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A solution to the Rectangle Puzzle of size n is an arrangement of n rectangles into a larger rectangle, such that no smaller rectangle is formed by outlining 2 or more of the placed pieces.

For example, here are solutions to the Rectangle Puzzle of sizes 2 and 5:

2 rectangles

5 rectangles

And here is an arrangement that fails to solve the puzzle for size 6, because the two tiles in the middle form a rectangle:

6 rectangles

3 and 4 are not solvable. Among the numbers 6, 7, 8, 9, 10, for which are the Rectangle Puzzle solvable?

share|improve this question
    
Aren't 2 pieces forming a rectangle? – TTT 14 hours ago
    
@TTT in which one? – Owen 14 hours ago
    
In the rectangle of Size 2. – TTT 14 hours ago
    
@TTT "that no smaller rectangle is formed by 2 or more of the pieces." – Lukas Rotter 14 hours ago
    
Right, got it. Thx. – TTT 14 hours ago
up vote 13 down vote

Here is a general solution for n>6.

enter image description here

Explanation:

The rectangles spiral around. To get the solution for a particular n, stop when you have placed the nth rectangle and truncate it so that the whole figure becomes a rectangle. Sadly this solution does not work for n=6, because then rectangles 2 and 6 together form a rectangle.

share|improve this answer
    
Oh wow, I was in the process of redoing my answer with this solution. Beat me to it. I also came to the conclusion that 6 doesn't work. 6 may not be possible at all. – TTT 13 hours ago
    
I assume you need to truncate #9? Otherwise, this fails to meet the "larger rectangle" requirement. It would be nice to mention so explicitly. – jpmc26 6 hours ago
    
@jpmc26 - truncation is mentioned in the explanation. – TTT 4 hours ago
up vote 4 down vote

Here are ones for

7 and 9 rectangles. 8 and 10 covered by comments. And a general solution greater then 6 by Jaap.

Pic1

enter image description here

pic2

enter image description here

Some general rules

Can't have an outer rectangle with a full side length
Can't have a square in a corner

share|improve this answer
    
In Pic 1, if you extend rect 1 up by 1, you can place an 8th horizontal next to it to get 8. – TTT 14 hours ago
    
You can do the same with Pic 2 to get 10. – TTT 13 hours ago
up vote 2 down vote

I (am pretty sure) I found the answer for n = 6. Please excuse the meager paint skillsenter image description here

Please feel free to let me know of the rectangle in here if I'm missing it, or to update the picture to make it prettier.

share|improve this answer
5  
The five rectangles on the right form a rectangle, and are basically the n=5 solution. – Jaap Scherphuis 13 hours ago
    
Ahhhh I see it now. Unlucky. Was only thinking of two smaller rectangles, not more than 2 – Avik Mohan 13 hours ago
up vote 1 down vote

Note: I misread the question. Leaving this here (for now) in case it helps others.

The answer is (probably not):

All rectangles of size N >= 5 are solvable.

This can('t) be accomplished by:

Starting with size 5 as drawn in the question, and adding a rectangle to the right that is the same length as the right side of it to make 6. Then add another rectangle across the top that is the size of the entire top to make 7. Repeat as needed.

Example:

Demo of pattern

share|improve this answer
    
I feel like that only works up until 9, because then adding a rectangle to a side that is the size of one whole side would make a smaller rectangle. – Bcmonks 14 hours ago
1  
@Bcmonks - I'm working on a picture right now... – TTT 14 hours ago
3  
Unforunately, the original 5 pieces form a rectangle smaller than the whole. – Jaap Scherphuis 14 hours ago
2  
You won't be able to find a solution for more rectangles that builds upon an existing smaller solution without modifications; the smaller solution will still be present and fail the condition outlined by the puzzle. – Ian MacDonald 14 hours ago
    
@JaapScherphuis - oh wow. I have now misread the question twice. In retrospect, that would have been way too easy for an Owen puzzle. – TTT 14 hours ago

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