Discussion
A frontier holds \( 10^{9} \) nodes at 100 bytes each. What memory is needed?
Answer: A
Multiplying the node count by the size gives:
\[ 10^{9} \times 100 = 10^{11} \]
bytes, that is about 100 gigabytes, beyond most machines. The same search would take about seventeen minutes at a million nodes a second, which is why memory rather than time is the binding constraint. Comparing the two resources explicitly, in the units the machine actually has, is the calculation that decides which algorithm is usable.
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