#!/usr/local/bin/apl --script

⍝ Run with:  ./MOD_TEST.apl

⍝ !!! This test is totally bogus because the residues produced by MODJ and ∣ for
⍝ a given modulo basis belong to different complete residue classes. !!!


pwold ← ⎕pw
⎕pw ← 196

∇ CONT
  ""
  ⍞ ← "Press enter to continue..."
  ⍞
  ""
∇

∇ EXIT
  ""
  ⍞ ← "Press enter to exit..."
  ⍞
∇

∇ z ← rcl LABEL tbl
  z ← ⍉ ⌽ ( ⌽ ⍉ ¯1 ⌽ tbl , 1 ⊃ rcl ) , ¯99J¯99, 2 ⊃ rcl
∇


∇ z ← NORMJ m
  z ← m × + m
∇

⍝ Truncate toward negative infinity.
⍝ This is an array function.
∇z ← ROUND x
  z ← ⌊ x
∇

⍙J ← 0J1

∇ z ← CMPLX a; m; n
  ( m n ) ← a
  z ← m + ⍙J × n
∇

∇ z ← m CMPLX0 n
  z ← m + ⍙J × n
∇

∇ z ← ROUNDJ m
  z ← CMPLX0 / ROUND 9 11 ○ m
∇

⍝ Euclidean division algorithm for Gaussian integers.
⍝ DIVJ is a scalar function.
∇ z ← a DIVJ b; q
  → ( b ≠ 0J0 ) / L0
  z ← 0 a
  → 0
 L0:
  z ← q , a - b × q ← CMPLX ⌊ ( 9 11 ) ○ a ÷ b
∇

⍝ Modulo function for Gaussian integers.  A test replacement for current
⍝ gnu-apl ∣ function applied to Gaussian integers.
⍝ MODJ is a scalar function.
∇ z ← a MODJ b
  z ← ¯1 ↑ b DIVJ a
∇

∇REPORT; n; r; c; rsd0; rsd1; bad; cnt; cl; clr
  ⍝ Test is performed on a grid of complex numbers ranging from ¯nJ¯n in the
  ⍝ lower left corner to nJn in the upper right corner.
  n ← 6 ⍝ 50
  n2p1 ← 1 + 2 × n
  ⍝ r is a vector of Gaussian integer real parts
  r ← ( ¯1 - n ) + ⍳ n2p1
  ⍝ c is a vector of all the required Gaussian integers generated from r.
  c ← , r ∘.+ 0J1 × r
  ⍝ rsd0 is matrix of residue's for all possible pairings of the elements of
  ⍝ c using MODJ.
  rsd0 ← , ⊃ c ∘.MODJ c
  ⍝ rsd1 is matrix of residue's for all possible pairings of the elements of c
  ⍝ using APL mod operator (∣).
  rsd1 ← , ⊃ c ∘.∣ c
  ⍝ colarg is a ( n2p1 ⋆ 2 ) ⋆ 2 vector composed of the c vector repeated
  ⍝ n2p1 ⋆ 3 times.
  colarg ← ( ( n2p1 ⋆ 2 ) ⋆ 2 ) ⍴ c
  ⍝ rowarg is a ( n2p1 ⋆ 2 ) ⋆ 2 vector composed of the c vector elements
  ⍝ repeated n2p1 ⋆ 2 times each.
  rowarg ← ( ( n2p1 ⋆ 2 ) ⍴ n2p1 ⋆ 2 ) / c
  " "
  ⍝ Compare rsd0 and rsd1 to determine if there are discrepancies.
  bad ← rsd0 ≠ rsd1
  ⍝ Count the bad elements and return if cnt is zero.
  → ( 0 = cnt ← + / bad ) / 0
  ⍝ Otherwise, generate a table of the arguments and differing results
  '  LA    RA   MODJ   |'
  d ← cnt 1
  ( d ⍴ bad / , rowarg ) , ( d ⍴ bad / , colarg ) , ( d ⍴ bad / rsd0 ) , d ⍴ bad / rsd1
∇

REPORT

"FINISHED"