algorithm PrintingNeatly ( M; l1 , . . . , ln )
 pre-cond : l1 , . . . , ln are the lengths of the words, and M is the length of each line.
 post-cond : opt Sol splits the text into lines in an optimal way, and cost is its cost.
begin
    % Table: optSol[i] would store an optimal way to print the first i words of the input,
    but actually we store only the bird’s advice for the subinstance and the cost of its
    solution.
    table[0..n] birdAdvice, cost
    % Base case: The only base case is for the best printing of the first zero words.
    Its solution is the empty printing with cost zero.
    % optSol[0] = ∅
    cost[0] = 0
    birdAdvice[0] = ∅
    % General cases: Loop over subinstances in the table.
    for i = 1 to n
          % Solve instance M; l1 , . . . , li and fill in table entry i .
          K = maximum number k such that the words of length li−k+1 , . . . , li fit
          on a single line.
          % Try each possible bird answers.
          for k = 1 to K
               % The bird-and-friend algorithm: The bird tells us to put k words on the
               last line. We ask the friend for an optimal printing of the first i − k words.
               He gives us optSol[i − k], which he had stored in the table. To this we add
               the bird’s k words on a new last line. This gives us optSolk , which is a best
               printing of the first i words from amongst those printings consistent with
               the bird’s answer.
               % optSolk = optSub[i − k], k
               costk = cost[i − k] + (M − k + 1 − ij =i−k+1 l j )3
          end for
          % Having the best, optSolk , for each bird’s answer k, we keep the best of these
          best.
          k min = a k that minimizes costk
               birdAdvice[i] = k min
         % optSol[i] = optSolkmin
         % cost[i] = costkmin
    end for
    optSol = PrintingNeatlyWithAdvice
    return optSol, cost[n]
end algorithm