Discussion on optimal Boop theory

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BoopMaster
Posts: 3
Joined: 20 February 2024, 18:01

Re: Discussion on optimal Boop theory

Post by BoopMaster »

Here's an attempt of representing the board through Emoji:

Code: Select all

Empty Spot = 🟦
Orange Kitten = 🟨
Orange Cat = πŸŸ₯
Grey Kitten = πŸ”²
Grey Cat = ⬛

🟦🟦🟦🟦🟦🟦
🟦🟦🟦🟦🟦🟦
🟦🟦🟦🟦🟦🟦
🟦🟦🟦🟦🟦🟦
🟦🟦🟦🟦🟦🟦
🟦🟦🟦🟦🟦🟦
Let's experiment:

Code: Select all

Grey: (1 πŸ”² | 0 ⬛) / Orange: (3 🟨 | 0 πŸŸ₯) 
πŸ”²πŸŸ¦πŸ”²πŸŸ¨πŸŸ¦πŸŸ¦
πŸŸ¦πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²
🟦🟦🟨🟦🟨🟦
🟦🟦🟦🟦🟦🟦
πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²πŸŸ¦
πŸŸ¦πŸ”²πŸŸ¦πŸŸ¨πŸŸ¦πŸŸ¨
I often myself in situations where I will have all 8 felines on the board AND 3 in a row potential.

In the situation above, from a real game, I could get three in a row by playing (row 3, col 1). But I didn't do that.

But let's suppose I did promote all three kittens:

Code: Select all

Grey: (0 πŸ”² | 3 ⬛) / Orange: (3 🟨 | 0 πŸŸ₯) 
πŸ”²πŸŸ¦πŸŸ¦πŸŸ¨πŸŸ¦πŸŸ¦
πŸŸ¦πŸŸ¦πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²
🟦🟦🟨🟦🟨🟦
🟦🟦🟦🟦🟦🟦
πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²πŸŸ¦
πŸŸ¦πŸ”²πŸŸ¦πŸŸ¨πŸŸ¦πŸŸ¨
If I were Orange, I would play at row 5, col 5 to knock two of grey's kittens off the board. While orange is still losing, they have a lot of influence on the board. And I don't like that.

Code: Select all

Grey: (0 πŸ”² | 3 ⬛) / Orange: (3 🟨 | 0 πŸŸ₯) 
πŸ”²πŸŸ¦πŸŸ¦πŸŸ¨πŸŸ¦πŸŸ¦
πŸŸ¦πŸŸ¦πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²
🟦🟦🟨🟦🟨🟦
🟦🟦🟦🟦🟦🟦
πŸŸ¦πŸŸ¨πŸŸ¦πŸŸ¦πŸ”²πŸŸ¦
🟦🟦🟦🟨🟦🟨
So instead promoting all three kittens, I go with row 3, col 4:

Code: Select all

Grey: (0 πŸ”² | 1 ⬛) / Orange: (3 🟨 | 0 πŸŸ₯) 
πŸ”²πŸŸ¦πŸ”²πŸŸ¨πŸŸ¦πŸŸ¦
πŸŸ¦πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²
πŸŸ¦πŸŸ¨πŸŸ¦πŸ”²πŸŸ¦πŸŸ¨
🟦🟦🟦🟦🟦🟦
πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²πŸŸ¦
🟦🟦🟦🟨🟦🟨
My reasoning is as follows:
1) Center Control
2) Divide my opponent's pieces
3) Promote one of my kittens in the lower left corner. Otherwise, my opponent could play row 5, col 5 to push two of my kittens off the board. But now, if they play there, they can only boop one of my kittens off the board.
4) I maintain my 3-in-a-row threat in the upper left quadrant of the board. And if I play there, I immediately create another threat because that would boop my center kitten into a two-in-a-row.

