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Coursenotes
CS3114: Data Structures andAlgorithms*
Clifford A. ShafferDepartment of Computer Science
Virginia TechCopyright © 2008
*Temporarily listed as CS2984, this course replaces CS2606.
Goals of this Course
1. Reinforce the concept that costs and benefits exist for every data structure.
2. Learn the commonly used data structures.– These form a programmer's basic data
structure ``toolkit.'‘
3. Understand how to measure the cost of a data structure or program.– These techniques also allow you to judge the
merits of new data structures that you or others might invent.
The Need for Data Structures
Data structures organize data
more efficient programs.
More powerful computers
more complex applications.
More complex applications demand more calculations.
Complex computing tasks are unlike our everyday experience.
Organizing Data
Any organization for a collection of records can be searched, processed in any order, or modified.
The choice of data structure and algorithm can make the difference between a program running in a few seconds or many days.
Efficiency
A solution is said to be efficient if it solves the problem within its resource constraints.– Space– Time
• The cost of a solution is the amount of resources that the solution consumes.
Selecting a Data Structure
Select a data structure as follows:
1. Analyze the problem to determine the basic operations that must be supported.
2. Quantify the resource constraints for each operation.
3. Select the data structure that best meets these requirements.
Some Questions to Ask
• Are all data inserted into the data structure at the beginning, or are insertions interspersed with other operations?
• Can data be deleted?
• Are all data processed in some well-defined order, or is random access allowed?
Costs and Benefits
Each data structure has costs and benefits.
Rarely is one data structure better than another in all situations.
Any data structure requires:– space for each data item it stores,– time to perform each basic operation,– programming effort.
Costs and Benefits (cont)
Each problem has constraints on available space and time.
Only after a careful analysis of problem characteristics can we know the best data structure for the task.
Bank example:– Start account: a few minutes– Transactions: a few seconds– Close account: overnight
Example 1.2
Problem: Create a database containing information about cities and towns.
Tasks: Find by name or attribute or location
• Exact match, range query, spatial queryResource requirements: Times can be
from a few seconds for simple queries to a minute or two for complex queries
Scheduling
• Managing large-scale projects involves scheduling activities– It is human nature to work better toward
intermediate milestones.
• The same concepts can/should be applied to mid-sized projects encountered in class.– For any project that needs more than a week
of active work to complete, break into parts and design a schedule with milestones and deliverables.
Real Results #1
• CS2606, Fall 2006• 3-4 week projects• Kept schedule information:
– Estimated time required– Milestones, estimated times for each– Weekly estimates of time spent.
Real Results #2
Real Results #3
• Results were significant:– 90% of scores below median were students
who did less than 50% of the project prior to the last week.
– Few did poorly who put in > 50% time early– Some did well who didn’t put in >50% time
early, but most who did well put in the early time
Real Results #4
• Correlations:– Strong correlation between early time and
high score– No correlation between time spent and
score– No correlation between % early time and
total time
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What is the Mechanism?
• Correlations are not causal– Do they behave that way because they are good,
or does behaving that way make them good?
• Spreading projects over time allows the “sleep on it” heuristic to operate
• Avoiding the “zombie” effect makes people more productive (and cuts time requirements)
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Mathematical Background
Set concepts and notation
Logarithms
Recursion
Induction Proofs
Summations
Recurrence Relations
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Estimation Techniques
Known as “back of the envelope” or “back of the napkin” calculation
1. Determine the major parameters that effect the problem.
2. Derive an equation that relates the parameters to the problem.
3. Select values for the parameters, and apply the equation to yield and estimated solution.
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Estimation Example
How many library bookcases does it take to store books totaling one million pages?
Estimate:– Pages/inch– Feet/shelf– Shelves/bookcase
Abstract Data Types
Abstract Data Type (ADT): a definition for a data type solely in terms of a set of values and a set of operations on that data type.
Each ADT operation is defined by its inputs and outputs.
Encapsulation: Hide implementation details.
Data Structure
• A data structure is the physical implementation of an ADT.– Each operation associated with the ADT is
implemented by one or more subroutines in the implementation.
• Data structure usually refers to an organization for data in main memory.
• File structure: an organization for data on peripheral storage, such as a disk drive.
Metaphors
An ADT manages complexity through abstraction: metaphor.– Hierarchies of labels
Ex: transistors gates CPU.
In a program, implement an ADT, then think only about the ADT, not its implementation.
Logical vs. Physical Form
Data items have both a logical and a physical form.
Logical form: definition of the data item within an ADT.– Ex: Integers in mathematical sense: +, -
Physical form: implementation of the data item within a data structure.– Ex: 16/32 bit integers, overflow.
Data Type
ADT:TypeOperations
Data Items: Logical Form
Data Items: Physical Form
Data Structure:Storage SpaceSubroutines
Example 1.8
A typical database-style project will have many interacting parts.
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Algorithm Efficiency
There are often many approaches (algorithms) to solve a problem. How do we choose between them?
At the heart of computer program design are two (sometimes conflicting) goals.
1. To design an algorithm that is easy to understand, code, debug.
2. To design an algorithm that makes efficient use of the computer’s resources.
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Algorithm Efficiency (cont)
Goal (1) is the concern of Software Engineering.
Goal (2) is the concern of data structures and algorithm analysis.
When goal (2) is important, how do we measure an algorithm’s cost?
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How to Measure Efficiency?
1. Empirical comparison (run programs)2. Asymptotic Algorithm Analysis
Critical resources:
Factors affecting running time:
For most algorithms, running time depends on “size” of the input.
Running time is expressed as T(n) for some function T on input size n.
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Examples of Growth Rate
Example 1.
// Return position of largest value in "A"static int largest(int[] A) { int currlarge = 0; // Position of largest for (int i=1; i<A.length; i++) if (A[currlarge] < A[i]) currlarge = i; // Remember pos return currlarge; // Return largest pos}
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Examples (cont)
Example 2: Assignment statement.
Example 3:
sum = 0;for (i=1; i<=n; i++) for (j=1; j<n; j++) sum++;}
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Growth Rate Graph
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Best, Worst, Average Cases
Not all inputs of a given size take the same time to run.
Sequential search for K in an array of n integers:
• Begin at first element in array and look at each element in turn until K is found
Best case:
Worst case:
Average case:
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Which Analysis to Use?
While average time appears to be the fairest measure, it may be difficult to determine.
When is the worst case time important?
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Faster Computer or Algorithm?Suppose we buy a computer 10 times faster.n: size of input that can be processed in one second
on old computer (in 1000 computational units)n’: size of input that can be processed in one second
on new computer (in 10,000 computational units)
T(n) n n’ Change n’/n
10n 100 1,000 n’ = 10n 10
10n2 10 31.6 n’= 10n 3.16
10n 3 4 n’ = n + 1 1 + 1/n
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Asymptotic Analysis: Big-oh
Definition: For T(n) a non-negatively valued function, T(n) is in the set O(f(n)) if there exist two positive constants c and n0 such that T(n) <= cf(n) for all n > n0.
Use: The algorithm is in O(n2) in [best, average, worst] case.
Meaning: For all data sets big enough (i.e., n>n0), the algorithm always executes in less than cf(n) steps in [best, average, worst] case.
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Big-oh Notation (cont)
Big-oh notation indicates an upper bound.
Example: If T(n) = 3n2 then T(n) is in O(n2).
Look for the tightest upper bound:
While T(n) = 3n2 is in O(n3), we prefer O(n2).
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Big-Oh Examples
Example 1: Finding value X in an array (average cost).
Then T(n) = csn/2.
For all values of n > 1, csn/2 <= csn.
Therefore, the definition is satisfied for f(n)=n, n0 = 1, and c = cs.
Hence, T(n) is in O(n).
