1. Atlantis HKOI2005 Final Event (Senior Group)

2. 2 Background 9000 B.C. 9000 B.C. Atlantis Zeus Other gods Destroy!!

3. 3 Problem There are N pairs of (island + controller) There are N pairs of (island + controller) In each pair, the island is K meters above the controller for some fixed positive integer K In each pair, the island is K meters above the controller for some fixed positive integer K Given the heights of the 2N objects, which of them are controllers? What is K? Given the heights of the 2N objects, which of them are controllers? What is K?

4. 4 Model Two integer sequences Two integer sequences A = (a 1, a 2, …, a N ), B = (b 1, b 2, …, b N ) are parallel if for some positive integer K, a 1 = b 1 +K, a 2 = b 2 +K, …, a N = b N +K Given a collection C of 2N positive integers, partition it into 2 parallel sequences of size N, if possible Given a collection C of 2N positive integers, partition it into 2 parallel sequences of size N, if possible

5. 5 Smaller problems How to determine if two sequences are parallel? How to determine if two sequences are parallel? Trivial – O(N) Trivial – O(N) Given two collections of size N, is it possible to form a sequence of size N from each of the sets such that the two sequences are parallel? Given two collections of size N, is it possible to form a sequence of size N from each of the sets such that the two sequences are parallel? Sorting – O(NlgN) (O(N) for count sort) Sorting – O(NlgN) (O(N) for count sort) 1 35 8 9 73 5 1 3 5 8 3 5 7 10 +2

6. 6 Algorithm 1 Idea Idea Partition C into two collections of size N Partition C into two collections of size N Determine if two parallel sequences can be formed from the two collections (previous slide) Determine if two parallel sequences can be formed from the two collections (previous slide) 2N C N ways to partition C 2N C N ways to partition C Exponential Exponential Expected score: 50% Expected score: 50%

7. 7 Observations Out of the 2N C N different partitions, most of them should not be tested, for example Out of the 2N C N different partitions, most of them should not be tested, for example {1, 2, …, 2N}  {1, 4, ??, …, ??}, {2, 3, ??, …, ??} {1, 2, …, 2N}  {1, 4, ??, …, ??}, {2, 3, ??, …, ??} Why? Why? {1, 2, …, 2N}  {1, 2, 4, ??, …, ??}, {3, ??, …, ??} {1, 2, …, 2N}  {1, 2, 4, ??, …, ??}, {3, ??, …, ??} The smallest integer from one collection and the smallest integer from another collection form a pair The smallest integer from one collection and the smallest integer from another collection form a pair

8. 8 Almost there… Suppose we know the smallest pair, can we reconstruct the partition? Suppose we know the smallest pair, can we reconstruct the partition? Fact: The smallest pair must contain the smallest integer in the C Fact: The smallest pair must contain the smallest integer in the C

9. 9 6+2 = 87+2 = 9 67 An Example Given C = {2, 6, 1, 3, 7, 4, 8, 9} and suppose we are told that (1, 3) is the smallest pair Given C = {2, 6, 1, 3, 7, 4, 8, 9} and suppose we are told that (1, 3) is the smallest pair 2 6 1 3 7 4 8 9 (1, 3) Smallest remaining integer = Its partner = 2 2+2 = 4 (2, 4) (6, 8) (7, 9)

10. 10 Algorithm 2 Idea Idea Suppose (a, b) is the smallest pair; let d = b – a Suppose (a, b) is the smallest pair; let d = b – a Repeat N times Repeat N times Let s be the smallest integer in C Let s be the smallest integer in C If s+d is not in C, return failure If s+d is not in C, return failure Otherwise (s, s+d) is a pair, remove s and s+d from C Otherwise (s, s+d) is a pair, remove s and s+d from C For how many times do we need to carry out the above steps? For how many times do we need to carry out the above steps?

11. 11 Algorithm 2 - Analysis How many different “smallest pairs” should be tried? How many different “smallest pairs” should be tried? N (smallest, 2 nd smallest), (smallest, 3 rd smallest), …, (smallest, (n+1) th smallest) (smallest, 2 nd smallest), (smallest, 3 rd smallest), …, (smallest, (n+1) th smallest)

12. 12 Algorithm 2 – Analysis (2) The loop takes O(N 2 ) time The loop takes O(N 2 ) time Overall time complexity is O(N 3 ) Overall time complexity is O(N 3 ) Too slow when N = 1000 Too slow when N = 1000 Observation: The pairs we get are increasing Observation: The pairs we get are increasing Both coordinates are increasing Both coordinates are increasing (1, 3) ≤ (2, 4) ≤ (6, 8) ≤ (7, 9) (1, 3) ≤ (2, 4) ≤ (6, 8) ≤ (7, 9)

13. 13 2+2 = 4 2 6+2 = 8 6 7+2 = 9 7 Algorithm 2A Sort the integers in C Sort the integers in C Keep two pointers (integers) Keep two pointers (integers) 1 2 3 4 6 7 8 9 (1, 3) Smallest remaining integer = Its partner = (2, 4) (6, 8) (7, 9) AB

14. 14 Algorithm 2A The loop takes O(N) time The loop takes O(N) time Overall time complexity is O(N 2 ) Overall time complexity is O(N 2 ) Not a problem even when N = 10000 Not a problem even when N = 10000

15. 15 Algorithm 3 (Out of Syllabus) For trainees with graph theory background For trainees with graph theory background Fix a height difference, construct a graph, then find a maximum bipartite matching Fix a height difference, construct a graph, then find a maximum bipartite matching Example: Height difference = 2 Example: Height difference = 2 Perfect matching Perfect matching  Feasible Overall time complexity: Overall time complexity: O(N 3.5 ) or O(N 3 ) 1 2 4 3 6 7 8 9

16. 16 Extensions and Mutations Constraints: Constraints: Real numbers instead of integers Real numbers instead of integers Height sums instead of height differences Height sums instead of height differences Height differences can lie in a range instead of a fixed number Height differences can lie in a range instead of a fixed number Objectives: Objectives: Minimize sum of height differences Minimize sum of height differences Minimize maximum height difference Minimize maximum height difference