Square Room Tiles

Question

The square floor of a square room was covered with colored 1 square unit tiles.The covering tiles were designed to have different (solid) colored tiles that forms rectangular areas but none of these areas have the same dimension with another area. How many tiles are needed to cover the smallest floor?

Answer 1

As others have already said, the arrangement with the fewest rectangles has 5 rectangles. This is because:

No rectangle can use up the whole side of the floor. If one did, the rectangular arrangement used in the remainder of the floor could be scaled to fill the whole floor with one fewer rectangle.

You therefore need a different rectangle in each corner of the floor.

Each rectangle has at least 3 neighbours (otherwise on its two internal sides one of the neighbours has a matching dimension)

Four rectangles (one in each corner) cannot work as both diagonally opposite pairs would need to be neighbours.

So we need at least 5 rectangles. If the fifth rectangle were in the middle of a side, the two adjacent corner rectangles would still need to be adjacent to their diagonally opposite corners to get three neighbours, so that is not possible. The fifth rectangle must therefore be an interior rectangle like this:

With 5 rectangles, the smallest dimensions they could have are the numbers 1 to 10, so the square floor has dimensions of at least 11x11. I think there are only two solutions for 11x11.

Using the letters in the diagram, we clearly have:
b+c=11
d+e=11
f+g=11
h+a=11
a+e+i=11
c+g+j=11
It turns out that all further equations you can set up (e.g. for total area) are linearly dependent on these.

{b,c},{d,e},{f,g},{h,a} are four of the five pairs {1,10},{2,9},{3,8},{4,7},{5,6}. This means that {i,j} is the fifth unused pair, and we can deduce that i+j=11.

So a+e=j and c+g=i. Obviously i,j>=3. This rules out {i,j}={1,10} or {2,9}.

If j=5, then the only choices for {a,e} are {2,3} and {1,4}. So {a,h,d,e} is either {2,9,3,8} or {1,10,4,7}, and {b,c,f,g} is the other. In neither case can we get c+g=6. The same goes for i=5, so {i,j}={5,6} is not possible.

If j=3, then the only choice for {a,e} is {1,2}. So {a,h,d,e}={1,10,2,8}, and {b,c,f,g}={4,7,5,6}. From this we cannot get c+g=8. The same goes for i=3, so {i,j}={3,8} is not possible.

This leaves only {i,j}={4,7}. We can assume wlog that i=4, j=7 because we can rotate everything by 90 degrees to swap i and j. Then {c,g}={1,3}. We can rotate everything by 180 degrees to swap c and g, so we can assume wlog that c=1, g=3. We have {a,e}={2,5}, and a=2 leads to the solution above, and a=5 gives the following solution:

In text form, the two solutions are:
2x10,1x6,5x8,3x9,4x7
5x10,1x9,2x8,3x6,4x7

For completeness, suppose we allow some of the areas to be squares, but disallow the trivial solution consisting of a single area. Much of the above argument works, until the point where it was deduced that {i,j} is the remaining pair. It is now possible for i=j to equal only one of the numbers in the remaining pair. One that choice is made, similar arguments to before then either prove there is no solution, or produce 2 solutions. I won't go through the cases in detail, but just list the solutions:

1x8,3x7,4x9,2x10,6x6
1x9,2x7,4x8,3x10,6x6

1x5,6x4,7x9,2x10,3x3
1x9,2x4,7x5,6x10,3x3

1x6,5x3,8x7,4x10,2x2
1x7,4x3,8x6,5x10,2x2

2x7,4x3,8x5,6x9,1x1
2x5,6x3,8x7,4x9,1x1

I found no solution when i=j are equal to 4, 5 or greater than 6.

Answer 2

The smallest floor can be covered in

5 rectangles.

I know this because

You cannot have a square area covered in 4 rectangles because there is no way to put 4 rectangles in a square grid without at least 2 of them sharing an edge. I have found an example for 5 rectangles for an 11 by 11 unit room. In that example you have (in clockwise order) a 6*3 rectangle, a 2*8 rectangle, a 1*9 rectangle and a 5*10 rectangle on the edges. This leaves a 4*7 rectangle in the center. I know that an 11 by 11 unit room is the smallest room made up of 5 rectangles because for 5 rectangles you need 10 dimensions and they cannot be 0 or the width of the square.

Floor Tiles layout:

Answer 3

The first key observation is that

where two rectangles meet, they can never do so along a complete edge of both -- for then they would have to have a dimension in common.

This means that

the number of rectangles (what I called k in a clarification-seeking comment on the original question, and will continue to call k now) must be at least 5; and I think there is "combinatorially/topologically" exactly one pattern with only 5 rectangles.

Here is a pretty small way to do it with that layout. I don't have a proof that this makes the room size (what I called n in that comment) as small as possible, and indeed it may very well not be. I doubt it's possible to do much better, though.

A.A.A.A.A.B.B.B.B.B.B.B.B. A.A.A.A.A.B.B.B.B.B.B.B.B. A.A.A.A.A.B.B.B.B.B.B.B.B. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. A.A.A.A.A.C.C.C.C.C.C.D.D. E.E.E.E.E.E.E.E.E.E.E.D.D.

The dimensions are:

13x13 (for the room); 12x5, 8x3, 9x6, 10x2, 11x1 for the rectangles.

Answer 4

(Assuming I read this correctly; dimensions equivalent means both dimensions are equivalent) If square sub tiles are allowed, then $3*1$ can be tiled with $2*1$ and $1*1$. Otherwise, tile an $5*1$ with a $3*1$ and a $2*1$