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 00:16 It’s Solving Time 01:08 Story Behind This Puzzle 01:30 Crosshatching & Snyder Notation 02:46 BONUS Tip 04:11 Advanced Strategy Options 05:07 Overlooked Strategy 08:32 Hidden Pair 09:57 BONUS Strategy #1 12:07 BONUS Strategy #2
The 4:15 mark in this video is where Timberlake abandons Snyder and begins the search for the advanced strategy. When I reached that same point in my puzzle, I just kept going by filling in BVC's and Tri-value cells. That took me to 22 minutes, then I couldn't advance any more without just throwing darts and randomly attempting 2-string kites, x-wings, swordfishes or whatever. These random searches could take an incredible amount of time. So instead, I just clicked through the video until I saw where he pointed out the cells involved in the xyz wing. I didn't turn on the sound to see how it's solved, but I went back to my puzzle and was able to figure it out myself because I already had all those cells filled in with all possibilities. Then I cruised through the rest of the puzzle and finished in 39:36.
Neat, good show. How far was I from finding the key to unlock this great puzzle from Mr. Locke. My miserable option was bifurcation with the 6s. After trying to figure out whether there was a logival solution for them, the best I could do was to start with a 6 in r1c8 (rather than c9). Lucky no doubt, but how easy and promptly the puzzle was finished !
I did this my usual slow, leisurely way. In the lower band, I hit on an alternate inference chain. Row 8 almost had a 159 triple, but one of the cells was 1359. So I decided to chain on 3 vs. not3 in that cell. R9C4 had 79, and both cases forced 7. Later, I found an XYZ-wing: 139 in columns 5 and 6. I still hadn't centermarked the entire grid, so I still had some work to do, but it was straightforward. 4:10 Yes, that was approximately the state of the grid when I got stuck. I saw a couple five-digit rows to fill, and searched for some apparently restricted cells, and then began centermarking the grid. I can see the XYZ-wing in the grid now, with the blocks I filled. I didn't see it during the solve. Instead, the 159 almost-triple attracted me, and I got lucky and placed 7 in R9C4. I only filled out further cells until I saw the XYZ-wing. 4:30 Speaking of finned X-wings... I didn't find any here, although I didn't look too hard. But I've lost count of the number of times a finned X-wing substituted for the intended strategy. 7:20 I once found an XYZ-wing in an NYT hard puzzle. When Rangsk did the puzzle, he placed a naked 5 which broke up the XYZ-wing. He replied that NYT hard puzzles aren't supposed to use any of those advanced techniques, but I was welcome to apply them if I found them.
Ugh. I’m not giving up, but I’m struggling. I can do an easy puzzle in under 5 minutes. But I cannot get anywhere on a medium. So right off the bat …. Please tell me why you started with block 3. What drew you to that? I start with block 1.
Heh. I feel for you! Keep it up 👍🏼There is no "right" way to solving obviously, but to answer your question, it's not that he "picked" Block 3~it just so happens he is solving for 1s first (then 2s, 3s, 4s and so on) and the 1s (c1r3, c8r9) intersect in that block.
