Transcript Data Structure

Data Structure
Chapter 1 – Basic Concepts
1.5 Algorithm Specification
• Definition
– An algorithm is a finite set of instructions that,
if followed, accomplishes a particular task and
satisfies:
• Input: 0 or more quantities are externally supplied.
• Output: At least one quantity is produced.
• Definiteness:
– Each instruction must be clear and unambiguous.
• Finiteness:
– An algorithm terminates in a finite number of steps.
• Effectiveness:
– Every instruction must be basic enough to be carried out.
Example 1.2 Selection Sort
• Goal:
– To sort a collection of n≧1 integers in nonincreasing order.
• Idea:
– From those integers that are currently
unsorted, find the smallest and place it next in
the sorted list.
Selection Sort Algorithm
• An algorithm is often written partially in C++
and partially in English.
• Program 1.7:
Assume integers are initially stored in an array a, and the length of a is n.
SelectionSort (int a[], int n)
for (int i=0; i<n; i++) {
examine a[i] to a[n-1] and suppose the smallest integer is at a[j];
interchange a[i] and a[j];
}
C++ Implementation of Selection Sort
1.7.1 Performance Analysis
• Priori performance evaluation.
– Space complexity
• The required amount of memory to run a program
– Time complexity
• The required amount of computer time to run a
program.
• Instance characteristic
– The number/size/scale of data to be dealt with
in a problem.
1.7.1.1 Space Complexity
• Let S(P) be the space requirement of any
program P.
– S(P) = c + Sp(n)
• c is constant; n denotes instance characteristic.
– When analyzing S(P), we concentrate solely
on estimating Sp(n).
• Program 1.16
float Abc(float a, float b, float c)
{
return a+b+b*c+(a+b-c)/(a+b)+4.0;
}
• The fixed part:
– The space to store a, b, and c, and the return
value.
• Sp(n) = 0.
• Program 1.17
float Sum(float *a, const int n)
{
float s = 0;
for (int i=0; i<n; i++)
s += a[i];
return s;
}
– The instance characteristic is n.
– Since a is actually the address of the first
element of a[], and n is passed by value, the
space needed by Sum() is constant
(Ssum(n)=0).
• Program 1.18
float RSum(float *a, const int n)
{
if (n <= 0) return 0;
else return (RSum(a, n-1) + a[n-1]);
}
n=997
n=998
n=999
n=1000
– Each call requires at least 4 words
• The values of n, a, return value and return address.
– The depth of the recursion is n+1.
• The stack space needed is 4(n+1).
1.7.1.2 Time Complexity
• Let T(P) be the time requirement of any
program T (compile time+run time).
– We shall concern ourselves with run time
(tp(n)), where n denote instance characteristic.
 Count a particular operation
 Count program steps
 Asymptotic complexity
 Complexity in terms of O, Ω, and Θ.
1.7.1.3 Asymptotic Notation
• Comparing the time complexity of two programs,
it is too difficult to determine exact step count.
– a step could compound several steps.
– Consider 2n+2 and 1000n+3
• When n is very large, the factors of coefficients and constants
become less important.
– Consider 1000n+3 and 2n2+4
• As n increases, 2n2+4 grows relatively faster than 1000n+3.
• We shall concern ourselves with the relationship
of program to the instance characteristic n.
2n+2
1000n+3
2n2+4
Definition [Big “oh”]
• f(n) = O(g(n)) iff there exist positive
constants c and n0 such that f(n)≦cg(n) for
all n where n≧n0.
– Example 1:
• Prove 3n+2=O(n)
Suppose f(n) = 3n+2, g(n) = n.
∵Let c=4, n0=2, and then for all n≧2, 3n+2≦4n.
∴3n+2=O(n)
Definition [Big “oh”]
– Example 2:
• Prove 1000n2+10n-6=O(n2)
Suppose f(n) = 1000n2+10n-6, g(n) = n2.
∵Let c=2000, n0=1, and then for all n≧n0=1, 1000n2+10n-6 ≦ n2.
∴ 1000n2+10n-6=O(n2)
– In other words, O(．) denotes an upper bound
for f(n).
Note
• The coefficient of g(n) is usually 1.
• g(n) should be as the smallest function for
f(n) = O(g(n)).
• It is usually written as “3n+2=O(n)” rather
than “O(n) = 3n+2”.
Theorem 1.2
If f (n)  am n m  ...  a1n  a0 , then f (n)  O(n m ).
P roof:
m
m
f (n)   ai n   ai n i
i
i 0
i 0
m
 n m  ai n i  m
i 0
m
 n m  ai , for n  1.
i 0
m
T herefore,let c   ai , n0  1, we have
i 0
f (n)  cnm , for n  n0 . T hes, f (n)  O(n m ).
Definition [Omega]
• f(n) = Ω(g(n)) iff there exist positive
constants c and n0 such that f(n)≧cg(n) for
all n where n≧ n0.
– Example 1:
• 3n+2=Ω(n)
Suppose f(n) = 3n+2, g(n) = n.
∵Let c=4, n0=1, and then we have 3n+2≧cn=2n for all n≧n0=1.
∴3n+2=Ω(n).
