Nurikabe II is a puzzle type made by Sam Griffiths-Jones, based on nikoli’s Nurikabe.
Just like the original Nurikabe; in an unsolved puzzle, there seems to be only one type of cell, white cells, though some white cells have numbers in them. However, in the solved puzzle, a new type of cell appears: black cells. White cells still exist in the solution; in fact, all of the cells with numbers are white.
The black cells divide the grid into separate parts of white cells. Each of these parts is called an “island”.
An island must have exactly one number, and its area is equal to the number. And no island may border another island. They may touch at corners, though.
The black cells also never form a group of 2×2 black squares, and all black cells are connected, not counting diagonally adjacent as connected. So if R4C2 was white, there will be two groups of black cells which is forbidden.
However, this type introduces extra restrictions… There must be no cycle of black cells; the no 2×2 black cells is then included within this rule. A cycle is defined as having two black squares on which one can be visited from another by at least two ways. For example, if R3C3 was black, R1C1 can visit R3C3 by either going down first or going right first.
Moreover, the white cells have the same restrictions. There must be no cycle of white cells. However, as it’s clear that a cycle of white cells that is larger than 2×2 is either composed of 2×2 cycles or containing isolated black cells which has been disallowed above, this restriction is simplified: there must be no 2×2 block of white cells.
Shade some cells to be black such that the above conditions are satisfied.
Walkthrough to the example
Begin with R4C1 and R2C2. An island with a 1 is always “full”, so we can shade R3C1, R4C2, R1C2, R2C1, R2C3, and R3C2. This also makes R1C1 black, since it is obviously unreachable by any number. (In particular, isolated white cells must be black.)
R1C3 is also unreachable by the 4, so it must be black…and now here comes the problem. R3C3 must be white; otherwise, a cycle of black cells is formed. Moreover, R3C4 must also be white, otherwise there’s not enough space for the 4.
Finally, if R4C3 is white, a block of 2×2 white cells is formed, so R4C3 must be black. This trivially makes the other two cells white to satisfy the 5 condition.