JPH0410179A - printing device - Google Patents

Abstract

PURPOSE:To shorten time for executing algorithm by turning on all the bits, where a straight line is passed through, and replacing one part, which is painted out, with an electronic circuit. CONSTITUTION:When a data is inputted to a data input/output means 1 and a curve is included in this data, the data is approximated to the straight line by a straight line approximation part 2. Next, a paint-out data calculation part 3 calculates a data to be required for painting-out so that all the bits, where the straight line is passed through, can be turned on. The data required for painting-out at high speed is sent to a paint-out range decision circuit 4 so as to decide a range to be painted out. This data decided by the paint-out range decision circular for the range to be painted-out is sent to a paper sheet transfer device 5 and based on the data, a bit map data is prepared and transferred to the sheets of paper.

Description

[Detailed description of the invention] [Industrial application field] The present invention relates to a printing device.

[Prior Art] Taking an example of a conventional printer, 2 in Fig. 2(a)
4. When displaying parallelograms shown in 25 and 26.27.

For example, IBM System Jo published in 1965.
umal, vol, 4°no, 1. 3 from page 25 of
It is written on page 0. J.E.

Using Mr. Bresenham's algorithm (hereinafter referred to as the Presentham algorithm), Figure 2 (a)
30, 31, 32, 33, 34, indicated by dotted arrows.
For each integer X coordinate of 35, 36.37, calculate which bit the intersection point between the horizontal line determined by the X coordinate and the actual straight line is on, and use the method shown in Figure 2 (
In a), as shown in 28, the bits closest to the actual straight line, which are colored gray, are turned on. Furthermore, all of these processes were performed by software. In addition.

Here, 21 and 22 indicate the y-axis and the y-axis, respectively.

23 indicates a coordinate point where both X and y are integers.

[Problems to be Solved by the Invention] However, in such a conventional method, when displaying the elongated shapes indicated by 42.43 and 44.45, as shown in FIG. Even though the figures are continuous, they may be interrupted when displayed twice. In addition.

Here, 40.41 are the y-axis and y-axis, respectively, and 4
6.47°4g, 49,50,51.52 are horizontal lines indicating integer X coordinates.

The present invention has been made in view of these problems, and its purpose is to further develop the conventional algorithm without changing its essential parts.

Turn on all the bits that the straight line passes through so that any straight line will be displayed continuously.

By replacing part of the filling part with an electronic circuit, the execution time of the algorithm can be significantly shortened.

[Means for Solving the Problems] The printing device of the present invention treats one character or figure as xy coordinate data, which is an integer representing two outlines, and prints it on paper. In the printing device, a data input/output means inputs and outputs the data of the characters and figures, and data of two curves included in the xy coordinate data, which are integers representing the two outlines of the figures and characters, is input and outputted as a straight line. Approximate to integer xy coordinate data. With the straight line approximation part.

Data of one curve was approximated by the linear approximation section. Using the integer xy coordinate data of the straight line and the integer xy coordinate data of the straight line as input, the straight line passes through when the straight line is made to correspond to the discrete bits of the printing device. data necessary for determining the range of filling in the figure or character made up of the straight line is calculated and output so that the bit corresponding to the straight line corresponds to the straight line. A range to be filled is determined using a fill data calculation section and the output of the fill data calculation section. The present invention is characterized by comprising a filling range determining circuit and a paper transfer device for creating bitmap digital data and transferring it to paper according to the output of the filling range determining circuit.

[Effect] The operation of the printing apparatus of the present invention will be explained.

As shown in the functional block diagram of Figure 1, when data is input to the data input/output means 1, in the linear approximation section 2, if the input data includes a curve, it is approximated to a straight line. Second, in the filling data calculation section 3, data necessary for filling is obtained so that the bits through which the straight line passes are turned on. Note that these data will be described later. Here, we will explain how to select bits when filling.

Basically, filling is turning on all the bits between two opposite sides.

In addition, in a printing device, the surface to be printed is made up of discrete bits, so at any integer X coordinate, it is sufficient to find the integer X coordinate of each of the two opposing sides. The situation is shown in Figure 3. Here, Figure 3 (
a) is the left side of the opposing sides (hereinafter referred to as
(referred to as the left side), at each integer X coordinate, and
3o2 in the figure 1 shows whether the bit corresponding to the coordinate is selected as the leftmost bit for filling.
The bit colored gray as shown in is the bit determined to be the leftmost bit in the X coordinate of the integer. 303 indicates the coordinates of the center of an arbitrary bit. Furthermore, the coordinates indicated by 303 are +
Both X+'l are integers. 304 and 305 are the y-axis and the y-axis, respectively.

FIG. 3(b) similarly shows the side on the right side (hereinafter referred to as the right side). Here, the bit colored gray is the bit determined to be the rightmost bit in the y-coordinate of each integer.

In this way, the data necessary for performing the filling at high speed is sent to the filling range determination circuit 4, which determines the range to be filled.

The data of the range to be filled determined by the filling range determination circuit is sent to the paper transfer device 5, and based on the data, bitmap data is created and transferred onto the paper.

[Example] Details of a printing apparatus of the present invention will be described below based on an illustrated example.

FIG. 10 is a system configuration diagram showing an embodiment of the present invention.

Reference numeral 111 in the figure indicates a microprocessor 112 (hereinafter referred to as "microprocessor").

CPU), 113 read-only memory
(below.

ROM), 114 random access memories (hereinafter referred to as work memories), and 115 ILOs (hereinafter referred to as ILOs), which are connected to 116 paper transfer devices and 117 fill-in range determination circuits. ing.

The microcomputer 111 has a data input/output function with the paper transfer device 116, a linear approximation function, and a filling data calculation function, and these functions are performed according to the program contained in the ROM 113.

This is achieved by operating the CPU 112 and operating the Ilo 115 and work memory 114.

FIG. 16 is a block diagram of the filling range determination circuit 117, and also shows the connection state between the CPU and the paper transfer device. A portion surrounded by a dotted line indicated by 91O is a filling range determining circuit. 912° and 913 are necessary for high-speed filling.

This is a circuit block that calculates the data on the left side.

914 and 915 are circuit blocks that calculate data on the right side, which is necessary for high-speed filling. These data will be described later.

911 is a CPU, and 916 is a paper transfer device.

In this embodiment, the coordinates are expressed as 8-bit signed integers, but the coordinates may have any number of bits.

Next, the operation of the device configured in this way is shown in Figures 11 and 12.
13, 14, and 15, circuit diagrams shown in FIGS. 17, 18, 19, and 20, and FIGS. 3 and 4. The explanation will be based on the conceptual diagrams shown in FIGS. 5, 6, 7, and 8.

