09-04-88/T\OSB21

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09-04-88/T\OSB21

This website contains an archive of files for the Acorn Electron, BBC Micro, Acorn Archimedes, Commodore 16 and Commodore 64 computers, which Dominic Ford has rescued from his private collection of floppy disks and cassettes.

Some of these files were originally commercial releases in the 1980s and 1990s, but they are now widely available online. I assume that copyright over them is no longer being asserted. If you own the copyright and would like files to be removed, please contact me.

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OSBITS - An Exploration of the BBC Micro at Machine Level

By Programmer

..........................................................


Part 21: Real Numbers Explore the Mandelbrot Set


In the last module I introduced the idea of splitting up a
word (in that case it was 4 bytes as used by BASIC integers)
such that the lower few bytes were in fact representing
fractional parts of 1.  This module exploits that idea in
order to perform some pretty hairy computation in order to
plot a crude diagram showing the Mandelbrot Set.

The Mandelbrot Set illustrates the performance of the
function z=z+c where both z and c are complex numbers.  This
means they have a real part and an imaginary part and the
imaginary part is a multiple (or fraction) of the square
root of -1.  Mathematicians call that quantity i while
engineers usually call it j.  (i to an electrical engineer
denotes current, as in Ohm's Law v=ir.)

So essentially we take z=z+c, starting with z=0 and keep
putting it into the equation.  Obviously the first time
around z will equal c.  If we keep doing this so called
iterative process and check on the size of z as we do it we
can see how the function behaves.

In order to draw the Mandelbrot Set we simply carry out this
iteration using screen co-ordinates to represent values of
c.  Since c has these real and imaginary parts we can plot
real values along the x axis and imaginary ones up the y
axis.  The colour of the pixel plotted at that point is used
to tell us how the size of the function behaves.

We are interested in whether or not the size of the function
increases without limit.  For our purposes 'without limit'
means only that it reaches 2 because if it reaches 2 it
will, I am reliably informed, reach infinity!  Different
colour values tell us how many iterations it took for the
function to reach 2 in size.  If after a certain number of
iterations it has not yet reached 2 then the pixel is
coloured black and that value of c is part of the Mandelbrot
Set.  This is all named after a Belgian mathematician named
Benoit B Mandelbrot who discovered this particular behaviour
of complex numbers.

Mandelbrot also discovered fractals and the boundary of the
black area in the middle of our eventual diagram is a
fractal.  If we blow it up ever larger we reveal more and
more detail.  Now you can imagine that since many iterations
are taking place at each pixel on the screen the whole
process can take a long time.  I have drawn blow-ups of
areas of the boundary that took over 48 hours to plot on a
BBC micro in BASIC.  (I've also done it in FORTRAN on a
32016 second processor in an hour but that's down to the
power of the machinery and of FORTRAN rather than any fancy
coding of mine.)

The algorithm is like this:

Map out values of c covering the range -2.25-1.5i to
0.75+1.5i across the screen area.  This means that the
graphics origin (0,0) represents -2.25-1.5i and (1024,1024)
represents 0.75+1.5i and each pixel on the screen would
represent a value of c.

Work your way over the screen and for each pixel:

Set a counter at zero, set z equal to zero (0+0i)

Loop start

Calculate size of z, if > 2 then colour the pixel according
to the value in the counter and move on to next pixel

If the counter has reached a preset value colour the pixel
black and move on to the next pixel

If neither condition is met then recalculate z=z+c using the
previous value of z and go back to the start of the loop

In practice the tendency is to calculate the new value of
the size squared, which is the sum of the real part squared
and the imaginary part squared.  There is no point in taking
a spurious square root.  So the size variable being used is
the square and it is therefore compared with 4 rather than
2.  Also the actual order of carrying out the tests will
vary from the straight algorithm.  In the module program,
for example, the counter is decremented at the end of the
loop rather than the beginning so the test is carried out
there.  Also the size is tested as soon after it is
calculated as feasible bearing in mind the length of the
branch needed after the test.  Because of this branch length
problem the code here actually tests in a rather odd place,
during the calculation of the new value of y.

The Mandelbrot set calculation is suited to limited
fixed-point arithmetic because it works within a limited
range of numbers and because, being a graphic, it is perhaps
more tolerant of poor accuracy in the calculations.  I will
admit this in advance, my fixed-point routine is not
amazingly accurate, but your trade-off is speed against
precision.  So, before we look at the code I shall explain
the fixed-point method used here.

