/usr/lib/pyrite-publisher/PyritePublisher/doc_compress.py is in pyrite-publisher 2.1.1-11.
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# $Id: doc_compress.py,v 1.1 2001/03/29 21:15:44 rob Exp $
#
# Copyright 1999-2001 Rob Tillotson <rob@pyrite.org>
# All Rights Reserved
#
# Permission to use, copy, modify, and distribute this software and
# its documentation for any purpose and without fee or royalty is
# hereby granted, provided that the above copyright notice appear in
# all copies and that both the copyright notice and this permission
# notice appear in supporting documentation or portions thereof,
# including modifications, that you you make.
#
# THE AUTHOR ROB TILLOTSON DISCLAIMS ALL WARRANTIES WITH REGARD TO
# THIS SOFTWARE, INCLUDING ALL IMPLIED WARRANTIES OF MERCHANTABILITY
# AND FITNESS. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR ANY
# SPECIAL, INDIRECT OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES WHATSOEVER
# RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN ACTION OF
# CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF OR IN
# CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE!
#
"""Doc compression in pure Python.
"""
__version__ = '$Id: doc_compress.py,v 1.1 2001/03/29 21:15:44 rob Exp $'
__copyright__ = 'Copyright 1999-2001 Rob Tillotson <rob@pyrite.org>'
import string
COUNT_BITS = 3
DISP_BITS = 11
def compress(s):
# optimizations
# this cut off about 0.1 sec/call
_find = string.find
out = []
space = 0
# first phase: sliding window
sstart = 0
i = 0
imax = len(s)
while 1:
if i >= imax: break
if (i - sstart) > 2047: sstart = i - 2047
#ts = s[i:i+10] # substring to search for
#while len(ts) >= 3:
# f = _find(s, ts, sstart, i)
# if f >= 0: break
# ts = ts[:-1]
# see that code above? it's the obvious way, but it's slow.
# what it does is basically do string.find on the data ahead
# of the current position, on an ever-shrinking buffer. With
# that code, this function took around 1.6 seconds per call
# (on a Cyrix M2-266 [207 MHz/83 MHz bus/1MB cache]).
#
# The below is a bit different: basically, it takes advantage
# of the fact that if we don't have a length 3 substring, we
# can't possibly have a length 4 substring, and so on. Thus,
# it first looks for a length 3 substring (we don't care about
# anything shorter). If it finds one, it then attempts to
# find a length 4 substring between that position and the
# current location, and so on.
#
# I suspect the primary benefit of this is to avoid 6 out of 7
# calls to string.find every time we aren't looking at a
# compressible string. At any rate, changing this code cuts
# the time to about 0.38 seconds per call.
e = 3
ns = sstart
ts = s[i:i+e]
while (imax - i) >= e and e <= 10:
f = _find(s, ts, ns, i)
if f < 0: break
e = e + 1
ts = s[i:i+e]
ns = f
e = e - 1
#if len(ts) >= 3: #we have a match, f is the location and len(ts) is the length
# l = len(ts)
# ns = f
if e >= 3:
dist = i - ns
byte = (dist << 3) | (e-3)
if space:
out.append(32)
space = 0
out.append(0x80 | (byte >> 8))
out.append(byte & 0xff)
i = i + e
else:
c = ord(s[i])
i = i + 1
if space:
if c >= 0x40 and c <= 0x7f: out.append(c | 0x80)
else:
out.append(32)
if c < 0x80 and (c == 0 or c > 8):
out.append(c)
else:
out.append(1)
out.append(c)
space = 0
else:
if c == 32: space = 1
else:
if c < 0x80 and (c == 0 or c > 8):
out.append(c)
else:
out.append(1)
out.append(c)
if space: out.append(32)
# second phase: look for repetitions of '1 <x>' and combine up to 8 of them.
# interestingly enough, in regular text this hardly makes a difference.
return string.join(map(chr, out),'')
def uncompress(s):
s = map(ord, s)
x = 0
o = []
try:
while 1:
c = s[x]
x = x + 1
if c > 0 and c < 9: # just copy that many bytes
for y in range(0, c):
o.append(s[x])
x = x + 1
elif c < 128: # a regular ascii character
o.append(c)
elif c >= 0xc0: # a regular ascii character with a space before it
o.append(32)
o.append(c & 0x7f)
else: # a compressed sequence
c = c << 8
c = c | s[x]
x = x + 1
m = (c & 0x3fff) >> COUNT_BITS
n = (c & ((1 << COUNT_BITS)-1)) + 3
for y in range(0, n):
o.append(o[len(o)-m])
except IndexError:
pass
return string.join(map(chr, o), '')
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