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I am currently looking into Big O notation and computational complexity.

Problem 1.1 in CLRS asks what seems a basic question, which is to get an intuition about how different algorithmic complexities grow with the size of the input.

The question asks:

For each function $f(n)$ and time $t$ in the following table, determine the largest size $n$ of a problem that can be solved in time $t$, assuming that the algorithm to solve the problem takes $f(n)$ microseconds.

The time periods are 1 second, 1 minute, 1 hour, 1 day, 1 month, 1 year, 1 century.

The functions $f(n)$ are seemingly common time complexities that arise in algorithms frequently, the list being:

$$ \log_2n, \quad \sqrt{n}, \quad n, \quad n \log_2 n, \quad n^2, \quad n^3, \quad 2^n \quad \text{and} \quad n!$$

Most are fairly straightforward algebraic manipulations. I am struggling with two of these, and both for the same reason:

If $c$ is the time in microseconds, the two I am struggling with are $$ n \log_2 n = c$$ $$ n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n = c$$

For $n!$ I thought of using Stirling's Approximation.

These both require the ability to solve $n \log_2 n = c$, with Stirling require a little more manipulation.

Questions

  1. As $n \log_2 n$ is not solvable using elementary functions (only Lambert W), what are some good ways to approximate $n log_2 n$? Or how do we implement Lambert W?
  2. How do we solve n! = c, necessarily approximately as n grows large. Is Stirling the right way to go, and if so how to solve $\sqrt{2\pi n} \left(\frac{n}{e}\right)^n = c$

Here is some python code I put together to complete the table with my current output:

EDIT: Based on a couple of answers, I have used a binary search method (except for lg n). I have edited the code below to reflect this:

+---------+-------------+-------------+-------------+-------------+-------------+-------------+-------------+
| f(n)    |    1 sec    |    1 min    |    1 Hour   |    1 Day    |   1 Month   |    1 Year   |  1 Century  |
+---------+-------------+-------------+-------------+-------------+-------------+-------------+-------------+
| lg n    | 2^(1.0E+06) | 2^(6.0E+07) | 2^(3.6E+09) | 2^(8.6E+10) | 2^(2.6E+12) | 2^(3.2E+13) | 2^(3.2E+15) |
| sqrt(n) |   1.0E+12   |   3.6E+15   |   1.3E+19   |   7.5E+21   |   6.7E+24   |   9.9E+26   |   9.9E+30   |
| n       |   1.0E+06   |   6.0E+07   |   3.6E+09   |   8.6E+10   |   2.6E+12   |   3.2E+13   |   3.2E+15   |
| n log n |    62746    |   2.8E+06   |   1.3E+08   |   2.8E+09   |   7.2E+10   |   8.0E+11   |   6.9E+13   |
| n^2     |     1000    |     7745    |    60000    |    293938   |   1.6E+06   |   5.6E+06   |   5.6E+07   |
| n^3     |     100     |     391     |     1532    |     4420    |    13736    |    31593    |    146645   |
| 2^n     |      19     |      25     |      31     |      36     |      41     |      44     |      51     |
| n!      |      9      |      11     |      12     |      13     |      15     |      16     |      17     |
+---------+-------------+-------------+-------------+-------------+-------------+-------------+-------------+

Python code:

import math
import decimal
from prettytable import PrettyTable

def binary_search_guess(f, t, last=1000):
    for i in range(0, last):
        guess = pow(2,i)
        if f(guess) > t:
            return binary_search_function(f, pow(2,i-1), guess, t)

    return -1 

def binary_search_function(f, first, last, target):
    found = False

    while first<=last and not found:
        midpoint = (first + last)//2
            if f(midpoint) <= target and f(midpoint+1) > target:
                found = True
            else:
                if target < f(midpoint):
                    last = midpoint-1
                else:
                    first = midpoint+1
    best_guess = midpoint

    return best_guess

def int_or_sci(x):
    if x >= math.pow(10,6):
        x = '%.1E' % decimal.Decimal(x)
    else:
        x = int(x)

    return x

def input_size_calc():
    #Create Pretty Table Header
    tbl = PrettyTable(["f(n)", "1 sec", "1 min", "1 Hour", "1 Day", "1 Month", "1 Year", "1 Century"])
    tbl.align["f(n)"] = "l" # Left align city names
    tbl.padding_width = 1 # One space between column edges and contents (default)

    #Each Time Interval in Microseconds
    tsec = pow(10,6)
    tmin = 60 * tsec
    thour = 3600 * tsec
    tday = 86400 * tsec
    tmonth = 30 * tday
    tyear = 365 * tday
    tcentury = 100 * tyear

    tlist = [tsec,tmin,thour,tday,tmonth,tyear,tcentury]
    #print tlist

    #Add rows   
    #lg n
    f = lambda x : math.log(x,2)
    fn_list = []
    for t in tlist:
        #This would take too long for binary search method
        ans = int_or_sci(t)
        fn_list.append("2^(%s)" % ans)
    tbl.add_row(["lg n",fn_list[0], fn_list[1], fn_list[2], fn_list[3], fn_list[4], fn_list[5], fn_list[6]])

    #sqrt(n)
    f = lambda x : math.pow(x,1/2.0)
    fn_list = []
    for t in tlist:
        fn_list.append(int_or_sci(binary_search_guess(f, t)))
    tbl.add_row(["sqrt(n)",fn_list[0], fn_list[1], fn_list[2], fn_list[3], fn_list[4], fn_list[5], fn_list[6]])

