Comment on page

13.4 Asymptotic Behavior

Be on your best behavior!
In most applications, we are most concerned about what happens for very large values of
NN
. This is known as the asymptotic behavior. We want to learn what types of algorithms are able to handle large amounts of data. Some examples of applications that require highly efficient algorithms are:
  • Simulating the interactions of billions of particles
  • Maintaining a social network with billions of users
  • Encoding billions of bytes of video data
Algorithms that scale well have better asymptotic runtime behavior than algorithms that scale poorly.
Let us return to the original problem of characterizing the runtimes of our dup functions. Recall that we desire characterizations that have the following features:
  • Simple and mathematically rigorous.
  • Clearly demonstrate the superiority of dup2 over dup1.
We’ve accomplished the second task! We are able to clearly see that dup2 performed better than dup1. However, we didn’t do it in a very simple or mathematically rigorous way. Luckily, we can apply a series of simplifications to solve these issues.

Simplification 1: Consider Only the Worst Case

When comparing algorithms, we often only care about the worst case. The worst case is often where we see the most interesting effects, so we can usually ignore all other cases but the worst case.

Example:

Consider the operation counts of some algorithm below. What do you expect will be the order of growth of the runtime for the algorithm?
  • NN
    (linear)
  • N2N^2
    (quadratic)
  • N3N^3
    (cubic)
  • N6N^6
    (sextic)
Operation
Count
less than (<)
100N2+3N100N^2 + 3N
greater than (>)
N3+1N^3+1
and (&&)
50005000
Answer:
N3N^3
(cubic)
Intuitively,
N3N^3
grows faster than
N2N^2
, so it would "dominate." To help further convince you that this is the case, consider the following argument:
  • Suppose the < operator takes
    α\alpha
    nanoseconds, the > operator takes
    β\beta
    nanoseconds, and the && takes
    γ\gamma
    nanoseconds.
  • The total time is
    α(100N2+3)+β(2N3+1)+5000γ\alpha (100N^2 + 3)+\beta (2N^3+1)+5000\gamma
    nanoseconds.
  • For very large
    NN
    , the
    2βN32\beta N^3
    term is much larger than others.
  • It can help to think of it in terms of calculus. What happens as
    NN
    approaches infinity? Which term ends up dominating?

Simplification 2: Restrict Attention to One Operation

Pick some representative operation to act as a proxy for overall runtime. From our dup example:
  • Good choice: increment, or less than or equals or array accesses.
  • Bad choice: assignment of j = i + 1, or i = 0.
The operation we choose is called the “cost model.”

Simplification 3: Eliminate Low Order Terms

Ignore lower order terms.
Sanity check: Why does this make sense? (Related to the checkpoint above!)

Simplification 4: Eliminate Multiplicative Constants

Ignore multiplicative constants.
  • Why? No real meaning!
  • By choosing a single representative operation, we already “threw away” some information.
  • Some operations had counts of
    3N23N^2
    , ​​
    N2/2N^2/2
    , etc. In general, they are all in the family/shape of
    N2N^2
    .

Checkpoint Exercise:

Apply our four simplification rules to the dup2 table.
Operation
Symbolic Count
i = 0
1
j = i+1
0 to
NN
<
0 to
N1N-1
==
1 to
N1N-1
array accesses
2 to
2N22N-2
Example answer: array accesses with order of growth
NN
.
<, ==, and j=i+1 would be fine answers as well.

Simplification Summary

  • Only consider the worst case.
  • Pick a representative operation (aka: cost model)
  • Ignore lower order terms
  • Ignore multiplicative constants.