JPS59152782A - Picture compressing system - Google Patents

Abstract

PURPOSE:To store a picture with less capacity by dividing a picture until dots constituting the divided partial picture become all the identical value and representing the entire picture by the value of the dots constituting the partial picture. CONSTITUTION:A picture 1 of 64-bit constitution is divided into four, partial pictures of 21-24. Since the bits of the partial picture 21 are all ''0'', the partial picture 21 is represented by ''00''. The partial pictures 22-24 are divided further into four to obtain partial pictures 221-224, 231-234 and 241-244. The dot value of the partial pictures is identical except the pictures 223 and 232 and the partial pictures are represented by ''00'' or ''11''. The partial picture 223, 232 are divided further into four and they are represented by the dot ''00'' or ''11'' finally. Thus, the picture 1 of 64-bit constitution is represented by 36-bit and the storage capacity is saved remarkably.

Description

[Detailed description of the invention] Technical field The present invention relates to image compression.

[Prior art] Images handled by computers are generally expressed by dividing them into dot grids or matrices. In the simplest method, each divided square corresponds to black and white, and the information is 0.
It is expressed by whether it is N(1) or OFF(0). In other words, the simplest method for storing images has traditionally been to associate each dot with one bit of memory, and turn on the corresponding bit depending on whether each dot is ON or OFF.
It has been expressed by setting it to (1) or OF F (0). In this case, regardless of the contents of the image to be stored, it was necessary to store as many bits e3t as the number of dots in the image.

Furthermore, in the case of a color image, if we assume that 8 types of one color are used, as a color manager, you can calculate the numbers from 0 to 7, and one dot is 3 dots) (from o to 7). value).

In this way, the entire screen including colors can be stored as a collection of dots. In this case, as well as the white image,
Regardless of the content of the image, it is necessary to store the number of bits equal to 3 x (the number of dots in the image). However, in many black and white images, in some parts of the image, all dots are ON (1), and in other parts, all dots are OFF.
F (0) in many cases. For color images,
All parts are often the same color.

Purpose The purpose of the present invention is to divide an image into parts until the contents are constant (all ON or OFF or the same color) by taking advantage of the above-mentioned image characteristics, and to record the entire image as a collection of B1 (partial images). This provides a compressive force formula for tQ images that allows the image to be stored in a smaller storage capacity.

Example Examples will be described below.

In the embodiment, it is assumed that the image is composed of 2''×2n dots and is black and white.

FIG. 1 shows an example of the concept of image division. The image is constant (
When all dots are not ON or OFF, the image is divided into 4 parts (the divided images are called partial images).If each divided partial image is not constant, it is further divided into 4 parts, until the dots are constant. This partial image division is repeated, and when the partial image becomes constant, that partial image is turned ON or OFF.
Represent it as F and memorize it. In the case of FIG. 1, if the division into one partial image is performed 111iiJ times, one dot is reached, so a constant partial image can always be obtained by dividing the image up to n times.

Now, we decided to describe the image information in the form pat(n), and according to BUF (/<Tsukasnauer notation),
Super symbol +1. Using =l+ and ``1''', it is determined that the image on the left side of =゛° is represented by one of the images separated by 1°'' on the right side, and a certain image pat(n) is expressed as the first Let us express it as follows.

pat(n)+=Ooffl 0onl 1pat(n
+11) pat (n+12) pat (n+13)
pat(n+14)−=(1)(off=o, o
n=1) That is, pat(n) means that all are off, all are on, or a set of four partial images divided into four.

Expressing equation (1) using actual bint information, equation (2)
It becomes like the expression.

pat (1): = 00 l 01 I 1 pat
(11:1 nat (12) pat (13) pat (14
)...(2) can be done. That is, if the initial focus is 0, the partial image is a constant image in which all dots have the value of 0 or 1 of the next bit, and if the first bit is l, then the partial image is a constant image with the value of 0 or 1 of the next bit. express.

