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Connor 2018-09-10 12:08:23 -07:00 committed by GitHub
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## Arithmetic coding compressor and uncompressor for binary source.
## This is a cleaned-up version of AEncode.py
BETA0=1;BETA1=1 ## default prior distribution
M = 30 ; ONE = (1<<M) ; HALF = (1<<(M-1))
QUARTER = (1<<(M-2)) ; THREEQU = HALF+QUARTER
def clear (c,charstack):
## print out character c, and other queued characters
a = `c`+`(1-c)`*charstack[0]
charstack[0]=0
return a
pass
def encode (string, c0=BETA0, c1=BETA1, adaptive=1,verbose=0):
b=ONE; a=0; tot0=0;tot1=0; assert c0>0; assert c1>0
if adaptive==0:
p0 = c0*1.0/(c0+c1)
pass
ans="";
charstack=[0] ## how many undecided characters remain to print
for c in string:
w=b-a
if adaptive :
cT = c0+c1
p0 = c0*1.0/cT
pass
boundary = a + int(p0*w)
if (boundary == a): boundary += 1; print "warningA"; pass # these warnings mean that some of the probabilities
if (boundary == b): boundary -= 1; print "warningB"; pass # requested by the probabilistic model
## are so small (compared to our integers) that we had to round them up to bigger values
if (c=='1') :
a = boundary
tot1 += 1 ; if adaptive: c1 += 1.0 ; pass
elif (c=='0'):
b = boundary
tot0 +=1 ; if adaptive: c0 += 1.0 ; pass
pass ## ignore other characters
while ( (a>=HALF) or (b<=HALF) ) : ## output bits
if (a>=HALF) :
ans = ans + clear(1,charstack)
a = a-HALF ;
b = b-HALF ;
else :
ans = ans + clear(0,charstack)
pass
a *= 2 ; b *= 2
pass
assert a<=HALF; assert b>=HALF; assert a>=0; assert b<=ONE
## if the gap a-b is getting small, rescale it
while ( (a>QUARTER) and (b<THREEQU) ):
charstack[0] += 1
a = 2*a-HALF
b = 2*b-HALF
pass
assert a<=HALF; assert b>=HALF; assert a>=0; assert b<=ONE
pass
# terminate
if ( (HALF-a) > (b-HALF) ) :
w = (HALF-a) ;
ans = ans + clear(0,charstack)
while ( w < HALF ) :
ans = ans + clear(1,charstack)
w *=2
pass
pass
else :
w = (b-HALF) ;
ans = ans + clear(1,charstack)
while ( w < HALF ) :
ans = ans + clear(0,charstack)
w *=2
pass
pass
return ans
pass
def decode (string, N=10000, c0=BETA0, c1=BETA1, adaptive=1,verbose=0):
## must supply N, the number of source characters remaining.
b=ONE ; a=0 ; tot0=0;tot1=0 ; assert c0>0 ; assert c1>0
model_needs_updating = 1
if adaptive==0:
p0 = c0*1.0/(c0+c1)
pass
ans=""
u=0 ; v=ONE
for c in string :
if N<=0 :
break ## break out of the string-reading loop
assert N>0
## // (u,v) is the current "encoded alphabet" binary interval, and halfway is its midpoint.
## // (a,b) is the current "source alphabet" interval, and boundary is the "midpoint"
assert u>=0 ; assert v<=ONE
halfway = u + (v-u)/2
if( c == '1' ) :
u = halfway
elif ( c=='0' ):
v = halfway
else:
pass
## // Read bits until we can decide what the source symbol was.
## // Then emulate the encoder's computations, and tie (u,v) to tag along for the ride.
while (1): ## condition at end
firsttime = 0
if(model_needs_updating):
w = b-a
if adaptive :
cT = c0 + c1 ; p0 = c0 *1.0/cT
pass
boundary = a + int(p0*w)
if (boundary == a): boundary += 1; print "warningA"; pass
if (boundary == b): boundary -= 1; print "warningB"; pass
model_needs_updating = 0
pass
if ( boundary <= u ) :
ans = ans + "1"; tot1 +=1 ; if adaptive: c1 += 1.0 ; pass
a = boundary ; model_needs_updating = 1 ; N-=1
elif ( boundary >= v ) :
ans = ans + "0"; tot0 +=1 ; if adaptive: c0 += 1.0 ; pass
b = boundary ; model_needs_updating = 1 ; N-=1
## // every time we discover a source bit, implement exactly the
## // computations that were done by the encoder (below).
else :
## // not enough bits have yet been read to know the decision.
pass
## // emulate outputting of bits by the encoder, and tie (u,v) to tag along for the ride.
while ( (a>=HALF) or (b<=HALF) ) :
if (a>=HALF) :
a = a-HALF ; b = b-HALF ; u = u-HALF ; v = v-HALF
pass
else :
pass
a *= 2 ; b *= 2 ; u *= 2 ; v *= 2 ;
model_needs_updating = 1
pass
assert a<=HALF; assert b>=HALF; assert a>=0; assert b<=ONE
## if the gap a-b is getting small, rescale it
while ( (a>QUARTER) and (b<THREEQU) ):
a = 2*a-HALF; b = 2*b-HALF ; u = 2*u-HALF ; v = 2*v-HALF
pass
if not (N>0 and model_needs_updating) : ## this is the "while" for this "do" loop
break
pass
pass
return ans
pass
def test():
sl=["1010"]
sl=["1010", "111", "00001000000000000000",\
"1", "10" , "01" , "0" ,"0000000", \
"000000000000000100000000000000000000000000000000100000000000000000011000000" ]
for s in sl:
print "encoding", s
N=len(s)
e = encode(s,10,1)
print "decoding", e
ds = decode(e,N,10,1)
print ds
if (ds != s) :
print s
print "ERR@"
pass
else:
print "ok ---------- "
pass
pass
if __name__ == '__main__': test()