Transcript Pruning
Pruning
26-Jul-16
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
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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 3
saves 7
saves 15
The higher up in the tree we can prune, the more effort we
can save
The advantage is exponential
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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
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
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0
0
0
0
0
0
0
0
0
0
0
0
0
5
0
0
0
0
0
0
0
1
1
1
1
1
1
7
0
0
0
0
1
1
1
0
0
0
0
1
1
8 13
0 0 0
0 1 13
1 0 8
1 1 21
0 1 20
1 0 15
1 1 28
0 0 5
0 1 18
1 0 13
1 1 26
0 0 12
0 1 25
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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
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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
As we search the binary tree,
3 (yes or no)
8 (yes or no)
9 (yes or no)
17 (yes or no)
etc.
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?
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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 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 0 0 0 0 0 0 1 x numbers
When we get to
1 1 1 1 1 1 0 0 101
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 1 1 1 1 1 1 x x numbers
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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
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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
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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!
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The End
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