dr. alagoz chapter 2 analysis of algorithms algorithm input output an algorithm is a step-by-step...
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Dr. Alagoz
CHAPTER 2Analysis of Algorithms
AlgorithmInput Output
An algorithm is a step-by-step procedure forsolving a problem in a finite amount of time.
Analysis of Algorithms 2
Dr. AlagozAlgorithm Analysis: Do NOT worry !!!
Analysis of Algorithms 3
Dr. Alagoz
properties of logarithms:logb(xy) = logbx + logby
logb (x/y) = logbx - logby
logbxa = alogbx
logba = logxa/logxbproperties of exponentials:a(b+c) = aba c
abc = (ab)c
ab /ac = a(b-c)
b = a logab
bc = a c*logab
Series summationsLogarithms and Exponents
Proof techniquesBasic probability
Math you need to Review
Analysis of Algorithms 4
Dr. Alagoz
Assumptions for the computational model
Basic computer with sequentially executed instructions
Infinite memory
Has standard operations; addition, multiplication, comparison in 1 time unit unless stated otherwise
Has fixed-size (32bits) integers such that no-fancy operations are allowed!!! eg.matrix inversion,
Proof techniquesBasic probability
Analysis of Algorithms 5
Dr. Alagoz
Running Time Most algorithms transform input objects into output objects.The running time of an algorithm typically grows with the input size.Average case time is often difficult to determine.We focus on the worst case running time.
Easier to analyze Crucial to applications
such as games, finance, image,robotics, AI, etc.
0
20
40
60
80
100
120
Runnin
g T
ime
1000 2000 3000 4000
Input Size
best caseaverage caseworst case
Analysis of Algorithms 6
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Experimental Studies
Write a program implementing the algorithmRun the program with inputs of varying size and compositionAssume a subprogram is written to get an accurate measure of the actual running timePlot the results 0
1000
2000
3000
4000
5000
6000
7000
8000
9000
0 50 100
Input Size
Tim
e (
ms)
Analysis of Algorithms 7
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Limitations of Experiments
It is necessary to implement the algorithm, which may be difficultResults may not be indicative of the running time on other inputs not included in the experiment. In order to compare two algorithms, the same hardware and software environments must be used
Analysis of Algorithms 8
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Theoretical AnalysisUses a high-level description of the algorithm instead of an implementationCharacterizes running time as a function of the input size, n.Takes into account all possible inputsAllows us to evaluate the speed of an algorithm independent of the hardware/software environment
Analysis of Algorithms 9
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Pseudocode
High-level description of an algorithmMore structured than English proseLess detailed than a programPreferred notation for describing algorithmsHides program design issues
Algorithm arrayMax(A, n)Input array A of n integersOutput maximum element of A
currentMax A[0]for i 1 to n 1 do
if A[i] currentMax thencurrentMax A[i]
return currentMax
Example: find max element of an array
Analysis of Algorithms 10
Dr. Alagoz
Pseudocode Details
Control flow if … then … [else …] while … do … repeat … until … for … do … Indentation replaces
braces
Method declarationAlgorithm method (arg [, arg…])
Input …Output …
Method callvar.method (arg [, arg…])
Return valuereturn expression
Expressions Assignment
(like in Java) Equality testing
(like in Java)n2 Superscripts and
other mathematical formatting allowed
Analysis of Algorithms 11
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Important Functions Seven functions that often appear in algorithm analysis:
Constant 1 Logarithmic log n Linear n N-Log-N n log n Quadratic n2
Cubic n3
Exponential 2n
In a log-log chart, the slope of the line corresponds to the growth rate of the function
1E+01E+21E+41E+61E+8
1E+101E+121E+141E+161E+181E+201E+221E+241E+261E+281E+30
1E+0 1E+2 1E+4 1E+6 1E+8 1E+10n
T(n
)
Cubic
Quadratic
Linear
Analysis of Algorithms 12
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Primitive Operations
Basic computations performed by an algorithm
Identifiable in pseudocode
Largely independent from the programming language
Exact definition not important (we will see why later)
Assumed to take a constant amount of time in the RAM model
Examples: Evaluating an
expression Assigning a
value to a variable
Indexing into an array
Calling a method Returning from a
method
Analysis of Algorithms 13
Dr. AlagozDetailed Model 4 counting operations-1
fetchto fetch an integer operant from memory
storeto store an integer result in memory
+to add two integers
-to subtract two integers
xto multiply two integers
/to divide two integers
<to comparison of two integers
The time require for the following operations are all constants.
