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Exhaustion, Branch and Bound Gary Wong If you have any question, please ask via Email: garywong612@gmail.com MSN: gary_wong612@hotmail.com

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Welcome! Your 1 st training on advanced topics this year I assume you have sufficient basic knowledge

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Pre-requisites Recursion Basic concepts of permutation and combination Graph Searching (Optional)

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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

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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

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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

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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

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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

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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(2 NxN ) A more efficient algorithm actually exists! Hint: its time complexity is O(2 N x N 2 )

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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

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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

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Magic triangle – 2 nd attempt Observation: – Numbers in lower layers can be used to derive the numbers above! – Why not just exhausting the lowest layer? Time complexity: O( T P N ), where T = N(N+1)/2 15 P 5 = 15! / 10! = 360360

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What have you learnt? Smart exhaustion vs Stupid exhaustion Now I would teach you what “smart” means…

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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

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Solution 1 S = Sum of volumes of all objects C = Capacity of a container Answer = ceil (S / V) Wrong! Why?

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Solution 2 Sort the objects in ascending order of their volumes Put the smaller objects first Wrong! Why?

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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

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Search tree DFS For N objects, O(N!) Empty bins B1(10) B1(50)B1(20) B1(10,20) B1(50); B2(10) … … …

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The cruel reality N <= 40 Performance: – input1.txt: Accepted – input2.txt: Accepted – input3.txt: TLE How to speed up?

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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 know here that at most 2 bins are needed

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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?

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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?

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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?

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Equivalent configurations = =

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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

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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

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Summary Tree Pruning A* Avoid equivalent configurations

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Exercises 1049 Chocolate 1050 Bin Packing 2005 Magic Triangle 2065 Toggle 3031 Trishade

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