CMU 15-112: Fundamentals of Programming and Computer Science
Class Notes: Efficiency
- Big-Oh
- Describes asymptotic behavior of a function
- Informally (for 15112): ignore all lower-order terms and constants
- Formally (after 15112): see here
- A few examples:
- 3n^{2} - 2n + 25 is O(n^{2})
- 30000n^{2} + 2n - 25 is O(n^{2})
- 0.00000000001n^{2} + 123456789n is O(n^{2})
- 10nlog_{17}n + 25n - 17 is O(nlogn)
- Common Function Families
- Constant O(1)
- Logarithmic O(logn)
- Square-Root O(n^{0.5})
- Linear O(n)
- Linearithmic, Loglinear, or quasilinear O(nlogn)
- Quadratic O(n^{2})
- Exponential O(k^{n})
- Efficiency
When we say the program runs in O(f(N)), we mean...- N is the size of our input
- For a string s, N = len(s)
- For a list L, N = len(L) (also true for sets, dictionaries, and other collections)
- For an integer n, N = numberOfDigits(n) = log_{b}(n), so n = b^{N} (where b is the base, and you can use any base b >= 2).
- In the literature, N is often written in lowercase n, but we use that often to represent an integer n, which is different from the size of that integer. So in 112, we use uppercase N for the size of the input.
- f(N) = resource consumption of our program
- Resource can be time, space, bandwidth, ...
- For 15112, we mainly care about time
- For time, we usually measure algorithmic steps rather than elapsed time (These share the same big-oh, but algorithmic steps are easier to precisely describe and reason over)
- Note that you can measure worst-case or average case, or even other
cases such as best case (which often is trivial to compute and not very useful in practice). For 15-112, we often omit this term (which is a notable simplification
that you will not see in future courses), and we nearly always mean worst-case,
which is quite useful and generally easier to compute than average-case.
- Count steps in a written algorithm, or comparisons and swaps in a list, etc
- Can verify by timing your code's execution with: time.time()
- N is the size of our input
- The Big Idea
- Each function family grows much faster than the one before it!
- And: on modern computers, any function family is usually efficient enough on small n, so we only care about large n
- So... Constants do not matter nearly as much as function families
- Practically...
- Do not prematurely or overly optimize your code
- Instead: think algorithmically!!!
- Examples
- Sequences, Nesting, and Composition
- Sequencing
# what is the total cost here? L = [ 52, 83, 78, 9, 12, 4 ] # assume L is an arbitrary list of length N L.sort() # This is O(NlogN) L.sort(reverse=True) # This is O(NlogN) L[0] -= 5 # This is O(1) print(L.count(L[0]) + sum(L)) # This is O(N) + O(N) - Nesting
# what is the total cost here? L = [ 52, 83, 78, 9, 12, 4 ] # assume L is an arbitrary list of length N for c in L: # This loop's body is executed O(N) times L[0] += c # This is O(1) L.sort() # This is O(NlogN) print(L) # This is O(N) (why?) - Composition
# what is the total cost here? def f(L): # assume L is an arbitrary list of length N L1 = sorted(L) # This is O(NlogN) return L1 # This is O(1) def g(L): # assume L is an arbitrary list of length N L1 = L * len(L) # This is O(N**2) (why?) return L1 # This is O(1) L = [ 52, 83, 78, 9, 12, 4 ] # assume L is an arbitrary list of length N L = f(g(L)) # What is the big-oh of this? print(L) # This is O(N**2) (why?)
- Sequencing
- Python Builtins
Here we use S for a set and L for a list:- Some are O(1), including len(L), (val in S), L.append(item)
- Some are O(N), including max(L), min(L), (val in L), L.count(val), set(L)
- Sorting is O(NlogN)
- For a more complete list, see here
- isPrime vs fasterIsPrime
- From these examples, we see that isPrime tests O(n) factors whereas fasterIsPrime tests O(n^{0.5}) factors.
- But the size of the input is N=log_{2}(n), so we substitute n=2^{N}.
- And we conclude that our isPrime is O(2^{N}) and fasterIsPrime is O((2^{N})^{0.5})) = O(2^{N/2}).
- And so: isPrime(n) is exponential, and hopelessly slow on large inputs.
- And: fasterIsPrime, while much faster, is also exponential and so also hopelessly slow on large inputs.