Analysis of Algorithms 14
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returnto return from a method
callto call a method
storeto pass an integer argument to a method
[.]for the address calculation (not including the computation of the subscription)
newto allocate a fixed amount of storage from the heap using new
Detailed Model 4 counting operations-2
The time require for the following operations are all constants.
Analysis of Algorithms 15
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Statement Time requiredy = x; fetch + store
y = 1; fetch + store
y = y + 1; 2 fetch + + + store
y += 1; 2 fetch + + + store
++y; 2 fetch + + + store
y++; 2 fetch + + + store
Detailed Model 4 counting operations-3
Analysis of Algorithms 16
Dr. AlagozDetailed Model... Example1:Sum
...public static int sum(int n) {
int result = 0;for (int i = 1; i <= n; ++i){
result += i;}return result;
}...
Statement Time requiredint result = 0; fetch + store
int i = 1 fetch + store
i <= n (2fetch + < ) (n+1)++i (2fetch + + + store)
nresult += i (2fetch + + + store)
nreturn result fetch + return
n
i
i1
Analysis of Algorithms 17
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Detailed Model... Example2
y = a [i]
3fetch + [.] + store
operations fetch a (the base address of the
array) fetch i (the index into the array) address calculation fetch array element a[i] store the result
Analysis of Algorithms 18
Dr. AlagozDetailed Model...Example3: Horner (*)
public class Horner {
public static void main(String[] args) {Horner h = new Horner();int[] a = { 1, 3, 5 };System.out.println("a(1)=" + h.horner(a, a.length - 1,
1));System.out.println("a(2)=" + h.horner(a, a.length - 1,
2));}
int horner(int[] a, int n, int x) {int result = a[n];for (int i = n - 1; i >= 0; --i) {
result = result * x + a[i];/**/System.out.println("i=" + i + " result" +
result);}return result;
}}
i=1 result8i=0 result9a(1)=9i=1 result13i=0 result27a(2)=27
outputoutput
in
ii xa
0
Analysis of Algorithms 19
Dr. Alagoz
Detailed Model...Example-4 (*)public class FindMaximum {
public static void main(String[] args) {FindMaximum h = new FindMaximum();int[] a = { 1, 3, 5 };System.out.println("max=" + h.findMaximum(a));
}
int findMaximum(int[] a) {int result = a[0];for (int i = 0; i < a.length; ++i) {
if (result < a[i]) {result = a[i];
}System.out.println("i=" + i + " result=" + result);
}return result;
}}
i=0 result=1i=1 result=3i=2 result=5max=5
outputoutput
3fetch + [.] + store
fetch + store(2fetch + <) n(2fetch + + + store) (n-1)
(4fetch + [.] + <) (n-1)
(3fetch + [.] + store) ?
fetch + store
Analysis of Algorithms 20
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Simplified Model … More Simplification
All timing parameters are expressed in units of clock cycles. In effect, T=1. The proportionality constant, k, for all timing parameters is assumed to be the same: k=1.
Analysis of Algorithms 21
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Simplified Model …Example_FindMax public class FindMaximum {
public static void main(String[] args) {FindMaximum h = new FindMaximum();int[] a = { 1, 3, 5 };System.out.println("max=" + h.findMaximum(a));
}
int findMaximum(int[] a) {int result = a[0];for (int i = 0; i < a.length; ++i) {
if (result < a[i]) {result = a[i];
}System.out.println("i=" + i + " result=" + result);
}return result;
}}
i=0 result=1i=1 result=3i=2 result=5max=5
outout
detailed2 3fetch + [.] + store
3a fetch + store3b (2fetch + <) n3c (2fetch + + + store) (n-1)
4 (4fetch + [.] + <) (n-1)
6 (3fetch + [.] + store) ?