I would say that these are videos that feature the best path to solve each puzzle quickly; that path may or may not be obvious to the average sudoku player. This uploader must already know the solution at the time of filming the video, for he often has important cells highlighted from the very start of the video, and there are other signs as well that he already knows exactly what to focus on. Cracking the Cryptic channel is different in that they work to try to solve the puzzle themselves as they go along in the video, thus their video lengths are often longer than the ones here showing the same puzzles. What keeps me going is I give little regard to the times posted by other commenters, for there is no way to verify the validity of those claims. I finished this puzzle in 39:36 using one big jolt of help from the video, and that's okay with me. (there is additional time where I paused the puzzle timer to click on the video). I suppose you might think that I'm writing false completion times for my puzzles as well, but if I was going to give fake times, I'd come up with better times than I claim. 😐
Presently,I’m having a go at rated “expert” on my app.I can do the “hard” without too much problem.I know most strategic methods BUT spotting them requires a lot of practise.Good to see the “mallet” explained 👍
This is readily solved by hitting the jackpot in picking the right candidate to be the CLOSURE!!! First the preliminary sweep across the board : Instant takeaway for 5 on c6@(86), 5 splits on r9B7, in B3. 7 on c8@(88), splits on r9B8, on c6 @(26)&(46)(highly relevant)! Looking for bivalued cells by direct point counting, we have (35)=[37] (relevant!), (76)=[19], (99)=[23]; 8 on r7@(77), on c9B6. 9 splits in B2. 1 splits in B3. More significantly, 68 given on c1&r3, both outside of B1 ==> [68] cp(conjugate pairs) in B1@(22)&(13). Now 6 given on c3 out of B1=> 6@(22)&8@(13)!! => 6 on r1B3 and 8 on r2@(24)(see how?)!!=> 8 in B5 on c5!! plus 8 on r5 @(52), by row 5(in view of 8 given on cols 6&8, and uncovered on cols 347, all five off r5! OR simply by noticing 8 in Boxes 5&6 are on cols 5&9, forming an X-Wing with both off r5 => 8 in B4 on r5!). Now both 3&7 in cell(35) can force a sustaining chain reaction. Consider first 7@(37)! It would lead to 7 on r2@(21)& on c6@(46)===> 7 on r6@(63) !! Note 7 uncovered @(21)B1 ==> 4 in B1 on r3& 4 on r2@(28)!!(see how?)==> 4 on r5@(53)===> 5 on col 3@(33)!!! (See how?)+=> 4 in B1 @(32) , immediately followed by 9 on r3@(34) , and also followed by 2 in B2@(26), leaving 1 on r2@(27), plus leaving [23] tcp on r3B3 and thereby leaving [56] tcp on r1B3. With 5 given @(79), 6@(19)& with 9 given @(29), together ==> [69] cp in B9@(97)&(78), leaving [23] tcp on r9B9!! ====> cell (39)= ZERO!!! So it’s 3@cell(35)!!! It would be the true closure . That is from now onwards every step taken will be by pure logical deduction, no fancy name calling, no more second guessing, just due diligence and patience. What follows is to help the ‘novice’ in solving : To be continued in the reply comment box below :
So 3&6 uncovered in B1 on r2 and in B2 on r3 ==> [36] tcp on r1B3 !! This leads to several ramifications !! Right away 5 in B3 is @(37) by cc(crisscrossing) of the two 5s given on r2&c9, both outside of B3, plus 5 on r1@(11), thereby leaving [19] tcp on r1B2! The next takeaway is not so obvious but with eyes wide open, 18 given on r3, 1 given in B9, 8s given on rows 8&9, all off c9 ==> [18] tcp on c9B6 !!! to be immediately followed by 4 on c9@(39) and next by 2 on c9@(99)(hope u see how!) 4 uncovered @(39) leaves [12] tcp on r2B3,and leads to 2 on r3@(34)!, whereby leaving 7 in B2 @(26), and also leaving [79] tcp on r3B1, plus 4 on r2@(21). So except for the [36] tcp in B3, all the open cells are isolated! (how so?). Note 3&6 cp on c9@(19)&(89)! Now in hot pursuit of the uncovering of 4&5 on c1 , both off r9 ==> 5 on r9@(92), followed by 4 on r9@(95), and 7 on r9@(94), leaving 3&9 cp on r9@(91)&(97)!! Note an extra bonus: [24] tcp on r7B7, by virtue of 24 given on r8& uncovered on r9, both outside of B7. Now turning our attention to 2 uncovered on r2@(34) and 7 uncovered on r9@(94), they lead to 9 on c4@(84)!!!, resulting in four repercussions : leaving [16] tcp on c4B5 ! and leaving [16] tcp in B8 on r7 !! plus 9 on r7@(78)! Last but not least, 6 on r8@(89), leaving [13] tcp on r8B7 and leaving 3 in B9@(97) ==> 9 on r9@(91) !! Upon close examination, all open cells in the Bottom Block are isolated! Now time to look at the middle row Block : 7&8 uncovered on cols 4&6 outside of B5 ==> [78] tcp on c5B5, leaving [23] tcp on c6B5 !! Also 4 uncovered @(73) and 1 uncovered @(83) ==> [14] tcp on c2 btw rows 4&6! Now 6&9 uncovered on cols 8&9 and on r5, all off B6 ===> [69] tcp on c7B6 btw rows 4&6! leaving 2 on c7@(57) ==> in hot pursuit with 9@(55), [29] tcp on c3B4 btw rows 4&6, leaving 5 in B4@(53)!! Note 4 uncovered @(42), together with 5 given @(44) ==> [45] tcp on c8B6 btw rows 5&6 !! , leaving 3 in B6@(48)! That should do it!!! All the remaining open TCPs are to be readily broken up by their respective danglers!! Note on c4, 6@(64)&1@(54). For example, 3 uncovered @(48) breaks up the [23] tcp on c6B4 with 2 uncovered @(57), and also together with 4 given @(66), breaks up the four TCPs between rows 4&6 : [37], [78],[81],[14].
