copy of The 11th Hour and have another go at it.

  
  
  

  Baseline
  The game consists of 4 CD-ROMs. Each disc has a media/ directory that has a series of files bearing the extension .gjd, likely the initials of one Graeme J. Devine. These are resource files which are merely headerless concatenations of other files. Thus, at first glance, one file might appear to be a single RoQ file. So that’s the source of some of the difficulty : Sending an apparent RoQ .gjd file through a RoQ player will often cause the program to complain when it encounters the header of another RoQ file.

  I have uploaded some samples to the usual place.

  However, even the frames that a player can decode (before encountering a file boundary within the resource file) look wrong.

  Investigating Codebooks Using dreamroq
  I wrote dreamroq last year– an independent RoQ playback library targeted towards embedded systems. I aimed it at a gjd file and quickly hit a codebook error.

  RoQ is a vector quantizer video codec that maintains a codebook of 256 2×2 pixel vectors. In the Quake III and later RoQ files, these are transported using a YUV 4:2:0 colorspace– 4 Y samples, a U sample, and a V sample to represent 4 pixels. This totals 6 bytes per vector. A RoQ codebook chunk contains a field that indicates the number of 2×2 vectors as well as the number of 4×4 vectors. The latter vectors are each comprised of 4 2×2 vectors.

  Thus, the total size of a codebook chunk ought to be (# of 2×2 vectors) * 6 + (# of 4×4 vectors) * 4.

  However, this is not the case with The 11th Hour RoQ files.

  Longer Codebooks And Mystery Colorspace
  Juggling the numbers for a few of the codebook chunks, I empirically determined that the 2×2 vectors are represented by 10 bytes instead of 6. Now I need to determine what exactly these 10 bytes represent.

  I should note that I suspect that everything else about these files lines up with successive generations of the format. For example if a file has 640×320 resolution, that amounts to 40×20 macroblocks. dreamroq iterates through 40×20 8×8 blocks and precisely exhausts the VQ bitstream. So that all looks valid. I’m just puzzled on the codebook format.

  Here is an example codebook dump :
  

  ID 0x1002, len = 0x0000014C, args = 0x1C0D
    0 : 00 00 00 00 00 00 00 00 80 80
    1 : 08 07 00 00 1F 5B 00 00 7E 81
    2 : 00 00 15 0F 00 00 40 3B 7F 84
    3 : 00 00 00 00 3A 5F 18 13 7E 84
    4 : 00 00 00 00 3B 63 1B 17 7E 85
    5 : 18 13 00 00 3C 63 00 00 7E 88
    6 : 00 00 00 00 00 00 59 3B 7F 81
    7 : 00 00 56 23 00 00 61 2B 80 80
    8 : 00 00 2F 13 00 00 79 63 81 83
    9 : 00 00 00 00 5E 3F AC 9B 7E 81
    10 : 1B 17 00 00 B6 EF 77 AB 7E 85
    11 : 2E 43 00 00 C1 F7 75 AF 7D 88
    12 : 6A AB 28 5F B6 B3 8C B3 80 8A
    13 : 86 BF 0A 03 D5 FF 3A 5F 7C 8C
    14 : 00 00 9E 6B AB 97 F5 EF 7F 80
    15 : 86 73 C8 CB B6 B7 B7 B7 85 8B
    16 : 31 17 84 6B E7 EF FF FF 7E 81
    17 : 79 AF 3B 5F FC FF E2 FF 7D 87
    18 : DC FF AE EF B3 B3 B8 B3 85 8B
    19 : EF FF F5 FF BA B7 B6 B7 88 8B
    20 : F8 FF F7 FF B3 B7 B7 B7 88 8B
    21 : FB FF FB FF B8 B3 B4 B3 85 88
    22 : F7 FF F7 FF B7 B7 B9 B7 87 8B
    23 : FD FF FE FF B9 B7 BB B7 85 8A
    24 : E4 FF B7 EF FF FF FF FF 7F 83
    25 : FF FF AC EB FF FF FC FF 7F 83
    26 : CC C7 F7 FF FF FF FF FF 7F 81
    27 : FF FF FE FF FF FF FF FF 80 80
  

  Note that 0x14C (the chunk size) = 332, 0x1C and 0x0D (the chunk arguments — count of 2×2 and 4×4 vectors, respectively) are 28 and 13. 28 * 10 + 13 * 4 = 332, so the numbers check out.

  Do you see any patterns in the codebook ? Here are some things I tried :

  • Treating the last 2 bytes as U & V and treating the first 4 as the 4 Y samples :
    
    
    
  • Treating the last 2 bytes as U & V and treating the first 8 as 4 16-bit little-endian Y samples :
    
    
    
  • Disregarding the final 2 bytes and treating the first 8 bytes as 4 RGB565 pixels (both little- and big-endian, respectively, shown here) :
    
    
    
  • Based on the type of data I’m seeing in these movies (which appears to be intended as overlays), I figured that some of these bits might indicate transparency ; here is 15-bit big-endian RGB which disregards the top bit of each pixel :
    
    
    

  These images are taken from the uploaded sample bdpuz.gjd, apparently a component of the puzzle represented in this screenshot.

