It is important to be able to measure, or at least make educated statements about, the space and time complexity of an algorithm. This will allow you to compare the merits of two alternative approaches to a problem you need to solve, and also to determine whether a proposed solution will meet required resource constraints before you invest money and time coding.
Analysis is done before coding. Profiling (a.k.a. benchmarking) is done after the code is written.
The absolute running time of an algorithm cannot be predicted, since this depends on the programming language used to implement the algorithm, the computer the program runs on, other programs running at the same time, the quality of the operating system, and many other factors. We need a machine-independent notion of an algorithm’s running time.
The current state-of-the-art in analysis is finding a measure of an algorithm’s relative running time, as a function of how many items there are in the input, i.e., the number of symbols required to reasonably encode the input, which we call $n$. The $n$ could be:
The first three only make sense when each of the elements (items, characters, coefficients/exponents) have a bounded-size. For example: if all items are 64-bit integers, then the number of items can be used for $n$, but if numeric values are unbounded, you have to count bits!
We count the number of abstract operations as a function of $n$.
for (int i = 0; i < a.length; i++) { System.out.println(a[i]); }
Here $n = a.\mathtt{length}$ (provided we know that all of the items in the array have a fixed size, which is often the case). We have
System.out.println
so we write $T(n) = 4n + 1$.
Or should we?
All operations are not created equal. The printing completely overwhelms the increments, compares and indexing operations. We might as well talk about having $n$ compare-index-print-increment operations. Then $T(n) = n + 1$. While we’re at it, we might as well realize that initialization doesn’t mean much, so we’re safe to write $T(n) = n$.
for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { double sum = 0; for (int k = 0; k < n; k++) { sum += a[i][k] * b[k][j]; } c[i][j] = sum; } }
Working from the inside-out we see each iteration of the $k$-loop has
4 array indexing operations, one multiplication, one addition, and
one assignment. Or, heck, one big sum += a[i][k]*b[k][j]
operation.
Inside the $j$-loop there are n of these, together with an initialization
of sum and an assignment to c[i][j]
. Does this mean
$n+2$? Hardly. Those other two operations
don’t count for much. By a similar argument we can ignore all the
simple for-loop initializing, testing, and incrementing operations.
They are fast. All we need to do is count the number of
sum += a[i][k]*b[k][j]
operations — that’s really what
counts.
So we have $T(n) = n^{3}$.
But is this a fair measure of the complexity? We counted operations relative to the width of the matrices. We can also count operations relative to the number of items in the two matrices, which is $2n^2$. Under this measure we’d say $T(n) = n^{1.5}$.
for (int i = 1; i <= n; i *= 2) { for (int j = 0; j < n; j++) { count++; } }
Here the outer loop is done $\log n$ times and the inner loop is done $n$ times, so $T(n) = n \log n$. (Yes, the default base for logarithms in Computer Science is 2.)
if (a.length > 0) { return a[a.length - 1]; } else { throw new NoSuchElementException(); }
Here $n = a\mathtt{.length}$, and $T(n) = 1$.
x + y
If $x$ and $y$ are primitive int values that each fit in a word
of memory, any computer can add them in one step: $T(n) = 1$, But if these are
BigInteger
values, then what matters is how many digits (or bits)
there are in each value. Since we just have to add the digits
in each place and propagate carries, we get
$T(n) = n$.
for (int i = 0; i < a.length; i++) { if (a[i] == x) { return Optional.of(i); } } return Optional.empty();
Here we need to introduce the $B$ (best-case) and $W$ (worst-case) functions: $B(n) = 1$ and $W(n) = n$. (What about the average case? It’s often difficult and sometimes impossible to compute an average case. You have to know something about the expected distributions of items.)
