R Under development (unstable) (2024-11-22 r87365 ucrt) -- "Unsuffered Consequences"
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> require(DEoptimR)
Loading required package: DEoptimR
> 
> c.time <- function(...) cat('Time elapsed: ', ..., '\n')
> S.time <- function(expr) c.time(system.time(expr))
> (doExtras <- DEoptimR:::doExtras())
[1] FALSE
> 
> set.seed(2345)
> # Bound-constrained test problems ----------------------------------------------
> 
> bl <- function(x) {
+ #   Becker and Lago problem
+ #
+ #   -10 <= x1, x2 <= 10
+ #   The function has four minima located at (+-5, +-5), all with f(x*) = 0.
+ #
+ #   Source:
+ #     Ali, M. Montaz, Khompatraporn, Charoenchai, and Zabinsky, Zelda B. (2005).
+ #     A numerical evaluation of several stochastic algorithms on selected
+ #     continuous global optimization test problems.
+ #     Journal of Global Optimization 31, 635-672.
+ 
+     sum((abs(x) - 5)^2)
+ }
> 
> S.time(bl_ <- NCDEoptim(-c(10, 10), c(10, 10),
+                         bl,
+                         niche_radius = 5,
+                         maxiter = 100))
Time elapsed:  1.22 0.05 1.27 NA NA 
> # Only inequality constraints --------------------------------------------------
> 
> #   Function F1
> #
> #   f(x) = x^2
> #   subject to:
> #   g(x) = 1 - x^2 <= 0
> #
> #   -2 <= x <= 2
> #   The two global optima are (x1*, x2*; f*) = (1, -1; 1).
> #
> #   Source:
> #     Poole, Daniel J. and Allen, Christian B. (2019).
> #     Constrained niching using differential evolution.
> #     Swarm and Evolutionary Computation 44, 74-100.
> 
> S.time(F1_ <- NCDEoptim(-2, 2,
+                         function(x) x^2,
+                         function(x) 1 - x^2,
+                         niche_radius = 1,
+                         maxiter = 200))
Time elapsed:  2.4 0 2.41 NA NA 
> 
> # Expected optimal values ------------------------------------------------------
> 
> stopifnot(
+   all.equal( as.vector(abs(bl_$solution_arch)), rep(5, 8), tolerance = 1e-3 ),
+   all.equal( as.vector(abs(F1_$solution_arch)), c(1, 1), tolerance = 1e-2 )
+ )
> 
> c.time(proc.time())
Time elapsed:  3.79 0.06 3.84 NA NA 
> 
> proc.time()
   user  system elapsed 
   3.79    0.06    3.84