# George Orwell Sudoku

Here is a puzzle I created today. The title is "George Orwell Sudoku"

Normal Sudoku rules apply. In every coloured group of three cells the middle digit must equal the sum of the other two digits – and remember that (yes you guessed it) two plus two equals FIVE not four 😊

Good luck 😊 • Both $2+3=5$ and $2+2=5$? Jun 29 '20 at 13:10
• Is the group of three white cells at the bottom of the first column a "coloured group of three cells"? Jun 29 '20 at 14:46
• Culver Kwan is correct, 2+3 = 2+2 = 5. The three white cells at the bottom of the first column is not a coloured group of three cells so don't have to obey the sum Jun 30 '20 at 12:13
• You made George Orwell proud. Aug 8 '20 at 11:00

The Solved Grid

Also:

Per prompting from the OP, there is an Easter egg in the puzzle. The first four digits of the top row are 1984, referencing the title of Orwell's most famous book.

Solution Method

A general observation:

In any colored group of three squares, the sum of the entries is even, unless it is an instance of 2,2,5. Otherwise, the sum is twice the middle entry.

The first deduction we can make is in:

the middle box. The sum of the yellow squares is even, and the sum of the blue squares is even, and 6 is even, so since the sum of all squares in the middle box must be 45, the sum of the two green squares in the middle box must be odd. This forces the sum of the three green squares in this group to be odd, implying it must be 2,2,5.

Continuing:

In the middle box, the remaining unplaced numbers are 1, 3, 4, 7, 8, 9. The 9 must be the middle of either the blue or yellow squares, and it must be paired with 1 and 8. Thus the remaining triplet must be 3, 4, 7, which must be blue since there is already a 4 in the top row of the middle box. This gives: Next:

Moving to the blue group with 4 in the center at right, it has to be 1,3,4, since it cannot be 2,2,4 as 2+2 = 5. There is already a 1 in row 4 from the top, so we can position the 3 and 1. Moving down to the bottom-center box, the 6 must be in the bottom-left or bottom-center square. But were it in the bottom-center it would force the center to be 7, a contradiction. This gives: Working in row 6 from the top:

In the middle-right box, the lower-left gold square cannot be 7 or greater, for if it were it would force the upper-right gold square in the lower-middle box to be 1 or 2, which is a contradiction. Thus this box must be 1 as 2-6 are forbidden by normal Sudoku rules. Next, the lower-right square in the middle-right box must be 7, 8 or 9 by normal Sudoku rules, and cannot be 9 since it is a non-center piece of the green group. So it must be 7 or 8, implying the center of that green group is either 8 or 9, and the other end of the green group is either 1 or 2. Since there is already a 1 in this box, it must be a 2, which forces the remainder of the green group to be 7 and 9.

Continuing in this row:

The middle-left box must contain the 3, 4 and 8 of row 6 from the top. The 3 and 4 cannot both be in the green squares in this box, since it would force the bottom green square of that group to be a 1. Thus the lower-middle square in this box must be the 8, and the other two squares contain the 3 and 4 in some order. This gives: Some small deductions from here:

In the top-left box, there are two complete coloured groups, whose sum must be even, the 7, and two yellow-coloured blocks. Since the sum of all squares in the box is odd, the sum of the two yellow boxes must be even. This precludes the possibility of 2+2 = 5, and allows us to conclude that the upper-left yellow square in the middle-left box is also even. It cannot be 4 or 8 by normal Sudoku rules, so it must be 2 or 6. In the bottom-left box, the top-middle green square must be either 4 or 5 to make the sum work out. The bottom-middle square must be either 7 or 9, since it must be greater than 6 and cannot be 8. This forces the bottom-right square in this box to be either 1 or 3. This gives: Now let's try to place the 9's:

Looking in column 6 from the left, and noting that a 9 can only be in the middle of a group of three coloured squares, the only possible locations for the 9 are the top row or the bottom. Suppose first it is in the top row. Then we get the chain of inferences: bottom-middle box's 9 in its left-middle square; bottom-right box's 9 in its upper-left square, upper-right box's 9 in its right-middle square; upper-left box's 9 in its bottom-left square; bottom-left box's 9 in its bottom-middle square. Once this is set, the bottom-right square of the bottom-left box must be 3, which forces the bottom-right square of the left-middle box to be 4, the bottom-left square of the left-middle box to be 3, and finally the upper-middle square of the bottom-left box to be 5. This creates a contradiction in the upper-right square of the bottom-left box: it has to be greater than 4 for the blue group, cannot be 5, 6, or 9 since those are already present in the box, and cannot be 7 or 8 since summing the blue group would create another 3 or 4 in column three from the left. The end state is shown here: Moving on:

