Algorithm to generate mountain ranges with upstrokes and down-strokes (java)

Question

I tried to do the classical problem to implement an algorithm to print all valid combinations of n pairs of parentheses. And I found this program (which works perfectly) :

Answer 1

There are more possible approaches for using the available parentheses generation code.

• Use it exactly as it is, and convert the resulting set of parentheses strings into mountain representations.

• Update it in order to directly generate the mountain strings. This is the alternative detailed in this answer.

Modifications

• Update the recursive function to work with a char matrix instead of a char array.

This prevents dealing with complications of inserting newlines while constructing the solution. A new string will be generated from this matrix once a solution is complete.

• Generate a string from the char matrix when a solution is complete.

Concatenate the strings associated to each row of the matrix, adding newlines after each row. Additionally (not implemented in the solution below), the trailing spaces of each row could be removed.

• Update the signature of the recursive function in order to accept now two position parameters, instead of a single one.

We use two position parameters, denoted row and col, as we are moving in two dimensions now, and they are the correspondents of the count parameter in the old code. The row and col indicate the corner where the mountain line has led us so far. The col (column) parameter increases by 1 after each character we add. The row parameter is changed according to whether the current character corresponds to climbing or descending.

• Clear (replace with space) any character that we have added as soon as they are not anymore being part of the currently investigated solution.

This was implicit in the 1D case, because we always ended up with strings of a fixed length and every new solution overwrote the previous ones. However, in the 2D case, if we don't clean up the path generated for a solution, we might see parts of it in the following solutions.

• Initialize the char matrix before the first recursive call.

The size of the matrix is count rows (because this is the maximum height of a solution that will be generated) and 2 * count columns (because this is the length when using count pairs of strokes). The matrix is initially filled with white spaces.

Java code

Below is the Java code that was updated according to the ideas above. Despite the modifications enumerated, the core logic is the same (the recursive structure is the same - the decision of whether to attempt adding an up-stroke / down-stroke and the termination criteria are not altered).

public static void addParen(ArrayList list, int leftRem, int rightRem, char[][] str, int row, int col) {
    if (leftRem < 0 || rightRem < leftRem) return; // invalid state

    if (leftRem == 0 && rightRem == 0) { /* all out of left and right parentheses */
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < str.length; i++) {
            sb.append(String.copyValueOf(str[i]));
            sb.append(System.lineSeparator());
        }
        list.add(sb.toString());
    } else {
        if (leftRem > 0) { // try a left paren, if there are some available
            str[row][col] = '/';
            addParen(list, leftRem - 1, rightRem, str, row - 1, col + 1);
            str[row][col] = ' ';
        }
        if (rightRem > leftRem) { // try a right paren, if there’s a matching left
            str[row + 1][col] = '\\';
            addParen(list, leftRem, rightRem - 1, str, row + 1, col + 1);
            str[row + 1][col] = ' ';
        }
    }
}

public static ArrayList generateParens(int count) {
    char[][] str = new char[count][count * 2];
    for (int i = 0; i < str.length; i++) {
        Arrays.fill(str[i], ' ');
    }

    ArrayList list = new ArrayList<>();
    addParen(list, count, count, str, count - 1, 0);
    return list;
}

Results

Below are the resulting mountains when the input is 3 (i.e. the width of the string is 6, because we have 3 up-strokes and 3 down-strokes):

  /\  
 /  \ 
/    \


 /\/\ 
/    \


 /\   
/  \/\


   /\ 
/\/  \



/\/\/\

Analysis

There are a few interesting questions that now can be answered about these strings.

(Q1) how many valid strings are there for a specific width?

(Q2) what is the probability of a random sequence of '/' and '\' to be a valid mountain?

(Q3) what is the probability of a random sequence containing equal number of '/' and '\' to be a valid mountain?

The table below answers to these questions for various string lengths:

 Length           Valid           Total        Prob. Q2   Equal / and \        Prob. Q3
      2               1               4        25.0000%               2        50.0000%
      4               2              16        12.5000%               6        33.3333%
      6               5              64         7.8125%              20        25.0000%
      8              14             256         5.4688%              70        20.0000%
     10              42           1,024         4.1016%             252        16.6667%
     12             132           4,096         3.2227%             924        14.2857%
     14             429          16,384         2.6184%           3,432        12.5000%
     16           1,430          65,536         2.1820%          12,870        11.1111%
     18           4,862         262,144         1.8547%          48,620        10.0000%
     20          16,796       1,048,576         1.6018%         184,756         9.0909%
     22          58,786       4,194,304         1.4016%         705,432         8.3333%
     24         208,012      16,777,216         1.2398%       2,704,156         7.6923%
     26         742,900      67,108,864         1.1070%      10,400,600         7.1429%
     28       2,674,440     268,435,456         0.9963%      40,116,600         6.6667%
     30       9,694,845   1,073,741,824         0.9029%     155,117,520         6.2500%