9 fetch + store
simple2 5
3a 2
3b (3)n
3c (4)(n-1)
4 (6)(n-1)
6 (5)?
9 2
12345678910
Analysis of Algorithms 22
Dr. AlagozSimplified Model > Algorithm1…Geometric Series Sum
public class GeometrikSeriesSum {
public static void main(String[] args) {System.out.println("1, 4: " + geometricSeriesSum(1, 4));System.out.println("2, 4: " + geometricSeriesSum(2, 4));
}
public static int geometricSeriesSum(int x, int n) {int sum = 0;for (int i = 0; i <= n; ++i) {
int prod = 1;for (int j = 0; j < i; ++j) {
prod *= x;}sum += prod;
}return sum;
}} 1, 4: 5
2, 4: 31
outputoutput
simple12 23a 23b 3(n+2)3c 4(n+1)4 2(n+1)5a 2(n+1)5b 3 (i+1)5c 4 i6 4 i78 4(n+1)910 21112
Total 11/2 n2 + 47/2 n +27
123456789101112
n
i 0
n
i 0
n
i 0
Analysis of Algorithms 23
Dr. AlagozSimplified Model > Algorithm_SumHornerGeometric Series Sum …public class GeometricSeriesSumHorner {
public static void main(String[] args) {System.out.println("1, 4: " + geometricSeriesSum(1, 4));System.out.println("2, 4: " + geometricSeriesSum(2, 4));
}
public static int geometricSeriesSum(int x, int n) {int sum = 0;for (int i = 0; i <= n; ++i) {
sum = sum * x + 1;}return sum;
}}
1, 4: 52, 4: 31
outputoutput
2 23a 23b 3(n+2)3c 4(n+1)4 6(n+1)6 2
Total 13 n + 22
1234567
Analysis of Algorithms 24
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Simplified Model > Algorithm_SumPower Geometric Series Sum …
public class GeometrikSeriesSumPower {
public static void main(String[] args) { System.out.println("1, 4: " + powerA(1, 4)); System.out.println("1, 4: " + powerB(1, 4)); System.out.println("2, 4: " + powerA(2, 4)); System.out.println("2, 4: " + powerB(2, 4));
}
...public static int powerA(int x, int n) {
int result = 1;for (int i = 1; i <= n; ++i) {
result *= x;}return result;
}
public static int powerB(int x, int n) {if (n == 0) {
return 1;} else if (n % 2 == 0) { // n is even
return powerB(x * x, n / 2);} else { // n is odd
return x * powerB(x * x, n / 2);}
}}
1, 4: 11, 4: 12, 4: 162, 4: 16
outputoutput
powerB 0<n 0<n n=0 n even n odd
10 3 3 311 2 - -12 - 5 513 - 10+T(n/2) -15 - - 12+T(n/2)
Total 5 18+T(n/2) 20+T(n/2)
1234567891011121314151617
powerB 1 n=0
xn = (x2)n/2 0<n, n is even x(x2)n/2 0<n, n is odd
powerB 5 n=0
xn = 18+T(n/2) 0<n, n is even 20+T(n/2) 0<n, n is odd
Analysis of Algorithms 25
Dr. AlagozSimplified Model > Calculation of PowerB …Geometric Series Sum … (*)
Let n = 2k for some k>0.Since n is even, n/2 = n/2 = 2k-1.
For n = 2k, T(2k) = 18 + T(2k-1).
