Default hopboxes have a 9 in the center cell. Box 1 (red) has its center seen by the 9 in box 7, and so does Box 9 (blue). Therefore hopboxes must be green. r5c8 = 9
Magic squares have their even numbers in the corners, but all the other cells of box 9 are seen by the 9s in either box 7 or box 6, therefore the blues must be young.
r7c7 = 1 r9c9 = 9 r1c4 = 1 r3c6 = 9
That leaves red as magic squares. r2c2 = 5 r4c4 = 5
Step 2 : Pencil mark red and blue
Use the 9s in other boxes to orient the magic squares, but otherwise just pencil in each blue and red box by what's known about magic and young.
Hidden single in row 6; R6C2 → 8 Hidden single in row 6; R6C1 → 7 Hidden single in column 1; R5C1 → 2 Hidden single in column 2; R4C2 → 4 Hidden single in row 4; R4C3 → 3
Step 10 : Row 2
Only place for 2&3 are c4 and c9, so nothing else can go in those spots.
Step 11 : Skyscraper on 5s
in rows 3 and 7; 5 removed as a candidate from R1C6, R8C4 and R9C4
From uniqueness, if r9c3 is not an 8 (and thus removing the possibility of 1 and 6 from anywhere in that box except r9c3 r9c2) then you'd have a deadly pair with r5c2 r5c3 in box 4.
![[Home]](/img/text_logo.jpg){kind=logo}
SudokuSolves/MagicYoungHopbox