witzar

It is not that I'm a busy person, it just didn't occur to me that there could be any doubt about what n stands for. Anyway, my apologies to everyone confused.

You start and finish the n-th cut at some two points at the edge of the pizza. Ont it's way this cut crosses each of n-1 previous cuts. When the n-th cut crosses first previous cut, a new piece appears. When it crosses second cut we have second piece, and so on. When we cross last previous cut we have (n-1) new pieces, and finally we reach the edge of the pizza which produces one more piece. This makes total of n new pieces with n-th cut.

Thanks, k-man. You have explained it all perfectly.

Wrong.

Why this works? Suppose chessboard squares have horizontal and diagonal coordinates ranging from 1 to 8, so that the Kings starts at square (1,1) and finishes at square (8,8). Observe that the above strategy ensures, that after every move of the first player the King is on the square with both coordinates being even numbers, and every move of his opponent the King is on the square with at least one coordinate being odd. The final square of the King journey (8,8) has both coordinates even, so only the first player can get there.

There is no way to put number 6 in 3rd column.

Will just one isosceles trapezoid do?

What you actually ask is to arrange given vectors into an ordered (rooted) tree that has the highest probability of construction according to given rules and vectors out of all possible ordered trees. But observe, that when creating set of vectors according to given rules it is equally likely to produce vector W from vector V and to produce vector V from vector W. Therefore changing the root of the tree to any other node will not alter the probability of the tree, but will alter the relation of precession. Finding the (non-ordered) tree with highest probability is definitely doable, but choosing the root is just guessing. Maybe asking about non-ordered tree would make the contest more interesting.

I totally agree. This is why I wrote "sketch of a proof" instead of "proof". Besides, I've left a case when there is one large circle (L) and a number of small circles, small circles are pair-wise externally tangent to each other and each small circle is internally tangent to the big circle. I just said, that "it is easy to see" that at most 3 small circles are possible, so I just want to complete it. Suppose we have 4 or more inner circles. Let's number them 1, 2, 3, 4... in the following way: number 1 is assigned freely, and subsequent numbers are assigned subsequently clockwise according to tangent points to the circle L. Since circle 2 is tangent to circle 4 and both are tangent to the circle L, they divide interior of circle L in two parts. But circles 1 and 3 are in different parts, so these circles cannot be tangent. This contradiction shows that 4 or more small circles are not possible.

There are two types of tangency: internal and external. In case when only external tangency exists, 4 is a maximum: we start with 3 circles, then the only option is to put the fourth one "in the middle", and that's it. Now suppose there are two internally tangent circles L (larger) and S (smaller). Now all other circles have to go inside L (if not, they will not be able to be tangent with S) and have to be externally tangent with S (if not, they will not be able to be tangent with L). So in this case we have one "big" circle and a number of "small" circles, small circles are pair-wise externally tangent with each other and each small circle is internally tangent with the big circle. In this case at most 3 small circles are possible which is easier "to see", than to prove. (I'm not suggesting that the prove is hard.) So 4 is the maximum and only 2 configurations are possible. PS I believe this reasoning can be extended for 3-dimensional case (mutually touching spheres).

I see. Then you must be asking about http://en.wikipedia.org/wiki/Descartes%27_theorem The simple part is a special case with one of the circles replaced by a straight line.

It is clearly required for each pair of circles to be tangent, which is not the case in your example. For example the top-left circle is not tangent with bottom-right circle.

You didn't get it. Move "4U" means: move 4th column Up (just one cell). (If you repeat this move 4 times, then you will return to the same position.)

I've played for a while with Anza Power's simulator and I've managed to swap "1" and "2" in original position. This means that it is possible to swap any two horizontally adjacent cells, you just have to: 1. Make a couple of obvious moves to bring these two cells in place where "1" and "2" originally are. 2. Apply the procedure that swaps cells "1" and "2". 3. Undo moves made in step 1 (perform "opposite" moves in reversed order). Exactly the same can be done for every vertically adjacent cells due to the symmetry of the puzzle, you can easily transform procedure swapping "1" and "2" into procedure swapping "1" and "4". Now when you know how to swap any two adjacent (horizontally or vertically) cells, it is easy to solve every possible puzzle: you can just put numbers into their cells one by one. First put 1 in place, then 2 without touching 1, then 3 without touching 1 and 2, etc. Alternatively you can solve first three rows using method described by Anza Power and use above for the last row. All you have to do is to find a sequence of moves that swaps "1" and "2" (possibly by trial and error). And such a sequence exists, I've checked it.

Think for example about a graph in a shape of a cross (or letter X): four arms, each arm made of 100 segments. This graph has 400 edges (segments) and only 4 odd nodes (one at the end of each arm). You can start and finish where You want. Can you traverse it in 402 steps? Can You do better then 600? PS If You don't like "dead ends" (vertices with only one segment)in graph, You can add a triangle (loop made of 3 segments) at the end of each arm.