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Big-Oh Examples
Example 2: Suppose T(n) = c1n2 + c2n, where c1 and c2 are positive.
c1n2 + c2n <= c1n2 + c2n2 <= (c1 + c2)n2 for all n > 1.
Then T(n) <= cn2 whenever n > n0, for c = c1 + c2 and n0 = 1.
Therefore, T(n) is in O(n2) by definition.
Example 3: T(n) = c. Then T(n) is in O(1).
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A Common Misunderstanding
“The best case for my algorithm is n=1 because that is the fastest.” WRONG!
Big-oh refers to a growth rate as n grows to .
Best case is defined for the input of size n that is cheapest among all inputs of size n.
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Big-Omega
Definition: For T(n) a non-negatively valued function, T(n) is in the set (g(n)) if there exist two positive constants c and n0 such that T(n) >= cg(n) for all n > n0.
Meaning: For all data sets big enough (i.e., n > n0), the algorithm always requires more than cg(n) steps.
Lower bound.
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Big-Omega Example
T(n) = c1n2 + c2n.
c1n2 + c2n >= c1n2 for all n > 1.
T(n) >= cn2 for c = c1 and n0 = 1.
Therefore, T(n) is in (n2) by the definition.
We want the greatest lower bound.
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Theta Notation
When big-Oh and coincide, we indicate this by using (big-Theta) notation.
Definition: An algorithm is said to be in (h(n)) if it is in O(h(n)) and it is in (h(n)).
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A Common Misunderstanding
Confusing worst case with upper bound.
Upper bound refers to a growth rate.
Worst case refers to the worst input from among the choices for possible inputs of a given size.
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Simplifying Rules
1. If f(n) is in O(g(n)) and g(n) is in O(h(n)), then f(n) is in O(h(n)).
2. If f(n) is in O(kg(n)) for some constant k > 0, then f(n) is in O(g(n)).
3. If f1(n) is in O(g1(n)) and f2(n) is in O(g2(n)), then (f1 + f2)(n) is in O(max(g1(n), g2(n))).
4. If f1(n) is in O(g1(n)) and f2(n) is in O(g2(n)) then f1(n)f2(n) is in O(g1(n)g2(n)).
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Time Complexity Examples (1)
Example 3.9: a = b;
This assignment takes constant time, so it is (1).
Example 3.10:sum = 0;for (i=1; i<=n; i++) sum += n;
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Time Complexity Examples (2)
Example 3.11:sum = 0;for (j=1; j<=n; j++) for (i=1; i<=j; i++) sum++;for (k=0; k<n; k++) A[k] = k;
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Time Complexity Examples (3)
Example 3.12:sum1 = 0;for (i=1; i<=n; i++) for (j=1; j<=n; j++) sum1++;
sum2 = 0;for (i=1; i<=n; i++) for (j=1; j<=i; j++) sum2++;
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Time Complexity Examples (4)
Example 3.13:sum1 = 0;for (k=1; k<=n; k*=2) for (j=1; j<=n; j++) sum1++;
sum2 = 0;for (k=1; k<=n; k*=2) for (j=1; j<=k; j++) sum2++;
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Binary Search
How many elements are examined in worst case?
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Binary Search// Return the position of an element in "A"// with value "K". If "K" is not in "A",// return A.length.static int binary(int[] A, int K) { int l = -1; // Set l and r int r = A.length; // beyond array bounds while (l+1 != r) { // Stop when l, r meet int i = (l+r)/2; // Check middle if (K < A[i]) r = i; // In left half if (K == A[i]) return i; // Found it if (K > A[i]) l = i; // In right half } return A.length; // Search value not in A}
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Other Control Statements
while loop: Analyze like a for loop.
if statement: Take greater complexity of then/else clauses.
switch statement: Take complexity of most expensive case.
Subroutine call: Complexity of the subroutine.
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Problems
• Problem: a task to be performed.– Best thought of as inputs and matching
outputs.– Problem definition should include constraints
on the resources that may be consumed by any acceptable solution.
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Problems (cont)
• Problems mathematical functions– A function is a matching between inputs (the
domain) and outputs (the range).– An input to a function may be single number,
or a collection of information.– The values making up an input are called the
parameters of the function.– A particular input must always result in the
same output every time the function is computed.
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Algorithms and ProgramsAlgorithm: a method or a process followed to
solve a problem.– A recipe.
An algorithm takes the input to a problem (function) and transforms it to the output.– A mapping of input to output.
A problem can have many algorithms.
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Analyzing Problems
Upper bound: Upper bound of best known algorithm.
Lower bound: Lower bound for every possible algorithm.
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Space/Time Tradeoff Principle
One can often reduce time if one is willing to sacrifice space, or vice versa.
• Encoding or packing informationBoolean flags
• Table lookupFactorials
Disk-based Space/Time Tradeoff Principle: The smaller you make the disk storage requirements, the faster your program will run.
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Analyzing Problems: Example
May or may not be able to obtain matching upper and lower bounds.
Example of imperfect knowledge: Sorting
1. Cost of I/O: (n).
2. Bubble or insertion sort: O(n2).
3. A better sort (Quicksort, Mergesort, Heapsort, etc.): O(n log n).
4. We prove later that sorting is in (n log n).
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Multiple Parameters
Compute the rank ordering for all C pixel values in a picture of P pixels.
for (i=0; i<C; i++) // Initialize count count[i] = 0;for (i=0; i<P; i++) // Look at all pixels count[value(i)]++; // Increment countsort(count); // Sort pixel counts
If we use P as the measure, then time is (P log P).
More accurate is (P + C log C).
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Space Complexity
Space complexity can also be analyzed with asymptotic complexity analysis.
Time: AlgorithmSpace: Data Structure
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Lists
A list is a finite, ordered sequence of data items.
Important concept: List elements have a position.
Notation: <a0, a1, …, an-1>
What operations should we implement?
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List Implementation Concepts
Our list implementation will support the concept of a current position.
Operations will act relative to the current position.
