1. Pruning
24-Feb-19

2. Exponential growth
How many leaves are there in a complete binary tree of depth N?
depth = 0, count = depth = 1, count = depth = 3, count = depth = 4, count = depth = 5, count = 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
saves 15
saves 7
saves 3
The higher up in the tree we can prune, the more effort we can save
The advantage is exponential

4. Sum of subsets Problem: Example:
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
             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 are 3, 8, 9, 17, 26, 39, 43, 56 and our goal is 100
We can describe this as a binary tree search
3 (yes or no) (yes or no) (yes or no) (yes or no) etc.
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 can’t 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
That is, we try things in the order
When we get to  43 we notice that even if we include all the remaining numbers (in this case, there is only one), we can’t get to 100
There is no need to try the x numbers
When we get to  101 we notice that we have overshot, hence no solution is possible with what we have so far
We don’t need to try any of the 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 much—exponential growth is fast
Bottom line: Avoid these problems if at all possible!

11. The End