Using repeated substitutionT(2k) = 18 + T(2k-1)
= 18 + 18 + T(2k-2) = 18 + 18 + 18 + T(2k-3) …= 18j + T(2k-j)
Substitution stops when k = jT(2k) = 18k + T(1)
= 18k + 20 + T(0)= 18k + 20 + 5= 18k + 25.
n = 2k then k = log2 n
T(2k) = 18 log2 n + 25
powerBxn = 18+T(n/2) 0<n
n is even
Analysis of Algorithms 26
Dr. Alagoz
Suppose n = 2k–1 for some k>0.Since n is odd, n/2 = (2k–1)/2 = (2k–2)/2= 2k-1
For n = 2k-1, T(2k-1) = 20 + T(2k-1-1), k>1.
Using repeated substitutionT(2k-1) = 20 + T(2k-1-1)
= 20 + 20 + T(2k-2-1) = 20 + 20 + 20 + T(2k-3-1) …= 20j + T(2k-j-1)
Substitution stops when k = jT(2k-1) = 20k + T(20-1)
= 20k + T(0)= 20k + 5.
n = 2k-1 then k = log2 (n+1)
T(n) = 20 log2 (n+1) + 5
Therefore, powerB 5 n=0
xn = 18+T(n/2) 0<n, n is even 20+T(n/2) 0<n, n is odd
Average: 19(log2(n+1) + 1) + 18
Simplified Model > Calculation of PowerB Geometric Series Sum … (*)
n is odd
Analysis of Algorithms 27
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public class GeometrikSeriesSumPower {
public static void main(String[] args) {System.out.println(“s 2, 4: " + geometrikSeriesSumPower
(2, 4));}
...public static int geometrikSeriesSumPower (int x, int n) {
return powerB(x, n + 1) – 1 / (x - 1);}
public static int powerB(int x, int n) {if (n == 0) {
return 1;} else if (n % 2 == 0) { // n is even
return powerB(x * x, n / 2);} else { // n is odd
return x * powerB(x * x, n / 2);}
}}
s 2, 4: 31
outputoutput
Simplified Model > Algorithm_SumPower Geometric Series Sum …
Analysis of Algorithms 28
Dr. AlagozComparison of 3 Algorithms
algorithm T(n) Sum 11/2 n2 + 47/2 n + 27Horner 13n + 22Power 19(log2(n+1) + 1) + 18
sum
Horner
Power
Analysis of Algorithms 29
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Counting Primitive Operations for
Pseudocodes By inspecting the pseudocode, we can determine the maximum number of primitive operations executed by an algorithm, as a function of the input size
Algorithm arrayMax(A, n)
# operations
currentMax A[0] 2for i 1 to n 1 do 2n
if A[i] currentMax then 2(n 1)currentMax A[i] 2(n 1)
{ increment counter i } 2(n 1)return currentMax 1
Total 8n 2
Analysis of Algorithms 30
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Estimating Running Time
Algorithm arrayMax executes 8n 2 primitive operations in the worst case. Define:a = Time taken by the fastest primitive operationb = Time taken by the slowest primitive
operation
Let T(n) be worst-case time of arrayMax. Then
a (8n 2) T(n) b(8n 2)
Hence, the running time T(n) is bounded by two linear functions
Analysis of Algorithms 31
Dr. AlagozAsymptotic Algorithm Analysis
The asymptotic analysis of an algorithm determines the running time in big-Oh notationTo perform the asymptotic analysis
We find the worst-case number of primitive operations executed as a function of the input size
We express this function with big-Oh notation
Example: We determine that algorithm arrayMax executes at
most 8n 2 primitive operations We say that algorithm arrayMax “runs in O(n) time”
Since constant factors and lower-order terms are eventually dropped anyhow, we can disregard them when counting primitive operations
Analysis of Algorithms 32
Dr. AlagozGrowth Rate of Running Time
Changing the hardware/ software environment Affects T(n) by a constant factor, but Does not alter the growth rate of T(n)
The linear growth rate of the running time T(n) is an intrinsic property of algorithm arrayMax
Analysis of Algorithms 33
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Constant Factors
The growth rate is not affected by
constant factors or
lower-order terms
Examples 102n 105 is a
linear function 105n2 108n is a
quadratic function 1E+01E+21E+41E+61E+8
1E+101E+121E+141E+161E+181E+201E+221E+241E+26
1E+0 1E+2 1E+4 1E+6 1E+8 1E+10n
T(n
)
Quadratic
Quadratic
Linear
Linear
Analysis of Algorithms 34
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Big-Oh Notation
