It is given that none of the streets has more than one team traveling along it in any direction at any point in time (point 1), which implies at
, all 4 teams have chosen different roots from the starting point.
It is also known that Teams 2 and 3 are the only ones in stations
and
respectively at 10:00 hrs, and Team 1 and Team 4 are the only teams that patrol the street connecting stations
and
.
It is only possible when Team 2 traveled (A-E) via F, and Team 3 reached station D via station C.
It is also known that Teams 1 and 3 are the only ones in Station
at 10:30 hrs, and Team 4 never passes through Stations B, D, or F. Hence, Team 1 must have chosen the (A-B) root at the starting point, and Team 4 has chosen the
root at 9.00 hrs.
Hence, Team 1 will reach
at 9.30 , and come to
at 10.00 hrs. After that, they will go to
at 10.30 hrs.
Since Team 4 never passes through stations B, D, or F. Team 4 only can pass through stations
, and
.
Hence, the roots of team 4 to reach station
at 11.30 will be (A-E-A-C-A-E) or (A-E-A-E-A-E).
Since team 1 is already traveling to
from
at
, at that time team 4 can't choose the same route. Hence, the final route for team 4 to reach
at 11.30 is (A-E-A-C-A-E), and at
, team 4 will come back to station
.
Hence, the complete route diagram for team 4 is (A-E-A-C-A-E-A)
We can see that team 1 is at station
at
, and they will reach station
at
, which is only possible when they travel to
via
.
Hence, the complete route diagram for team 1 is (A-B-A-E-A-B-A). It is also known that Teams 1 and 3 are the only ones in station E at 10:30 hrs.
The only possible root for Team 2 at 10.00 hrs is from
to
since they can't choose
to
because Team 3 is already on this route. Since team 3 has to reach A at 12.00 . The only possible combination for team 3 is E-D-C-A
Now the roots for team 2 going back to
is from
at 10.30 hrs (F-A-F-A) or (F-E-F-A).
Hence, the final table is given below:
From the table, we can see that the teams have passed through B2 times in this given period