Rules and how to solve RippleEffect.
Rules
  1. Enter numbers in each cell to meet the following conditions.
  2. One or more continuous cells surrounded by a thick frame are in the same group, and the same group contains numbers from 1 to the number of cells in the group, but the same number does not belong to the same group.
  3. The numbers may be arranged as problem numbers in advance. If there is the same number in the left, right, up, and down direction of the number to enter the cell, there must be an interval of the same number or more. For example, when 3 is in another cell in the same row, there must be at least 3 cells with other numbers between the cells containing 3 cells.


jump to RippleEffect problem.
Tips for solving
解くポイントはまず太線で囲まれたいくつかの枡からなるグループの中で数字の決まっていない枡が一つなら残り一つの数字が決まります。
またあるグループの中にある数字の決まっていない枡では そこにはこのグループの枡数までの数字しか入りませんが上下左右に間隔が十分ある数字が一つしかないならその数字に決まります。
さらにグループで既に決まっていない数字・上下左右に間隔が十分ある数字が一つならその数字に決まります。
数字が未定の枡に 入る可能性のある候補数字をメモ書きし、周囲の数字が決まった時に候補メモを修正していくのも良いでしょう。
The point to be solved is that if there is one unnumbered cell in a group consisting of several cells enclosed by a thick line, the remaining number will be determined.
Also, if a cell in a group has an undetermined number, it will only contain numbers up to the number of cells in this group, but if there is only one number that has sufficient space above, below, left, and right, that number will be determined.
It is a good idea to write down the candidate numbers that may enter the unspecified cell, and then modify the candidate notes when the surrounding numbers are determined.
Let's explain with a real problem. The following figure is the problem. The problem cell is a large numbered cell with a net. (The problem numbers are shaded here to make them clear, but sometimes they are bold.) ex0
First, notice that there is only one square in the same group whose number is not determined.
The pink framed cell on the left end of the first row is a group of one cell, so 1 is entered.
The cell with the light blue frame at the bottom is a group of two cells. The cell on the left is already filled with 1, so 2 will be entered.
Next, consider the cell with a yellow-green frame on the fourth line. Since this square is a group of four squares , the numbers 1 to 4 will be included, but 2 is already in the same group and 1 is in the lower area (interval 0) so 1 and 2 do'nt enter. Furthermore, there is 4 in the second on the right side, so 4 can not be entered unless 4 spaces are left. After all, the number entered is 3.
ex1
Let's consider a group of three inverted L-shaped cells with the upper left 1. In the previous figure, 3 was placed in the yellow-green frame of the fourth cell from the top and the second cell from the left, so 3 could not be entered in the three cells above this cell. Therefore, in a group of three inverted L-shaped cells, only 3 can be inserted into the red framed cell at the left end and second raw from the top. It is thought that only 3 can be inserted because there is 1 above this cell and 2 below it. The remaining two cells of the group of inverse L will be filled with 1 and 2,but in the orange frame of the second cell from the left at the top, 1 is on the left, so 1 cannot be entered and the remaining 2 will be entered.
Next, let's consider a group of four cells that are 3rd to 5th cells from the top left (including one cellwith a green frame). This group of four cells contains 1-4, but already has 2 and 3. There are 4 in the 4th cell from the top and the 4th cell from the left (the 3rd cell on the right of the green frame 1), so 4 is not inserted in the 4 cells on the left (actually 3 cells). So 4 can only be placed in the 5th cell from the top, the cell with the green frame at the left end. The remaining 1 goes into the fourth cell from the top and left end,which is vacant and green framed.
Now consider a group of three inverted L-shaped cells, including a cell with a blue frame in the lower right corner. This group contains 1-3. Consider a cell of this group which is a blue framed cell of the fifth cell from the top and the fifth cell from the left at the lower right corner.
There are 2 in the 2nd cell on the left and 3 in the 3rd cell on the top, so neither 2 nor 3 can be put in this cell, so the remaining 1 will be inserted. (Since 2 is determined for the cell on the top and 3 for the cell on the left, it is considered that the remaining 1 enters the cell with the blue frame.)
ex2
Consider again the group of three inverted L-shaped cells in the lower right corner. There is 2 next to the cell on the left of 1 in the yellow-green frame of the 5th cell from the top and the 4th cell from the left, so the remaining 3 will be entered without entering 2.
The remaining cells with a green frame above the cell 1 in the fifth cell from the right and the fourth cell from the top contain the remaining number 2. The remaining number 1 is entered in the pink frame of the second cell from the top and the second cell from the left of the group of three inverted L-shaped cells in the upper left.
ex3
This time, we will consider a group of five red framed cells which are 3rd to 4th squares from the top and 2nd to 4th cells from the left. These five cells are filled with numbers from 1 to 5, but 4 is already in the fourth cell from the top and left of the same group.
If you look at the red cell of the third cell from the top and the second cell from the left, there are 1 on the upper cell, 2 on the left, and 3 on the bottom, so 1-3 are not included. Since 4 is already used by another cell in the same group, only 5 can be placed in this cell.
Next, consider a group of three I-shaped cells located in the first to third squares from the top right. 3 is already in the second cell from the top. Considering the green cell of the third cell from the top, the cell next to it contains 2 immediately, so 2 is not included and the remaining 1 is included.
There is a 1 in the bottom cell of the fifth cell from the top, but a different number 2 in one cell between them, so put a 1 in the third cell from the top.
ex4
If you come to this point, you can see the figure afterwards. ex5
ex6 ex7
This is the correct solution. ex8
The following are some tips.
In the figure below, one cell in the light blue frame at the bottom center contains 1, so the two cells in contact with the top contain 2,1 as blue numbers.
1 is entered in one cell of the red frame at the lower left of the figure, so 1 cannot be entered in the two adjacent pink cells. Then there is no choice but to put (red) 1 in the remaining cell.
Consider a group of three cells sandwiched between one cell group and two cell group at the top of the figure. One cell in the yellow-green frame is filled with 1, and no matter which of the two cells in the green frame is filled with 2, neither the cell 1 nor the cell 2 can't be filled in the center of the three cells, so this cell can be filled with 3.
In the light purple frame on the right side of the figure, there is a 3 somewhere. For this reason, 3 is not put in the purple frame of the cell in contact with the three cell groups, and 1, 2, and 4 remain as candidate numbers of this cell.
ex10
In the following figure, where two cells group are in contact with the one cell group, they are determined as green / yellow green 2, blue 1, etc. Think in a group of four cell on the left of them. In the green frame, there are blue 1 and green 2 on the right side, and 1 and 2 are not included, so 3, 4 remains as candidates. In the yellow-green framed cell, there are blue 1 and yellow-green 2 on the upper side, and 1 and 2 do not enter, so 3, 4 will remain as candidates. In the four cell group, candidates of two cells are 3 and 4, so 3 and 4 cannot be entered in the remaining cell. Therefore, 1 and 2 remain as candidate numbers in the two cells in the blue frame.
Where the lower two cells group meet the four cells group, there are only two light blue frames at 2, two yellow frames at 1, or two light blue frames at 1, and two yellow frames at 2. Then, 1 and 2 can not be entered in the two red frame cells sandwiched between the light blue frame and the yellow frame, so 3, 4 will remain as candidate numbers of them. 3 and 4 reserved in two places cannot be entered in the orange frame, so 1,2 will remain as candidate numbers in the two orange frames.
There are many other patterns.
ex11



2020.2.28 Modified
2010.6.13 First edition
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