Therefore the 9 in column 6 from the left must be in the bottom row. This immediately gives the bottom-center and bottom-right squares in the bottom-left box as 7 and 1 respectively. We also determine with standard Sudoku reasoning that the center square of the bottom-right box is 1 and the upper-right square of the left-middle box is 7. We can also determine that the lower-right square of the left-middle box is 3: were it 4, the upper-right square of the bottom-left box would have to be greater than 4, but not 6 or 7; 5 forces two 1's in the column, 8 forces two 4's in the column, and 9 forces both the upper-middle green square and the right-middle blue square to be 5. Filling in a couple of places where our possibilities have been reduced to 2, our grid is: Let's tackle the upper-left box:

Focus in on this box, assigning some letters to particular squares, as in this image: With these variables, we must have 2A + 2B + C + D = 38 and C-D = ether 2 or 6. We also have the following constraints on each individual value: A must be one of 3, 5 or 9 (1 and 2 are too small, and 4 and 6-8 already appear in the column); B must be one of 6, 8 or 9 (1 and 2 are too small; 3, 5 and 7 already appear in the column [3 and 7 explicitly, and one of the two blue squares in the bottom-left box must be a 5], and 4 would force there to be two 1s or 3s in the column) and C is one of 5, 8 or 9 (it must be at least 3; 3 would force two 1s in the row, 4 and 7 are already in the column, and 6 would force the square below C to be 4, making two 4s in the column). If C is 9, D must be 3 since it cannot be 7; if C is 8, D must be 2 because there is already a 6 in D's column; if C is 5, D must be 3. By enumeration, there are exactly three possible solutions for (A,B,C,D) which meet these constraints: (5,8,9,3), (5,9,8,2) and (9,6,5,3). Each of these can be realized as a candidate for the upper-left box in exactly one way consistent with the rest of the grid: Notice two important properties of these solutions:

In all of them, there is a 1 in the upper-left square of the upper-left box, and the 2 is in the middle column of the upper-left box. The former fact forces the center square of the left-middle box to be a 1, while the latter forces the two blue squares in the lower-left box to be 5 and 2. But also notice that the first two solutions do not have a 9 in the middle column. This now forces the 9 to be in the top-middle square of the left-middle box, which is a contradiction as there is already a 9 in that row. Hence the third grid must be the correct solution, we get some easy fill-ins with this revelation, yielding: Some more deduction:

In the third row from top, the 2 must be in the 6th column from the left, since the only other candidate is column 7, and we cannot have a two as the center of a coloured group. In the same row, the 9 must be in the 4th column from the left, as again column 7 is the only other candidate, and putting 9 there would force a repeated 7 in the second row. This allows us to place a 9 in the right-middle square of the upper-right box as well, since 9 cannot be in a non-central square of a coloured group. Back in the third row, we now can put a 7 in the seventh column (and the corresponding 5 above it), since there is already a 6 in the column, and putting an 8 there would create another 6 in that column (square above). The grid at this stage is: Working the second row from top:

The middle square in the upper-right box must be a 4. Only 1, 3, 4, and 8 are possibilities in the row, the column excludes 1 and 3, and an 8 would force a second 9 in the box. Continuing in this box, we know the 2 and 3 must be in the upper-left and upper-right corners, while the 6 must be in the top-middle square and the 8 in the bottom-middle square. This lets us finish off the third row with a 6 in the bottom-middle square of the top-middle box. Back in the second row, the middle square of the upper-middle box must also be an 8, since the 8 for that column must be in this box, and it cannot be the top-middle square. A 1 in the right-middle square of the top-middle box would force two 7s in the 5th column, and at this point we have enough of the grid that the remainder is easy deduction.

• Your solution is correct but there is one further point of interest: <rot13> Gur gbc ebj pbagnvaf na rnfgre rtt </rot13> Jul 5 '20 at 12:39
• Got it...will mod the answer. Jul 5 '20 at 13:30