Amb
You are encouraged to solve this task according to the task description, using any language you may know.
Define and give an example of the Amb operator.
The Amb operator takes some number of expressions (or values if that's simpler in the language) and nondeterministically yields the one or fails if given no parameter, amb returns the value that doesn't lead to failure.
The example is using amb to choose four words from the following strings:
set 1: "the" "that" "a"
set 2: "frog" "elephant" "thing"
set 3: "walked" "treaded" "grows"
set 4: "slowly" "quickly"
It is a failure if the last character of word 1 is not equal to the first character of word 2, and similarly with word 2 and word 3, as well as word 3 and word 4. (the only successful sentence is "that thing grows slowly").
Ada
<lang ada> with Ada.Strings.Unbounded; use Ada.Strings.Unbounded; with Ada.Text_IO; use Ada.Text_IO;
procedure Test_Amb is
type Alternatives is array (Positive range <>) of Unbounded_String;
type Amb (Count : Positive) is record
This : Positive := 1;
Left : access Amb;
List : Alternatives (1..Count);
end record;
function Image (L : Amb) return String is
begin
return To_String (L.List (L.This));
end Image;
function "/" (L, R : String) return Amb is
Result : Amb (2);
begin
Append (Result.List (1), L);
Append (Result.List (2), R);
return Result;
end "/";
function "/" (L : Amb; R : String) return Amb is
Result : Amb (L.Count + 1);
begin
Result.List (1..L.Count) := L.List ;
Append (Result.List (Result.Count), R);
return Result;
end "/";
function "=" (L, R : Amb) return Boolean is
Left : Unbounded_String renames L.List (L.This);
begin
return Element (Left, Length (Left)) = Element (R.List (R.This), 1);
end "=";
procedure Failure (L : in out Amb) is
begin
loop
if L.This < L.Count then
L.This := L.This + 1;
else
L.This := 1;
Failure (L.Left.all);
end if;
exit when L.Left = null or else L.Left.all = L;
end loop;
end Failure;
procedure Join (L : access Amb; R : in out Amb) is
begin
R.Left := L;
while L.all /= R loop
Failure (R);
end loop;
end Join;
W_1 : aliased Amb := "the" / "that" / "a"; W_2 : aliased Amb := "frog" / "elephant" / "thing"; W_3 : aliased Amb := "walked" / "treaded" / "grows"; W_4 : aliased Amb := "slowly" / "quickly";
begin
Join (W_1'Access, W_2); Join (W_2'Access, W_3); Join (W_3'Access, W_4); Put_Line (Image (W_1) & ' ' & Image (W_2) & ' ' & Image (W_3) & ' ' & Image (W_4));
end Test_Amb; </lang> The type Amb is implemented with the operations "/" to construct it from strings. Each instance keeps its state. The operation Failure performs back tracing. Join connects two elements into a chain. The implementation propagates Constraint_Error when matching fails. Sample output:
that thing grows slowly
ALGOL 68
Note: This program violates ALGOL 68's scoping rules when a locally scoped procedure is returned to a more global scope. ELLA ALGOL 68RS misses this violation, but ALGOL 68 Genie spots it at run time and then produces an assert. However ELLA ALGOL 68RS does produce the desired result, but may potentially suffer from "mysterious" stack problems. <lang algol>MODE STRINGS = [0][0]CHAR; MODE YIELDSTRINGS = PROC(STRINGS)VOID; MODE ITERSTRINGS = PROC(YIELDSTRINGS)VOID;
OP INITITERSTRINGS = (STRINGS self)ITERSTRINGS:
(YIELDSTRINGS yield)VOID: # scope violation #
FOR i TO UPB self DO
yield(self[i])
OD;
OP + = (ITERSTRINGS for strings, STRINGS b)ITERSTRINGS:
(YIELDSTRINGS yield)VOID: # scope violation #
for strings((STRINGS amb)VOID:(
[UPB amb + 1]STRING joined;
joined[:UPB amb] := amb;
STRING last string := amb[UPB amb];
CHAR last char := last string[UPB last string];
FOR i TO UPB b DO
IF last char = b[i][1] THEN
joined[UPB joined] := b[i];
yield(joined)
FI
OD
));
OP + = (STRINGS a, STRINGS b)ITERSTRINGS: INITITERSTRINGS a + b;
ITERSTRINGS for amb :=
STRINGS("the", "that", "a") +
STRINGS("frog", "elephant", "thing") +
STRINGS("walked", "treaded", "grows") +
STRINGS("slowly", "quickly");
STRINGS sep;
- FOR amb IN for amb DO#
for amb((STRINGS amb)VOID:( print((amb[1]," ",amb[2]," ",amb[3]," ",amb[4], new line)) ))
- OD#</lang>
Output:
that thing grows slowly
C
Note: This uses the continuations code from http://homepage.mac.com/sigfpe/Computing/continuations.html <lang c> typedef char * amb_t;
