In a $m\times{n}$ grid are there are token. Every  token [i]dominates [/i] every square on its same row ($\leftrightarrow$), its same column ($\updownarrow$), and diagonal ($\searrow\hspace{-4.45mm}\nwarrow$)(Note that the token does not \emph{dominate} the diagonal ($\nearrow\hspace{-4.45mm}\swarrow$), determine the lowest number of tokens that must be on the board to [i]dominate [/i] all the squares on the board.
1. **Define the problem and assumptions:**
   We need to determine the minimum number of tokens required to dominate all squares on an \(m \times n\) grid. A token dominates all squares in its row, column, and one of the diagonals (\(\searrow\hspace{-4.45mm}\nwarrow\)).

2. **Identify "naughty" squares:**
   Let \(k\) be the number of tokens. "Naughty" squares are those not in the same row or column as any token, meaning they are only dominated diagonally. The number of naughty squares is \((m - k)(n - k)\).

3. **Distribute naughty squares in diagonals:**
   We distribute these naughty squares along the \(\searrow\hspace{-4.45mm}\nwarrow\) diagonals. The first and last diagonals contain at most one square each. If a diagonal contained more than one square, the square at the intersection of the row of one and the column of the other would also be naughty, contradicting the assumption.

4. **Count squares in diagonals:**
   The second and next-to-last diagonals contain at most two naughty squares each, and so on. Thus, the naughty squares are covered diagonally in bunches of \(1, 1, 2, 2, 3, 3, \ldots\).

5. **Calculate the number of diagonals:**
   The first \(2n - 2k - 2\) diagonals dominate at most \((n - k)(n - k - 1)\) naughty squares. 

6. **Symmetry consideration:**
   Assume \(n \leq m\). If \(m < n\), we can perform a symmetry around the \(\searrow\hspace{-4.45mm}\nwarrow\) diagonal, reversing \(m\) and \(n\).

7. **Inequality for naughty squares:**
   We need \((n - k)(n - k - 1) < (m - k)(n - k)\). This implies we need additional \((m - k) - (n - k - 1) = m - n + 1\) tokens to dominate each \(n - k\) naughty squares.

8. **Final inequality:**
   The number of tokens \(k\) must satisfy:
   \[
   k \geq 2n - 2k - 2 + m - n + 1
   \]
   Simplifying this, we get:
   \[
   3k \geq m + n - 1
   \]
   Therefore:
   \[
   k \geq \left\lceil \frac{m + n - 1}{3} \right\rceil
   \]

9. **Special cases:**
   If \(m\) or \(n\) are less than this value, we need just one token per row or column to dominate the entire board, and no naughty squares exist.

The final answer is \( \boxed{ \left\lceil \frac{m + n - 1}{3} \right\rceil } \).