If we take a four byte word and put the 'binary' point in
the middle we then have a fixed point number.  If you look
at the algorithms for arithmetic I listed in the last module
you will find that a constant factor was involved in
changing the numbers into the fixed point representation. 
It was 10000 last time but in the program in this module I
will use 65536.  This puts the binary point exactly in the
middle of the number so that the top two bytes are the
integer part and the lower two bytes are the fractional
part.  In this way we can represent numbers up to 65536 and
as small as 1.526E-5 (which is 0.00001526).

In order to carry out the necessary multiplications I have
modified the multi-byte multiplication routine used in
module 13 to produce a five byte result instead of four. 
This is because we have to divide the result of a fixed
point multiplication by the relevant factor.  It looks like
this:

Here is a six byte number. The binary point is represented
by a plus sign.

  ------------------------------------+------------------
  | byte 5 | byte 4 | byte 3 | byte 2 | byte 1 | byte 0 |
  ------------------------------------+------------------

If we multiply a four byte number x

  ------------------------------------+------------------
  |        |        | xxxxxx | xxxxxx | xxxxxx | xxxxxx |
  ------------------------------------+------------------

by a four byte number y

  ------------------------------------+------------------
  |        |        | yyyyyy | yyyyyy | yyyyyy | yyyyyy |
  ------------------------------------+------------------

we can get a eight byte number xy, part of which is here.

  ------------------------------------+------------------
  | xyxyxy | xyxyxy | xyxyxy | xyxyxy | xyxyxy | xyxyxy |
  ------------------------------------+------------------

To divide by 65536 we simply discard bytes 0 and 1 and read
from byte 2 to byte 5.

  ------------------+------------------
  | byte 5 | byte 4 | byte 3 | byte 2 |
  ------------------+------------------

And this essentially is how the multiplication routine
works.  Now I know from experience that we will not be
multiplying any numbers to give products greater than 255 so
we can use a 5 byte representation.  This results in one
byte above the binary point in the result, like this.

           ---------+------------------
           | byte 4 | byte 3 | byte 2 |
           ---------+------------------

In order to cope with 2's complement negatives you have to
also pad out the higher bytes in the numbers to be
multiplied and in the result with &FF if necessary.  That is
why you see checks for negative and &FFs being put into 5th
bytes in the fixed point multiplication routine.

The routine lies between lines 3780 and 4580 in the module
program.

And so to the code itself.

You will find that this code is almost a straight line, with
only two subroutines, the Mandelbrot routine itself and the
multiplying routine.  This was done partly for speed, but
also partly because by its nature this is a linear process
rather than a multi-branched one, and so subroutines might
not have clarified much.

At the start of the assembly I test to see if your program
is running in a second processor or in the BBC Micro itself,
the I/O processor.  This is because the program uses Mode 2
graphics and will not fit in memory easily.  With a second
processor that is no problem but in the micro itself you
must be in Mode 7 to assemble and must clear the BASIC
variable workspace (with a NEW) before you can run the final
code.  The first few lines of the program take care of that
and will tell you what is going on.  The code is executed by
pressing function key 0: function key 1 is available to list
the program in Mode 3.  In the micro the code is assembled
at &1300, in a 'safe' part of the disc workspace.  In a
second processor it can just be put alongside the
BASIC/Assembler using DIM.  If you have a BBC Master without
a second processor you will have to change this part of the
code to force the use of DIM because the value of PAGE is
&E00.  [You can also use shadow memory by using Mode 130
instead of Mode 2.]

We set the mode as 2 and this gives us increments of 8
graphics units per pixel on the x axis and 4 on the y.  If
you want to change to Mode 0 or Mode 1 you have to modify
these variables accordingly.  The horizontal increments will
be 2 and 4 for Modes 0 and 1 respectively.

For speed I am using zero page workspace for the
multiplying routines, and that is set next.

In the assembler itself Mode 2 is set up followed by a VDU5
equivalent (both done using OSWRCH) so that the text cursor
is removed.  I know that there is a VDU23 instruction to
switch off the cursor, but VDU5 is quicker and easier.