amb_t amb(size_t argc, ...) {
amb_t *choices;
va_list ap;
int i;
if(argc) {
choices = malloc(argc*sizeof(amb_t));
va_start(ap, argc);
i = 0;
do { choices[i] = va_arg(ap, amb_t); } while(++i < argc);
va_end(ap);
i = 0;
do { TRY(choices[i]); } while(++i < argc);
free(choices);
}
FAIL;
}
int joins(char *left, char *right) { return left[strlen(left)-1] == right[0]; }
int _main() {
char *w1,*w2,*w3,*w4;
w1 = amb(3, "the", "that", "a");
w2 = amb(3, "frog", "elephant", "thing");
w3 = amb(3, "walked", "treaded", "grows");
w4 = amb(2, "slowly", "quickly");
if(!joins(w1, w2)) amb(0);
if(!joins(w2, w3)) amb(0);
if(!joins(w3, w4)) amb(0);
printf("%s %s %s %s\n", w1, w2, w3, w4);
return EXIT_SUCCESS;
} </lang>
Haskell
Haskell's List monad returns all the possible choices. Use the "head" function on the result if you just want one. <lang haskell> import Control.Monad
amb = id
joins left right = last left == head right
example = do
w1 <- amb ["the", "that", "a"] w2 <- amb ["frog", "elephant", "thing"] w3 <- amb ["walked", "treaded", "grows"] w4 <- amb ["slowly", "quickly"] unless (joins w1 w2) (amb []) unless (joins w2 w3) (amb []) unless (joins w3 w4) (amb []) return (unwords [w1, w2, w3, w4])
</lang>
Prolog
<lang prolog> amb(E, [E|_]). amb(E, [_|ES]) :- amb(E, ES).
joins(Left, Right) :-
append(_, [T], Left), append([R], _, Right), ( T \= R -> amb(_, []) % (explicitly using amb fail as required) ; true ).
amb_example([Word1, Word2, Word3, Word4]) :-
amb(Word1, ["the","that","a"]), amb(Word2, ["frog","elephant","thing"]), amb(Word3, ["walked","treaded","grows"]), amb(Word4, ["slowly","quickly"]), joins(Word1, Word2), joins(Word2, Word3), joins(Word3, Word4).
</lang>
Python
Python does not have the amb function, but, in the spirit of the task, here is an implementation in Python (version 2.6) that uses un-ordered sets of words; the itertools.product function to loop through all the word sets lazily; and a generator comprehension to lazily give the first answer: <lang python> >>> from itertools import product >>> sets = [ set('the that a'.split()), set('frog elephant thing'.split()), set('walked treaded grows'.split()), set('slowly quickly'.split()) ] >>> success = ( sentence for sentence in product(*sets)
if all(sentence[word][-1]==sentence[word+1][0]
for word in range(3))
)
>>> success.next() ('that', 'thing', 'grows', 'slowly') >>> </lang>
The following is inspired by Haskell. For loops in a generator kind of act as an amb operator. Of course the indenting won't be right because for-blocks have to be indented. I will try to replicate the "amb with empty list" here faithfully but it is really awkward:. <lang python> def amb(*args): return args
def joins(left, right): return left[-1] == right[0]
def example():
for w1 in amb("the", "that", "a"):
for w2 in amb("frog", "elephant", "thing"):
for w3 in amb("walked", "treaded", "grows"):
for w4 in amb("slowly", "quickly"):
for _ in joins(w1,w2) and amb(42) or amb(): # this is really just "if joins(w1,w2):"
for _ in joins(w2,w3) and amb(42) or amb(): # this is really just "if joins(w2,w3):"
for _ in joins(w3,w4) and amb(42) or amb(): # this is really just "if joins(w3,w4):"
yield "%s %s %s %s" % (w1,w2,w3,w4)
</lang>
<lang python> >>> list(example()) ['that thing grows slowly'] </lang>
Ruby
<lang ruby> class Amb
class ExhaustedError < RuntimeError; end
def initialize
@fail = proc { fail ExhaustedError, "amb tree exhausted" }
end
def choose(*choices)
prev_fail = @fail
callcc { |sk|
choices.each { |choice|
callcc { |fk| @fail = proc { @fail = prev_fail fk.call(:fail) } if choice.respond_to? :call sk.call(choice.call) else sk.call(choice) end }
}
@fail.call
}
end
def failure choose end
def assert(cond) failure unless cond end
end
A = Amb.new w1 = A.choose("the", "that", "a") w2 = A.choose("frog", "elephant", "thing") w3 = A.choose("walked", "treaded", "grows") w4 = A.choose("slowly", "quickly")
A.choose() if not w1[-1] == w2[0] A.choose() if not w2[-1] == w3[0] A.choose() if not w3[-1] == w4[0]
puts w1, w2, w3, w4 </lang>
Scheme
<lang scheme> (define fail
(lambda () (error "Amb tree exhausted")))
(define-syntax amb
(syntax-rules ()
((AMB) (FAIL)) ; Two shortcuts.
