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 Match Editorial
2004 TopCoder Collegiate Challenge
Online Round 3

Wednesday, March 10, 2004

Summary

It must be tournament time. I've never seen so many reds open and abandon multiple problems. Of the fifty advancers, at least fourteen abandoned either the easy or the medium problem. And at the end of the night, seven reds and yellows were left with all three problems open, but unsubmitted. Congratulations to bstanescu, who rode two problems and a challenge to victory! The surprise of the night came when tomek resubmitted the medium, dropping him to 19th for the night, and 2nd in the overall rankings.

# The Problems

Fences
Used as: Division One - Level One:
 Value 250 Submission Rate 70 / 95 (73.68%) Success Rate 68 / 70 (97.14%) High Score Ruberik for 241.06 points (5 mins 30 secs) Average Score 193.18 (for 68 correct submissions)

"Arghh!", you may have thought. "Intersections? Inclusion-exclusion to avoid over counting?" Then you took a deep breath and said to yourself, "No way, it's only an Easy!"

The use of integer coordinates means that the smallest units of pasture are 1 unit on each side. And there are only a million such units under consideration. Just make a 1000-by-1000 array of booleans that say whether each unit square is inside a rectangle or not. For each rectangle, loop through all its unit squares and set them all to true. Note that if a unit square is inside several rectangles it will be set to true several times, but that does no harm. At the end, loop through all the unit squares and, for each unit square that is inside a rectangle, count how many of its four borders touch either an outside square or an edge of the 1000-by-1000 array. Return the total count at the end.

```  for each rectangle do // (x1,y1) and (x2,y2) are opposite corners
for x in min(x1,x2) to max(x1,x2)-1 do
for y in min(y1,y2) to max(y1,y2)-1 do
set inside[x,y] to true
count = 0
for x in 0 to 999 do
for y in 0 to 999 do
if inside[x,y] then
if x==0 or !inside[x-1,y] then count++
if x==999 or !inside[x+1,y] then count++
if y==0 or !inside[x,y-1] then count++
if y==999 or !inside[x,y+1] then count++
return count
```
Note that inside[x,y] represents the unit square whose upper left corner is at coordinates (x,y). Also, notice the use of min and max to take care of the problem that we don't know which corners of the rectangle we have been given, only that they are opposites.

A nifty trick that's worth remembering anytime you have arrays representing these kinds of maps is that you can avoid all the edge checks by putting an extra layer of cells around all edges of the array. In this case, you could use a 1002-by-1002 array and add 1 to all the coordinates. Then the four inner if-statements become

```           if !inside[x-1,y] then count++
if !inside[x+1,y] then count++
if !inside[x,y-1] then count++
if !inside[x,y+1] then count++
```

Now, there can be up to fifty rectangles, each of which can contain up to a million unit squares. That's 50 million unit squares we might touch. And the final loop that counts the fence segments can touch nearly 5 million more unit squares. Doing something 55 million times should raise your timeout antennae, because it is often the case that a few million operations is enough to fill 8 seconds. Ah, but that is a few million complex operations. These are 55 million simple operations and they run in about a second—or even less if you're in a language that doesn't do bounds checks on arrays.

ImageCompress
Used as: Division One - Level Two:
 Value 500 Submission Rate 50 / 95 (52.63%) Success Rate 29 / 50 (58.00%) High Score m00tz for 385.47 points (16 mins 32 secs) Average Score 277.28 (for 29 correct submissions)

It was easy to time out on this problem. The examples did not include any large test cases, not out of deviousness but because the output would not have fit comfortably on the screen. It would have been straightforward to test a large case manually, but many coders (myself included) have gotten into the bad habit of submitting as soon as they pass all the example cases. Tonight, that was fatal.

The simplest implementation is a recursive function that tries all possible decomposition patterns and keeps the best, with memoization to keep it efficient. A key question is how to represent subimages. Although you could actually construct each subimage as strings, it is much simpler and faster to store a single copy of the original image and specify subimages with six numbers:

• row: the topmost row in the subimage
• col: the leftmost column in the subimage
• rowCount: the number of rows in the subimage
• colCount: the number of columns in the subimage
• rowStep: the distance between adjacent rows in the subimage, relative to the original image
• colStep: the distance between adjacent columns in the subimage, relative to the original image
The last two of these numbers allow us to handle the patterns that decompose the image into even and odd rows/columns. Initially, row and col are the top left corner of the original image, and both step sizes are 1. There are four decomposition patterns:
• Left-Right: Sets colCount to half for the left subimage and colCount-half for the right subimage, where half is (colCount+1)/2. Also sets col to col+half*colStep for the right subimage. All other values stay the same.
• Upper-Lower: Sets rowCount to half for the upper subimage and rowCount-half for the lower subimage, where half is (rowCount+1)/2. Also sets row to row+half*rowStep for the lower subimage. All other values stay the same.
• Even-Odd Columns: Sets colCount to half for the even subimage and colCount-half for the odd subimage, where half is (colCount+1)/2. Also sets col to col+colStep for the odd subimage, and colStep to 2*colStep for both subimages. All other values stay the same.
• Even-Odd Rows: Sets rowCount to half for the even subimage and rowCount-half for the odd subimage, where half is (rowCount+1)/2. Also sets row to row+rowStep for the odd subimage, and rowStep to 2*rowStep for both subimages. All other values stay the same.
For each image, you find the best encodings for each of the eight subimages, combine them into the best encodings for each of the four decomposition patterns, and pick the best of these as the encoding for the whole image. Because you can reach the same subimage by different paths—for example, splitting horizontally and then vertically produces the same subimages as splitting vertically and then horizontally—you memoize the results for each subimage to avoid recomputing them over and over again.