In this embodiment, it is assumed that the positions at which figures and characters are printed are expressed in terms of x and y coordinates.

When the printing apparatus starts operating, it waits for input from the paper transfer device 116 according to the program stored in the ROM 113 of the microcomputer 111.

(Step 1) From 116 paper transfer devices to 115 work/○.

When an execution request and text or graphic data are input.

(Step 1) First, check whether there are two curves. (Step 2) If there is a curve, approximate it to a straight line.

(Step 3) In step 4, data necessary for high-speed filling is calculated. These data will be explained in detail. Note that the coordinates of the starting point of the 9th side are (Xl.

yl) and the coordinates of the end point are (x2.y2).

Also, let yl be y2.

First, a method for calculating data on the left side and a method for selecting bits will be explained based on the conceptual diagram shown in FIG. 4. In FIG. 4(a), the left side indicated by 401 and the bit determined to be the leftmost position at each y-coordinate are shown colored gray as indicated by 402, and 404 indicates y-axis, 4o5
is the y-axis ・Figure 4(b) is 403 in Figure 4(a)
This is an enlarged view of the inside of the circle indicated by . here,
421°422 are bit boundaries in the X direction, 451,
452, 453, and 454 are bit boundaries in the X direction.

Then, the distance from the bit boundary in the X direction indicated by 410, 411, and 412 (hereinafter).

err) is the criterion for bit selection. Viewed from the bit boundary in the X direction, if err is negative, the left bit.

If positive, select the right bit. In the case of Figure 4(b),
When we check the err of 410 at the y-coordinate of 431, it is negative, so the err of the 0th order with the bit of 441 as the leftmost bit is (x2-x 1) / (y2-y
1) at the bit boundary in the X direction of 452. The err indicated by 411 is determined. Since this err is positive, the bit indicated by 442 is selected.

Then, the reference bit boundary in the X direction is moved to 422.

1- (x2-xl)/(y2
Find the err of 412 by adding yl),
Repeat the bit selection decision.

eTing calculates the initial value only once before performing this procedure, and then uses the constant c2 = (x2-xl)/(y
2-yl) or cl = 1-(x2-xi
)/(y2-yl).

Next, how to obtain the initial value of err on the left side will be explained.

FIG. 5 is a conceptual diagram showing an example of calculating the initial value of err. FIG. 5(a) shows the case where the slope of the side indicated by 501 is positive, and FIG. 5(b) shows the case where the slope of the side indicated by 521 is negative.

A portion surrounded by a square indicated by 505 and 526 corresponds to one bit, and 506 and 527 indicate the center coordinates of the bit.

Note that the center coordinates of the bits are integers, and the starting point of the straight line is (X 1 + y 1) and the ending point is (x2.y2). 6 Here, it is assumed that yl<y2.

First, FIG. 5(a) will be explained.

FIG. 5(a) shows the vicinity of the starting point of the straight line indicated by 501. The distance from the boundary in the X direction of 507 to the starting point of a straight line as shown in 502 is determined. Since the xy coordinates of the starting point of the straight line are integers, this distance is -0,5. Add the distance shown in 503 to this value and get the err shown in 504.
Find the initial value of. Here, the distance shown at 503 is y
Set 2-yl as deltay and x2-xl as
If we set deltaX, the distance shown in 503 is.

0.5 x deltax/deltay. Therefore.

err = 0.5 x deltax / del
tay -0.5. However, as it is, there will be division by 2 and decimal numbers, so we will make it possible to perform calculations using only integers. In other words, 2 X deltay on both sides
Apply.

2 X deltay X err= delt
Let ax −deltay. err only makes positive and negative decisions, so it's new.

2 X deltay X err, err
I'll keep it that way.

err = deltax - deltay.

Also, the value added to err mentioned above.

c2 = (x2-xl)/(y2-yl), ie.

c2 = deltax/deltay or.

cl = 1-(x2-xl)/(y2-yl), ie.

cl = 1-deltax/delta
Because y must also be multiplied by 2 X deltay.

c2 = 2 x deltax cl = 2 x (deltay - delta
x).

Now we have found the data for the left side with a positive slope.

Next, Figure 5 (b
) to explain.

FIG. 5(b) shows the vicinity of the starting point of the straight line indicated by 521. The distance from the boundary in the X direction of 524 to the starting point of a straight line as shown in 521 is determined. Since the xy coordinates of the starting point of the straight line are integers, this distance is 0.5. The distance (negative value) shown at 523 is added to this value to obtain the initial value of en- shown at 524. Here, the distance shown in 523 is set as deltay for y2-yl, and x2
If −xi is set as deltax, the distance shown in 523 is.

0.5 x deltax/deltay. Therefore.

err = 0.5 x deltax / del
tay + 0.5. However, as it is, there will be division by 2 and decimal numbers, so we will make it possible to perform calculations using only integers. In other words, 2 X deltay on both sides
Go over it.

2 X deltay X err= delt
Let ax + deltay. err only makes positive and negative decisions, so it's new.

2 X deltay X err, err
I'll keep it that way.

err = deltax + deltay.

moreover.

Find m=deltax div deltay. Here, div indicates integer division, and.

ml = m + 1 ml=m seek.

Now we have found the data on the left side. (Step next,
The method of calculating the data on the right side and the method of selecting bits will be explained based on the conceptual diagram shown in FIG. Figure 6(a)
, the right side indicated by 601 and the bit determined to be the rightmost at each X coordinate are colored gray as indicated by 602, 6o4 is the y axis, and 605 is the X coordinate. It is the axis.

FIG. 6(b) is an enlarged view of the inside of the circle indicated by 603 in FIG. 6(a). Here, 621°62
2 is the bit boundary in the X direction, 651, 652, 653
, 654 are bit boundaries in the X direction. And 61
Distance from the bit boundary in the X direction indicated by 0.611, 612 (hereinafter).

err) is the criterion for bit selection. Viewed from the bit boundary in the X direction, if err is negative, the left bit.

If positive, select the right bit. In the case of Figure 6(b),
When we check the err of 610 at the X coordinate of 631, it is negative, so we set the bit of 641 as the leftmost bit to the 0th order err: (x2-xl)/(y2 yl)
By adding the value 652 at the bit boundary in the X direction. The err indicated by 611 is determined. This e
Since the value is positive, the bit indicated by 642 is selected. Then, the reference bit boundary in the X direction is moved to 622.

