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2.1 Computational Tractability
"For me, great algorithms are the poetry of computation. Just like verse, they can be terse, allusive, dense, and even mysterious. But once unlocked, they cast a brilliant new light on some aspect of computing." - Francis Sullivan
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Computational Tractability
As soon as an Analytic Engine exists, it will necessarily guide the future course of the science. Whenever any result is sought by its aid, the question will arise - By what course of calculation can these results be arrived at by the machine in the shortest time? - Charles Babbage Charles Babbage (1864) Analytic Engine (schematic)
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Polynomial-Time Brute force. For many non-trivial problems, there is a natural brute force search algorithm that checks every possible solution. Typically takes 2N time or worse for inputs of size N. Unacceptable in practice. Desirable scaling property. When the input size doubles, the algorithm should only slow down by some constant factor C. Def. An algorithm is poly-time if the above scaling property holds. n ! for stable matching with n men and n women focusing on worst-case poly-time is a subtle modeling decision - we try to justify it here There exists constants c > 0 and d > 0 such that on every input of size N, its running time is bounded by c Nd steps. choose C = 2d
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Worst-Case Analysis Worst case running time. Obtain bound on largest possible running time of algorithm on input of a given size N. Generally captures efficiency in practice. Draconian view, but hard to find effective alternative. Average case running time. Obtain bound on running time of algorithm on random input as a function of input size N. Hard (or impossible) to accurately model real instances by random distributions. Algorithm tuned for a certain distribution may perform poorly on other inputs.
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Worst-Case Polynomial-Time
Def. An algorithm is efficient if its running time is polynomial. Justification: It really works in practice! Although 6.02 1023 N20 is technically poly-time, it would be useless in practice. In practice, the poly-time algorithms that people develop almost always have low constants and low exponents. Breaking through the exponential barrier of brute force typically exposes some crucial structure of the problem. Exceptions. Some poly-time algorithms do have high constants and/or exponents, and are useless in practice. Some exponential-time (or worse) algorithms are widely used because the worst-case instances seem to be rare. simplex method Unix grep
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Why It Matters table taken from More Programming Pearls, p. 82, 400MhZ Pentium II scaled up one step takes one nanosecond msec = millisecond
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2.2 Asymptotic Order of Growth
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Complexity The complexity of an algorithm associates a number T(n), the worst-case time the algorithm takes, with each problem size n. Mathematically, T: N+ → R+ that is T is a function that maps positive integers (giving problem sizes) to positive real numbers (giving number of steps).
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Complexity Measures Problem size n
Worst-case complexity: max # steps algorithm takes on any input of size n Best-case complexity: min # steps algorithm takes on any input of size n Average-case complexity: avg # steps algorithm takes on inputs of size n Best-case : unrealistic Average-case : over what probability distribution?, analysis often hard Worst-case : a fast algorithm has a comforting guarantee maybe too pessimistic
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Asymptotic Order of Growth
Upper bounds. T(n) is O(f(n)) if there exist constants c > 0 and n0 0 such that for all n n0 we have T(n) c · f(n). Lower bounds. T(n) is (f(n)) if there exist constants c > 0 and n0 0 such that for all n n0 we have T(n) c · f(n). Tight bounds. T(n) is (f(n)) if T(n) is both O(f(n)) and (f(n)). Ex: T(n) = 32n2 + 17n + 32. T(n) is O(n2), O(n3), (n2), (n), and (n2) . T(n) is not O(n), (n3), (n), or (n3).
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Asymptotic Order of Growth in words
A way of comparing functions that ignores constant factors and small input sizes O(g(n)): class of functions f(n) that grow no faster than g(n) Θ(g(n)): class of functions f(n) that grow at same rate as g(n) Ω(g(n)): class of functions f(n) that grow at least as fast as g(n)
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Notation Slight abuse of notation. T(n) = O(f(n)). Asymmetric:
f(n) = 5n3; g(n) = 3n2 f(n) = O(n3) = g(n) but f(n) g(n). Better notation: T(n) O(f(n)). Meaningless statement. Any comparison-based sorting algorithm requires at least O(n log n) comparisons. Statement doesn't "type-check." Use for lower bounds.
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Properties Transitivity. If f = O(g) and g = O(h) then f = O(h).
If f = (g) and g = (h) then f = (h). If f = (g) and g = (h) then f = (h). Additivity. If f = O(h) and g = O(h) then f + g = O(h). If f = (h) and g = (h) then f + g = (h). If f = (h) and g = O(h) then f + g = (h).
