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 00:18 It’s Solving Time 00:49 Puzzle Story 01:13 Finding Snyder Restrictions 03:40 What To Do After Snyder 04:54 Finding Bivalue Cells 06:56 Powerful BVC Strategy 08:50 Unlocking More Cells 11:25 Powerful Single Candidate Strategy 15:19 Searching For Next Strategy 16:15 BONUS Tip
A nice way to find the 3 in r8c8 is to notice the empty rectangle for 1 in box 1. r8c2 and r2c8 cannot both be 1 because they nullify the possibilities of 1 in box 1. Therefore at least one of the two cells is a 2, so the cell that they see in common (r8c8) cannot be 2, therefore it is 3. The 3 in r8c8 concludes the 3 in r7c6 so the rest is the same.
14:09 I've done this a few times. My name for it is "direct exclusion", because your chain ends in the same house it starts in. But now I know it's a discontinuous loop, so I'll start calling it that. I've found those to be quite rare, but they are very powerful--because if you start from a BVC and land on ANY digit on the same house, that means that the OTHER digit from your STARTING cell can be eliminated from the cell you LANDED in. For example, you started with 23@IX5 (r8c8 in standard notation). Then you found that 2@IX5 would lead to 4@IX2 (r7c8), and since the other digit at your starting point is a 3, which would clearly also eliminate a 3 from IX2, you can eliminate a 3 from IX2! Also, check out block I at the beginning. There's that "under-locked" set thing I was talking about the last time I commented. You have 1 and 3 looking in from row 1, and 1 and 3 looking in from column 1, so you need to find one more digit in r1 or c1 outside of block I to get a naked triple. Here the other digit turned out to be a 4. I haven't quite figured out what to do with these yet, other than using them as a link in a chain, but they are fascinating!
Hi Jon! Welcome back. It’s great to hear from you. I understand about being busy this time of year. I hope you get a chance to catch up on the videos. 👍🏻
12:31 Proud of myseld for finding the 12 """pair"""at R2C8 and R8C2!!! (It can't be both 1) You solved in another way, nice! I couldn't see this path! 😊
Bi valued cells 12 in boxes 3 and 7 and bivalued cell 23 in box 9 . Let us focus our attention on these 3 cells . If R8C2(12) is 2 then R8C8(23) becomes 3 and if R8C2 is 1 then R2C3 is 1 and R2C8 is 2 which again leads to the same result making R8C8 as 3. Thus R8C8 is 3 in each case. This is sufficient to solve the puzzle completely.
I think that I placed, three 2s, two 8s, and a 9 before filling out the grid. Block 9 had a 23 cell facing two 12 cells, and I showed that at least one of the 12 cells was a 2, placing the 3. That enabled me to place all the 3s. But that's all I got for a while. I decided that something was up with the 14 cell in block 1. My mind kept jumping around, and I kept losing what I thought I saw. So I placed 4 in the corner of the cell, 1 in the center -- graying the cells to remind me that the usual convention was contravened. Centermarking went one way, cornermarking another. Both gave me 7 in R1C7, which gave me a couple other digits -- and a centermarking null. I selected the grays, removed the centermarks, and placed the cornermarks as digits (or pairs). The remainder was unraveling the centermarks. 7:40 Ack! I hope that I learn one thing: pay more attention to bivalue cells. I saw it with the 12 and 23 cells, but not the 49 cell and its chain. It removes the 4 from R2C6 (my centermarking null cell), and places the 4 in R2C3 (my bifurcating cell). Once that happens, it's all unraveling the centermarks. Or just follow a better chain from where I began: 1 in R2C3 leads to 4 in R5C6, eliminating the same 4. I can see it once something draws my attention to it, but otherwise... 8:30 Once the 4 in R2C6 is eliminated, I get a 68 pair, and the rest of the centermarks unravel. Something nice may be said for filling out the grid. 11:30 I'd already earlier used the 12s to place the 3 in R8C8. Therefore, nothing remained to get stuck on. I remember somewhere, you said something about changing mental states when going from one type to another. It seems to me that using the 12s is easier to see than what you did. And you're using the same method as before. (Maybe it's me. Something about identical bi-value cells attracts me to them.)
