**Undead**TerminologyPaths and SidesEach number or

**quota** around the grid is at the same time an entrance and an exit of a path to and from another number. Each path may be divided by zero or more mirrors, indicated by a slash in a cell. We call a number's

**side** for every cell between the number and the first mirror when traversing the path starting from that number. Conversely, the

**other side** of the path in this direction are every cell from the last mirror to the other number at the end of the path. In other words, each number owns a side that goes no further than the first mirror closest to the number.

Sides will be important for advanced strategies, like parity.

Moreover, everything between the first mirrors from both sides are shared cells. They have nothing to do with parity, and will count the same towards both number's quota. For shared cells, a Vampire counts as 0, and a Zombie or Ghost counts as +1 for both sides.

ParityParity in Undead means zero-sum. When counting just the cells for each side (not the shared cells), Zombie is worth 0 points, Vampire worth +1, and Ghost -1. So two Zombies on the same side is worth the same as one Vampire and one Ghost. In other words, the Vampire and Ghost cancel each other. Conversely, these symbols on the other side of a path is worth the inverse amount. That is, a Zombie on the other side is worth 0 points, a Vampire on the opposite side is worth -1, and a Ghost is worth +1. Thus, the same symbol appearing on both sides will cancel each other. A Vampire on one side is +1 and a Vampire on the other side is -1, thus they cancel each other.

Quota DifferenceThe shortcut to solving the quotas is to get the difference of the numbers at both ends of the path. In other words, subtract the two numbers. Then using parity, find out how many extra Vampires and Ghosts are at either side. Because of parity concept explained above, Vampires are worth +1 for that side and Ghosts -1; and vice versa if they appear on the other side. Generally, the higher number has more Vampires and fewer Ghosts (not always true, but that's the basic idea.) Use Vampires if you need to raise a side to match the difference and Ghosts to lower a side.

Example 1: Suppose the two sides of a path are 3 and 1. The difference is 2. That could mean: the 3 side has two more Vampires, or the 1 side has two more Ghosts, or the 3 side has a Vampire and the 1 side has a Ghost. This is the simple way to look at it if only two cells are involved.

Example 2: Suppose the two sides of a path are 3 and 1. The difference is 2. But the 3 side already has a Vampire, which is worth +1. That means we need either another Vampire at the 3 side or a Ghost at the 1 side. Both solutions give +1 to the 3 side.

Example 3: Suppose the two sides of a path are 3 and 1. The difference is 2. But the 3 side already has a Ghost, which is worth -1. We need to raise +3 in order to match the difference of 2. That means we need either three more Vampires at the 3 side or three more Ghosts at the 1 side, or some combination in between (i.e. 1 Vampire 2 Ghosts, 2 Vampires 1 Ghost).

2 for 1A Ghost and a Vampire on the same side or both in shared cells will do two things: cancel each other to make zero parity, and cause two cells to provide only one toward the quota. This is useful if you have just enough cells to fill a quota or you need to eliminate extra cells.

Example 1: Suppose two sides of a path are both 2, and exactly two cells and one mirror in the path. Then you know both cells are Zombies, filling 2 to both quotas. A Vampire or Zombie would give 1 less toward their quotas.

Example 2: Suppose two sides of a path are both 2, and exactly three cells and one mirror in the path. The number of cells in the path is greater than the quota of one or both sides. Thus you know two of those cells must be Vampires and/or Ghosts, and the third cell is Zombie. The Vampire/Ghost will take up two cells to fill 1 quota.

Zombie CountAnother form of parity is odd/even parity (in general mathematics.) This may be useful to find whether there are odd or even number of Zombies on both sides of a path.

- Let E = number of cells along the path that are not shared.
- Let D = the quota difference for this path.
- E - D = Zp, the odd/even parity of Zombies for this path.
- Zp = odd means odd number of Zombies.
- Zp = even means even number of Zombies.

0 (Zero) QuotaA path that involves a 0 (zero) endpoint or quota is the most trivial to solve and will help fill out the cells. There are two simple rules to remember:

- All cells on the 0's side (before the first mirror) are Ghosts.
- All cells beyond the first mirror are Vampires.

No MirrorsA path with no mirrors is only possible if a path runs straight across the same row or column. The quotas or numbers on the endpoints will be identical. The number of Ghosts on this path is equal to number of cells minus either quota.

Effective Single MirrorOne or more consecutive mirrors with no empty cells in between effectively act as a single mirror, unless the entrance and exit of a possible effective single mirror crosses the same cell. A special case is a path between two endpoints with only a true effective single mirror and no other mirrors (or if all shared cells are Vampires). Then the following are true.

- Let E = the number of non mirror cells along this path.
- Let P and Q be the quotas of the two endpoints of this path.
- 0 <= P <= E and 0 <= Q <= E.
- P and Q may or may not be equal.
- Let T = P + Q, the total quota of both endpoints.
- Let M = 2*E, the maximum total quota along this path.
- E is also the minimum total quota along this path.
- Then E <= T <= M must be true.
- M - T = G + V, the number of non Zombies along this path.
- T - E = Z, the number of Zombies along this path.
- When T = M, then all cells along this path are Zombies. In other words, P = Q = Z = E.
- When T = E, then all cells are non Zombies.

Multi-MirrorMulti-mirror paths can be reduced to effective single mirror paths, if

**all** shared cells in the path are

**known**. Then subtract values as follows, and solve like an effective single mirror path.

- Let E = cells in the path that are not shared.
- For each Ghost and Zombie in shared cells, subtract 1 (one) from both P and Q.
- Solve like an effective single mirror path, except using these new values.