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Asymptotic Bounds for Some Common Functions
Polynomials. a0 + a1n + … + adnd is (nd) if ad > 0. Polynomial time. Running time is O(nd) for some constant d independent of the input size n. Logarithms. O(log a n) = O(log b n) for any constants a, b > 0. Logarithms. For every x > 0, log n = O(nx). Exponentials. For every r > 1 and every d > 0, nd = O(rn). can avoid specifying the base log grows slower than every polynomial every exponential grows faster than every polynomial
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More Examples 10n2-16n+100 is O(n2) also O(n3) 10n2-16n+100 ≤ 11n2 for all n ≥ 10 10n2-16n+100 is Ω (n2) also Ω (n) 10n2-16n+100 ≥ 9n2 for all n ≥16 Therefore also 10n2-16n+100 is Θ (n2) 10n2-16n+100 is not O(n) also not Ω (n3)
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Time efficiency of nonrecursive algorithms
General Plan for Analysis - Decide on parameter n indicating input size - Identify algorithm’s basic operation - Determine worst, average, and best cases for input of size n - Set up a sum for the number of times the basic operation is executed - Simplify the sum using standard formulas and rules
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Useful summation formulas and rules
liu1 = 1+1+…+1 = u - l + 1 In particular, liu1 = n = n (n) 1in i = 1+2+…+n = n(n+1)/2 n2/2 (n2) 1in i2 = …+n2 = n(n+1)(2n+1)/6 n3/3 (n3) 0in ai = 1 + a +…+ an = (an+1 - 1)/(a - 1) for any a 1 In particular, 0in 2i = …+ 2n = 2n (2n ) (ai ± bi ) = ai ± bi cai = cai liuai = limai + m+1iuai
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Example Problem Problem: Given N integers stored in an array X (int X[N]), find the sum of the numbers How can you design an algorithm for this problem? Iterative (Non-Recursive) Solution Use a for or while loop and add the numbers one by one Recursive Solution A solution that calls itself on smaller problems to solve the big problem
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Finding the sum of a set of numbers: Non-Recursive Algorithm
Int X[N]; Sum = 0; For (int i = 0; i < N; i++) Sum = Sum + X[i];
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Finding the sum of a set of numbers: Recursive Algorithm
int sum ( int A[], int N) { if (N == 0) return 0; Stopping rule else return sum(A, N-1) + A[N-1]; -- Key Step } /* end-sum */ Why recursion? Simplifies the code drastically
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Analyzing Running Time
RT: the amount of time it takes for the algorithm to finish execution on a particular input size More precisely, the RT of an algorithm on a particular input is the number of primitive operations or steps executed. We define a step to be a unit of work that can be executed in constant amount of time in a machine.
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Finding the sum of a set of numbers: Iterative Algorithm and its analysis
Assume int X[N] of integer is our data set Cost Times Sum = 0; C For (int i = 0; i < N; i++) C N Sum = Sum + X[i]; C N T(n) = C0 + C1*N + C2*N Since C0, C1 and C2 are constants, T(n) can be expressed as a linear function n, i.e., T(n) = a + b*n, for some constants a, b
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Another Example: Searching for a number in an array of numbers
Assume int X[N] of integer is our data set and we are searching for “key” Cost Times Found = 0; C I = 0; C while (!found && i < N){ C <= L < N If (key ==X[I]) found = 1; C <= L <= N I++; C <= L <= N } T(n) = C0 + C1 + L*(C2 + C3 + C4), where 1 <= L <= N is the number of times that the loop is iterated.
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Example2: Searching for a number in an array of numbers (continued)
What’s the best case? Loop iterates just once => T(n) = C0 + C1 + C2 + C3 + C4 What’s the average (expected) case? Loop iterates N/2 times => T(n) = C0 + C1 + N/2 * (C2 + C3 + C4) Notice that this can be written as T(n) = a + b*n where a, b are constants What’s the worst case? Loop iterates N times => T(n) = C0 + C1 + N * (C2 + C3 + C4)
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Worst Case Analysis of Algorithms
We will only look at WORST CASE running time of an algorithm. Why? Worst case is an upper bound on the running time. It gives us a guarantee that the algorithm will never take any longer For some algorithms, the worst case happens fairly often. As in this search example, the searched item is typically not in the array, so the loop will iterate N times The “average case” is often roughly as bad as the “worst case”. In our search algorithm, both the average case and the worst case are linear functions of the input size “n”
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Asymptotic Notation We will study the asymptotic efficiency of algorithms To do so, we look at input sizes large enough to make only the order of growth of the running time relevant That is, we are concerned with how the running time of an algorithm increases with the size of the input in the limit as the size of the input increases without bound. Usually an algorithm that is asymptotically more efficient will be the best choice for all but very small inputs. 3 asymptotic notations Big O, Q, W Notations
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Big-Oh Notation: Asymptotic Upper Bound
