<>(()){<>((([][][][][])<(((({}){})(({})({}))[])({}(({})({}({})({}{}(<>)))))[])>{()<{}>}{})<{{}}{}>())}{}<>(<(({()(((<>))<>)}{}{<({}(([][][])((({})({}))[]{})){})>((){[]<({}{})((){[]<({}{}<>((({})({})){}{}){})(<>)>}{}){{}{}<>(<({}{}())>)(<>)}>}{}){(<{}{}{}((<>))<>>)}{}}<>)<{({}[]<({}<>)<>{(<{}>)<>{<>({}[])}{}<>({}<>)(<>)}{}>)}{}<>>)>)<>{(({}[])(){(<{}>)<><(({})[])>[][][][]{()()()()(<{}>)}{}<>}{}<>)<>}<>{}{(({})<({()<<>({}<>)>}{})>([]))((){[](<(({}()()(<>))()()()){(<{}>)<>}>)}{}<>){{}((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[](<{}<>{({}<>)<>}{}(({}))({<{}({}<>)<>>{}(<<>({}[]<>)>)}<><{({}<>)<>}>{})>)}{}){{}{}(<([])>)}>}{}){{}<>{({}<>)<>}{}((({})())<{({}[]<({}<>)<>>)}>{}){({}[]<><({}<><({()<({}[]<({}<>)<>>)>}{}<>)><>)<>({()<({}[]<({}<>)<>>)>}{}<>)>)}<>(<{({}<>)<>}>)}>}{}){{}{}(<(())>)}>}{}){(<{}{}>)<>{({}<>)<>}{}(({}))({<{}({}<>)<>>({})(<<>({}<>)>)}<><{({}<>)<>}>){{}([][][])<>(((<{}>)<>))}}>}{}){{}(<([{}])>)}>}{}){{}((<{}>))}>}{}){{}(({})(<()>)<<>{({}<>)<>}{}({}()<>)<>>)<>(<({}<>)>)<>{({}<>)<>}}{}(<({}<({}<>)<>>{})<>({}<>)>)<>(<({}())>)}{}({}<{({}[]<({}<>)<>>)}{}>){((({}[]<>){(<{}({}<>)>)}{}())<{({}()<({}<>)<>(({})[])>{[][](<{}>)}{})}{}>()){{}(<>)}}{}}{}{({}[]<[{}]>)}{}{({}[]<{}>)}{}
Pruébalo en línea!
+4 bytes de arreglar un error con la condición en el {...} mónada, y -36 bytes de varios campos de golf.
1238 bytes de código, +1 byte para el -a indicador (que se puede combinar con el indicador de idioma).
Esto ahora evalúa {...} como cero según la especificación de desafío. Tenga en cuenta que Brain-Flak se ha evaluado {...}como la suma de todas las ejecuciones desde la corrección de errores del 7 de mayo de 2016 dos días antes de que se publicara este desafío.
El siguiente código interpreta Brain-Flak Classic correctamente, con {...}la suma de todas las ejecuciones. La única diferencia entre los dos intérpretes es la colocación de una {}nilad.
<>(()){<>((([][][][][])<(((({}){})(({})({}))[])({}(({})({}({})({}{}(<>)))))[])>{()<{}>}{})<{{}}{}>())}{}<>(<(({()(((<>))<>)}{}{<({}(([][][])((({})({}))[]{})){})>((){[]<({}{})((){[]<({}{}<>((({})({})){}{}){})(<>)>}{}){{}{}<>(<({}{}())>)(<>)}>}{}){(<{}{}{}((<>))<>>)}{}}<>)<{({}[]<({}<>)<>{(<{}>)<>{<>({}[])}{}<>({}<>)(<>)}{}>)}{}<>>)>)<>{(({}[])(){(<{}>)<><(({})[])>[][][][]{()()()()(<{}>)}{}<>}{}<>)<>}<>{}{(({})<({()<<>({}<>)>}{})>([]))((){[](<(({}()()(<>))()()()){(<{}>)<>}>)}{}<>){{}((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[]<({}())((){[](<{}<>{({}<>)<>}{}(({}))({<{}({}<>)<>>{}(<<>({}[]<>)>)}<><{({}<>)<>}>{})>)}{}){{}{}(<([])>)}>}{}){{}<>{({}<>)<>}{}((({})())<{({}[]<({}<>)<>>)}>{}){({}[]<><({}<><({()<({}[]<({}<>)<>>)>}{}<>)><>)<>({()<({}[]<({}<>)<>>)>}{}<>)>)}<>(<{({}<>)<>}>)}>}{}){{}{}(<(())>)}>}{}){(<{}>)<>{({}<>)<>}{}(({}))({<{}({}<>)<>>({})(<<>({}<>)>)}<><{({}<>)<>}>{}){{}([][][])<>(((<{}>)<>))}}>}{}){{}(<([{}])>)}>}{}){{}((<{}>))}>}{}){{}(({})(<()>)<<>{({}<>)<>}{}({}()<>)<>>)<>(<({}<>)>)<>{({}<>)<>}}{}(<({}<({}<>)<>>{})<>({}<>)>)<>(<({}())>)}{}({}<{({}[]<({}<>)<>>)}{}>){((({}[]<>){(<{}({}<>)>)}{}())<{({}()<({}<>)<>(({})[])>{[][](<{}>)}{})}{}>()){{}(<>)}}{}}{}{({}[]<[{}]>)}{}{({}[]<{}>)}{}
Pruébalo en línea!