611 err 1- (x2-xl)/(y2-y
1), the err of 612 is obtained and the bit selection judgment is repeated.

In this way, the data on the right side can be calculated in exactly the same way as the data on the left side. (Step 6) To summarize the bit selection method explained above, focusing on the difference between the left side and the right side, as shown in the conceptual diagram of Fig. 7, the difference in processing between the left side and the right side is as follows.

When determining the leftmost bit or the rightmost bit, if the slope is positive, as shown in Figure 7(a),
The left side looks at elT and determines the bit above it, as shown in FIG. 7(c).

The right side looks at err and determines the bit below it. Also, if the slope is negative, as shown in Figure 7(b), the left side looks at err and determines the bit below it, while the right side looks at err and determines the bit below it, as shown in Figure 7(d). e
Look at rr and determine the bit above it.

2.For convenience, err on the left side is errl and cl is ell.
, c2 to c21. Let ml be ml and m2 be m12.

Similarly, for the right side, err on the right side is errr, c
l is clr, c2 is c2r.

Let ml be mrl and m2 be mr2.

In this way, errl, cll, c on the left side
21, mll.

m12. and errr, clr, c2 on the right side
After calculating r, mrl, and mr2, these data are sent to the filling range determining circuit. (Step 7) This will be explained based on the flowchart shown in FIG. 12 and the conceptual diagram shown in FIG.

In FIG. 8, the thick line indicates the outline of the polygon to be filled in. For convenience, the upper left corner indicated by 803 is the coordinate axis origin, 801 is the X axis, and 802 is the y axis. Also,
The point indicated by 810 is the conceptual center coordinate of the bit, and its value is an integer. Furthermore, bits are conceptually expressed as squares lined up without gaps, and bits that are turned on are colored gray, 804', 805, 80.
6°807, 808, 809 are each side of the polygon. Furthermore, 841,842,843,844°845.
The scales marked 846, 847, 848, 849 represent integer X coordinates, 851,852°853.854,8
56,857,858,859゜The scale shown at 860 indicates the integer X coordinate, and 821,822,823
The white arrow is.

Indicates the err of each side of 805, 831°832
．． The white arrows shown at 833 and 834 are.

The err of each side of 804 is shown.

First, arrange them in descending order of the X coordinate of the starting point of each side.

The sides with the same X coordinate of the starting point are not considered here. (Step 18) Next, the X coordinate value of the side having the smallest starting point or end point is taken out and set as the current X coordinate. here,
The X coordinate is indicated by 851.

(Step 19) In step 20, a filling range determining circuit is provided.

Initialize by sending an initialization signal.

In step 21, it is checked whether processing has been completed for all sides, that is, whether filling has been completed. If filling is completed, go to step 30, otherwise,
Go to step 22.

In step 22, all sides whose starting point is at the current X coordinate are rearranged in descending order of X coordinate. X
If the coordinates are the same, deltax / deltay
Arrange the sides in descending order of slope, expressed by . The X coordinate of 851 is 804. ．． Side 805 has the starting point.

Since they have the same X coordinate, they are ordered in descending order of slope.

Arrange in the order of 804 and 805.

Then, let the side on the left side be the left side, and the side on the right side be the right side. 804 is the left side, and 805 is the right side.

(Step 23) In step 24, a reset signal is sent to the reset terminal of the filling range determining circuit to reset the two circuits.

In step 25, the filling data calculation unit calculates the data. errl, cll, c21, m
ll, m12 and.

The X coordinate of the starting point on the left side and the X coordinate of the ending point (hereinafter, X5EN
D), the value obtained by subtracting delt on the left side from 127 (hereinafter referred to as ENDYI), and the sign bit of deltax that represents the slope of the left side (hereinafter referred to as s 5
lopel), errr, clr
, c2r, mrl, mr2, the X coordinate of the starting point on the right side, the X coordinate of the ending point (hereinafter referred to as XeEND), and the value obtained by subtracting the deltay on the right side from 127 (hereinafter referred to as E
ND Yr) and de where the slope of the right side is expressed.
The sign bit of ltax (hereinafter referred to as s5loper), the current X coordinate, and the end point X of the left and right sides.
Of the coordinates.

A value obtained by adding 1 to the smaller value (hereinafter referred to as "Yend") and subtracting the current X coordinate is set in the filling range determining circuit, and a load signal is sent. Regarding this, the first
The flowchart shown in Fig. 3, Fig. 17, Fig. 18, and Fig. 1
This will be explained in detail with reference to the circuit diagrams shown in FIGS. 9 and 20.

FIG. 17 shows a circuit for determining the coordinates representing the bits on the left side. The initialization signal sent in step 20 is applied to the INITIAL terminal of 216. This signal is 217
221 through the or gate.

Each flip-flop except 227 and 238.

and is applied to the CLR terminal of the latch. This initializes the two circuits.

The left side and the right side of each data set performed in step 25 will be explained.

First, the left side of FIG. 17 will be explained.

Put s 5lopel on the 201 terminal, errl on the 202 terminal, all on the 204 terminal, c21 on the 205 terminal, and Xs end on the 207 terminal.

mll to the 208 terminal, 2 to the 209 terminal, 21
Send Xs to the 0 terminal and END Yl to the 211 terminal. The terminal 201 is connected to latches 221 and 222. 221 and 222 are asynchronous latches, and C
The data is held until the LR signal is input or new data is input. 202,20
Terminals 4, 205, 207° 208, 209, 210 are 227° 218.219, 220, 232, 2, respectively.
It is connected to a 34,238° latch. 22
7,218°219.220,232,234,238
Is it an asynchronous latch and receives the CLR signal?

Alternatively, the data is held until new data is input. The terminal 211 is connected to a counter 247 having a data holding function. The counter of 247 temporarily stores input data and starts counting according to the input data when a load signal (hereinafter referred to as LD) is input.

Next, the right side of FIG. 18 will be explained.

s 5loper to the 251 terminal, errr to the 252 terminal, clr to the 254 terminal, c2r to the 255 terminal, Xe end to the 257 terminal, mrl to the 258 terminal, mr2 to the 259 terminal. , sends Xe to the terminal 260, and sends END Yr to the terminal 261.

The terminal 251 is connected to latches 271 and 272. 271 and 272 are asynchronous latches, and CLR
Data is held until a signal is input or new data is input. 252, 254,
255, 257, 258° 259.260 terminals are 277.269° 270.293, 282, 284, respectively.
, 288. connected with the latch. 277.269
, 270° 293, 282, 284, and 288 are asynchronous latches to which the CLR signal is input, or.