Dynamic programming and memoization are usually interchangable, so it is instructive to consider why dynamic programming does not work very well for this problem. In dynamic programming, you usually use a few loops to generate all the subproblems from smallest to largest. But what should those loops look like here?

Consider the rowCount value. How should a loop generate the possible values of rowCount? It would be easy to write a loop from 1 to the number of rows in the original image. However, this would be wasteful because not all of those values are actually needed. For example, suppose the original image had 21 rows. Then various subimages have 1, 2, 3, 5, 6, 10, 11, or 21 rows, but not 4, 7, 8, 9, or 12-20 rows. A dynamic programming solution would have to work very hard indeed to avoid computing answers for subimages with invalid numbers of rows. (Here is a challenge for the round tables: write a simple loop that takes a number of rows and produces the sequence of the numbers of rows that occur in subimages, from smallest to largest!) On the other hand, a recursive function augmented with memoization naturally considers only those values that are valid. It is no extra work to filter out the unwanted values—in fact, it would take substantial extra work to include the unwanted values if we decided we wanted them for some perverse reason.

And yes, the last example case was indeed a black heart, which some took as evidence of the writer's character.

Decaffeinated
Used as: Division One - Level Three:
 Value 1000 Submission Rate 2 / 95 (2.11%) Success Rate 0 / 2 (0.00%) High Score null for null points (NONE) Average Score No correct submissions

Don't be fooled. This problem was hard, but not as hard as the submission stats suggest. People spent so much time on the other two problems that they ran out of time on this one.

In this problem, you had to find the shortest path through a state space. I guess you could break out the big shortest-path guns, like Dijkstra's algorithm, but that would be silly when breadth-first search is all you need.

A state includes at most 5 numbers: the amount of coffee in each serving cup and the amount (and type) of coffee in each measuring cup. For the measuring cups, I used negative numbers to indicate regular coffee and positive numbers to indicate decaffeinated coffee. You could also use a boolean flag to distinguish between regular and decaffeinated, but then you would have to be very careful how you handled an empty cup. For example, you would not want to avoid pouring regular coffee into an empty measuring cup just because its boolean flag said decaffeinated.

The tricky part about this problem is handling all the possible state transitions, because a single transition can involve up to three simultaneous pouring actions. Handling all the possible cases here (to and from the reservoirs, to the serving cups, or between measuring cups, with one, two, or three simultaneous actions) probably accounted for most of the time and code spent on this problem.

In breadth-first search, you usually use a queue. You begin by putting the initial state (all cups empty) into the queue, and then go into a loop where you repeatedly pull a state out of the queue and put all the new states reachable from that state at the end of the queue, stopping when either you find the end state (both serving cups full) or the queue becomes empty, in which case there is no way to get to the end state.

To keep the search efficient, it is important to avoid putting states in the queue if you've already seen them before. There might be several hundred thousand to a few million states, so you have to be careful how you keep track of what states you've seen. A hashtable or tree runs the risk of taking up too much memory and/or being too slow, although the constraints were small enough that you could probably get away with it. A big bitvector works well, because even the maximum of 4.5 million possible states can be represented in about half a megabyte.

Once you find the end state, you still need to return the number of seconds it took to get there. You could add this information to each state, but an easier approach is to rethink the use of a queue. Instead of one queue, you can use two queues: one that contains the states reachable in the current second, and one that contains the states reachable in the next second. You then take states out of the queue for the current second, and add new states to the queue for the next second. When the queue for the current second becomes empty, you increment a clock variable and swap the two queues so that the queue for the next second becomes the queue for the current second, and the queue for the current second (which is empty) becomes the queue for the next second. When you find the end state, you simply return the clock variable. One pleasant feature of this approach is that the queues don't need to be queues anymore. Stacks would work just as well—the order in which states are removed from the data structure no longer matters.

Finally, there is a cute trick for determining that the end state is going to be unreachable without searching at all. Take the GCD of the measuring cup sizes and if it does not divide evenly into both serving cup sizes, then the end state is unreachable.

By vorthys
TopCoder Member