Problem: Prove that an a × b rectangle can be covered by 1 × n rectangles iff na or nb.
Solution: The proof is given by means of the following 2 claims.
Claim 1:
Given a
rectangular area is covered by tiles of dimension 1xn placed in horizontal or
vertical orientation,
it is
possible to remove the
tiles
one at a time in such a way that at any point of time, the covered region’s top
border consists of a sequence of steps increasing from left to right as shown
in diagram 1.
Proof:
Define step
blocks to
be rectangular regions defined by the steplike shape of the covered area. In
diagram 1, step
blocks are
represented by the areas shaded blue. Start from the left most step block and
consider the top left tile in that block. This tile must fall entirely within the current
block (call it block A) or must extend beyond the
right of the current
step block
(it cannot beyond the bottom of the current block since the first block is also
the bottommost). In the former
case, removing the
tile
will leave behind a step shape,
thereby proving the claim. In the latter case, inspect the next step block for
the possibility of removing a tile. The top left tile of this next block (call
it block B) cannot extend beyond the bottom of that block since the block A’s
top left tile occupies that portion as per the previous statement. Hence the
top left tile of block B too must either lie entirely within B or extend beyond
the right of B. As before, in the former case, removing the tile will leave a
step shape whereas in the latter case, the search for a tile to remove can be
repeated with the block to the right of C. Note that this procedure can be
repeated at most till the rightmost block is
reached, in which case the top left tile cannot extend beyond the right of the
block (and must therefore lie entirely within that block), thereby becoming
eligible for removal while retaining the step shape.
Claim 2: When
tiles are removed as per the procedure of claim 1 and neither a not b are
divisible by n, then at any point of time at least one of the step regions on
the top border will have a height and a width (as defined by diagram 2) that are
both indivisible by n.
Proof: The
proof is by induction on r, the number of tiles removed so far under the
procedure of claim 1. When r is 0, the covered region and also the only step
region is the original axb
rectangle. The height and width, namely a and b, are both indivisible by n as
per the assumption. Suppose the claim is true for values of r upto R.
Now consider r = (R + 1). Suppose the tile being removed at the current
iteration lies in a step region whose width is W and height Is H. If either W
or H is divisible by n, then there must be some other step region with width
and height indivisible by n and that region will remain after iteration (R +
1) as well. But if say W and H are both
indivisible by n. After removing the tile in this iteration, the 2 new step
regions would have dimensions (a) (W, H  1) and (W  n, H) or (b) (W, H
 n) and (W  1, H), depending on the orientation of the tile; in either case,
one of the new step regions would have its width and height indivisible by n.
Hence proved.
Diagram 1: Step
blocks denoted
by shaded blue regions

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