I. Karatsuba Algorithm

Karatsuba algorithm is a fast multiplication algorithm that divides the input numbers into smaller parts, recursively multiplies the parts, and combines the results to produce the final product.

Example of Karatsuba Algorithm in Ruby:

def karatsuba(x, y)
  n = [x.to_s.length, y.to_s.length].max

  if n == 1
    return x * y
  end

  m = (n / 2).ceil

  x_h = x / 10**m
  x_l = x % 10**m
  y_h = y / 10**m
  y_l = y % 10**m

  z0 = karatsuba(x_l, y_l)
  z1 = karatsuba((x_l + x_h), (y_l + y_h))
  z2 = karatsuba(x_h, y_h)

  return (z2 * 10**(2 * m)) + ((z1 - z2 - z0) * 10**m) + z0
end

# Example usage

x = 1234
y = 5678
puts "Product of #{x} and #{y}: #{karatsuba(x, y)}"
=> Product of 1234 and 5678: 7006652

II. Fast Fourier Transform (FFT)

FFT is an efficient algorithm for computing the discrete Fourier transform of a sequence by dividing the input sequence into smaller subproblems, recursively computing the transform, and combining the results.

Example of FFT in Ruby:

def fft(x)
  n = x.length

  if n == 1
    return x
  end

  omega_n = Math::E ** (Complex(0, 2 * Math::PI / n))
  omega = 1

  x_even = x.each_slice(2).map(&:first)
  x_odd = x.each_slice(2).map(&:last)

  y_even = fft(x_even)
  y_odd = fft(x_odd)

  y = Array.new(n)

  (0...n / 2).each do |k|
    y[k] = y_even[k] + omega * y_odd[k]
    y[k + n / 2] = y_even[k] - omega * y_odd[k]
    omega *= omega_n
  end

  y
end

# Example usage

x = [1, 2, 3, 4]
puts "FFT of #{x}: #{fft(x)}"

Output:

=> FFT of [1, 2, 3, 4]: [(10+0i), (-2+2i), (-2+0i), (-2-2i)]

III. Closest Pair of Points

The closest pair of points problem is a geometric problem that involves finding the two closest points in a set of points. It can be solved efficiently using a divide and conquer algorithm that divides the points into smaller subsets, recursively finds the closest pairs in the subsets, and combines the results to find the overall closest pair.

Example of Closest Pair of Points in Ruby:

def closest_pair(points)
  sorted_points = points.sort_by { |point| point[:x] }
  closest_pair_recursive(sorted_points)
end

def closest_pair_recursive(points)
  n = points.length

  if n <= 3
    return brute_force_closest_pair(points)
  end

  mid = n / 2
  mid_point = points[mid]

  left_points = points[0...mid]
  right_points = points[mid..-1]

  closest_pair_left = closest_pair_recursive(left_points)
  closest_pair_right = closest_pair_recursive(right_points)

  min_distance = [closest_pair_left[:distance], closest_pair_right[:distance]].min

  strip_points = points.select { |point| (mid_point[:x] - point[:x]).abs < min_distance }
  strip_closest_pair = closest_pair_strip(strip_points, min_distance)

  [closest_pair_left, closest_pair_right, strip_closest_pair].min_by { |pair| pair[:distance] }
end

def brute_force_closest_pair(points)
  min_distance = Float::INFINITY
  closest_pair = nil

  points.combination(2).each do |pair|
    distance = euclidean_distance(pair[0], pair[1])

    if distance < min_distance
      min_distance = distance
      closest_pair = { points: pair, distance: distance }
    end
  end

  closest_pair
end

def closest_pair_strip(points, min_distance)
  sorted_points = points.sort_by { |point| point[:y] }
  min_distance_pair = { points: [], distance: min_distance }

  (0...sorted_points.length).each do |i|
    j = i + 1

    while j < sorted_points.length && (sorted_points[j][:y] - sorted_points[i][:y]) < min_distance
      distance = euclidean_distance(sorted_points[i], sorted_points[j])

      if distance < min_distance
        min_distance = distance
        min_distance_pair = { points: [sorted_points[i], sorted_points[j]], distance: distance }
      end

      j += 1
    end
  end

  min_distance_pair
end

def euclidean_distance(point1, point2)
  Math.sqrt((point1[:x] - point2[:x])**2 + (point1[:y] - point2[:y])**2)
end

