R Under development (unstable) (2023-11-26 r85638 ucrt) -- "Unsuffered Consequences"
Copyright (C) 2023 The R Foundation for Statistical Computing
Platform: x86_64-w64-mingw32/x64

R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.

R is a collaborative project with many contributors.
Type 'contributors()' for more information and
'citation()' on how to cite R or R packages in publications.

Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.

> library(diptest)
> ## These are from
> ## the 217-readme.doc file that explains the bug fixed by
> ## Ferenc Mechler (fmechler@med.cornell.edu). [5/Sep/2002]
> ##
> ex1 <- c(0.0198, 0.0198, 0.1961, 0.2898, 0.3184, 0.3687,
+          0.4336, 0.4987, 0.5661, 0.6530, 0.7476, 0.8555)
> 
> ex2 <- c(0.0198, 0.1961, 0.2898, 0.3184, 0.3687, 0.4336,
+          0.4987, 0.5661, 0.6530, 0.7476, 0.8555, 0.9912)
> 
> ## Multiply them by 10000 here:
> 
> (D1 <- dip(10000*ex1, full=TRUE, debug=2))
dip() in C: n = 12; starting with  2N*dip = 1.
'dip': LOOP-BEGIN: 2n*D= 1         [low,high] = [  1, 12]; l_lcm/gcm = ( 6, 3)
  while(gcm[ix] != lcm[iv]) :
 L(3,2) --> ix = 2, iv = 3
 G(2,3) --> ix = 1, iv = 3
 --> ix = 1, iv = 4
 --> ix = 1, iv = 5
 --> ix = 1, iv = 6
  calculating dip .. (dip_l, dip_u) = (2, 1) -> new larger dip 2 (j_best = 2)
'dip': LOOP-BEGIN: 2n*D= 2         [low,high] = [  4,  9]; l_lcm/gcm = ( 6, 2)
  while(gcm[ix] != lcm[iv]) :
 L(2,2) --> ix = 1, iv = 3
 L(2,3) --> ix = 1, iv = 4
 --> ix = 1, iv = 5
 --> ix = 1, iv = 6

Call:
dip(x = 10000 * ex1, full.result = TRUE, debug = 2)

n = 12.  Dip statistic, D_n = 0.08333333 = 2/(2n)
 Modal interval [xL, xU] = [x[4], x[9]] = [2898, 5661]
 GCM and LCM have 2 and 6 nodes inside [xL, xU], respectively.
> str(D1, digits = 10, vec.len = 12)
List of 15
 $ call       : language dip(x = 10000 * ex1, full.result = TRUE, debug = 2)
 $ x          : num [1:12] 198 198 1961 2898 3184 3687 4336 4987 5661 6530 7476 8555
 $ n          : int 12
 $ dip        : num 0.08333333333
 $ lo.hi      : int [1:2] 4 9
 $ ifault     : int 0
 $ gcm        : int [1:2] 9 4
 $ lcm        : int [1:6] 4 5 6 7 8 9
 $ mn         : int [1:12] 1 1 1 1 4 4 4 4 4 4 4 4
 $ mj         : int [1:12] 2 9 6 5 6 7 8 9 10 11 12 12
 $ min.is.0   : logi FALSE
 $ debug      : int 2
 $ xl         : num 2898
 $ xu         : num 5661
 $ full.result: logi TRUE
 - attr(*, "class")= chr "dip"
> 
> (D2 <- dip(10000*ex2, full=TRUE, debug=2))
dip() in C: n = 12; starting with  2N*dip = 1.
'dip': LOOP-BEGIN: 2n*D= 1         [low,high] = [  1, 12]; l_lcm/gcm = ( 6, 4)
  while(gcm[ix] != lcm[iv]) :
 G(3,2) --> ix = 2, iv = 2
 G(2,2) --> ix = 1, iv = 2
 --> ix = 1, iv = 3
 --> ix = 1, iv = 4
 --> ix = 1, iv = 5
 --> ix = 1, iv = 6
  calculating dip .. (dip_l, dip_u) = (1, 1)
'dip': LOOP-BEGIN: 2n*D= 1         [low,high] = [  3,  8]; l_lcm/gcm = ( 6, 2)
  while(gcm[ix] != lcm[iv]) :
 L(2,2) --> ix = 1, iv = 3
 L(2,3) --> ix = 1, iv = 4
 --> ix = 1, iv = 5
 --> ix = 1, iv = 6
  calculating dip .. (dip_l, dip_u) = (0, 1)
'dip': LOOP-BEGIN: 2n*D= 1         [low,high] = [  3,  5]; l_lcm/gcm = ( 3, 2)
  while(gcm[ix] != lcm[iv]) :
 L(2,2) --> ix = 1, iv = 3
  calculating dip .. (dip_l, dip_u) = (0, 1)
'dip': LOOP-BEGIN: 2n*D= 1         [low,high] = [  3,  4]; l_lcm/gcm = ( 2, 2)
  ** (l_lcm,l_gcm) = (2,2) ==> d := 1
  calculating dip .. (dip_l, dip_u) = (0, 0)
No improvement in  low = 3  nor  high = 4 --> END

Call:
dip(x = 10000 * ex2, full.result = TRUE, debug = 2)

n = 12.  Dip statistic, D_n = 0.04166667 = 1/(2n)
 Modal interval [xL, xU] = [x[3], x[4]] = [2898, 3184]
 GCM and LCM have 2 and 2 nodes inside [xL, xU], respectively.
> str(D2, digits = 10, vec.len = 12)
List of 15
 $ call       : language dip(x = 10000 * ex2, full.result = TRUE, debug = 2)
 $ x          : num [1:12] 198 1961 2898 3184 3687 4336 4987 5661 6530 7476 8555 9912
 $ n          : int 12
 $ dip        : num 0.04166666667
 $ lo.hi      : int [1:2] 3 4
 $ ifault     : int 0
 $ gcm        : int [1:2] 4 3
 $ lcm        : int [1:2] 3 4
 $ mn         : int [1:12] 1 1 2 3 3 3 3 3 3 3 3 3
 $ mj         : int [1:12] 8 5 4 5 6 7 8 9 10 11 12 12
 $ min.is.0   : logi FALSE
 $ debug      : int 2
 $ xl         : num 2898
 $ xu         : num 3184
 $ full.result: logi TRUE
 - attr(*, "class")= chr "dip"
> 
> proc.time()
   user  system elapsed 
   0.18    0.12    0.25