Counting bars in Code 128

I was trying to decode a generated barcode by hand and it was just too hard. So why not use my favorite new language, darkbasic classic.

I load the bitmap and then just display the bar widths and manually look it up in the table at https://en.wikipedia.org/wiki/Code_128.

Quick and dirty but so much easier than trying to count pixels by hand...

`load bitmap "code128_gif_saved_to_png.png"
load bitmap "code128.bmp"

ink rgb(255,0,0),0

ink rgb(255,0,0),0
xcord = 0
ycord = 2

`quick test of point and getr,getg,getb routines
set cursor 0,100
black=1
for i = 1 to 10
print point(xcord,ycord)," ",getr(point(xcord,ycord))," ",getg(point(xcord,ycord))," ",getb(point(xcord,ycord))
inc xcord
next i

xcord = 0
ycord = 2
unitthickness=2
WHILE getr(point(xcord,ycord))>128
  inc xcord
endwhile

do
currentcolor=getr(point(xcord,ycord))
thick = 0
while currentcolor = getr(point(xcord,ycord))
line xcord,ycord+90,xcord,ycord+100
   inc thick
`   if xcord>165 then print "xcord=",xcord," getr=",getr(point(xcord,ycord))
   if thick > 5 then exit
   if xcord > 320 then exit
`this exit only exits the inner while
  for i = 1 to unitthickness: inc xcord:  next i
endwhile
`print "xcord=",xcord," thick=",thick," color=",currentcolor
inc group
printc thick
if mod(group,6)=0 then printc " ":printc " >"
currentcolor = getr(point(xcord,ycord))
   if thick > 20 then print "TOO THICK":exit
   if xcord > 320 then print "XCORD TOO BIG":exit
loop

`more testing of our getr,getg,getb routines

ink rgb(255,128,2),0
dot xcord,ycord

print getr(point(xcord,ycord))
print getg(point(xcord,ycord))
print getb(point(xcord,ycord))


end

`mod is seriously messy because of conversions...
`0.166666*6 gives you less than one,
`int conversion truncates
function mod(x,y)
x# = x
y# = y
frac# = (x#/y#) - int(x#/y#)
`print frac#,modval
modval = frac# * y#
modval# = frac# * y#
modval = modval#+0.00001
`print "frac=",frac#,"y=",y#,"modval=",modval#,modval,int(frac#*y#)
endfunction modval

function getbyte(x,bytenum)
retval=(x & (255*(256^bytenum)))/(256^bytenum)
endfunction retval

function getb(x)
endfunction getbyte(x,0)

function getg(x)
endfunction getbyte(x,1)

function getr(x)
endfunction getbyte(x,2)

So why only 107 symbols in Code 128?

Sometimes you just wonder why things have a certain number of possibilities. Like code 128 barcodes. Why does it have a certain # of symbols?

From https://en.wikipedia.org/wiki/Code_128

It's got a constant 11 bit width with 3 bars and 3 spaces. Each bar can be from 1 to 4 units thick.

So you have 6 bars and they can have 4 possibilities each.

You'd think it would be 4 * 4 * 4 * 4 * 4 = 4096 possibilities, but the number of units all has to add up to 11.

Let's write some javascript to do some counting for us:

<script>
count=0;

for (a1=1;a1<5;a1++)
for (a2=1;a2<5;a2++)
for (a3=1;a3<5;a3++)
for (a4=1;a4<5;a4++)
for (a5=1;a5<5;a5++)
for (a6=1;a6<5;a6++)
if ((a1+a2+a3+a4+a5+a6)==11) 
{count++; 
 mykey=a1.toString()+a2.toString()+a3.toString()+a4.toString()+a5.toString()+a6.toString();
 document.write(mykey+" count="+count+"<BR>");
}
document.write("count="+count+"<BR>");
</script>

and it dutifully gives us 216 possibilities.

If we remove half of those 216 taking away the reverse bit patterns, we get 108 which is pretty close to the 107 symbols that you can have in code 128.