Given functions f(n) and g(n), we say that f(n) is O(g(n)) if there are positive constantsc and n0 such that
f(n) cg(n) for n n0
Example: 2n 10 is O(n) 2n 10 cn (c 2) n 10 n 10(c 2) Pick c 3 and n0 10
1
10
100
1.000
10.000
1 10 100 1.000n
3n
2n+10
n
Analysis of Algorithms 35
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Big-Oh Example
Example: the function n2 is not O(n)
n2 cn n c The above
inequality cannot be satisfied since c must be a constant
1
10
100
1.000
10.000
100.000
1.000.000
1 10 100 1.000n
n 2̂
100n
10n
n
Analysis of Algorithms 36
Dr. Alagoz
More Big-Oh Examples
7n-2
7n-2 is O(n)need c > 0 and n0 1 such that 7n-2 c•n for n n0
this is true for c = 7 and n0 = 1
3n3 + 20n2 + 53n3 + 20n2 + 5 is O(n3)need c > 0 and n0 1 such that 3n3 + 20n2 + 5 c•n3 for n
n0
this is true for c = 4 and n0 = 21 3 log n + 53 log n + 5 is O(log n)need c > 0 and n0 1 such that 3 log n + 5 c•log n for n
n0
this is true for c = 8 and n0 = 2
Analysis of Algorithms 37
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Big-Oh and Growth RateThe big-Oh notation gives an upper bound on the growth rate of a functionThe statement “f(n) is O(g(n))” means that the growth rate of f(n) is no more than the growth rate of g(n)
We can use the big-Oh notation to rank functions according to their growth rate
f(n) is O(g(n)) g(n) is O(f(n))
g(n) grows more
Yes No
f(n) grows more No Yes
Same growth Yes Yes
Analysis of Algorithms 38
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Big-Oh Rules
Use the smallest possible class of functions Say “2n is O(n)” instead of “2n is O(n2)”
Use the simplest expression of the class Say “3n 5 is O(n)” instead of “3n 5 is
O(3n)”
Theorem
Consider polynomial where am>0.
Then f(n) = O(nm). i.e.,1. Drop lower-order terms2. Drop constant factors
m
i
iinanf
0
)(
Analysis of Algorithms 39
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Big-Oh Rules-2 logk n = O(n) for any constant kZ+
f(n) = O (f(n))
c O(f(n)) = O (f(n))
O(f(n)) + O(f(n)) = O(f(n))
O(O(f(n))) = O(f(n))
O(f(n)) O(g(n)) = O(f(n) g(n))
Analysis of Algorithms 40
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Relatives of Big-Oh
big-Omega f(n) is (g(n)) if there is a constant c > 0
and an integer constant n0 1 such that
f(n) c•g(n) for n n0
big-Theta f(n) is (g(n)) if there are constants c’ > 0
and c’’ > 0 and an integer constant n0 1 such that c’•g(n) f(n) c’’•g(n) for n n0
Analysis of Algorithms 41
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Intuition for Asymptotic Notation
Big-Oh f(n) is O(g(n)) if f(n) is
asymptotically less than or equal to g(n)
big-Omega f(n) is (g(n)) if f(n) is
asymptotically greater than or equal to g(n)
big-Theta f(n) is (g(n)) if f(n) is
asymptotically equal to g(n)
Analysis of Algorithms 42
Dr. AlagozExample Uses of the Relatives of Big-Oh
f(n) is (g(n)) if it is (n2) and O(n2). We have already seen the former, for the latter recall that f(n) is O(g(n)) if there is a constant c > 0 and an integer constant n0 1 such that f(n) < c•g(n) for n n0
Let c = 5 and n0 = 1
5n2 is (n2)
f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0 1 such that f(n) c•g(n) for n n0
let c = 1 and n0 = 1
5n2 is (n)
f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0 1 such that f(n) c•g(n) for n n0
let c = 5 and n0 = 1
5n2 is (n2)
Analysis of Algorithms 43
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Comparison of OrdersO(1) < O(log n) < O(n) < O(n log n) < O(n2) < O(n3) <
O(2n)
Analysis of Algorithms 44
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O(log n)
Comparison of OrdersO(1) < O(log n) < O(n) < O(n log n) < O(n2) < O(n3) <
O(2n)
O(1)
O(2n)
O(n3)
O(n2)
O(n log n)O(n)
Analysis of Algorithms 45
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O(n) Analysis of Running Time
Example Worst-case running time
if statement 5 executes all the time…
Best-case running time if statement 5 never executes…
On-the-average running time if statement 5 executes half of the time????
