This week, guest puzzle master Aaron Brager has a scavenger hunt for you.
You'll use SQL to find Carmen Sandiego! She could be anywhere in the world.
How to Play
To get started, fork and clone this GitHub project. This project uses PostgreSQL, a popular database system. You almost certainly already have it installed.
Follow the eight clues in the project's
readme.md file to find Carmen. For each clue, you'll need to write a SQL query to hone in on her location.
To win, send your SQL queries, their output, and where you think Carmen is, to email@example.com by March 4th 9am PST.
For bonus points, determine Carmen's location using the fewest number of queries, or using interesting SQL syntax.
Last Week's Result Snail Sort
Last week the challenge was to sort an associative array in the pattern of a snail:
The "Who's on First" award
Goes to Daniel Rassiner who completed the puzzle in 2 hours.
def snail(array) array_amount = array.size - 1 total_loops = (array.size + array.size) - 2 answer = array x = 0 y = array_amount index = 1 what_to_pick = ["r", "d", "l", "u"] total_loops.times do case what_to_pick[index] when "r" array_amount.times do y += 1 answer << array[x][y] end when "d" array_amount.times do x += 1 answer << array[x][y] end when "l" array_amount.times do y -= 1 answer << array[x][y] end when "u" array_amount.times do x -= 1 answer << array[x][y] end index = -1 end if (index % 2 == 0) array_amount -= 1 end index += 1 end answer end
The "Loopty Loop" award
Goes to Fong Fan for his multi-loop based solution:
def snail(array) final_array =  xMin = 0 yMin = 0 xMax = array.length - 1 yMax = xMax # Base Cases return final_array if array.length == 0 if array.length == 1 final_array << array return final_array end # Clockwise Loop Over All Four Sides while (xMin <= xMax) && (yMin <= yMax) # Traverse top side for y in yMin..yMax final_array << array[xMin][y] end xMin = xMin + 1 # Traverse right side for x in xMin..xMax final_array << array[x][yMax] end yMax = yMax - 1 # Traverse bottom side for y in (yMax).downto(yMin) final_array << array[xMax][y] end xMax = xMax - 1 # Traverse left side for x in (xMax).downto(xMin) final_array << array[x][yMin] end yMin = yMin + 1 end return final_array end
The "Jony Ive Minimalism" award
goes to Aaron Brager for his tiny solution.
def snail(array) result =  while !array.empty? do result << array.shift array = array.transpose.reverse end result.flatten end
The key to this solution is the ruby
transpose method which allows you to rotate the entire dataset by turning the columns into rows.