syntacticsugar

(define (abs x)
    (cond ((= 0 x) 0)
          ((< 0 x) x)
          ((> 0 x)(- 0 x))))

(define abs
    (lambda (x)
    (cond ((= 0 x) 0)
          ((< 0 x) x)
          ((> 0 x)(- 0 x)))))

syntacticsugar

// Find the sum of all the multiples of 3 or 5 below 1000.
//
//
// ruby version: (inspired by Lfborjas' Ruby Fuckery)
//multi = ->(x) { (x % 3).zero? or (x % 5).zero? }
//puts (1...1000).select{ |x| multi[x] }.inject(:+).to_s
//233168

function arrayFromRange(lo, hi){

syntacticsugar

class Array
  def insert_every!(skip, str)
    orig_ary_length = self.length
    new_ary_length = orig_ary_length + ((orig_ary_length-1) / skip)
    i = skip
    while(i<new_ary_length) do self.insert(i,str); i+=(skip+1) end
    self
  end
end

syntacticsugar

=begin
The sum of the squares of the first ten natural numbers is,

  1**2 + 2**2 + ... + 102 = 385
The square of the sum of the first ten natural numbers is,

  (1 + 2 + ... + 10)**2 = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025  385 = 2640.

  Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

syntacticsugar

# Euclid's formula yields Pythagorean triples for integers m and n with m < n: 
# a = m**2 - n**2 ; b = 2*m*n ; c = m**2 + n**2

x = 1000

def euclids upto 
  result = []
  (2..upto).each do |m| # Start at 2 as 1 results nothing anyway
    (1...m).each do |n| # Euclid's formula only works for m > n
      result << [m**2 - n**2, 2*m*n, m**2 + n**2]

syntacticsugar

# goal: get a number like 25164150
# and transform it to: 25,164,150

# =begin
# 1) number.to_s.split("").length
# 2) start from end, count multiples of 3
# 3) insert comma
# 4) turn array back into a string
# .join(",")
# x.unshift(",")unless (x % 3 != 0)

syntacticsugar

def modulo_3?(number)
  # (number % 3).zero? 
  number.modulo(3).zero? # i guess this is more idiomatic Ruby, though I like the previous better
end

def commify(number)
  reversed = number.to_s.reverse
  number_size = number.to_s.size
  
  result = ""

syntacticsugar

# Find the greatest product of five consecutive digits in the 1000-digit number.

madness=7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315

syntacticsugar

class Celery
  def initialize(grid)
    @grid = grid
    @height = grid.length
    @width = grid[0].length
  end
  def self.mini
     [[ 0, 1, 2, 3, 4],
      [10,11,12,13,14],
      [20,21,22,23,24],

syntacticsugar

# proccc

def one proccc, x
  proccc.(x)
end

def two proccc, x
  proccc.(proccc.(x))
end