The number of turns is related to the sum of logarithms of the dimensions ($\lceil\log_2(n')\rceil + \lceil\log_2(m')\rceil$), not their product $n' \times m'$. a smaller area doesn't always mean fewer steps in the game:

i.e.,
$n_1 \times m_1 < n_2 \times m_2 \;\;\;\not\Rightarrow\;\;\; T(n_1, m_1) < T(n_2, m_2)$
or more simply,
$n_1 \times m_1 < n_2 \times m_2 \;\;\;\not\equiv\;\;\; T(n_1, m_1) < T(n_2, m_2)$

egg::
the dude has to decide between these two grid options (ts after first cut):

**option a**: a grid of size $31 \times 1$
- area: $31 \times 1 = 31$
- turns other guy can force:
- for 31: $(31+1)/2 = 16$, then 8, 4, 2, 1 → that’s 5 steps
- for 1: nothing to cut → 0 steps
- total: $5 + 0 = 5$ turns

**option b**: a grid of size $5 \times 5$
- area: $5 \times 5 = 25$
- turns other guy can force:
- for 5: $(5+1)/2 = 3$, then 2, then 1 → 3 steps
- same for the other 5 → another 3 steps
- total: $3 + 3 = 6$ turns

so even though option b has a smaller area (25), it results in more turns (6) than option a (5 turns with area 31).

if mouf only looked at area, he’d pick option b and end up with a longer game. but if he calculates the number of future turns, he sees option a is actually better

**∴** to minimize the total number of turns,be checking all 4 poss outcomes from the init cut compute how many turns fouad can force in each and pick the one with the fewest