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Beau­tiful Racket / tuto­rials

Imagine a language: wires

First, we’ll test our language with our sample circuit. All we do is add a #lang wires desig­na­tion to the top:

test.rkt
#lang wires
x AND y -> d
x OR y -> e
x LSHIFT 2 -> f
y RSHIFT 2 -> g
NOT x -> h
NOT y -> i
123 -> x
456 -> y
#lang wires
x AND y -> d
x OR y -> e
x LSHIFT 2 -> f
y RSHIFT 2 -> g
NOT x -> h
NOT y -> i
123 -> x
456 -> y
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When we run this in DrRacket, we get the result:

d: 72
e: 507
f: 492
g: 114
h: 65412
i: 65079
x: 123
y: 456
d: 72
e: 507
f: 492
g: 114
h: 65412
i: 65079
x: 123
y: 456
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Exactly right. Because we’ve modeled our wires as Racket func­tions, we can fiddle with them on the REPL just like any other func­tions:

x
#<procedure:x>
(x)
123
(y)
456
(d)
72
> x
#<proce­dure:x>
> (x)
123
> (y)
456
> (d)
72
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Because our wires expander exports the 16-bit oper­a­tions, we can use those on the REPL too:

(AND (x) (y)) ; same as definition of `d`
72
(LSHIFT (x) 2) ; same as definition of `f`
492
> (AND (x) (y)) ; same as defi­n­i­tion of `d`
72
> (LSHIFT (x) 2) ; same as defi­n­i­tion of `f`
492
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Now we can try the main puzzle file, which is a 339-wire circuit. As before, we just add #lang wires to the top (excerpt below, but the whole thing is avail­able in the source listing):

puzzle.rkt
#lang wires
bn RSHIFT 2 -> bo
lf RSHIFT 1 -> ly
fo RSHIFT 3 -> fq
cj OR cp -> cq
fo OR fz -> ga
t OR s -> u
lx -> a
NOT ax -> ay
he RSHIFT 2 -> hf
lf OR lq -> lr
lr AND lt -> lu
dy OR ej -> ek
1 AND cx -> cy
hb LSHIFT 1 -> hv
1 AND bh -> bi
ih AND ij -> ik
···
#lang wires
bn RSHIFT 2 -> bo
lf RSHIFT 1 -> ly
fo RSHIFT 3 -> fq
cj OR cp -> cq
fo OR fz -> ga
t OR s -> u
lx -> a
NOT ax -> ay
he RSHIFT 2 -> hf
lf OR lq -> lr
lr AND lt -> lu
dy OR ej -> ek
1 AND cx -> cy
hb LSHIFT 1 -> hv
1 AND bh -> bi
ih AND ij -> ik
···
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This will print a list of all the wire values:

bo: 10292
ly: 4045
fq: 6560
cq: 15518
ga: 60815
u: 1
a: 46065
ay: 65531
hf: 15568
lr: 8187
lu: 7027
ek: 15935
cy: 0
hv: 1438
bi: 1
ik: 9873
···
bo: 10292
ly: 4045
fq: 6560
cq: 15518
ga: 60815
u: 1
a: 46065
ay: 65531
hf: 15568
lr: 8187
lu: 7027
ek: 15935
cy: 0
hv: 1438
bi: 1
ik: 9873
···
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Since we might not notice the value of wire a as it goes whizzing by, we have two choices. We can update our define/display macro in the expander to only print the value of the a wire. (An exer­cise left to the reader.) Or we can just eval­uate a on the REPL:

(a)
46065
> (a)
46065
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Puzzle solved. (If you sign up at Advent of Code, you can get your own puzzle file and feed it to #lang wires.)

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