5 BEST Coloring Hacks For Hard Sudoku 

Join the Smartie Party now 🥳to get EXCLUSIVE reward puzzle packs, ad free content, discord access, and so much more👉👉www.buymeacoffee.com/timberlakeB/membership Timestamps 0:00 Intro 00:24 It’s Solving Time 01:10 Hack #1 02:00 Puzzle Story 04:53 Hack #2 07:32 Hack #3 09:04 Hack #4 10:21 Hack #5 11:10 How to Get Puzzle Pack 11:57 Following The Snyder

Shortcut Synder notation by catching the impact of 1 & 9 on block 4 and then 2 & 8 on block 2 : two pairs of claiming pairs. Then 5 & 9 on block 2 leaving the 6 & 7 pair in block 2.

so rewarding to solve a puzzle with a special trick you found and then watch the video and realize that was the intended path. I did a bit more snyder before coloring those 57s, and used them first to unwind a 67 in R3C2 into a naked 6, but that didnt break the puzzle so i then unwound the 257 into a naked 2 in R1C9 as you did and finished it.

Used 57 remote pairs coloring to place 2 at R1C9. Smooth sailing thereafter.

I managed to solve the puzzle without getting stuck. I didn't understand why the cells were colored starting out. I used temporary colors to visualize pairs and triples -- something I do often -- but those were the only colors I used. I went with letters to deal with 57 cells. My two final entries were changing the two letters to the digits.

I didn't understand the colors added to this at all, and I found them distracting. I know that content creators sometimes make changes just for the sake of changing, and I don't always think it's a good choice. I sometimes add colors myself, and I always pick the same color to represent any given number (e.g., 1 is always yellow, like the pool ball.) This grid made any of my coloring virtually impossible. My colors were blending in with the default colors, and it made my coloring very had to see. Anyway, the chain around the outer loop was obvious, and I got the puzzle done in 24:56, with minimal help from the video.

A fine puzzle, which is elegant even without strict symmetry. Oddly, I didn't use either simple or multiple colouring for it. I found what perhaps counts as an XY-Chain or Remote Pair, but followed in my own way (21 = row 2 column 1): 21, 17, 91, 88, 81, 94, 54, 42, 97 (hidden pair in box 3), 18 (busy box 2), 38, 12, 77, 92, 79, 99, 73, 32, 14, 34, 93, 19 (XY-Chain, or Remote Pair, followed from R1C1)*, 15 (order pretty much arbitrary from here), 16, 45, 65, 47, 76, 96, 89, 98, 75, 82, 71, 62, 48, 27, 29, 36, 35, 11, 13, 23, 43, 57, 67, 53, 63, 56, 66, 95, 52, 68, 58. * R1C1 = 5 → R8C2 = 5 → R9C8 = 5 → R2C9 = 5 (→ R2C = 7**) → R1C9 = 2. R1C1 = 7 → R8C2 = 7 → R9C8 = 7 → R2C9 = 7 (→ R2C = 5**) → R1C9 = 2. ** Remote Pair.

What software or app are you using to solve these puzzles? You have gotten me interested in Sudoku!