The next section is a plotting loop that uses two words of
memory labelled 'x_coord' and 'y_coord' to set the position
of the pixel being plotted currently.  The Mandelbrot
subroutine uses a GCOL equivalent to set the colour before
the pixel is plotted using a PLOT 69 equivalent (plot a
point).  For convenience I am using a 1024 units square area
of the screen.

Since the numerical area covered is 3 units along both axes
the increment per graphics unit is 3/1024.  Since our fixed
point representation effectively multiplies that by 65536
(which is 256*256) the increment in fixed point is in fact
&C0 in both directions.  This figure is put into memory at
labels h_inc and v_inc during the assembly.

I have in fact eased the difficulty of this program greatly
by assuming it will only be plotting the whole set.  You
could modify it to plot a smaller section at the risk of
greater inacuracy with very small numbers but you will have
to change the increments and the plot starting points
manually, as it were, to do this.  Simply, to find the four
byte fixed point equivalent of a number, just multiply it by
65536.

The complex numbers are broken down into real and imaginary
parts.  Their labels here are x and y for the components of
z (since z is usually written x+iy) and c_real and c_imag
for c.  These are the calculations we must make:

    size = x^2 + y^2  (remember size is itself a square)

    New value of x = x^2 - y^2 + c_real

    New value of y = 2 * x * y + c_imag

From line 1050 the code computes the value of c_real, based
upon the current position on the screen.  By using the X
register counting down from 4 and offset against 'h_inc-1'
we can use a BNE to test the loop during the transfer of
bytes into multiplication workspace.  You will see that most
of the transfers are done this way.

In line 1310 -2.25 is added in the calculation of c_real. 
-2.25 is the starting point of that axis.  In our fixed
point representation the number to be added is -2.25 * 65536
and that (in 2's complement negative) is &FFFDC000.

From line 1420 c_imag is calculated and you will note that
in line 1680 -1.5 is added in, this is &FFFE8000 in our
fixed point.

Just prior to the start of the main iteration loop the
counter is initialised (we count down as usual in order to
use a BNE/BEQ trap at the end of the loop) and x and y are
initialised at zero, because z is zero as we start
iterating.

x squared and y squared are calculated simply by multiplying
each with itself.  The size is then calculated by adding the
two results.  Ideally we should test the size now, but there
is too much code between here and the nearest point to put
the other end of the branch.  Remember a branch can only be
+127 or -128 bytes.  We will find the test shortly.

In the calculations for the new x and y you will notice that
while x needs x^2 and y^2, y needs x and y, so we must
calculate the new y before we calculate the new x.  From
line 2540 the code calculates the new y.  In the middle of
this there is the earliest spot for the size test, which is
why it appears at line 2850.  The reason for putting the
test as early as possible is that it will trap on size over
most of the screen so by trapping here, rather than at the
end of the loop, we save executing code unnecessarily, which
saves time albeit only a small amount.

We finish calculating the new y and then calculate the new
x ready for the next iteration.

The counter is decremented for the next iteration, but if
that will take us past our iterations limit, in this case
32, we exit instead, using a BEQ in line 3410.

At 'size_out' is the code that uses GCOL (VDU 18) to set the
pixel colour.  Rather than waste time and effort on a
division I have simply masked off all but the last three
bits of the size, which gives values from 0 to 7.  If the
iterations limit is exceeded then we reach 'count_out' and
set the colour to black (GCOL 0,0).

After some variable and workspace setting we finally come to
the fixed point multiplication code.  You will notice that
we test for a negative number and if we find one we make it
up to five bytes before the multiplication, since we want a
good five byte result.  The same applies to the result.  The
loop is executed 40 times because 5 bytes is 40 bits.

Plotting the Mandelbrot set on a 3MHz 6502 second processor
from BASIC using floating point takes me over 45 minutes,
this routine using fixed point machine code arithmetic takes
only 22 minutes.  If I could speed up the multiplications I
could probably get it even faster.  Fair takes your breath
away dunnit?

The alternative to fixed point, a little slower but more
precise, is floating point.  That's for next time.

