Amortized analysis refers to determining the time-averaged running time for a sequence (not an individual) operation. It is different from average case analysis because here, we don’t assume that the data arranged in average (not very bad) fashion like we do for average case analysis for quick sort. That is, amortized analysis is worst case analysis but for a sequence of operation rather than an individual one. It applies to the method that consists of the sequence of operation, where a vast majority of operations are cheap but some of the operations are expensive. This can be visualized with the help of binary counter which is implemented below.
Let’s see this by implementing an increment counter in C. First, let’s see how counter increment works.
Let a variable i contains a value 0 and we performs i++ many time. Since on hardware, every operation is performed in binary form. Let binary number stored in 8 bit. So, value is 00000000. Let’s increment many time. So, the pattern we find are as :
00000000, 00000001, 00000010, 00000011, 00000100, 00000101, 00000110, 00000111, 00001000 and so on …..
1. Iterate from rightmost and make all one to zero until finds first zero.
2. After iteration, if index is greater than or equal to zero, then make zero lie on that position to one.
On a simple look on program or algorithm, its running cost looks proportional to the number of bits but in real, it is not proportional to a number of bits. Let’s see how!
Let’s assume that increment operation is performed k time. We see that in every increment, its rightmost bit is getting flipped. So, the number of flipping for LSB is k. For, second rightmost is flipped after a gap, i.e., 1 time in 2 increments. 3rd rightmost – 1 time in 4 increments. 4th rightmost – 1 time in 8 increments. So, the number of flipping is k/2 for 2nd rightmost bit, k/4 for 3rd rightmost bit, k/8 for 4th rightmost bit and so on …
Total cost will be the total number of flipping, that is,
C(k) = k + k/2 + k/4 + k/8 + k/16 + …… which is Geometric Progression series and also,
C(k) < k + k/2 + k/4 + k/8 + k/16 + k/32 + …… up to infinity
So, C(k) < k/(1-1/2)
and so, C(k) < 2k
So, C(k)/k < 2
Hence, we find that average cost for increment a counter for one time is constant and it does not depend on the number of bit. We conclude that increment of a counter is constant cost operation.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
- Analysis of Algorithm | Set 5 (Amortized Analysis Introduction)
- Difference between Straight Ring Counter and Twisted Ring Counter
- Analysis of Algorithms | Set 4 (Analysis of Loops)
- Analysis of Algorithms | Big-O analysis
- Analysis of Algorithms | Set 1 (Asymptotic Analysis)
- Ripple Counter in Digital Logic
- n-bit Johnson Counter in Digital Logic
- Design counter for given sequence
- Differences between Synchronous and Asynchronous Counter
- Mod 6 Johnson Counter (with D flip-flop)
- Mod 2 Ring Counter (with D flip-flop)
- Non-binary Counter in Digital Logic
- Synchronous Series Carry Counter
- Self Starting Counter
- Free Running Counter
- Ring Counter in Digital Logic
- Minimum increment or decrement operations required to make the array sorted
- Analysis of Algorithms | Set 3 (Asymptotic Notations)
- Analysis of Algorithm | Set 4 (Solving Recurrences)
- Asymptotic Analysis and comparison of sorting algorithms
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.