<20, 23 | 12, 15>
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List ADTpublic interface List<E> { public void clear(); public void insert(E item); public void append(E item); public E remove(); public void moveToStart(); public void moveToEnd(); public void prev(); public void next(); public int length(); public int currPos(); public void moveToPos(int pos); public E getValue();}
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List ADT Examples
List: <12 | 32, 15>
L.insert(99);
Result: <12 | 99, 32, 15>
Iterate through the whole list:
for (L.moveToStart(); L.currPos()<L.length(); L.next()) { it = L.getValue(); doSomething(it);}
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List Find Function
/** @return True if "K" is in list "L", false otherwise */public static boolean find(List<Integer> L, int K) {
int it; for (L.moveToStart();
L.currPos()<L.length(); L.next()) { it = L.getValue(); if (K == it) return true; // Found K } return false; // K not found}
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Array-Based List Insert
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Array-Based List Class (1)class AList<E> implements List<E> { private static final int defaultSize = 10;
private int maxSize; private int listSize; private int curr; private E[] listArray;
// Constructors AList() { this(defaultSize); } @SuppressWarnings("unchecked") AList(int size) { maxSize = size; listSize = curr = 0; listArray = (E[])new Object[size]; }
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Array-Based List Class (2)
public void clear() { listSize = curr = 0; }public void moveToStart() { curr = 0; }public void moveToEnd() { curr = listSize; }public void prev() { if (curr != 0) curr--; }public void next() { if (curr < listSize) curr++; }public int length() { return listSize; }public int currPos() { return curr; }
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Array-Based List Class (3)
public void moveToPos(int pos) { assert (pos>=0) && (pos<=listSize) : "Position out of range";
curr = pos;}
public E getValue() { assert (curr >= 0) && (curr < listSize) : "No current element";
return listArray[curr];}
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Insert
// Insert "it" at current position */public void insert(E it) { assert listSize < maxSize : "List capacity exceeded"; for (int i=listSize; i>curr; i--) listArray[i] = listArray[i-1]; listArray[curr] = it; listSize++;}
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Appendpublic void append(E it) { // Append "it" assert listSize < maxSize : "List capacity exceeded"; listArray[listSize++] = it;}
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Remove
// Remove and return the current element.public E remove() { assert (curr >= 0) && (curr < listSize) :
"No current element"; E it = listArray[curr]; for(int i=curr; i<listSize-1; i++) listArray[i] = listArray[i+1]; listSize--; return it;}
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Link ClassDynamic allocation of new list elements.class Link<E> { private E element; private Link<E> next;
// Constructors Link(E it, Link<E> nextval) { element = it; next = nextval; } Link(Link<E> nextval) { next = nextval; }
Link<E> next() { return next; } Link<E> setNext(Link<E> nextval) { return next = nextval; } E element() { return element; } E setElement(E it) { return element = it; }}
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Linked List Position (1)
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Linked List Position (2)
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Linked List Class (1)
class LList<E> implements List<E> { private Link<E> head; private Link<E> tail; protected Link<E> curr; int cnt;
//Constructors LList(int size) { this(); } LList() { curr = tail = head = new Link<E>(null); cnt = 0; }
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Linked List Class (2)public void clear() { head.setNext(null); curr = tail = head = new Link<E>(null); cnt = 0;}public void moveToStart() { curr = head; }public void moveToEnd() { curr = tail; }public int length() { return cnt; }public void next() { if (curr != tail) { curr = curr.next(); }}public E getValue() { assert curr.next() != null : "Nothing to get"; return curr.next().element();}
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Insertion
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Insert/Append// Insert "it" at current positionpublic void insert(E it) { curr.setNext(new Link<E>(it, curr.next()));
if (tail == curr) tail = curr.next(); cnt++;}
public void append(E it) { tail = tail.setNext(new Link<E>(it, null)); cnt++;}
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Removal
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Remove/** Remove and return current element */public E remove() { if (curr.next() == null) return null; E it = curr.next().element(); if (tail == curr.next()) tail = curr; curr.setNext(curr.next().next()); cnt--; return it; }
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Prev/** Move curr one step left; no change if already at front */public void prev() { if (curr == head) return; Link<E> temp = head; // March down list until we find the // previous element while (temp.next() != curr) temp = temp.next(); curr = temp;}
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Get/Set Position/** Return position of the current element */public int currPos() { Link<E> temp = head; int i; for (i=0; curr != temp; i++) temp = temp.next(); return i;}
/** Move down list to "pos" position */public void moveToPos(int pos) { assert (pos>=0) && (pos<=cnt) : "Position out of range"; curr = head; for(int i=0; i<pos; i++) curr = curr.next();}
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Comparison of Implementations
Array-Based Lists:• Insertion and deletion are (n).• Prev and direct access are (1).• Array must be allocated in advance.• No overhead if all array positions are full.
Linked Lists:• Insertion and deletion are (1).• Prev and direct access are (n).• Space grows with number of elements.• Every element requires overhead.
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Space Comparison
“Break-even” point:
DE = n(P + E);
n = DE P + E
E: Space for data value.P: Space for pointer.D: Number of elements in array.
Space Example
• Array-based list: Overhead is one pointer (4 bytes) per position in array – whether used or not.
• Linked list: Overhead is two pointers per link node– one to the element, one to the next link
• Data is the same for both.• When is the space the same?
– When the array is half full
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Freelists
System new and garbage collection are slow.
• Add freelist support to the Link class.
Link Class Extensionsstatic Link freelist = null;
static <E> Link<E> get(E it, Link<E> nextval) { if (freelist == null) return new Link<E>(it, nextval); Link<E> temp = freelist; freelist = freelist.next(); temp.setElement(it); temp.setNext(nextval); return temp;}
void release() { // Return to freelist element = null; next = freelist; freelist = this;}
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Using Freelistpublic void insert(E it) { curr.setNext(Link.get(it, curr.next())); if (tail == curr) tail = curr.next(); cnt++;}
public E remove() { if (curr.next() == null) return null; E it = curr.next().element(); if (tail == curr.next()) tail = curr; Link<E> tempptr = curr.next(); curr.setNext(curr.next().next()); tempptr.release(); cnt--; return it; }
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Doubly Linked Listsclass DLink<E> { private E element; private DLink<E> next; private DLink<E> prev;
DLink(E it, DLink<E> n, DLink<E> p) { element = it; next = n; prev = p; } DLink(DLink<E> n, DLink<E> p) { next = n; prev = p; }
DLink<E> next() { return next; } DLink<E> setNext(DLink<E> nextval) { return next = nextval; } DLink<E> prev() { return prev; } DLink<E> setPrev(DLink<E> prevval) { return prev = prevval; } E element() { return element; } E setElement(E it) { return element = it; }}
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Doubly Linked Lists
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Doubly Linked Insert
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Doubly Linked Insert
public void insert(E it) { curr.setNext(new DLink<E>(it, curr.next(), curr));
if (curr.next().next() != null) curr.next().next().setPrev(curr.next());
if (tail == curr) tail = curr.next(); cnt++;}
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Doubly Linked Remove
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Doubly Linked Remove
public E remove() { if (curr.next() == null) return null; E it = curr.next().element(); if (curr.next().next() != null) curr.next().next().setPrev(curr); else tail = curr; curr.setNext(curr.next().next()); cnt--; return it; }
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Stacks
LIFO: Last In, First Out.
Restricted form of list: Insert and remove only at front of list.
Notation:• Insert: PUSH• Remove: POP• The accessible element is called TOP.
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Stack ADTpublic interface Stack<E> { /** Reinitialize the stack. */ public void clear();
/** Push an element onto the top of the stack. @param it Element being pushed onto the stack.*/ public void push(E it);
/** Remove and return top element. @return The element at the top of the stack.*/ public E pop();
/** @return A copy of the top element. */ public E topValue();
/** @return Number of elements in the stack. */ public int length();};
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Array-Based Stack
// Array-based stack implementationprivate int maxSize; // Max size of stackprivate int top; // Index for topprivate E [] listArray;
Issues:• Which end is the top?• Where does “top” point to?• What are the costs of the operations?
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Linked Stackclass LStack<E> implements Stack<E> { private Link<E> top; private int size;
What are the costs of the operations?
How do space requirements compare to the array-based stack implementation?
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Queues
FIFO: First in, First Out
Restricted form of list: Insert at one end, remove from the other.
Notation:• Insert: Enqueue• Delete: Dequeue• First element: Front• Last element: Rear
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Queue Implementation (1)
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Queue Implementation (2)
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Dictionary
Often want to insert records, delete records, search for records.
Required concepts:• Search key: Describe what we are looking
for• Key comparison
– Equality: sequential search– Relative order: sorting
Records and Keys
• Problem: How do we extract the key from a record?
• Records can have multiple keys.
• Fundamentally, the key is not a property of the record, but of the context.
• Solution: We will explicitly store the key with the record.
