Cheap online selection with weighted comparisons
Brute-force case analysis reveals that the optimal competitive ratio for the special case  $n=3$, with no other restrictions on the cost matrix, is the golden ratio $\phi = (\sqrt{5}+1)/2$.  Thus, no online algorithm can achieve a competitive ratio better than $\phi$.

Suppose $C_{1,2}=0$, $C_{1,3}=1$, and $C_{2,3}=\phi$.


Without loss of generality, the algorithm starts by comparing $S_1$ to $S_2$, at cost zero.
The adversary declares $S_1 &gt; S_2$.
If the algorithm compares $S_1$ to $S_3$:


The adversary declares $S_1 &gt; S_3$.
The algorithm must compare $S_2$ and $S_3$.
The adversary declares $S_2 &lt; S_3$, so $S_2$ is the minimum.
The total cost of the algorithm's comparisons is $1+\phi$.
The adversary reveals the total order $S_2 &lt; S_3 &lt; S_1$.
The total cost of the optimal comparisons ($S_1&gt;S_2$ and $S_2&lt;S_3$) is $\phi$.

If the algorithm compares $S_2$ to $S_3$:


The adversary declares $S_3 &gt; S_2$, so $S_2$ is the minimum.
The total cost of the algorithm's comparisons is $\phi$.
The adversary reveals the total order $S_2 &lt; S_1 &lt; S_3$.
The total cost of the optimal comparisons ($S_1&gt;S_2$ and $S_1&lt;S_3$) is $1$.

In either case, the algorithm's comparisons cost a factor of $\frac{1+\phi}{\phi} = \frac{\phi}{1} = \phi$ more than the optimal set of comparisons for the revealed total order.


More generally, the competitive ratio is $\min\{\frac{a+c}{a+b} , \frac{a+b+c}{a+c}\}$, where $a\le b\le c$ are the three comparison costs.  (There are more cases to consider here, because there are optimal algorithms that do not perform the cheapest comparison first, but the case analysis is still elementary.)  Tedious calculations imply that the expression $\min\{\frac{a+c}{a+b} , \frac{a+b+c}{a+c}\}$ is maximized when $a=0$, $b=1$, and $c=\phi$.

In particular, if $\frac{a+c}{a+b} &gt; \frac{a+b+c}{a+c}$, the best possible algorithm can be forced to perform all three comparisons.