The data is retained until new data is input. The terminal of 261 is connected to a counter of 297 having a data holding function. The counter 297 temporarily stores the input data and starts counting according to the input data when the LD is input.

Furthermore, to the terminal 922 in FIG.

Set Yend+1-current y. This value indicates the y-coordinate range to be filled in. A terminal 922 is connected to a counter 926. The counter 926 has the same function as the counter 297 in FIG.

Then, connect the current y to the terminal 951 in Figure 20.
Set. The terminal of 951 is connected to the latch of 955. The latch 955 has a similar function to the latch 288 in FIG.

After setting the data in the filling range determination circuit in this manner, the CPU sets these data to 0 and proceeds to step 26. In step 26, it is determined whether the current processing is processing of the first side. In Figure 8, 80
The processing of edges 4 and 805 is the processing of the first edge. If it is determined in step 26 that the first edge is to be processed, step 2
Go to 7 and enter INIT L in the filling range determination circuit.
Send OAD signal and LOAD signal. If not, go to step 28 and send a LOAD signal to the fill range determining circuit. The filling range determining circuit is LOAI
) signal, sends a halt signal to the CPU,
The CPU operation is temporarily stopped until the filling process is completed.

Here, regarding the operation of the filling range determining circuit, the first
This will be explained in detail with reference to the flowchart shown in FIG. 2 and the circuit diagrams shown in FIGS. 17, 18, 19, and 20.

First, the overall function of the circuit diagram shown in FIG. 17 will be explained.

FIG. 17 is a circuit diagram for determining coordinate points on the left side. 20
errl applied to the terminal 2 is temporarily stored in the latch 227 and sent to the flip-flop 229 through the OR gate 228.

In addition, the latch 227 sends errl to the flip-flop 229, and then clear is applied to the terminal 203.
It is reset by the r signal. This will be discussed later. Added to 229 flip-flops, 206
The sampled errl is added to the adder of 230 by the clock applied to the terminal of all
,or.

The sum with c21 is calculated.

ell and c21 are applied to terminals 204 and 205, respectively, and temporarily stored in latches 218 and 219, respectively. Then, it is sent to selectors 223 and 224, and err
Should cll be added to err 1 when l is positive, and c21 be added when errl is negative?

Alternatively, whether to add c21 when errl is positive and add cll when errl is negative can be added to the terminal of 201 and 222
2 is selected by the value of s5lopel applied to the SEL terminals of selectors 223 and 224 through the latch. Furthermore, it is sent to the selector of 225, where er
Depending on the sign bit of rl, output ell or c2
It is selected whether to output 1 and sent to 226 flip-flops.

Here, after being sampled with a clock, it is sent to an adder at 230.

230 adders, errl, all, or c
After calculating the sum of 21, the calculation result passes through the flip-flop of 231 and passes through the OR gate of 228 to 229.
Go back to the flip-flop and repeat the calculation.

The XS applied to the terminal 210 is temporarily stored in the latch 238 and sent to the flip-flop 240 through the OR gate 239. In the flip-flop 240, the slope of the left side, or err
The sum with 9m11 or m12 is determined depending on the value of l.

m11 and m12 are applied to terminals 208 and 209, respectively, and temporarily stored in latches 232 and 234, respectively. Then, it is sent to selectors 233 and 235, and er
Add mll to XS when rl is positive and add m12 when errl is negative, or m when errl is positive.
Add L2 and add mll when it is negative to the 201 terminal.

SE of selector 233, 235 via latch 222
By the value of s5lopel applied to the L terminal.

selected. Furthermore, it is sent to the selector of 236.

Here, the sign bit of errl selects whether to output mll or m12.

237 flip-flops. here.

After being sampled by the clock, it is sent to a 241 adder.

241 adder, xs, mll, or.

After calculating the sum of m12, the calculation result is sent to 246 flip-flops and 242 selectors. 246
The output of the flip-flop is sent to the OR gate 228 and returns to the flip-flop 229 to repeat the calculation.

In the selector of 242, Xs which is the X coordinate of the end point of the left side
A selection is made between END and XS. This is to correct the X coordinate of the end point on the left side.

The end point of the left side is detected by the counter 247. This will be discussed later.

The output of the selector 242 is input to the flip-flop 245, and after being sampled with the clock, the output of the selector 215
is sent to the paper transfer device from the output terminal of.

Note that the counter 247 starts counting from END Yl, and when all outputs become 1, it outputs a pull carry, but due to the ripple carry output of the counter 247,
The LOAD signal and reset signal are 249 and 25, respectively.
It is gated with an AND gate of 0. For this reason, 247
The ripple carry signal of the counter is not output, that is, 1
The LO and AD double signal reset signals are not accepted until the processing of the left side of the book is completed. Also.

The clear signal is also gated with a 991 AND gate.

FIG. 18 is a circuit diagram for determining coordinate points on the right side. 25
errr applied to the terminal 2 is temporarily stored in the latch 277 and sent to the flip-flop 279 through the OR gate 278.

In addition, the latch 277 sends errr to the flip-flop 279, and then clear is applied to the terminal 253.
It is reset by the r signal. This will be discussed later. Added to 279 flip-flops, 256
With the clock being applied to the terminal of clr, the sampled errr is added to the adder of 280 and clr
,or.

The sum with c2r is calculated.

clr and c2r are applied to terminals 254 and 255, respectively, and temporarily stored in latches 269 and 270, respectively. Then, it is sent to selectors 273 and 274, and err
Should clr be added to errr when r is positive, and c2r be added when errr is negative?

Alternatively, whether to add c2r when errr is positive and add clr when errr is negative can be added to the terminal of 251 and 272
2 is selected by the value of s5loper applied to the SEL terminals of selectors 273 and 274 through latches. Furthermore, it is sent to the selector of 275, where er
Depending on the sign bit of rr, output clr or output c2
It is selected whether to output r and sends it to 276 flip-flops.

Here, after being sampled with a clock, it is sent to a 280 adder.

280 adder, errr, clr, or c
After calculating the sum of 2r, the calculation result passes through the flip-flop of 281 and passes through the OR gate of 278 to 279.
Go back to the flip-flop and repeat the calculation.

The xe applied to the terminal of 260 is temporarily stored in the latch of 288, and sent to the flip-flop of 290 through the OR gate of 289. In the flip-flop 290, it is sampled by the clock and added to the adder 291. Here, the slope of the right side or err
Depending on the value of r, the sum with mrl or mr2 is determined.

mrl and m r 2 are applied to terminals 258 and 259, respectively, and temporarily stored in latches 282 and 284, respectively. Then, it is sent to selectors 283 and 285, and e
Add mrl to xe when rrr is positive and mr2 when errr is negative, or add mr2 when errr is positive and mrl when it is negative, to the terminal of 251.

