JP2783221B2 - Decryption method - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION

[0001]

The present invention relates to a method for decoding image information and the like.

[0002]

2. Description of the Related Art For encoding a Markov information source, a symbol sequence is mapped on a number line from 0.0 to 1.0, and its coordinates are encoded, for example, in binary notation as a code word. A linear display encoding method is known. FIG. 2 is a conceptual diagram showing a non-memory information source in which the appearance probability of the symbol "1" is r and the appearance probability of "0" is 1-r as a binary information source for simplicity. Assuming that the output sequence length is 3, the coordinates of each of C (000) to C (111) at the right end are binary-displayed, and if each codeword is truncated to a digit that can be distinguished, the receiving side will transmit. Decoding can be performed by ending the process similar to.

[0003] In such a sequence, the mapping range A _i of the symbol sequence at the i-th point, and its smaller coordinate is C _i
Means that the output symbol a _i is 0 (dominant symbol: M
PS)

[0004]

(Equation 1)

[0005]

(Equation 2)

When the output symbol a _i is 1 (inferior symbol: hereinafter referred to as LPS)

[0007]

(Equation 3)

[0008]

(Equation 4)

## EQU1 ##

[0010] Also, An overview of
he basic princes of the th
e Q-Coder adaptivebinary
Arithmetic coder (IBM joint
al of Reserchand Development
ent VOL. 32 No. 6 NOV. 1988)
As described in, in order to reduce the number of arithmetic circuits for multiplication and the like, it is conceivable to prepare a plurality of fixed values and select a certain value from among them without necessarily calculating rA _i-1. I have.

That is, assuming that rA _i-1 in the above equation is S, when a _i = 0

[0012]

(Equation 5)

[0013]

(Equation 6)

When a _i = 1

[0015]

(Equation 7)

[0016]

(Equation 8)

## EQU1 ##

[0018] However, in this case A _i-1 is day become sequentially smaller as the successive output of the symbol must be S also sequentially reduced, the A _i-1 in order to maintain the calculation accuracy 2
In the actual code word, the fixed value is always assumed to be the same, and the value is raised to a power of 2 (ie, a binary number shift) during the operation. Processing.

[0019]

When a constant value is used as S as described above, there is a problem particularly when S is large and normalized A _i-1 is relatively small.

An example is shown below. Assuming that A _i-1 is slightly larger than 0.5, A _i is extremely small when a _i is MPS and may be smaller than the range S when a _i is LPS. That is, although the appearance probability of the MPS is higher, the area assigned to the MPS is smaller than the area assigned to the LPS, which leads to a decrease in decoding efficiency. If the area allocated to the MPS is always larger than the area allocated to the LPS, S needs to be 0.25 or less from A _i-1 > 0.5. Therefore, assuming that S = 0.25, when the range A _i-1 of the previous symbol is 1.0, r = 0.25, and A
_{When i-1} is close to 0.5, almost r = 0.5, so that the appearance ratio of LPS after decoding becomes 1/4 to 1 /
Therefore, if the fluctuation width can be reduced, an area can be allocated according to the appearance ratio, and improvement in decoding efficiency can be expected. The present invention has been made to solve the above-described problems, and particularly when the appearance ratio of LPS (inferior symbol) is close to 1/2, the efficiency is remarkably improved.

[0021]

According to a first aspect of the present invention, there is provided a decoding method comprising: comparing means for comparing the range on the number line between a superior symbol and an inferior symbol; Control means is provided for controlling the range of the inferior symbol and the range of the superior symbol on the number line so that the symbol range does not exceed the range of the superior symbol.

In the decoding method according to a second aspect of the present invention, a first step of comparing a range on a number line assigned to a superior symbol with a predetermined value, and a range given to the superior symbol on a number line If the value is less than the predetermined value, the range on the number line between the superior symbol and the inferior symbol is controlled so that half of the value below the predetermined value of the superior symbol range is shifted from the inferior symbol range to the superior symbol range. And steps.

[0023]

FIG. 1 shows an embodiment of the present invention. Reference numeral 1 denotes an adder which adds the input S value to the immediately preceding output of the offset operation unit 3 to calculate the upper address of the LPS. The comparator of 2 compares this value with 0.5. If the value is 0.5 or less and the generated symbol is the dominant symbol (MPS), the processing of the offset calculation unit of 3 stops at the addition of S. Similarly, in the comparator of 2,.
5 or less, and the generated symbol is the inferior symbol (LP
If S), the base operation is performed by the base operation device 4 and the base coordinates are output as a code. Further, a multiple (2 ⁿ times) necessary to normalize (effective range is set to 0.5 to 1.0) from the value of S is obtained and output as a shift digit is a shift digit calculation of 5 Department.