@@SmartHobbies Thanx for your kind words . In watching your clip, your key breakthrough is the detection of 139 forming an XYZ system at cells (15),(16)&(76). This results in 1 not at (26) looking at 3&9 in (15)&(76) at same time , but 1 on r2@(27), leading immediately to [18] tcp on c9B6, to be followed by 4 on c9@(39) (and 2 on c9@(99) ==> 4 on r2@(21) ==> 7 in B1@(33) !! ===> 3@(35) by point counting!!! and the rest is history!!! So I choose an alternative route by zeroing in on a closure, for that purpose digging out bivalued cells are important steps during the preliminary sweep. Again the saying is : all roads lead to Rome. And there are more than one way to skin a cat!!
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
00:16 It’s Solving Time
01:08 Story Behind This Puzzle
01:30 Crosshatching & Snyder Notation
02:46 BONUS Tip
04:11 Advanced Strategy Options
05:07 Overlooked Strategy
08:32 Hidden Pair
09:57 BONUS Strategy #1
12:07 BONUS Strategy #2
Nice puzzle and solve. This stumped me, took me almost an hour. Great trick!
Thank you for the feedback, Laszlo. I am glad the strategy helped you with the solving.
The 4:15 mark in this video is where Timberlake abandons Snyder and begins the search for the advanced strategy. When I reached that same point in my puzzle, I just kept going by filling in BVC's and Tri-value cells. That took me to 22 minutes, then I couldn't advance any more without just throwing darts and randomly attempting 2-string kites, x-wings, swordfishes or whatever. These random searches could take an incredible amount of time. So instead, I just clicked through the video until I saw where he pointed out the cells involved in the xyz wing. I didn't turn on the sound to see how it's solved, but I went back to my puzzle and was able to figure it out myself because I already had all those cells filled in with all possibilities. Then I cruised through the rest of the puzzle and finished in 39:36.
Neat, good show. How far was I from finding the key to unlock this great puzzle from Mr. Locke. My miserable option was bifurcation with the 6s. After trying to figure out whether there was a logival solution for them, the best I could do was to start with a 6 in r1c8 (rather than c9). Lucky no doubt, but how easy and promptly the puzzle was finished !
I am sorry to hear that you went down the wrong track Georges. Thought you might appreciate the strategy that unlocked this one.
I did this my usual slow, leisurely way. In the lower band, I hit on an alternate inference chain. Row 8 almost had a 159 triple, but one of the cells was 1359. So I decided to chain on 3 vs. not3 in that cell. R9C4 had 79, and both cases forced 7. Later, I found an XYZ-wing: 139 in columns 5 and 6. I still hadn't centermarked the entire grid, so I still had some work to do, but it was straightforward.
4:10 Yes, that was approximately the state of the grid when I got stuck. I saw a couple five-digit rows to fill, and searched for some apparently restricted cells, and then began centermarking the grid. I can see the XYZ-wing in the grid now, with the blocks I filled. I didn't see it during the solve. Instead, the 159 almost-triple attracted me, and I got lucky and placed 7 in R9C4. I only filled out further cells until I saw the XYZ-wing.
4:30 Speaking of finned X-wings... I didn't find any here, although I didn't look too hard. But I've lost count of the number of times a finned X-wing substituted for the intended strategy.