SE of selector 283, 285 via latch 272
By the value of s 5loper applied to the L terminal.

selected. Furthermore, it is sent to the 286 selector.

Here, depending on the sign bit of errr, it is selected whether to output mrl or mr2.

287 flip-flop. here.

After being sampled by the clock, it is sent to the adder 291.

291 adder, xe, mrl, or.

After calculating the sum of mr2, the calculation result is sent to flip-flop 296 and selector 292. 296
The output of the flip-flop is sent to the OR gate 289 and returns to the flip-flop 290 to repeat the calculation.

In the selector of 292, Xe which is the X coordinate of the end point of the right side
A selection is made between END and xe. This is to correct the X coordinate of the end point on the right side.

The end point of the right side is detected by the counter 297. This will be discussed later.

The output of the selector 292 is input to the flip-flop 295, and after being sampled by the clock, the output of the selector 292 is
is sent to the paper transfer device from the output terminal of.

Note that the counter 297 starts counting from END Yr and outputs a pull carry when all outputs become 1, but due to the ripple carry output of the counter 297,
The LOAD signal and reset signal are 299 and 26, respectively.
It is gated with an AND gate of 5. For this reason, 297
The ripple carry signal of the counter is not output, that is, 1
Neither the LOAD signal nor the reset signal is accepted until the processing of the right side of the book is completed. Also.

The clear signal is similarly gated with a 992 AND gate.

The circuits shown in FIGS. 17 and 18 both operate in synchronization with a clock. The circuit for controlling the clock for this purpose is the circuit shown in FIG.

921 is a terminal for inputting the LOAD signal, 922 is a Yen
d+1-current y input terminal, 923
is a terminal for inputting MASTER CLOCK, and 924 is a reset terminal. here.

MASTERCLOCK is the output of a rectangular wave oscillation circuit for generating a clock. The LOAD signal is output in step 27. Alternatively, it is the signal sent in step 28. Yend+1-current y is data set in step 25. The reset signal is the signal sent in step 24.

When the LOAD signal is sent to the 926 counter,
Count the clock from Yend+1-Current-y.

When all outputs become l, a ripple carry is output. The ripple carry is input to the ENABLE terminal of 926, and when the ripple carry is output, the counter stops operating.

Also, ripple carry passes through the or gate of 938, and
It is inverted by 27 inverters and input to 929 flip-flops. In the 929 flip-flop, M
The input signal is sampled using a clock obtained by inverting ASTERCLOCK using a 928 inverter. 929
The non-inverted output of the flip-flop is ANDed with MASTERCLOCK by the AND gate of 930.

It is output from the output terminal of 931 as CLOCK.

This CLOCK is input to each circuit block. In this way, since MASTERCLOCK is gated by ripple carry, each circuit is (Yend+1-c
The inverted output of the flip-flop 929 is sent as is to the CPU as a HALT signal. While this signal is active, the CPU is not operating. In other words,
While the CPU is not operating, the filling range determining circuit is operating.

While the CPU is operating, the filling range determining circuit is not operating. In this way, synchronization with the CPU is achieved.

In the above manner, XS and xe can be obtained.

In addition to this, the value of the y coordinate must also be determined.

The circuit for calculating this y-coordinate is the circuit shown in FIG.

In 951, current y is set, and.

It is temporarily stored in the latch of 955. The output of latch 955 is sent to flip-flop 956. Since the flip-flop of 956 does not output data unless CLOCK is input manually, no data is sent to the adder until CLOCK is input.
An AD signal is added.

The LOAD signal is transmitted to the 959 counter.

When the LOAD signal is input, the counter 959 takes in data, and this data is O. 953
is a terminal to which CLOCK is input manually. The CLOCK input here is the CLOCK controlled by the circuit in Figure 19.
It is. Therefore, CLOCK is input only when one circuit is operating, and 954 is a reset terminal.

The current y stored in the latch of 955 is 95.
It is sent to the flip-flop No. 6, and when CLOCK is input, it is sent to the adder No. 957. The output of the counter at 959 is added to the flip-flop at 960, and the output from the counter at 960 is
The output of the flip-flop is .

957 adder. With 957 adders.

The sum of current y and the output of the counter is calculated, and the value of y is output from the terminal 961.

Note that the circuit configuration from when current y is manually input until it is output is similar to the circuit for calculating XS or xe. Therefore.

Data at each output terminal is synchronized.

Next, let's talk about the operation of the filling range determining circuit.

The flowcharts shown in FIGS. 12, 13, 14, and 15, and the conceptual diagram shown in FIG. 8.

This will be explained in detail with reference to the circuit diagrams shown in FIGS. 17, 18, 19, and 20.

In step 31, if there is a new side, its starting point is corrected. This will be explained in detail with reference to the flowchart shown in FIG.

In step 51, it is determined whether the left side is a new side.

Here, a new side is a side whose starting point's y-coordinate is the current y-coordinate. In FIG. 8, 804 is the new left side. If it is a new edge, go to step 52,
If not, go to step 56<, In the circuit of FIG. 17, this judgment is the ripple carry output of the counter 247.

INITIAL signal or INIT performed by INITIAL signal and INIT LOAD signal
When the LOAD signal is input, since the first side is being processed, the left side is determined to be a new side. Also.

Immediately after the LOAD signal is input, a new edge is processed. However, if the ripple carry signal is not output, processing of one left edge has not yet been completed, so the LOAD signal is input.
AD signal, CLEAR signal.

It does not accept the RESET signal, so no new edges are processed.

In step 52, it is checked whether the slope of the left side is positive. If the slope is positive, go to step 56; otherwise, go to step 53. Since the slope of the side 804 is negative,
Go to step 53, in FIG. 17, by the output of the latch 221.

This is realized by inputting the CLR signal to the flip-flops 226 and 237. That is,
If the slope is positive, the sign bit of the slope, s 5lop
Since el is 0, the inverted output of the latch 2221 is 1. Therefore, flip-flops 226 and 237 are cleared and 0 is output.

Therefore, for 230 and 241 adders, err 1 and X
O is added to S. This implements the route from step 52 to step 56.

In step 53, it is checked whether errl is positive.

