A look at the Chocolate Bar Problem

• Definition
  • A bar is divided (by lines) into mxn squares.
  • We wish to break it into m*n individual squares
  • No parallel breaking
• A brute force approach
  • Just snap along all of one direction
  • Then break each sub piece.

  • BRUTE_FORCE(bar)
    // bar is mxn
    // BREAKR(bar, row) breaks the bar along the given row.
    // BREAKC(bar, col) breaks the bar along the given column
    
     for i ← 1 to m-1 do
          (bari,n,bar BREAKR(bar,1)
     for i ← 1 to m do 
         for j ← 1 to n-1 do
          (chunk, bari,n) = BREAKC(bari,n,1)
    
    	 

  • Input size : m is the number of rows in the bar, n is the number of columns in the bar.
  • Critical operations: lines 2 and 5
  • Performance:

    
    T(m,n) = Σ1m-11 + Σ1m Σ 1n-1 1
           = m-1 -1+1 + Σ1m n-1-1+1
           = m-1 + (n-1)Σ1m 1
           = m-1 +  (n-1)(m-1+1)
           = m-1 +mn -m
           = mn - 1
           ∈ Θ(mn)
    	 

  • Best Worst or Average?
  • Do we believe that this works?
• The problem, however, in mentioning binary trees suggests that some other form may be faster.
• QuickBreak
  • A divide and conquer attack
  • Break along the major axis the recursively break the resulting bars.

  • QUICK_BREAK(bar)
    
     if m =1  and  n = 1 then
          return
     elseif m ≥ n then
          (a,b) = BREAKR(m/2,n)
          QUICK_BREAK(a);
          QUICK_BREAK(b);
          return
     else
          (a,b) = BREAKC(m,n/2)
          QUICK_BREAK(a);
          QUICK_BREAK(b);
          return
    
    	 

  • Performance:

             | 1  n = 0 and m = 1   (We are counting breaks, none needed here)
    T(m,n) = | 2T(m/2,n) +1 m ≥ n
             | 2T(m,n/2) + 1 otherwise
    
           WLOG assume m=2a and n = 2b
           Also assume the bar is roughly square, but this really doesn't matter.
    
           T(m,n) = 2T(m/2),n)  + 1,      T(m/2,n) = 2T(m/2,n/2)+1
                                                  or 2T(m/22,n)+1
           Since we assumed the bar is roughly square
           T(m,n) = 2[2T(m/2,n/2) + 1] + 1
                = 22T(m/2,n/2)+21 + 20
    	                              T(m/2,n/2) = 2T(m/22,n/2)+1
    				               or  2T(m/2,n/s2)+1 
           T(m,n) = 22[2T(m/22,n/2) + 1] + Σ012i
                  = 23T(m/22,n/2)  Σ022i
    
           at step k,   k = c+d
           T(m,n) = 2kT(m/2c,n/2d) + Σ0k-1 2i
                  = 2kT(m/2c,n/2d)  + 2k-1
           when c = a and d = b
           T(m,n) = 2a+bT(1,1) + 2a+b-1
                  = 2a+b*0 + 2a+b-1
                  = 2a2b-1
                  = mn-1
                  ∈ O(mn)
    	 

  • Basic questions?
• Ok, so is one better, Let's try our new toy, Dynamic Programming
  • This is an optimization problem, find the fewest breaks.
  • And it consists of sub problems, (optimal strategy for breaking smaller bars)
  • F(m,n) is the minimal cost to break a bar of size m,n

  • 
    F(m,n) = 0 if m=1 and n=1
             min(min row breaks, min col breaks) + 1
    	 min row breaks = min1⌊m/2⌋ (F(i,n)+F(m-i,n))
    	 min col breaks = min1⌊n/2⌋ (F(m,i)+F(m,n-i))
    	

  • For example
  • 

    +	F(3,1) + F(3,3)
    +	F(3,2) + F(3,2)

  • Note, there is no need to do F(3,3)+ F(3,1)

  • 
    F n  1   2   3   4    
    m 1  0                    F(1,2) = F(1,1) + F(1,1) + 1  = 1
      2                       F(2,1) = F(1,1) + F(1,1) + 1 = 1
      3
      4
      5
    
    F n  1   2   3   4       F(1,3) = F(1,1) + F(1,2) + 1 = 0 + 1 + 1 = 2
    m 1  0   1               F(3,1) = F(1,1) + F(2,1) + 1 = 0 + 1 + 1 = 2
      2  1                   F(2,2) = min[ F(2,1)+F(2,1), F(1,2)+F(1,2)] + 1
      3                             = mim(2+2, 2+2) + 1 = 3
      4
      5
    
    F n  1   2   3   4       F(1,4) = min(F(1,1)+F(1,3),F(1,2)+F(1,2)]+1
    m 1  0   1   2                  = min(0+2,1+1) + 1 = 3 
      2  1   3               Same for F(4,1)
      3  2                   F(3,2) = F(1,2)+F(2,2)+1 = 1+3+1 = 5
      4                      F(2,3) = F(2,1)+F(2,2) + 1 = 5
      5
    
    F n  1   2   3   4
    m 1  0   1   2   3
      2  1   3   5
      3  2   5
      4  3
      5
    
    F n  1   2   3   4
    m 1  0   1   2   3
      2  1   3   5   4
      3  2   5   8   9
      4  3   8  11  14
      5  4   9  14  19