I see it now. Thanks for the clarification!

If I call [1,2,3].reduce( f, nil ) it should do f( f( f(nil, 1), 2), 3) - Your function seems to treat nil as the test for an undefined starting point, but that shouldn't be the case.

For [1,2,3].reduce( f, nil ):

• My solution: f.( f.( f.( nil, 1), 2), 3)
• Your solution:f.( f.( 1, 2), 3)