Get Time
statistics_w  Match Editorial
2005 TopCoder Collegiate Challenge
Online Round 3

January 26, 2005


Round 3 proved to be leaps and bounds harder than round 2. Only 3 coders submitted the easy problems in the first 10 minutes. In fact, it was so hard that tomek had to resubmit the easy problem once, and the medium problem twice. reid, a regular at most onsite events didn't make it past the first problem during the coding phase. After an agonizing 75 minutes, however, 90 coders had scraped together solutions for the easy, while 62 had solutions for the medium.

However, about two thirds of all the submissions couldn't make it past the system tests, and no one was able to solve all three problems successfully. Im2Good ended up with the highest score, based solely on his submission for the hard problem. Eryx was closest to solving all three, but his medium was challenged, and he ended up in second. In third place, John Dethridge had the highest score from the first two problems.

tomek, the defending champ, had a good showing, taking a respectable 8th place. reid the third highest rated coder in the round, was the only target not to advance, but was in good company as antimatter, m00tz, pparys, and a slew of other red coders didn't make the cut. From the other end of the spectrum, kindloaf was the lowest rated coder to advance, while dexy was the coder who advanced with the worst odds (9.07% according to radeye). One final noteworthy event was that this was the first time the tiebreaker rules were applied to determine advancers, as there was a 6-way tie for 46th place - with 50 points.

I guess that the lesson from this round, if one was inclined to take one, was that when the problems are hard, it's often worth a bit of extra time to make sure you get them right. With harder problems, correct solutions are more important that fast solutions.

The Problems

AutoAdjust discuss it
Used as: Division One - Level One:
Value 250
Submission Rate 90 / 97 (92.78%)
Success Rate 30 / 90 (33.33%)
High Score misof for 214.37 points (11 mins 59 secs)
Average Score 162.54 (for 30 correct submissions)

The algorithm for adjusting the brightness and contrast of a particular image boiled down to a fairly simple formula:

    newColor = (oldColor - 63.5 + brightness) * contrast / 100 + 63.5
The idea is that you are first adjusting the brightness, and then pushing the color values away from grey so that there ends up being more of a difference between the different colors (or shades of grey) in the image. This problem asks you to find values for the brightness and constrast that cause the darkest colors to be black, and the brightest colors to be white. The idea here is that you want to use the full range of available colors, but not lose any details by adjusting the contrast too much.

If you understand what adjusting the brightness and contrast is doing, it should be clear that the brightest pixels will remain the brightest pixels after making the adjustment, and that the darkest pixels will also remain the same. So, to find the values of the brightness and contrast that provide the adjustment you need, you only need to know the color of the brightest pixel, and the color of the darkest pixel. From here, you could determine the values of brightness and contrast by solving a system of two equations with two unknowns:
    (darkest - 63.5 + brightness) * contrast / 100 + 63.5 = 32
    (brightest - 63.5 + brightness) * contrast / 100 + 63.5 = 95
Rather than doing the algebra, the symmetry of the problem dictates that the brightness should be selected so that after adjusting the brightness, the darkest and brightest color values are equidistant from 63.5. Once you know the brightness, its easy to figure out the constrast:
    brightness = 63.5 - (darkest + brightest) / 2
    contrast = (32 - 63.5) * 100 / (darkest - 63.5 + brightness)
One difficulty with this is that solving these equations will not necessarily give you integers for brightness and contrast. Additionally, while a contrast of 200, for example, might be the value needed to give 32 and 95, a contrast of 199 might give 32.4 and 94.6, which then round to 32 and 95. The problem statement indicates that, in this case, you should use the smaller contrast - 199. While you can work around all of these issues with a bit of cleverness, it's a lot harder than the problem needs to be.

Instead of doing all this math, most coders noticed that the brightness only ranges from -100 to 100, and the contrast only ranges from 100 to 20,000, which means that there are only about 4,000,000 different combinations, few enough to just try them all, and see which ones have the desired properties. In fact, there is no reason to use a brightness less than -31 or greater than 31, or a contrast greater than 6201. So, we can simply iterate over all contrasts, starting at 100, and for each contrast iterate over all brightnesses. By designing our loops properly, we can assure that we find the smallest contrast that works first, and with that the smallest absolute brightness. Then all that remains is to adjust each pixel in the image, and return the result.

