Arithmetic Coding
Arithmetic coding is the most efficient method to code symbols according to the probability
of their occurrence. The average code length corresponds exactly to the possible minimum
given by information theory. Deviations which are caused by the bit-resolution of binary
code trees do not exist.
In contrast to a binary Huffman code tree the arithmetic coding offers a clearly better
compression rate. Its implementation is more complex on the other hand.
In arithmetic coding, a message is encoded as a real number in an interval from one to zero.
Arithmetic coding typically has a better compression ratio than Huffman coding, as it
produces a single symbol rather than several separate codewords.
Arithmetic coding differs from other forms of entropy encoding such as Huffman coding in
that rather than separating the input into component symbols and replacing each with a code,
arithmetic coding encodes the entire message into a single number, a fraction n where (0.0
n < 1.0)
Arithmetic coding is a lossless coding technique. There are a few disadvantages of
arithmetic coding. One is that the whole codeword must be received to start decoding the
symbols, and if there is a corrupt bit in the codeword, the entire message could become
corrupt. Another is that there is a limit to the precision of the number which can be
encoded, thus limiting the number of symbols to encode within a codeword. There also
exist many patents upon arithmetic coding, so the use of some of the algorithms also call
upon royalty fees.
Arithmetic coding is part of the JPEG data format. Alternative to Huffman coding it will be
used for final entropy coding. In spite of its less efficiency Huffman coding remains the
standard due to the legal restrictions mentioned above.
Arithmetic Coding Algorithm:
The arithmetic coding algorithm works from leaves to the root in the opposite direction.
1. Start with an interval [0, 1), divided into subintervals of all possible symbols to appear
within a message. Make the size of each subinterval proportional to the frequency at
which it appears in the message.
2. When encoding a symbol, "zoom" into the current interval, and divide it into
subintervals like in step one with the new range.
3. Repeat the process until the maximum precision of the machine is reached, or all
symbols are encoded.
4. Transmit some number within the latest interval to send the codeword. The number
of symbols encoded will be stated in the protocol of the image format.
Example 1:
The source of information A generates the symbols {A0, A1, A2, A3 and A4} with the
corresponding probabilities {0.4, 0.3, 0.2 and 0.1}. Encoding the source symbols using
Huffman encoder gives:
Source Symbol P
i
Binary Code Huffman
A0 0.4 00 0
A1 0.3 01 10
A2 0.2 10 110
A3 0.1 10 111
L
avg
H = 1.846 2 1.9
The Entropy of the source is
Since we have 4 symbols (4=2
2
), we need 2 bits at least to represent each symbol in binary
(fixed-length code). Hence the average length of the binary code is
Thus the efficiency of the binary code is
The average length of the Huffman code is
Thus the efficiency of the Huffman code is
The Huffman encoder has the closest efficiency to the entropy that can be obtained using a
prefix code. Higher efficiency can be yielded with the arithmetic coding.
Dividing into Intervals
On the basis of a well-known alphabet the probability of all symbols has to be determined
and converted into intervals. The size of the interval depends linearly on the symbol's
probability. If this is 50% for example, then the associated sub-interval covers the half of the
current interval. Usually the initial interval is [0; 1) for the encoding of the first symbol.
Source Symbol P
i
Sub-interval
A0 0.4 [0.0;0.4)
A1 0.3 [0.4;0.7)
A2 0.2 [0.7;0.9)
A3 0.1 [0.9;1.0)
Assume that the message to be encoded is A0A0A3A1A2. The first symbol to be encoded is
A0. We "zoom" into the interval corresponding to "A0", and divide up that interval into
smaller subintervals like before. We now use this new interval as the basis of the next
symbol encoding step.
Source Symbol New A0 Interval
A0 [0.0;0.16)
A1 [0.16;0.28)
A2 [0.28;0.36)
A3 [0.36;0.4)
To encode the next character "A0", we use the "A0" interval created before, and zoom into
the subinterval "A0", and use that for the next step. This produces
Source Symbol New A0 Interval
A0 [0.0;0.064)
A1 [0.064;0.112)
A2 [0.112;0.144)
A3 [0.144;0.16)
To encode the next character "A3", we use the "A0" interval created before, and zoom into
the subinterval "A3", and use that for the next step. This produces
Source Symbol New A3 Interval
A0 [0.144;0.1504)
A1 [0.1504;0.1552)
A2 [0.1552;0.1584)
A3 [0.1584;0.16)
And lastly, the final result is
Source Symbol New A0 Interval
A0 [0.1504;0.15232)
A1 [0.15232;0.15376)
A2 [0.15376;0.15472)
A3 [0.15472;0.1552)
Transmit some number within the latest interval to send the codeword. The number of
symbols encoded will be stated in the protocol of the image format, so any number within
[0.15376, 0.15472) will be acceptable.
Lets choose the number 0.1543. The binary representation of this number is 0.001001111.
We need 10 bits to encode the message (9 bits and the floating point). The minimum number
of bits needed to fully encode the message is
H*N = 1.846*5= 9.23 bit
Using Huffman code, the message is encoded to 0 0 111 10 110 which need also 10 bits. The
larger is the number of symbols, the wider is the gap in efficiency.
Decoding the code is a reverse approach. Lets assume the number 0.1543 has been received
at the decoder. 0.1543 lies in the interval [0; 0.4), then the first symbol of the message is A0.
Then, we "zoom" into the interval corresponding to "A0", and divide up that interval into
smaller subintervals. We now use this new interval as the basis of the next symbol decoding
step.
Source Symbol New A0 Interval
A0 [0.0;0.16)
A1 [0.16;0.28)
A2 [0.28;0.36)
A3 [0.36;0.4)
0.1543 lies in the interval [0; 0.16), then the second symbol of the message is A0. Zoom into
the subinterval "A0", and use that for the next step. This produces
Source Symbol New A0 Interval
A0 [0.0;0.064)
A1 [0.064;0.112)
A2 [0.112;0.144)
A3 [0.144;0.16)
0.1543 lies in the interval [0.144; 0.16), then the third symbol of the message is A3. Zoom
into the subinterval "A3", and use that for the next step. This produces
Source Symbol New A3 Interval
A0 [0.144;0.1504)
A1 [0.1504;0.1552)
A2 [0.1552;0.1584)
A3 [0.1584;0.16)
0.1543 lies in the interval [0.1504; 0.1552), then the fourth symbol of the message is A1.
Zoom into the subinterval "A1", and use that for the next step. This produces
Source Symbol New A0 Interval
A0 [0.1504;0.15232)
A1 [0.15232;0.15376)
A2 [0.15376;0.15472)
A3 [0.15472;0.1552)
0.1543 lies in the interval [0.15376; 0.15472), then the last symbol of the message is A2. The
decoder stops after this as the number of symbols encoded will be stated in the protocol of the
message.
Exercise 1:
The source of information A generates the symbols {A0, A1, A2, A3 and A4} with the
probabilities shown in the table below. Encode the source symbols using Arithmetic encoder
and Huffman encoder. The message is A4A1A0A3A2
Source Symbol P
i
A0 0.4
A1 0.4
A2 0.12
A3 0.06
A4 0.02
Compare the efficiency of both codes and comment on the results.
Exercise 2:
The source of information A generates the symbols {A0, A1, A2 and A3} with the
probabilities shown in the table below. Encode the source symbols arithmetic encoder if the
message is A2A0A2A3
Source Symbol P
i
A0 0.5
A1 0.3
A2 0.2
Compare the efficiency of both codes and comment on the results.
