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• Encoding a video in FFmpeg to X264 and have it playable in Quicktime

  7 mai 2013, par illu

  I am wondering which command line settings i need to explicitly set (or avoid) to make a video encoded into x264 (in the mp4 format) using ffmpeg by default playable in Quicktime. I find that a number of the predefined preset files work for me but some of them won't, for example I can't get any of the lossless ones to work and I'm interested in those ones as well. For example libx264-lossless_max.ffpreset will encode my video but it's only playable in VLC, not in Quicktime. In Quicktime the video stays black. I know Perian is an option but I want my file to be playable without installing Perian. Thanks for your help.

• Basic Video Palette Conversion

  20 août 2011, par Multimedia Mike — General, Python

  How do you take a 24-bit RGB image and convert it to an 8-bit paletted image for the purpose of compression using a codec that requires 8-bit input images ? Seems simple enough and that’s what I’m tackling in this post.

  Ask FFmpeg/Libav To Do It
  Ideally, FFmpeg / Libav should be able to handle this automatically. Indeed, FFmpeg used to be able to, at least at the time I wrote this post about ZMBV and was unhappy with FFmpeg’s default results. Somewhere along the line, FFmpeg and Libav lost the ability to do this. I suspect it got removed during some swscale refactoring.

  Still, there’s no telling if the old system would have computed palettes correctly for QuickTime files.

  Distance Approach
  When I started writing my SMC video encoder, I needed to convert RGB (from PNG files) to PAL8 colorspace. The path of least resistance was to match the pixels in the input image to the default 256-color palette that QuickTime assumes (and is hardcoded into FFmpeg/Libav).

  How to perform the matching ? Find the palette entry that is closest to a given input pixel, where "closest" is the minimum distance as computed by the usual distance formula (square root of the sum of the squares of the diffs of all the components).

  
  
  

  That means for each pixel in an image, check the pixel against 256 palette entries (early termination is possible if an acceptable threshold is met). As you might imagine, this can be a bit time-consuming. I wondered about a faster approach...

  Lookup Table
  I think this is the approach that FFmpeg used to use, but I went and derived it for myself after studying the default QuickTime palette table. There’s a pattern there— all of the RGB entries are comprised of combinations of 6 values — 0x00, 0x33, 0x66, 0x99, 0xCC, and 0xFF. If you mix and match these for red, green, and blue values, you come up with 6 * 6 * 6 = 216 different colors. This happens to be identical to the web-safe color palette.

  The first (0th) entry in the table is (FF, FF, FF), followed by (FF, FF, CC), (FF, FF, 99), and on down to (FF, FF, 00) when the green component gets knocked down and step and the next color is (FF, CC, FF). The first 36 palette entries in the table all have a red component of 0xFF. Thus, if an input RGB pixel has a red color closest to 0xFF, it must map to one of those first 36 entries.

  I created a table which maps indices 0..215 to values from 5..0. Each of the R, G, and B components of an input pixel are used to index into this table and derive 3 indices ri, gi, and bi. Finally, the index into the palette table is given by :

    index = ri * 36 + gi * 6 + bi
  

  For example, the pixel (0xFE, 0xFE, 0x01) would yield ri, gi, and bi values of 0, 0, and 5. Therefore :

    index = 0 * 36 + 0 * 6 + 5
  

  The palette index is 5, which maps to color (0xFF, 0xFF, 0x00).

  Validation
  So I was pretty pleased with myself for coming up with that. Now, ideally, swapping out one algorithm for another in my SMC encoder should yield identical results. That wasn’t the case, initially.

  One problem is that the regulation QuickTime palette actually has 40 more entries above and beyond the typical 216-entry color cube (rounding out the grand total of 256 colors). Thus, using the distance approach with the full default table provides for a little more accuracy.

  However, there still seems to be a problem. Let’s check our old standby, the Big Buck Bunny logo image :

  
  
  Distance approach using the full 256-color QuickTime default palette
  
  
  Distance approach using the 216-color palette
  
  
  Table lookup approach using the 216-color palette
  

  I can’t quite account for that big red splotch there. That’s the most notable difference between images 1 and 2 and the only visible difference between images 2 and 3.

