Sudoku solution - The Guardian 23 May 2024 Hard level

Here's today's Guardian Hard level sudoku: www.theguardian.com/lifeandst...
I used sudokuexchange.com to play this.

[@enjoyalaugh133]

I enjoyed this. 9 mins.

[@ckoh010]

The instant takeaways are 1) For hidden singles 8 in Box 4 is @(42) and 8 in B1 on col3; 1 on row 6 is @(64); 3 in B6@(58) and on c7B3; 7 on c7@(67),followed by 8 on r6@(68)( see how?), thereby leaving [46] tcp (true coupled pairs) on c9B6 !! plus in hot pursuit 8 on r9@(94)(?), followed by 8 on c6@(16)!! 2) Now we turn our attention to the Right Hand Block : Note 1&4 given on c7&r9, both outside of B9 => 1&4 cp in B9@(78)&(89), with 1@(89) followed by 4 @(78); Also note right away [19] tcp on c8B3 !! by virtue of 1&9 given in B6, 9 given in B9,plus 1s given on rows 7&9, all off c8!! to be followed by a) 6 on c8@(98), and concurrently [28] tcp on c7B9 with 8@(77) & 2@(87)==> 2 on c5@(95)!, to be followed by 2 on c3@(13) !! (see how? Note the 2s given in B4& on r3, and uncovered on rows 3,7&9, all off c3!) b) [36] tcp on c7B3, leaving [28] tcp on c9B, with 2@(29)&8@(39) ==> 8 on r2/c3 @23), thus completing all 8s in grid !! Note in hot pursuit, 6 uncovered @(98), leaves [37] tcp on r9B7, plus with 4&6 uncovered in Boxes 8&9, both off r8 ==> [46] tcp on r8B7 !!, thereby leaving in B7 [29] tcp on r7B7 AND on r8 [39] tcp on r8B8; So at this stage of Midgame, we have breakdowns of Right Hand Block and Bottom Block with Boxes 6&9 completed , and the other Boxes 367&8, each left with a tcp yet to be resolved. Now focusing on c3, we have with 1 uncovered @(38)&off c3 ==> 1 on c3@(51), to be followed by 4 on c3@(83), leaving 6 on c3@(33)!! (Indeed cell (33)= [168] by direct point count is reduced to [6] preemptively by the uncovering of I @(83)&8@(93)) Northbound we note that 9 uncovered @(28) with the help of 1 uncovered @(64) and the three 9s given on rows 45&9, all off c4 ==> 9 on c4@(14), to be immediately followed by 9 on r3@(31) ==> in hot pursuit, 3 on r3@(35), by virtue of 3 uncovered @(92), off r3!!, resulting in four ramifications: a) leaving 5 on r3@(32) ==> [56] tcp on c5B2 btw rows 1&2, (Hint : by crisscrossing of 56 uncovered on r3& given on c4, both outside of B2), leaving 4 in B3@(24) AND 4 on c5@(56); b) 3 on c4@(44), leaving 4 on c4@(24) (thereby leaving [56] tcp in B2 on c5) c) 9 on r6@(66), leaving [56] tcp on c6B5 btw rows 4&5!! In hot pursuit, 1&4 uncovered on rows 2&3 => [14] tcp on r1B1,btw cols 1&2, leaving [37] tcp in B1 on r2 btw cols1&2 (as indeed verified by 3 uncovered @(17)!!) ===> 4@(11) and together with 8 given @(81) ==> [48] tcp on c2B4, leaving [57] tcp on c1B4 btw rows 4&5!! And we are all done!! All that’s left are open TCPs that would be broken up by their respective danglers!!