Lecture 03-17-17

Matrix Chain Multiplication

• Greedy algorithms
  • Fast
  • Easy to understand
  • Have to prove if correct
• Goal: minimize total operations
• Take advantage of subproblems - adjacent multiplication
• As always: go from easiest
• M1, M2, M3, … Mn
• Mij - represents matrix multiplication of the matrices at indices i, j
• M11, M22, M33, … Mnn
• M12, M23, M34, M(n-1)n
• M13, M24, M25, .. M2N
• Eventually M1N
• Most trivial subproblem is when i = j
• Theta of N² problems to solve
• The shape eventually looks like a staircase (common for DP problems)
• Algorithm
  • Store in a table the results of Mij
  • Row 0 (all cost 0) and Row 1 can be calculated easily
  • Subsequent rows check the ways to do it, get the optimal value, and keep track of where the “breakpoint” is
    • The breakpoint is where the decision was made
  • For each M_ij:
    • Calculates every combination of adjacent sequences
    • For example M14 = min(m11 + m24, m12 + m34, m13 + m44)
• Analysis
  • O(n³)
  • Each row (descending from M1N) takes n - 1, n - 2, etc time
  • Say only half is being calculated:
    • Each row checks n - 1 breakpoints, each entry takes >= n/2 - 1
    • For 1 + 2 + n/2 rows = N^2
  • N³ where N is the number of different matrices
  • No one knows how to do this correctly in a greedy manner

Optimal Binary Search Tree Problem

• Almost the same algorithm as matrix chain multiplication but twist at the end that applies to this but not matrix chain
• Should always look for this particular technique
• Dynamic set: add and remove from dictionary
  • Also called the dictionary problem
• Static dictionary: words are all fixed
  • No insertion or deletion
  • Space efficient spell checker
    • Some words are more common so should be at the root
    • For each word: frequency
    • Assumption: every lookup is in the dictionary
    • Evaluate cost of two trees: compare cost = sum(weight(i) * level(i))
• To be continued