u/Automatic_Loan8312

The solution strategy for a Killer Sudoku titled 7.5 on Sudoku Coach by a user named 333 is discussed here.

Original puzzle

Constraints:

Killer cage:- Within a cage, the digits must add up to the sum shown in the top left-hand corner of the cage. Digits must not repeat within a cage.

Here is the solution strategy.

Using Rule of 45 for boxes 1 and 2, we get that r1c7 = 5. Similarly, for boxes 7 and 8, we get that r9c7 = 2. Likewise, for boxes 3 and 9, we get that r4c8 = 3 and r6c8 = 6. Thus, we reach the first checkpoint.

First checkpoint

Here, it can be inferred that 4 is invalid in each of the cells r13c89, by Kakuro logic, because then, in order to satisfy a total of 9 in each of the two-cell cages r1c89 and r3c89, the other cell would have to be a 5. This is impossible, as 5 already exists in the same box as r13c89.

Thus, 4 is restricted to r23c7 (yellow), which means that r23c7 is a naked pair {3,4}. Also, the two-cell cages r1c89 and r3c89 both add up to 9 each therefore, without 3 or 4 in them, the valid combos can be {1,8} or {2,7}. Thus, {7,8} is invalid combo in the two-cell cage r2c89 with a total of 15. Therefore, r2c89 is a {6,9}, further, because of the 6 in r6c8, we have r2c8 = 9 and r2c9 = 6.

Next, using rule of 45 for box 5, we get that r46c7 adds up to 10. Further, using rule of 45 in column 7, we get that r5c7 is 8.

As r78c7 is a naked pair {6,7}, by Kakuro logic, for the two-cell cage r8c89, the combos {2,8}, {3,7}, and {4,6} are all invalid. Thus, the only valid combo for r8c89 is {1,9}. Further, we have r8c8 = 1 and r8c9 = 9.

Using Kakuro logic for r7c89, which adds up to 8, the only valid combo for r7c89 is {3,5} (no 1 or 2, because they’re in the same box as r7c89). Further, r7c9 is a 3 while r7c8 is a 5. We now reach the second checkpoint.

https://preview.redd.it/x1c7b1d9zg1h1.png?width=750&format=png&auto=webp&s=9072cc367ecbfb9c4d700ebb37efe41471bbd976

Using rule of 45 for columns 1 through 4 (gray), we get that the orange cells r46c5 add up to 12. Notice that the yellow cells r78c5 add up to 14, and can have either {5,9} or {6,8}. Further, if r78c5 be {5,9}, then the orange cells r46c5 will have to be {4,8}, which means that the pink cells r23c5, which add up to 6, can have neither 5 nor 4. This is an invalid state, as neither combo {1,5} nor {2,4} is possible for r23c5. Therefore, r78c5 isn’t {5,9}. Thus, r78c5 is a naked pair {6,8}. This means that r46c5 can be either {3,9} or {5,7}.

https://preview.redd.it/sxmff6rd3i1h1.png?width=750&format=png&auto=webp&s=044e02bce608303e3497f66534cbcbd35627bfd9

The above figure is useful for deducing a combo in r46c5 using Kakuro logic, and it is demonstrated as follows:

Assuming that r46c5 be the combo {5,7} (orange), we get that r23c5 (pink cells) form a pair {2,4}. Now, notice that by the rule of 45, we get that r123c4 adds up to 17, but will also have a 5, but not a 4 in them. Thus, blue cells form a triple {3,5,9}.

Likewise, using rule of 45 for box 7, we get that r789c4 adds up to 14. We see that because r46c5 is {5,7}, 7 is now restricted to r789c4 in column 4. Further, 2 is restricted to r78c2 (see above), which means that in order to satisfy a total of 14, r789c4 must be {2,5,7}. But, 5 is already present in r123c4 (as a triple {3,5,9}), which means that our assumption that r46c5 being {5,7} was wrong.

