Pruning

1
Pruning
2
Exponential growth

• How many leaves are there in a complete binary
  tree of depth N?

depth 0, count 1 depth 1,
count 2 depth 3, count
4 depth 4, count 8
depth 5, count 16
depth N, count 2N

• This is easy to demonstrate
• Count going left as a 0
• Count going right as a 1
• Each leaf represents one of the 2N possible N-bit
  binary numbers
• This observation turns out to be very useful in
  certain kinds of problems

3
Pruning

• Suppose the binary tree represents a problem that
  we have to explore to find a solution (or goal
  node)
• If we can prune (decide we can ignore) a part of
  the tree, we save effort

• The higher up in the tree we can prune, the more
  effort we can save
• The advantage is exponential

4
Sum of subsets

• Problem
• There are n positive integers, and a positive
  integer W
• Find a subset of the integers that sum to exactly
  W
• Example
• The numbers are 2, 5, 7, 8, 13
• Find a subset of numbers that sum to exactly 25
• We can multiply each number by 1 if it is in the
  sum, 0 if it is not

• 2 5 7 8 130 0 0 0 0 ? 00 0 0 0 1 ?
  130 0 0 1 0 ? 80 0 0 1 1 ? 210 0 1
  0 1 ? 200 0 1 1 0 ? 150 0 1 1 1 ? 280
  1 0 0 0 ? 50 1 0 0 1 ? 180 1 0 1 0
  ? 130 1 0 1 1 ? 260 1 1 0 0 ? 120 1
  1 0 1 ? 25

5
Brute force

• We have a brute-force method for solving the sum
  of subsets problem
• For N numbers, count in binary from 0 to 2N
• For each 1, include the corresponding number for
  each 0, exclude the corresponding number
• Stop if we get lucky
• This is clearly an exponential-time algorithm
• It seems like, with a little cleverness, we could
  do better
• It turns out that we can use pruning to do
  somewhat better
• But we are still left with an exponential-time
  algorithm

6
Binary tree representation

• Suppose our numbers are3, 8, 9, 17, 26, 39, 43,
  56and our goal is 100
• We can describe thisas a binary tree search

• As we search the binary tree,
• A node is promising if we might be able to get to
  a solution from it
• A node is nonpromising if we know we cant get to
  a solution
• When we detect a nonpromising node, we can prune
  (ignore) the entire subtree rooted at that node
• How do we detect nonpromising nodes?

7
Detecting nonpromising nodes

• Suppose we work from left to right in the
  sequence 3 8 9 17 26 39 43 56
• That is, we try things in the order 0 0 0 0
  0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0
  0 0 0 1 1 0 0 0 0 0 0 ...
• When we get to 0 0 0 0 0 0 1 0 ? 43we
  notice that even if we include all the remaining
  numbers (in this case, there is only one), we
  cant get to 100
• There is no need to try the 0 0 0 0 0 0 1 x
  numbers
• When we get to 1 1 1 1 1 1 0 0 ? 101we
  notice that we have overshot, hence no solution
  is possible with what we have so far
• We dont need to try any of the 1 1 1 1 1 1 x x
  numbers

8
Still exponential

• Even with pruning, the sum of subsets typically
  requires exponential time
• However, in some cases, pruning can save
  significant amounts of time
• Consider trying to find a subset of 23, 29, 35,
  41, 43, 46, 48, 51 that sums to 100
• Here, pruning can save substantial effort
• Sometimes, common sense can be a big help
• Consider trying to find a subset of 16, 20, 28,
  34, 44, 48 that sums to 75

9
Boolean satisfaction

• Suppose you have n boolean variables, a, b, c,
  ..., that occur in a logical expression such as
  (a or c or not f) and (not b or not d or a) and
  ...
• The problem is to assign true/false values to
  each of the boolean variables in such a way as to
  satisfy (make true) the logical expression
• The brute-force algorithm is the same as before
  (0 is false, 1 is true, try all n binary numbers)
• Again, you can do significant pruning, if you
  think about the problem
• Anything you do, you will still end up with a
  solution that is exponential in n

10
Intractable problems

• The technical term for a problem that takes
  exponential time is intractable
• Intractable problems can only be solved for small
  input sizes
• Faster computer speeds will not help
  muchexponential growth is fast
• Bottom line Avoid these problems if at all
  possible!

11
The End