PackageShipping discuss it
Used as: Division One - Level Two:
Value 500
Submission Rate 62 / 97 (63.92%)
Success Rate 23 / 62 (37.10%)
High Score overwise for 328.52 points (23 mins 15 secs)
Average Score 240.35 (for 23 correct submissions)

There are a couple of very different ways to approach this problem. One way requires dynamic programming or memoization, coupled with a breadth first search, or something similar. The second is simple brute force. I'm going to start with the harder solution, which scales quite a bit better, and come back to the brute force approach at the end. As in any dynamic programming problem, the question you should ask is, what are my states? In this case, a state is a combination of a location, a time, and a cost. For each such triple, you want to find the route to that state that has the minimum probability of damaging the package. To compute this minimum, you should perform a breadth first search. Starting at the start state, you should calculate all states that are reachable from the start state - these correspond to the various routes leaving the start location. Given the probability, p, corresponding to a particular (location, cost, time) and a route to (location -> destination, route_cost, route_time, probability), you can then reach the state (destination, cost+route_cost, time+route_time) with a damage probability of 1 - (1-p) * (1-probability). So, if this probability is better than the current probability for that state (all states are initialized to 1), then update the probability for that state and add the state to a list of states that still need to be processed. To make this really fast, you should use a priority queue so that you process the states in the queue with the better probabilities earlier. As in Dijkstra's shortest path algorithm, this will guarantee that you never look at a state more than once. However, in this problem, that was not necessary, and a simple first in first out queue was sufficient.

  queue q
  map<state -> probability> best
  start = state(origin,0,0,0) 
  insert start into q
  best(start) = 0
  while(q is not empty)
    s = q.remove_first()
    if(s.prob != best(s)) continue//this helps to avoid doing extra work
    foreach (route r with r.origin = s.location)
      state next(r.destination, s.cost + r.cost, s.time + r.time, 1-(1-s.prob)*(1-r.prob))
      if(next.time ≤ MAX_TIME and next.prob < best(s))
        insert next into q
Once you have the best probabilities for all of the states, you should consider each state that is at the destination, and compute the expected cost to reach that state: the cost of the state plus the package cost times the probability of damaging the package. Once you find the minimum, simply return that value. There are a couple of caveats to this problem. One thing to watch for is that you don't run out of memory, depending on your implementation details. If you remove routes that you know you won't use, it can help you for some pathological cases.

Now, for the promised brute force solution. Since there are only 50 routes, there are not that many different paths from the origin to the destination. It seems like there could be a lot, and in fact there are an exponential number of them, but with only 50 routes there are few enough paths that you can consider them all - so long as you don't go around in circles. So, the basic algorithm is a quite simple recursive function:
  map<location -> bool> visited
  best = INFINITY
  recurse(location, cost, time, prob)
    if(location == FINAL_DESTINATION)
      total = cost + prob * PACKAGE_COST
      if(total < best)
        best = total
    foreach (route r with r.origin = location)
      if(!visited(r.destination) and time + r.time ≤ MAX_TIME)
        visited(r.destination) = true
        recurse(r.destination, cost+r.cost,time+r.time,1-(1-prob)*(1-r.prob))
        visited(r.destination) = false
Surprisingly, the most recursive calls you will need to make with this algorithm is comfortably under a million, and in Java it runs in under a second on all inputs. However, the runtime is at least as bad as O(φN/2), where N is the number of routes, and φ is the golden ratio. There are also certainly cases that have a worse runtime, but their runtimes are harder to evaluate, and I'm not sure what the worst case is. I'd be interested to see how bad a case someone can construct with 50 edges.

DFAReversal discuss it
Used as: Division One - Level Three:
Value 1000
Submission Rate 3 / 97 (3.09%)
Success Rate 2 / 3 (66.67%)
High Score Im2Good for 895.25 points (9 mins 57 secs)
Average Score 728.81 (for 2 correct submissions)

First, a bit of background for this problem. There are two different kinds of finite automata - deterministic and non-deterministic. The determinisitc ones are described in the problem statement, and for every state there is exactly one outgoing edge corresponding to each symbol in the alphabet. In a non-deterministic finite automata, this is not the case. A state may have multiple outgoing edges corresponding to the same symbol, or 0 outgoing edges for a symbol. In the case of multiple edges, when we reach a state, we can follow any one of the edges corresponding to the next symbol in the input sequence. If there are no outgoing edges, then the sequence is rejected. The NFA accepts a sequence if and only if there is some way to choose which edge to follow at each point where a decision must be made such that it ends up in an accept state. For example, consider an NFA with a single symbol in its alphabet, with states 0 and 1, and with two edges - one from 0 to 0 and another from 0 to 1. If the start state is 0, and 1 is an accept state then this NFA will accept any sequence of 1 or more symbols. So long as we choose to follow the loop edge from 0 to 0 for every symbol except the last, we will end up in state 1 at the end - an accept state.

