🚀 go-pugleaf

In article <1992Nov19.102634.20745@vpnet.chi.il.us> cgordon@vpnet.chi.il.us (gordon hlavenka) writes:

>
>A puzzle?  We ran this little ditty on almost a daily basis.
>0000 ADDR LD
>DCA I 0010
>JMP 0000
>0000 ADDR LD CLEAR CONT
>
>2 instructions.  Actually, we'd run it with the data field set to 1,
>then 2, and finally zero.  The program autoincrements itself right
>through the field, finally (in DF0) overwriting the DCA with a zero.
>(AND 0000)  Then the loop was a lot tighter -- you could see the
>difference in the panel lights.  You do have to make sure that
>location 10 has a 10 in it when you start, though...

When the DF is 10-70, it doesn't much matter what's in 00010, since you just
let it run more than enough times to clobber everything many times over.

With the DF 00, when the DCA I 10 is destroyed, the JMP isn't destroyed.
However, since the switches are zero from the last load address function, the
switches are thus conveniently set to 0000.  Hitting deposit will put a 0000
into the latest location, which since the running program is now:

	*0000

	AND	0
	JMP	0

means that the PC is either 00000 or 00001, so worst case you can either
deposit a 0000 over 00000 and 00001 if you hit deposit twice, or a 0000 over
00001 and 00002 if you hit deposit twice.  This can be done without looking at
the lights if you're in a hurry.  An alternative is to do ADDR LOAD then
DEP DEP but that might take most people a few extra milliseconds.

Here's one for the fans of the (nearly) cleared-out memory PDP-8:

Assume that all of field 0 is all 0000 as the above might accomplish.  ADDR
LOAD 0000, change the switches to 1777, press DEP, CONT.  Describe what the
AC lights are doing, and why.

(On a difficulty scale of 1-10, this problem is rated 2.)

cjl