We can group complexity functions by their “growth rates” (something we’ll formalize in a minute). For example:
Class | Example | Description |
---|---|---|
Constant | $\lambda n . 1$ | Fixed number of steps, regardless of input size |
Logarithmic | $\lambda n . \log n$ | Number of steps proportional to log of input size |
Polynomial | $\lambda n . n^k$, $k \geq 1$ | Number of steps a polynomial of input size |
Exponential | $\lambda n . c^n$, $c > 1$ | Number of steps exponential of input size |
Those are four useful families to be sure, but we can define lots more:
Class | Example ($\lambda n .$) | Description |
---|---|---|
Constant | $1$ | Doing a single task at most a fixed number of times. Example: retrieving the first element in a list. |
Inverse Ackermann | $\alpha(n)$ | |
Iterated Log | $\log^{*} n$ | |
Loglogarithmic | $\log \log n$ | |
Logarithmic | $\log n$ | Breaking down a large problem by cutting its size by some fraction. Example: Binary Search. |
Polylogarithmic | $(\log n)^c$, $c>1$ | |
Fractional Power | $n^c$, $0<c<1$ | |
Linear | $n$ | "Touches" each element in the input. Example: printing a list. |
N-log-star-n | $n \log^{*} n$ | |
Linearithmic | $n \log n$ | Breaking up a large problem into smaller problems, solving them independently, and combining the solutions. Example: Mergesort. |
Quadratic | $n^2$ | "Touches" all pairs of input items. Example: Insertion Sort. |
Cubic | $n^3$ | |
Polynomial | $n^c$, $c\geq 1$ | |
Quasipolynomial | $n^{\log n}$ | |
Subexponential | $2^{n^\varepsilon}$, $0<\varepsilon<1$ | |
Exponential | $c^n$, $c>1$ | Often arises in brute-force search where you are looking for subsets of a collection of items that satisfy a condition. |
Factorial | $n!$ | Often arises in brute-force search where you are looking for permutations of a collection of items that satisfy a condition. |
N-to-the-n | $n^n$ | |
Double Exponential | $2^{2^n}$ | |
Ackerman | $A(n)$ | |
Runs Forever | $\infty$ |
To get a feel for the complexity classes consider a computer that performed one million abstract operations per second:
$T(n)$ | 10 | 20 | 50 | 100 | 1000 | 1000000 |
---|---|---|---|---|---|---|
1 | 1 μs | 1 μs | 1 μs | 1 μs | 1 μs | 1 μs |
$\log n$ | 3.32 μs | 4.32 μs | 5.64 μs | 6.64 μs | 9.97 μs | 19.9 μs |
$n$ | 10 μs | 20 μs | 50 μs | 100 μs | 1 msec | 1 second |
$n \log n$ | 33.2 μs | 86.4 μs | 282 μs | 664 μs | 9.97 msec | 19.9 seconds |
$n^2$ | 100 μs | 400 μs | 2.5 msec | 10 msec | 1 second | 11.57 days |
$1000 n^2$ | 100 msec | 400 msec | 2.5 seconds | 10 seconds | 16.7 minutes | 31.7 years |
$n^3$ | 1 msec | 8 msec | 125 msec | 1 second | 16.7 minutes | 317 centuries |
$n^{\log n}$ | 2.1 ms | 420 ms | 1.08 hours | 224 days | 2.5×10^{7} Ga | 1.23×10^{97} Ga |
$1.01^n$ | 1.10 μs | 1.22 μs | 1.64 μs | 2.7 μs | 20.9 ms | 7.49×10^{4298} Ga |
$0.000001 \times 2^n$ | 1.02 ns | 1.05 msec | 18.8 minutes | 40.2 Ga | 3.40×10^{272} Ga | 3.14×10^{301001} Ga |
$2^n$ | 1.02 ms | 1.05 seconds | 35.7 years | 4.02×10^{7} Ga | 3.40×10^{278} Ga | 3.14×10^{301007} Ga |
$n!$ | 3.63 seconds | 771 centuries | 9.64×10^{41} Ga | 2.96×10^{135} Ga | 1.28×10^{2545} Ga | — |
$n^n$ | 2.78 hours | 3323 Ga | 2.81×10^{62} Ga | 3.17×10^{177} Ga | 3.17×10^{2977} Ga | — |
$2^{2^n}$ | 5.70×10^{285} Ga | 2.14×10^{315630} Ga | — | — | — | — |
By the way, Ga is a gigayear, or one billion years.
