Exhaustion, Branch and BoundGary Wong

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Email: garywong612@gmail.com

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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

• Time limit: 1s

6

4

2

3

1

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

10

20

30

50

20

Capacity: 50

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 foundpick a bin Bfor each object Odo if volume of O does not exceed remaining space in B

then Put O into BBIN PACKING()Remove O from B

Search tree

• DFS• For N objects, O(N!)

Empty bins

B1(10) B1(50) B1(20)

B1(10,20)

B1(50); B2(10)

… … …

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(10) B1(50) B1(20)

B1(10,20)

B1(50); B2(10)

… … …

Suppose you knowhere that at most2 bins are needed

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 non-

increasing 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