If the entire image is pat(n), pat(n) = 00 means an image with the entire screen OFF (white), and pat(n) = '01 means an image with the entire screen ON (black).
means a statue of

Figure 2 is an example of an image with a small number of dots (dragon number 8 x 8)
It is. Expressing this using equation (1) and introducing parentheses to assist in delimiting results in equations (3) 2 to (28).
It becomes like the expression.

pat (1) = 1 pat (21) pat (22)
pat(23) pat(24)... (3) However, pat(21) = 00
...(4) pat(22) = Ipat(
22+) pa't (222) pat (223) pat (
224)... (5) pat(22j) -00... (6) pat(22
2) = 00...
・ (7) pat (223') = Ipat (2231
,) pat (2232) pat (2233) pat (2
234) ... (8) pat
(2231) = 00 ・
... (9) pat (2232) = 01
... (10) pat(2233)=
01...(11)p
at(2234) = 01
... (12) pat (224) = 00
... (13) pat(23
) = Ipat(231)pat(232)pat
(233) pat (234)... (14) pat (23+) = 00
... (15) pat (232) -1 pat (2
321) pat (2322) pat (2323) pat
(2324) ... (16) pat
(232+)=00-
(17) pat(2322) = 01
- (18) pat (2323') = 01
・” (19) pat(
2324) = 01...
・(20)pat(233)=00
... (21) pat (234) =
01... (22
) pat (24) = 1 pat (241) pat (
242) pat (243) pat (244)... (
23) pat(241)=01
... (24) pat (242) = 00
... (25) pat(2
43)=01...
(26) pat(244) = 00
... (27) Therefore, equation (3) becomes as follows.

... (28)-(3) = 100100001000+0101001001
000101010001101000100
... (28) Furthermore, (28) −
The numbers written in parentheses below (1) to (28) - (3) correspond to the numbers in each formula above.

Therefore, although diagram S2 conventionally required 8×8=64 bits, it can be expressed using 44 bits according to the embodiment. Note that this effect becomes greater as the number of dots on the screen increases.

FIG. 3 shows an expression method that is more effective in saving memory. The image to be handled is the same 2n x 2n black and white model as in FIG. Here, the final divided images are defined as p a, t as follows.

pat(i+1):=OOl 01 l 1pat(i
) pat(i) pat(i) pat(i)
...(29) However, i=0.1, ..., n-
1 patO:=Ol ] ・= (
30) That is, if the divided image is one dot in the final division, the first one bit is saved by taking the 0 key 171 to indicate that the dot information is uniform.

When the image shown in FIG. 2 is expressed using this method, it becomes as shown in FIG. 3, which can be expressed using 36 bits. That is, pat n= 1
0010000101110010010111000
1101000100...
(3) Next, let's consider the efficiency of memory saving with reference to Figure 2.

First, let's consider the worst case. Now, if we assume that all image divisions do not take a constant value until they reach one dot, the image will be dotted with dots or completely scattered,
This is the worst case.

In reality, images like this are rare.

Here, let us assume that the storage section of p a t (i) is D1 bit.

'3 ('32) formula is obtained. Solving this recurrence formula, we get
The number of bits of the n×n overall image is as shown in equation (32).

Dn=4, -1 (
32) Therefore, since the capacity required to store all dots in bit correspondence is 21 x 2n ~ 41, calculating this, formula (32) requires 473 times the normal storage capacity. become.

However, the purpose of the present invention is not to store such images, and the value of equation (33)'' is obtained only in the case of very special images.

Now, let us consider the process of dividing a general image n times. In particular, consider the average case in which 174 partial images take a constant value among the partial images in the process of dividing each image. In this case as well, equation (34) is obtained as before.

Solving this recurrence equation yields equation (35).

Dn=3”-'X(45/8)-5/8-,・(
35).

Therefore, the ratio On/4'' with the normal storage capacity is f5 (
36) The equation is as follows.

Now, if we consider the case of n-8 (the number of dots is 25 [I x 256)], from equation 55 (3B), the value is 0°18
It will be 77. That is, according to the present invention, it is possible to store an image in which 174 partial images take a constant 4i with a storage capacity of about 115.3 compared to the conventional one. Furthermore, the 36th
) As shown by the equation (3/4)''-', the larger n is, that is, the larger the number of dots to represent the target image, the greater the memory saving effect.

Next, we will discuss the hardware configuration that realizes the image compression method.

FIG. 4 is an example of a hardware configuration for realizing the present invention. GM in the figure is an image memory that stores one dot of an image (pixel) corresponding to one bit. This memory is addressed one bit at a time. The horizontal address signal line (0~2''-1) of the image memory GM is the address signal line (0~2''-1)
2n-1) is Av.

Bit information can be read from and written to the image memory GM through the signal line Dg.

PM in the figure is a compression memory that stores image information compressed according to the present invention. This memory stores data in a stacked manner, and can be read and written one bit at a time through the signal line Dg. A write operation adds a new bit to the top of the stack. Also, a read operation takes out the most significant bit of the stack. In the figure, PUi±, the above image memory GM
PAC that compresses the image information inside and stores it in compressed memory PM
This is the processing flow of the K operation and the UNPA (:) operation that returns the compressed information in the compression memory PM to the conventional representation and stores it in the image memory GM. This is a processing flow of the operation.Next, the memory in the processing device PU shown in FIG. 4 used in the above processing will be explained.

