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statistics_w  Match Editorial
SRM 178
Wednesday, January 7, 2004

Match summary

Last night's dynamic programming course began with an excursion into the land of typesetting. After coders completed the Div 1 easy (Div 2 medium), and were quite happy to be through with it, they moved on to a sequence of DP/Memoization problems. Div 2 coders faced a short, but tricky problem on strings, with interesting mathematical properties. Div 1 coders faced a sequence of DP problems, the second of which required 2 levels of DP. radeye and John Dethridge, clearly experts at all things DP, lead the pack. Each of them conducted a challenging lecture later on, preying on many of the Div 1 easy submissions. In Division 2, the newcomer matmis proved he wasn't new to programming. Newcomers win division 2 so frequently, other competitors must be cautious competing with them. After all, tomek and SnapDragon were newcomers once.

The Problems

SimpleCalculator discuss it
Used as: Division Two - Level One:
Value 250
Submission Rate 214 / 236 (90.68%)
Success Rate 208 / 214 (97.20%)
High Score Krumble for 248.75 points (2 mins 1 secs)
Average Score 202.90 (for 208 correct submissions)

The major issue with this problem, was correctly parsing the input. Once complete, the actual computation was trivial. The C++-ish solution uses a single sscanf statement, or a few stringstreams. A Java-ish solution uses either StringTokenizer, or split combined with a replaceAll regular expression:

       input.replaceAll("([-+*/])"," $1 ").split(" ");
Understanding the above code, gives some good insight into Java regular expressions.

TeXLeX discuss it
Used as: Division Two - Level Two:
Value 600
Submission Rate 80 / 236 (33.90%)
Success Rate 23 / 80 (28.75%)
High Score BradAustin for 400.41 points (22 mins 34 secs)
Average Score 282.71 (for 23 correct submissions)
Used as: Division One - Level One:
Value 350
Submission Rate 173 / 200 (86.50%)
Success Rate 73 / 173 (42.20%)
High Score radeye for 316.08 points (9 mins 30 secs)
Average Score 190.75 (for 73 correct submissions)

This problem was nasty. TeX, Donald Knuth's ubiquitous typesetting software, has many intracacies that even diehard TeX/LaTeX users don't know about. In order to support many different input mechanisms, TeX gives users the ability to easily access all characters using caret sequences. These are particularly useful when trying to access characters outside of the printable range. The code that goes into handling this convenience was the point of this problem.

To implement the primitive TeX lexer required for this problem, we read in characters from the left side of the input. At any point in time, we are only concerned with the leftmost 4 characters. If the leftmost character is not a caret, we simply remove it from the input and append its value to the returned list. Otherwise, we try to match one of the caret processing rules to the 4-character block. In this process, we need be careful that the string doesn't end prematurely, and that we store the values in an int or unsigned character, to avoid negatives. The last 2 pitfalls accounted for most of the incorrect submissions. The character produced by the caret rule should be returned to the input, to be processed again. The characters used to produce the new character are removed.

BadSubstring discuss it
Used as: Division Two - Level Three:
Value 1000
Submission Rate 40 / 236 (16.95%)
Success Rate 20 / 40 (50.00%)
High Score matmis for 934.41 points (7 mins 38 secs)
Average Score 701.16 (for 20 correct submissions)

This problem is a bit tricky conceptually, but requires very little code. There are many ways of understanding how to solve this problem. If you happen to be familiar with DFAs (Deterministic Finite Automata), build a 3 state DFA that accepts the set of strings which don't have "ab" as a substring ( L((b+a*c)*a*) ) for those savvy with regular expressions). Consider the expression for how many strings of a given length will end up on either of the final states. An analogous approach is to consider, for any given length, how many strings (which do not have an "ab" substring) end with an 'a', and how many do not. The key here, is that these values can be defined inductively. Let A(k) and N(k) denote how many strings of length k end with 'a', or do not respectively. Again, we are only referring to the strings that do not contain "ab". We then have the following recurrence:

         A(k) = A(k-1) + N(k-1) for k > 0       (1)                            
         N(k) = 2*N(k-1) +A(k-1) for k > 0     (2)                             
These recurrence relations are staring at you when you build the DFA. The reasoning is as follows:
  • 1) Any string that does or doesn't end with an 'a', can be made into a string that does by appending an 'a'.
  • 2) Any string that ends with an 'a' can be made into one that doesn't by appending a 'c'. We cannot append a 'b' here, since we would create an "ab" substring. For the ones that don't end with an 'a', 'b' or 'c' can be appended (the 2 choices are what cause the multiplier of 2 in the formula for N(k)).
Iterating this recurrence, the answer is computed quite quickly. For those math buffs out there, try to prove that the answer is just the fibonacci sequence in disguise.