  Unseen Types
  It has long been rumored that early RoQ files could contain JPEG images. I finally found one such specimen. One of the files bundled early in the uploaded fhpuz.gjd sample contains a JPEG frame. It’s a standard JFIF file and can easily be decoded after separating the bytes from the resource using ‘dd’. JPEGs serve as intraframes in the coding scheme, with successive RoQ frames moving objects on top.

  However, a new chunk type showed up as well, one identified by 0×1030. I have never encountered this type. Where could I possibly find data about this ? Fortunately, iD Games recently posted all of their open sourced games at Github. Reading through the code for their official RoQ decoder, I see that this is called a RoQ_PACKET. The name and the code behind it are both supremely unhelpful. The code is basically a no-op. The payloads of the various RoQ_PACKETs from one sample are observed to be either 8784, 14752, or 14760 bytes in length. It’s very likely that this serves the same purpose as the JPEG intraframes.

  Other Tidbits
  I read through the readme.txt on the first game disc and found this nugget :

          g)      Animations displayed normally or in SPOOKY MODE
                  SPOOKY MODE is blue-tinted grayscale with color cursors, puzzle
  
                  and game pieces.  It is the preferred display setting of the
  
                  developers at Trilobyte.  Just for fun, try out the SPOOKY
  
                  MODE.
  

  The MobyGames screenshot page has a number of screenshots labeled as being captured in spooky mode. Color tricks ?

  Meanwhile, another twist arose as I kept tweaking dreamroq to deal with more RoQ weirdness : After modifying my dreamroq code to handle these 10-byte vectors, it eventually chokes on another codebook. These codebooks happen to have 6-byte vectors again ! Fortunately, I was already working on a scheme to automatically detect which codebook is in play (plugging the numbers into a formula and seeing which vector size checks out).

• CD-R Read Speed Experiments

  21 mai 2011, par Multimedia Mike — Science Projects, Sega Dreamcast

  I want to know how fast I can really read data from a CD-R. Pursuant to my previous musings on this subject, I was informed that it is inadequate to profile reading just any file from a CD-R since data might be read faster or slower depending on whether the data is closer to the inside or the outside of the disc.

  Conclusion / Executive Summary
  It is 100% true that reading data from the outside of a CD-R is faster than reading data from the inside. Read on if you care to know the details of how I arrived at this conclusion, and to find out just how much speed advantage there is to reading from the outside rather than the inside.

  Science Project Outline

  • Create some sample CD-Rs with various properties
  • Get a variety of optical drives
  • Write a custom program that profiles the read speed

  Creating The Test Media
  It’s my understanding that not all CD-Rs are created equal. Fortunately, I have 3 spindles of media handy : Some plain-looking Memorex discs, some rather flamboyant Maxell discs, and those 80mm TDK discs :

  
  
  

  My approach for burning is to create a single file to be burned into a standard ISO-9660 filesystem. The size of the file will be the advertised length of the CD-R minus 1 megabyte for overhead— so, 699 MB for the 120mm discs, 209 MB for the 80mm disc. The file will contain a repeating sequence of 0..0xFF bytes.

  Profiling
  I don’t want to leave this to the vagaries of any filesystem handling layer so I will conduct this experiment at the sector level. Profiling program outline :

  • Read the CD-ROM TOC and get the number of sectors that comprise the data track
  • Profile reading the first 20 MB of sectors
  • Profile reading 20 MB of sectors in the middle of the track
  • Profile reading the last 20 MB of sectors

  Unfortunately, I couldn’t figure out the raw sector reading on modern Linux incarnations (which is annoying since I remember it being pretty straightforward years ago). So I left it to the filesystem after all. New algorithm :

  • Open the single, large file on the CD-R and query the file length
  • Profile reading the first 20 MB of data, 512 kbytes at a time
  • Profile reading 20 MB of sectors in the middle of the track (starting from filesize / 2 - 10 MB), 512 kbytes at a time
  • Profile reading the last 20 MB of sectors (starting from filesize - 20MB), 512 kbytes at a time

  Empirical Data
  I tested the program in Linux using an LG Slim external multi-drive (seen at the top of the pile in this post) and one of my Sega Dreamcast units. I gathered the median value of 3 runs for each area (inner, middle, and outer). I also conducted a buffer flush in between Linux runs (as root : 'sync; echo 3 > /proc/sys/vm/drop_caches').

  LG Slim external multi-drive (reading from inner, middle, and outer areas in kbytes/sec) :

  • TDK-80mm : 721, 897, 1048
  • Memorex-120mm : 1601, 2805, 3623
  • Maxell-120mm : 1660, 2806, 3624

  So the 120mm discs can range from about 10.5X all the way up to a full 24X on this drive. For whatever reason, the 80mm disc fares a bit worse — even at the inner track — with a range of 4.8X - 7X.

