26 Aug 06
Originally posted by Fat LadyI didn't think that the first person to reply would get it right.
There are:
8*8 squares of side 1.
7*7 squares of side 2.
6*6 squares of side 3.
5*5 squares of side 4.
4*4 squares of side 5.
3*3 squares of side 6.
2*2 squares of side 7.
1 square of side 8.
Making a total of 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204.
27 Aug 06
Here's a slightly trickier one, one of Martin Gardner's.
Imagine you have an empty chess board in front of you and 32 2x1 rectangle shaped pieces of wood, each one composed of a black square glued onto a white square. The rectangles are such that each one will exactly cover two adjacent squares of the chess board.
With your 32 rectangular pieces it is easy to see that you can cover the chess board so that every black square on the chess board has a black side of a rectangle over it, and every white square on the board has a white side of a rectangle over it.
This is the puzzle - suppose I take away one of your rectangles, leaving you with 31, and obliterate the a1 and h8 squares on the chess board. Is it possible to cover the board (except for a1 and h8) in the same way as before?
27 Aug 06
2 edits
Originally posted by Fat LadySorry, this was posted just the other day (in the correct forum). 🙄
Here's a slightly trickier one, one of Martin Gardner's.
http://www.timeforchess.com/board/showthread.php?threadid=50465
So before this thread dies, how many rectangles are there on a chessboard?
29 Aug 06
Originally posted by ThudanBlunder1x2=2
So before this thread dies, how many rectangles are there on a chessboard?
3x2=6
4x2=8
5x2=10
6x2=12
7x2=14
8x2=16
Total: 68
1x3=3
2x3=6
4x3=12
5x3=15
6x3=18
7x3=21
8x3=24
Total: 99
1x4=4
2x4=8
3x4=12
5x4=20
6x4=24
7x4=28
8x4=32
Total: 128
1x5=5
2x5=10
3x5=15
4x5=20
6x5=30
7x5=35
8x5=40
Total: 155
1x6=6
2x6=12
3x6=18
4x6=24
5x6=30
7x6=42
8x6=48
Total: 180
1x7=7
2x7=14
3x7=21
4x7=28
5x7=35
6x7=42
8x7=56
Total: 203
1x8=8
2x8=16
3x8=24
4x8=32
5x8=40
6x8=48
7x8=56
Total: 224
224 + 203 + 180 + 155 + 128 + 99 + 68 = 1057
29 Aug 06
Originally posted by lauseyI have no idea what you are doing here. This problem can be bruteforced (there are 2*(9-m)*(9-n) mxn rectangles on a chessboard where m!=n [if so then remove the 2*]).
1x2=2
3x2=6
4x2=8
5x2=10
6x2=12
7x2=14
8x2=16
Total: 68
1x3=3
2x3=6
4x3=12
5x3=15
6x3=18
7x3=21
8x3=24
Total: 99
1x4=4
2x4=8
3x4=12
5x4=20
6x4=24
7x4=28
8x4=32
Total: 128
1x5=5
2x5=10
3x5=15
4x5=20
6x5=30
7x5=35
8x5=40
Total: 155
1x6=6
2x6=12
3x6=18
4x6=24
5x6=30
7x6=42
8x6=48
Total: 180
1x7=7
2x7=1 ...[text shortened]... 24
4x8=32
5x8=40
6x8=48
7x8=56
Total: 224
224 + 203 + 180 + 155 + 128 + 99 + 68 = 1057
However a more elegant solution is that a chessboard has 9 vertical boundaries and 9 horizontal boundaries and to form a rectangle we choose any two of each.
There are 36 ways of choosing 2 from 9 (9C2) and therefore there are 36^2 rectangles.
36^2 = 1296
29 Aug 06
2 edits
Originally posted by XanthosNZI realised I made a mistake in my bruteforce approach. Wrote an algorithm that did indeed get 1296. If you are talking about oblongs, it is 1092 (previously assumed that rectangles aren't squares).
I have no idea what you are doing here. This problem can be bruteforced (there are 2*(9-m)*(9-n) mxn rectangles on a chessboard where m!=n [if so then remove the 2*]).
However a more elegant solution is that a chessboard has 9 vertical boundaries and 9 horizontal boundaries and to form a rectangle we choose any two of each.
There are 36 ways of choosing 2 from 9 (9C2) and therefore there are 36^2 rectangles.
36^2 = 1296
I like your elegant solution though. 🙂
EDIT (short delphi code of bruteforce if anyone is interested):
procedure TForm1.Button1Click(Sender: TObject);
var
n, m, total: integer;
begin
total := 0;
for n := 1 to 8 do
for m := 1 to 8 do
if m n then // this is just for oblongs. For all rectangles, remove this line
total := total + (m*n);
showmessage(inttostr(total));
end;
EDIT2: meant to be an "m greater or less than n" in code, but did not render here, as well as indentation.