# Least amount of (n)x(n+1) tiles to perfectly cover a 7x7 floor with two irregular tiles

What is the least amount, $\mathscr{L}$, of $(n)\times(n+1)$ tiles that can perfectly cover a $7\times 7$ floor of unit tiles,

$\hskip2in$,

while still using these two red tiles:

$\hskip2in$

$$\tiny\text{NOT TO SCALE}$$

These two red tiles can be rotated or reflected. They are special and are comprised of 3 and 4 unit tiles respectively. Just to be clear, you put these two red tiles on the grid, and then try to find the least amount, $\mathscr{L}$, of $(n)\times(n+1)$ tiles, which themselves can also be rotated/reflected, to cover the grid perfectly with no overlaps. So you're going to have a certain number of $(n)\times(n+1)$ tiles and the two irregular tiles to cover the floor. "Cover" doesn't necessarily mean "contain within" the grid, but the two irregular pieces, and any tile you generate, needs to cover the grid orthogonally to the sides of the grid.

For an interesting wrinkle to the original problem post your answer for the least amount, $\mathscr{M}$, of $(n_i)\times(n_i+1)$ tiles for any number of $i=1,2,3,4,...$ which are contained in the $7\times7$ grid. $n_i=n_j$ for $i\neq j$ is allowed. That is, you can have $n_1=n_2=2$, for example.

So if you choose $n_1$ to be $2$, then you'd have your first tile as $(n_1)\times(n_1+1)=2\times3$, then your second tile could be $3\times4$ corresponding to an $n_2$ equal to $3$. In other words, any number of different sized $(n)\times(n+1)$ tiles.

I hope this turns out to be fun and less confusing than my first question. If it's not clear, then leave a constructive comment, and I'll answer it.

• lol to my surprise, the weirdo pieces make the solution easier not harder! – Tony Ennis Aug 10 '17 at 16:19
• The sentence starting "Just to be clear" seems to me to make things less clear, not more. Until reading that I was certain that I just had to find one placement of the red tiles which gives a minimal covering, but after reading it I'm not sure whether I have to find a minimal set of rectangles which works for any placement of the red tiles. – Peter Taylor Aug 10 '17 at 18:01
• @PeterTaylor, naturally. – user39732 Aug 10 '17 at 18:14
• The edit has reduced clarity further. What does "perfectly" mean in "cover the grid perfectly" if not that the tiles may not go outside the grid? – Peter Taylor Aug 10 '17 at 18:48
• @PeterTaylor, as discussed in the other answers' respective comment section, "cover" could mean "covering the grid with some non-irregular tiles on the outside" OR it could mean "contain within the grid". Either is accepted, and there is now a new, perhaps more interesting, third approach highlighted in the yellow box, which allows for different $(n)\times(n+1)$ tiles. – user39732 Aug 10 '17 at 19:46

The minimum is

7 tiles of 2 by 3

Explanation

The tiles have to be 1x2=2, 2x3=6, 3x4=12, 5x6=30, or 6x7=42, and 6x7 is impossible by inspection. They have to fill an area of 7x7-(3+4) = 42 tiles. 42 is not divisible by 12 or 30, so the rectangles have to be 1x2 or 2x3. Since the goal is to minimize the number of rectangles, 2x3 would be ideal if possible, which it is:

• That's great. Good job. – user39732 Aug 10 '17 at 5:59
• you've got 3 3*4's in there – JMP Aug 10 '17 at 5:59
• I still don't get What to do, even after i see solution – Jan Ivan Aug 10 '17 at 6:00
• Oh whoa, @JonMarkPerry, are you color blind? – user39732 Aug 10 '17 at 6:00
• Is it not allowed to mix different sizes? You can combine some of those into 3x4 tiles. I think that's what JonMark Perry meant too. – Kruga Aug 10 '17 at 7:48

Just to be annoying and pedantic, I've done it with

3 6x5 tiles.

As seen here, the grid is covered perfectly with no overlaps.

• Ha, ha, ha. Your answer is also correct. Nothing prevents you from doing this because I said "cover" not "contain". Good job. – user39732 Aug 10 '17 at 6:14
• 3 < 7. Winner winner! :P – Alpha Aug 10 '17 at 6:17
• But, Alpha, think of this: If cover is the aim, then I could do just one $7\times8$with the two irregular tiles just dangling on the edge. It still covers, right? – user39732 Aug 10 '17 at 6:17
• I did think of that, but the question does specify "you put these two tiles on the grid". That would be an efficient way to cover a floor though. – Alpha Aug 10 '17 at 6:25
• True. I did say that. Anyway, bravo. – user39732 Aug 10 '17 at 6:27

You can cover the 7x7 square perfectly (with no overlap and no excess) with a 5x4, a 6x3 (4x3 + 2x3), two 2x1 and the prescribed irregular tiles, for a total of 5 n x (n+1) tiles.

• You should show how, not merely assert that one can do so. – Rubio Aug 10 '17 at 15:50
• There exists a smaller solution for different sized $(n)\times(n+1)$ tiles. – user39732 Aug 10 '17 at 19:06
• @Rubio, once you fit the 5x4 and the 6x3 which really only works one way, the rest basically follows... – hkBst Aug 11 '17 at 7:56

Nobody seems to have answered part 2 yet, which is to tile the 7x7 with the given tetromino plus tromino plus all-different-sized n by n+1 rectangles. Here is a complete set of solutions of all the ways you can tile a 7x7 with the two given pieces plus any number of n by n+1 rectangles, allowing repeats. As you can see it is impossible. There are 97 100 440 tilings, with just those 28 sets of pieces.