There is something. Compare the $2^n$ row with the $0.000001\cdot 2^n$ row. The latter represents something running one million times faster than the former, but still, even for an input of size 50, requires a run time in the thousands of centuries.
Since we are really measuring growth rates, we usually ignore:
so for example
The justification for doing this is
How can we formalize this? That is, how do we formalize things like “the set of all quadratic functions” or the “set of all cubic functions” or “the set of all logarithmic functions”?
A function $f$ is in $O(g)$ whenever there exist constants $c$ and $N$ such that for every $n > N$, $f(n)$ is bounded above by a constant times $g(n)$. Formally:
$$ O(g) =_{def} \{ f \mid \exists c\,N . \forall n > N . |f(n)| \leq c|g(n)| \} $$This means that $O(g)$ is the set of all functions for which $g$ is an asymptotic upper bound of $f$.
By convention, people usually drop the lambdas when writing expressions involving Big-O, so you will see things like $O(n^2)$.
Well let’s see why we should be able to, but only argue with one special case. (You can do the formal proof on your own.) Let’s show that $7n^4 + 2n^2 + n + 20 \in O(n^4)$. We have to choose $c$ and $N$. Let’s choose $c=30$ and $N=1$. So since $n \geq 1$: $$ \begin{eqnarray} |7n^4 + 2n^2 + n + 20| & \leq & 7n^4 + 2n^2 + n + 20 \\ & \leq & 7n^4 + 2n^4 + n^4 + 20n^4 \\ & \leq & 30n^4 \end{eqnarray} $$
There is a Big-$\Omega$ notation for lower bounds.
A function $f$ is in $\Omega(g)$ whenever there exist constants $c$ and $N$ such that for every $n > N$, $f(n)$ is bounded below by a constant times $g(n)$.
$$ \Omega(g) =_{def} \{ f \mid \exists c\,N . \forall n > N . |f(n)| \geq c|g(n)| \} $$Lower bounds are useful because they say that an algorithm requires at least so much time. For example, you can say that printing an array is $O(2^n)$ because $2^n$ really is an upper bound, it’s just not a very tight one! But saying printing an array is $\Omega(n)$ means it requires at least linear time which is more accurate. Of course just about everything is $\Omega(1)$.
When the lower and upper bounds are the same, we can use Big-Theta notation.
$$\Theta(g) =_{def} \{ f \mid \exists c_1\,c_2\,N. \forall n > N . c_{1} |g(n)| \leq |f(n)| \leq c_2 |g(n)| \}$$Example: printing each element of an array is $\Theta(n)$. Now that’s useful!
Most complexity functions that arise in practice are in Big-Theta of something. An example of a function that isn’t in Big-$\Theta$ of anything is $\lambda n . n^2 \cos n$.