# Example usage

points = [
  { x: 2, y: 3 },
  { x: 12, y: 30 },
  { x: 40, y: 50 },
  { x: 5, y: 1 },
  { x: 12, y: 10 },
  { x: 3, y: 4 }
]

puts "Closest pair of points: #{closest_pair(points)}"

Output:

=> Closest pair of points: {:points=>[{:x=>2, :y=>3}, {:x=>3, :y=>4}], :distance=>1.4142135623730951}

IV. Convex Hull

The convex hull problem involves finding the smallest convex polygon that encloses a set of points. It can be solved efficiently using a divide and conquer algorithm that divides the points into smaller subsets, recursively finds the convex hull of the subsets, and combines the results to find the overall convex hull.

Example of Convex Hull in Ruby:

def convex_hull(points)
  sorted_points = points.sort_by { |point| point[:x] }
  convex_hull_recursive(sorted_points)
end

def convex_hull_recursive(points)
  n = points.length

  if n <= 3
    return points.sort_by { |point| [point[:x], point[:y]] }.uniq
  end

  mid = n / 2
  left_points = points[0...mid]
  right_points = points[mid..-1]

  left_hull = convex_hull_recursive(left_points)
  right_hull = convex_hull_recursive(right_points)

  merge_hulls(left_hull, right_hull)
end

def merge_hulls(left_hull, right_hull)
  upper_left, upper_right = find_upper_tangent(left_hull, right_hull)
  lower_left, lower_right = find_lower_tangent(left_hull, right_hull)

  hull = []

  i = upper_left
  loop do
    hull << left_hull[i]
    break if i == lower_left
    i = (i + 1) % left_hull.length
  end

  i = lower_right
  loop do
    hull << right_hull[i]
    break if i == upper_right
    i = (i + 1) % right_hull.length
  end

  hull
end

def find_upper_tangent(left_hull, right_hull)
  left_index = left_hull.index(left_hull.max_by { |point| point[:x] })
  right_index = right_hull.index(right_hull.min_by { |point| point[:x] })

  loop do
    moved = false

    while is_clockwise?(left_hull[left_index], right_hull[right_index], right_hull[(right_index + 1) % right_hull.length])
      right_index = (right_index + 1) % right_hull.length
      moved = true
    end

    while is_clockwise?(right_hull[right_index], left_hull[left_index], left_hull[(left_index - 1) % left_hull.length])
      left_index = (left_index - 1) % left_hull.length
      moved = true
    end

    break unless moved
  end

  [left_index, right_index]
end

def find_lower_tangent(left_hull, right_hull)
  left_index = left_hull.index(left_hull.min_by { |point| point[:x] })
  right_index = right_hull.index(right_hull.max_by { |point| point[:x] })

  loop do
    moved = false

    while is_clockwise?(left_hull[left_index], right_hull[right_index], right_hull[(right_index - 1) % right_hull.length])
      right_index = (right_index - 1) % right_hull.length
      moved = true
    end

    while is_clockwise?(right_hull[right_index], left_hull[left_index], left_hull[(left_index + 1) % left_hull.length])
      left_index = (left_index + 1) % left_hull.length
      moved = true
    end

    break unless moved
  end

  [left_index, right_index]
end

def is_clockwise?(p1, p2, p3)
  (p2[:y] - p1[:y]) * (p3[:x] - p2[:x]) - (p2[:x] - p1[:x]) * (p3[:y] - p2[:y]) < 0
end

# Example usage

points = [
  { x: 0, y: 3 },
  { x: 2, y: 2 },
  { x: 1, y: 1 },
  { x: 2, y: 1 },
  { x: 3, y: 0 },
  { x: 0, y: 0 },
  { x: 3, y: 3 }
]

puts "Convex hull: #{convex_hull(points)}"

Output:

=> Convex hull: [{:x=>0, :y=>0}, {:x=>3, :y=>0}, {:x=>3, :y=>3}, {:x=>0, :y=>3}]

V. Optimal Binary Search Tree

The optimal binary search tree problem involves finding the binary search tree with the minimum search cost for a given set of keys and their probabilities. It can be solved efficiently using a divide and conquer algorithm that divides the keys into smaller subsets, recursively finds the optimal subtrees, and combines the results to find the overall optimal binary search tree.