@@SmartHobbies I don't know, I'm sorry to say. Typically when something attracts my attention. (The two 12 cells attracted me.) I tried figuring out whether they were the same, opposite, or other. All I found was that if R8C2 was 1, R2C8 had to be 2 -- and that was enough to place the 3.
That was exactly the intended break in nicely done! I definitely still get choked up on the basics often so I can relate to the struggle with the aftersolve. I can say for my puzzles at least, that I try my best to make the first few digits after a difficult break-in as obvious as possible and chained together in a way such that the rest of the solve is a breeze ;)
After getting few no's and XY chain, One of the bifurcation R2C3 (14) is 1 or 4 R8C8 mustbe 3. If it is 1 R2C8 is 2, then R8C8 is 3. If it is 4 1s in R23C2, R2C3 is 2, then R8C8 is 3, puzzle solved.
Puzzle solved other ways :- W wing 12s, conjugate 1s in Col 3, effect eliminating 2 in R8C8 (23=3) puzzle solved. I found this one after i saw in comment already commented. The bifurcation ways., 14 R2C3 12 R2C8 13 R3C2 47 R7C2 12 R8C2 39 R7C6 79 R7C9 These all cell no's targetting one cell this or that R8C8 mustbe 3. Puzzle soved more other 7 ways. Nice puzzle from SULPHER, I watched vedio because stuck after finding some no's and XY chain. Bifurcation gives me idea.
This took me 65:41 to solve, but I'm not too upset about it because I came up with a way that helped me to find all the important cells quickly. If I had seen the xy chain, I could have solved much easier because everything else important was already filled in.
@@SmartHobbies I'd been losing too much time just randomly looking around for either BVC's or naked singles. Sometimes I will run Snyder all the way from 1-9, and still not have much to work with. So then I tried clicking on each number individually to see which ones have been solved the LEAST amount of times. (If the entire puzzle were filled in with all possible candidates, these numbers will show up as candidates most often.) Look at this puzzle: after Snyder, the 4,6& 7 have all been solved twice, so they are represented the least. Then I focus on looking at the cells that fall within the houses of those numbers, and it's even better if a cell is covered by 2 or all 3 of those numbers. So in this puzzle, we immediately get a 59 BVC in block 8. Then we get a 23 in block 9, and a 12 in block 3. Then we get a 12 and a 15 in block 7. All this combined is enough for a major solve, because when you look at how the 12's interact with block 1, you know that they both can't be 1's because there would be no room for a 1 in block 1. This fact makes at least one of the 12's a 2, and that puts a 3 in the 23 cell in block 9. This solves every 3 in the entire puzzle, and gives you a 14 in block 1. If you mention this technique in a video, please give me some credit for it because I'd never heard of it being done before.
At around 6:30 I we had the same solve but I looked at R2C8, R8C2 and R8C8 if R8C2 isn't a 2 then R2C8 is so R8C8 is a 3 and the puzzle solves easily from there.
Just now I watched ur video. At 14:40 u discussed abt discont loop relating to two weak links starting frm the cell R8C8 (23)with a strong link frm 3 to 2 within the cell and then ending at cell R8C8 . That's fine . Instead if we had stopped at cell R7C8 with a strong link at 4 in this cell. Then we can treat this as an AIC type 2( as it starts with a strong link frm 3 and ends with a strong link at 4 and both the cells R7C8 and R8C8 see each other, therefore the start candidate '3' can be dismissed frm end cell R7C8 ) Am I right? I hv never applied discont / cont loop. Thanks for repeating this concept .
Great puzzle. Instead of using the single candidate strategy on the 3's, I noticed if cell R8C8 = 2, then the BVC's along both row 2 and col 2 leave cell R2C2 with no value, so 3 can be placed in R8C8. What's the name of the strategy there?
It does seem that way. The actual strategy is a Discontinuous Loop, which uses a series of strong and weak links to get to the same conclusion. Do you like to solve with bifurcation or do you try to avoid it?
Any kind of chaining is bifurcation. In fact, all advanced positions are based on bifurcation. Not2 in R8C2 makes it 1, forces 1 in R2C3, and 2 in R2C8. One of R2C8 and R8C2 must be 2, and R8C8 sees both. (This was my tactic, not necessarily the video's.) EDIT: I replied first before seeing how the video treated the 3. I see that it was completely different.