- Much faster primality tests exist. For example, the AKS Primality Test is polynomial not exponential!
- Linear Search vs Binary Search
- Linear search
- Basic idea: check each element in turn
- Use: find an element in an unsorted list
- Cost: O(N)
- Binary search
- Basic idea: in a sorted list, check middle element, eliminate half on each pass
- Uses:
- Find an element in a sorted list
- Number-guessing game (eg: guess a random number between 1 and 1000)
- Find a root (zero) of a function with bisection (adapted binary search)
- Cost: O(logN)
- Linear search
- Sorting
- Sorting Examples
See here. - SelectionSort vs MergeSort
- Definitions
- selectionsort: repeatedly select largest remaining element and swap it into sorted position
- mergesort: sort blocks of 1's into 2's, 2's into 4's, etc, on each pass merging sorted blocks into sorted larger blocks
- Sorting Links
- Wikipedia page on Sorting
- David Eck's xSortLab applet (or you might try this jar file )
- Our sorting sample code (You need to be able to write all of this (except bubblesort) from scratch!)
- Excellent sorting animation website
- Sorting algorithm animation video (15 algorithms in 6 minutes)
- Even more sorting algorithm animations
- Analysis
This is mostly informal, and all you need to know for a 112-level analysis of these algorithms. You can easily find much more detailed and rigorous proofs on the web.- selectionsort
On the first pass, we need N compares and swaps (N-1 compares and 1 swap).
On the second pass, we need only N-1 (since one value is already sorted).
On the third pass, only N-2.
So, total steps are about 1 + 2 + ... + (N-1) + N = N(N+1)/2 = O(N^{2}). - mergesort
On each pass, we need about 3N compares and copies (N compares, N copies down, N copies back).
So total cost = (3N steps per pass) x (# of passes)
After pass 0, we have sorted lists of size 2^{0} (1)
After pass 1, we have sorted lists of size 2^{1} (2)
After pass 2, we have sorted lists of size 2^{2} (4)
After pass k, we have sorted lists of size 2^{k}
So we need k passes, where N = 2^{k}
So # of passes = k = log_{2}N
Recall that total cost = (3N steps per pass) x (# of passes)
So total cost = (3N)(log_{2}N) = O(NlogN).
Note: This is the theoretical best-possible O() for comparison-based sorting!
- selectionsort
- Definitions
- Sorting Examples
- sumOfSquares Examples
Note: Run this code in Standard Python, as it will timeout if you run it in brython.# The following functions all solve the same problem: # Given a non-negative integer n, return True if n is the sum # of the squares of two non-negative integers, and False otherwise. def f1(n): for x in range(n+1): for y in range(n+1): if (x**2 + y**2 == n): return True return False def f2(n): for x in range(n+1): for y in range(x,n+1): if (x**2 + y**2 == n): return True return False def f3(n): xmax = int(n**0.5) for x in range(xmax+1): for y in range(x,n+1): if (x**2 + y**2 == n): return True return False def f4(n): xyMax = int(n**0.5) for x in range(xyMax+1): for y in range(x,xyMax+1): if (x**2 + y**2 == n): return True return False def f5(n): xyMax = int(n**0.5) for x in range(xyMax+1): y = int((n - x**2)**0.5) if (x**2 + y**2 == n): return True return False def testFunctionsMatch(maxToCheck): # first, verify that all 5 functions work the same print("Verifying that all functions work the same...") for n in range(maxToCheck): assert(f1(n) == f2(n) == f3(n) == f4(n) == f5(n)) print("All functions match up to n =", maxToCheck) testFunctionsMatch(100) # use larger number to be more confident import time def timeFnCall(f, n): # call f(n) and return time in ms # Actually, since one call may require less than 1 ms, # we'll keep calling until we get at least 1 secs, # then divide by # of calls we had to make calls = 0 start = end = time.time() while (end - start < 1): f(n) calls += 1 end = time.time() return float(end - start)/calls*1000 #(convert to ms) def timeFnCalls(n): print("***************") for f in [f1, f2, f3, f4, f5]: print ("%s(%d) takes %8.3f milliseconds" % (f.__name__, n, timeFnCall(f, n))) timeFnCalls(1001) # use larger number, say 3000, to get more accurate timing
- Sequences, Nesting, and Composition
Graphically (Images borrowed from here):