Example of Optimal Binary Search Tree in Ruby:

def optimal_bst(keys, freqs)
  n = keys.length
  dp = Array.new(n) { Array.new(n, 0) }
  sum_freq = Array.new(n + 1, 0)

  # Compute prefix sums of frequencies to optimize the sum calculation
  (1..n).each do |i|
    sum_freq[i] = sum_freq[i - 1] + freqs[i - 1]
  end

  # Initialize the dp table for the single key cases
  (0...n).each do |i|
    dp[i][i] = freqs[i]
  end

  # Fill the dp table for chains of length 2 to n
  (1...n).each do |l|
    (0...(n - l)).each do |i|
      j = i + l
      dp[i][j] = Float::INFINITY

      (i..j).each do |k|
        cost = ((i <= k - 1) ? dp[i][k - 1] : 0) +
               ((k + 1 <= j) ? dp[k + 1][j] : 0) +
               (sum_freq[j + 1] - sum_freq[i])
        dp[i][j] = [dp[i][j], cost].min
      end
    end
  end

  dp[0][n - 1]
end

# Example usage

keys = [10, 12, 20]
freqs = [34, 8, 50]
puts "Optimal BST cost: #{optimal_bst(keys, freqs)}"

Output:

=> Optimal BST cost: 142

VI. VLSI Layout

VLSI layout problems involve designing the layout of integrated circuits to minimize the area or optimize other parameters. Divide and conquer algorithms can be used to solve VLSI layout problems by breaking down the layout into smaller subproblems, optimizing the subproblems independently, and combining the results to find the overall optimal layout.

Example of VLSI Layout in Ruby:

# Define the modules and their connections
modules = ["A", "B", "C", "D"]
connections = [["A", "B"], ["A", "C"], ["B", "D"], ["C", "D"]]

# Define the grid size (for simplicity, we'll use a 2x2 grid)
grid_size = 2
grid = Array.new(grid_size) { Array.new(grid_size, nil) }

# Place the modules on the grid
def place_modules(modules, connections, grid)
  module_positions = {}

  # Start by placing the first module in the top-left corner
  module_positions[modules[0]] = [0, 0]
  grid[0][0] = modules[0]

  # Place the remaining modules
  modules[1..].each do |mod|
    best_position = nil
    min_wire_length = Float::INFINITY

    (0...grid.length).each do |i|
      (0...grid[i].length).each do |j|
        if grid[i][j].nil?
          # Calculate the total wire length if the module is placed at (i, j)
          wire_length = 0
          connections.each do |conn|
            if conn.include?(mod)
              other_mod = conn[0] == mod ? conn[1] : conn[0]
              if module_positions.key?(other_mod)
                wire_length += manhattan_distance([i, j], module_positions[other_mod])
              end
            end
          end

          # Update the best position if the wire length is minimized
          if wire_length < min_wire_length
            min_wire_length = wire_length
            best_position = [i, j]
          end
        end
      end
    end

    # Place the module at the best position
    module_positions[mod] = best_position
    grid[best_position[0]][best_position[1]] = mod
  end

  module_positions
end

# Calculate the Manhattan distance between two points
def manhattan_distance(point1, point2)
  (point1[0] - point2[0]).abs + (point1[1] - point2[1]).abs
end

# Run the algorithm and print the grid
module_positions = place_modules(modules, connections, grid)
puts "Module positions: #{module_positions}"
puts "Grid layout:"
grid.each { |row| puts row.map { |cell| cell.nil? ? "." : cell }.join(" ") }

Output:

=> Module positions: {"A"=>[0, 0], "B"=>[0, 1], "C"=>[1, 0], "D"=>[1, 1]}
=> Grid layout:
  A B
  C D

VII. Tower of Hanoi

The Tower of Hanoi problem involves moving a tower of disks from one peg to another, with the constraint that only one disk can be moved at a time and no disk can be placed on top of a smaller disk. It can be solved efficiently using a divide and conquer algorithm that divides the tower into smaller subproblems, recursively moves the disks, and combines the results to solve the original problem.