Orange might consider playing row 4, col 2, creating their own threat:

Code: Select all

Grey: (0 πŸ”² | 1 ⬛) / Orange: (2 🟨 | 0 πŸŸ₯) 
πŸ”²πŸŸ¦πŸ”²πŸŸ¨πŸŸ¦πŸŸ¦
πŸŸ¦πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²
πŸŸ¦πŸŸ¨πŸŸ¦πŸ”²πŸŸ¦πŸŸ¨
🟦🟨🟦🟦🟦🟦
πŸŸ¦πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²πŸŸ¦
🟦🟦🟦🟨🟦🟨
But I don't think that's a good move. Grey has 1 piece left. On their next turn, they can easily ruin Orange's threat and promote their own kitten into a cat AGAIN! The promotion of one feline at a time can have a nasty rhythm going on.

So sadly, orange's best move might be to play row 1, col 2 for the simple reason of depriving Grey from getting into that rhythm:

Code: Select all

Grey: (0 πŸ”² | 2 ⬛) / Orange: (2 🟨 | 0 πŸŸ₯) 
πŸŸ¦πŸŸ¨πŸ”²πŸŸ¨πŸŸ¦πŸŸ¦
πŸŸ¦πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²
πŸŸ¦πŸŸ¨πŸŸ¦πŸ”²πŸŸ¦πŸŸ¨
🟦🟦🟦🟦🟦🟦
πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²πŸŸ¦
🟦🟦🟦🟨🟦🟨
Although grey can get three in a row here, the silver lining for orange is that grey has to boop an orange piece towards the center.

Often, in situations of having all eight felines on the board and three-in-a-row potential, I tend to get 4 Cats instead of 3 Cats.

That is because I promote 1 feline at first which can ONLY occur during the condition that I have all eight pieces on the board. And I do so in such a way that my three-in-a-row threat is still present. So I end up getting 4 cats in the end.

If I promoted three kittens at first, I will "only" have 3 cats and I deprived myself of the option to leverage the all-8 feline condition.

This is the "slow and steady wins the race". You promote 1 to get 4 Cats. If you're greedy, you only get 3 Cats.
User avatar
Ishmokin
Posts: 3
Joined: 10 April 2024, 12:48

Re: Discussion on optimal Boop theory

Post by Ishmokin »

Honorarius wrote: ↑11 April 2024, 00:02 Welcome. I'm hoping to find some more on boop strategy too.
thanks! Glad to be here! :)
User avatar
Ishmokin
Posts: 3
Joined: 10 April 2024, 12:48

Re: Discussion on optimal Boop theory

Post by Ishmokin »

BoopMaster wrote: ↑11 April 2024, 18:12 Here's an attempt of representing the board through Emoji:

Code: Select all

Empty Spot = 🟦
Orange Kitten = 🟨
Orange Cat = πŸŸ₯
Grey Kitten = πŸ”²
Grey Cat = ⬛

🟦🟦🟦🟦🟦🟦
🟦🟦🟦🟦🟦🟦
🟦🟦🟦🟦🟦🟦
🟦🟦🟦🟦🟦🟦
🟦🟦🟦🟦🟦🟦
🟦🟦🟦🟦🟦🟦
Let's experiment:

Code: Select all

Grey: (1 πŸ”² | 0 ⬛) / Orange: (3 🟨 | 0 πŸŸ₯) 
πŸ”²πŸŸ¦πŸ”²πŸŸ¨πŸŸ¦πŸŸ¦
πŸŸ¦πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²
🟦🟦🟨🟦🟨🟦
🟦🟦🟦🟦🟦🟦
πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²πŸŸ¦
πŸŸ¦πŸ”²πŸŸ¦πŸŸ¨πŸŸ¦πŸŸ¨
I often myself in situations where I will have all 8 felines on the board AND 3 in a row potential.

In the situation above, from a real game, I could get three in a row by playing (row 3, col 1). But I didn't do that.

But let's suppose I did promote all three kittens:

Code: Select all

Grey: (0 πŸ”² | 3 ⬛) / Orange: (3 🟨 | 0 πŸŸ₯) 
πŸ”²πŸŸ¦πŸŸ¦πŸŸ¨πŸŸ¦πŸŸ¦
πŸŸ¦πŸŸ¦πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²
🟦🟦🟨🟦🟨🟦
🟦🟦🟦🟦🟦🟦
πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²πŸŸ¦
πŸŸ¦πŸ”²πŸŸ¦πŸŸ¨πŸŸ¦πŸŸ¨
If I were Orange, I would play at row 5, col 5 to knock two of grey's kittens off the board. While orange is still losing, they have a lot of influence on the board. And I don't like that.