La entrada (a cualquiera de los intérpretes) es el programa Brain-Flak Classic para interpretar, luego una nueva línea, luego una lista de enteros separados por espacios. No se realiza validación en la entrada. Se requiere la nueva línea, incluso si el programa o la entrada están en blanco.
El primer paso es analizar toda la entrada, comenzando con los corchetes:
# Move to right stack, and push 1 to allow loop to start
<>(())
{
# While keeping -5 on third stack:
<>((([][][][][])<
# Pop bracket or newline k from left stack, and push 0, k-10, k-40, k-60, k-91, k-123 on right stack
(((({}){})(({})({}))[])({}(({})({}({})({}{}(<>)))))[])
# Search this list for a zero, and push the number of nonzero entries popped minus 5
# (thus replacing the 0 if it was destroyed)
>{()<{}>}{})
# Remove rest of list, and push the same number plus 1
# Result is -4 for {, -3 for [, -2 for <, -1 for (, 0 for newline, or 1 for everything else (assumed closing bracket)
<{{}}{}>())
# Repeat until newline found
}{}<>
Luego se analizan los enteros. Esto normalmente no sería necesario, pero la entrada se tomó como ASCII. Sin embargo, esto tiene un lado positivo: el ingreso de texto nos permite determinar la altura de la pila, lo que simplifica las cosas cuando no tenemos acceso a la altura de la pila nilad.
Los enteros se analizan en dos números en la segunda pila: uno para el valor absoluto y otro para el signo. Estos se mueven de nuevo a la primera pila.
Las pilas interpretadas se almacenan debajo del código en la primera pila en el siguiente orden: altura de pila actual, pila actual, otra altura de pila, otra pila. El 0 para la otra altura de la pila no necesita ser empujado en este punto, ya que será un cero implícito la primera vez que se lea.
(<((
# If stack nonempty, register first stack entry.
{()(((<>))<>)}{}
# For each byte k of input:
{
# Push -3, -13, and k-32
<({}(([][][])((({})({}))[]{})){})>
# Evaluate to 1 if space
# If not space (32):
((){[]<
# If not minus (45):
({}{})((){[]<
# Replace top of right stack (n) with 10*n + (k-48)
({}{}<>((({})({})){}{}){})(<>)
# Else (i.e., if minus):
>}{}){
# Remove excess "else" entry and -3
{}{}
# Set sign to negative (and destroy magnitude that shouldn't even be there yet)
<>(<({}{}())>)(<>)}
# Else (i.e., if space):
>}{}){
# Remove working data for byte, and push two more 0s onto right stack
(<{}{}{}((<>))<>>)
# Push number of integers found
}{}}<>)
# For each integer:
<{({}[]<
# Move magnitude back to left stack
({}<>)<>
# If sign is negative, negate
{(<{}>)<>{<>({}[])}{}<>({}<>)(<>)}{}
>)}{}
# Push stack height onto stack
<>>)
# Push 0
>)
La representación del código ahora se mueve de regreso a la pila izquierda. Para facilitar las cosas más adelante, restamos 4 de los corchetes de apertura de nilads, para que cada operación tenga un número entero único de -1 a -8.
# For each bracket in the code:
<>{
# Push k-1 and evaluate to k
(({}[])()
# If not closing bracket:
{
# Check next bracket (previously checked, since we started at the end here)
(<{}>)<><(({})[])>
# Subtract 4 if next bracket is closing bracket
# Inverting this condition would save 8 bytes here, but cost 12 bytes later.
[][][][]{()()()()(<{}>)}{}
<>}{}
# Push result onto left stack
<>)
<>}<>{}
La parte principal del programa es interpretar las instrucciones. Al comienzo de cada iteración del bucle principal, la instrucción actual está en la parte superior de la pila izquierda, todo lo que está debajo está en la misma pila, y todo lo que está antes en la pila derecha. Tiendo a visualizar esto como si tuviera un libro abierto en una página determinada.
{
(
# Get current instruction
({})
# Move all code to left stack, and track the current position in code
<({()<<>({}<>)>}{})>
# Push -1, signifying that the code will move forward to just before a matching }.
# In most cases, this will become 0 (do nothing special) before it is acted upon
([])
# Push instruction minus 1
)
# If opening bracket:
((){[](<
# Push instruction+1 and instruction+4
(({}()()(<>))()()())
