Let a, b, c, and d denote constants
In order of dominance (top is most dominant). Realize that algorithms of higher dominance are bad!
Function 
condition 
common name 
N^{n} 


N! 

N factorial 
a^{n} 
dominates b^{n} if a > b 
exponential 
N^{c} 
dominates N^{d} if c > d 
polynomial (cubic, quadratic, etc.) 
n log n 

n log n 
n 

linear 
log n 

logarithmic (regardless of base) 
1 

constant 
Linear 
logarithmic 
n*log_{2}N 
quadratic 
cubic 
exponential 
exponential 
factorial 
1 
0 
0 
1 
1 
2 
3 
1 
2 
1 
2 
4 
8 
4 
9 
2 
4 
2 
8 
16 
64 
16 
81 
24 
8 
3 
24 
64 
512 
256 
6561 
40320 
16 
4 
64 
256 
4096 
65,536 
43,046,721 
2.09E+013 
32 
5 
160 
1024 
322,768 
4,294,967,296 
…1.85E+15 
2.63E+035 
64 
6 
384 
4096 
262,144 
1.84E+17 
…3.43E+30 
1.27E+089 
128 
7 
896 
16,384 
2,097,152 
3.4E+38 
…1.18E+61 
3.86E+215 
256 
8 
2048 
65,536 
1,677,216 
1.16E+77 ??? 
…1.39E+122 
Find other calculator 
Note 1: The value here is approximately the number of machine instructions executed by a 1 gigaflop computer in 5000 years, or 5 years on a current supercomputer (teraflop computer)
Note 2: The value here is about 500 billion times the age of the universe in nanoseconds, assuming a universe age of 20 billion years.log_{2}n 
n 
nlog_{2}n 
n^{2} 
2^{n} 

Current
Computers 
N1 
N2 
N3 
N4 
N5 
Ten
times
faster 
N1 x 30 
N2
x
10 
N3 x 3 
N4 x 3 
N5 + 3 
Thousand
times
faster 
N1 x 9,000 
N2
x
1,000 
N3 x 111 
N4 x 31 
N5 + 10 
Million
times
faster 
N1 x 19E+6 
N2
x
1E+6 
N3 x 5,360 
N4 x 1,000 
N5 + 20 
array int sorta; // this comes from somewhere in a messy state
int temp; // for the swap
int outIx, inIx; // these are indecies
// outer loop over the sorted array  when everything's sorted we're done
for (outIx gets the values sorta.length down to 1) {
// from the beginning of the list to the end of the unsorted bit
for(inIx = 1; inIx < outIx1; inIx++) {
// If adjacent items are out of order, swap them!
if (sorta[inIx1]>sorta[inIx]) {
temp = sorta[inIx1];
sorta[inIx1] = sorta[inIx];
sorta[inIx] = temp;
}
}