0. Overview =========== 1) Work out what the three important digits are (a) work out their values (b) work out their posistions 2) Solve the rest of the puzzles The slog is front loaded, as you need to dismiss a bunch of alternatives, some require deeper probing than others. Fortunately, while some paths are deepish, they aren't branchy, if you've got decent short-term memory you can do it in your head: "so that's a 6, so these are a 4 and 5, which makes these 2 and 3, which means those aren't 2, and then we've got 4 cells that need to be [145]" kind of thing. Several of the dependency chains overlap, so they get easier when you recognise them. 1. Work out what the three important digits are =============================================== They're the final branching points of the tree, the zipper middles in row 3. Let's call them A,B, and C in columns 2, 5, and 8. They partner in the cages with (11-A), (11-B), (11-C) They merge at cells (22-A-B) and (22-B-C) Which partner in the cages with (A+B-11) and (B+C-11) Which creates the root node in r9c5 with value A+2B+C-22 This is better viewed as: root = ((A+B+C)-22) + B. 1a. Work out their values ------------------------- The 6 branches sum to at least 21+21=42. Therefore the three values sum to at least 21. 21=6+7+8, 21=5+7+9, 22=6+7+9, 22=5+8+9, 23=6+8+9, 24=7+8+9. We can start with an absolute hack and slay move: if the sum A+B+C is 22, then B=root, r3c5=r9c5, Fail. Two down, four to go. Looking for weaknesses, high values with positive feedback seems likely to be fragile. The extreme case 24=7+8+9 has (A+B+C)-22=2, so 7 must go in the middle, putting 9 in the root. However, you can't put 8 into box 2 because of the zippers and [89] pair in rown 3. Fail. Three down, three to go. Next most extreme, 23=6+8+9, requires a little more attention. B can't be 9, as that makes the root 10. It also can't be 8, as there's nowhere for 9 in its box. 6 in B requires a bit more care: Fortunately, it's so constraining that you can instantly place 2,4,9, and 8, leaving a [15] pair in row 2 and a [37] pair in row 3. Wherever you now put the 8 in A or C, it must be accompanied by a [26] pair in row 2. If A is 8, the 4 in row 1 means that you can't fulfil the 11-cage, you've run out of evens. If the 8 is in C, the [15] and [26] pairs mean you can't create the sum to 9 in box 1. Fail. Four down, two to go. So A+B+C=21. The zippers are populated with [123456], sandwiching two from [789]. From the possible A,B,C values, the 5 in 5+7+9 can also be considered "extreme", which might show a weakness, and indeed it immediately does: [579] in cages pair with [642], and therefore to make a zipper the 2 must chaperone the 4 and 6, so the 9 would have to be in B, making the root 8. 7 is needed in both cells between the zipper. Fail. Five down, only one possibility remains. So A,B,C are 6,7,8, and part 1 of the slog is completed. 1b. Work out their positions ---------------------------- There are 6 ways of permuting 6,7, and 8, but the cracks we exploited earlier inform us what's most likely to be easily rejectable. But first we should familiarise ourselves with some simple patterns. We know that the zipper cells in rows 1 and 2 must pair the following ways with row 3's 6,7, and 8: 6 7 8 5+1 3+4 2+6 4+2 6+1 5+3 We also know that in 2 of the boxes there's an 11-cage, which forces other values into the box: If 6 (inducing a +5) is in a box with cage, there's 274 or 472 in row 1, both with 8 on row 2. Too fast? Step by step: If 6 is in a box with a cage, we know that box also has 1,2,4, and 5 in its zippers. The cell in the cage pairs with a value (11-6)=5 higher than its zipper partner's