# If instruction+4 is nonzero (not loop monad), replace the earlier -1 with 0 to cancel forward seek
# This would be clearer as {(<{}>)<>(<{}>)<>}, but that would be unnecessarily verbose
{(<{}>)<>}
# Else (i.e., if closing bracket):
>)}{}<>){
# If closing bracket, parse command
# Post-condition for all: if not moving to {, pop two and push evaluation, 0.
# (For nilads, can assume second from top is 0.)
# If moving to {, pop one, push -3, 0, 0.
# Seven nested if/else statements, corresponding to eight possible instruction.
# The "else" statements end with 0 already on the stack, so no need to push a 0 except in the innermost if.
# Each one beyond the first increments the instruction by 1 to compare the result with 0
# Each instruction will pop the instruction, leaving only its evaluation (with a 0 on top).
{}((){[]<
({}())((){[]<
({}())((){[]<
({}())((){[]<
({}())((){[]<
({}())((){[]<
({}())((){[](<
# -7: pop
# Pop instruction to reveal existing 0 evaluation
{}
# Move code out of the way to access stack
<>{({}<>)<>}{}
# Duplicate stack height (only useful if stack height is zero)
(({}))
(
# If stack height nonzero
{
# Save stack height on second stack
<{}({}<>)<>>
# Pop stack
{}
# Move stack height back and subtract 1
(<<>({}[]<>)>)
}
# Move code back to normal position
<><{({}<>)<>}>{}
# Evaluate as popped entry (0 if nothing popped)
)
# (else)
>)}{}){
# -6: -1 nilad
# Just evaluate as -1
{}{}(<([])>)
# (else)
}>}{}){
# -5: swap nilad
# Move code out of the way to access stack
{}<>{({}<>)<>}{}
# Number of integers to move: stack height + 1 (namely, the stack height and every entry in the stack)
((({})())
# Move to second stack
<{({}[]<({}<>)<>>)}>{}
# Do (stack height + 1) times again
){({}[]<><
# Get stack element
({}<><
# Move alternate (interpreted) stack to second (real) stack, and push length on top of it
({()<({}[]<({}<>)<>>)>}{}<>)
# Push current stack element below alternate stack
><>)
# Move alternate stack back above newly pushed element
<>({()<({}[]<({}<>)<>>)>}{}<>)
>)}
# Move code back to normal position
<>(<{({}<>)<>}>)
# (else)
}>}{}){
# -4: 1
# Just evaluate to 1
{}{}(<(())>)
# (else)
}>}{}){
# -3: loop
# Create zero on stack while keeping existing evaluation
# This becomes (<{}{}>) in the version that meets the challenge spec
(<{}>)
# Move code out of the way to access stack
<>{({}<>)<>}{}
# Duplicate stack height
(({}))
(
# If stack height nonzero
{
# Save stack height on second stack
<{}({}<>)<>>
# Peek at top of stack
({})
# Move stack height back
(<<>({}<>)>)
}
# Move code back to normal position
<><{({}<>)<>}>
# Look at peeked entry
# Remove the {} in the version meeting the challenge spec
{})
# If peeked entry is nonzero
{
# Replace -3 instruction on third stack
{}([][][])
# Replace loop indicator to 0 (to be incremented later to 1)
<>(((<{}>)
# Create dummy third stack entry to pop
<>))
}
# (else)
}>}{}){
# -2: print
# Just print evaluation without modifying it
{}(<([{}])>)
# (else)
}>}{}){
# -1: evaluate as zero
# Just change evaluation to 0
{}((<{}>))
# else
}>}{}){
# 0: push
# Get current evaluation (without modifying it)
{}(({})
# Create zero on stack as barrier
(<()>)
# Move code out of the way to access stack
<<>{({}<>)<>}{}
# Increment stack height and save on other stack
({}()<>)<>
# Push evaluation
>)
# Move stack height back (and push zero)
<>(<({}<>)>)
# Move code back to normal position
<>{({}<>)<>}
}{}
# Update third stack by adding evaluation to previous entry's evaluation
# Previous entry's instruction is saved temporarily on left stack
(<({}<({}<>)<>>{})<>({}<>)>)
# Increment loop indicator
# If instruction was loop monad and top of stack was nonzero, this increments 0 to 1 (search backward)
# Otherwise, this increments -1 to 0 (do nothing)
<>(<({}())>)
}{}
# While holding onto loop indicator
({}<
# Go to immediately after executed symbol
{({}[]<({}<>)<>>)}{}
>)
# If looping behavior:
{
# Switch stack and check if searching forward
((({}[]<>)
# If so:
{
# Move just-executed { back to left stack, and move with it
(<{}({}<>)>)
}{}
# Either way, we are currently looking at the just-executed bracket.
# In addition, the position we wish to move to is on the current stack.
# Push unmodified loop indicator as initial value in search
())
# While value is nonzero:
<{
# Add 1
({}()
# Move current instruction to other stack
<({}<>)<>
# Check whether next instruction is closing bracket
(({})[])>
# If opening bracket, subtract 2 from value
{[][](<{}>)}{}
)
}{}>
# If searching backward, move back to left stack
()){{}(<>)}
}{}
}
Después de salir del bucle principal, todo el código está en la pila correcta. Las únicas cosas en la pila izquierda son un cero y las dos pilas interpretadas. Producir la salida correcta es una cuestión simple.
# Pop the zero
{}
# Output current stack
{({}[]<[{}]>)}{}
# Discard other stack to avoid implicit printing
{({}[]<{}>)}{}