7:20 I once found an XYZ-wing in an NYT hard puzzle. When Rangsk did the puzzle, he placed a naked 5 which broke up the XYZ-wing. He replied that NYT hard puzzles aren't supposed to use any of those advanced techniques, but I was welcome to apply them if I found them.
Rangsk makes a good point. Neat that you found that in a NYT Hard.
11:34 to solve. I didn’t need a xyz chain so will take that! Happy Sunday Timberlake and I hope you have had an awesome weekend?
I did, Jon. How about you? Great job with the solve!
@@SmartHobbies very good thanks Timberlake. Travelling to Lincolnshire this week for work purposes!
@@jonbrowne I have never been there. Is that along the coast?
@@SmartHobbies it’s the east of England about 40 minutes from Nottingham!
Ugh. I’m not giving up, but I’m struggling. I can do an easy puzzle in under 5 minutes. But I cannot get anywhere on a medium. So right off the bat …. Please tell me why you started with block 3. What drew you to that? I start with block 1.
Block 1 is a good start here, as one can place two digits. But each to his own. I began with 1s, the digit.
Heh. I feel for you! Keep it up 👍🏼There is no "right" way to solving obviously, but to answer your question, it's not that he "picked" Block 3~it just so happens he is solving for 1s first (then 2s, 3s, 4s and so on) and the 1s (c1r3, c8r9) intersect in that block.
I would say that these are videos that feature the best path to solve each puzzle quickly; that path may or may not be obvious to the average sudoku player. This uploader must already know the solution at the time of filming the video, for he often has important cells highlighted from the very start of the video, and there are other signs as well that he already knows exactly what to focus on. Cracking the Cryptic channel is different in that they work to try to solve the puzzle themselves as they go along in the video, thus their video lengths are often longer than the ones here showing the same puzzles. What keeps me going is I give little regard to the times posted by other commenters, for there is no way to verify the validity of those claims. I finished this puzzle in 39:36 using one big jolt of help from the video, and that's okay with me. (there is additional time where I paused the puzzle timer to click on the video). I suppose you might think that I'm writing false completion times for my puzzles as well, but if I was going to give fake times, I'd come up with better times than I claim. 😐
Block I already has a 1
Presently,I’m having a go at rated “expert” on my app.I can do the “hard” without too much problem.I know most strategic methods BUT spotting them requires a lot of practise.Good to see the “mallet” explained 👍
This is readily solved by hitting the jackpot in picking the right candidate to be the CLOSURE!!!
First the preliminary sweep across the board :
Instant takeaway for 5 on c6@(86), 5 splits on r9B7, in B3.
7 on c8@(88), splits on r9B8, on c6 @(26)&(46)(highly relevant)!
Looking for bivalued cells by direct point counting, we have
(35)=[37] (relevant!),
(76)=[19], (99)=[23];
8 on r7@(77), on c9B6.
9 splits in B2.
1 splits in B3.
More significantly,
68 given on c1&r3, both outside of B1 ==> [68] cp(conjugate pairs) in B1@(22)&(13).
Now 6 given on c3 out of B1=> 6@(22)&8@(13)!! => 6 on r1B3 and 8 on r2@(24)(see how?)!!=> 8 in B5 on c5!! plus 8 on r5 @(52), by row 5(in view of 8 given on cols 6&8, and uncovered on cols 347, all five off r5! OR simply by noticing 8 in Boxes 5&6 are on cols 5&9, forming an X-Wing with both off r5 => 8 in B4 on r5!).
Now both 3&7 in cell(35) can force a sustaining chain reaction.
Consider first 7@(37)!
It would lead to 7 on r2@(21)& on c6@(46)===> 7 on r6@(63) !!
Note 7 uncovered @(21)B1 ==> 4 in B1 on r3& 4 on r2@(28)!!(see how?)==> 4 on r5@(53)===> 5 on col 3@(33)!!! (See how?)+=> 4 in B1 @(32) , immediately followed by 9 on r3@(34) , and also followed by 2 in B2@(26), leaving 1 on r2@(27), plus leaving [23] tcp on r3B3 and thereby leaving [56] tcp on r1B3.
With 5 given @(79), 6@(19)& with 9 given @(29), together ==> [69] cp in B9@(97)&(78), leaving [23] tcp on r9B9!! ====> cell (39)= ZERO!!!
