the following is true: j + m ≤ 1. This means that the variable i must be 1. Since i = 1 and the variable i is the neighbor to the clue cell with number 1, the variables a, b, c, e, f, h, and j must be zero. The same result can be obtained by replacing i = 1 into the second equation as follows: a + b + c + e + f + h + j = 0. This is equivalent to a = 0, b = 0, c = 0, e = 0, f = 0, h = 0, j = 0. Figure 3 is obtained after Step 1 and Step 2. The grayed cells with '–' are the variables with value 0. The cell with the symbol Δ corresponds to the variable with value 1. The variable k is the only neighbor of the left most clue cell with value 2. This clue cell has one neighbor with an object and only one remaining cell with variable k. Therefore, k must be 1. Similarly, the variable m must be 1 too because it is the only remaining variable neighbor to the right most clue cell with value 2. Since k = 1, m = 1 and i = 1, we complete the marking of three hidden objects therefore d = 0, and g = 0. The final solution is given in Figure 4. Use of uniqueness In the example above (Figure 2), the variables a, b, c, and e are the neighbors of the clue cell 1 and they are not neighbors of any other cell. It is obvious that the followings are possible solutions: a = 1, b = 0, c = 0, e = 0 a = 0, b = 1, c = 0, e = 0 a = 0, b = 0, c = 1, e = 0 a = 0, b = 0, c = 0, e = 1 However, if the puzzle is prepared so that we should have one only one (unique) solution, we can set that all these variables a, b, c, and e must be 0. Otherwise there become more than one solutions. Use of partitioning Some puzzle configurations may allow the player to use partitioning for complexity reduction. An example is given in Figure 5. Each partition corresponds to a number of the objects hidden. The