There are also little-oh and little-omega. $$\omicron(g) =_{def} \{ f \mid \lim_{x\to\infty} \frac{f(x)}{g(x)} = 0 \}$$ $$\omega(g) =_{def} \{ f \mid \lim_{x\to\infty} \frac{f(x)}{g(x)} = \infty \}$$
There’s also the Soft-O:
$$\tilde{O}(g) =_{def} O(\lambda n.\,g(n) (\log{g(n)})^k)$$Since a log of something to a power is bounded above by that something to a power, we have
$$\tilde{O}(g) \subseteq O(\lambda n.\,g(n)^{1+\varepsilon})$$Normally you can just look at a code fragment and immediately "see" that it is $\Theta(1)$, $\Theta(\log n)$, $\Theta(n)$, $\Theta(n \log n)$, or whatever. But when you can’t tell by inspection, you can write code to count operations for given input sizes, obtaining $T(n)$. Then you "guess" different values of f for which $T \in \Theta(f)$. To do this, generate values of $\frac{T(n)}{f(n)}$ for lots of different $f$s and $n$s, looking for the $f$ for which the ratio is nearly constant. Example:
package edu.lmu.cs.algorithms; import java.util.Arrays; /** * An application class that illustrates how to do an empirical * analysis for determining time complexity. */ public class PrimeTimer { /** * A method that computes all the primes up to n, and returns the * number of "primitive operations" performed by the algorithm. */ public static long computePrimes(int n) { boolean[] sieve = new boolean[n]; Arrays.fill(sieve, true); sieve[0] = sieve[1] = false; long ticks = 0; for (int i = 2; i * i < sieve.length; i++) { ticks++; if (sieve[i]) { ticks++; for (int j = i + i; j < sieve.length; j += i) { ticks++; sieve[j] = false; } } } return ticks; } /** * Runs the computePrimes() methods several times and displays the * ratio of the number of primitive operations to n, n*log2(log2(n)), * n*log2(n), and n*n, for each run. The idea is that if the ratio is * nearly constant for any one of the expressions, that expression is * probably the asymptotic time complexity. */ public static void main(String[] args) { System.out.println(" n T(n)/n T(n)/nloglogn" + " T(n)/nlogn T(n)/n^2"); for (int n = 100000000; true; n += 10000000) { double time = (double)computePrimes(n); double log2n = Math.log(n) / Math.log(2.0); double log2log2n = Math.log(log2n) / Math.log(2.0); System.out.printf("%9d%15.9f%15.9f%15.9f%18.12f\n", n, time / n, time / (n * log2log2n), time / (n * log2n), time / ((double)n * n)); } } }
$ javac edu/lmu/cs/algorithms/PrimeTimer.java && java edu.lmu.cs.algorithms.PrimeTimer n T(n)/n T(n)/nloglogn T(n)/nlogn T(n)/n^2 100000000 2.483153670 0.524755438 0.093437967 0.000000024832 110000000 2.488421536 0.525042572 0.093154203 0.000000022622 120000000 2.492800475 0.525216963 0.092881654 0.000000020773 130000000 2.496912269 0.525397614 0.092636275 0.000000019207 140000000 2.500607893 0.525543667 0.092406844 0.000000017861 150000000 2.504261540 0.525726296 0.092202718 0.000000016695 160000000 2.508113469 0.525989754 0.092029052 0.000000015676 170000000 2.511378312 0.526164370 0.091854066 0.000000014773 180000000 2.514174300 0.526271114 0.091679818 0.000000013968 190000000 2.517111653 0.526434419 0.091526593 0.000000013248 200000000 2.519476410 0.526502103 0.091366731 0.000000012597 210000000 2.521918771 0.526607744 0.091222446 0.000000012009 220000000 2.524643468 0.526791900 0.091099845 0.000000011476 230000000 2.526841665 0.526883963 0.090968655 0.000000010986 240000000 2.529254646 0.527037032 0.090854692 0.000000010539 250000000 2.531615772 0.527194102 0.090747527 0.000000010126 260000000 2.533554869 0.527276931 0.090633206 0.000000009744 270000000 2.535273511 0.527326521 0.090518378 0.000000009390 280000000 2.537081721 0.527406431 0.090413568 0.000000009061 290000000 2.538856217 0.527490156 0.090313866 0.000000008755 300000000 2.540369853 0.527529768 0.090210757 0.000000008468 . . . 580000000 2.573933419 0.529233035 0.088416447 0.000000004438 590000000 2.574801915 0.529278658 0.088371416 0.000000004364 600000000 2.575498323 0.529291236 0.088321815 0.000000004292 610000000 2.576229731 0.529313262 0.088274708 0.000000004223 620000000 2.576874421 0.529319651 0.088225880 0.000000004156 630000000 2.577634338 0.529351816 0.088182205 0.000000004091 640000000 2.578348789 0.529376678 0.088138140 0.000000004029 650000000 2.579215052 0.529434671 0.088100389 0.000000003968 660000000 2.579956808 0.529469006 0.088059472 0.000000003909 670000000 2.580654654 0.529496175 0.088018114 0.000000003852 680000000 2.581385994 0.529532005 0.087978922 0.000000003796 690000000 2.582226029 0.529591860 0.087944424 0.000000003742 700000000 2.582908279 0.529621034 0.087905512 0.000000003690 710000000 2.583473030 0.529627754 0.087863538 0.000000003639 . . .