If errl is positive, go to step 54, otherwise go to step 55, step 54゜step 55
Then, as mentioned above, fix errl and xs
err 1 on side 804 is 8 in FIG.
As shown in 31, since it is positive, go to step 54.
Here, correct errl.

errl indicated by 832 is obtained. Moreover, xs is the X coordinate of the starting point of the side 804. (Step 54) Step 53. Step 54. A method for realizing the function of step 55 will be described later.

In step 56, it is determined whether the right side is a new side. If the right side is a new side, go to step 57, otherwise go to step 61<, In the circuit of FIG.
This judgment is based on the ripple carry output of the counter 297,
INITIAL signal.

INITI performed by INIT LOAD signal
When the AL signal or INIT LOAD signal is input.

Since this is the first side to be processed, the right side is determined to be a new side. Also, immediately after the LOAD signal is input, a new side is processed.However, if the ripple carry signal is not output, the processing of one right side has not yet been completed, so the LOAD signal, CLEAR signal , RESE
The T signal is not accepted, so no new edge processing is performed.

In step 57, it is checked whether the slope of the right side is positive. If the slope of the right side is positive, the process goes to step 58; otherwise, the process goes to step 59.

Since the slope of the right side indicated by 805 is positive, the process goes to step 58. In FIG. Realize. That is, when the slope is negative, the sign bit of the slope, s5loper, is 1, so 1 is output to the non-inverting output of the latch 271. Therefore, 276.287
The flip-flop is cleared and O is output. Therefore, in the 280°291 adder, errr and xe
O is added to. This implements the route from step 57 to step 61.

In step 58, it is checked whether errr is positive. er
If rr is positive, go to step 60; otherwise,
Go to step 59.

Step 59. In step 60, as described above, errr is corrected and xe is determined, as indicated by 821. Since errr on the right side of 805 is O, go to step 59 and add clr to em of 821 to find errr of 822. xe becomes the starting point of side 805. Step 58. Step 59. A method for implementing the function of step 60 will be described later.

In step 61, the Xs obtained in FIG.

The values of Xe determined in FIG. 18 and y determined in FIG. 20 are sent to the paper transfer device. Based on this value.

The paper transfer device has xs and xe at the X coordinate of 851.
Turn on the bits between. In Figure 8.

Turn on the bit between (X coordinate of 845, X coordinate of 851) and (X coordinate of 845, X coordinate of 851), that is, (X coordinate of 845, X coordinate of 851). The current X coordinate is.

The counter of 959 increases by 1.

Automatically increases by 1. In addition, here, Yend is 854
is the X coordinate of

In this way, once the new side has been corrected.

Go to step 32 in the flowchart of FIG.

In step 32, the current X coordinate (hereinafter referred to as

y) is smaller than Yend. If it is smaller, go to step 33, otherwise go to step 48
Since the X coordinate currently being processed is the X coordinate of 825, the process goes to step 33. This function is realized by the circuit shown in FIG. That is, the counter of 926 is
Count up to yena+t-current y,
By outputting a ripple carry, it can be seen that processing up to Yend has been completed.

Then, due to ripple carry output, CLOCK
stops, and the filling range determining circuit stops operating.

Then, by releasing HALT, CP
U starts working.

In step 33, it is checked whether the slope of the left side is positive. If it is positive, go to step 34; otherwise, go to step 35. Since the slope of the side 804, which is the left side currently being processed, is negative, go to step 35.

Step 34. Step 36. Step 37 shows the procedure of the bit selection method described above for the left side with a positive slope. Also, step 35° and step 38. Step 39 shows the procedure of the bit selection method described above for the left side with a negative slope.

In step 35, on the left side of 804.

When the value of errl indicated by 832 is checked, it is negative, so the process goes to step 39.

In step 39, errl of 832 is changed to errl of 833.
Then, xs, which was at the X coordinate of 845, is changed to the X coordinate of 844.

Step 33. Step 34. Step 35° Step 36. Step 37. Step 38. A method for implementing step 39 will be explained. Thinking about the difference in the judgment made in step 33.

If the slope of the side is positive or negative, the determination of the magnitude of errl and O is reversed.

In FIG. 17, the determination of the size of errl is fixed.

This is achieved by replacing the values added to errl or XS. cll, c21.

223°224 through latches 218 and 219, respectively.
is sent to the selector of The selectors 223, 224 and 225 send the data input to the A terminal to the output Y when the input to the SEL terminal is 0.

When the slope of the side is positive, s 5lopel is O, so
The selectors 223 and 224 output the input of the A terminal.

Therefore, the A input terminal of the selector 225 has all
However, c21 is sent to the input terminal of B.

The selector of 225 also outputs A when SEL is O, and outputs S.
Since B is output when EL is 1, if the slope is positive, e
If rrl is non-negative, cll is sent through flip-flops 226 to adder 230, and if errl is negative, c21 is sent to the adder as well. mll, m12
It works exactly the same way.

In step 40, it is checked whether the slope of the right side is positive. If it is positive, go to step 41; otherwise, go to step 42. Since the slope of the right side currently being processed, 805, is positive, go to step 41.

Step 41. Step 43. Step 44 shows the procedure of the bit selection method described above for the right side with a positive slope. Also, step 42° and step 45. Step 46 shows the procedure of the bit selection method described above for the right side with a negative slope.

Step 41 is indicated by 822. When the value of errr on the right side of 805 is checked, it is a dog than 0, so the process goes to step 43.

In step 43, 822 errr is changed to 823 errr.
Then, xe, which was at the X coordinate of 845, is changed to the X coordinate of 846.

Step 40. Step 41. Step 42゜Step 43. Step 44. Step 45. The method for implementing step 46 is exactly the same as the case on the left side.

Then, the values of XS, xe, and y are sent to the paper transfer device.
(Step 47) After that, return to step 32 and change the current y to Ye.
nd, i.e., 247. Alternatively, until the ripple carry of the counter 297 is output, step 3
Repeat steps 2 to 47.

Current y is no longer smaller than Yend.

When the ripple carry of the counter 247 or 297 is output, the process goes to step 48.For sides 804 and 805 in FIG. 8, when y becomes the X coordinate of 845, the process goes to step 48.

In step 48, the side on which processing is to be completed at the X coordinate,
In other words, the end point of the side whose end point is at the X coordinate is corrected.

This will be explained according to the flowchart in FIG. 15.

First, check whether processing on the left side has finished. (Step 7
1) If the processing on the left side has not been completed, go to step 72. In step 72, normal processing is performed. The normal processing here refers to the processing shown in steps 33 to 29 in the flowchart of FIG. The process is exactly the same. After performing this normal processing, go to step 77. If the line is <, on the left side of 804, the processing has not yet ended.

