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Message: [QUOTE="Matthew Moss, post: 4619929"] 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, :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. [/QUOTE]

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