3 resistors?  I've tried to fit that, 4 is the only number I come up
with:  ## is an anti-parallel pair of LED's, below, and \/\/ a resistor;

(0)--\/\/--+-----+-----+
           #     |     |
           #     |     |
(1)--\/\/--+--+--(--+  |
              #  #  #  #
              #  #  #  #
(2)--\/\/--+--+--+  |  |
           #        |  |
           #        |  |
(3)--\/\/--+--------+--+

I've tried to reduce the number of resistors - no fewer resistors would
work, IMO.  If you omit the resistor in lead (0) above, then 3 of the
pairs of LEDs would run at increased brightness - the LEDs connected
between pins 1 and 2, those connected from pins 1 to 3, and those from 2
to 3, would run at the same brightness, though.

Of course, you could use a pair of 2.5V diodes in series for each diode,
or just using short, high current pulses for your "on" time, let the PIC
pin current limit the LED's current - which are quite do-able (test your
code with resistors, tho <G>)

  Mark

Craig Lee wrote:
>
> I'll CAD the schematic later, but my technique will give you
> 12 LEDs with 4 PIC I/O and 3 resistors.  Z denotes I/O set to
> input.
>
> The scheme demands that only one LED be on at one time.  However,
> with a little slight of micro, you can easily make more than one
> appear on at once.
>
> I worked out a formula for muxing more LEDs by adding more I/O,
> but didn't write it down. Argghh!
>
> Heres the truth table for 12:
>
> ABCD  LED ON
> ----  ------
> 0000    none
> 1111  none
> zzzz  none
> 01zz   1
> 10zz   2
> 0z1z   3
> 1z0z   4
> z01z   5
> z10z   6
> 0zz1   7
> 1zz0   8
> zz01   9
> zz10  10
> z0z1  11
> z1z0  12
>
> Craig

--
I re-ship for small US & overseas businesses, world-wide.
(For private individuals at cost; ask.)