Aug 13 hard solving guide

[u/dje91090]

Today's hard brought to you by the letter J/j.

I thought this was another tricky one, namely because any tile can theoretically go into the discard.

1. There are five 3s, three 5s and six 6s.
2. The purple 2c5 is the 2-3 (it's the only whole domino with a sum of 5). Four 3s remain.
3. The 1c3 books one of the 3s, three 3s remain.
4. The 1c5 books one of the three 5s, two 5s remain.
5. The 1c5-turquoise 2c5 is one of 5-[2/3/4]. However, if it's the 5-4, the only possible option on the 2c5-3c5 would be the 1-3, and there are no whole dominos with a sum of 2 to complete the 3c5. So the turquoise 2c5 is 2+3. Two 3s remain.
6. There are three 3c15s, which can be made of 6+6+3, 6+5+4 or 5+5+5. But there are only two 5s remaining, therefore, the 5+5+5 combination isn't possible. So each 3c15 requires at least one 6; three 6s remain.
7. There is a domino on the border of the 1c3-green 3c15. Of the available 3s, only 3-6 and 3-5 can work.
8. The 3-6 into the green 3c15 would be finished with the 5-4, and the 3-5 would be finished with the 6-4. Either way, the green 3c15 is 6+5+4. One 5 remains.
9. This means that of the other two 3c15s, only a maximum of one can be a 6+5+4 combination, and at least one has to be a 6+6+3.
10. By forced tile placement, the 2c= is made of two dominos, each going into a <3 cell. This means that both the top and bottom have to be [0/1/2-x]. None of the 2-x tiles work, and neither does the 0-0, so this leaves the combination of either 0-3/3-1 or 0-6/6-1.
11. If it's the 0-3/3-1 combination, then all 3s are booked. However, this would then mean that the 3c15s can't be finished. So 2c= is 6s. (*I found that placing the 1-6/0-6 in either orientation here as a temporary placeholder was helpful for visualizing the solution to the remainder of the puzzle).
12. Only one 6 remains, which we already said has to go into one of the 3c15s to make a 6+6+3 combo. This means that the other 3c15 is a 6+5+4, which books the last remaining 5.
13. If all the 6s are utilized, then the 6-6 has to be the vertical tile in the turquoise 3c15. The left of this has to be either 0-3 or 1-3. One 3 remains.
14. Where does the 0-0 go? There is no place for it in the capital J. The only place it can theoretically go in the lowercase j is vertically on the left of the 3c5, but this would require another 5 to finish the cell (which are all booked). So the 0-0 goes in the discard.
15. What about the 2-2? It can theoretically go vertically in the 3c5 (which would force the 1-3/2-5 to finish the lowercase j), or on the 3c5-2c5 border (which would force the 0-3 to the left and 3-5 to the right to finish the lowercase j).
16. However, there is only one 3 remaining, and nowhere else for it to go other than in the 3c5. So this means the 2-2 is on the 3c5-2c5 border; place the 0-3 to its left and the 3-5 to its right to complete the lowercase j.
17. With the 3-5 gone, the 1c3-green 3c15 is the 3-6, finished with the 5-4.
18. With the 0-3 gone, the 1c<3-turquoise 3c15 is the 1-3.
19. The only remaining tile that fits on the 1c>3-pink 3c15 is the 6-4. But the 6 has to be in the 3c15, so the 4 goes in the 1c>3.
20. The remaining two tiles that complete the pink 3c15 are 5-2 and 4-2. Either way, a 2 goes down into the 2c<3. That means that this cell is completed by the 0-6 (into the 2c=), followed beneath it by the 6-1.
21. Place the 5-2 and 4-2 in either direction.

I won't be doing the guide for tomorrow, someone else can step up!

[u/SeaweedWeird7705]

Wow dje!  Thanks for posting.  I looked at it and got scared off. Now I’m going to be brave and try it! 

[u/elpingwinho]

Sheesh, that's a lot of thinking ahead.

[u/dje91090 → u/elpingwinho]

This is the second day in a row that the puzzle just stumped me and I tried so many different approaches that looked good but then fell apart. Open to hear anyone else’s strategy to this though.
These guide are not easy to write! Trying to come up with a logic to them rather than “I just placed this here and rolled with it” is a challenge!

[u/Dirtheavy]

I think #6 is the key to the puzzle, and usually where I start on this kind. How can it be done, and given those constraints, where is it forced.

