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 Problem Statement for RandomPaintingOnABoard

Problem Statement


You are given a board with N row and M columns. We will use (i,j) to denote the j-th cell in the i-th row (0-based indices). Each cell is initially white. You goal is to have a board that has at least one black cell in each row, and at the same time at least one black cell in each column.

You will paint the cells black in turns. In each turn, you select one cell (in a specific way that is given below) and color it black. You stop as soon as your goal is achieved.

You are given a String[] prob consisting of N elements, each M characters long. All characters in prob are digits ('0'-'9'). The digit represented by the j-th character in the i-th row (0-based indices) will be denoted p[i][j]. This digit is the relative probability of cell (i,j) being chosen in each turn.

Formally, let s be the sum of all p[i][j]. Then in each turn the cell (i,j) is chosen with probability p[i][j]/s. Note that the same cell may be chosen multiple times: we may sometimes choose a cell that is already black.

The constraints guarantee that achieving your goal is possible. Return the expected number of turns until your goal is achieved.



Method signature:double expectedSteps(String[] prob)
(be sure your method is public)


-Your return value must have a relative or an absolute error of less than 1e-9.


-prob will contain between 1 and 21 elements, inclusive.
-Each element of prob will be between 1 and 21 characters long, inclusive.
-The elements of prob will be of the same length.
-Each element of prob will consist of digits only ('0'-'9').
-The total number of cells (i.e., the total number of digits in prob) will be between 1 and 150, inclusive.
-In each row there will be at least one non-zero digit.
-In each column there will be at least one non-zero digit.


Returns: 3.0

Each of the cells (0,0) and (1,1) has probability 1/2 to be chosen in each turn. One of those cells will be colored black in the first turn.

Then there will be some additional turns until the other cell is chosen. As that happens with probability 1/2, the expected number of additional turns is 2.

Thus the total expected number of turns is 1+2 = 3.

Returns: 3.6666666666666665

This board is symmetric, so we can just assume that the cell (0,0) was chosen in the first turn.

Then there will be some additional turns until some other cell is chosen for the first time. The probability of choosing a different cell is 3/4, thus the expected number of these additional turns is 4/3.

Now, what is the second cell to be colored black?

  • With probability 1/3 it is the cell (1,1) and we are done.
  • With probability 2/3 it is one of the other two cells. In either of these two cases, we still need to color any one of the other two white cells black, and the expected number of turns until that happens is 2.

Thus the answer is 1 + 4/3 + (2/3 * 2) = 11/3.

Returns: 3.9166666666666665
Returns: 11.214334077031307
Returns: 1028.7662876159634

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This problem was used for:
       Single Round Match 583 Round 1 - Division I, Level Three