The normal process of step 72 is performed, and the value of the X coordinate of 842 is substituted for XS.

If it is determined in step 71 that the processing of the left side has been completed, the process proceeds to step 73 to check whether the slope of the left side is positive. If the slope of the left side is positive, the process goes to step 75; otherwise, the process goes to step 74.

In step 74, the X coordinate of the end point of the left side is substituted for XS.

In step 75, it is determined whether errl is positive. er
If rl is positive, go to step 74.

Otherwise, go to step 77.

Next, check whether the processing on the right side has finished. (Step 7
7) If the processing on the right side has not been completed, go to step 78. In step 78, normal processing is performed. The normal processing here refers to the processing shown in steps 40 to 46 in the flowchart of FIG. The process is exactly the same. After performing this normal processing, the process proceeds to step 83.

If it is determined in step 77 that the processing of the right side has been completed, the process proceeds to step 79 to check whether the slope of the right side is positive. If the slope of the right side is positive, go to step 80; otherwise, go to step 81.

In step 80, the X coordinate of the end point of the right side is substituted for xe.

In step 81, it is checked whether errr is positive. er
If rr is positive, go to step 80.

Otherwise, go to step 83.

On the right side of 805, the process ends, so go to step 79, and since the slope is positive, go to step 8o, and
Assign the X coordinate of 848 of the end point of side 805 to x6.

In step 83, XS, xe, and y are sent to the paper transfer device. In Figure 8, at the X coordinate of 854, from xs to x
e, that is, (X coordinate of 842, X coordinate of 854) and (
848 X coordinate, 854 X coordinate).

A method for realizing this operation will be explained with reference to FIGS. 17 and 18.

The end point of the left side is detected by the counter 247 as described above. When the end point of the left side is reached, the counter 247 outputs a ripple carry. This is.

Because it is input to the AND gate of 244 and 993.

Each gate allows nine signals other than pull-carry to pass through. At this time, the AND gate 244 outputs 1 when the slope of the left side is negative. This output is inverted by the NOR gate 994 to become O, and is input to the SEL terminal of the selector 242. When the SEL terminal becomes O, the selector 242 outputs the input of A. As a result, Xs END is output, and step 71.
The flow of steps 73 and 74 can be realized. Steve 71. Step 73. Step 75. Step 74
To realize the flow of 243 inverter, when the slope on the left side is positive, input 1 to 993 AND gate,
When errl is positive in the 995 inverter, 1 is input to the 993 AND gate. Then, by inverting the output of the AND gate 993 with the NOR gate 994 and inputting it to the SEL terminal of the selector 242,
Output end.

Detection of the end point of the right side is similar to detection of the end point of the left side.
When the right side reaches the end point performed by the counter 7, a ripple carry is output from the counter 297. This is 294
and 997 AND gates, each gate can allow two signals other than pull-carry to pass through.

At this time, the AND gate 294 receives 1 from the inverter 996 and outputs 1 when the slope of the right side is positive. This output is inverted by the NOR gate 998 to become O, and is input to the SEL terminal of the selector 292. When the SEL terminal becomes 0, the selector 292 outputs the input of A. As a result, Xe end is output, and steps 77, 79 . The flow of step 80 can be realized. Step 77, Step 79. Step 81. To implement the flow of step 80, when the slope of the right side is negative, 21 is input to the AND gate of 997, and when em is negative, 1 is input to the AND gate of 997.

Then, the output of the AND gate 997 is inverted by the NOR gate 998 and inputted to the SEL terminal of the selector 292, thereby outputting Xe end.

Processing up to Yend as described above.

The ripple carry of the counter 926 in FIG. 19 is output, and counting is stopped. And this ripple carry is 9
After 38 OR gates and 927 inverters, 929
CLOCK is stopped when applied from the flip-flop to the AND gate of 930. HALT is also released and the CPU resumes program execution. C
The PU executes step 29 in FIG. In step 29, the YendO value is set as the current y-coordinate value, and the process returns to step 21.

Steps 21 to 29 are then repeated until all edges have been processed.

After processing all edges, return to step 1 in FIG. 11 and wait for input.

[Effects of the Invention] As explained above, according to the present invention, all the bits through which the sides of the figure to be filled can be turned on, so that the second
The figure shown in the figure can also be filled in correctly, as shown in FIG. Furthermore, since complex repetitive processing is implemented as a circuit, processing speed is significantly increased. Also 2 circuits.

- It is easy to realize two circuits because it is composed of only basic circuit elements, and it is easy to implement LSI such as gate array.
The effects of the present invention are significant.