Figure 5(a) shows the main flow, step (1)
HP is the horizontal address of the image dot (0~2°-
1) is stored. Therefore, it corresponds to the horizontal address Ah of the image memory GM in FIG. Similarly, VP in step (1) of FIG. 5 indicates that the vertical address (0 to 2n-1) of the image dot is stored. Therefore, it corresponds to the vertical address Av of the image memory GM in FIG. The address in the image memo 50M is (HP).

VP). The number of dots (2°, 21, . . . 2n) currently targeted is stored in the storage unit DC. When the main flow of the 51st Δ(a) starts, in step (1), the horizontal address HP and vertical atsis of the image dot are set to O (zero), and the current target driver l-
Set the number DC to 21).

The operation performed in step (2) (see (b)) stores the compression information in order starting from the first bit, and determines whether it can be compressed or not.
Or judge whether it is compressed.

Furthermore, the Dots and Dots that appear in Figures 5 to 7
1-11ot4 is a 1-bit memory that temporarily stores information about image dots.

Step (1) is a step for performing initial settings, and settings are made for the entire image. In step (2), a subroutine UNPAC:KS that performs actual processing is called. Note that the subroutine UNPACKS is a recursive program.

Step (3) of the subroutine UNPACKS is a branch of special processing when one dot is compressed at the final stage.
Inspect the storage unit DC for the number of dots currently targeted, and l
Otherwise, proceed to step (4). In step (4), one bit is extracted from the compression memory PM and stored in the memory Te5t, and in step (5) it is determined whether or not the currently targeted partial image has a constant value. If it is not a constant value, Te5
Since t is 1, the process proceeds to step (6) 5, where the target dot e is further halved so as to reproduce a partial image divided into four parts. UNPACKS is performed on each partial image divided into four by step (7L(8), (+3), (to)).

In step (11), the horizontal address HP, the vertical address vP, and the storage section DC for the number of dots to be processed are returned to their original values, and the process ends. On the other hand, if the partial image has a constant value in step (5) and if it is determined that the partial image is one dot in step (3), proceed to step (12);
The value of the constant value is taken out from the compression memory PM and in step (13) the information is reproduced in the image memory GM. Thus, the processing of tlNPAcXs is completed.

As described above, when the program finally returns to END (14), the compressed image of PM has been reproduced on GM.

FIG. 6 shows the procedure for compressing the image on the image memory GM and storing it on the compression memory PM.

In step (1), initial settings are made for the entire image. In step (2), a subroutine PACKS that performs actual processing is called. Note that PACKS is a recursive program. When the subroutine PACKS is called, it is determined in step (3) whether the currently targeted partial image is in the final stage of 2×2. If it is not the final stage, proceed to step (4) and further halve the target driver hgl so that the partial image divided into four parts is handled. PAC:KS is performed on each of the four divided partial images in steps (5), (8), (?), and (8). At this time, it can be noted that the four partial images are handled in the reverse order as in the case of image re-presentation in FIG. That is, this is due to the fact that the structure of the compressed memory PM is a stack structure. The storage unit DC for the number of dots targeted in step (S) is returned to its original state. At this time, the horizontal address HP,
Vertical address VPLt has already returned to its original value.

In step (10), it is determined whether the set of four 81 (minute images) stored in the compression memories P and M can be further compressed, and if it can be further compressed, it is compressed.

This will be described later using FIG. 7. On the other hand, if the partial image becomes 2x2 at the final stage in step (3) of Fig. 86, the information of these 4 dots is read out from the image memory GM in step (11), and in the next step (12). It is determined whether all the four dots called up match, and if they do not match, the process moves to step (13), the information of these four dots is packed into the compression memory PM, and the bit "" indicating that compression could not be performed is stored in step (14). 1″ further compressed memory PM
Fill it in. On the other hand, in step (12) all 4 dots are 12
If it is, the process moves to step (15), and only the dots showing the constant value are stored in the compressed memory PM, and step (15) is performed.
In G), a bit "0" indicating that the data has been further compressed is packed into the compression memory PM.In this way, PAC:KS ends.

As described above, when the program finally returns to END (17), the image in the image memory GM is compressed and stored on the compression memory PM.

