I'd like to play pong on my computer and I plan to use an old scope in the X/Y mode for the screen. So I need a digital-to-analog converter. Naturally, resistor ladders come into mind, however R-2R would not work directly.

Let us try to find resistor values. Current intensities are tied with the voltages in the following way:

You can try the following code snippet in the sage calculator:

var("R,R1,R2,R3,R4,R5,R6,V1,V2,V3")
A=matrix([[0,0,1,-1,0,0],[0,1,0,1,-1,0],[1,0,0,0,1,-1],[0,0,R3,R4,R5,R6],[0,R2,0,0,R5,R6],[R1,0,0,0,0,R6]])
b=matrix([[0],[0],[0],[V3],[V2],[V1]])
I=(A.inverse()*b).simplify_full()

Vo=(I[5][0]*R6+I[4][0]*R5+I[3][0]*R4).simplify_full()
Vo=Vo.substitute(R1==R,R2==R,R3==R)

eq13=(26/27==Vo.substitute(V1= 1,V2= 1,V3= 1))
eq12=(24/27==Vo.substitute(V1= 0,V2= 1,V3= 1))
eq11=(22/27==Vo.substitute(V1=-1,V2= 1,V3= 1))
eq10=(20/27==Vo.substitute(V1= 1,V2= 0,V3= 1))
eq09=(18/27==Vo.substitute(V1= 0,V2= 0,V3= 1))
eq08=(16/27==Vo.substitute(V1=-1,V2= 0,V3= 1))
eq07=(14/27==Vo.substitute(V1= 1,V2=-1,V3= 1))
eq06=(12/27==Vo.substitute(V1= 0,V2=-1,V3= 1))
eq05=(10/27==Vo.substitute(V1=-1,V2=-1,V3= 1))
eq04=( 8/27==Vo.substitute(V1= 1,V2= 1,V3= 0))
eq03=( 6/27==Vo.substitute(V1= 0,V2= 1,V3= 0))
eq02=( 4/27==Vo.substitute(V1=-1,V2= 1,V3= 0))
eq01=( 2/27==Vo.substitute(V1= 1,V2= 0,V3= 0))

sln=solve([eq01,eq02,eq03,eq04,eq05,eq06,eq07,eq08,eq09,eq10,eq11,eq12,eq13],R4,R5,R6)
show(sln)

Turns out that it is not R-2R. It is R-2R-4/3R! Let us plug my 3-trit counter to the resistor ladder:

The sawtooth wave clocks the counter, and we have a nice staircase output. With few spikes, but those are easy to filter out.