|
Stick-Ken
Stick-Ken is a Sudoku-like puzzle where you fill a grid with the digits 1–9 (for a 9x9 Kendoku) so that each row and column has all digits once and only once.
In addition, each shape within the grid must fulfill the arithmetic result given for that shape. Individual shapes may contain a digit twice, but within each row and column a digit can be used only once.
The four arithmetic operations used are Addition (+),Subtraction (-), Multiplication ( ), and Division (÷).
A sample 6x6 Kendoku is given here. In a 6x6 Kendoku, you need to use the digits 1–6.
Solving a Puzzle:
The clues in single-box shapes are direct values for that box. Fill them in first. For example, in the 6x6 grid below, the fourth cell in the bottom row and the second cell in the fifth row are single-box shapes.
Next, try to find possibilities for a small shape. The ÷6 shape in the last column has only one possible combination of numbers (i.e. 1 and 6). Given that a number 1 is already found in the fifth row (in the second column), the number 1 that fits this equation can only go in the fourth row, leaving the number 6 to go in the fifth row.
People get intimidated by the fact that a shape may have a digit twice. But that sometimes helps us solve the puzzle quicker.
For example, the 150 in the last two columns can have only one possible combination of numbers (i.e. 6 5 5). Given that 5 is used twice, and that the 5 has to be in different rows and different columns, there is only one way to place these three numbers. The 6 goes in the third row, fifth column, and 5s go in the other cells.
In the fifth row, 30 has two possible number combinations: 1 6 5 or 2 3 5. Given that 1 and 6 are already used in this row, we can eliminate that possibility. This means that we have to use 2 3 5. Although we don’t know which digit goes where, this does leave the first column of the fifth row with only one remaining possibility—4.
The 8 in the first two rows (last two columns) has only one possible combination: 1 2 4. Given that digits 1 and 2 are already used in the last column, the cell on the top row, last column must be 4. The fifth cell in the top row must be 2 (because 2 is already used in the second row and cannot be repeated). The remaining cell in the second row, fifth column is therefore 1.
Now you can solve the rest of the puzzle using similar techniques. The solution for this example is:
|
