Board Game Arena

Hi everyone and welcome to a new weekly Azul puzzle!

First of all, huge thanks to Propaganda_Panda for the great answer to last week's puzzle! Keeping the spirit of the last answer, here are the notations that make the most sense if you want to detail which moves would be made:
- Trays are numbered 1 to 5 (the upper "Northern" tray is 1, and from there you go clockwise). The centre is 0.
- Colours are abbreviated as such: BL (Blue), Y (Yellow), R (Red), BK (Black), W (White)
- Rows 1 to 5 are numbered 1 to 5, obviously. The floor line is 0.
- Top player is Player A, bottom player is Player B.
- You note a move by naming, in that order: 1) The tray you take from. 2) The colour you take. 3) The row you place it. So, for example: 3-Y-2 simply means: You take yellow from tray 3 and put them in row 2. 0-BK-0 means: You take black from the centre and put them in your floor line. And so on.

In all the puzzles, it is your turn to play and you are top player. Relative info such as previously lost tiles and/or score will be added when relevant. If it's the final round, these aren't really necessary as you just need to maximize your score with what you have available!

With that out of the way, on to the new puzzle!
Not as elaborate as last week's but I think a good opportunity for the more experienced players to share some wisdom with lower ELO players on how to finish a round strong.

Do you go for the red but provide your opponent to complete first two rows with the grouping of the whites or try something else?

Take 1 white (doesn’t matter from where in this situation). Denying your opponent the 1st or 2nd row in any round is extremely strong. Haven’t decided yet if I play it on row 1 or trash it lol

Bit confused, but if it's the upper players' turn then taking the one red and placing it on the bottom seems like the best move to me.

If the lower player picks two whites, then they can't finish the row 1 because the the upper player will just pick the black tile and the yellow one will remain.
Same goes if he takes the black one for the row 1 instead, then the opponent will just place those 2 whites on the top.

And why not place the red one on the top but instead on the bottom? The adjacency bonus is negligible (just 1 extra point + who knows how much will it matter in the next round), would have to put more to the floor, and it's much easier to finish the reds during the next turn.

I use the suggested notation, with exception of using B for Blue and K for blacK. I like using only one letter per color...
I separate player 1 and player 2 moves by a semicolon (,) and seperate the next move sequence by two slashes (//).

Options, ordered by optimal moves by both sides (in my opinion):

1-R-5, 0-W-2 // 0-K-1, 0-Y-0 ... Player 1 takes red into row 5, player 2 finishes row 2 with white (+2 points) // player 1 takes black (+1 point), and player 2 discards yellow (-1 point), because starting a new row with yellow is suboptimal. Total +-0 points.

1-R-5, 0-K-1 // 0-W-1, 0-Y-0 ... Player 1 takes red into row 5, player 2 finishes row 1 with black (+1 point) // player 1 takes white in row 1 while discarding one of the two whites (+2-2 = 0 points). Player 2 then again discards the last yellow for -1 points. Total +-0 points.

Bonus question for professionals: Which of these 2 responses should player 2 choose? As they are identical pointswise I have no idea which I would chose...

Now to the suboptimal sequences from player 1's perspective:

1-R-1, 0-K-1 // 0-W-0, 0-Y-0 ... If player 1 wrongly puts the red into row 1 (+2 points), player 2 will take black for row 1 (+1 point) // Since player 1 cannot let player 2 get the whites (he cannot allow player 2 to complete row 1 AND row 2!) he now has to discard white (-4 points). Player 2 then discards yellow (-1 point). Total: -2 points (from player 1's perspective)

1-W-1, 0-R-1 // 1-K-0, 1-W-0 // 1-Y-0 ... If player 1 starts with white (0-W-1 is identical to 1-W-1 as Priptonite already noted) (+2 points), then player 2 will take the red for his row 1 (+2 points) // and all the remaining tiles will be discarded since starting a new incomplete row is not a good idea. Since player 1 has 2 moves to play he will discard for (-4 points) and player 2 only discards (-1 point). Total: -3 points (from player 1's perspective)

Please note that 1-W-1 will not be answered by 0-R-4! Putting the red in row 4 loses one point in row 1 and gains one point because of no discarding. Because it is the same point-wise Player 2 should play a compact position in row 1 (red neighboring yellow) and thus answer 1-W-1 by 0-R-1 and not 0-R-4.

