I have come up with an interesting puzzle but I can't for the life of me figure out how to solve it.
It follows like this:
You have 2 types of squires that are congruent with one another. Let's call them A-squares and B-squares. They both have an area of n²
The rules are:
You can place as many B-squares on a square board with a side length of N as you want. (imagine placing Black squares on an empty chess board)
A-squares must be placed next to at least 1 B-square (they can be placed next to each other as long as both squares are touching at least 1 B-square)
The question is: What is the Maximum number of A-squares that is possible for any n-length square board?
If this question can't easily be answered I want to at least know if it's possible to calculate the maximum number of A-Squares that can fit into a 7x7 board.
This is an image of the two best configurations that i could come up with manually for a 7x7 board.
(The green squares are A, blue squares are B and the the yellow/black squares outlines the board)
For context, I came up with this puzzle a long time ago when I wanted to create the most space-efficient sugarcane farm i could in minecraft seeing as how sugarcanes could only be planted adjacent to water blocks. I don't play minecraft anymore but I like the mathy-ness of many of it's aspects.
I'm still in high-school and i'm not good at programming. Sorry for all the trouble.
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pentominoes it takes to cover your board completely. $\endgroup$