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Dictionary ADTpublic interface Dictionary<K, E> {
public void clear(); public void insert(K k, E e); public E remove(K k); // Null if none public E removeAny(); // Null if none public E find(K k); // Null if none public int size();};
Payroll Class// Simple payroll entry: ID, name, addressclass Payroll { private Integer ID; private String name; private String address;
Payroll(int inID, String inname, String inaddr) { ID = inID; name = inname; address = inaddr; }
public Integer getID() { return ID; } public String getname() { return name; } public String getaddr() { return address; }}
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Using Dictionary// IDdict organizes Payroll records by IDDictionary<Integer, Payroll> IDdict = new UALdictionary<Integer, Payroll>();
// namedict organizes Payroll records by nameDictionary<String, Payroll> namedict = new UALdictionary<String, Payroll>();
Payroll foo1 = new Payroll(5, "Joe", "Anytown");Payroll foo2 = new Payroll(10, "John", "Mytown");IDdict.insert(foo1.getID(), foo1);IDdict.insert(foo2.getID(), foo2);namedict.insert(foo1.getname(), foo1);namedict.insert(foo2.getname(), foo2);Payroll findfoo1 = IDdict.find(5);Payroll findfoo2 = namedict.find("John");
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Unsorted List Dictionaryclass UALdictionary<K, E> implements Dictionary<K, E> {
private static final int defaultSize = 10; private AList<KVpair<K,E>> list;
// Constructors UALdictionary() { this(defaultSize); } UALdictionary(int sz) { list = new AList<KVpair<K, E>>(sz); }
public void clear() { list.clear(); }
/** Insert an element: append to list */ public void insert(K k, E e) { KVpair<K,E> temp = new KVpair<K,E>(k, e); list.append(temp); }
Sorted vs. Unsorted List Dictionaries
• If list were sorted– Could use binary search to speed search– Would need to insert in order, slowing
insert
• Which is better?– If lots of searches, sorted list is good– If inserts are as likely as searches, then
sorting is no benefit.
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Binary Trees
A binary tree is made up of a finite set of nodes that is either empty or consists of a node called the root together with two binary trees, called the left and right subtrees, which are disjoint from each other and from the root.
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Binary Tree Example
Notation: Node, children, edge, parent, ancestor, descendant, path, depth, height, level, leaf node, internal node, subtree.
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Full and Complete Binary Trees
Full binary tree: Each node is either a leaf or internal node with exactly two non-empty children.
Complete binary tree: If the height of the tree is d, then all leaves except possibly level d are completely full. The bottom level has all nodes to the left side.
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Full Binary Tree Theorem (1)
Theorem: The number of leaves in a non-empty full binary tree is one more than the number of internal nodes.
Proof (by Mathematical Induction):
Base case: A full binary tree with 1 internal node must have two leaf nodes.
Induction Hypothesis: Assume any full binary tree T containing n-1 internal nodes has n leaves.
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Full Binary Tree Theorem (2)
Induction Step: Given tree T with n internal nodes, pick internal node I with two leaf children. Remove I’s children, call resulting tree T’.
By induction hypothesis, T’ is a full binary tree with n leaves.
Restore I’s two children. The number of internal nodes has now gone up by 1 to reach n. The number of leaves has also gone up by 1.
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Full Binary Tree Corollary
Theorem: The number of null pointers in a non-empty tree is one more than the number of nodes in the tree.
Proof: Replace all null pointers with a pointer to an empty leaf node. This is a full binary tree.
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Binary Tree Node Class/** ADT for binary tree nodes */public interface BinNode<E> { /** Return and set the element value */ public E element(); public E setElement(E v);
/** Return the left child */ public BinNode<E> left();
/** Return the right child */ public BinNode<E> right();
/** Return true if this is a leaf node */ public boolean isLeaf();}
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Traversals (1)
Any process for visiting the nodes in some order is called a traversal.
Any traversal that lists every node in the tree exactly once is called an enumeration of the tree’s nodes.
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Traversals (2)
• Preorder traversal: Visit each node before visiting its children.
• Postorder traversal: Visit each node after visiting its children.
• Inorder traversal: Visit the left subtree, then the node, then the right subtree.
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Traversals (3)/** @param rt The root of the subtree */void preorder(BinNode rt){ if (rt == null) return; // Empty subtree visit(rt); preorder(rt.left()); preorder(rt.right());}
void preorder2(BinNode rt) // Not so good{ visit(rt); if (rt.left() != null) preorder(rt.left()); if (rt.right() != null) preorder(rt.right());}
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Recursion Examplesint count(BinNode rt) { if (rt == null) return 0; return 1 + count(rt.left()) + count(rt.right());}
boolean checkBST(BSTNode<Integer,Integer> root, Integer low, Integer high) { if (root == null) return true; Integer rootkey = root.key(); if ((rootkey < low) || (rootkey > high)) return false; // Out of range if (!checkBST(root.left(), low, rootkey)) return false; // Left side failed return checkBST(root.right(), rootkey, high);}
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Binary Tree Implementation (1)
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Binary Tree Implementation (2)
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Inheritance (1)public interface VarBinNode { public boolean isLeaf();}
class VarLeafNode implements VarBinNode { private String operand; public VarLeafNode(String val) { operand = val; } public boolean isLeaf() { return true; } public String value() { return operand; }};
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Inheritance (2)
/** Internal node */class VarIntlNode implements VarBinNode { private VarBinNode left; private VarBinNode right; private Character operator;
public VarIntlNode(Character op, VarBinNode l, VarBinNode r) { operator = op; left = l; right = r; } public boolean isLeaf() { return false; } public VarBinNode leftchild() { return left; } public VarBinNode rightchild(){ return right; } public Character value() { return operator; }}
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Inheritance (3)
/** Preorder traversal */public static void traverse(VarBinNode rt) { if (rt == null) return; if (rt.isLeaf()) Visit.VisitLeafNode(((VarLeafNode)rt).value()); else { Visit.VisitInternalNode( ((VarIntlNode)rt).value()); traverse(((VarIntlNode)rt).leftchild()); traverse(((VarIntlNode)rt).rightchild()); }}
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Composition (1)
public interface VarBinNode { public boolean isLeaf(); public void traverse();}
class VarLeafNode implements VarBinNode { private String operand;
public VarLeafNode(String val) { operand = val; } public boolean isLeaf() { return true; } public String value() { return operand; }
public void traverse() { Visit.VisitLeafNode(operand); }}
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Composition (2)
class VarIntlNode implements VarBinNode { private VarBinNode left; private VarBinNode right; private Character operator;
public VarIntlNode(Character op, VarBinNode l, VarBinNode r) { operator = op; left = l; right = r; } public boolean isLeaf() { return false; } public VarBinNode leftchild() { return left; } public VarBinNode rightchild() { return right; } public Character value() { return operator; }
public void traverse() { Visit.VisitInternalNode(operator); if (left != null) left.traverse(); if (right != null) right.traverse(); }}
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Composition (3)
/** Preorder traversal */public static void traverse(VarBinNode rt) { if (rt != null) rt.traverse();}
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Space Overhead (1)
From the Full Binary Tree Theorem:• Half of the pointers are null.
If leaves store only data, then overhead depends on whether the tree is full.
Ex: Full tree, all nodes the same, with two pointers to children and one to element:
• Total space required is (3p + d)n• Overhead: 3pn• If p = d, this means 3p/(3p + d) = 3/4 overhead.
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Space Overhead (2)
Eliminate pointers from the leaf nodes:
n/2(2p) + np 2p n/2(2p) + np +dn 2p + d
This is 2/3 if p = d.
(2p +p)/(2p + d + p) if data only at leaves 3/4 overhead.
Note that some method is needed to distinguish leaves from internal nodes.
=
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Array Implementation (1)
Position 0 1 2 3 4 5 6 7 8 9 10 11
Parent -- 0 0 1 1 2 2 3 3 4 4 5
Left Child 1 3 5 7 9 11 -- -- -- -- -- --
Right Child 2 4 6 8 10 -- -- -- -- -- -- --
Left Sibling -- -- 1 -- 3 -- 5 -- 7 -- 9 --
Right Sibling -- 2 -- 4 -- 6 -- 8 -- 10 -- --
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Array Implementation (1)
Parent (r) =
Leftchild(r) =
Rightchild(r) =
Leftsibling(r) =
Rightsibling(r) =
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Binary Search Trees
BST Property: All elements stored in the left subtree of a node with value K have values < K. All elements stored in the right subtree of a node with value K have values >= K.
