I am reading CLRS 3rd Ed, chapter 17.1 (Aggregate analysis pg453) and I came across this statement.

Let us analyze a sequence of n PUSH, POP, and MULTIPOP operations on an initially empty stack.

I am confused as in:

  1. Do the push, pop, and multipop cumulatively add up to n operations.
    (there is a total of n operations which consists of x pushes, y pops, and z multipops where x+y+z = n)
  2. Are they talking about n push, n pop, and n multipop operations?
  3. 1st operation(push, pop, Multipop), 2nd operation(push, pop, Multipop), ..., nth operation(push, pop, Multipop)

The question is does the statement from the book imply 1 or 2 or 3 or something I did not mention above. Thanks

  • 5
    $\begingroup$ I don't have the book with me so I am not posting this as an answer but in all likelihood $n$ refers to the number of operations and "pop, push and multipop" to the kind of operations, so number 1 of your suggestions. If the intended meaning was different the phrase would (probably) have a different structure. $\endgroup$
    – phan801
    Commented Nov 3, 2021 at 16:32
  • $\begingroup$ @phan801 thanks that makes sense. $\endgroup$ Commented Nov 3, 2021 at 19:12

1 Answer 1


As phan801 commented, the first interpretation, a sequence of $n$ operations, each of which is either push or pop or multipop, is correct.

Either one of the other two interpretations might stand a small chance without surrounding context or with a very different context. However, had it been the intended meaning, "the phrase would probably have a different structure", such as, "n push operations, n pop operations, and n multipop operations" or "n operations, each being a push operation followed by a pop operation followed by a multipop operation".

A strong indication of the actual meaning comes from the introductory sentence of this section, "17.1 Aggregate analysis".

In aggregate analysis, we show that for all $n$, a sequence of $n$ operations takes worst-case time $T(n)$ in total.

The interpretation 2 means three sequences of $n$ operations instead of "a sequence of $n$ operations".

The interpretation 3 could have been more possible if push-then-pop-then multipop is a reasonable combination. However, after push-then-pop, multipop operation will always be a non-operation, since the stack is assumed empty initially. There is nothing interesting to analyze for this interpretation. (Even if the stack is not empty initially, we would probably use "push-then-multipop" since we can combine a pop operation followed by a multipop operation into another multipop operation".) So this combination of three operations as one operation does not make much sense.

  • 2
    $\begingroup$ By the way, the first interpretation fits the ensuing discussion in the textbook perfectly. $\endgroup$
    – John L.
    Commented Nov 3, 2021 at 19:45

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