Apologies for being a bit late... Here's this week's summary.

The primary point of this week's quiz was to highlight the `?`

pseudo-operator... Simply, a character preceded by `?` is converted

to its ASCII value. (I hesitate to call it a true operator, since it

is almost certainly not defined nor implemented in such manner.

Rather, I expect, it is part of Ruby's lexer and so forms part of the

definition of a number. But, in some ways, it _looks_ like a unary

operator.) Type `?(` into irb and the interpreter will return the

value 40.

Knowing this, the five quiz-permitted characters give easy access to

five integer values.

> [?(, ?), ?*, ?-, ??]
=> [40, 41, 42, 45, 63]

But I specifically permitted four characters (aside from the `?`) that

can be used for a number of mathematical operations.

Grouping/precedence, addition, subtraction, multiplication and

exponentiation are all possible using those four characters, and so we

can express values aside from the five shown above.

> ?) - ?( # 41 - 40
=> 1

> (?* - ?() ** ?( # 2 ** 40
=> 1099511627776

Using these techniques, it is possible to form an expression that

evaluates to any number, though perhaps not compactly. Let's look at a

couple solutions.

Our first solution is from _Alex LeDonne_:

def symbolify(i)

"??-??" + "-?(--?)" * i

end

Remember, it's the string output from `symbolify` that, by request of

the quiz specification, must contain only the five permitted

characters. Alex begins with `??-??`. As shove above, `??` evaluates

to 63, so this expression is `63-63`: that is, zero.

I'm going to examine the next string in a few steps, just to make it

clear what's going on. First, add in some whitespace.

"- ?( - - ?)"

Second, replace the `?` pseudo-operators with appropriate ASCII values.

"- 40 - - 41"

Third, realize that one of those `-` characters now represent unary

negation operators.

"- 40 - -41"

Fourth, simply the mathematical identity that subtracting a negative

is the same as adding the positive.

"- 40 + 41"

Finally, do the math.

"+ 1"

So now let's look at Alex's solution again, using these equivalent

strings, to understand the intent.

def symbolify(i)

"0" + "+1" * i

end

So the call `symbolify(5)` results in a string that is effectively

`0+1+1+1+1+1`. Which, when evaluated, equals five. It's not the most

efficient way to count, but it is the simplest. Of course, this

conversion was done just to show how Alex's solution produces the

correct values on evaluation. The _actual_ output of `symbolify(5)`

is:

=> "??-??-?(--?)-?(--?)-?(--?)-?(--?)-?(--?)"

I apologize for a bit long-winded explanation, but I wanted to make it

clear for anyone who might be confused about all this. For the rest of

the solutions I look at here, I'll skip the string and symbol details

and look specifically at the algorithm.

As programmers, I certainly expected someone to provide a simple

powers-of-two solution, and _Bill Kelly_ provides one. (Bill golfed

this to a single line, but it is shown here with additional whitespace

for readability.)

def symbolify(i)

x = "(?(-?()"

i.to_s(2).each_byte { |d|

x="(#{x}*(?*-?()--(?#{(d-

.chr}-?())"

}

x

end

The first line assigns a default value of zero. The second line

enumerates over the digits of the binary representation of the input

(as provided by `to_s(2)`). For each digit, the preceding value of `x`

is doubled (multiplication by `(?*-?()`), and then the next digit is

added in. This is basic binary counting.

What might be a little confusing here is why eight is subtracted from

`d`. Realize that enumerating the binary string using `each_byte` sets

`d` to the ASCII value: in this case, either 48 or 49, the ASCII

values of "0" and "1". Subtracting eight gives us 40 or 41, equivalent

to `?(` or `?)`.

As a final non-cheating solution, I want to look at one from _Dana Merrick_.

require 'mathn'

def symbolify(i)

i.zero?? "??-??":"-"+("(?(-?))-"*i).chop

end

def symbolify_short(i)

return "??-??" if i.zero?

factors = i.prime_division

factors.inject("") do |string,pair|

string << "(" << symbolify(pair.first) << ")**("

<< symbolify(pair.last) << ")*"

end.chop

end

Dana's `symbolify` function is very similar to Alex's solution above:

counting by ones. But in the added function, `symbolify_short`, Dana

makes use of `prime_division` from the `mathn` module, which will

calculate the prime factorization of an integer. For example:

> 360.prime_division
=> [[2, 3], [3, 2], [5, 1]

> (2 ** 3) * (3 ** 2) * (5 ** 1)
=> 360

This provides an efficient way to represent the inputs while still

adhering to the permitted characters. `symbolify_short` alone isn't

enough, since it can't provide output for the prime numbers, but it

calls on the simpler `symbolify` to take care of that.

Now, I figured there might be some other representations possible,

except that `eval` was going to be too limiting. So I allowed

cheating. And cheats we got, of a few different varieties.

_Chris Shea_ redefines the multiplication operator for numbers like so:

def symbolify(i)

Fixnum.class_eval <<-EOD

def *(other)

#{i}

end

EOD

'?**?*'

end

This isn't, of course, a general purpose solution, and fails the

delayed evaluation test I provided. Still, Chris managed to return a

string (the same string every time) and have that evaluate to the

number passed in.

Then we have one _Ryan Davis_, which converts the number to base-5,

using the five symbols as the digits:

FROM, TO = '01234', '?*()-'

def symbolify n

n.to_s(5).tr(FROM, TO)

end

Of course, using `eval` on the result of `symbolify` won't work, so

`eval` had to be redefined as well.

alias :real_eval :eval

def eval s

return real_eval(s) unless s =~ /^[#{Regexp.escape TO}]+$/

s.tr(TO, FROM).to_i(5)

end

Another method redefinition comes from _Lucas_ (and done in similar

fashion by a few other Rubyists), which was written solely to pass the

automated tests provided.

def symbolify (n)

s = "#{n}"

def s.delete(*args)

""

end

s

end

Note that here Lucas adds a method directly to instance `s` rather

than the String class, so it won't interfere with other code. The sole

intent of this `delete` method is to fool the automated tests.

It didn't fool _me_, as the output clearly contains characters other

than those permitted, but that's okay... I'll let it slide. They're

ugly, but I like 'em.

Finally, I present to you my own disturbingly ugly cheat:

def symbolify(num)

def eval(str) $num end; $num = num; "()"

end

Not only did I redefine `eval` (bad) in a poor manner (badder), but I

started using globals (baddest). Now that you've seen it, I recommend

you never look at it again. Just learn from the experience.

Thanks for all the submissions! I applaud your techniques and

occasional deviousness...

--

Matthew Moss <(E-Mail Removed)>