I am resigned to the ugly reality that I'll never, ever be able to do a grid like this by conventional means. Although I found an XY-Wing along the way, my main strategy was using several virtual chains. I'll watch this out of curiosity, but I am certain that even if I can understand the logic, I could never spot it in the wild. Follow-up while watching the video: Interestingly, I found that 4 in box 2 by different means, but it was my first write-in after the few easy cells at the beginning. Oh, I actually found the 4 in box 5 before the one in box 2. In any case, it led to the same sequence of easy 4s elsewhere. Well, I was right. I would never have spotted the Discontinuous Loop as such, but some of my virtual chains are not unlike it.
Heh. I know what you mean! But hopefully you meant, "XY Chains"- as XY Wings (aka Y Wings) are separate solving strategies. For me, finding them is easier when (unfortunately) I fill out the grid entirely which does take hard work & experience because you want to be accurate in your Snyder notation. Timberlake can do it the way he does it because he is soo experienced (and the Man! 🙌🏼). After all that, then I start looking for BVCs and any commonalities. Just keep playing (and watching this channel 👍🏼) eventually you'll spot XY Chains and other solving strategies. Don't be bogged down by trying to understand any new strategy you learn~just keep grinding and like I said, you'll eventually begin to see it and implement the strategy. Hope that helps 🤙🏼
@@deaahimm6334 No, I meant an XY-Wing. I really did find one, but only after following two or three virtual chains, by which I mean chains based on a starting hypothesis (i.e., mental bifurcation).
Oh ok, my bad. Well, my advice still holds~don't trip out on trying to "find" them... eventually they'll find you. Heh. For years I'd been solving without any knowledge of strategies, until about 3-4 years ago (during the Pandemic) I started to actually watch Sudoku channels and realized I'd been using some strategies already (X-wings, Y-wings, XY-chains, URs) although crudely. I'm still grinding away, and even though I'm far better than I was back then, I sometimes go back to an old puzzle that gave me fitz (heh that took me days to solve) and look for how I could've solved it more simply & efficiently. And I'm like, "geez it was right there!" Anyways, keep grinding bro 🤙🏼
(1,3)=9 (1,5)=2; tercia(4,6,7) en 1a fila y en 1a col ent (7,1) =8 (5,1)=2 (6,6)=2 (5,3)=8; Tanteo Indirecto: si (5,6)=4 ent (1,4)=4 o si (5,6)=9 ent (5,7)=7 (1,7)=6 y (1,4)=9, o sea en ambos casos se llega a (1,4)=4 formándose el par (6,7) en 1a fila y en 1a col dentro del bloque 1 ent (4,1)=4 (6,2)=6 (5,6)=4 (6,9)=4; par (5,9) en 5a col quedando la tercia(1,6,8) en dicha col ent (6,5)=9 (7,5)=5; Tanteo por Combinaciones: primero (8,8)=2 con (8,3)=1 se llega a un Error! en la casilla (8,2) = (1,2) quedaría vacia y luego (8,8)=2 con (8,3)=5 Error! se duplica el #1, por lo tanto en (8,8)=3 (6,7)=3 (6,3)=7 (4,3)=3 (7,6)=3 (3,5)=6 (3,1)=7 (1,1)=6 (1,7)=7 (5,7)=9 (4,9)=5 (4,6)=6 (4,5)=1 (4,8)=7 (5,9)=1 (5,4)=7 (6,4)=5 (3,4)=3 (3,2)=1 (2,2)=3 (2,3)=4 (8,2)=2...
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Timestamps
0:00 Intro
00:18 It’s Solving Time
00:49 Puzzle Story
01:13 Finding Snyder Restrictions
03:40 What To Do After Snyder
04:54 Finding Bivalue Cells
06:56 Powerful BVC Strategy
08:50 Unlocking More Cells
11:25 Powerful Single Candidate Strategy
15:19 Searching For Next Strategy
16:15 BONUS Tip
A nice way to find the 3 in r8c8 is to notice the empty rectangle for 1 in box 1. r8c2 and r2c8 cannot both be 1 because they nullify the possibilities of 1 in box 1. Therefore at least one of the two cells is a 2, so the cell that they see in common (r8c8) cannot be 2, therefore it is 3. The 3 in r8c8 concludes the 3 in r7c6 so the rest is the same.