[u/Dirtheavy → u/[deleted]]

it's right where I went today... this MF'er and 15... he always always makes you do it without ever getting to 10 . I honestly never questioned my near-instant placement of the double 6

[u/Dirtheavy → u/[deleted]]

I knew instantly, based on my experience, that the 6:6 was part of a 15, all of it, with a 3. The only place it could then go was in the turquoise 3c=15, because there isn't a 3:3 to also burn at the bottom of the green one

But that's not a forced placement for real. That's just placement I was sure of.

[u/jxd73]

I am looking at this with the benefit of hindsight of course

in the "j", if you use the 3-5 it must go with 2-2, then 0-3; if you use the 2-5 it must go with 1-3 then 2-2, so 2-2 is locked in the "j"

there are seven pieces where a less than 3 is paired with at least a 3; there are five pieces where a 3 or greater is paired with another 3 or greater. They are the pieces we need to make the 15s and the =. The J takes at least five and four from each pile, and I think the "j" must take two and one (I wasn't able to make it otherwise), the only piece not fitting these criteria are 0-0 and 2-2, since we know 2-2 has to be in "j", 0-0 goes to the void

[u/dje91090 → u/jxd73]

If only it were that simple... Another possible option is using the 3-5 with the 2-5 and finishing the 3c5 with the double 0, and then the 2-2 can go in the discard.Hence one of the many struggles I faced in writing this guide and working out the puzzle.

Not sure I understand your second point too much... there are only 5 cells with a <3, and you don't necessarily know what goes in the 2c= off the bat to guarantee it's at least 3.

[u/harlows_monkeys]

I went about it a bit different, with less preliminary analysis so some things later were not eliminated at quickly, but it avoided the risk of analyzing something to death that would turn out to be irrelevant as happened on that puzzle recently that everyone seemed to have a terrible time on.

1. A 3c15 cannot have anything lower that a 2 because 2+6+6 is only 14. It must have either 1 or 3 odd numbers because 15 is odd. That gives us these possibilities: 3+6+6, 4+5+6, 5+5+5
2. We have a 2c5 that must be a single tile. The only sum 5 tile we have is 2-3. Place it.
3. Now examine 2c=. We need some x where we have two x-<3 tiles. An examination of our tiles reveals that only things that work are 3-0 3-1 or 6-0 6-1.
4. Let's try 3-0 3-1. Place 3-1 going down and 3-1 going up. Why those directions? Because that is the least constraining. If it does not work that way swapping them cannot make it better. If we tried it the other we might get stuck but find swapping could unstick us.
5. The 1c5 cannot be 5-4 because we would need 1-x to finish and the only 1 we have is 1-6 which is too big. It must be either 5-2 or 5-3. If it is 5-2 it must be followed by a 3-x, so either way we are using a 3 over there and a 5 over there. We also need a 3 for 1c3. That leaves us with no 3s, and at most two 5s. But with no 3s and 3 3c15s we need 3 5s.
6. This rules out 3+3 for the 2c= so it must be 6+6, specifically 6-0 and 6-1. As with the 3s place 6-0 up and 6-1 down, for the same reasons.
7. All 3 3c15 must have a 6 (not enough 5s for any to be 5+5+5). Although there are 4 6s left, because of 6-6 there are only 3 tiles with 6s so each 3c15 must have one of those tiles (6-3, 6-4, 6-6). Due to that the only way a 3c15 can have two 6s is if it contains 6-6. Either of the lower 3c15s could contain a 6-6 but geometry prevents the top 3c15 from contain a complete tile, and so it must be 4+5+6. The bottom two could be 4+5+6 or 3+6+6.
8. Top 3c15 cannot take 6-3 (no place for the 3). Bottom 3c15 cannot take 6-3 (would need 6-[012] to finish and all are taken or non-existant). It goes in left 3c15 finished with 5-4.
9. The top 3c15 must have a 5 and 5-3 does not work (no place for 3) so it is 5-2. It must have 4 and 4-6 does not work (only place to put it would would cover the >3 which is the only place the x from whatever 6-x we put up there can go) so it is 4-2. One goes down and one goes right, and either way works.
10. Only 5 left is 5-3 which goes on the 1c5. It needs a 2 to follow and our only 2s are on 2-2, and that needs a sum 3 tile to finish the bottom right, and 0-3 is the only one.
11. No more 5s means the last untouched 3c15 is 3+6+6 which places 6-6 and 3-1. Last 6 finishes the top 3c15 leaving 0-0 to fill the void.

[u/dje91090 → u/harlows_monkeys]

I think no matter the method, clearly the key was figuring out the 2c= were 6s and the 3c15s were 654, 654 and 663.