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Dan said:
Could you show us what you mean? I for one do not know what a mutually
recursive function is.
I have written it all out. Now, since I'm fresh to this list, I dunno if
attachments show. Indulge me, please. I'll paste all the code (it's a
bit literate') and also attach it (as mutrec.rb).
By the way, ruby -v is: 1.8.5
#! /usr/bin/ruby
# Two examples of a mutually-recursive function.
# These are functions that depend on each other - recursively.
# even and odd:
# If a number is zero, then it is even. Otherwise, it is even if the one before it is odd.
# A number is odd if it is not zero. Otherwise, it is odd if it the one before it is even.
)
# This is the canonical example.
def my_even(x)
x == 0 or odd(x - 1)
end
def my_odd(x)
x != 0 and even(x - 1)
end
# That is, of course, a silly way to implemet even and odd. So, let me give you the problem
# I was trying to solve.
#
# To generate the lettering similar to that above the spreadsheet columns, where the first
# column has 'A', the second 'B' ... the twenty-sixth 'Z'. Then, the twenty-seventh has
# 'AA', the twenty-eight has 'AB', and so on. This is recursive by nature. Here is what
# should work:
Letters = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J',
'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T',
'U', 'V', 'W', 'X', 'Y', 'Z']
def alphabetise(pos)
pile_letters(0, pos) # LABEL: 1
end
def pile_chars(thus_far, pos)
len = Letters.length
if pos < len then
Letters[pos]
else
alphabeticise(thus_far) + pile_chars(thus_far + 1, pos - len)
end
end
# Labels:
# 1. That's where it fails. I guess 'tis because the scope is searched backwards
# from that point.
# OCaml, for example, requires me to put the `and' keyword between mutually-recursive
# functions. I'd not mind doing that in Ruby, but I don't know how. n00b. ;-)
# Also, it works in That Other Language.
#
# Now, the (dirty) solution I found. I think you will help me with how to get it done
# without all this voodoo:
Funcs = [lambda do |pos|
Funcs[1].call(0, pos)
end,
lambda do |thus_far, pos|
len = Letters.length
if pos < len then
Letters[pos]
else
Funcs[0].call(thus_far) + Funcs[1].call(thus_far + 1, pos - len)
end
end]
def main(args)
puts Funcs[0].call((26 * 26) - 1) # Prints A to Z. But see how un-Rubyistic it is!
# puts alphabetise(675) # This fails.
# if my_even 4 and my_odd 5 then 0 else 1 end # This also fails
0
end
exit(main(ARGV))
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filename="mutrec.rb"
#! /usr/bin/ruby
# Two examples of a mutually-recursive function.
# These are functions that depend on each other - recursively.
# even and odd:
# If a number is zero, then it is even. Otherwise, it is even if the one before it is odd.
# A number is odd if it is not zero. Otherwise, it is odd if it the one before it is even.
)
# This is the canonical example.
def my_even(x)
x == 0 or odd(x - 1)
end
def my_odd(x)
x != 0 and even(x - 1)
end
# That is, of course, a silly way to implemet even and odd. So, let me give you the problem
# I was trying to solve.
#
# To generate the lettering similar to that above the spreadsheet columns, where the first
# column has 'A', the second 'B' ... the twenty-sixth 'Z'. Then, the twenty-seventh has
# 'AA', the twenty-eight has 'AB', and so on. This is recursive by nature. Here is what
# should work:
Letters = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J',
'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T',
'U', 'V', 'W', 'X', 'Y', 'Z']
def alphabetise(pos)
pile_letters(0, pos) # LABEL: 1
end
def pile_chars(thus_far, pos)
len = Letters.length
if pos < len then
Letters[pos]
else
alphabeticise(thus_far) + pile_chars(thus_far + 1, pos - len)
end
end
# Labels:
# 1. That's where it fails. I guess 'tis because the scope is searched backwards
# from that point.
# OCaml, for example, requires me to put the `and' keyword between mutually-recursive
# functions. I'd not mind doing that in Ruby, but I don't know how. n00b. ;-)
# Also, it works in That Other Language.
#
# Now, the (dirty) solution I found. I think you will help me with how to get it done
# without all this voodoo:
Funcs = [lambda do |pos|
Funcs[1].call(0, pos)
end,
lambda do |thus_far, pos|
len = Letters.length
if pos < len then
Letters[pos]
else
Funcs[0].call(thus_far) + Funcs[1].call(thus_far + 1, pos - len)
end
end]
def main(args)
puts Funcs[0].call((26 * 26) - 1) # Prints A to Z. But see how un-Rubyistic it is!
# puts alphabetise(675) # This fails.
# if my_even 4 and my_odd 5 then 0 else 1 end # This also fails
0
end
exit(main(ARGV))
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