This is the logic I saw too 👍
Hello it is I Sulphur. I can confirm that this indeed was the intended solution :)
@@drkn0ckers715 Great, a very cool puzzle :)
14:09 I've done this a few times. My name for it is "direct exclusion", because your chain ends in the same house it starts in. But now I know it's a discontinuous loop, so I'll start calling it that. I've found those to be quite rare, but they are very powerful--because if you start from a BVC and land on ANY digit on the same house, that means that the OTHER digit from your STARTING cell can be eliminated from the cell you LANDED in. For example, you started with 23@IX5 (r8c8 in standard notation). Then you found that 2@IX5 would lead to 4@IX2 (r7c8), and since the other digit at your starting point is a 3, which would clearly also eliminate a 3 from IX2, you can eliminate a 3 from IX2!
Also, check out block I at the beginning. There's that "under-locked" set thing I was talking about the last time I commented. You have 1 and 3 looking in from row 1, and 1 and 3 looking in from column 1, so you need to find one more digit in r1 or c1 outside of block I to get a naked triple. Here the other digit turned out to be a 4. I haven't quite figured out what to do with these yet, other than using them as a link in a chain, but they are fascinating!
Awesome. Thank you for sharing.
Great puzzle but very hard to solve. Thanks Timberlake. Been a very busy past 6 weeks and apologies if I have not seen all of your great videos!
Hi Jon! Welcome back. It’s great to hear from you. I understand about being busy this time of year. I hope you get a chance to catch up on the videos. 👍🏻
12:31
Proud of myseld for finding the 12 """pair"""at R2C8 and R8C2!!! (It can't be both 1)
You solved in another way, nice! I couldn't see this path! 😊
Nice job!!!!
It's called W wing conjugate 1s in Col 3, effect R8C8 is 3.
Thats how I solved. Definitely a cleaner, more logical approach than in the video and probably what Sulphur intented
@@Ramakrishnagm Yes! Thanks for the info!
Bi valued cells 12 in boxes 3 and 7 and bivalued cell 23 in box 9 .
Let us focus our attention on these 3 cells .
If R8C2(12) is 2 then R8C8(23) becomes 3 and if R8C2 is 1 then R2C3 is 1 and R2C8 is 2 which again leads to the same result making R8C8 as 3.
Thus R8C8 is 3 in each case.
This is sufficient to solve the puzzle completely.
I think that I placed, three 2s, two 8s, and a 9 before filling out the grid. Block 9 had a 23 cell facing two 12 cells, and I showed that at least one of the 12 cells was a 2, placing the 3. That enabled me to place all the 3s.
But that's all I got for a while. I decided that something was up with the 14 cell in block 1. My mind kept jumping around, and I kept losing what I thought I saw. So I placed 4 in the corner of the cell, 1 in the center -- graying the cells to remind me that the usual convention was contravened. Centermarking went one way, cornermarking another. Both gave me 7 in R1C7, which gave me a couple other digits -- and a centermarking null. I selected the grays, removed the centermarks, and placed the cornermarks as digits (or pairs).
The remainder was unraveling the centermarks.
7:40 Ack! I hope that I learn one thing: pay more attention to bivalue cells. I saw it with the 12 and 23 cells, but not the 49 cell and its chain. It removes the 4 from R2C6 (my centermarking null cell), and places the 4 in R2C3 (my bifurcating cell). Once that happens, it's all unraveling the centermarks. Or just follow a better chain from where I began: 1 in R2C3 leads to 4 in R5C6, eliminating the same 4.
I can see it once something draws my attention to it, but otherwise...
8:30 Once the 4 in R2C6 is eliminated, I get a 68 pair, and the rest of the centermarks unravel. Something nice may be said for filling out the grid.
11:30 I'd already earlier used the 12s to place the 3 in R8C8. Therefore, nothing remained to get stuck on.
I remember somewhere, you said something about changing mental states when going from one type to another. It seems to me that using the 12s is easier to see than what you did. And you're using the same method as before. (Maybe it's me. Something about identical bi-value cells attracts me to them.)
Great observations, John. This was not an easy puzzle. How often do you look for XY-Chains in your solving?
@@SmartHobbies I don't know, I'm sorry to say. Typically when something attracts my attention. (The two 12 cells attracted me.) I tried figuring out whether they were the same, opposite, or other. All I found was that if R8C2 was 1, R2C8 had to be 2 -- and that was enough to place the 3.
