What I did with the Python challenge was try and solve the puzzles without python. It turned out to be pretty easy to do the first 4 or 5 with bourne shell command line tools such as tr(1), sed(1), awk(1) and grep(1). BTW, if you are a grep(1) fan, you should check out Bill Tannenbaum's cgrep tool. It is typically 1 to 2 orders of magnitude (10 to 100 times) faster than any other grep I've ever seen. Try it on really huge files (10M or larger) and see what you think.

I've blogged before about how many of these dynamic languages aren't providing any new tools, just recasting old things in new ways. Now, I am a proponent of recasting APIs to speak the language of a problem domain so that you can easily solve the problem and so that people reading the code can follow easily.

[2, 4, 3], {'bob':34.3, 'jerry':12.0}

sum(x*x for x in range(10))
def qsort(l):
  if len(l) <= 1:
    return l
  return (qsort([x for x in l[1:] if x < l[0]])
          + l[0:1]
          + qsort([x for x in l[1:] if x >= l[0]]))

def make_adder(n):
  def add(x):
    return x+n
  return add

signal.signal(signal.SIGALRM,
  lambda n,f: sys.stderr.write('signal_handler\n'))

class EnvironmentVariables(object): 
   def __getattr__(self, name):
      return os.environ.get(name,'')

env = EnvironmentVariables()

[2, 4, 3], {'bob':34.3, 'jerry':12.0}

sum(x*x for x in range(10))

One can define a method called range(int) that returns <List<Integer>> and then use some of the loop abreviations in JDK1.5.  But, you will have some extra variables.

What I notice the most about these clever, notational shortcuts is that they are typically used in line, in places that I would be compelled to write a simple method to do the same so that there was a name for the operation that made it more obvious what I wanted to happen.

> def qsort(l):
>   if len(l) <= 1:
>     return l
>   return (qsort([x for x in l[1:] if x < l[0]])
>           + l[0:1]
>           + qsort([x for x in l[1:] if x >= l[0]]))