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3.12 ≠ π

3.12 ≠ π. What an inspiring title. PotW Solution - Bovinekiin.

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3.12 ≠ π

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  1. 3.12 ≠ π What an inspiring title

  2. PotW Solution - Bovinekiin int stall = 0;for (inti = 0; i < 90001; i++) {inta = rand.nextInt(n);intb = rand.nextInt(n); Item temp = cur.get(a);cur.set(a, cur.get(b));cur.set(b, temp);if(calcGreedy(cur) > calcGreedy(best)) { best = newArrayList<Item>(cur); // update best } else { cur = newArrayList<Item>(best); // restore best stall++;if(stall == 9001) Collections.shuffle(cur); }}printSolution(best); • Full solution at http://ideone.com/yHPi6

  3. About Harker • Deadline to register was last Saturday • You probably received an email with info if you’re registered • Read it! • Schedule • 8:30-9:30 Check-in • 9:30-10:30 Speaker • 10:30-10:40 Break • 10:40-11:00 Diagnostics/Introduction to competition infrastructure • 11:00-12:30 Competition • 12:30-1:45 Lunch • 1:45-2:00 Set-up for Challenge Round • 2:00-2:30 Challenge Round • 2:30-3:00 Awards and Raffle

  4. March USACO I know this title is off-center, but that’s the way this slide theme works.

  5. Bronze – Times 17 • Given a number N, print 17 * N. • The catch: Input and output are in binary, and N may have up to 1000 digits • The solution: 17N = 16N + N • 16N in binary is just N followed by four 0s • Simple binary addition to add 16N to N

  6. Silver – Flowerpot • Given the locations of N raindrops in the 2D plane, compute the minimum width of a flowerpot along the x-axis such that the vertical distance between the first and last raindrop to hit the flowerpot is at least some value D. • 1 <= N <= 100000 • 1 <= D <= 1000000

  7. Flowerpot – Solution • Sort all points by x-value, “sweep” a pair of vertical lines from left to right • Store y-values of points between sweep lines in a data structure that can easily find min and max • Such as pair of priority queues • Whenever difference btwn. min and max is at least D, check if this is best (min.) flowerpot width so far and advance left sweep line • If not at least D, advance right sweep line

  8. Gold – Skyscraper • Given an elevator with a weight capacity, what is the minimal number of trips that will be needed to deliver n < 18 cows with certain weights from one floor of a skyscraper to the next. n < 18, so the solution is probably exponential in time • Use dynamic programming • The state is a bitmask of n bits: O(2n) • ith bit on: ith cow delivered, off: ith cow not delivered • The DP function maps the bitmask to a pair of integers: • First, what's the minimum # of occupied elevators that must be used to carry these cows • Second, what's the minimum amount of weight in the last elevator given this minimal # of elevators

  9. Skyscraper – Dynamic Programming • This DP function only takes into account the weight of last elevator (weights in the other elevators are ignored) • Minimize # of occupied elevators first • Then minimize amount of weight in last elevator • Basically just find the minimal pair lexicographically • Base Case: • DP(all 0s) = (0, 0) • Recurrence: • Try adding each non-used cow (of weight W) to the last elevator (adds insignificant factor of O(n)) • If it overflows the capacity, add a new elevator, and let the last elevator have weight W • If it doesn't, add W to the last elevator

  10. No PotW this week! • Have fun at Harkerthis Saturday!

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