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00000010  6f 72 61 74 69 6f 6e 20  6f 66 20 74 68 65 20 42  |oration of the B|
00000020  42 43 20 4d 69 63 72 6f  20 61 74 20 4d 61 63 68  |BC Micro at Mach|
00000030  69 6e 65 20 4c 65 76 65  6c 0d 0d 42 79 20 50 72  |ine Level..By Pr|
00000040  6f 67 72 61 6d 6d 65 72  0d 0d 2e 2e 2e 2e 2e 2e  |ogrammer........|
00000050  2e 2e 2e 2e 2e 2e 2e 2e  2e 2e 2e 2e 2e 2e 2e 2e  |................|
*
00000080  2e 2e 2e 2e 0d 0d 0d 50  61 72 74 20 32 31 3a 20  |.......Part 21: |
00000090  52 65 61 6c 20 4e 75 6d  62 65 72 73 20 45 78 70  |Real Numbers Exp|
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000000b0  72 6f 74 20 53 65 74 0d  0d 0d 49 6e 20 74 68 65  |rot Set...In the|
000000c0  20 6c 61 73 74 20 6d 6f  64 75 6c 65 20 49 20 69  | last module I i|
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00000740  20 63 65 72 74 61 69 6e  20 6e 75 6d 62 65 72 20  | certain number |
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000032c0  20 69 6e 73 74 65 61 64  2c 20 75 73 69 6e 67 20  | instead, using |
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00003330  20 52 61 74 68 65 72 20  74 68 61 6e 20 77 61 73  | Rather than was|
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00003380  74 20 74 68 65 20 6c 61  73 74 20 74 68 72 65 65  |t the last three|
00003390  0d 62 69 74 73 20 6f 66  20 74 68 65 20 73 69 7a  |.bits of the siz|
000033a0  65 2c 20 77 68 69 63 68  20 67 69 76 65 73 20 76  |e, which gives v|
000033b0  61 6c 75 65 73 20 66 72  6f 6d 20 30 20 74 6f 20  |alues from 0 to |
000033c0  37 2e 20 20 49 66 20 74  68 65 0d 69 74 65 72 61  |7.  If the.itera|
000033d0  74 69 6f 6e 73 20 6c 69  6d 69 74 20 69 73 20 65  |tions limit is e|
000033e0  78 63 65 65 64 65 64 20  74 68 65 6e 20 77 65 20  |xceeded then we |
000033f0  72 65 61 63 68 20 27 63  6f 75 6e 74 5f 6f 75 74  |reach 'count_out|
00003400  27 20 61 6e 64 0d 73 65  74 20 74 68 65 20 63 6f  |' and.set the co|
00003410  6c 6f 75 72 20 74 6f 20  62 6c 61 63 6b 20 28 47  |lour to black (G|
00003420  43 4f 4c 20 30 2c 30 29  2e 0d 0d 41 66 74 65 72  |COL 0,0)...After|
00003430  20 73 6f 6d 65 20 76 61  72 69 61 62 6c 65 20 61  | some variable a|
00003440  6e 64 20 77 6f 72 6b 73  70 61 63 65 20 73 65 74  |nd workspace set|
00003450  74 69 6e 67 20 77 65 20  66 69 6e 61 6c 6c 79 20  |ting we finally |
00003460  63 6f 6d 65 20 74 6f 0d  74 68 65 20 66 69 78 65  |come to.the fixe|
00003470  64 20 70 6f 69 6e 74 20  6d 75 6c 74 69 70 6c 69  |d point multipli|
00003480  63 61 74 69 6f 6e 20 63  6f 64 65 2e 20 20 59 6f  |cation code.  Yo|
00003490  75 20 77 69 6c 6c 20 6e  6f 74 69 63 65 20 74 68  |u will notice th|
000034a0  61 74 0d 77 65 20 74 65  73 74 20 66 6f 72 20 61  |at.we test for a|
000034b0  20 6e 65 67 61 74 69 76  65 20 6e 75 6d 62 65 72  | negative number|
000034c0  20 61 6e 64 20 69 66 20  77 65 20 66 69 6e 64 20  | and if we find |
000034d0  6f 6e 65 20 77 65 20 6d  61 6b 65 20 69 74 0d 75  |one we make it.u|
