The tag bits in direct mapping can be given by

Key idea: In direct mapping, every main-memory block maps to exactly one cache line.

• Let the number of main-memory blocks be m.

• Let the number of cache lines be n.

• For a fixed cache line, the number of possible memory blocks that can map there is m/n.

• The tag must identify which one of these m/n blocks is currently present.

Therefore: tag bits = log2(m/n).

Final answer: log2(m/n).