Definition [Omega]
• Ω(．) denotes an lower bound for f(n).
– Like O(．), g(n) has to be the largest function and
usually has coefficient 1.
• Theorem 1.3
If f (n)  amnm  ... a1n  a0 and am  0, then f (n)  (nm ).
Comparison
• Constant time: O(1)
• Polynomial time:
– Linear: O(n)
– Quadratic: O(n2)
– Cubic: O(n3)
• Exponential time: O(2n)
O(1)<O(logn)<O(n)<O(nlogn)<O(n2)<O(n3)<O(2n)<O(n!)
Definition [Theta]
• f(n) = Θ(g(n)) iff there exist positive constants c1, c2 and
n0 such that c1g(n)≦ f(n)≦c2g(n) for all n where n≧ n0.
– Example 1:
• 3n+2=Θ(n)
∵Let c1=3, c2=4, n0=2, and then we have 3n+2≦3n and
3n+2≧4n all n≧n0=2.
• ∴3n+2=Θ(n).
6  2n  n2  (2n )
Definition [Theta]
• Θ(g(n)) means f(n) will be bounded around g(n).
– g(n) is an upper and lower bound on f(n).
• Theorem 1.4
If f (n)  amnm  ... a1n  a0 and am  0, then f (n)  (nm ).
Asymptotic Analysis
• Asymptotic notations are used to evaluate
computation time of an algorithm.
– O and Ω correspond to computation time in
worst- and best case for an algorithm.
– Usually, we consider computation time in
worst case (O(．)) rather than that in average
case.
• O(．) provides an upper bound for the entire
execution.
• Average time is hard to define.
A Simple Example
• Program 1.17
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float Sum(float *a, const int n)
{
float s = 0;
for (int i=0; i<n; i++)
s += a[i];
return s;
}
– Line 1: O(1).
– Line 2-3: Line 3 executes in O(1) and repeat n times. Therefore,
totally O(n) time.
– Line 4: O(1).
Overall, the computation time of Program 1.17 is
O(1) + O(n) + O(1) = O(n)
An Recursive Example
float RSum(float *a, const int n)
{
1
if (n <= 0) return 0;
2
else return (RSum(a, n-1) + a[n-1]);
}
Let TRSum(n) denote the computation time to execute RSum(n).
Therefore,
Line 1: O(1).
Line 2: TRSum(n-1) + O(1).
n
Therefore,
TRSum(n) = O(1) + TRSum(n-1)
= O(1) + O(1) + TRSum(n-2) = O(1) + O(1) + … + TRSum(0)
= (n+1)O(1) = O(n)
Example 1.3: Binary Search
• Assume there are n≧1 distinct integers
that are sorted and stored in the array a[0],
a[1], …, a[n-1].
• Objective: to determine if the integer x is in
the array.
– If so, return j if x = a[j];
– Otherwise, return -1.
Idea
• Let left and right denote the left and right ends of
the list to be searched.
– Initially, left=0, right=n-1.
• Let middle = (left+right) / 2.
• Three cases:
– x < a[middle]: search x in the left half. right = middle -1.
– x == a[middle]: x is found, return middle.
– x > a[middle]: search x in the right half. left = middle + 1.
• if left > right, it means x is not found.
Iterative Binary Search
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int BinarySearch(int *a, const int x, const in n)
{
Initialize left and right;
while (left <= right)
{
middle = (left + right) / 2;
if (x < a[middle])
right = middle - 1;
else if (x > a[middle]) left = middle + 1;
else Return middle;
}
Return -1;
}
Performance Analysis of Iterative
Binary Search
• Choose n as the instance characteristic.
• Space Complexity:
– The memory space is used to store the values
of x, n, return value and the start address of
a[], which is independent of n. Therefore,
space complexity is O(1).
Performance Analysis of Iterative
Binary Search
• Time complexity:
– Line 1 and 9: O(1).
– Line 2 to 8:
• a while-loop whose body (Line 4 to 7) executes
totally in O(1) time.
• The loop ends when the length of the list to be
searched is equal to 0 and the length is decreased
about one-half in each iteration. Therefore, there
are log2 n iterations.
• Totally, O(1)xO(log2n)= O(log2n)
– Overall, the time complexity is O(log2n).
Recursive Binary Search
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int BinarySearch(int *a, const int x, const in left, const in right)
{
If (left <= right) {
middle = (left + right) / 2;
if (x < a[middle]) Return BinarySearch(a, x, left, middle – 1);
if (x > a[middle]) Return BinarySearch(a, x, middle+1, right);
Return middle;
}
Return -1;
}
Performance Analysis of Recursive
Binary Search
• Choose n as the length of the list (rightleft+1) to be the instance characteristic.
• Space Complexity:
– O(1) memory space is used when the function
is invoked each time. In worst cast (x is not
found), the size of the stack space is O(log2n).
Therefore, the space complexity is O(log2n).
Performance Analysis of Iterative
Binary Search
• Time complexity:
Let T (n) denote the computation time to execute BinarySearch(). Therefore,
log2 n
T (n) = O(1) + T (n/2)
= O(1) + O(1) + T (n/4) = O(1) + O(1) + … + T (0)
= O(log2n)O(1) = O(log2n)