BRIEF DESCRIPTION OF THE DRAWINGS FIG. 1 is a functional block diagram showing the configuration of a printing apparatus according to the present invention. FIG. 2 is a conceptual diagram showing problems in the conventional example. FIG. 3 shows one example of filling in the present invention. A conceptual diagram showing the basic policy of bit selection method. FIG. 4 is a conceptual diagram showing data necessary for filling in the left side according to the present invention. FIG. 5 is a conceptual diagram showing how to obtain data necessary for filling in the left side according to the present invention. FIG. 6 is a conceptual diagram showing data necessary for filling in the right side according to the present invention. FIG. 7 is a conceptual diagram showing the difference in the bit selection method for filling according to the present invention depending on the slope of each of the left side and right side. FIG. 8 is a conceptual diagram showing how to fill in a polygon according to the present invention. FIG. 9 is a conceptual diagram showing that the problems of the conventional example shown in FIG. 2 are solved by the present invention. FIG. 10 is a system configuration diagram showing two embodiments of the present invention. FIG. 11 is a flowchart showing two embodiments of the printing apparatus of the present invention. FIG. 12 shows one embodiment of the printing apparatus of the present invention. Flowchart showing the overall flow of creating bit data. FIG. 13 shows two embodiments of the printing apparatus of the present invention. A flowchart showing the detailed flow of filling in and creating bit data. FIG. 14 is a flowchart showing the flow of correcting the coordinate points of a new side among the flowcharts showing the detailed flow of filling in and creating bit data in an embodiment of the printing apparatus of the present invention. FIG. 15 is a flowchart showing the flow of correcting the coordinate point of the side where the process ends, among the flowcharts showing the detailed flow of the portion of filling in and creating bit data in the second embodiment of the printing apparatus of the present invention. . FIG. 16 shows two embodiments of the printing apparatus of the present invention. FIG. 2 is a block diagram showing the configuration of a filling range determining circuit and connections with a CPU and a paper transfer device. FIG. 17 shows nine embodiments of the printing apparatus of the present invention. FIG. 3 is a circuit diagram showing an implementation example of a circuit that processes the left side of a filling range determining circuit. FIG. 18 shows two embodiments of the printing apparatus of the present invention. FIG. 3 is a circuit diagram illustrating an implementation example of a circuit that processes the right side of a filling range determining circuit. FIG. 19 shows two embodiments of the printing apparatus of the present invention. FIG. 3 is a circuit diagram showing an implementation example of a circuit that controls the operation of a filling range determining circuit. FIG. 20 shows nine embodiments of the printing apparatus of the present invention. FIG. 3 is a circuit diagram showing an implementation example of a circuit that calculates a y-coordinate value of a filling range determining circuit. 1 ... 2 ... 3 ... 4 ... 5 ... Data person. Output means Linear approximation section Fill data calculation section Bit data creation section Paper transfer device 21... Square 28...Example of filled bits 30.31, 32, 33, 34, 35, 36, 37
. . . Straight line 40 indicating an integer y-coordinate. . . Polygons to be filled 46.47, 48, 49, 50, 51.52...
Straight line showing the y-coordinate of an integer 301...302...303.307 304.309 305.310 306...308... Straight line showing the left side Example showing the bits selected as the left side... Center coordinates...Y axis...Example showing the bits selected as the right side of the X axis Straight line 401 showing the right side...Straight line 402 showing the left side...Pitch 403 selected as the left side...Enlarged view Circle indicating range 404...Y axis 405...X axis 410.411,412 421.422...431.432,433 441.442,443 451.452,453...Indicating err on the left side Arrow Bit boundary in the X-axis direction...Arrow indicating the y-coordinate of an integer...Bit selected as the left side...Example 501 showing the bit boundary in the y-axis direction...Slope indicating the vicinity of the starting point of the left side is a positive straight line 50
2...Distance from the bit boundary in the X-axis direction 503...
Distance in the X-axis direction between the intersection of the left side with a positive slope and the bit boundary in the y-axis direction and the start point of the left side with a positive slope 504... Initial value of err on the left side with a positive slope 505.
526... Legend 506.527 indicating one bit
...Bit center coordinates 507.524...Bit boundary in the X-axis direction 521...A straight line with a negative slope indicating the vicinity of the starting point of the left side 522...Distance from the bit boundary in the X-axis direction 523...・Distance in the X-axis direction between the intersection of the left side with a negative slope and the bit boundary in the y-axis direction and the starting point of the left side with a negative slope 525 ... Initial value of err on the left side with a negative slope 601 ・
... Straight line 602 indicating the right side...Example 60 indicating the bits selected as the right side
3...Circle 604 indicating the range of the enlarged view...Y axis 605...X axis 610, 611, 612...Arrow indicating err on the right side 621, 622...Bit boundary in the X axis direction 631 ．．
632, 633 ...Arrow 64 indicating the y-coordinate of an integer
1.642,643...Bits selected as the right side 651.652,653,654...Bit boundary in the y-axis direction 801...X-axis 802...y-axis 803...Coordinate axis origin 804. 805, 806, 807, 808, 809
...Polygon 810 to be filled represented by sides...Center coordinates of bits 821, 822, 823 er of each side of 805
r831.832,833,834 err841.842,843,844,845 for each side of 804
, 846, 847, 848, 849... Integer coordinates of the y-axis 851.852, 853, 854, 855, 856, 8
57,858,859,860...integer coordinates on the y-axis 901,902,903,904...A problem occurs with the conventional method shown in FIG. Polygon to be filled 910... 911... 912... 913... 914... Filling range determination circuit CPU errl calculation circuit xs calculation circuit errr calculation circuit 915... xe calculation circuit 916... paper Transfer device 201-ss1ope1 input terminal 202...
・ errl input terminal 203.253-cl e a r input terminal 204
...all input terminal 205...c21 input terminal 206,256,953...CLOCK input terminal 207...Xs END input terminal 208...m11 input terminal 209...m12 input terminal 210... Xs input terminal 211...END Yl input terminal 212.262... INIT LOAD input terminal 213.263,921,952...LOAD
Input terminal 214.264, 924, 954...R
ESET input terminal 215...Xs output terminal 216.267...INITIAL input terminal 251
... 252 ... 254 ... 255 ... 257 ... 258 ... 259 ... 260 ... 261 ... 266 ... 922 ... 923 ... 931 ...・ 936 ... 933 ... 951 ... 961 ... s 5loper input terminal errr input terminal clr input terminal c2r input terminal Xe END input terminal mr1 input terminal mr2 input terminal Xe input terminal END Yr input terminal Xe output terminal Yend+1-current y input terminal MASTE
R-CLOCK input terminal CLOCK output terminal HALT output terminal c, l e a r output terminal Current-y input terminal y output terminal 221.222, 227, 218.219.220, 2
32,234,238.271.277. 272.269,270,293,282,284,2
88,925,937,955...Latch 229.226,231,237,240,245,2
46,279,276.281.287゜290.29
6,295,929,932,956,960,958
...Flip-flop 223.224, 225.233.235.236, 2
42,273.274,275.283. 285.286,292...Selector 230.2
41,280,291,957...Adder 247
．． 297,926,959...Counter 228.
239, 217, 248, 278, 289, 268, 2
98,938...OR gate 243.994,996,927,928...Inverter 244.249,250,265.294,2
99,991.992,993.997,930°93
4...AND gate 994.998...Noah gate 111...Microcomputer 112...CPU 113...ROM 114...Work memory 116...117...O Paper transfer device filling range determination More than a circuit

Claims (1)

[Scope of Claims] a) A printing device that handles characters and figures as xy coordinate data, which are integers representing outlines, and prints them on paper; b) A data input device that inputs and outputs the data of the characters and figures. output means; c) a linear approximation unit that approximates curve data included in the integer xy coordinate data representing the contour of the figure or character to the integer xy coordinate data of a straight line; d) The linear approximation unit uses the integer xy coordinate data of the straight line that approximates the curve data and the integer xy coordinate data of the straight line as input, and converts the straight line into a discrete image of the printing device. When making the bits that the straight line passes correspond to the bits that the straight line passes through, calculate the data necessary to determine the range of filling in the figure or character that is made up of the straight line. e) a fill range determination circuit that determines a range to be filled using the output of the fill data calculation unit; and f) a bitmap digital output according to the output of the fill range determination circuit. A printing device characterized by having a paper transfer device that creates data and transfers it onto paper.