Next, when it is judged by the comparator 2 that it exceeds 0.5 (decimal), the LPS upper address corrector 6 makes L
Based on whether the high-order address of the PS is corrected and is MPS or LPS, a base operation is performed at 4 and a shift digit operation is performed at 5 and a base coordinate and a shift digit number are output. As described above, the fact that the upper address of the LPS calculated by adding the value of the input S exceeds 0.5 means that the range of the inferior symbol on the number line is equal to the range of the superior symbol on the number line. May be exceeded. The output codeword is added in the addition register to become a codeword. The number of shift digits indicates how many digits to shift the position to add the next output codeword. FIGS. 3 and 4 show flowcharts of encoding and decoding in order to more accurately express the above processing. In each case, S is selected to be a power of 1/2.

Next, a specific coding example will be described. In FIG. 5, it is assumed that the coordinates are displayed in binary. S is 1/8 or 1
/ 4. First S from Markov state at Markov source
= 1/8 is known. This time. From 000.
001 is assigned 1 (LPS). 0 (MPS) is assigned from 001 to 1.000. Now suppose that 0 symbols are generated. The range is limited from 001 to 1.000. At this time, the offset is. 001.
Since it is known for both transmission and reception to use S = 1/4 from the appearance probability of 1 for the next symbol. From 001. One is assigned to 011. When 0 appears here, the range of the number line is. It changes from 011 to 1.000.
Next, if S = 1/4, the upper limit of the LPS allocation range is. 0
11+. 01 =. Since the value exceeds 0.1 (0.5 in decimal) at 101, when the correction exceeding 0.1 is halved. 1001.

Here, LPS has appeared, and the size of the LPS region is. 1001-. 011 =. If then 2 ² multiplied by because it is 0011 more than 0.1 (0.5 in decimal). Therefore, the number of shift digits is two. The base is. 1001
−. 01 =. 0101 and outputs this value as a code word. The new offset is. 011-. 0101 =. 00
Since 01 is shifted by two digits, it becomes 0.01. Then S = 1
/ 8. 01+. 001 =. 011 is the boundary between 0 and 1.

Here, when 0 occurs, the offset becomes. 01
Go up to 1. Here, assuming that S = 1/4. 011+. 0
Since 1 = 0.101> 0.1 and exceeds 0.1, halving the amount exceeding 0.1 is used. 1001. If the symbol is 0, the area of 0 is less than 0.1, so the base value. 10
It is necessary to output 1000 corresponding to 00 as a codeword output and to normalize it by a factor of two. . 1000 because base new offset becomes 0.001 in 2 ¹ times the (0.1001-0.1). The next state is S = 1/8 and MPS
Is assumed to be 0.001 + 0.001
= 0.010. Further, assuming that S = 1/4, if 1 (LPS) appears, the offset value. 0100 is output as a codeword.

The final code word is the one calculated based on the code word issued above and the shift digits. (Shown in Figure 5 below)

The value of S is 1/2, 1/4, 1 /
If a value set to a power of 1/2 such as 8 is selected, the assigned area of the MPS becomes 0.5 on the normalized number line.
Is convenient because the power-of-two multiple for normalization does not change even if the value of S fluctuates due to correction when the value is less than.

Now, when areas are given to 0 (MPS) and 1 (LPS) according to the above-described method, the relationship between the value of S and the assumed appearance ratio r of LPS when S is determined is as follows.

[0031]

(Equation 9)

## EQU1 ##

Therefore, if S = 1/2, r≡1 / 2, which is stable.

If S = 1/4,

[0035]

(Equation 10)

Is as follows.

On the other hand, if S is fixed as before, the assumed appearance ratio r of LPS is

[0038]

[Equation 11]

If S = 1/2

[0040]

(Equation 12)

If S = 1/4

[0042]

(Equation 13)

Is as follows. That is, since the variation range of r is larger in the conventional method, the proposed method is more efficient.

Embodiment 1 of the Invention Next, a specific decoding example will be described with reference to FIGS. 4 and 5. FIG. In FIG. 5, it is assumed that the coordinates are displayed in binary. S is 1/8 or 1/4. It is assumed that S = 1/8 is first known from the Markov state in the Markov information source. This time. From 000. 001 is assigned 1 (LPS). 0 (MPS) is assigned from 001 to 1.000.

Now, the code word. .. and boundary. 0
Compared with 01, since codeword ≧ boundary and the codeword is a region of 0 symbols, 0 symbol (MPS) is decoded and. The range is limited from 001 to 1.000. That is, the offset is. 001. Since it is known for both transmission and reception to use S = 1/4 from the appearance probability of 1 for the next symbol. From 001. One is assigned to 011. Where codeword. .. and boundary. 0
When 11 is compared, codeword ≧ boundary, the codeword is a region of 0, and 0 symbols are decoded.

Next, the range of the number line is. 011 to 1.0
Change to 00. Next, if S = 1/4, the upper limit of the LPS allocation range is. 011+. 01 =. 0.1 (1
Since it exceeds 0.5 in 0-base, if you make a correction to halve the amount exceeding 0.1, 1001.

