Does Prolog have an alias “operator” like Haskell?

Question

In Haskell, there is a language feature called "as"-operator (sometimes known as alias). The idea is the following: given you have a function that for instance takes as input a list and wants to return all tails, you can implement this as:

tails a@(_:xs) = a : tails xs
tails [] = [[]]

The @ ensures that you have a reference to both the entire argument as well as a reference to some parts of the structure of the parameter. This is intelligent performance-wise (it is more a performance hack, because reconstructing an array (x:xs) ) in the body of the first line, will - if not optimized by the compiler - result in allocating a new object, modifying fields, etc. See here for more information.

I was wondering if Prolog has something equivalent: for instance if you want to implement tails in Prolog, it can be done the following way:

tails([H|T],[[H|T]|TA]) :-
    tails(T,TA).
tails([],[[]]).

But it could be more efficient if there was an "as"-operator like:

tails(L@[_|T],[L|TA]) :-  %This does not compile
    tails(T,TA).
tails([],[[]]).

Is there any such construct, or a language extension?

Answer 1

TL;DR: Good idea ¹ ! The speedup appears to be limited to ~20% (for most list sizes).

In this answer, we compare three different predicates that differ regarding @ -like data reuse:

list_tails([], [[]]).                % (1) like `tails/2` given by the OP ...
list_tails([E|Es], [[E|Es]|Ess]) :-  %     ....... but with a better name :-)
   list_tails(Es, Ess).

list_sfxs1(Es, [Es|Ess]) :-          % (2) "re-use, mutual recursion"
   aux_list_sfxs1(Es, Ess).          %     "sfxs" is short for "suffixes"

aux_list_sfxs1([], []).
aux_list_sfxs1([_|Es], Ess) :-
   list_sfxs1(Es, Ess).

list_sfxs2([], [[]]).                % (3) "re-use, direct recursion"
list_sfxs2(Es0, [Es0|Ess]) :-
   Es0 = [_|Es],
   list_sfxs2(Es, Ess).

To measure runtime, we use the following code:

:-( dif(D,sicstus), current_prolog_flag(dialect,D)
  ; use_module(library(between))
  ).

run_benchs(P_2s, P_2, L, N, T_ms) :-
   between(1, 6, I),
   L is 10^I,
   N is 10^(8-I),
   length(Xs, L),
   member(P_2, P_2s),
   garbage_collect,
   call_walltime(run_bench_core(P_2,Xs,N), T_ms).

run_bench_core(P_2, Xs, N) :-
   between(1, N, _),
   call(P_2, Xs, _),
   false.
run_bench_core(_, _, _).

To measure wall-time ² , we utilize call_ walltime /2 —a variation of call_time/2 :

call_walltime(G, T_ms) :-
   statistics(walltime, [T0|_]),
   G,
   statistics(walltime, [T1|_]),
   T_ms is T1 - T0.

Let's put above code variations to the test...

• ...using different list lengths L ...
• ...and running each test a number of times N (for better accuracy).

First, we use swi-prolog version 7.3.14 (64-bit):

?- run_benchs([list_sfxs1,list_sfxs2,list_tails], P_2, L, N, T_ms).
   P_2 = list_sfxs1, L*N = 10*10000000, T_ms =  7925
;  P_2 = list_sfxs2, L*N = 10*10000000, T_ms =  7524
;  P_2 = list_tails, L*N = 10*10000000, T_ms =  6936
;
   P_2 = list_sfxs1, L*N = 100*1000000, T_ms =  6502
;  P_2 = list_sfxs2, L*N = 100*1000000, T_ms =  5861
;  P_2 = list_tails, L*N = 100*1000000, T_ms =  5618
;
   P_2 = list_sfxs1, L*N = 1000*100000, T_ms =  6434
;  P_2 = list_sfxs2, L*N = 1000*100000, T_ms =  5817
;  P_2 = list_tails, L*N = 1000*100000, T_ms =  9916
;
   P_2 = list_sfxs1, L*N = 10000*10000, T_ms =  6328
;  P_2 = list_sfxs2, L*N = 10000*10000, T_ms =  5688
;  P_2 = list_tails, L*N = 10000*10000, T_ms =  9442
;
   P_2 = list_sfxs1, L*N = 100000*1000, T_ms = 10255
;  P_2 = list_sfxs2, L*N = 100000*1000, T_ms = 10296
;  P_2 = list_tails, L*N = 100000*1000, T_ms = 14592
;
   P_2 = list_sfxs1, L*N = 1000000*100, T_ms =  6955
;  P_2 = list_sfxs2, L*N = 1000000*100, T_ms =  6534
;  P_2 = list_tails, L*N = 1000000*100, T_ms =  9738.

Then, we repeat the previous query ³ using sicstus-prolog version 4.3.2 (64-bit):

?- run_benchs([list_sfxs1,list_sfxs2,list_tails], P_2, L, N, T_ms).
   P_2 = list_sfxs1, L*N = 10*10000000, T_ms =  1580
;  P_2 = list_sfxs2, L*N = 10*10000000, T_ms =  1610
;  P_2 = list_tails, L*N = 10*10000000, T_ms =  1580
;
   P_2 = list_sfxs1, L*N = 100*1000000, T_ms =   710
;  P_2 = list_sfxs2, L*N = 100*1000000, T_ms =   750
;  P_2 = list_tails, L*N = 100*1000000, T_ms =   840
;
   P_2 = list_sfxs1, L*N = 1000*100000, T_ms =   650 
;  P_2 = list_sfxs2, L*N = 1000*100000, T_ms =   660
;  P_2 = list_tails, L*N = 1000*100000, T_ms =   740
;  
   P_2 = list_sfxs1, L*N = 10000*10000, T_ms =   620
;  P_2 = list_sfxs2, L*N = 10000*10000, T_ms =   650
;  P_2 = list_tails, L*N = 10000*10000, T_ms =   740
;
   P_2 = list_sfxs1, L*N = 100000*1000, T_ms =   670
;  P_2 = list_sfxs2, L*N = 100000*1000, T_ms =   650
;  P_2 = list_tails, L*N = 100000*1000, T_ms =   750
;
   P_2 = list_sfxs1, L*N = 1000000*100, T_ms = 12610
;  P_2 = list_sfxs2, L*N = 1000000*100, T_ms = 12560
;  P_2 = list_tails, L*N = 1000000*100, T_ms = 33460.

Summary:

• The alias-thingy can and does improve performance significantly.
• In above tests, the SICStus Prolog jit ⁴ gives 10X speedup , compared to SWI-Prolog!

---

^{Why do the stunt of putting (@)/2 in the rule head?} 为什么把(@)/2放在规则头上的噱头？ ^{To end up with non- idiomatic Prolog code?
We are interested in the total runtime.} 我们对总运行时间感兴趣。 ^{Why? Because garbage-collection costs show with larger data sizes!
The answer sequence has been post-processed for the sake of readability.} 为了便于阅读，答案序列已经过后处理。
^{Available since release 4.3.0 .} 自4.3.0版以来可用。 ^{Current target architectures include IA-32 and AMD64 .}