A 5x5 grid has been colored in such a way that any group of cells consisting of a cell and its two, three or four neighboring cells has each of its cells being of a different color. This has been done in such a way as to use the minimum possible number of colors that will accomplish that. Two of the red cells have been shown in the diagram below. Which of the other numbered cells are also red?
| 1 |
2 |
3 |
4 |
5 |
| 6 |
7 |
8 |
9 |
10 |
| 11 |
12 |
13 |
14 |
15 |
| 16 |
17 |
18 |
19 |
20 |
| 21 |
22 |
23 |
24 |
25 |
Cells 9, 12 and 20.
By neighboring cells, I assume that the puzzle refers to horizontally and vertically neighboring, namely those with a common border.
In that case, a minimum of 5 colors is required, because a cells and its 4 neighbors must all be colored differently. And this minimum is achievable in two basic ways (not counting reflections, etc).
ABCDE or ABCDE
BCDEA CDEAB
CDEAB EABCD
DEABC BCDEA
EABCD DEABC
In the first way, all the same colored cells in effect share one common "diagonal" (imagine forming a cylinder by taping the top row to the bottom, or the first column to the last).
In the second way (where matching colors are separated by "knight" moves), each of the 10 "diagonals" have each of the 5 colors.
At any rate, the two reds are not on a common "diagonal", so the arrangement bust of the 2nd type.
It's Greek to me (spoiler)