If we had followed the wrong edge at any point, we would not have ended up in an accept state. However, by definition the NFA accepts the input as long as there is some way to choose which edges to follow that lead to acceptance. The problem, it might seem, is that if you have to consider all possible paths, the time it takes to determine whether a sequence is accepted or not would be painfully long. However, a bit of dynamic programming can help us out here. Rather than considering all paths, we can keep track of the set of states that we could be in after some sequence of symbols. For example, if the start state is 0, then after 0 symbols, our set is {0} - we can only be in one state at the start of execution. If, there are edges from state 0 to states 0, 1 and 3 corresponding to the first symbol in a sequence, then after the first symbol in the sequence, we could be in any of those three states, which we represent as the set {0,1,3}. Now, lets say that there are edges from state 1 to state 2 and from state 3 to state 4 corresponding to the next symbol in the input. That would mean that from the states {0,1,3} we could reach states {2,4}, so after two symbols, considering all possible paths, we could be in either state 2 or state 4. In general, we are keeping track of the set of states that we could be in after the first K symbols. To compute the set for the first K+1 symbols, we need only see which states are reachable on symbol K+1 from one of the states reachable after the first K symbols. This is the basis of any regular expression matcher. The regular expression is typically converted to an NFA, and then the input sequence is evaluated using sets of states as described here.

    //Code to compute whether a sequence is accepted by an NFA
    current = {start}
    for(K = 1 to length(sequence))
        s = sequence[K]
        next = ∅
        foreach (state u ∈ current)
            foreach (edge (u,v) corresponding to s)
                next = next ∪ {v}
        current = next
    if(current ∩ accept ≠ ∅)
        return true
        return false
This suggests an algorithm to convert an NFA into a DFA. First, create a new DFA where each state in the DFA corresponds to a set of states in the NFA. Then, for each set of states, U, and each symbol, s, compute the set of states, V, that are reachable from a state in U on s, and add an edge from U to V on s. The start state in the DFA is {start}, and each state U in the DFA where U ∩ accept ≠ ∅ is an accept state. Unfortunately, there are 2N states in this DFA, where N is the number of states in the NFA, but this is the best we can do in some cases. The reason this works is that traversing the states in the constructed DFA is exactly the same as traversing sets of states in the NFA, as described above.

Now, back to the problem (finally!). You are given a DFA, M, and want to find a DFA, M', that accepts the reverse of each sequence accepted by M. The simplest way to do this is simply to reverse all the edges in M. This gives you an NFA. You then set your start state to the set of all accept states in M (we're going to be converting the NFA to a DFA later, so its ok to have multiple start states for now). The accept state in the NFA is identical to the start state in M. If a sequence is accepted by M, then there is a path from the start state in M to one of the accept states. Therefore, with all the edges reversed, there is a path from one of the accept states in M back to the start state in M on a sequence s if and only if there was a path from the start state in M to an accept state on reverse(s).

Now, from the NFA, we can construct a new DFA, using the subset construction described. The only problem is that we need to generate a minimal DFA - one with as few states as possible. It turns out that there is a very simple way to do that in this case. Starting from the start state in the reversed DFA (which is the set of all accept states in the input) do a simple depth first search, to find all the sets of states in the reversed DFA that are reachable from the start state. Clearly, if a state is not reachable from the start state, it can be removed, but proving that this DFA is minimal requires a bit more insight. The first step in this proof is to note that any DFA can be made minimal by first finding all pairs of states such that for every sequence s, the outcome (accept or reject) will be the same regardless of which of the two states you start from. Any such pair of states can be merged into a single state, while any pair of states for which this is not true may not be merged.

Now, consider two sets of states in the input DFA (M), u and v, representing states in the reversed DFA (M'). u and v can be merged in M' if and only if every sequence in M' has the same outcome (accept or reject) when leaving either u or v. What this means in relation to M is that for every sequence s, there is a path backwards from one of the states in u and also backwards from one of the states in v that goes to the start state in M. However, if there is a path backwards from state x ∈ u to the start state on s, that means that M goes from the start state to x on rev(s), as it is deterministic. Therefore, it must be true that x ∈ v also, since there must be a path backwards from a state in v to the start state on s. This leads to the conclusion that u and v must in fact be the same (assuming all states in M are reachable).

At long last, we can implement an algorithm to solve the problem, which turns out to be quite simple:
    map<state -> bool> visited
    dfs(DFA M, state cur)
        if(visited(cur))return 0;
        visited(cur) = 0
        ret = 1
        foreach (symbol s)
            state next = ∅
            foreach (u ∈ cur)
                foreach (edge (u,v) corresponding to s)
                    next = next ∪ {v}
            ret = ret + dfs(M,next)
        return ret
    reverse(DFA M)
        start = M.accept
        return dfs(M,start)
As an implementation detail, the state can best be implemented as a bitmask, in which case visited can be a simple array. The solution can be under 25 lines long, without doing anything too clever.

By lbackstrom
TopCoder Member