Embedded Files
New Site https://blog.stimim.tw (under construction)
step 1
step 1
By the power of SET (link), we know that green cells and red cells must contain the same digits.
This also mean that green cells and red cells sum to the same number.
There are some circle arrows whose arrow and circle are in the different colors (r3c6, r8c3, r8c8). If we remove those cells on arrow and circle at the same time, the sum of remaining green cells and remaining red cells will still be the same.
Now the only arrows on the colored regions are two digits arrows. Their tens digits can only be 1 or 2, so they form a 12 pair in row 3.
More importantly, the digits on the arrow sum to the "number in the pill". Therefore, the digits sum of those green cells on the arrows, and the digits sum of those red cells on the pills are differ by: (10+20+r4c1+r4c9)-(1+2+r4c1+r4c9)=27.
Now, let's consider the rest of the green cells and red cells that are not on the arrows, the sum of those red cells need to be 27 greater than the sum of those green cells.
The max difference we can make is:
digits on row3 + digits on column 1 - digits on row 1 - digits on row 8 - digits on row 9.
(8+9) + (7+8+9) - (1) - (1+2) - (1+2+3+4) = 27
Therefore, the remaining green cells must be minimum values, the remaining red cells must be maximum values.
step 2
step 2
Now we get a grid like this.
The only cell that could be 1 on column 1 is r4c1.
8 and 9 must be on the arrow in box 1, therefore, the 3rd digit on the arrow is 4.
On column 1, 4 must be in box 7, so r9c2 must be 3. r8c1 must be 4.
There is no 1 on the arrow of r8c8, so it must be 9=2+3+4. And we know 3 must be in r7c9.
On row 7, 124 must be on the arrow, so the arrow is 7=1+2+4.
step 3
step 3
7 in box 1 must be in r3c2.
The rest of the cells in box 1 are 356, r3c3 can only be 5 or 6.
r3c6 is on the circle of length 3 arrow, the minimum sum of the arrow is 5, so r3c6 can only be 5 or 6.
Therefore r3c7, r3c8 is a 34 pair.
1 in box 6 is can only be in r6c8.
The minimum value of r5c8 is 8=3+5, and it's also the maximum value it could be.
The 35 pair on the arrow disambiguate r2c7 and r3c7.
step 4
step 4
1 must be on the r3c6 arrow, so it must be in r5c4.
We can also deduct some more digits by doing regular Sudoku.
step 5
step 5
The minimum value of r4c5, r6c3 are 2, so they form a banded 23 pair.
Because r4c5, r6c3 is a 23 pair, 23 in box 4 must be in r5c3, r6c3.
The minimum value of arrow on r5c2 is 4+5=9. And we can fill in more digits.
step 6
step 6
6 cannot be on the arrow of r9c6, so the arrow can only be 3+5=8.
We can fill in more digits by doing regular Sudoku.
Finally, the arrow in r1c4 can only be 3=1+2.
Page updated
Google Sites
Report abuse