Overview ======== 5 nodes - find the king High numbers highs and lows in box 2 sums to 9 in box 6 5 nodes ------- It can't be alternating high/low for the between lines - there's a middle There has to be a high node that connects to a low node and a middle node. Likewise, there's a low node that connects to a middle node and a high node. We call these the king node and the peon node. Start by finding minimal values for the nodes based on the arrows Only box 2 and box 6's nodes are even vague candidates for the king node High numbers ------------ Row 6 can only have a 9 in r6c4, everything else is between or arrow shaft Column 6 can only have a 9 in r3c6, for the same reason box 2's 8 can only be in r2c[56] if it's on the node, it's the king node, and the peon node must have value at most 3, but neither candidate can be that low. => r2c6 is the 8 => r4c8 is above 8, so is 9, and the king node => box 2's 7 can only go in r2c5 => r3c7 is also 8 => box 2 has the middle node, and box 4 has the peon node => r2c5 is 7, r4c2 is [345], r7c7 is [456], and r7c3 is [89]. => r7c7 is actually [45], because of the between line from the [89] In column 7, r1c7 is 9. Highs and lows in box 2 ----------------------- With the 3-way 7-arrow, we can start to partition the remaining 6 cells into higher (actually mid) and low entropies. The between line must be high because of the min-3 node it connects to The wholely in-box arrow is one of each. If r3c5 was high, then both r1c6 and r4c5 are low, and that puts too much pressure on r7c7's arrows - at least 1+5 and 4+? Therefore r3c6 is low, those two are high, and r3c4 is low => r2c7 is also low sums to 9 in box 6 ------------------ With 4+5 excluded by r7c7, and 8+1 impossible, column 7's arrow is either 2+7 or 3+6 WIth 2 from [678] on the between line, and 1 from [67] on the column 7 arrow, the only sum to 9 available for the column 9 arrow is 4+5. If r7c7 was 4, the [123] triple (including a virtual pair) in column 7 pushes both 1 and 2 into r4c9 => r7c7 is 5 => the between line is a [678] triple, and r7c3 is 9 => r9c5, r8c9, r5c1, and r2c2 are 9 => r9c4 is box 8's 5 Even with r7c7 a 5, there's still a [123] triple (inc virt) in column 7 => r8c7 is 4 and r9c6 is 1 => the arrow in box 9 is a [23] pair Do some sudoku -------------- A 4 in r1c6 bounces off r3c6 to put a 5 in r6c5, which prevents box 2 from having a 5. => r1c6 is [56] and r2c7 is [12]. This limits column 7s 3 to box 6, making r4c9 a [12]. Between peon and mid -------------------- THe peon hasn't been inspected much at all, and if it's low it applies some real pressure to box 2. If peon is 3, then it has two 1+2 arrows. If the 2s are in rows 2 & 3, then 2, 3, and 7 are forced into the box, and the arrow fails If the 1s are in rows 2 & 3, then 1, 3, and 5 are forced into the box, and the arrow fails => r4c2 is not 3 And that's huge! It removes 4 from the between line => r2c4 is 6 => r1c6 is 5, and r2c7 is 2 => the r1c[45] arrow now must be a 4+3 pair => r3c5 is 1, and r3c4 is 2 => r[56]c7 is a [36] pair => r4c7 is 1 and r4c9 is 2 => r4c3 is [34], but sums with 2 to make [45], therefore is 3, summing to 5 => r2c3 is 1 and r3c2 is 4 In column 4, the only place for the 1 is r5c4 => r6c3 is 8 => r6c8 is 7 and r5c8 is 8 In row 2, the only place for a 4 is r2c8 => r3c8 is box 3's 5 => r2c1 is box 1's 5 => box 9's 9-arrow needs 4, from 1 above 3 => r3c9 is 7, r1c8 is 6, and r3c1 is 3 => the [68] pair in box 9 makes r7c8 the 1 and r9c7 the 7 Row 6's 1 can only be in r6c1 => r8c2 is the final 1 Box 4's 2 can only be in r5c3 r8c3 is [457] but not 4 or [678], so is 5 r9c3 is 4 completing the column The between line in box 4 must be 7 above 6 This resolves teh r6c5 and r6c7 arrows r5c6 can now only be 4 (has been for a while) THe [678] between line has been available as 8-6-7 for ages too To be honest, it's cracked wide open, everuytjong resolves trivially now, and its' just a speedrun.