[Guest]

Here is a riddle i came up with.It is my first one so i hope you like it

If we have 10 cards, 5 with x's and 5 with +'s on a table, like below,and in each move, we can switch 2 cards.What is the minimum amount of movements needed so that we have first 5 +'s and then 5 x's ?

+ x + x + x + x + x

Now what if we have 10 x's and 10 +'s?

+ x + x + x + x + x + x + x + x + x + x

Grab a pen!

Edited December 30, 2010 by Tsopi

[Guest]

Oh and another question i forgot to post.

Question 3 : Can you make a mathematical equation that give us the amount of movements needed, when we put in the number of cards?Supposing that the total of the cards is always a 2,4,6,8,10,12....number so that there are half X and half +

Edited December 30, 2010 by Tsopi

[unreality]

if you take the second half and put it underneath like this:

+ x + x +

x + x + x

2 switches will do the trick. I was going to just count the displacement (3/5 of each set of five is already complete, so 2 changes are needed in both, which amounts to 2 swaps) but not only that I can tell you which to swap: the two columns where x is on top and + on bottom.

I think this generalizes to floor(n/2) if 2n is the total size

[Guest]

I would say that it takes 2 moves on the 5,5 set, rotate 2 cards 45 degrees.

Same logic follows for the 10 cards, 5 moves.

Edit: Didn't re-read the question. It states that you must "switch" cards.

Edited December 30, 2010 by adamjgp

[Guest]

Two switches.

1. Switch 2 and 9. You get:

+ + + x + x + x x x

2. Switch 4 and 7. You get:

+ + + + + x x x x x

Edited December 31, 2010 by slashdot

[Guest]

Two switches.

1. Switch 2 and 9. You get:

+ + + x + x + x x x

2. Switch 4 and 7. You get:

+ + + + + x x x x x

Five switches.

2 and 19 giving +++x+x+x+x+x+x+x+xxx

4 and 17 giving +++++x+x+x+x+x+xxxxx

6 and 15 giving +++++++x+x+x+xxxxxxx

8 and 13 giving +++++++++x+xxxxxxxxx

10 and 11 giving ++++++++++xxxxxxxxxx

Which leads me to speculate:

For the general case with 2n total cards, the maximum number of switches would be the n/2 rounded down.

[Guest]

Five switches.

2 and 19 giving +++x+x+x+x+x+x+x+xxx

4 and 17 giving +++++x+x+x+x+x+xxxxx

6 and 15 giving +++++++x+x+x+xxxxxxx

8 and 13 giving +++++++++x+xxxxxxxxx

10 and 11 giving ++++++++++xxxxxxxxxx

Which leads me to speculate:

For the general case with 2n total cards, the maximum number of switches would be the n/2 rounded down.

Nice one :

My answer to the third question is : Moves = Cards/ 4 , Moves belong to Z numbers

Edited December 31, 2010 by Tsopi

[Guest]

I would say that it takes 2 moves on the 5,5 set, rotate 2 cards 45 degrees.

Same logic follows for the 10 cards, 5 moves.

Edit: Didn't re-read the question. It states that you must "switch" cards.

That's something i didn't think of, but it requires the same amount of moves if you do normally!