Weekender #1
Update Added at the bottom of the post.
We used to have weekender puzzles in BITS. Every weekend, there would be puzzles posted on mess notice board along with the names of the winners of the last week. Fun times!
We do it sporadically on our mailing list. I hope to be more regular as finding newer and cool puzzles is not that easy. I found one though:
You have a chess-board *grin*. You have a rectangle whose longest side is shorter than a side of this board. You place this rectangle randomly on this chess board with just one condition: Sides of the rectangle have to be parallel to the sides of the chess-board.
Now the question:
What is the condition that needs to be satisfied to ensure that the rectangle has equal amounts of both colors(black and white) inside it?
Update:
Maybe this was a little confusing puzzle for weekender #1. Here is a little clarification and what exactly I am looking for.
Ok. There are two types of rectangles that satisfy this.
Type A rectangle can be placed anywhere with sides parallel to chessboard grid and it will satisfy the condition.
Type B rectangle needs to be placed in a certain way on the board as randomly placing it cannot include equal amount of black and white in all cases.
What is a Type A rectangle?
What is a Type B rectangle and what is the extra condition for Type B?
I should not have used the word random in the original post as it is true only for half of the rectangles. I hope this clarifies it.
We used to have weekender puzzles in BITS. Every weekend, there would be puzzles posted on mess notice board along with the names of the winners of the last week. Fun times!
We do it sporadically on our mailing list. I hope to be more regular as finding newer and cool puzzles is not that easy. I found one though:
You have a chess-board *grin*. You have a rectangle whose longest side is shorter than a side of this board. You place this rectangle randomly on this chess board with just one condition: Sides of the rectangle have to be parallel to the sides of the chess-board.
Now the question:
What is the condition that needs to be satisfied to ensure that the rectangle has equal amounts of both colors(black and white) inside it?
Update:
Maybe this was a little confusing puzzle for weekender #1. Here is a little clarification and what exactly I am looking for.
Ok. There are two types of rectangles that satisfy this.
Type A rectangle can be placed anywhere with sides parallel to chessboard grid and it will satisfy the condition.
Type B rectangle needs to be placed in a certain way on the board as randomly placing it cannot include equal amount of black and white in all cases.
What is a Type A rectangle?
What is a Type B rectangle and what is the extra condition for Type B?
I should not have used the word random in the original post as it is true only for half of the rectangles. I hope this clarifies it.
posted by Nikhil at 10:40 AM
11 Comments:
Call me simple ... but isn't a square a rectangle? So ... having equal sides would be a convenient solution, no?
Problem solved! Tea & buns for everyone!
Simple,
Have you tried to put a 3 X 3 square on a chessboard so that it resembles a tic-tac-toe board?
Or for that matter what about a 1 X 1 square, the smallest square on the chessboard? Anything more would be a hint :)
And BTW, I should stress that the sides of this rectangle need to be parallel to the grid lines, need not be coinciding.
Everyone, please return your tea and yes, you can keep your buns.
OK OK, a square whose side (measured in grid squares) is an even number.
Solved! Tea & jam tarts for everyone!
Again.. this time people can keep the tea but have to return the jam tarts as this is only 1/4th solved.
Our Algebra book started with some witty comments by mathmaticians. One chapter started with: "Be wise, generalize." You are being too specific.
Do you mean to say that only squares with even sides will satisfy this?
What about a 1 x 4 rectangle that has 2 black and 2 white squares inside?
What about a 1 X 3 rectangle? Can it never achieve this?
You are on the right track but I need a rock-solid solution that considers all cases. My questions above should be enough to reach a perfect solution. And this time, don't be in a hurry to throw a tea party. :)
if length of each square is 1 unit, and sides of rectangle have length 'a' and 'b'
then at least one of ceil(a) or ceil(b) should be even.
no ... what i said was completely wrong. the condition should be: 'a' and 'b' are the number of squares that the sides pass through.
Hey Nikhil, one clarification is necessary. Can the condition that needs to be satisifed contain something other than rectange sides etc? For example, a 1x1 rectangle can satisfy the property you need if you place it such that the center of the rectangle coincides with a vertex of the board. (Or does randomly placing mean that this has to work for all possible placings?)
Ok. There are two types of rectangles that satisfy this.
Type A rectangle can be placed anywhere with sides parallel to chessboard grid and it will satisfy the condition.
Type B rectangle needs to be placed in a certain way on the board as randomly placing it cannot include equal amount of black and white in all cases.
What is Type A? What is Type B and what is the extra condition for Type B?
I should not have used the word random as it is true only for half of the rectangles. I hope this clarifies it.
Maybe this was a little confusing puzzle for weekender #1.
Not really...its the way you stated it that was confusing :-)
Anyway, assume that the squares on the chessboard are 1x1. Then, the type-A rectangles need to have one even side, and the other side integral. For proving this, you can show it for all 1x(even) rectangles, and then use induction to extend that to any Yx(even).
For type-B, my guess is that integral sides should be enough, and the condition for placing it is that one of the "lines" of the chessboard passes through the centre of the rectangle.
Actually, the integrality constraints are not needed. Type-A rectangles are the ones with even side, and type-B contains all the other rectangles. The placement condition for type-B is the same.
Anon,
You have cracked it. Congratulations! :)
Type A have at least one side even.
Type B have both sides odd and the center lies on a grid line (excluding edges).
Well done!
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