000034e0  70 20 74 6f 20 66 69 76  65 20 62 79 74 65 73 20  |p to five bytes |
000034f0  62 65 66 6f 72 65 20 74  68 65 20 6d 75 6c 74 69  |before the multi|
00003500  70 6c 69 63 61 74 69 6f  6e 2c 20 73 69 6e 63 65  |plication, since|
00003510  20 77 65 20 77 61 6e 74  20 61 0d 67 6f 6f 64 20  | we want a.good |
00003520  66 69 76 65 20 62 79 74  65 20 72 65 73 75 6c 74  |five byte result|
00003530  2e 20 20 54 68 65 20 73  61 6d 65 20 61 70 70 6c  |.  The same appl|
00003540  69 65 73 20 74 6f 20 74  68 65 20 72 65 73 75 6c  |ies to the resul|
00003550  74 2e 20 20 54 68 65 0d  6c 6f 6f 70 20 69 73 20  |t.  The.loop is |
00003560  65 78 65 63 75 74 65 64  20 34 30 20 74 69 6d 65  |executed 40 time|
00003570  73 20 62 65 63 61 75 73  65 20 35 20 62 79 74 65  |s because 5 byte|
00003580  73 20 69 73 20 34 30 20  62 69 74 73 2e 0d 0d 50  |s is 40 bits...P|
00003590  6c 6f 74 74 69 6e 67 20  74 68 65 20 4d 61 6e 64  |lotting the Mand|
000035a0  65 6c 62 72 6f 74 20 73  65 74 20 6f 6e 20 61 20  |elbrot set on a |
000035b0  33 4d 48 7a 20 36 35 30  32 20 73 65 63 6f 6e 64  |3MHz 6502 second|
000035c0  20 70 72 6f 63 65 73 73  6f 72 0d 66 72 6f 6d 20  | processor.from |
000035d0  42 41 53 49 43 20 75 73  69 6e 67 20 66 6c 6f 61  |BASIC using floa|
000035e0  74 69 6e 67 20 70 6f 69  6e 74 20 74 61 6b 65 73  |ting point takes|
000035f0  20 6d 65 20 6f 76 65 72  20 34 35 20 6d 69 6e 75  | me over 45 minu|
00003600  74 65 73 2c 0d 74 68 69  73 20 72 6f 75 74 69 6e  |tes,.this routin|
00003610  65 20 75 73 69 6e 67 20  66 69 78 65 64 20 70 6f  |e using fixed po|
00003620  69 6e 74 20 6d 61 63 68  69 6e 65 20 63 6f 64 65  |int machine code|
00003630  20 61 72 69 74 68 6d 65  74 69 63 20 74 61 6b 65  | arithmetic take|
00003640  73 0d 6f 6e 6c 79 20 32  32 20 6d 69 6e 75 74 65  |s.only 22 minute|
00003650  73 2e 20 20 49 66 20 49  20 63 6f 75 6c 64 20 73  |s.  If I could s|
00003660  70 65 65 64 20 75 70 20  74 68 65 20 6d 75 6c 74  |peed up the mult|
00003670  69 70 6c 69 63 61 74 69  6f 6e 73 20 49 0d 63 6f  |iplications I.co|
00003680  75 6c 64 20 70 72 6f 62  61 62 6c 79 20 67 65 74  |uld probably get|
00003690  20 69 74 20 65 76 65 6e  20 66 61 73 74 65 72 2e  | it even faster.|
000036a0  20 20 46 61 69 72 20 74  61 6b 65 73 20 79 6f 75  |  Fair takes you|
000036b0  72 20 62 72 65 61 74 68  0d 61 77 61 79 20 64 75  |r breath.away du|
000036c0  6e 6e 69 74 3f 0d 0d 54  68 65 20 61 6c 74 65 72  |nnit?..The alter|
000036d0  6e 61 74 69 76 65 20 74  6f 20 66 69 78 65 64 20  |native to fixed |
000036e0  70 6f 69 6e 74 2c 20 61  20 6c 69 74 74 6c 65 20  |point, a little |
000036f0  73 6c 6f 77 65 72 20 62  75 74 20 6d 6f 72 65 0d  |slower but more.|
00003700  70 72 65 63 69 73 65 2c  20 69 73 20 66 6c 6f 61  |precise, is floa|
00003710  74 69 6e 67 20 70 6f 69  6e 74 2e 20 20 54 68 61  |ting point.  Tha|
00003720  74 27 73 20 66 6f 72 20  6e 65 78 74 20 74 69 6d  |t's for next tim|
00003730  65 2e 0d                                          |e..|
00003733

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