RandomFA discuss it
Used as: Division One - Level Two:
Value 500
Submission Rate 52 / 200 (26.00%)
Success Rate 43 / 52 (82.69%)
High Score mathijs for 407.93 points (14 mins 10 secs)
Average Score 273.06 (for 43 correct submissions)

This was one of those DP-Probability questions. Since this was a medium, the bounds on maxLength were low enough to allow brute force, but the DP solution is far cleaner in my opinion. We maintain an array of states that will hold the probability of being on each state, for a given length of string. The probability of being on state p for strings of length n is directly computable from the probabilities for length n-1. This property allows DP to work. The only remaining trick is to properly weight the given probabilities. Since there are 3 times more strings of length k than k+1, we need to weight its affect on the final probability three times more. There are many ways to accomplish this task. Java code for one such method follows:

public double getProbability(String[] rulesa, String[] rulesb, 
                             String[] rulesc, int finalState, int maxLength) {
   double[] ret = new double[rulesa.length+1], actual = new double[ret.length];
   double[][][] probs = new double[ret.length][ret.length][3];
   String[][] arrs = {rulesa, rulesb, rulesc};
   //Parse the input, and store it in probs
   //The value ret.length-1 is used for the state 999
   for (int a = 0; a < 3; a++) {
      for (int r = 0; r < arrs[a].length; r++) {
         String toks[] = arrs[a][r].split(" ");
         int left = 100;
         if (arrs[a][r].length()==0) {
            probs[r][ret.length-1][a] = left/100.0;
         for (int c = 0; c < toks.length; c++) {
            String[] pair = toks[c].split(":");
            int st = Integer.parseInt(pair[0]), 
                prob = Integer.parseInt(pair[1]);
            probs[r][st][a] = prob/100.0;
         } probs[r][ret.length-1][a] = left/100.0;
      } probs[ret.length-1][ret.length-1][a] = 1;
   //Now the actual good stuff begins
   double num = 1, denom = 4, adder = 9;
   ret[0] = actual[0] = 1;
   for (int i = 0; i < maxLength; i++) {
      double[] newprob = new double[ret.length];
      for (int a = 0; a < 3; a++) 
         for (int s = 0; s < newprob.length; s++) 
            for (int t = 0; t < newprob.length; t++) 
               newprob[t] += actual[s]*probs[s][t][a]/3.0;
      for (int s = 0; s < ret.length; s++) 
         ret[s] = (num*ret[s] + (denom-num)*newprob[s]) / denom;
      actual = newprob;
      num = denom;
      denom+= adder;
   return ret[finalState==999?ret.length-1:finalState];

MiniPaint discuss it
Used as: Division One - Level Three:
Value 1000
Submission Rate 13 / 200 (6.50%)
Success Rate 9 / 13 (69.23%)
High Score dgarthur for 830.04 points (13 mins 26 secs)
Average Score 591.46 (for 9 correct submissions)

This problem was DP in disguise. At first glance, the solution appears to be greedy. Will arranging the contiguous sections of paint in order of length work? Unfortunately, greedy methods do not seem applicable. It turns out, you probably need to use DP twice! Consider the following strip:

Let's consider the least number of mispainted squares, for each possible number of strokes:
   0 Strokes : 14 mispainted
   1 Stroke  : 6 mispainted
   2 Strokes : 3 mispainted
   3 Strokes : 3 mispainted
   4 Strokes : 0 mispainted
The greedy methods I have seen have a hard time producing these results, and completely breakdown for more complex strips. Using DP/Memoization we compute the cost for the whole strip, using the costs for smaller substrips. Java code implementing memoization follows:
int[][][] cache = new int[3][50][51];
int BIG = 3000;
int stripScore(int[][] row, int[][][] cache, int start, int needed, int prev) {
   if (needed > 0 && start >= row[0].length) return BIG;
   if (needed == 0 || start >= row[0].length) return 0;
   if (cache[prev][start][needed] != -1) return cache[prev][start][needed];
   int best = BIG;
   for (int i = 0; i < 3; i++) {
      int newneeded = needed, adder = 0;
      if (i == 0 || i == prev) newneeded--;
      if (i == 0 || i!=row[1][start]) adder += row[0][start];
      best = Math.min(best,stripScore(row,rowid,start+1,newneeded,i)+adder);
   } return cache[prev][start][needed] = Math.min(BIG,best);
In the code above, prev denotes the color we used to paint the region immediately to the left of the current region. prev values of 0,1 and 2 denote not painted, white, and black respectively. row[0][k] denotes the length of the kth contiguous region of the strip, and row[1][k] denotes its color, using the same numbering system as prev. start is the number of the currently processed region, starting from the left. needed denotes the number of strokes that should be used. For example, passing 3,5,and 1 for start, needed and prev is like asking: "Starting at the 3rd region from the left, with the last color used being white, and using exactly 5 strokes, what is the least number of mispainted spaces possible?" To obtain the best score for a row using exactly k strokes, we pass 0, k, and 0 to start needed and prev. The key to this algorithm, is that the best score for the area to the right of the kth region, is uniquely determined by the number of strokes needed, and the color used to paint the (k-1)th region.

stripScore allows us to find the least number of mispainted spaces, for any row, given a particular number of strokes. Using this row data, we now have to find the best score for the whole picture. Once again, this process requires DP/Memoization. We build an array that stores, for each number of strokes, the least number of mispainted spaces. Then, using the same process that is used to solve 0-1 knapsack or coin changing, we try each stroke amount for each row, and compute the best score.

By brett1479
TopCoder Member