Chessboard puzzle
 

Okay, nothing quite so sophisticated as an enigma puzzle (as they have become, it seems) but perhaps worth the while of the FOFC puzzling community, as it were.


Imagine you have a chessboard. Standard 8x8 layout of squares.

You also have 21 "bars" that will cover an area of 1 square by 3 squares on the chessboard.

With your keen math instincts, you determine that the 21 tiles each cover 3 squares, totalin 63 squares covered -- leaving only one square uncovered on the chessboard.

So far, so good.



The puzzle:

How many squares on the chessboard may be the "one uncovered square" after you have placed all 21 of your 3-tile bars onto the chessboard? And where are they?

OK--

If we start in the top left corner and number squares 1, 2, 3 , 1, 2, 3, 1, 2, 3--continuing with the row below starting at the left (so that the 9th number starts row 2, column 1), you will have 21 sets of 1,2,3 and one extra "1." We see that we have created a board such that it is impossible to lay down a piece that does not cover a 1,2,3 combination and ONLY a 1,2,3 combination. That means that the 21 pieces must cover 21 sets of 1,2,3, leaving a "1" standing alone.

Therefore, the empty square must be one that is designated with the number "1" using the numbering system outlined above.

There are probably some other restrictions that limit the number of empty squares even more, but I think that the above will be a good start.

I'm not sure I understand albionmoonlight's theory. a lot of those 123's overlap across rows.

I understand these to all be 1x3 unbendable bars... so they can't go across rows, etc. and can't bend.

Using that as my starting point, I haven't figure out how to get them all on the board at all yet.


edit: rows, not columns, dumbass.

One huge caveat--I am working on the assumption that it is possible at all to fit the pieces on the board--not an assumption that I have proven in the slightest.

I was working on the unbendable bar assumption as well.

Bishop to King's Rook 3. Checkmate.

Just to illustrate (if possible) what I mean

1--2--3--1--2--3--1--2
3--1--2--3--1--2--3--1
2--3--1--2--3--1--2--3
1--2--3--1--2--3--1--2
3--1--2--3--1--2--3--1
2--3--1--2--3--1--2--3
1--2--3--1--2--3--1--2
3--1--2--3--1--2--3--1

You see that there are no three consecutive numbers in a row or column that do not include and only include a 1, 2 and 3.

a	a	a	b	b	b	g	h
c	u	t	s	s	s	g	h
c	u	t	q	r		g	h
c	u	t	q	r	k	k	k
d	n	o	q	r	l	l	l
d	n	o	p	p	p	j	i
d	n	o	m	m	m	j	i
e	e	e	f	f	f	j	i

That's what I came up with for one pattern. I would figure that you can then "reflect" the results so that the empty square would be in 4 different parts of the board.

a	a	a	b	b	b	g	h
c	u	t	s	s	s	g	h
c	u	t	q	r		g	h
c	u	t	q	r	k	k	k
d	n	o	q	r	l	l	l
d	n	o	p	p	p	j	i
d	n	o	m	m	m	j	i
e	e	e	f	f	f	j	i

Those bolded letters will be the new places for the empty space if you started putting the bars differently.

There's probably more answers.

This is the only one I have so far, as well.

I've got a solution to the problem, but I'm helped by the fact that I studied these sort of problems back in college. I won't post it yet.

EDIT: Whoops, already solved. Nice job, Taz

Really????

Woo Hoo!!!!!!!

Well, there's a couple of loose ends, like showing that those are the only four squares, but you've got the gist of it.

Works for me. albion's 1-2-3 approach is correct, and as it turns out, it's only those four spots which can end up as the uncovered square. Well done.