[edit] I just realized that by using B for black, I cannot use player A and player B anymore. Perhaps using BL and BK for blue and black is not such a bad idea.

LL-azul wrote: ↑11 August 2023, 18:17 If the lower player picks two whites, then they can't finish the row 1 because the the upper player will just pick the black tile and the yellow one will remain.
Same goes if he takes the black one for the row 1 instead, then the opponent will just place those 2 whites on the top.

Yup good point, missed this. I am still not sure red is better given that our opponent requires rows 2 and 3 white. Straining the resource may be optimal.

1) Player 1 should take the red and place in the bottom row. I don't see why this is even a matter of argument. The reason it's good is because it makes it VERY hard for Player 2 to both complete column 2 and row 4 anytime in the near future--Player 2 needs to collect 6 red.

2) Player 2 has a few options but the one I like most, because it looks towards the next round, is taking the single black and putting it on the top row. You could take the 2 white and place on row 2, of course, but why do it? Player 1 has placed whites in rows 3 and 4, and row 5 is under construction, so if they take the whites then will have only one row free for whites for a while. There are about 11 whites left in the bag and in the next few rounds Player 2 can probably take them with little interference, or force Player 1 to eat negative points.

3) Then of course it plays out as follows: Player 1 takes 2 white, smashes 1 and places 1 on row 1; Player 2 either places yellow on row 2 or junks it. Junking it is probably the best option, though I feel that players are often too ready to trash singletons rather than look at the board and consider the gestalt: it's not necessarily bad to get started on yellow in row 2, as surely at some point soon you're going to want to fill that spot (it's the central spot in row 2) and a lot of yellow tiles have already come and gone.

nandblock wrote: ↑12 August 2023, 16:20 1) Player 1 should take the red and place in the bottom row. I don't see why this is even a matter of argument. The reason it's good is because it makes it VERY hard for Player 2 to both complete column 2 and row 4 anytime in the near future--Player 2 needs to collect 6 red.

2) Player 2 has a few options but the one I like most, because it looks towards the next round, is taking the single black and putting it on the top row. You could take the 2 white and place on row 2, of course, but why do it? Player 1 has placed whites in rows 3 and 4, and row 5 is under construction, so if they take the whites then will have only one row free for whites for a while. There are about 11 whites left in the bag and in the next few rounds Player 2 can probably take them with little interference, or force Player 1 to eat negative points.

3) Then of course it plays out as follows: Player 1 takes 2 white, smashes 1 and places 1 on row 1; Player 2 either places yellow on row 2 or junks it. Junking it is probably the best option, though I feel that players are often too ready to trash singletons rather than look at the board and consider the gestalt: it's not necessarily bad to get started on yellow in row 2, as surely at some point soon you're going to want to fill that spot (it's the central spot in row 2) and a lot of yellow tiles have already come and gone.

Agree! I think this would be the optimal sequence for both players. Discarding the yellow tile or placing it in row 2 is a difficult question, I would probably just place it, but it shouldn't be a big deal either way.

Also, red really is the contested colour here. Our opponent has a realistic path to two columns, he will need to sacrifice a completed 2nd row but it would be totally worth it. Given that, he will want at least 8 red tiles (1 for row 1, 2 for row 4, 5 for row 5), ideally even 10 because the red in row 2 is fairly central and would give good adjacency bonuses. We want at least 6 (1 for row 1, 2 for row 2, 3 for row 5). Including the red tile on display, there are 16 red tiles left, so there will be a hard fight for the remaining red.