Here, the code word. .. and boundary.
When 1001 is compared, codeword <boundary, and 1 (LPS) is decoded in the area of codeword 1. That is, LP
S has appeared and the size of the LPS region is. 1001
−. 011 =. If then 2 ² multiplied by because it is 0011 more than 0.1 (0.5 in decimal). Therefore, the number of shift digits is two. The base is. 1001-. 01 =. 0101
And subtract this value from the codeword. The result of subtraction is. 00
10100 ... (= 0.111100 ...- 0.010)
1).

The new offset is. 011-. 0101
=. Since 0001 is shifted by two digits, it becomes 0.01.

The code word is also. 0010100 ... 2
Digit shift, so. 10100 ... Then S =
As 1/8. 01+. 001 =. 011 is the boundary between 0 and 1. Here, first, the codeword. 10100 ... and boundary. 011, codeword> boundary, and the codeword is determined to be a 0 region. Here, since 0 is decoded, the offset is. Go up to 011. Next, let S = 1/4. 011+. Since 01 = 0.101> 0.1 and exceeds 0.1, if the amount exceeding 0.1 is halved, the boundary becomes. 10
01.

Here, the code word. 10100 ... and boundary.
1001 and codeword> boundary and codeword is 0
Area. Therefore, 0 is decoded here. Next, since this 0 area is less than 0.1, it is the base value. 1
It is necessary to subtract 1000 corresponding to 000 from the codeword and normalize (shift by one digit) twice.

[0051] 1000 because base new offset becomes 0.001 in 2 ¹ times the (0.1001-0.1). Subtracting this value from the codeword will also result in. 001
0 (= .10100... -1000). 1
If you shift by digits. 010 ...

The next state is S = 1/8 and the boundary is 0.
001 + 0.001 = 0.010. Codeword. 01
0 ... and boundary. Compared with 010, codeword ≧ boundary, and the codeword is an area of 0. In the next state, S = 1 /
If it is 4, the boundary is 0.010 + 0.01 = 0.1.

Here, the code word. 010 ... and boundary. 10
When compared with 0, the codeword <boundary, the codeword is an area of 1 and 1 is decoded. Thereafter, decoding is performed in the same manner.

[0054]

As described above, according to the present invention, the area allocated to the LPS in the number line coding can be controlled in accordance with the LPS appearance probability, so that more efficient decoding can be realized.

It is needless to say that the present invention can be applied to a multi-valued information source because a multi-valued information source can be easily converted to a binary information source by tree expansion.

[Brief description of the drawings]

FIG. 1 is a diagram showing an encoding device according to an embodiment of the present invention.

FIG. 2 is a conventional diagram showing the concept of number line coding.

FIG. 3 is an encoding flowchart in one embodiment of the present invention.

FIG. 4 is a decoding flowchart according to an embodiment of the present invention.

FIG. 5 is an operation example in one embodiment of the present invention.

[Explanation of symbols]

1 adder, 2 comparator, 3 offset calculator, 4
Base arithmetic unit, 5 shift digit number arithmetic unit, 6 Upper address correction device.

──────────────────────────────────────────────────続 き Continuing on the front page (72) Inventor Tomohiro Kimura 5-1-1, Ofuna, Kamakura City Mitsubishi Electric Corporation Communication Systems Laboratory (56) References Proc. Of the IEICE Spring National Convention (1989) , Ono, Kimura, Kino, Yoshida, P.S. 1-152 (58) Field surveyed (Int. Cl. ⁶ , DB name) H03M 7/40

Claims (2)

(57) [Claims] When decoding a coded signal, a dominant symbol which is a symbol having a high appearance probability and a inferior symbol which is a symbol having a low appearance probability are divided into a range (A _i-1 ) on the number line up to the previous symbol and A decoding method for outputting a decoded signal from a range correspondence result on the number line and a coded signal input based on the predetermined range information on the number line based on the predetermined range information (S) on the number line of the inferior symbol. A first step of comparing the range on the number line between the superior symbol and the inferior symbol; and, based on the comparison result of the first step, the range of the inferior symbol on the number line does not exceed the range of the superior symbol on the number line. And a second step of controlling the range of the inferior symbol and the range of the superior symbol on the number line, and performing decoding based on the controlled range. 2. In decoding a coded signal, a dominant symbol having a high appearance probability and a inferior symbol having a low appearance probability are represented by a number line range (A _i-1 ) up to the previous symbol and A decoding method for outputting a decoded signal based on a predetermined range on the number line based on a predetermined range (S) on the number line of the inferior symbol and a result of the range correspondence on the number line and an input coded signal, A first step of comparing the range on the number line assigned to the dominant symbol with a predetermined value; and, when the range given to the dominant symbol is below the predetermined value on the number line, half of the value below the predetermined value in the range of the dominant symbol. A second step of controlling the range on the number line between the superior symbol and the inferior symbol so as to shift the value from the inferior symbol range to the inferior symbol range. Decoding method and performing come decoding.