    #n
    f = lambda x : x
    fn_list = []
    for t in tlist:
        fn_list.append(int_or_sci(binary_search_guess(f, t)))
    tbl.add_row(["n",fn_list[0], fn_list[1], fn_list[2], fn_list[3], fn_list[4], fn_list[5], fn_list[6]])

    #n log n
    f = lambda x : x * math.log(x,2)
    fn_list = []
    for t in tlist:
        fn_list.append(int_or_sci(binary_search_guess(f, t)))
    tbl.add_row(["n log n",fn_list[0], fn_list[1], fn_list[2], fn_list[3], fn_list[4], fn_list[5], fn_list[6]])

    #n^2
    f = lambda x : math.pow(x,2)
    fn_list = []
    for t in tlist:
        fn_list.append(int_or_sci(binary_search_guess(f, t)))
    tbl.add_row(["n^2",fn_list[0], fn_list[1], fn_list[2], fn_list[3], fn_list[4], fn_list[5], fn_list[6]])

    #n^3
    f = lambda x : math.pow(x,3)
    fn_list = []
    for t in tlist:
        fn_list.append(int_or_sci(binary_search_guess(f, t)))
    tbl.add_row(["n^3",fn_list[0], fn_list[1], fn_list[2], fn_list[3], fn_list[4], fn_list[5], fn_list[6]])

    #2^n
    f = lambda x : math.pow(2,x)
    fn_list = []
    for t in tlist:
        fn_list.append(int_or_sci(binary_search_guess(f, t)))
    tbl.add_row(["2^n",fn_list[0], fn_list[1], fn_list[2], fn_list[3], fn_list[4], fn_list[5], fn_list[6]])

    #n!
    f = lambda x : math.factorial(x)
    fn_list = []
    for t in tlist:
        fn_list.append(int_or_sci(binary_search_guess(f, t)))
    tbl.add_row(["n!",fn_list[0], fn_list[1], fn_list[2], fn_list[3], fn_list[4], fn_list[5], fn_list[6]])

    print tbl

#PROGRAM BEGIN
input_size_calc()
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  • 3
    $\begingroup$ You could approximate $W_n$ by simply doing binary search on the value of $n$ or using a Taylor series. You can extend the factorial function to be continuous, to the gamma function and you can probably find some information on its inverse - but your approach using the Stirling approximation seems fine as well. $\endgroup$ – Tom van der Zanden Aug 1 '15 at 17:49
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    $\begingroup$ check out this pdf: cs-people.bu.edu/lapets/resource/nlogn.pdf $\endgroup$ – Sagnik Aug 1 '15 at 19:03
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    $\begingroup$ @TomvanderZanden Forget $W_n$ -- you can approximate the whole question just by doing a binary search! $\endgroup$ – David Richerby Aug 1 '15 at 21:07
  • $\begingroup$ Be careful about fractions... using three significant digits is reasonable when n is large, but for small n, make sure to round down to the nearest integer. $\endgroup$ – Ben Voigt Aug 2 '15 at 3:21
  • $\begingroup$ @DavidRicherby Can you expand on this please? $\endgroup$ – stats_novice_123 Aug 2 '15 at 14:55
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The approximate inverse of $c = n\log n$ is $n = c/\log c$. Indeed, for this value of $n$ we have $$ n\log n = \frac{c}{\log c} \log \frac{c}{\log c} = \frac{c}{\log c} (\log c - \log\log c) = c \left(1 - \frac{\log\log c}{\log c}\right). $$ This approximation is usually good enough.

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You don't need to approximate anything to solve the exercise. All the functions you're given are monotone so you can just use binary search. That is, to solve $f(n)=c$ for $n$, just guess $n=1, 2, 4, 8, \dots$ until you find the first $k$ that $2^k \log 2^k > c$. Then, do ordinary binary search between $2^{k-1}$ and $2^k$. If the solution is $x$, this takes roughly $2\log x$ evaluations of $f$.

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  • $\begingroup$ Wouldn't a binary search on lg n take an infeasible amount of time? $\endgroup$ – stats_novice_123 Aug 2 '15 at 18:29
  • $\begingroup$ I have edited my code in my question and answer table to reflect this approach. It seems impractical for lg n however, would you agree? $\endgroup$ – stats_novice_123 Aug 2 '15 at 18:50
  • $\begingroup$ No, it's fine for solving $\log n = c$. Once you've doubled $\lceil c\rceil$ times, you've overshot, and then you do a binary search between $2^{\lceil c\rceil}$ and $2^{\lfloor c\rfloor}$. That range is about $2^c$ wide, so it takes about $c$ iterations of binary search to find the correct value. $\endgroup$ – David Richerby Aug 2 '15 at 19:10
  • $\begingroup$ But just for 1 second, c = 10^6 ? 1 year c = 3.2 x 10^13 ? as f(n) is in microseconds according to the question $\endgroup$ – stats_novice_123 Aug 2 '15 at 19:14
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    $\begingroup$ OK, I see your point. But, still, you'd only need about $2c$ iterations of the search and, with an efficient bignum library, iterations wouldn't take too long. (And, yeah, as you say, $\log$ is trivially invertible anyway so you can side-step the issue.) $\endgroup$ – David Richerby Aug 2 '15 at 19:23

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