We wish to estimate the surprise number $(2$nd moment$)$ of a data stream, using the method of

AMS. It happens that our stream consists of ten different values, which we'll call $1,2,$dots,$10,$ that

cycle repeatedly. That is$,$ at timestamps $1$ through $10,$ the element of the stream equals the

timestamp, at timestamps $11$ through $20,$ the element is the timestamp minus $10,$ and so on$.$ It

is now timestamp $75($it is the last timestamp in the stream$),$ and a $5$ has just been read from

the stream.

For our estimate of the surprise number, we shall choose three timestamps at random, and

estimate the surprise number from each, using the AMS approach $($length of the stream times

$2m-1,$ where $m$ is the number of occurrences of the element of the stream at that timestamp,

considering all times from that timestamp on$,$ to the current time $75).$ Then, our estimate will

be the median of the three resulting values.

Identify from the list below the set of three "random" timestamps that give the closest

estimate, and explain why.

a$)\{24,44,65\}$

b$)\{31,48,50\}$

c$)\{37,46,55\}$

d$)\{25,34,47\}$