@@JohnRandomness105 That works for me. Thanks John!
That was exactly the intended break in nicely done! I definitely still get choked up on the basics often so I can relate to the struggle with the aftersolve. I can say for my puzzles at least, that I try my best to make the first few digits after a difficult break-in as obvious as possible and chained together in a way such that the rest of the solve is a breeze ;)
After getting few no's and XY chain,
One of the bifurcation
R2C3 (14) is 1 or 4 R8C8 mustbe 3.
If it is 1 R2C8 is 2, then R8C8 is 3.
If it is 4 1s in R23C2, R2C3 is 2, then
R8C8 is 3, puzzle solved.
Puzzle solved other ways :-
W wing 12s, conjugate 1s in Col 3, effect eliminating 2 in R8C8 (23=3) puzzle solved. I found this one after i saw in comment already commented.
The bifurcation ways.,
14 R2C3
12 R2C8
13 R3C2
47 R7C2
12 R8C2
39 R7C6
79 R7C9
These all cell no's targetting one cell this or that R8C8 mustbe 3. Puzzle soved more other 7 ways.
Nice puzzle from SULPHER, I watched vedio because stuck after finding some no's and XY chain.
Bifurcation gives me idea.
This took me 65:41 to solve, but I'm not too upset about it because I came up with a way that helped me to find all the important cells quickly. If I had seen the xy chain, I could have solved much easier because everything else important was already filled in.
That is great to hear, Brad. Can you explain a little more about the method you used?
@@SmartHobbies I'd been losing too much time just randomly looking around for either BVC's or naked singles. Sometimes I will run Snyder all the way from 1-9, and still not have much to work with. So then I tried clicking on each number individually to see which ones have been solved the LEAST amount of times. (If the entire puzzle were filled in with all possible candidates, these numbers will show up as candidates most often.) Look at this puzzle: after Snyder, the 4,6& 7 have all been solved twice, so they are represented the least. Then I focus on looking at the cells that fall within the houses of those numbers, and it's even better if a cell is covered by 2 or all 3 of those numbers. So in this puzzle, we immediately get a 59 BVC in block 8. Then we get a 23 in block 9, and a 12 in block 3. Then we get a 12 and a 15 in block 7. All this combined is enough for a major solve, because when you look at how the 12's interact with block 1, you know that they both can't be 1's because there would be no room for a 1 in block 1. This fact makes at least one of the 12's a 2, and that puts a 3 in the 23 cell in block 9. This solves every 3 in the entire puzzle, and gives you a 14 in block 1. If you mention this technique in a video, please give me some credit for it because I'd never heard of it being done before.
At around 6:30 I we had the same solve but I looked at R2C8, R8C2 and R8C8 if R8C2 isn't a 2 then R2C8 is so R8C8 is a 3 and the puzzle solves easily from there.
Very nice find :3
Just now I watched ur video.
At 14:40 u discussed abt discont loop relating to two weak links starting frm the cell R8C8 (23)with a strong link frm 3 to 2 within the cell and then ending at cell R8C8 . That's fine .
Instead if we had stopped at cell R7C8 with a strong link at 4 in this cell. Then we can treat this as an AIC type 2( as it starts with a strong link frm 3 and ends with a strong link at 4 and both the cells R7C8 and R8C8 see each other, therefore the start candidate '3' can be dismissed frm end cell R7C8 )
Am I right?
I hv never applied discont / cont loop.
Thanks for repeating this concept .
Nice. Thank you for bringing up AIC Type 2. I’m glad you found that. They are closely related to DNL.
Great puzzle. Instead of using the single candidate strategy on the 3's, I noticed if cell R8C8 = 2, then the BVC's along both row 2 and col 2 leave cell R2C2 with no value, so 3 can be placed in R8C8. What's the name of the strategy there?
Thank you for your comment. According to Sulphur, this is a type of W-Wing that makes the solve happen.
Eliminating 3 from column 8 row 7 is almost like solving by bifurcation
It does seem that way. The actual strategy is a Discontinuous Loop, which uses a series of strong and weak links to get to the same conclusion. Do you like to solve with bifurcation or do you try to avoid it?