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BSTNode (1)class BSTNode<K,E> implements BinNode<E> { private K key; private E element; private BSTNode<K,E> left; private BSTNode<K,E> right;
public BSTNode() {left = right = null; } public BSTNode(K k, E val) { left = right = null; key = k; element = val; } public BSTNode(K k, E val, BSTNode<K,E> l, BSTNode<K,E> r) { left = l; right = r; key = k; element = val; }
public K key() { return key; } public K setKey(K k) { return key = k; }
public E element() { return element; } public E setElement(E v) { return element = v; }
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BSTNode (2) public BSTNode<K,E> left() { return left; } public BSTNode<K,E> setLeft(BSTNode<K,E> p) { return left = p; }
public BSTNode<K,E> right() { return right; } public BSTNode<K,E> setRight(BSTNode<K,E> p) { return right = p; }
public boolean isLeaf() { return (left == null) && (right == null); }}
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BST (1)/** BST implementation for Dictionary ADT */class BST<K extends Comparable<? super K>, E> implements Dictionary<K, E> { private BSTNode<K,E> root; // Root of BST int nodecount; // Size of BST
/** Constructor */ BST() { root = null; nodecount = 0; }
/** Reinitialize tree */ public void clear() { root = null; nodecount = 0; }
/** Insert a record into the tree. @param k Key value of the record. @param e The record to insert. */ public void insert(K k, E e) { root = inserthelp(root, k, e); nodecount++; }
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BST (2) /** Remove a record from the tree. @param k Key value of record to remove. @return Record removed, or null if there is none. */ public E remove(K k) { E temp = findhelp(root, k); // find it if (temp != null) { root = removehelp(root, k); // remove it nodecount--; } return temp; }
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BST (3) /** Remove/return root node from dictionary. @return The record removed, null if empty. */ public E removeAny() { if (root != null) { E temp = root.element(); root = removehelp(root, root.key()); nodecount--; return temp; } else return null; }
/** @return Record with key k, null if none. @param k The key value to find. */ public E find(K k) { return findhelp(root, k); }
/** @return Number of records in dictionary. */ public int size() { return nodecount; }}
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BST Searchprivate E findhelp(BSTNode<K,E> rt, K k) { if (rt == null) return null; if (rt.key().compareTo(k) > 0) return findhelp(rt.left(), k); else if (rt.key().compareTo(k) == 0) return rt.element(); else return findhelp(rt.right(), k);}
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BST Insert (1)
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BST Insert (2)private BSTNode<K,E> inserthelp(BSTNode<K,E> rt, K k, E e) { if (rt == null) return new BSTNode<K,E>(k, e); if (rt.key().compareTo(k) > 0) rt.setLeft(inserthelp(rt.left(), k, e)); else rt.setRight(inserthelp(rt.right(), k, e)); return rt;}
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Get/Remove Minimum Valueprivate BSTNode<K,E> getmin(BSTNode<K,E> rt) { if (rt.left() == null) return rt; else return getmin(rt.left());}
private BSTNode<K,E> deletemin(BSTNode<K,E> rt) { if (rt.left() == null) return rt.right(); else { rt.setLeft(deletemin(rt.left())); return rt; }}
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BST Remove (1)
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BST Remove (2)/** Remove a node with key value k @return The tree with the node removed */private BSTNode<K,E> removehelp(BSTNode<K,E> rt, K k) { if (rt == null) return null; if (rt.key().compareTo(k) > 0) rt.setLeft(removehelp(rt.left(), k)); else if (rt.key().compareTo(k) < 0) rt.setRight(removehelp(rt.right(), k));
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BST Remove (3) else { // Found it, remove it if (rt.left() == null) return rt.right(); else if (rt.right() == null) return rt.left(); else { // Two children BSTNode<K,E> temp = getmin(rt.right()); rt.setElement(temp.element()); rt.setKey(temp.key()); rt.setRight(deletemin(rt.right())); } } return rt;}
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Time Complexity of BST Operations
Find: O(d)
Insert: O(d)
Delete: O(d)
d = depth of the tree
d is O(log n) if tree is balanced. What is the worst case?
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Priority Queues (1)
Problem: We want a data structure that stores records as they come (insert), but on request, releases the record with the greatest value (removemax)
Example: Scheduling jobs in a multi-tasking operating system.
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Priority Queues (2)
Possible Solutions:- insert appends to an array or a linked list ( O(1) )
and then removemax determines the maximum by scanning the list ( O(n) )
- A linked list is used and is in decreasing order; insert places an element in its correct position ( O(n) ) and removemax simply removes the head of the list( O(1) ).
- Use a heap – both insert and removemax areO( log n ) operations
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Heaps
Heap: Complete binary tree with the heap property:
• Min-heap: All values less than child values.• Max-heap: All values greater than child values.
The values are partially ordered.
Heap representation: Normally the array-based complete binary tree representation.
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Max Heap Example
88 85 83 72 73 42 57 6 48 60
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Max Heap Implementation (1)public class MaxHeap<K extends Comparable<? super K>, E> { private KVpair<K,E>[] Heap; // Pointer to heap array private int size; // Maximum size of heap private int n; // # of things in heap
public MaxHeap(KVpair<K,E>[] h, int num, int max){ Heap = h; n = num; size = max; buildheap(); }
public int heapsize() { return n; }
public boolean isLeaf(int pos) // Is pos a leaf position?{ return (pos >= n/2) && (pos < n); }
public int leftchild(int pos) { // Leftchild position assert pos < n/2 : "Position has no left child"; return 2*pos + 1;}
public int rightchild(int pos) { // Rightchild position assert pos < (n-1)/2 : "Position has no right child"; return 2*pos + 2;}public int parent(int pos) { assert pos > 0 : "Position has no parent"; return (pos-1)/2;}
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Sift Downpublic void buildheap() // Heapify contents { for (int i=n/2-1; i>=0; i--) siftdown(i); }
private void siftdown(int pos) { assert (pos >= 0) && (pos < n) : "Illegal heap position"; while (!isLeaf(pos)) { int j = leftchild(pos); if ((j<(n-1)) && (Heap[j].key().compareTo(Heap[j+1].key()) < 0)) j++; // index of child w/ greater value if (Heap[pos].key().compareTo(Heap[j].key()) >= 0) return; DSutil.swap(Heap, pos, j); pos = j; // Move down }}
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RemoveMax, Insert
public KVpair<K,E> removemax() { assert n > 0 : "Removing from empty heap"; DSutil.swap(Heap, 0, --n); if (n != 0) siftdown(0); return Heap[n];}
public void insert(KVpair<K,E> val) { assert n < size : "Heap is full"; int curr = n++; Heap[curr] = val; // Siftup until curr parent's key > curr key while ((curr != 0) && (Heap[curr].key(). compareTo(Heap[parent(curr)].key()) > 0)) { DSutil.swap(Heap, curr, parent(curr)); curr = parent(curr); }}
Heap Building Analysis
• Insert into the heap one value at a time: – Push each new value down the tree from the root
to where it belongs– log i = (n log n)
• Starting with full array, work from bottom up– Since nodes below form a heap, just need to push
current node down (at worst, go to bottom)– Most nodes are at the bottom, so not far to go– (i-1) n/2i = (n)
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Huffman Coding Trees
ASCII codes: 8 bits per character.• Fixed-length coding.
Can take advantage of relative frequency of letters to save space.
• Variable-length coding
Build the tree with minimum external path weight.
Z K M C U D L E
2 7 24 32 37 42 42 120
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Huffman Tree Construction (1)
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Huffman Tree Construction (2)
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Assigning Codes
Letter Freq Code Bits
C 32
D 42
E 120
M 24
K 7
L 42
U 37
Z 2
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Coding and Decoding
A set of codes is said to meet the prefix property if no code in the set is the prefix of another.