  To prove to myself that the distance approach is equivalent to the table approach, I wrote a Python script to iterate through all possible RGB combinations and verify the equivalence. If you’re not up on your base 2 math, that’s 2²⁴ or 16,777,216 colors to run through. I used Python’s multiprocessing module to great effect and really maximized a Core i7 CPU with 8 hardware threads.

  So I’m confident that the palette conversion techniques are sound. The red spot is probably attributable to a bug in my WIP SMC encoder.

  Source Code
  Update August 23, 2011 : Here’s the Python code I used for proving equivalence between the 2 approaches. In terms of leveraging multiple CPUs, it’s possibly the best program I have written to date.

  PLAIN TEXT

  PYTHON :

  1. # !/usr/bin/python
  2.  
  3. from multiprocessing import Pool
  4.  
  5. palette = []
  6. pal8_table = []
  7.  
  8. def process_r(r) :
  9.  counts = []
  10.  
  11.  for i in xrange(216) :
  12.   counts.append(0)
  13.  
  14.  print "r = %d" % (r)
  15.  for g in xrange(256) :
  16.   for b in xrange(256) :
  17.    min_dsqrd = 0xFFFFFFFF
  18.    best_index = 0
  19.    for i in xrange(len(palette)) :
  20.     dr = palette[i][0] - r
  21.     dg = palette[i][1] - g
  22.     db = palette[i][2] - b
  23.     dsqrd = dr * dr + dg * dg + db * db
  24.     if dsqrd <min_dsqrd :
  25.      min_dsqrd = dsqrd
  26.      best_index = i
  27.    counts[best_index] += 1
  28.  
  29.    # check if the distance approach deviates from the table-based approach
  30.    i = best_index
  31.    r = palette[i][0]
  32.    g = palette[i][1]
  33.    b = palette[i][2]
  34.    ri = pal8_table[r]
  35.    gi = pal8_table[g]
  36.    bi = pal8_table[b]
  37.    table_index = ri * 36 + gi * 6 + bi ;
  38.    if table_index != best_index :
  39.     print "(0x%02X 0x%02X 0x%02X) : distance index = %d, table index = %d" % (r, g, b, best_index, table_index)
  40.  
  41.  return counts
  42.  
  43. if __name__ == ’__main__’ :
  44.  counts = []
  45.  for i in xrange(216) :
  46.   counts.append(0)
  47.  
  48.  # initialize reference palette
  49.  color_steps = [ 0xFF, 0xCC, 0x99, 0x66, 0x33, 0x00 ]
  50.  for r in color_steps :
  51.   for g in color_steps :
  52.    for b in color_steps :
  53.     palette.append([r, g, b])
  54.  
  55.  # initialize palette conversion table
  56.  for i in range(0, 26) :
  57.   pal8_table.append(5)
  58.  for i in range(26, 77) :
  59.   pal8_table.append(4)
  60.  for i in range(77, 128) :
  61.   pal8_table.append(3)
  62.  for i in range(128, 179) :
  63.   pal8_table.append(2)
  64.  for i in range(179, 230) :
  65.   pal8_table.append(1)
  66.  for i in range(230, 256) :
  67.   pal8_table.append(0)
  68.  
  69.  # create a pool of worker threads and break up the overall job
  70.  pool = Pool()
  71.  it = pool.imap_unordered(process_r, range(256))
  72.  try :
  73.   while 1 :
  74.    partial_counts = it.next()
  75.    for i in xrange(216) :
  76.     counts[i] += partial_counts[i]
  77.  except StopIteration :
  78.   pass
  79.  
  80.  print "index, count, red, green, blue"
  81.  for i in xrange(len(counts)) :
  82.   print "%d, %d, %d, %d, %d" % (i, counts[i], palette[i][0], palette[i][1], palette[i][2])

• 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. }