A very photogenic couple

[Dave Britten]

Here's my Tower of Hanoi solver that I adapted from an fx-7000G version I wrote. Instead of a graphical display, it uses this output format:

FT.XYZ

Where F is the stack to move from, T is the stack to move to, and X, Y, and Z indicate the total number of disks on the left, middle, and right stacks respectively.

This uses extensive register packing to store all three stacks in just three variables, as well as a small array to store pointers to the tops of each stack. (Note that there are some uses of variable O; be careful not to mistake this for zero.)

Code:

==Prog 0==
Fix 3
Mcl
"DISCS":?→N
2Frac (N÷2→O
2^N-1→M
9→P~R
N→U
9→V
18→W
U→I
Lbl 0
N-I+1→D
Prg 2
Dsz I
Goto 0
1→B
Prg 3
Lbl 1
B→F
B+1+O→G
G>3⇒G-3→G
B+2-O→B
B>3⇒B-3→B
T[F→I
Prg 1
D→C
T[G→I
Prg 1
C<D⇒Goto 2
D→C
F→S
G→F
S→G
Lbl 2
0→D
T[F→I
Prg 2
Dsz T[F]:Deg (no-op)
Isz T[G]
C→D
T[G→I
Prg 2
Prg 3
Dsz M
Goto 1
Norm

==Prog 1==
P[Int (I÷9→Z
9Frac (I÷9→E
10Frac (.1Int (Z÷(10^)E→D

==Prog 2==
D→H
Prg 1
Z-D(10^)E+H(10^)E→P[Int(I÷9)]

==Prog 3==
10F+G+.1U+.01(V-9)+.001(W-18◢