Code for DEED:
Decode 1011001110111101:
Expected cost per letter:
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General Trees
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General Tree Nodeinterface GTNode<E> { public E value(); public boolean isLeaf(); public GTNode<E> parent(); public GTNode<E> leftmostChild(); public GTNode<E> rightSibling(); public void setValue(E value); public void setParent(GTNode<E> par); public void insertFirst(GTNode<E> n); public void insertNext(GTNode<E> n); public void removeFirst(); public void removeNext();}
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General Tree Traversal/** Preorder traversal for general trees */static <E> void preorder(GTNode<E> rt) { PrintNode(rt); if (!rt.isLeaf()) { GTNode<E> temp = rt.leftmostChild(); while (temp != null) { preorder(temp); temp = temp.rightSibling(); } }}
Parent Pointer Implementation
Equivalence Class Problem
The parent pointer representation is good for answering:
– Are two elements in the same tree?
/** Determine if nodes in different trees */public boolean differ(int a, int b) { Integer root1 = FIND(array[a]); Integer root2 = FIND(array[b]); return root1 != root2;}
Union/Find/** Merge two subtrees */public void UNION(int a, int b) { Integer root1 = FIND(a); // Find a’s root Integer root2 = FIND(b); // Find b’s root if (root1 != root2) array[root2] = root1;}
public Integer FIND(Integer curr) { if (array[curr] == null) return curr; while (array[curr] != null) curr = array[curr]; return curr;}
Want to keep the depth small.
Weighted union rule: Join the tree with fewer nodes to the tree with more nodes.
Equiv Class Processing (1)
Equiv Class Processing (2)
Path Compressionpublic Integer FIND(Integer curr) { if (array[curr] == null) return curr; array[curr] = FIND(array[curr]); return array[curr];}
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Lists of Children
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Leftmost Child/Right Sibling (1)
170
Leftmost Child/Right Sibling (2)
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Linked Implementations (1)
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Linked Implementations (2)
Efficient Linked Implementation
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Sequential Implementations (1)
List node values in the order they would be visited by a preorder traversal.
Saves space, but allows only sequential access.
Need to retain tree structure for reconstruction.
Example: For binary trees, us a symbol to mark null links.
AB/D//CEG///FH//I//
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Sequential Implementations (2)
Example: For general trees, mark the end of each subtree.
RAC)D)E))BF)))
Sorting
Each record contains a field called the key.– Linear order: comparison.
Measures of cost:– Comparisons– Swaps
Insertion Sort (1)
Insertion Sort (2)static <E extends Comparable<? super E>>void Sort(E[] A) { for (int i=1; i<A.length; i) for (int j=i; (j>0) && (A[j].compareTo(A[j-1])<0); j--) DSutil.swap(A, j, j-1);}
Best Case:Worst Case:Average Case:
Bubble Sort (1)
Bubble Sort (2)static <E extends Comparable<? super E>> void Sort(E[] A) { for (int i=0; i<A.length-1; i++) for (int j=A.length-1; j>i; j--) if ((A[j].compareTo(A[j-1]) < 0)) DSutil.swap(A, j, j-1);}
Best Case:Worst Case:Average Case:
Selection Sort (1)
Selection Sort (2)static <E extends Comparable<? super E>>void Sort(E[] A) { for (int i=0; i<A.length-1; i++) { int lowindex = i; for (int j=A.length-1; j>i; j--) if (A[j].compareTo(A[lowindex]) < 0) lowindex = j; DSutil.swap(A, i, lowindex); }}
Best Case:Worst Case:Average Case:
Pointer Swapping
Summary
Insertion Bubble Selection
Comparisons:
Best Case (n) (n2) (n2)Average Case (n2) (n2) (n2)Worst Case (n2) (n2) (n2)
SwapsBest Case 0 0 (n)Average Case (n2) (n2) (n)Worst Case (n2) (n2) (n)
Exchange Sorting
All of the sorts so far rely on exchanges of adjacent records.
What is the average number of exchanges required?– There are n! permutations– Consider permuation X and its reverse, X’– Together, every pair requires n(n-1)/2
exchanges.
Shellsort
Shellsortstatic <E extends Comparable<? super E>> void Sort(E[] A) { for (int i=A.length/2; i>2; i/=2) for (int j=0; j<i; j++) inssort2(A, j, i); inssort2(A, 0, 1);}
/** Modified version of Insertion Sort for varying increments */static <E extends Comparable<? super E>>void inssort2(E[] A, int start, int incr) { for (int i=start+incr; i<A.length; i+=incr) for (int j=i;(j>=incr)&& (A[j].compareTo(A[j-incr])<0); j-=incr) DSutil.swap(A, j, j-incr);}
Quicksortstatic <E extends Comparable<? super E>>void qsort(E[] A, int i, int j) { int pivotindex = findpivot(A, i, j); DSutil.swap(A, pivotindex, j); // k will be first position in right subarray int k = partition(A, i-1, j, A[j]); DSutil.swap(A, k, j); if ((k-i) > 1) qsort(A, i, k-1); if ((j-k) > 1) qsort(A, k+1, j);}
static <E extends Comparable<? super E>>int findpivot(E[] A, int i, int j) { return (i+j)/2; }
Quicksort Partitionstatic <E extends Comparable<? super E>>int partition(E[] A, int l, int r, E pivot) { do { // Move bounds inward until they meet while (A[++l].compareTo(pivot)<0); while ((r!=0) && (A[--r].compareTo(pivot)>0)); DSutil.swap(A, l, r); } while (l < r); DSutil.swap(A, l, r); return l;}
The cost for partition is (n).
Partition Example
Quicksort Example
Cost of Quicksort
Best case: Always partition in half.Worst case: Bad partition.Average case:
T(n) = n + 1 + 1/(n-1) (T(k) + T(n-k))
Optimizations for Quicksort:– Better Pivot– Better algorithm for small sublists– Eliminate recursion
k=1
n-1
MergesortList mergesort(List inlist) { if (inlist.length() <= 1)return inlist; List l1 = half of the items from inlist; List l2 = other half of items from inlist;
return merge(mergesort(l1), mergesort(l2));}
Mergesort Implementationstatic <E extends Comparable<? super E>>void mergesort(E[] A, E[] temp, int l, int r) { int mid = (l+r)/2; if (l == r) return; mergesort(A, temp, l, mid); mergesort(A, temp, mid+1, r); for (int i=l; i<=r; i++) // Copy subarray temp[i] = A[i]; // Do the merge operation back to A int i1 = l; int i2 = mid + 1; for (int curr=l; curr<=r; curr++) { if (i1 == mid+1) // Left sublist exhausted A[curr] = temp[i2++]; else if (i2 > r) // Right sublist exhausted A[curr] = temp[i1++]; else if (temp[i1].compareTo(temp[i2])<0) A[curr] = temp[i1++]; else A[curr] = temp[i2++]; }}
Optimized Mergesortvoid mergesort(E[] A, E[] temp, int l, int r) { int i, j, k, mid = (l+r)/2; if (l == r) return; // List has one element if ((mid-l) >= THRESHOLD) mergesort(A, temp, l, mid); else inssort(A, l, mid-l+1); if ((r-mid) > THRESHOLD) mergesort(A, temp, mid+1, r); else inssort(A, mid+1, r-mid); // Do merge. First, copy 2 halves to temp. for (i=l; i<=mid; i++) temp[i] = A[i]; for (j=1; j<=r-mid; j++) temp[r-j+1] = A[j+mid]; // Merge sublists back to array for (i=l,j=r,k=l; k<=r; k++) if (temp[i].compareTo(temp[j])<0) A[k] = temp[i++]; else A[k] = temp[j--];}
Mergesort Cost
Mergesort cost:
Mergsort is also good for sorting linked lists.
Mergesort requires twice the space.
