Prove or disprove that all checkerboards of these shapes can...

Key Concepts

Inductive and Recursive Construction Methods

Many tiling proofs, particularly those dealing with boards whose dimensions are powers of some base (like 2^n or 3^n), use induction or recursive decomposition. In these constructions, a tiling for a board of a smaller size is used as a building block to tile a larger board. This method not only shows the existence of a tiling for all values of n but also provides a constructive way of building up a solution through self-similar patterns or by partitioning the board into smaller, manageable sections that can each be tiled independently.

Tile Geometry and Orientation

Understanding the shape and orientation of the tile is essential in tiling problems. Right triominoes have a specific geometry – they are ‘L’-shaped and can be rotated or reflected. Recognizing how these tiles can be placed on the board, how they cover adjacent squares, and how their orientation affects the tiling configuration is crucial in constructing a valid tiling or proving that one cannot exist.

Tiling Invariants and Coloring Arguments

A common method used in proving tiling results is the invariant or coloring technique, where the board is colored in a specific pattern (often with two or three colors) to detect parity or residue imbalances that might prevent a legal tiling. This technique helps to show why in some geometries a tiling may or may not be possible by ensuring that each tile covers a fixed configuration of colors. In tiling with right triominoes, such arguments can be crucial in providing necessary conditions for the tiling, or conversely, to demonstrate an impossibility when such invariant conditions fail.

Divisibility and Area Conditions

A fundamental key concept in tiling problems is ensuring that the total area of the board is a multiple of the area of a single tile. For triominoes, which cover three unit squares, the board’s total number of squares must be divisible by three. This arithmetic condition (often derived from calculating areas such as 3 × 2^n, 6 × 2^n, 3^n × 3^n, or 6^n × 6^n) acts as a necessary condition for a complete tiling, and it is the first step in determining the feasibility of tiling a given board with the chosen pieces.

Transcript

00:01 High in this question.

00:01 We are asked to prove off this proof whether the following sugar bowl uh sugar boss can be feel with tree only know so the block in our shaped like this we have four boards to consider so let's go with them one by one.

00:19 The 1st 1 is three times two to the end when aeneas positive integer if you look at the 1st 1 tree by two, you can see that is clearly can't be feel and now we are we can use an argument light that on induction suppose if we can view the vault at at some in as some integer in i argue that we can use that to construct the next iteration.

00:56 So at in plus one now how do we do that? well, since to to hear means the block on the wedding hall like this on dde we does had this block all three by two everyone everyone folks on both three by four we can just add more off them, right and this will become three by four everyone the next iteration ah, tree by 16.

01:37 Then we just cool cook this as one object and add um, tree mauled him.

01:47 Right.

01:49 So you're gonna be 44 by four, which is 16.

01:52 And we can keep, like, repeating this process.

01:56 And that is the same with this in and in plus one.

02:02 So we just add the 1st 1 to get the until we get the 2nd 1 and it can always be done because it just power up two.

02:14 Okay, i hope this convince you that for any in this bought, this kind of bought can be few.

02:21 And now we go on to the next.

02:24 We can use the pre areas, uh, answer to help with this body as well.

02:31 Because six by two to the end is actually just pre by two to the end act together.

02:42 I shouldn't i shouldn't use at here.

02:44 But i mean, what i mean by that is attached them to get the woody cali like this.

02:53 So this is treat by 23 by two.

02:56 Then we get six by to write the first.

02:59 The first iteration, like a basic step on dde with with similar argument from the previous kiss.

03:07 When we have six by two to the end, then the next one can be created by just i can add what the holly until until we get they hide, we won.

03:24 So there's nothing complicated that we can always field this ball as well.

03:31 Okay, now, next one ese different.

03:38 If you are familiar with this fuel, check about between the nose kind of questions.

03:45 You're gonna run into this eventually.

03:47 So first, first off them the tree by tree by itself can be few completely with tri tree only knows no matter how you try.

04:01 So to this progress, you can just go through every possible way to feel like try every possible way and you go, it will not be complete.

04:14 There will be some there'll be some square left or there will be some, like something pushed out off the boat like this.

04:23 So, given cal, i you can start by saying that the only three ways to feel this corner like this is a carnival.

04:35 So buy it bt the in one in off the criminals.

04:42 Or you'd be the corner and you go further from thes tree and you gotta fight it anyway.

04:49 We did it...

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