Matthew Moss
9/26/2008 2:18:00 AM
Thanks to everyone who joined in on the one-liners mashup. I think
part of the challenge to the mashup was coming up with new problems.
It seems that, unless you've already solved the problem, it takes a
bit of intuition to come up with a problem that isn't trivial, but
also that can be done within a typical line or two. All the problems
presented seemed to fit right in.
I'm not going to do a regular summary here: the discussion on Ruby
Talk pretty much _is_ the summary. I will run through all the problems
very quickly, just pulling out the solution that appealed to me the
most in each case.
The problem:
Write a single line method for Array that does this:
> [:one, "two", 4].repeat(3)
=> [:one, :one, :one, "two", "two", "two", 4, 4, 4]
"The Nice" solution (I agree):
def repeat(n)
inject([]) { |a,e| a.concat [e]*n }
end
The problem:
Print out a Sierpinski carpet.
The shorter (but still kinda long) solution:
def carpet(n)
n==0?"#\n":carpet(n-1).map{|l| ['\0\0\0','\0
\0','\0\0\0'].map{|c| l.gsub(/#| /,c)}}.join
end
The problem:
Given the class:
class Data2D
def initialize
@data = [ ] # in row major form
end
def add_row(*row)
@data << row
end
end
And this setup for an object:
data = Data2D.new
data.add_row(1, 2, 3, 4)
data.add_row(5, 6, 7, 8)
define a [] method that makes this form of access possible:
data[2][1] # => 7
The tricksy call-you-later solution:
class Data2D
def [](x) lambda { |y| @data[y][x] }
end
The problem:
Write an []= method to solve the following:
obj = YourClass.new
obj['answer'] = 42
obj.answer # => 42
The what-the-heck-with-all-the-eval solution:
class YourClass
def []=(f, v)
class << self; self end.instance_eval{ attr_accessor f };
instance_eval "@#{f}=v"
end
end
The problem:
Given one or more input filenames on the command line, report the
number of unique IP addresses
found in all the data.
The command-line solution:
ruby -e 'p ARGF.read.scan(/\d{1,3}(\.\d{1,3}){3}/).uniq.size'
The problem:
Here's something simple: define method roll(n, s) to roll a s-sided
die n times and display a bar graph of the results. So roll(20, 4)
might look like this:
1|#####
2|#####
3|######
4|####
A random solution:
def roll(n,s)
(1..n).map { |x| "#{x}|#{'#' * (1 + rand(s))}" } * "\n"
end
The problem:
Provide a one-liner that can wrap a body of text at a requested
maximum length.
The "Didn't we do this before?" solution:
text.gsub(/(.{1,80})\s|(\w{80})/, "\\1\\2\n")
The problem:
Sort an array of words by the words' values where a word's value is the
sum of the values of its letters and a letter's value is its position in
the alphabet. So the value of "Hello" is 8+5+12+12+15.
The "I mutated two solutions into one I like the bestest" solution:
class Array
def sortval()
sort_by { |x| x.upcase.sum - x.length * ?@ }
end
end
The problem:
Write a function per(n) which returns the periodicity of 1/n, i.e.
per(3) => 1
per(4) => 0
per(7) => 6
per(11) => 2
The "Ummm... yeah... how do these work?" solution:
def per(n, b=10)
i=1;x=b;h={};loop {x=x%n*b;break 0 if x==0;h[x]?(break
i-h[x]):h[x]=i; i+=1}
end
The problem:
Return the power set of an Array's elements.
The "Don't get up, I'll take care of it" solution:
class Array
def powerset
inject([[]]) { |a,e| a + a.map { |b| b+[e] } }
end
end
The problem:
Assuming you have an array of numeric data, write a method that
returns an array of progressive
sums. That is:
prog_sum( [1, 5, 13, -6, 20] ) => [1, 6, 19, 13, 33]
The obligatory inject solution:
def prog_sum(ary)
ary.inject([0, []]) {|(s, a), i| [s+i, a<<(s+i)]}.last
end
The problem:
Given an s-expression, print it out as a tree, where [:a,
:b, :c, :d] is the node with parent a and children b, c and d
[:foo, [:bar, [:baz, :quux], :hello, :world], :done] #=>
foo
| -- bar
| | -- baz
| | | -- quux
| | -- hello
| | -- world
| -- done
The last solution of the summary... solution:
class Array
def to_s
collect{|x| x.to_s.gsub(/\n/,"\n| ")}.join("\n|-- ")
end
end
Any unsolved problems? You bet! Here are the problems that didn't get
answered, in case you've got an itch for more.
1. Given a text from STDIN and a regexp, print each line which (in part)
matches regexp with two preceding and two successional lines. Do not output
a line more than once.
2. Starting with an array, find the first permutation of the elements of
that array that is lexicographically greater than (i.e. sorts after)
the given array.
3. Write a oneliner next_fib(n) which gives the smallest Fibonacci number
greater than n.
4. Given an epsilon, compute PI to that precision.
--
Matthew Moss <matthew.moss@gmail.com>