Heapsortstatic <E extends Comparable<? super E>>void heapsort(E[] A) { // Heapsort MaxHeap<E> H = new MaxHeap<E>(A, A.length, A.length); for (int i=0; i<A.length; i++) // Now sort H.removemax(); // Put max at end of heap}
Use a max-heap, so that elements end up sorted within the array.
Cost of heapsort:
Cost of finding K largest elements:
Heapsort Example (1)
Heapsort Example (2)
Binsort (1)
A simple, efficient sort:for (i=0; i<n; i++) B[A[i]] = A[i];
Ways to generalize:– Make each bin the head of a list.– Allow more keys than records.
Binsort (2)static void binsort(Integer A[]) { List<Integer>[] B = (LList<Integer>[])new LList[MaxKey]; Integer item; for (int i=0; i<MaxKey; i++) B[i] = new LList<Integer>(); for (int i=0; i<A.length; i++) B[A[i]].append(A[i]); for (int i=0; i<MaxKey; i++) for (B[i].moveToStart(); (item = B[i].getValue()) != null; B[i].next()) output(item);}
Cost:
Radix Sort (1)
Radix Sort (2)static void radix(Integer[] A, Integer[] B, int k, int r, int[] count) { int i, j, rtok;
for (i=0, rtok=1; i<k; i++, rtok*=r) { for (j=0; j<r; j++) count[j] = 0; // Count # of recs for each bin on this pass for (j=0; j<A.length; j++) count[(A[j]/rtok)%r]++; // count[j] is index in B for last slot of j for (j=1; j<r; j++) count[j] = count[j-1] + count[j]; for (j=A.length-1; j>=0; j--) B[--count[(A[j]/rtok)%r]] = A[j]; for (j=0; j<A.length; j++) A[j] = B[j]; }}
Radix Sort Example
Radix Sort Cost
Cost: (nk + rk)
How do n, k, and r relate?
If key range is small, then this can be (n).
If there are n distinct keys, then the length of a key must be at least log n.– Thus, Radix Sort is (n log n) in general case
Empirical ComparisonSort 10 100 1K 10K 100K 1M Up Down
Insertion .00023 .007 0.66 64.98 7281.0 674420 0.04 129.05
Bubble .00035 .020 2.25 277.94 27691.0 2820680 70.64 108.69
Selection .00039 .012 0.69 72.47 7356.0 780000 69.76 69.58
Shell .00034 .008 0.14 1.99 30.2 554 0.44 0.79
Shell/O .00034 .008 0.12 1.91 29.0 530 0.36 0.64
Merge .00050 .010 0.12 1.61 19.3 219 0.83 0.79
Merge/O .00024 .007 0.10 1.31 17.2 197 0.47 0.66
Quick .00048 .008 0.11 1.37 15.7 162 0.37 0.40
Quick/O .00031 .006 0.09 1.14 13.6 143 0.32 0.36
Heap .00050 .011 0.16 2.08 26.7 391 1.57 1.56
Heap/O .00033 .007 0.11 1.61 20.8 334 1.01 1.04
Radix/4 .00838 .081 0.79 7.99 79.9 808 7.97 7.97
Radix/8 .00799 .044 0.40 3.99 40.0 404 4.00 3.99
Sorting Lower Bound
We would like to know a lower bound for all possible sorting algorithms.
Sorting is O(n log n) (average, worst cases) because we know of algorithms with this upper bound.
Sorting I/O takes (n) time.
We will now prove (n log n) lower bound for sorting.
Decision Trees
Lower Bound Proof
• There are n! permutations.• A sorting algorithm can be viewed as
determining which permutation has been input.• Each leaf node of the decision tree corresponds
to one permutation.• A tree with n nodes has (log n) levels, so the
tree with n! leaves has (log n!) = (n log n) levels.
Which node in the decision tree corresponds to the worst case?
Primary vs. Secondary Storage
Primary storage: Main memory (RAM)
Secondary Storage: Peripheral devices– Disk drives– Tape drives– Flash drives
Comparisons
RAM is usually volatile.
RAM is about 1/2 million times faster than disk.
Medium 1996 1997 2000 20004 2006 2007 2008
RAM $45.00 7.00 1.500 0.3500 0.1500 0.0742 0.0339
Disk 0.25 0.10 0.010 0.0010 0.0005 0.0004 0.0001
Flash ----- ----- ----- 0.1000 0.0900 0.0098 0.0029
Floppy 0.50 0.36 0.250 0.2500 ----- ----- -----
Tape 0.03 0.01 0.001 0.0003 ----- ----- -----
Golden Rule of File Processing
Minimize the number of disk accesses!1. Arrange information so that you get what you want
with few disk accesses.2. Arrange information to minimize future disk accesses.
An organization for data on disk is often called a file structure.
Disk-based space/time tradeoff: Compress information to save processing time by reducing disk accesses.
Disk Drives
Sectors
A sector is the basic unit of I/O.
Terms
Locality of Reference: When record is read from disk, next request is likely to come from near the same place on the disk.
Cluster: Smallest unit of file allocation, usually several sectors.
Extent: A group of physically contiguous clusters.
Internal fragmentation: Wasted space within sector if record size does not match sector size; wasted space within cluster if file size is not a multiple of cluster size.
Seek Time
Seek time: Time for I/O head to reach desired track. Largely determined by distance between I/O head and desired track.
Track-to-track time: Minimum time to move from one track to an adjacent track.
Average Access time: Average time to reach a track for random access.
Other Factors
Rotational Delay or Latency: Time for data to rotate under I/O head.
– One half of a rotation on average.– At 7200 rpm, this is 8.3/2 = 4.2ms.
Transfer time: Time for data to move under the I/O head.
– At 7200 rmp: Number of sectors read/Number of sectors per track * 8.3ms.
Disk Spec Example
16.8 GB disk on 10 platters = 1.68GB/platter13,085 tracks/platter256 sectors/track512 bytes/sectorTrack-to-track seek time: 2.2 msAverage seek time: 9.5ms4KB clusters, 32 clusters/track.5400RPM
Disk Access Cost Example (1)
Read a 1MB file divided into 2048 records of 512 bytes (1 sector) each.
Assume all records are on 8 contiguous tracks.
First track: 9.5 + (11.1)(1.5) = 26.2 ms
Remaining 7 tracks: 2.2 + (11.1)(1.5) = 18.9ms.
Total: 26.2 + 7 * 18.9 = 158.5ms
Disk Access Cost Example (2)
Read a 1MB file divided into 2048 records of 512 bytes (1 sector) each.
Assume all file clusters are randomly spread across the disk.
256 clusters. Cluster read time is 8/256 of a rotation for about 5.9ms for
both latency and read time.
256(9.5 + 5.9) is about 3942ms or nearly 4 sec.
How Much to Read?
Read time for one track:9.5 + (11.1)(1.5) = 26.2ms
Read time for one sector:9.5 + 11.1/2 + (1/256)11.1 = 15.1ms
Read time for one byte:9.5 + 11.1/2 = 15.05ms
Nearly all disk drives read/write one sector (or more) at every I/O access
– Also referred to as a page or block
Recent Drive Specs
• Samsung Spinpoint T166
• 500GB (nominal)
• 7200 RPM
• Track to track: 0.8 ms
• Average track access: 8.9 ms
• Bytes/sector 512
• 6 surfaces/heads222
Buffers
The information in a sector is stored in a buffer or cache.
If the next I/O access is to the same buffer, then no need to go to disk.
There are usually one or more input buffers and one or more output buffers.
Buffer Pools
A series of buffers used by an application to cache disk data is called a buffer pool.
Virtual memory uses a buffer pool to imitate greater RAM memory by actually storing information on disk and “swapping” between disk and RAM.
Buffer Pools
Organizing Buffer Pools
Which buffer should be replaced when new data must be read?
First-in, First-out: Use the first one on the queue.
Least Frequently Used (LFU): Count buffer accesses, reuse the least used.
