conditional test in D-Fns

[Veli-Matti]

Had some spare time, and tried to find different ways to solve the problem.

      foo1←(((2|⊢)×1+3×⊢)+(~2|⊢)×0.5×⊢) 
      foo2←(6÷⍨¯1+⊢)@(2|⊢)(1+3×⊢)
      foo3←{(b×1+3×⍵)+⍵×0.5×~b←2|⍵}
      foo4←{2÷⍨@(⍸~b)⊢(1+3×⊢)@(⍸b←2|⍵)⊢⍵}
      foo5←{(⊂⍋⍋b)⌷(0.5×(~b)/⍵),1+3×⍵/⍨b←2|⍵} 
      foo6←{b←2|v←⍵ ⋄ v[⍸b]←1+3×b/v ⋄ v[⍸~b]←0.5×(~b)/v ⋄ v} 
      foo7←{2|⍵:1+3×⍵ ⋄ ⍵÷2}¨
      foo8←{b←2|⍵ ⋄ (~b){2÷⍨⍣⍺⊢⍵}¨b{(1+3×⊢)⍣⍺⊢⍵}¨⍵}
      w←1e5?1e6 
      ]runtime -c "foo1 w" "foo2 w" "foo3 w" "foo4 w" "foo5 w" "foo6 w" "foo7 w" "foo8 w"
                                                                     
  foo1 w → 5.8E¯4 |      0% 
  foo2 w → 7.3E¯4 |    +25% 
  foo3 w → 8.9E¯4 |    +54% 
  foo4 w → 1.0E¯3 |    +81% ⎕ 
  foo5 w → 1.5E¯3 |   +154% ⎕ 
  foo6 w → 1.7E¯3 |   +190% ⎕ 
  foo7 w → 2.0E¯2 |  +3345% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕ 
  foo8 w → 7.2E¯2 | +12363% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕

-wm

[paulmansour]

Try a 3d array and watch everything change:

Code: Select all

 w←100 100 10⍴100000?1000000

]runtime -c "foo1 w" "foo2 w" "foo3 w" "foo4 w"  "foo7 w" "foo8 w" "foo9 w"
                                                                     
  foo1 w → 3.6E¯4 |      0%                                          
  foo2 w → 4.3E¯3 |  +1091% ⎕⎕⎕                                      
  foo3 w → 2.5E¯4 |    -31%                                          
  foo4 w → 8.1E¯3 |  +2143% ⎕⎕⎕⎕⎕                                    
  foo7 w → 1.6E¯2 |  +4469% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕                               
  foo8 w → 6.6E¯2 | +18334% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕ 
  foo9 w → 3.1E¯3 |   +750% ⎕⎕    

foo5 and foo6 omitted as they don't work on high rank arrays and one addition:

Code: Select all

foo9←{⊃(⊂⍤¯1)(2|⍵)⊖↑(⍵÷2)(1+3×⍵)}

[Veli-Matti]

Hi Paul, what a nice surprise!

I _thought_ that the original problem was meant for simple vectors, but you are right, higher ranks should be taken into account as well.

A couple of other possible (but not necessarily effective) algorithms:

      foo10←{b←2|⍵ ⋄ (~b){2÷⍨⍣⍺⊢⍵}¨b{(1+3×⊢)⍣⍺⊢⍵}¨⍵}
      foo11←{(⍴⍵)⍴1↑(2|⍵)⊖(⍵÷2),[0.1]1+3×⍵}

It's funny how some simple problems can have so many different approaches.

-wm (looking forward to Olhão)