Therefore, r46c5 is {3,9}, which means that r6c5 is a 3.

Third checkpoint

By Kakuro logic, r5c12 (in gray) has the combo {6,9}, which means that there is a hidden single 3 in r5c3 in box 4. Next, r46c3 add up to 8, but cannot have a 6 or 3. Therefore, by Kakuro logic, r46c3 is {1,7}, thus, r4c3 is 7, while r6c3 is 1. Further, we have a hidden single 1 in r5c4 (because of the 1 in the 6-cell cage with total 29, 1 is eliminated from r5c56 and r46c6, and the 1 in r4c7 and r6c3 further eliminate 1 from r46c4).

Fourth checkpoint

By Kakuro logic, r46c4 adds up to 13 and can have either {5,8} or {6,7} as valid combos. Next, we have r78c5 adding up to 14, which is {6,8}. Therefore, r789c4, which adds up to 14, and has 2 in it, has either {2,3,9} or {2,5,7} as valid combos. Whichever combo among {5,8} or {6,7} is valid for r46c4, r789c4 cannot have {2,5,7} in it. Thus, r789c4 is {2,3,9}.

Further, by Kakuro logic, r9c3 = r7c4, and because r9c3 cannot have 2 and 3, r7c4 also cannot have 2 or 3. Thus, r9c3 and r7c4 are both 9.

Fifth checkpoint

As can be deduced from the above figure, 4 is locked to r123c4, which means that r23c5 cannot have a 4, therefore, the only valid combo is {1,5} for r23c5. Furthermore, we have {4,6,7} as the only valid combo for r123c4 (adding to 17).

Next, by Kakuro logic, r1c3 = r3c4 and because r3c4 cannot have a 4 (because it is a part of the five-cell cage r23c23 and r3c4, totalling 35), r1c3 also cannot be a 4, and because of the 7 in r4c3, we cannot have 7 in r1c3 (and also r3c4), thus, there’s a naked single 6 in r1c3 (and also r3c4). Further, as previously deduced, r4c12 is {4,5}, which means that r4c4 is 8, and r6c4 is 5.

From this point on, using Kakuro logic and scanning for the placement of the remaining numbers, the puzzle is solved.

The solution strategy to Anti-Chess Region Sum 888 on Sudoku Coach by Emeccu shall be discussed here.

Original puzzle

Constraints:

1. Anti-knight: Cells a knight’s move away cannot contain the same digit.
2. Anti-king: Cells a king’s move away cannot contain the same digit.
3. Region-sum-line: For each line, the digits on the line have an equal sum N within each box it passes through. If a line passes through the same box more than once, each individual segment of such a line within that box sums to N separately.

Here is the solution strategy for this puzzle.

Because of the multiple constraints operating simultaneously, reducing search space for digits in a cell, the original puzzle itself is the first checkpoint.

First checkpoint

Using anti-knight and anti-king constraints for 2 in box 2, we deduce that r1c4 is a hidden single 2.

We now continue by deducing the sum of each line segment of the region-sum-line constraint.

Notice that the pink cells r45c8 and r6c7, the green cell r6c6, the yellow cells r7c45, and the orange cells r56c4, all add up to a certain number. Let that number (the sum of these cells) be x.

We figure out that because the green cell and the pink cells have the same sum, we know that the search space for x can only be {6,7,8,9}. Any number x greater than 9 violates the green cell, and any number x less than 6 violates the region sum rule for the pink cells.

Further, using anti-knight and anti-king constraints with the 9 in r3c5, we find that either r6c4 or r6c6 can be a 9. Now, the presence of 9 in r6c4 leads to a contradiction, that both the orange cells and the green cell sum constraints cannot be fulfilled simultaneously.

Let me elaborate this.

If r6c4 is a 9, then in order to satisfy the green cell, r5c4 has to be a negative number. Not possible. If r6c4 is a 9, then any number in r5c4 will make the green cell greater than 9, which is again impossible.