int findMaximum(int[] a) {int result = a[0];for (int i = 1; i < a.length; ++i) {
if (result < a[i]) {result = a[i];
}}return result;
}
123456789
Analysis of Algorithms 46
Dr. AlagozAlgorithm Complexity as a function of size n
Problem Input of size n Basic operation
Searching a list
Sorting a list
Multiplying two matrices
Prime factorization
Evaluating a polynomial
Traversing a tree
Towers of Hanoi
lists with n elements
lists with n elements
two n-by-n matrices
n-digit number
polynomial of degrees n
tree with n nodes
n disks
comparison
comparison
multiplication
division
multiplication
accessing a node
moving a disk
A comparison-based algorithm for searching or sorting a list is
based on
•making comparisons involving list elements
•then making decisions based on these comparisons.
• SOON U ll know all about these!!!
Analysis of Algorithms 47
Dr. Alagoz
HWLA
Solve Exercises in Chapter2 : 1, 7, 9 ,12, 25, 29, 31
Analysis of Algorithms 48
Dr. AlagozComputing Prefix Averages(*)
We further illustrate asymptotic analysis with two algorithms for prefix averagesThe i-th prefix average of an array X is average of the first (i 1) elements of X:
A[i] X[0] X[1] … X[i])/(i+1)
Computing the array A of prefix averages of another array X has applications to financial analysis such as ARMA, ARIMA, FARIMA, models.
0
5
10
15
20
25
30
35
1 2 3 4 5 6 7
X
A
Analysis of Algorithms 49
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Prefix Averages (Quadratic) (*)
The following algorithm computes prefix averages in quadratic time by applying the definition
Algorithm prefixAverages1(X, n)Input array X of n integersOutput array A of prefix averages of X #operations A new array of n integers nfor i 0 to n 1 do n
s X[0] nfor j 1 to i do 1 2 … (n 1)
s s X[j] 1 2 … (n 1)A[i] s (i 1) n
return A 1
Analysis of Algorithms 50
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Arithmetic Progression (*)
The running time of prefixAverages1 isO(1 2 …n)The sum of the first n integers is n(n 1) 2
There is a simple visual proof of this fact
Thus, algorithm prefixAverages1 runs in O(n2) time 0
1
2
3
4
5
6
7
1 2 3 4 5 6
Analysis of Algorithms 51
Dr. AlagozPrefix Averages (Linear) (*)
The following algorithm computes prefix averages in linear time by keeping a running sum
Algorithm prefixAverages2(X, n)Input array X of n integersOutput array A of prefix averages of X #operationsA new array of n integers ns 0 1for i 0 to n 1 do n
s s X[i] nA[i] s (i 1) n
return A 1Algorithm prefixAverages2 runs in O(n) time
Analysis of Algorithms 52
Dr. AlagozThe Random Access Machine (RAM) Model (*)
A CPU
A potentially unbounded bank of memory cells, each of which can hold an arbitrary number or character
01
2
Memory cells are numbered and accessing any cell in memory takes unit time.
Analysis of Algorithms 53
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Recurrence Relations
Analysis of Algorithms 54
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