What follows is a solution to the chessboard problem (1.3#11), submitted by a student in our class (black text), along with some remarks by me (colored text).
Problem: If CB is n by m such that n and m are both odd, and a white piece is removed, then the remainder of the board can be perfectly covered by dominoes.
The assumptions of the problem are not completely stated. In particular, it should be mentioned that one of the corner pieces is assumed to be white.
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Definitions:
These aren't definitions. I'd call them "preliminary lemmas and notation." I'd say something like this:
"Before solving the problem, we first fix some notation and state some easy lemmas without proof."
Thm: Any board of the form (even by x), where x is some positive integer can be covered.
We noted this simple fact in class. But I'd be more clear about it. For instance,
Lemma. Any n×m chessboard can be perfectly covered by dominoes if at least one of n or m is even.
Some people like to be more concise, even when it is less obvious, for example,
Lemma. Any n×m chessboard can be perfectly covered by dominoes if nm is even.
I'd actually prefer the following, for future use.
Definition. We call an n×m chessboard even if either n or m is even.
Lemma 1. Any even chessboard can be perfectly covered by dominoes.
m,n,k,l all correspond to some positive integer.
The symbols k, l, m and n will denote positive integers.
Column and rows shall be counted from left to right, and from top to bottom respectively.
I'd say "numbered" instead of "counted", but that's pretty picky. And put a comma before "respectively."
Assume that one corner is white, therefore all corners are white.
Some grammatical adjustments are needed, and more assumptions needed. (The oddness of the chessboard isn't mentioned.)
Lemma 2. If nm is odd, then all corners have the same color.
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Proof: CB is 1 by 3. Below are the two possible cases._ represents white piece that is removed,(* *) represents a domino.
- (* *) , (* *) _ Thus the statement holds for a 1 by 3.
This is pretty skimpy. I understand the desire to use diagrams, but a few words can help. (I like the domino representation, though.) I'd revise as follows.
"We are now ready to prove the result. First we note that the case of a 1×3 (or 3×1) chessboard is trivial, for if a corner is removed then the remaining two pieces are perfectly covered by one domino."
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CB is a 3 by 3. Below are all 5 possible cases.
A1= _ (* *), A2=(* *) _, A3=* (* *), A4=(* *) *, A5=(* *) *
* (* *) (* *) * * (* *) (* *) * * _ *
* (* *) (* *) * _ (* *) (* *) _ * (* *)
Thus statement holds for 3 by 3.
Unfortunately, although a picture could probably do a nice job of aiding the explanation, these things don't always come through properly in text files. Also, the "vertical" dominos are not represented in this way. Furthermore, there are really only two subcases here. Like so:
"In the case of a 3×3 chessboard, either a corner square or a middle square is removed. If a corner is removed, it is easy to see that what is left can be decomposed into two even chessboards of size 1×2 and 2×3, each of which, by Lemma 1, can be perfectly covered. If the middle square is removed, then it is clear that four dominos can be placed in a circular fashion around the border."
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If greater than 3 by 3 then the following cases occur:
(1) the white piece that is removed lies on the border.
(2) the white piece that is removed lies on the interior.
It isn't clear what "greater than 3 by 3" means. Is 1×35 greater than 3×3? Be clear, even if it takes a few more words:
"Finally, if either n > 3 or m > 3, then we consider two cases: the removed square is (1) a border square or (2) an interior square. (By an interior square we mean a square in the original chessboard that is surrounded on all sides by other squares; a square is a border square if it is not an interior square.) We suppose our chessboard has n rows and m columns. "
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1a-If the white piece removed lies on the border in the 2k+1 row on the first or last column, then the remainder of the board is reduced to a (1 by 2k), a (1 by (n-(2k+1))), and a (n by m-1). All easily covered by Thm. defined earlier.
1b-If the white piece removed lies on the border in the 2l+1 column on the first or last row, then the remainder of the board is reduced to a (1 by 2l), a (1 by (m-(2l+1))), and a (n-1 by m). All easily covered by Thm.
These two cases can be combined.
"In case (1), we first note that the the removed piece must be in an odd-numbered column or row. By considering ninety-degree rotations of the chessboard, we may assume, without loss of generality, that it is in the first column of an odd-numbered row, say row number 2k+1. Then the remaining squares can be partitioned into rectangles of (a) 2k rows and 1 column, (b) n-2k-1 rows and 1 column, and (3) n rows and m-1 columns, each of which are even and `covered' by Lemma 1. "
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2a-If the white piece removed is on the interior in the 2k row and 2l column, then place dominoes surrounding the piece removed in the same fashion as defined by A5. the remainder of the board is reduced to a (m by (2k-2)), a (m by (n-(2k+1))), a (3 by (2l-2)), and a (3 by (m-(2l+1))). All easily covered by Thm.
2b-If the white piece removed is on the interior in the 2k+1 row and 2l+1column, then place dominoes surrounding the piece removed in the fashion defined by A1. The rest of the board can be reduced to a (m by 2k), a (m by (n-(2k+3))), a (3 by (2l)), and a (3 by (m-(2l+3))). All of which can be covered by Thm.
I'd begin by mentioning that this last case (2) has two subcases, in which either (a) the missing square is on a square with even-numbered coordinates or (b) odd-numbered coordinates. After that, with some minor rewording, the above two paragraphs will work.
Thus the statement holds.
Thus the theorem is proved! Nice work, all the ideas are correct.
R. Mabry