Code: Select all

Grey: (0 πŸ”² | 3 ⬛) / Orange: (3 🟨 | 0 πŸŸ₯) 
πŸ”²πŸŸ¦πŸŸ¦πŸŸ¨πŸŸ¦πŸŸ¦
πŸŸ¦πŸŸ¦πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²
🟦🟦🟨🟦🟨🟦
🟦🟦🟦🟦🟦🟦
πŸŸ¦πŸŸ¨πŸŸ¦πŸŸ¦πŸ”²πŸŸ¦
🟦🟦🟦🟨🟦🟨
So instead promoting all three kittens, I go with row 3, col 4:

Code: Select all

Grey: (0 πŸ”² | 1 ⬛) / Orange: (3 🟨 | 0 πŸŸ₯) 
πŸ”²πŸŸ¦πŸ”²πŸŸ¨πŸŸ¦πŸŸ¦
πŸŸ¦πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²
πŸŸ¦πŸŸ¨πŸŸ¦πŸ”²πŸŸ¦πŸŸ¨
🟦🟦🟦🟦🟦🟦
πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²πŸŸ¦
🟦🟦🟦🟨🟦🟨
My reasoning is as follows:
1) Center Control
2) Divide my opponent's pieces
3) Promote one of my kittens in the lower left corner. Otherwise, my opponent could play row 5, col 5 to push two of my kittens off the board. But now, if they play there, they can only boop one of my kittens off the board.
4) I maintain my 3-in-a-row threat in the upper left quadrant of the board. And if I play there, I immediately create another threat because that would boop my center kitten into a two-in-a-row.

Orange might consider playing row 4, col 2, creating their own threat:

Code: Select all

Grey: (0 πŸ”² | 1 ⬛) / Orange: (2 🟨 | 0 πŸŸ₯) 
πŸ”²πŸŸ¦πŸ”²πŸŸ¨πŸŸ¦πŸŸ¦
πŸŸ¦πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²
πŸŸ¦πŸŸ¨πŸŸ¦πŸ”²πŸŸ¦πŸŸ¨
🟦🟨🟦🟦🟦🟦
πŸŸ¦πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²πŸŸ¦
🟦🟦🟦🟨🟦🟨
But I don't think that's a good move. Grey has 1 piece left. On their next turn, they can easily ruin Orange's threat and promote their own kitten into a cat AGAIN! The promotion of one feline at a time can have a nasty rhythm going on.

So sadly, orange's best move might be to play row 1, col 2 for the simple reason of depriving Grey from getting into that rhythm:

Code: Select all

Grey: (0 πŸ”² | 2 ⬛) / Orange: (2 🟨 | 0 πŸŸ₯) 
πŸŸ¦πŸŸ¨πŸ”²πŸŸ¨πŸŸ¦πŸŸ¦
πŸŸ¦πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²
πŸŸ¦πŸŸ¨πŸŸ¦πŸ”²πŸŸ¦πŸŸ¨
🟦🟦🟦🟦🟦🟦
πŸ”²πŸŸ¦πŸŸ¦πŸŸ¦πŸ”²πŸŸ¦
🟦🟦🟦🟨🟦🟨
Although grey can get three in a row here, the silver lining for orange is that grey has to boop an orange piece towards the center.

Often, in situations of having all eight felines on the board and three-in-a-row potential, I tend to get 4 Cats instead of 3 Cats.

That is because I promote 1 feline at first which can ONLY occur during the condition that I have all eight pieces on the board. And I do so in such a way that my three-in-a-row threat is still present. So I end up getting 4 cats in the end.

If I promoted three kittens at first, I will "only" have 3 cats and I deprived myself of the option to leverage the all-8 feline condition.

This is the "slow and steady wins the race". You promote 1 to get 4 Cats. If you're greedy, you only get 3 Cats.
thanks for this! learning a lot!
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