max-min algorithm: each move of white is one that maximizes the new difference in times after the following move of black that tries to minimizes the time difference. A better, but more expensive, strategy on the part of white is to look-ahead more than one step.

So, let us then consider a seemingly easier question: is $c({\mathfrak{L}}({{A}_d}), {\mathfrak{L}}({{B}_d}))$ finite or infinite? For this question we do not need to speed-up the clocks. An infinite conformal distance occurs in one of the following three situations.

1.  The untimed language of ${{A}_d}$ is not included in that of ${{B}_d}$: there exists a path $q_0 \xrightarrow{a_1} q_1 \xrightarrow{a_2} \cdots \xrightarrow{a_n} q_n$ in ${{A}_d}$, with $q_n$ an accepting location, which either cannot be realized in ${{B}_d}$ with the same sequence of actions, or all such paths in ${{B}_d}$ do not terminate in an accepting location.

2.  There exists a path in ${{A}_d}$ of the form $q_0 \xrightarrow{a_1} q_1 \xrightarrow{a_2} \cdots \xrightarrow{a_n} q_n$, where the transition $q_{n-1} \xrightarrow{a_n} q_n$ has guard $t \geq m$, whereas for any path in ${{B}_d}$ of the form $q'_0 \xrightarrow{a_1} q'_1 \xrightarrow{a_2} \cdots \xrightarrow{a_n} q'_n$ the guard of the last transition $q'_{n-1} \xrightarrow{a_n} q'_n$ bounds $t$ from above.

3.  For 