  Sega Dreamcast (reading from inner, middle, and outer areas in kbytes/sec) :

  • TDK-80mm : 502, 632, 749
  • Memorex-120mm : 499, 889, 1143
  • Maxell-120mm : 500, 890, 1156

  It’s interesting that the 80mm disc performed comparably to the 120mm discs in the Dreamcast, in contrast to the LG Slim drive. Also, the results are consistent with my previous profiling experiments, which largely only touched the inner area. The read speeds range from 3.3X - 7.7X. The middle of a 120mm disc reads at about 6X.

  Implications
  A few thoughts regarding these results :

  • Since the very definition of 1X is the minimum speed necessary to stream data from an audio CD, then presumably, original 1X CD-ROM drives would have needed to be capable of reading 1X from the inner area. I wonder what the max read speed at the outer edges was ? It’s unlikely I would be able to get a 1X drive working easily in this day and age since the earliest CD-ROM drives required custom controllers.
  • I think 24X is the max rated read speed for CD-Rs, at least for this drive. This implies that the marketing literature only cites the best possible numbers. I guess this is no surprise, similar to how monitors and TVs have always been measured by their diagonal dimension.
  • Given this data, how do you engineer an ISO-9660 filesystem image so that the timing-sensitive multimedia files live on the outermost track ? In the Dreamcast case, if you can guarantee your FMV files will live somewhere between the middle and the end of the disc, you should be able to count on a bitrate of at least 900 kbytes/sec.

  Source Code
  Here is the program I wrote for profiling. Note that the filename is hardcoded (#define FILENAME). Compiling for Linux is a simple 'gcc -Wall profile-cdr.c -o profile-cdr'. Compiling for Dreamcast is performed in the standard KallistiOS manner (people skilled in the art already know what they need to know) ; the only variation is to compile with the '-D_arch_dreamcast' flag, which the default KOS environment adds anyway.

  PLAIN TEXT

  C :

  1. #ifdef _arch_dreamcast
  2.   #include <kos .h>
  3.  
  4.   /* map I/O functions to their KOS equivalents */
  5.   #define open fs_open
  6.   #define lseek fs_seek
  7.   #define read fs_read
  8.   #define close fs_close
  9.  
  10.   #define FILENAME "/cd/bigfile"
  11. #else
  12.   #include <stdio .h>
  13.   #include <sys /types.h>
  14.   #include </sys><sys /stat.h>
  15.   #include </sys><sys /time.h>
  16.   #include <fcntl .h>
  17.   #include <unistd .h>
  18.  
  19.   #define FILENAME "/media/Full disc/bigfile"
  20. #endif
  21.  
  22. /* Get a current absolute millisecond count ; it doesn’t have to be in
  23. * reference to anything special. */
  24. unsigned int get_current_milliseconds()
  25. {
  26. #ifdef _arch_dreamcast
  27.   return timer_ms_gettime64() ;
  28. #else
  29.   struct timeval tv ;
  30.   gettimeofday(&tv, NULL) ;
  31.   return tv.tv_sec * 1000 + tv.tv_usec / 1000 ;
  32. #endif
  33. }
  34.  
  35. #define READ_SIZE (20 * 1024 * 1024)
  36. #define READ_BUFFER_SIZE (512 * 1024)
  37.  
  38. int main()
  39. {
  40.   int i, j ;
  41.   int fd ;
  42.   char read_buffer[READ_BUFFER_SIZE] ;
  43.   off_t filesize ;
  44.   unsigned int start_time, end_time ;
  45.  
  46.   fd = open(FILENAME, O_RDONLY) ;
  47.   if (fd == -1)
  48.   {
  49.     printf("could not open %s\n", FILENAME) ;
  50.     return 1 ;
  51.   }
  52.   filesize = lseek(fd, 0, SEEK_END) ;
  53.  
  54.   for (i = 0 ; i <3 ; i++)
  55.   {
  56.     if (i == 0)
  57.     {
  58.       printf("reading inner 20 MB...\n") ;
  59.       lseek(fd, 0, SEEK_SET) ;
  60.     }
  61.     else if (i == 1)
  62.     {
  63.       printf("reading middle 20 MB...\n") ;
  64.       lseek(fd, (filesize / 2) - (READ_SIZE / 2), SEEK_SET) ;
  65.     }
  66.     else
  67.     {
  68.       printf("reading outer 20 MB...\n") ;
  69.       lseek(fd, filesize - READ_SIZE, SEEK_SET) ;
  70.     }
  71.     /* read 20 MB ; 40 chunks of 1/2 MB */
  72.     start_time = get_current_milliseconds() ;
  73.     for (j = 0 ; j <(READ_SIZE / READ_BUFFER_SIZE) ; j++)
  74.       if (read(fd, read_buffer, READ_BUFFER_SIZE) != READ_BUFFER_SIZE)
  75.       {
  76.         printf("read error\n") ;
  77.         break ;
  78.       }
  79.     end_time = get_current_milliseconds() ;
  80.     printf("%d - %d = %d ms => %d kbytes/sec\n",
  81.       end_time, start_time, end_time - start_time,
  82.       READ_SIZE / (end_time - start_time)) ;
  83.   }
  84.  
  85.   close(fd) ;
  86.  
  87.   return 0 ;
  88. }