Any kind of chaining is bifurcation. In fact, all advanced positions are based on bifurcation. Not2 in R8C2 makes it 1, forces 1 in R2C3, and 2 in R2C8. One of R2C8 and R8C2 must be 2, and R8C8 sees both. (This was my tactic, not necessarily the video's.)
EDIT: I replied first before seeing how the video treated the 3. I see that it was completely different.
I am resigned to the ugly reality that I'll never, ever be able to do a grid like this by conventional means. Although I found an XY-Wing along the way, my main strategy was using several virtual chains. I'll watch this out of curiosity, but I am certain that even if I can understand the logic, I could never spot it in the wild.
Follow-up while watching the video:
Interestingly, I found that 4 in box 2 by different means, but it was my first write-in after the few easy cells at the beginning.
Oh, I actually found the 4 in box 5 before the one in box 2. In any case, it led to the same sequence of easy 4s elsewhere.
Well, I was right. I would never have spotted the Discontinuous Loop as such, but some of my virtual chains are not unlike it.
Heh. I know what you mean! But hopefully you meant, "XY Chains"- as XY Wings (aka Y Wings) are separate solving strategies. For me, finding them is easier when (unfortunately) I fill out the grid entirely which does take hard work & experience because you want to be accurate in your Snyder notation. Timberlake can do it the way he does it because he is soo experienced (and the Man! 🙌🏼). After all that, then I start looking for BVCs and any commonalities. Just keep playing (and watching this channel 👍🏼) eventually you'll spot XY Chains and other solving strategies. Don't be bogged down by trying to understand any new strategy you learn~just keep grinding and like I said, you'll eventually begin to see it and implement the strategy. Hope that helps 🤙🏼
@@deaahimm6334 No, I meant an XY-Wing. I really did find one, but only after following two or three virtual chains, by which I mean chains based on a starting hypothesis (i.e., mental bifurcation).
Oh ok, my bad. Well, my advice still holds~don't trip out on trying to "find" them... eventually they'll find you. Heh. For years I'd been solving without any knowledge of strategies, until about 3-4 years ago (during the Pandemic) I started to actually watch Sudoku channels and realized I'd been using some strategies already (X-wings, Y-wings, XY-chains, URs) although crudely. I'm still grinding away, and even though I'm far better than I was back then, I sometimes go back to an old puzzle that gave me fitz (heh that took me days to solve) and look for how I could've solved it more simply & efficiently. And I'm like, "geez it was right there!" Anyways, keep grinding bro 🤙🏼
@@deaahimm6334
Are you solved puzzle or watched vedio, if you solved mean how you solved?
@@Ramakrishnagm Hi. No, I just watch the videos. I never do the puzzles.
W-wing сразу решает эту головоломку.
Keen observation. Thank you for sharing. How good are you at spotting W-Wings?
@@SmartHobbies достаточно быстро. На дуплеты всегда обращаю внимание впервую очередь.
(1,3)=9 (1,5)=2; tercia(4,6,7) en 1a fila y en 1a col ent (7,1) =8 (5,1)=2 (6,6)=2 (5,3)=8; Tanteo Indirecto: si (5,6)=4 ent (1,4)=4 o si (5,6)=9 ent (5,7)=7 (1,7)=6 y (1,4)=9, o sea en ambos casos se llega a (1,4)=4 formándose el par (6,7) en 1a fila y en 1a col dentro del bloque 1 ent (4,1)=4 (6,2)=6 (5,6)=4 (6,9)=4; par (5,9) en 5a col quedando la tercia(1,6,8) en dicha col ent (6,5)=9 (7,5)=5; Tanteo por Combinaciones: primero (8,8)=2 con (8,3)=1 se llega a un Error! en la casilla (8,2) = (1,2) quedaría vacia y luego (8,8)=2 con (8,3)=5 Error! se duplica el #1, por lo tanto en (8,8)=3 (6,7)=3 (6,3)=7 (4,3)=3 (7,6)=3 (3,5)=6 (3,1)=7 (1,1)=6 (1,7)=7 (5,7)=9 (4,9)=5 (4,6)=6 (4,5)=1 (4,8)=7 (5,9)=1 (5,4)=7 (6,4)=5 (3,4)=3 (3,2)=1 (2,2)=3 (2,3)=4 (8,2)=2...
Wow. Thank you for sharing!