There are so many instant giveaways (galore) in this puzzle that it’ll take us awhile to whittle them out in an orderly fashion, as long as we keep due diligence, patience, always with eyes open wide, not missing out any details. 1) In Box7, we have right away [39] tcp on c1 btw rows 8&9 (do you see how? It’s quite obvious) AND concurrently, 268 triplet in the three open cells @(73),(92)&(93) (Hint: 268 given on c1&r8, both sets outside of B7), thereby leaving 57 cp in B7 @(71)&(82)!! ( as readily verified by direct point counting). 2) right away, 1 on c4@(94), followed by 3 @(54), leaving [67] tcp on c4B2. Note right away also 2598 given on c4 outside of B2 ==> they are quadruplet in the four open cells in B2 off c4, leaving [67] tcp on c4B2 (as readily verified by point counting). Further more, 2&8 given on r3&c4, both outside of B2 ==> [28] cp in B2@(16)&(25), to be followed immediately by [59] tcp on r3B2 btw cols 5&6!! (That’s a lot of information in one stretch!!) 3) right away 1 on r8@(88) (=> 1 on c7B6 btw rows 5&6), AND concurrently [68] tcp on c9B9 btw rows 7&9 (obvious). 4) right away, 2&8 on c8B6; 4 on r6B5 btw cols 5&6, on c2B1; 3 on r4@(42); 1 on c1@(21) ==> [57] cp on c1 @(11)&(71)!!; 5) right away, 943 given on r2&c9, both sets outside of B3==> 943 triplet in B3@ (17),(18)&(38), leaving cell(27)=[57](verifiable by direct count). Indeed 3 is at (17), both by row & column ==> [49] tcp on c8B3 !! In hot pursuit, 3 uncovered @(17) , together with 94 given on r2 and in B6, both off c7 ==> 4&9 are on c7B9, with 9 @(97), followed by 4@(77) => 4 in B8 on r9 btw cols 5&6 !! Also, 9 uncovered on r3B2 ==> 4@(38) ==> 4 on c2@(12)===> 8 in B1 on c3 AND on c2@(92)!!! (Hint : the two given 8s on rows 3&8 are off c2!), thereby leaving [26] tcp on c3B7 ==> 6 on c2@(32), followed by 2 in B4 on c2 btw 5&6, AND 2 in B8 on r7 !!(with 5 remaining in B8 undetermined!!) 6) last but not least, we have right away [19] tcp on c3B4 btw rows 5&6!!, by the crisscrossing of 1&9 given on c2&r4, both outside of B4, leaving 5&7 split in B4!! Now this is abt as far as the arduous but fruitful preliminary sweep will take us. As we enter the stage of mid game, it’s time to take a deep breath and a grand view across the grid and see how sort of development has brought us. We see a gamut of 5&7 cps on c1, in Boxes 7&8, splits in B4, as well expected as they are the only single clues given. So what’s the next step forward? Apparently a call for some educated/creative guess to be made somewhere, but where? The best bet is to look at those bivalued cells and search for candidates which can force a STRONG chain reaction towards a finish or a contradiction, knowing well one has to be true and the other false. Enough said, let’s look at the bivalued cell(98) = [57] and pick 7 !! 7@(97) would lead to 5@(89), leaving [27] tcp on c9B3 and at same time 7 on c6@(76) ==> 2 on c6@(16) and concurrently 7 on c1@(11). Jointly on the same row 1 , they REDUCE cell(19) = ZERO !!! So it’s 5@(98)&7@(89) !!! This will take us to the finish line. The readers are encouraged to try on their own and if they get stuck somewhere along the line, they can always check the tutorial below. So 5 uncovered @(98) ==> [25] tcp on r7B8, leaving [47] tcp on r9B8 btw cols 5&6. On the other hand, 7@(89) ==> 7 in B3@(27), leaving [25] tcp on c9B3, PLUS 7@(71) => 5 on c1@(11) and in B7 @(82) ==> 5 in B4@(43), bringing three ramifications : leaving [27] tcp in B4 on c2 btw rows 5&6 / leaving [78] tcp on r3B1 / together with 1 given @(46), and 1&5 uncovered on c8, both outside of B6, [15] tcp on c7B6 btw rows 5&6 !! Now in hot pursuit 5 uncovered @(11) and 7 uncovered @(27) lead to 2&(25)&8@(16) !! ==> [68] tcp on c5B5, leaving [49] tcp on c6B5 !! (as they should be) Now 8 given @(61) ==> 8@(45), and together with 2 given @(44), plus 28 given on c7, both pairs outside of B6 ==> [28] tcp on c8B6 btw rows 5&6, leaving [67] tcp on r4B6 !!! That should do it!!! Every cell has its own delegate and every candidate its own cell. In retrospect, each step after the next taken has been by pure logical deduction until the mid game where a maneuver of proof by contradiction is required to isolate the true candidate, and even then after that uncovering , the rest is smooth voyage. So to repeat no fuss no muss just a bump of contradiction trick in terms of technique.