You can see that the most connections any area has is of degree 2.īut the most efficient algorithm will only choose to flip the middle area. Take the 1-dimension example I commented earlier. So you can see flip operation has the effect of merging all adjacent nodes into the one being flipped.ĮDIT: in fact, greedily flip area with most connections isn't optimal. Then flip area 3 again because it has 2 connections: (3) Find riddles about anything and everything, all specially designed for kids Home Printable Worksheets. I'm not sure if greedily flip the area with most connections will be correct or not.įor example I'll flip area 3 first because it has 4 connections. Here are many rhyming riddles for kids that have a color theme - great for using for games and other activities with your children. To repeatedly perform flip operation to merge the areas. Which is better represented as: 1 - 2 4 - 5 The only location where your criteria would be true is the North Pole, where every direction outside your window is South. Transformed into area notation as: 1 2 3 4 5 Since the nodes are sequentially black and white in any path in these graphs, it's not difficult to prove that the minimal changes needed is exactly n (2 in this case) and can be achieved by changing always the central area.įirst, you are off that trying to model the problem as a connection graph of cells when in fact you should model the problem as a connection graph of areas When the eccentricity is n, there is at least one "maximal" path of length 2n-1 or 2n (of 2n or 2n+1 nodes). It's obvious from the diagram that for the above graph, there is exactly one central point and the "eccentricity" is 2. a node A where the greatest distance d(A,B) to other nodes B is minimal (and this minimal greatest distance is sometimes called "eccentricity"). Only one color, but not one size, Stuck at the bottom, yet easily flies. We now need to find the graph's center, or just one of the central points, e.g. My color changes from bright red to blue, The power Im using will dictate my. So, after you establish, the connected (black and white components), we have a graph of connections between these components (like in answer). You know there is one multicolor hat, but you don’t know how many green and black hats there are of each color (all of them must be wearing one). He can only figure it out after listening to C, not before. 4- And finally, D can figure out which hat he is wearing. So, the minimal is only 2 moves, not 4, in this case. 3- Only after listening to B can C figure out what hat he is wearing. Flip this to Black and all the board is black. As a result, the flipped one and the 4 white components are now connected into one component. First flip (to White) the central big Black component. The minimal solution is only 2 moves though. Consider this matrix, where there are 13 Black connected components and 4 White ones. This seems to be quite good but the answer is not so simple as it seems at first. Then one has to find the number of "black" and "white" connected components and choose the smallest of both. The answer is the number of connected components in the graph. The vertices are connected iff the correspondent matrix elements are "neighbors" and have the same color.
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