[Richard Suchenwirth] 2004-09-02 - Discussion in the [Tcl chatroom] brought me to propose this challenge for a Tcl contest: * Given a set of moderate-sized black&white GIF images, * write a pair of encoder/decoder between photo image and a binary string * so that the decoding of the encoded image is equal to the original (lossless) * and the average length of the encoded strings is minimal I prefer GIF here because it is lossless in itself, and can be read in by plain Tk. Bitmap images cannot be processed pixel-by-pixel, so they're excluded, although they're binary. Test images (feel free to add more. The GIF file size is the initial target to beat): courier12.gif - 1923 bytes: [http://mini.net/files/courier12.gif] times12i.gif - 1947 bytes: [http://mini.net/files/times12i.gif] castle.gif - 11598 bytes [http://www.100dollarhamburger.com/castle.gif] ouster.gif - 2775 bytes [http://tcl.sf.net/tct/kennykb/ouster.gif] ---- Links to solutions to this challenge: [PS] My you can find my attempt on [Binary image compression challenge - Pascal's Entry]. It compresses courier12.gif to 720 bytes and times12.gif to 834 bytes. Castle.gif is actually a compressed gif file! Mine 'compressed' to 13672... [PS] Update 4sept2004: Attempt number two (second code block on my entry page): now with Huffman(?) encoding for runlengths which occur 5 or more times. This actually compresses castle.gif! My new size is 11442 bytes, which is 156 bytes less than the original file! courier12 compresses to 671 bytes, times12i compresses to 718... ---- Little helpers to compare and tally photo images: proc photo'eq {im1 im2} { #-- returns 1 if both images are exactly equal, else 0 set h [image height $im1] if {[image height $im2] != $h} {return 0} set w [image width $im1] if {[image width $im2] != $w} {return 0} for {set y 0} {$y<$h} {incr y} { for {set x 0} {$x<$w} {incr x} { if {[$im1 get $x $y] ne [$im2 get $x $y]} {return 0} } } return 1 } proc photo'colors img { #-- return a list of {{r g b} n} tallies of pixel colors, #-- sorted decreasing by n (number of pixels of that color) set h [image height $img] set w [image width $img] for {set y 0} {$y<$h} {incr y} { for {set x 0} {$x<$w} {incr x} { set color [$img get $x $y] if {![info exists a($color)]} {set a($color) 0} incr a($color) } } foreach {color n} [array get a] {lappend tally [list $color $n]} lsort -decreasing -index 1 -integer $tally } ---- [JH]: Here is my solution, which doesn't compression at all (doesn't beat the GIF format, but is close, the 1923b becomes 2163b). It just represents the data in a small, raw form, but not compressed. It will hopefully provide someone with a starting point to apply compression. ''[MS]: castle.gif is 'compressed' to 15764b by this; gzip reduces that file to 10305b.'' proc binimg'encode img { set clrs [photo'colors $img] if {[llength $clrs] != 2} { return -code error "not a 2 color image" } set clr0 [lindex $clrs 0 0] set clr1 [lindex $clrs 1 0] set h [image height $img] set w [image width $img] set str "" for {set y 0} {$y<$h} {incr y} { for {set x 0} {$x<$w} {incr x} { set color [$img get $x $y] append str [string equal $color $clr1] } } foreach {r g b} $clr0 { set color0 [expr {$r<<16 | $g <<8 | $b}]; break } foreach {r g b} $clr1 { set color1 [expr {$r<<16 | $g <<8 | $b}]; break } # store image as where w and h are shorts set binstr [binary format ssiib* $w $h $color0 $color1 $str] return $binstr } proc binimg'decode data { binary scan $data ssiib* w h color0 color1 clrs set img [image create photo -width $w -height $h] set clr(0) \#[format %.6x $color0] set clr(1) \#[format %.6x $color1] set i 0 set data "" set line "" foreach c [split $clrs {}] { lappend line $clr($c) if {[incr i] eq $w} { set i 0 lappend data $line set line "" } } $img put $data -to 0 0 return $img } See [Binary image compression challenge - mig's Entry] for a slight variant of this, which ''should'' compress better in general. ---- [DKF]: The simplest cheat way is to use ''gzip'' to compress the GIF files themselves! Both images compress to somewhere in the region of 1600-1650 bytes. Can you do better? [MS]: Yes; gzip'ing Jeff's coded files reduces the thing to 640-654 bytes. ---- [kroc]: My attempt, based on RLE is here: [Binary image compression challenge - Kroc's Entry] and my results are: * 923 bytes for courier12.gif * 904 bytes for times12i.gif * 15421 bytes for castle.gif (this one is very hard to beat). ---- [KBK]: I'm putting another attempt over at [Binary image compression challenge - KBK's entry]. It outperforms [PS]'s entry on two of the three test cases, and does only slightly worse on the third. It's smaller than the GIF in all cases. It's not much worse than JH's solution postprocessed through gzip on the text images, and a good bit better on the dithered photograph and line art. * 685 bytes for courier12.gif (vs. 671 for PS's solution) * 682 bytes for times12i.gif (vs. 718) * 9260 bytes for castle.gif (vs. 11442) * 1856 bytes for ouster.gif The way it works is that it marches through the image along the [Hilbert curve]. The advantage to the Hilbert curve as opposed to a raster pattern is that areas of a solid color are likely to result in runs that are much longer than a scan line, even though the other color is present in the same row and column. The sky in the castle picture or the jacket in Ousterhout's portrait are converted to only a few long runs of 0 and 1. Once the data are ordered in Hilbert-curve sequence, they are converted to a set of run lengths; note that since runs of 0 and 1 will alternate, the run lengths are all that's needed; the colors are not stored explicitly. The run lengths are arbitrarily limited to 127; longer runs are broken into pieces with dummy 0-length runs in between. Finally, [Huffman coding] is used to make the run lengths into a string of bits. Decoding is the same process, only in reverse; the Huffman bit sequences are expanded into run lengths and then to a bit string, and the bit string is folded along the Hilbert curve to yield the image again. Compression is, of course, lossless. ---- [Arts and crafts of Tcl-Tk programming]