The $\emph{Stooge sort}$ is a particularly inefficient recursive sorting algorithm defined as follows: given an array $A$ of size $n$, we swap the first and last elements if they are out of order; we then (if $n\ge3$) Stooge sort the first $\lceil\tfrac{2n}3\rceil$ elements, then the last $\lceil\tfrac{2n}3\rceil$, then the first $\lceil\tfrac{2n}3\rceil$ elements again. Given that this runs in $O(n^\alpha)$, where $\alpha$ is minimal, find the value of $(243/32)^\alpha$.
1. **Define the running time function:**
   Let \( f(n) \) be the running time of Stooge sort. According to the problem, the recurrence relation for the running time is:
   \[
   f(n) = 3f\left(\left\lceil \frac{2n}{3} \right\rceil\right) + O(1)
   \]
   For simplicity, we can approximate the ceiling function and ignore the constant term, giving us:
   \[
   f(n) \approx 3f\left(\frac{2n}{3}\right)
   \]

2. **Assume a polynomial solution:**
   Assume \( f(n) = O(n^\alpha) \). Then we can write:
   \[
   n^\alpha = 3 \left(\frac{2n}{3}\right)^\alpha
   \]

3. **Simplify the equation:**
   Simplify the right-hand side:
   \[
   n^\alpha = 3 \left(\frac{2n}{3}\right)^\alpha = 3 \left(\frac{2^\alpha n^\alpha}{3^\alpha}\right) = 3 \cdot \frac{2^\alpha}{3^\alpha} \cdot n^\alpha
   \]
   \[
   n^\alpha = 3 \cdot \left(\frac{2}{3}\right)^\alpha \cdot n^\alpha
   \]

4. **Isolate the term involving \(\alpha\):**
   Divide both sides by \( n^\alpha \):
   \[
   1 = 3 \cdot \left(\frac{2}{3}\right)^\alpha
   \]

5. **Solve for \(\alpha\):**
   \[
   1 = 3 \cdot \left(\frac{2}{3}\right)^\alpha
   \]
   \[
   \left(\frac{2}{3}\right)^\alpha = \frac{1}{3}
   \]
   Taking the natural logarithm of both sides:
   \[
   \alpha \ln\left(\frac{2}{3}\right) = \ln\left(\frac{1}{3}\right)
   \]
   \[
   \alpha \ln\left(\frac{2}{3}\right) = -\ln(3)
   \]
   \[
   \alpha = \frac{-\ln(3)}{\ln\left(\frac{2}{3}\right)}
   \]
   \[
   \alpha = \frac{\ln(3)}{\ln\left(\frac{3}{2}\right)}
   \]

6. **Calculate \((243/32)^\alpha\):**
   \[
   (243/32)^\alpha = \left(\frac{3^5}{2^5}\right)^\alpha = \left(\frac{3}{2}\right)^{5\alpha}
   \]
   Substitute \(\alpha\):
   \[
   \left(\frac{3}{2}\right)^{5 \cdot \frac{\ln(3)}{\ln\left(\frac{3}{2}\right)}} = \left(\frac{3}{2}\right)^{5 \cdot \frac{\ln(3)}{\ln(3) - \ln(2)}}
   \]
   Simplify the exponent:
   \[
   5 \cdot \frac{\ln(3)}{\ln(3) - \ln(2)} = 5 \cdot \frac{\ln(3)}{\ln\left(\frac{3}{2}\right)}
   \]
   \[
   \left(\frac{3}{2}\right)^{5 \cdot \frac{\ln(3)}{\ln\left(\frac{3}{2}\right)}} = 3^5 = 243
   \]

The final answer is \(\boxed{243}\).