Exhaustion, Branch and Bound
Gary Wong
If you have any question, please ask via
Email: garywong612@gmail.com
MSN: gary_wong612@hotmail.com
Welcome!
• Your 1st training on advanced topics this year
• I assume you have sufficient basic knowledge
Pre-requisites
• Recursion
• Basic concepts of permutation and combination
• Graph Searching (Optional)
Exhaustion
• Check for (almost) all possible cases
• Advantages
– Simple
– Correct in nearly all circumstances
• Provided that you implement correctly
• Disadvantages
– Can be too slow if too many cases (e.g. exponential)
– Implementation is sometimes difficult
When do we use exhaustion?
• Not many possible cases
• No other better methods are in your mind
– Might not be your fault!
– Many computational problems in the world have no
known polynomial-time solutions
– P, NP, NP-hard, NP-completeness…
• Usually when the constraints are very small, you
should think about whether exhaustion is possible
Let me count the ways
• Now you have unlimited supply of coins of values 1,
5, 10, 25 and 50 cents
• Find the number of ways to make up n cents
• 0 <= n <= 99
Let me count the ways
• Have a rough (and conservative) estimation on
number of possible cases
• If only 1 cent can be used, at most how many coins?
• If only 5 cents can be used, at most how many coins?
• …
• A very rough upper bound:
– 100 x 20 x 10 x 4 x 2 = 1600000
• Surely acceptable within 1s
Grid
• Given an N x N white grids (N <= 5)
• You may toggle the grid colors like this (white ->
black; black -> white):
• Given the initial colors of the grids, find the minimum
number of toggling required to make all grids to
become black
Grid
• Observations:
– Toggle twice = Not toggle at all
– Sequence of toggling does not affect the final appearance
of the board
– How many possible sequences of toggling are there?
• Time Complexity: O(2NxN)
• A more efficient algorithm actually exists!
• Hint: its time complexity is O(2N x N2)
Magic triangle
• Given a magic triangle of N layers (1 <= N <= 5),
comprising of N(N+1)/2 nodes
• Integers 1 to N(N+1)/2 are assigned into each node
exactly once, such that:
– The number in a node = absolute difference of the
numbers in its two children nodes
• M numbers (0 <= M <= N(N+1)/2) have filled into
some nodes, fill in the remaining nodes
1
• Time limit: 1s
4
6
2
3
5
Magic triangle
• Let’s fill in the N(N+1)/2 numbers into the circles!
• O((N(N+1)/2 – M)!)
• Worst case?
– N=5
– M = 0, i.e. no numbers are filled
– 15! = 1307674368000 
Magic triangle – 2nd attempt
• Observation:
– Numbers in lower layers can be used to derive the
numbers above!
– Why not just exhausting the lowest layer?
• Time complexity: O(TPN), where T = N(N+1)/2
• 15P5 = 15! / 10! = 360360 
What have you learnt?
• Smart exhaustion vs Stupid exhaustion
• Now I would teach you what “smart” means…
A Classical Problem – Bin Packing
• A number of objects are to be put into some
identical bins
• Find the minimum number of bins required to pack
all objects
Capacity: 50
10
20
30
50
20
Solution 1
• S = Sum of volumes of all objects
• C = Capacity of a container
• Answer = ceil (S / V)
• Wrong!
• Why?
Solution 2
• Sort the objects in ascending order of their volumes
• Put the smaller objects first
• Wrong!
• Why?
The correct solution
• Exhaustion 
BIN PACKING()
if All objects have been put in some bins
then update the best solution if a better one is found
pick a bin B
for each object O
do if volume of O does not exceed remaining space in B
then Put O into B
BIN PACKING()
Remove O from B
Search tree
Empty bins
B1(50)
B1(10)
…
B1(20)
…
B1(10,20)
B1(50); B2(10)
• DFS
• For N objects, O(N!)
…
The cruel reality
• N <= 40
• Performance:
– input1.txt: Accepted
– input2.txt: Accepted
– input3.txt: TLE
• How to speed up?
Back to the search tree…
• Is it necessary to go further here? Why?
Empty bins
B1(50)
B1(10)
…
…
B1(10,20)
B1(50); B2(10)
Suppose you know
here that at most
2 bins are needed
B1(20)
…
Optimization
• Tree Pruning (Branch and Bound!)
• Suppose f is the solution at this moment
• If f >= cur_best, do not go further
• Performance:
– input1.txt: Accepted
– input2.txt: Accepted
– input3.txt: TLE
• Still too slow… 
• Can we do better?
A* Search
• Let h be an estimated lower bound for the number of
bins required to pack all unpacked objects
• If f + h >= best, do not go further
• h is also known as a heuristic function
• But what makes a good function h in this case?
The cruel reality, again
• Performance:
–
–
–
–
input1.txt: Accepted
input2.txt: Accepted
input3.txt: Accepted
input4.txt: TLE
• Further improved now
• Can we do even better?
Equivalent configurations
=
=
Avoid equivalent configurations
• Sort the objects by their volumes before searching
• When putting the objects, you should ensure that:
– The volumes of objects in each bin are in non-decreasing
order, or
– The volumes of smallest objects in all bins form a nonincreasing order
A happy ending
• Performance:
–
–
–
–
–
–
–
–
input1.txt: Accepted
input2.txt: Accepted
input3.txt: Accepted
input4.txt: Accepted
input5.txt: Accepted
input6.txt: Accepted
input7.txt: Accepted
input8.txt: Accepted
• Result: Accepted
Summary
• Tree Pruning
• A*
• Avoid equivalent configurations
Exercises
•
•
•
•
•
1049 Chocolate
1050 Bin Packing
2005 Magic Triangle
2065 Toggle
3031 Trishade