Therefore, we arrive at a contradiction with r6c4 = 9.

Thus, r6c6 is a 9, and the sum of the region-sum-line constraint is derived as such.

Second checkpoint

Now, consider the region-sum-line segment in box 6 (in green). The sum is 9, so, by Kakuro logic, the possible combos are {1,2,6} and {2,3,4} (not {1,3,5}, because of the 5 in r5c7). By anti-knight rule, r4c8 cannot be 2, while r6c7 cannot be 2 because of r2c7 being a 2.

Therefore, r5c8 is a 2.

Third checkpoint

Here, the use of locked candidates deduced by anti-knight and anti-king rules to deduce further eliminations and placements is illustrated.

As shown above, by anti-knight and anti-king eliminations created by the 2 in r1c4 and r2c7 (in green), we deduce a locked candidate r3c12 (in yellow). Using this locked candidate, we again use anti-king and anti-knight constraints to eliminate 2 from r4c123. This results in a hidden single 2 in r4c5 (aqua cell).

Deducing a placement using the same logic as above

Another demonstration of the same deduction is discussed. The pink cells r6c123, which are featuring the locked candidate 2 in box 6, eliminate r7c123 by anti-king and anti-knight constraints. Thus, there's another hidden single 2 in r7c6.

Fourth checkpoint

An anti-knight elimination for 1 is illustrated here as an intermediate step to deduce a hidden single later on.

Here, the 1 in r3c4 eliminates 1 from r456c4 and from r5c5 (by anti-knight constraint). This locks 1 in r5c6 and r6c5. This means that 1 can be further eliminated from r7c5 (because that cell is simultaneously at a knight’s distance from r5c6 and at a king’s distance from r6c5).

This means that both the two-cell-sums in box 5 (r56c4) and in box 8 (r7c45) can have either {3,6} or {4,5}. Further, we can deduce, using anti-knight placements, that 5 is a locked candidate in box 5, restricted to only r46c4. This eliminates 5 from r7c4, so r7c4 is either a 4 or 6.

Using another anti-king, anti-knight elimination to deduce a hidden single

As illustrated above, whichever among 4 or 6 be in r7c4, that number is eliminated from r456c4, and also from r56c5 (due to anti-king and anti-knight constraints), so r5c6 is the only cell to have 4 or 6. Thus, both cells have been highlighted in purple color. Thus, that cell r5c6 is not a 1 either way.

Therefore, r6c5 is a hidden single 1.

Using anti-knight and anti-king constraints, further eliminations for 1 can be deduced.

Using anti-king placement to deduce a naked single

Using the 3 in r8c2 (yellow), we eliminate 3 from r79c4 (by anti-knight rule), which means that r79c5 is a locked candidate 3. Thus, r5c5 cannot be a 3. As previously proved, r5c5 cannot be a 4 or 6 as well, so we have a naked single 7 in r5c5.

Using a region-sum-line combo to deduce a contradiction

This is the most complex part in the solution strategy, as a region-sum-line combo is used to deduce a contradiction. For the region-sum of 9 in box 6 (see above figure), by Kakuro logic, the combos are {1,2,6} and {2,3,4}. Let’s consider that the combo is {2,3,4}. If r4c8 is 3, r6c7 is 4, which means that r5c6 is a 6 (by anti-knight for 3 and anti-king for 4). As proved previously, r7c4 and r5c6 both shall have the same digit, therefore r7c4 is a 6 and r7c5 is a 3.

This means that r56c4 cannot be a 3 (by anti-knight and anti-king rules), and also, r4c4 cannot be a 3 (because of the 3 in r4c8). This leads to a contradiction, as there is no place for 3 in box 5.

Similarly, with r4c8 being a 4, we can prove a contradiction, because r7c5 and r6c7, both cells at a knight’s distance, will have to be a 3, which is impossible due to anti-king and anti-knight rules.

