Square $\mathcal S$ has vertices $(1,0)$, $(0,1)$, $(-1,0)$ and $(0,-1)$.  Points $P$ and $Q$ are independently selected, uniformly at random, from the perimeter of $\mathcal S$.  Determine, with proof, the probability that the slope of line $PQ$ is positive.

[i]Proposed by Isabella Grabski[/i]
1. **Define the vertices and segments:**
   The square $\mathcal{S}$ has vertices at $(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$. The perimeter of the square consists of four segments: $AB$, $BC$, $CD$, and $DA$ where:
   - $A = (1,0)$
   - $B = (0,1)$
   - $C = (-1,0)$
   - $D = (0,-1)$

2. **Select points $P$ and $Q$:**
   Points $P$ and $Q$ are selected independently and uniformly at random from the perimeter of $\mathcal{S}$. We need to determine the probability that the slope of the line $PQ$ is positive.

3. **Consider the segments:**
   Without loss of generality (WLOG), assume $P$ is on segment $AB$ or $BC$. We will analyze the case when $P$ is on $AB$ and then extend the result to other segments by symmetry.

4. **Case 1: $P$ on $AB$:**
   - Parametrize $P$ on $AB$: Let $P = (1-t, t)$ where $0 \leq t \leq 1$.
   - Determine the regions where $Q$ must lie for the slope of $PQ$ to be positive:
     - If $Q$ is on $BC$, $CD$, or $DA$, the slope of $PQ$ will be positive.
     - If $Q$ is on $AB$, the slope will be zero or negative.

5. **Calculate the length of favorable regions:**
   - The total perimeter of the square is $4$ units.
   - The length of $BC$, $CD$, and $DA$ combined is $3$ units.
   - Therefore, the probability that $Q$ lies on $BC$, $CD$, or $DA$ is $\frac{3}{4}$.

6. **Case 2: $P$ on $BC$:**
   - Parametrize $P$ on $BC$: Let $P = (-t, 1-t)$ where $0 \leq t \leq 1$.
   - Determine the regions where $Q$ must lie for the slope of $PQ$ to be positive:
     - If $Q$ is on $CD$, $DA$, or $AB$, the slope of $PQ$ will be positive.
     - If $Q$ is on $BC$, the slope will be zero or negative.

7. **Calculate the length of favorable regions:**
   - The total perimeter of the square is $4$ units.
   - The length of $CD$, $DA$, and $AB$ combined is $3$ units.
   - Therefore, the probability that $Q$ lies on $CD$, $DA$, or $AB$ is $\frac{3}{4}$.

8. **Combine the probabilities:**
   - The probability that $P$ is on $AB$ is $\frac{1}{4}$.
   - The probability that $P$ is on $BC$ is $\frac{1}{4}$.
   - The probability that $P$ is on $CD$ is $\frac{1}{4}$.
   - The probability that $P$ is on $DA$ is $\frac{1}{4}$.
   - For each segment, the probability that $Q$ lies in the favorable regions is $\frac{3}{4}$.

9. **Calculate the overall probability:**
   \[
   \text{Overall probability} = \frac{1}{4} \cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{3}{4} + \frac{1}{4} \cdot \frac{3}{4} = 4 \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{3}{4}
   \]

The final answer is $\boxed{\frac{1}{2}}$