value. caged 5 partnering with 1 outside the cage => 6 in the cage - clashes with the 6 caged 4 partnering with 2 outside the cage => 7 in the cage - fine caged 2 partnering with 4 outside the cage => 9 in the cage - fine caged 1 partnering with 5 outside the cage => 10 in the cage - say what? If 7 (inducing a +4) is in a cage, row 1 has 651 with 8 in row 2, or has 483. => 6 and 7 cannot coexist in the boxes with cages - either a 4 or an 8 clash. Two down, four to go. If 8 (inducing a +3) is in a cage, row 1 has 692 (2+3=5, 3+3=6, and 5+3=8) => 6 and 8 cannot coexist in the boxes with cages - the 2 will clash in row 1. => 6 is not in a box with a cage, it's in box 3, paired with 5 in its cage. The above might seem a bit deep, it's probably worth just trying one or two out in the grid to see the logic/arithmetic working. Anyway, four down, only two remain. Rejecting 8 from box 2 is simple - where would the 7 go? It can't go on the zippers (which are [2356], it can't go in column 7, as the root of the tree is 8+(21-22)=7, and it can't go in row 3. 7 doesn't have that problem, because root in that case is 6, and there's no pressure on [789]. => The three simportant boxes are 6, 7, and 6 in that order. 2. Solve the rest of the puzzle ------------------------------- Instantly, we can fill in two dozen cells just by adding and subtracting - 4 pairs are all that remain unresolved in boxes 1-3. The cage on row 9 can only be 2-9. With 6 and 4 excluded from the cage in column 1, and 2+9 also excluded, it's 8 above 3. This puts a naked 6 into r4c3, pushing 5 into r5c3, 3 into r2c3, 5 into r2c1, and 8 into r8c3 Box 4 is left with a [29] and a [14] pair, and box 7 with a [17] pair and a [56] pair. With all the boxes and sums, we have a bunch of 9s - can we find any more? In box 5, they're excluded from row 6, column 5, and the zipper, so are in row 4 => box 4's [29] pair is 2 above 9. The same logic in box 8 places them in row 7 => The 9 in box 9 is in r8c8. => r7c8 is a naked 1 => r7c5 is a naked 5, and box 7's [56] resolves 6 above 5. => The zipper in box 8 must now be 2-4 => box 2's [25] pair resolves 5-2. Similar logic applies to 8s: In box 5, the 8 can't be in row 6, on the zipper, or in column 5. So there's an [89] pair in row 4. => there's a [17] pair in row 4 in box 6. Box 5's 5 can only be in r6c6 Rows 6 and 7 have started to put serious pressure on the cage in column 9, it's [34] above [78]. This makes a [234] triple in box 6. This makes a [68] pair in box 6 on row 5 => r5c4 is 3 This puts a [12] pair in box 5, also in column 5. => in the column, r8c5 is 3 => in the box, r6c4 is 6, r2c3 is 1, r2c6 is 6 => 7 is forced into r5c6, pairing with 2 on the zipper, and creating an [89] pair in the column => r9c6 is 1, r9c1 is 7, r8c1 is 7 => r9c4's a naked 8, r7c6 is 9, r7c4 is 7, r4c4 is 9, r4c6 is 8 => r7c9 is 8, putting a 3 in r6c9, also placing the 4, and running 1,4,1,2 round boxes 4 and 6 => r5c9 is 6, r5c7 is 8, r8c9 is 7, r4c9 is 1, r4c7 is 7, r8c7 is 6 => r1c9 is 5, r1c7 is 1, r8c7 is 6, r9c7 is 5, and yadda yadda yadda, you've finished. Proof of uniqueness: https://sigh.github.io/Interactive-Sudoku-Solver/?freeform-input=&q=.Zipper%7ER1C1%7ER2C1%7ER3C2%7ER2C3%7ER1C3.Zipper%7ER1C4%7ER2C4%7ER3C5%7ER2C6%7ER1C6.Zipper%7ER1C7%7ER2C7%7ER3C8%7ER2C9%7ER1C9.Zipper%7ER4C2%7ER5C2%7ER6C3%7ER5C4%7ER4C5.Zipper%7ER4C8%7ER5C8%7ER6C7%7ER5C6%7ER4C5.Zipper%7ER7C3%7ER8C4%7ER9C5%7ER8C6%7ER7C7.Cage%7E11%7ER3C2%7ER4C2.Cage%7E11%7ER3C5%7ER4C5.Cage%7E11%7ER3C8%7ER4C8.Cage%7E11%7ER6C3%7ER7C3.Cage%7E11%7ER6C7%7ER7C7.Cage%7E11%7ER6C1%7ER7C1.Cage%7E11%7ER6C9%7ER7C9.Cage%7E11%7ER1C2%7ER1C3.Cage%7E11%7ER1C5%7ER1C6.Cage%7E11%7ER9C3%7ER9C2