Therefore, the combo {2,3,4} for box 6 for the region-sum is invalid, therefore, {1,2,6} is the combo. Further, r4c8 is a 1, because r6c7 cannot be a 1.

Further anti-knight and anti-king placements for 1 can be deduced likewise.

Fifth checkpoint

Using the 6 in r6c7, we can deduce some eliminations. Anti-king rule for 6 removes 6 from r5c6, which means that there is a locked candidate 6 in r45c4. Therefore, 6 can be eliminated from r7c4, thus, r7c4 is a naked single 4.

Using anti-knight and anti-king constraints to deduce the eliminations, likewise, the rest of the puzzle is solved.

The solution strategy to the puzzle titled Nonconsecutive XV on Sudoku Coach by Ce is discussed.

Original puzzle

Constraints:

Non-consecutive: Cells that are next to each other orthogonally must not be consecutive (have a difference of 1), unless they have a white dot between them.

Two-cell-sum: Two adjacent cells with a V or X between them must add up to 5 (V) or 10 (X).

Here's the solution strategy.

A couple of things to be highlighted for this non-consecutive two-cell-sum hybrid Sudoku:

First, the two-cell-sums V, X, and XV (5, 10, and 15, respectively) in this puzzle cannot contain a 5 in each of the two cells comprising the sum.

Second, for the odd-numbered two-cell-sums, the non-consecutive rule states that the valid combos for the two-cell-sum cannot have a combo that has two numbers arithmetically adjacent to each other.

Let me elaborate this.

Suppose that two cells add up to XI (11). Kakuro logic states that the valid combos are {2,9}; {3,8}; {4,7}; and {5,6}. Here, combining it with the non-consecutive rule, we can eliminate the {5,6} combo. So, the remaining combos are {2,9}; {3,8}; and {4,7}.

Now, back to the puzzle.

Because of the multiple two-cell-sums V, X, and XV lined up near or on row 5, the original puzzle itself is the first checkpoint.

First checkpoint

Notice that because of each of these two-cell-sums (highlighted in different shades), 5 is eliminated in all the cells from row 5, except r5c8 (not highlighted). Thus, we have a hidden single 5 in r5c8. Similarly, we can deduce that r2c4 is also a 5.

Further, the non-consecutive rule is applied for two-cell-sums as well. This means that in this particular puzzle, a two-cell-sum of 5 can only have a combo {1,4} (not {2,3}). Similarly, a two-cell-sum of 15 can only have the combo {6,9}.

Combining this with the fact that a 4, which is a part of the two-cell-sum of V, cannot be orthogonally adjacent to a 5, we deduce that r5c9 is 1. Similarly, r2c5 is also 1.

Second checkpoint

Notice that r89c2 is another two-cell-sum V (combo {1,4}), which means that because of the 6 in r6c3 (gray), r8c3 cannot be a 6. It can only be a 9. Thus, r8c2 is 1.

Next, by non-consecutive rule, r5c7 cannot be a 6, thus, r5c7 is a 9.

Third checkpoint

Due to the non-consecutive rule, 5 is eliminated from r79c3. Further, 5 is eliminated from r1c3, which is a part of the two-cell-sum X. Thus, we have a hidden single 5 in r3c3. Similarly, by cross-hatching, we have a hidden single 5 in r1c7 in box 3, and a hidden single 5 in r9c9.

Similarly, in box 8, because all cells except r7c5 are a part of some or the other two-cell-sum, or share the same row as r9c9, r7c5 is a hidden single 5.

Fourth checkpoint

Notice that because of the 4 in r5c3, the combo {4,6} is invalid for r5c56. So, 6 is eliminated from r5c56. Also, because of the 6 in r6c3, 6 is eliminated from r5c12. Thus, we have a hidden single 6 in r5c4.

Likewise, using the non-consecutive and two-cell-sum rules, this puzzle is solved.