FIG. 7 is a flowchart showing the process of compressing the postage of the partial image called at (10) in FIG. Processing 1/]E
1' (+The upper content of the memory PM stack is “00θ
oooooo” means that all of the four partial images are O, and this is replaced with °00゛, and when the content of the compressed memory PM is “01010101”, it means that all of the four partial images are 1, Processing is performed to replace this with 01'. - In cases other than quadrants, a process is performed to add "1°" to the compression memory PM, which indicates that the set of four partial images is not a constant value.

In FIG. 7, when the CHECK program is called, first, in step (1), the count is set to 0, and then in step (2), 1 bit Te5t is taken out from the compressed memory PM. In step (3), the bit of Test t is determined, and if it is O, the process proceeds to step (4). In step (4), 1-pi 1-Dot is taken out from the compressed memory PM. Next, in step (5), increase the count by 1,
Next, in step (6), l bit Te is extracted from the compressed memory PM.
Take out to 5t. The bit of Te5t is determined in step (7), and if it is 0, the compressed memory P is determined in step (8).
Take out 1 human 1-Dot'l from M. Next, in step (9), Dot,! =Dot 1 or ']' is determined, and if they are equal, proceed to step (10),
Determine whether the count value is 3 or not. Karan)・
If the value of is not 3, return to step (5) and step (5
) to step (10) are repeated. In step (10), if the count value is 3, it means that all the bits in the compressed memory are equal, so in step (11), the bit of Dot is transferred to the compressed memory PM by 8 years old and then step ('
12) Pip 1-.

is packed into the compressed memory PM and completed, and if Te5t is 1 in step (3), the process moves to step (14) and Te5
Pack the bits of t into compressed memory PM. Step (15
), if the count is not 0, the process moves to step (16).

In step (16), the bits of Dot are stored in the memory PM. At step (17), bit 0 is packed into the memory PM, and at step '(18), the count (+i:i) is counted, and the process moves to step (5).

If the count value is 0 in step (15), this means that the value in the compressed memory PM has been returned to its original value, so in step (19) it is filled in the compressed memory and returns to the called place.

If Te5t is 1 in step (7), it means that the pattern bits are equal to or less than 1, so step (14)
and perform the process described above. In step (8), Do
If t is not equal to Dot 1, it also indicates that the bits of the pattern are not equal, so the bits of Dot 1 are packed into the compressed memory PM and the process moves to step (14), where the above-mentioned processing is performed.

According to the above explanation, that is, according to the present invention, the number 2ηX2
Considering that it takes a constant value of 2 bits in the minimum case and a 1/4 partial image on average to store a black and white image of ``,'' compared to the conventional method of storing all dots. According to the present invention, the effect of saving storage capacity becomes even greater when storing a large image with a large number of dots representing the image.

According to the present invention, more images can be stored on a disk or the like with the same storage capacity than before. That is, the storage area for images can be saved and a large number of images can be stored in a limited storage capacity.

[Brief explanation of drawings]

The first [/ is the part of the image to be stored using the method of the present invention'+
'; II conceptual example, Figure 2 is a diagram for explaining the storage method according to the invention, Figure 3 is an example of simplified storage when the final divided partial image is 11'. 4 is a hardware configuration example of the present invention, FIG. 5 is a flowchart 1 showing processing for re-expressing a compressed image, FIG. 6 is a flowchart showing JFE compression processing, and FIG. FIG. 3 is a flowchart diagram illustrating bit operations for compression. Here, l...dot matrix of all images. 20... Indicates the division of the bow, 21, 22, 23, 24,...
・1 divided partial image, 22o...22 divided partial images, 221, 222.223.224...22 divided partial images, 2230...223;'! '
1, 2231.2232, 2233.2234.
...223 divided partial images, PU...processing J1 location, GM...image memory, PM...pressure"ii?
+Memory, Ah...Address in the horizontal direction of the image memory, AV...Address in the vertical force direction of the image memory, I)g
, I) P... is a reliable line. Patent applicant: Canon Co., Ltd. Figure 3, Figure 4, Figure 5 (a) (1))

Claims (1)

[Claims] 1. Divide a dot matrix representing an image into parts,
The image is further divided into partial images until all the dots that make up the divided partial image reach the same value, and when all the dots that make up the partial image reach the same value, the previous dividing procedure and the configuration of the partial image are repeated. An image compression method characterized in that the entire image is represented by the value of a dot. 2. The image compression method described in Section 4, Item 1 of the patent claim, characterized in that when the divided B1 image reaches one dot, one dot is treated as partial image information.