1.00 Tutorial 10
4/29/2002
PS#9 Corrections on web site
• Pruning rule #2 updated
(should be “<=“, not “<“)
• Diagram corrected
1
PS#9: What's the point?
Given (1) a series of stereo components of varying lengths,
2
3
5
10
15
and (2) a pallet of width M,
M=20
How many ways can we pack some subset of the stereos,
side-by-side, so they they exactly fit on the pallet?
PS#9: What's the point?
For this example, there are three solutions
2 3
5
10
20
2 3
15
5
20
15
20
But how do we solve this problem in general?
2
PS#9: Tree Algorithm
s=Sum of lengths in this solution
r=Sum of lengths to be considered
2 3 5 10 15
M=20
x0=1
(0,35)
x0=0
2 3 5 10 15
2 3 5 10 15
(2,33)
x1=1
2 3 5 10 15
(5,30)
x2=1
2 3 5 10 15
x2=0
2 3 5 10 15
(0,33)
x1=0
2 3 5 10 15
(2,30)
x2=1
2 3 5 10 15
x2=0
2 3 5 10 15
x1=1
x1=0
2 3 5 10 15
(3,30)
x2=1
2 3 5 10 15
x2=0
2 3 5 10 15
2 3 5 10 15
(0,30)
x2=1
2 3 5 10 15
x2=0
2 3 5 10 15
(10,25) (5,25) (7,25) (2,25) (8,25) (3,25) (2,25) (0,25)
plus two more levels
PS#9 Tree Algorithm
•
•
•
•
Enumerates all combinations
Grows exponentially: O(2n)
Solution found whenever s=M
x[ ] is the current solution, takes on values of 0 if
length Lk is not in solution, or 1 if it is.
• Pruning rules reduce number of nodes by
“pruning” subtrees. (Note that our “pruning
rules” are actually “subtree generation rules”,
since the action taken is to generate nodes, not
prune existing ones.)
3
Pruning Rule #1
• If (s + Lk = M), output solution (and don’t
follow any subtrees!)
M=20
x2=1
Lk=10
2 3 5 10 15
S=10
(10,25)
x3=1
x3=0
2 3 5 10 15
Solution! (20,15)
x4=1
2 3 5 10 15
(10,15)
x4=0 x4=1
x4=0
Prune subtree
Note that Lk is the
next length to be
considered, not the
current one.
Pruning Rule #2
• If (s + Lk + Lk+1 <= M), generate left child
and set xk = 1 (otherwise do not!)
Lk=3
M=20
S=2
Lk+1=5
2 3 5 10 15
(2,33)
x1=1
2 3 5 10 15
(5,30)
2+3+5 <= 20,
so we generate a left child
Note that Lk is the
next length to be
considered, not the
current one.
4
Pruning Rule #3
• If (s + r - Lk >= M) && (s + Lk+1 <= M) generate
right child and set xk = 0
• Part 1: if we were to generate a right node, then
the maximum lengths we could add to our
current solution would be r-Lk. If that amount is
< M, we can never fill the pallet, so we might as
well stop now.
• Part 2: If the following length (Lk+1) is greater
than the pallet width, we also cannot use it and
can stop now as well.
Note that Lk is the
next length to be
considered, not the
current one.
Exercise
• Apply the 3 pruning rules to decide
whether or not to generate left & right
nodes when M=25, L= {10, 12, 13, 30}
(1) For root node (k=0)
(2) When k = 2, x={0, 1}
(3) When k = 2, x={1, 1}
Solution on next slide…
5
Exercise Solution
1. Left and right nodes generated. (Rules 2
& 3)
2. Solution found via rule 1. No more
nodes generated
3. Neither right nor left generated. (Rules 2
& 3.) This is a dead end.
PS#9 Implementation
• Input read from command line
(String[ ] argv). E.g.,
M
L
DOS> java SubsetTree 20 2 3 5 10 15
• Write a SubsetTree class as your main class.
It should create the root node and call
root.visit()
• Suggested implementation is to build a tree
using a Node class.
– Each node should carry a solution array
(int x[ ]) of size L.length. How should it be initialized?
– What other variables/methods will you need?
6
Node’s visit() method
• visit() method does all the work:
– Implements pruning rules
– Creates left & right child nodes, and calls visit
recursively (left.visit(), right.visit()) as
appropriate
– Sets x[i] to 0 or 1
– Prints solutions when found (a
printSolution method is handy to have)
Other solutions
• Many solutions are possible
• You don’t even need to create nodes (can
use recursion alone to traverse an implicit
tree)
7