((AMB expression) expression)
((AMB expression ...)
(LET ((FAIL-SAVE FAIL))
((CALL-WITH-CURRENT-CONTINUATION ; Capture a continuation to
(LAMBDA (K-SUCCESS) ; which we return possibles.
(CALL-WITH-CURRENT-CONTINUATION
(LAMBDA (K-FAILURE) ; K-FAILURE will try the next
(SET! FAIL K-FAILURE) ; possible expression.
(K-SUCCESS ; Note that the expression is
(LAMBDA () ; evaluated in tail position
expression)))) ; with respect to AMB.
...
(SET! FAIL FAIL-SAVE) ; Finally, if this is reached,
FAIL-SAVE))))))) ; we restore the saved FAIL.
(let ((w-1 (amb "the" "that" "a"))
(w-2 (amb "frog" "elephant" "thing"))
(w-3 (amb "walked" "treaded" "grows"))
(w-4 (amb "slowly" "quickly")))
(define (joins? left right)
(equal? (string-ref left (- (string-length left) 1)) (string-ref right 0)))
(if (joins? w-1 w-2) '() (amb))
(if (joins? w-2 w-3) '() (amb))
(if (joins? w-3 w-4) '() (amb))
(list w-1 w-2 w-3 w-4))
</lang>
SETL
<lang SETL>program amb;
sets := unstr('[{the that a} {frog elephant thing} {walked treaded grows} {slowly quickly}]');
words := [amb(words): words in sets]; if exists lWord = words(i), rWord in {words(i+1)} |
lWord(#lWord) /= rWord(1) then fail;
end if;
proc amb(words);
return arb {word in words | ok};
end proc;
end program;</lang> Sadly ok and fail were only ever implemented in CIMS SETL, and are not in any compiler or interpreter that is available today, so this is not very useful as it stands.
Alternate version (avoids backtracking)
<lang SETL>program amb;
sets := unstr('[{the that a} {frog elephant thing} {walked treaded grows} {slowly quickly}]');
print(amb(sets));
proc amb(sets);
return amb1([], {}, sets);
end proc;
proc amb1(prev, mbLast, sets);
if sets = [] then
return prev;
else
words fromb sets;
if exists word in words |
(forall last in mbLast |
last(#last) = word(1)) and
(exists sentence in {amb1(prev with word, {word}, sets)} |
true) then
return sentence;
end if;
end if;
end proc;
end program;</lang> We cheat a bit here - this version of amb must be given the whole list of word sets, and that list is consumed recursively. It can't pick a word from an individual list.
Tcl
Brute force, with quick kill of failing attempts: <lang Tcl> proc amb list {set list}
proc fit {a b} {
expr {[string index $a end] eq [string index $b 0]}
}
foreach i [amb {the that a}] {
foreach j [amb {frog elephant thing}] {
if ![fit $i $j] continue
foreach k [amb {walked treaded grows}] {
if ![fit $j $k] continue
foreach l [amb {slowly quickly}] {
if [fit $k $l] {
puts [list $i $j $k $l]
}
}
}
}
} </lang>