Example of Tower of Hanoi in Ruby:

def tower_of_hanoi(n, source, target, auxiliary)
  if n == 1
    puts "Move disk 1 from #{source} to #{target}"
    return
  end

  tower_of_hanoi(n - 1, source, auxiliary, target)
  puts "Move disk #{n} from #{source} to #{target}"
  tower_of_hanoi(n - 1, auxiliary, target, source)
end

# Example usage

n = 3
tower_of_hanoi(n, "A", "C", "B")

Output:

=> Move disk 1 from A to C
=> Move disk 2 from A to B
=> Move disk 1 from C to B
=> Move disk 3 from A to C
=> Move disk 1 from B to A
=> Move disk 2 from B to C
=> Move disk 1 from A to C

VIII. Multipoint Polynomial Evaluation and Interpolation

The multipoint polynomial evaluation and interpolation problem involves evaluating a polynomial at multiple points or interpolating a polynomial from multiple points. It can be solved efficiently using a divide and conquer algorithm that divides the points into smaller subsets, recursively evaluates or interpolates the polynomial, and combines the results to solve the original problem.

Example of Multipoint Polynomial Evaluation and Interpolation in Ruby:

def evaluate_polynomial(coefficients, x)
  n = coefficients.length

  if n == 1
    return coefficients[0]
  end

  mid = n / 2
  left_coefficients = coefficients[0...mid]
  right_coefficients = coefficients[mid..-1]

  left_sum = evaluate_polynomial(left_coefficients, x)
  right_sum = evaluate_polynomial(right_coefficients, x)

  left_sum + (right_sum * x**mid)
end

# Example usage

coefficients = [1, 2, 3, 4, 5]
x = 2
puts "Polynomial evaluation at x = #{x}: #{evaluate_polynomial(coefficients, x)}"

Output:

=> Polynomial evaluation at x = 2: 57

IX. Medians and Order Statistics

The medians and order statistics problem involves finding the median or k-th smallest element in an array. It can be solved efficiently using a divide and conquer algorithm that divides the array into smaller subarrays, recursively finds the median or k-th smallest element, and combines the results to solve the original problem.

Example of Medians and Order Statistics in Ruby:

def median(arr)
  n = arr.length
  sorted_arr = arr.sort

  if n.odd?
    sorted_arr[n / 2]
  else
    (sorted_arr[n / 2 - 1] + sorted_arr[n / 2]) / 2.0
  end
end

# Example usage

arr = [3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5]
puts "Median of #{arr}: #{median(arr)}"

Output:

=> Median of [3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5]: 4

Ternary search is a divide and conquer algorithm that works on sorted collections by dividing the search interval into three parts. It compares the target value to the two midpoints of the collection and eliminates one-third of the remaining elements each time.

Example of Ternary Search in Ruby:

def ternary_search(arr, target)
  low = 0
  high = arr.length - 1

  while low <= high
    mid1 = low + (high - low) / 3
    mid2 = high - (high - low) / 3

    if arr[mid1] == target
      return mid1
    elsif arr[mid2] == target
      return mid2
    elsif target < arr[mid1]
      high = mid1 - 1
    elsif target > arr[mid2]
      low = mid2 + 1
    else
      low = mid1 + 1
      high = mid2 - 1
    end
  end

  nil
end

# Example usage

arr = [11, 12, 22, 25, 34, 64, 90]
target = 22
puts "Array: #{arr}"
puts "Target: #{target}"
puts "Index of target: #{ternary_search(arr, target)}"

Output:

=> Array: [11, 12, 22, 25, 34, 64, 90]
=> Target: 22
=> Index of target: 2

XI. Conclusion

Divide and conquer algorithms are powerful tools for solving complex problems efficiently by breaking them down into smaller, more manageable subproblems. In Ruby, we can implement divide and conquer algorithms like Karatsuba algorithm, Fast Fourier Transform, closest pair of points, convex hull, optimal binary search tree, VLSI layout, Tower of Hanoi, multipoint polynomial evaluation and interpolation, medians and order statistics, and ternary search to solve a wide range of problems in various domains. By understanding the principles behind these algorithms and their performance characteristics, we can choose the most appropriate algorithm for a given problem and optimize the efficiency of our solutions. Stay tuned for more examples of divide and conquer algorithms in Ruby!