What we’re seeing here is that the ratio $\frac{T(n)}{n}$ is increasing, so the complexity is probably more than linear. The ratios $\frac{T(n)}{n \log n}$ and $\frac{T(n)}{n^2}$ are decreasing so those functions are probably upper bounds. But $\frac{T(n)}{n \log{\log{n}}}$ is also increasing, but the rate of increase seems to be slowing and who knows, might even converge. We could bet that the complexity $\in \Theta(n \log{\log{n}})$, but we should really do a formal proof.
Suppressing leading constant factors hides implementation dependent details such as the speed of the computer which runs the algorithm. Still, you can some observations even without the constant factors.
For an algorithm of complexity | If the input size doubles, then the running time |
---|---|
$1$ | stays the same |
$\log n$ | increases by a constant |
$n$ | doubles |
$n^2$ | quadruples |
$n^3$ | increases eight fold |
$2^n$ | is left as an exercise for the reader |
Getting a faster computer allows to solve larger problem sets in a fixed amount of time, but for exponential time algorithms the improvement is pitifully small.
For an algorithm of complexity | If you can solve a problem of this size on your 100MHz PC | Then on a 500MHz PC you can solve a problem set of this size | And on a supercomputer one thousand times faster than your PC you can solve a problem set of this size |
---|---|---|---|
$n$ | 100 | 500 | 100000 |
$n^2$ | 100 | 223 | 3162 |
$n^3$ | 100 | 170 | 1000 |
$2^n$ | 100 | 102 | 109 |
More generally,
$T(n)$ | On Present Computer | On a computer 100 times faster | On a computer 1000 times faster | On a computer one BILLION times faster |
---|---|---|---|---|
$n$ | N | 100N | 1000N | 1000000000N |
$n^2$ | N | 10N | 31.6N | 31623N |
$n^3$ | N | 4.64N | 10N | 1000N |
$n^5$ | N | 2.5N | 3.9N | 63N |
$2^n$ | N | N + 6.64 | N + 9.97 | N + 30 |
$3^n$ | N | N + 4.19 | N + 6.3 | N + 19 |
What if we had a computer so fast it could do ONE TRILLION operations per second?
$T(n)$ | 20 | 40 | 50 | 60 | 70 | 80 |
---|---|---|---|---|---|---|
$n^5$ | 3.2 μs | 102 μs | 313 μs | 778 μs | 1.68 msec | 3.28 msec |
$2^n$ | 1.05 msec | 1.1 seconds | 18.8 minutes | 13.3 days | 37.4 years | 383 centuries |
$3^n$ | 3.5 msec | 4.7 months | 227 centuries |
1.3 × 10^{7} centuries | 7.93 × 10^{11} centuries | 4.68 × 10^{16} centuries |
As you can see, the gap between polynomial and exponential time is hugely significant. We can solve fairly large problem instances of high-order polynomial time algorithms on decent computers rather quickly, but for exponential time algorithms, we can network together thousands of the world’s fastest supercomputers and still be unable to deal with problem set sizes of over a few dozen. So the following terms have been used to characterize problems:
Polynomial-time Algorithms | Exponential-time Algorithms |
---|---|
Good | Bad |
Easy | Hard |
Tractable | Intractable |