The $1088$ words constitute $68$ blocks, occuping blocks $0$ to $67$ in the memory. The

cache has space for $64$ blocks. Hence, after blocks $0,1,2,...,63$ have been read

from the memory into the cache on the first pass, the cache is full. The next four

blocks, numbered $64$ to $67,$ map to sets $0,1,2,$ and $3.$ Each of them will replace

the least recently used cache block in its set, which is block $0.$ During the second

pass, memory block $0$ has to be reloaded into set $0$ of the cache, since it has been

overwritten by block $64.$ It will be placed in the least recently used block of set $0$ at

that point, which is block $1.$ Next, memory blocks $1,2,$ and $3$ will replace block $1$ of

sets $1,2$ and $3$ in the cache, respectively. Memory blocks $4$ to $15$ will be found in the

cache. Memory blocks $16$ to $19,$ which were in block location $1$ of sets $0$ to $3,$ have

now been overwritten, and will be reloaded in block location $2$ of these sets.

How do we know that there are four sets