So it’s 3@cell(35)!!!
It would be the true closure .
That is from now onwards every step taken will be by pure logical deduction, no fancy name calling, no more second guessing, just due diligence and patience.
What follows is to help the ‘novice’ in solving :
To be continued in the reply comment box below :
Great job! Thank you for sharing. You seem passionate about solving Sudoku!
So 3&6 uncovered in B1 on r2 and in B2 on r3 ==> [36] tcp on r1B3 !!
This leads to several ramifications !!
Right away 5 in B3 is @(37) by cc(crisscrossing) of the two 5s given on r2&c9, both outside of B3, plus 5 on r1@(11), thereby leaving [19] tcp on r1B2!
The next takeaway is not so obvious but with eyes wide open, 18 given on r3, 1 given in B9, 8s given on rows 8&9, all off c9 ==> [18] tcp on c9B6 !!! to be immediately followed by 4 on c9@(39) and next by 2 on c9@(99)(hope u see how!)
4 uncovered @(39) leaves [12] tcp on r2B3,and leads to 2 on r3@(34)!, whereby leaving 7 in B2 @(26), and also leaving [79] tcp on r3B1, plus 4 on r2@(21).
So except for the [36] tcp in B3, all the open cells are isolated! (how so?). Note 3&6 cp on c9@(19)&(89)!
Now in hot pursuit of the uncovering of 4&5 on c1 , both off r9 ==> 5 on r9@(92), followed by 4 on r9@(95), and 7 on r9@(94), leaving 3&9 cp on r9@(91)&(97)!!
Note an extra bonus:
[24] tcp on r7B7, by virtue of 24 given on r8& uncovered on r9, both outside of B7.
Now turning our attention to 2 uncovered on r2@(34) and 7 uncovered on r9@(94), they lead to 9 on c4@(84)!!!, resulting in four repercussions :
leaving [16] tcp on c4B5 ! and leaving [16] tcp in B8 on r7 !! plus 9 on r7@(78)!
Last but not least, 6 on r8@(89), leaving [13] tcp on r8B7 and leaving 3 in B9@(97) ==> 9 on r9@(91) !!
Upon close examination, all open cells in the Bottom Block are isolated!
Now time to look at the middle row Block :
7&8 uncovered on cols 4&6 outside of B5 ==> [78] tcp on c5B5, leaving [23] tcp on c6B5 !!
Also 4 uncovered @(73) and 1 uncovered @(83)
==> [14] tcp on c2 btw rows 4&6!
Now 6&9 uncovered on cols 8&9 and on r5, all off B6 ===> [69] tcp on c7B6 btw rows 4&6! leaving 2 on c7@(57) ==> in hot pursuit with 9@(55), [29] tcp on c3B4 btw rows 4&6, leaving 5 in B4@(53)!!
Note 4 uncovered @(42), together with 5 given @(44) ==> [45] tcp on c8B6 btw rows 5&6 !! , leaving 3 in B6@(48)!
That should do it!!!
All the remaining open TCPs are to be readily broken up by their respective danglers!!
Note on c4, 6@(64)&1@(54).
For example, 3 uncovered @(48) breaks up the [23] tcp on c6B4 with 2 uncovered @(57), and also together with 4 given @(66), breaks up the four TCPs between rows 4&6 : [37], [78],[81],[14].
@@SmartHobbies
Thanx for your kind words .
In watching your clip, your key breakthrough is the detection of 139 forming an XYZ system at cells (15),(16)&(76).
This results in 1 not at (26) looking at 3&9 in (15)&(76) at same time , but 1 on r2@(27), leading immediately to [18] tcp on c9B6, to be followed by 4 on c9@(39) (and 2 on c9@(99) ==> 4 on r2@(21) ==> 7 in B1@(33) !! ===> 3@(35) by point counting!!! and the rest is history!!!
So I choose an alternative route by zeroing in on a closure, for that purpose digging out bivalued cells are important steps during the preliminary sweep.
Again the saying is : all roads lead to Rome. And there are more than one way to skin a cat!!
@@ckoh010 I love it when viewers find and share different solve paths.