T(n) = f(n) = O(g(n)) if f(n) <= c*g(n) for all n > n0, where c & n0 are constants > 0 n c*g(n) f(n) n0 Example: T(n) = 2n + 5 is O(n). Why? 2n+5 <= 3n, for all n >= 5 T(n) = 5*n2 + 3*n + 15 is O(n2). Why? 5*n2 + 3*n + 15 <= 6*n2, for all n >= 6
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W Notation: Asymptotic Lower Bound
T(n) = f(n) = W(g(n)) if f(n) >= c*g(n) for all n > n0, where c and n0 are constants > 0 n f(n) c*g(n) n0 Example: T(n) = 2n + 5 is W(n). Why? 2n+5 >= 2n, for all n > 0 T(n) = 5*n2 - 3*n is W(n2). Why? 5*n2 - 3*n >= 4*n2, for all n >= 4
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Q Notation: Asymptotic Tight Bound
T(n) = f(n) = Q(g(n)) if c1*g(n) <= f(n) <= c2*g(n) for all n > n0, where c1, c2 and n0 are constants > 0 n f(n) c1*g(n) c2*g(n) n0 Example: T(n) = 2n + 5 is Q(n). Why? 2n <= 2n+5 <= 3n, for all n >= 5 T(n) = 5*n2 - 3*n is Q(n2). Why? 4*n2 <= 5*n2 - 3*n <= 5*n2, for all n >= 4
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Big-Oh, Theta, Omega Tips to guide your intuition:
Think of O(f(N)) as “less than or equal to” f(N) Upper bound: “grows slower than or same rate as” f(N) Think of Ω(f(N)) as “greater than or equal to” f(N) Lower bound: “grows faster than or same rate as” f(N) Think of Θ(f(N)) as “equal to” f(N) “Tight” bound: same growth rate (True for large N and ignoring constant factors)
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Common Functions we will encounter
Name Big-Oh Comment Constant O(1) Can’t beat it! Log log O(loglogN) Extrapolation search Logarithmic O(logN) Typical time for good searching algorithms Linear O(N) This is about the fastest that an algorithm can run given that we need O(n) just to read the input N logN O(NlogN) Most sorting algorithms Quadratic O(N2) Acceptable when the data size is small (N<1000) Cubic O(N3) Exponential O(2N) Only good for really small input sizes (n<=20) Increasing cost Polynomial time
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Time and Space Tradeoffs
In turns out in most algorithm design, there is a tradeoff between time and space To make an algorithm faster, you might have to use more space Trade space away (use less space), then the algorithm will run slower
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2.4 A Survey of Common Running Times
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Linear Time: O(n) Linear time. Running time is at most a constant factor times the size of the input. Computing the maximum. Compute maximum of n numbers a1, …, an. max a1 for i = 2 to n { if (ai > max) max ai }
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Linear Time: O(n) Merge. Combine two sorted lists A = a1,a2,…,an with B = b1,b2,…,bn into sorted whole. Claim. Merging two lists of size n takes O(n) time. Pf. After each comparison, the length of output list increases by 1. i = 1, j = 1 while (both lists are nonempty) { if (ai bj) append ai to output list and increment i else(ai bj)append bj to output list and increment j } append remainder of nonempty list to output list
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O(n log n) Time O(n log n) time. Arises in divide-and-conquer algorithms. Sorting. Mergesort and heapsort are sorting algorithms that perform O(n log n) comparisons. Largest empty interval. Given n time-stamps x1, …, xn on which copies of a file arrive at a server, what is largest interval of time when no copies of the file arrive? O(n log n) solution. Sort the time-stamps. Scan the sorted list in order, identifying the maximum gap between successive time-stamps. also referred to as linearithmic time
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Quadratic Time: O(n2) Quadratic time. Enumerate all pairs of elements.
Closest pair of points. Given a list of n points in the plane (x1, y1), …, (xn, yn), find the pair that is closest. O(n2) solution. Try all pairs of points. Remark. (n2) seems inevitable, but this is just an illusion. min (x1 - x2)2 + (y1 - y2)2 for i = 1 to n { for j = i+1 to n { d (xi - xj)2 + (yi - yj)2 if (d < min) min d } don't need to take square roots see chapter 5
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Cubic Time: O(n3) Cubic time. Enumerate all triples of elements.
Set disjointness. Given n sets S1, …, Sn each of which is a subset of 1, 2, …, n, is there some pair of these which are disjoint? O(n3) solution. For each pairs of sets, determine if they are disjoint. foreach set Si { foreach other set Sj { foreach element p of Si { determine whether p also belongs to Sj } if (no element of Si belongs to Sj) report that Si and Sj are disjoint
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Polynomial Time: O(nk) Time
Independent set of size k. Given a graph, are there k nodes such that no two are joined by an edge? O(nk) solution. Enumerate all subsets of k nodes. Check whether S is an independent set = O(k2). Number of k element subsets = O(k2 nk / k!) = O(nk). k is a constant foreach subset S of k nodes { check whether S in an independent set if (S is an independent set) report S is an independent set } poly-time for k=17, but not practical
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Exponential Time Independent set. Given a graph, what is maximum size of an independent set? O(n2 2n) solution. Enumerate all subsets. S* foreach subset S of nodes { check whether S in an independent set if (S is largest independent set seen so far) update S* S }
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