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
- 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)
- Optional: For a more complete list, see here
- isPrime vs fasterIsPrime
- From here, isPrime tests O(n) values and fasterIsPrime tests only O(n^{0.5}) values.
- So fasterIsPrime is much faster.
- Considerably faster primality tests exist. For example, the AKS Primality Test.
- 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 Algorithms
There are dozens of common sorting algorithms. We will study these in particular:- Selection Sort: repeatedly select the smallest (or largest) remaining element and swap it into sorted position.
- Merge Sort: sort blocks of 1's into 2's, 2's into 4's, etc, on each pass merging sorted blocks into sorted larger blocks.
- Using the Sorting Simulator (xSortLab)
Check out xSortLab for Selection Sort and Merge Sort. Be sure to understand the Visual Sort as well as the Timed Sort. - Sorting Code Examples and Timing
# sorting.py import time, random #################################################### # swap #################################################### def swap(a, i, j): (a[i], a[j]) = (a[j], a[i]) #################################################### # selectionSort #################################################### def selectionSort(a): n = len(a) for startIndex in range(n): minIndex = startIndex for i in range(startIndex+1, n): if (a[i] < a[minIndex]): minIndex = i swap(a, startIndex, minIndex) #################################################### # mergeSort #################################################### def merge(a, start1, start2, end): index1 = start1 index2 = start2 length = end - start1 aux = [None] * length for i in range(length): if ((index1 == start2) or ((index2 != end) and (a[index1] > a[index2]))): aux[i] = a[index2] index2 += 1 else: aux[i] = a[index1] index1 += 1 for i in range(start1, end): a[i] = aux[i - start1] def mergeSort(a): n = len(a) step = 1 while (step < n): for start1 in range(0, n, 2*step): start2 = min(start1 + step, n) end = min(start1 + 2*step, n) merge(a, start1, start2, end) step *= 2 #################################################### # builtinSort (wrapped as a function) #################################################### def builtinSort(a): a.sort() #################################################### # testSort #################################################### def testSort(sortFn, n): a = [random.randint(0,2**31) for i in range(n)] sortedA = sorted(a) startTime = time.time() sortFn(a) endTime = time.time() elapsedTime = endTime - startTime assert(a == sortedA) print(f'{sortFn.__name__:>20} n={n}, time={elapsedTime:6.3f}') def testSorts(): n = 2**8 # use 2**8 for Brython, use 2**12 or larger for Python for sortFn in [selectionSort, mergeSort, builtinSort]: testSort(sortFn, n) testSorts() - Some Sorting Links
- Sorting 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
- Sorting Algorithms
- Optional: sumOfSquares Examples
Note: this section is optional. Also: 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(f'{f.__name__}({n}) takes {timeFnCall(f, n):8.3f} milliseconds') timeFnCalls(1001) # use larger number, say 3000, to get more accurate timing
- Python Builtins
Graphically (Images borrowed from here):