Least Recently used (LRU): Keep buffers on a linked list. When buffer is accessed, bring it to front. Reuse the one at end.
Bufferpool ADT: Message Passing
/** Buffer pool: message-passing style */public interface BufferPoolADT { /** Copy "sz" bytes from "space" to position "pos" in the buffered storage */ public void insert(byte[] space, int sz, int pos); /** Copy "sz" bytes from position "pos" of the buffered storage to "space". */ public void getbytes(byte[] space, int sz, int pos);}
Bufferpool ADT: Buffer Passing
/** Buffer pool: buffer-passing style */public interface BufferPoolADT { /** Return pointer to requested block */ public byte[] getblock(int block);
/** Set the dirty bit for the buffer holding "block" */ public void dirtyblock(int block);
/** Tell the size of a buffer */ public int blocksize();}
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Design Issues
Disadvantage of message passing:• Messages are copied and passed back and
forth. Disadvantages of buffer passing:• The user is given access to system memory (the
buffer itself)• The user must explicitly tell the buffer pool when
buffer contents have been modified, so that modified data can be rewritten to disk when the buffer is flushed.
• The pointer might become stale when the bufferpool replaces the contents of a buffer.
Some Goals
• Be able to avoid reading data when the block contents will be replaced.
• Be able to support multiple users accessing a buffer, and indpendantly releasing a buffer.
• Don’t make an active buffer stale.
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Improved Interface
public interface BufferPoolADT { Buffer acquireBuffer(int block);}
public interface BufferADT { // Read the block from disk public byte[] readBlock(); // Just get pointer to space, no read public byte[] getDataPointer(); // Contents have changed public void markDirty(); // Release access to the block public void release();}
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Programmer’s View of Files
Logical view of files:– An a array of bytes.– A file pointer marks the current position.
Three fundamental operations:– Read bytes from current position (move file
pointer)– Write bytes to current position (move file
pointer)– Set file pointer to specified byte position.
Java File Functions
RandomAccessFile(String name, String mode)
close()
read(byte[] b)
write(byte[] b)
seek(long pos)
External Sorting
Problem: Sorting data sets too large to fit into main memory.
– Assume data are stored on disk drive.
To sort, portions of the data must be brought into main memory, processed, and returned to disk.
An external sort should minimize disk accesses.
Model of External Computation
Secondary memory is divided into equal-sized blocks (512, 1024, etc…)
A basic I/O operation transfers the contents of one disk block to/from main memory.
Under certain circumstances, reading blocks of a file in sequential order is more efficient. (When?)
Primary goal is to minimize I/O operations.
Assume only one disk drive is available.
Key Sorting
Often, records are large, keys are small.– Ex: Payroll entries keyed on ID number
Approach 1: Read in entire records, sort them, then write them out again.
Approach 2: Read only the key values, store with each key the location on disk of its associated record.
After keys are sorted the records can be read and rewritten in sorted order.
Simple External Mergesort (1)
Quicksort requires random access to the entire set of records.
Better: Modified Mergesort algorithm.– Process n elements in (log n) passes.
A group of sorted records is called a run.
Simple External Mergesort (2)
• Split the file into two files.• Read in a block from each file.• Take first record from each block, output them in
sorted order.• Take next record from each block, output them
to a second file in sorted order.• Repeat until finished, alternating between output
files. Read new input blocks as needed.• Repeat steps 2-5, except this time input files
have runs of two sorted records that are merged together.
• Each pass through the files provides larger runs.
Simple External Mergesort (3)
Problems with Simple Mergesort
Is each pass through input and output files sequential?
What happens if all work is done on a single disk drive?
How can we reduce the number of Mergesort passes?
In general, external sorting consists of two phases:– Break the files into initial runs– Merge the runs together into a single run.
Breaking a File into Runs
General approach:– Read as much of the file into memory as
possible.– Perform an in-memory sort.– Output this group of records as a single run.
Replacement Selection (1)
• Break available memory into an array for the heap, an input buffer, and an output buffer.
• Fill the array from disk.• Make a min-heap.• Send the smallest value (root) to the
output buffer.
Replacement Selection (2)
• If the next key in the file is greater than the last value output, then
– Replace the root with this keyelse– Replace the root with the last key in the
arrayAdd the next record in the file to a new heap
(actually, stick it at the end of the array).
RS Example
Snowplow Analogy (1)
Imagine a snowplow moving around a circular track on which snow falls at a steady rate.
At any instant, there is a certain amount of snow S on the track. Some falling snow comes in front of the plow, some behind.
During the next revolution of the plow, all of this is removed, plus 1/2 of what falls during that revolution.
Thus, the plow removes 2S amount of snow.
Snowplow Analogy (2)
Problems with Simple Merge
Simple mergesort: Place runs into two files.– Merge the first two runs to output file, then
next two runs, etc.
Repeat process until only one run remains.– How many passes for r initial runs?
Is there benefit from sequential reading?Is working memory well used?Need a way to reduce the number of
passes.
Multiway Merge (1)
With replacement selection, each initial run is several blocks long.
Assume each run is placed in separate file.
Read the first block from each file into memory and perform an r-way merge.
When a buffer becomes empty, read a block from the appropriate run file.
Each record is read only once from disk during the merge process.
Multiway Merge (2)
In practice, use only one file and seek to appropriate block.
Limits to Multiway Merge (1)
Assume working memory is b blocks in size.
How many runs can be processed at one time?
The runs are 2b blocks long (on average).
How big a file can be merged in one pass?
Limits to Multiway Merge (2)
Larger files will need more passes -- but the run size grows quickly!
This approach trades (log b) (possibly) sequential passes for a single or very few random (block) access passes.
General Principles
A good external sorting algorithm will seek to do the following:
– Make the initial runs as long as possible.– At all stages, overlap input, processing and
output as much as possible.– Use as much working memory as possible.
Applying more memory usually speeds processing.
– If possible, use additional disk drives for more overlapping of processing with I/O, and allow for more sequential file processing.
Search
Given: Distinct keys k1, k2, …, kn and collection L of n records of the form
(k1, I1), (k2, I2), …, (kn, In)where Ij is the information associated with key
kj for 1 <= j <= n.
Search Problem: For key value K, locate the record (kj, Ij) in L such that kj = K.
Searching is a systematic method for locating the record(s) with key value kj = K.
Successful vs. Unsuccessful
A successful search is one in which a record with key kj = K is found.
An unsuccessful search is one in which no record with kj = K is found (and presumably no such record exists).
Approaches to Search
1. Sequential and list methods (lists, tables, arrays).
2. Direct access by key value (hashing)
3. Tree indexing methods.
Average Cost for Sequential Search
• How many comparisons does sequential search do on average?
• We must know the probability of occurrence for each possible input.
• Must K be in L?• For analysis, ignore everything except the
position of K in L. Why?• What are the n + 1 events?
Average Cost (cont)• Let ki = I+1 be the number of comparisons
when X = L[i].
• Let kn = n be the number of comparisons when X is not in L.
• Let pi be the probability that X = L[i].
• Let pn be the probability that X is not in L[i] for any I.
n
ii
n
iii ipnppkpkn
10
100)(T
Generalizing Average Cost
What happens to the equation if we assume all pi's are equal (except p0)?
Depending on the value of p0,
(n+1)/2 < T(n) < n.
2
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1
npnippn
n
i
T
Searching Ordered Arrays• Change the model: Assume that the elements are
in ascending order.• Is linear search still optimal? Why not?• Optimization: Use linear search, but test if the
element is greater than K. Why?• Observation: If we look at L[5] and find that K is
bigger, then we rule out L[1] to L[4] as well.• More is Better: If K > L[n], then we know in one
test that K is not in L. – What is wrong here?
Jump Search
• What is the right amount to jump?
• Algorithm:– Check every k'th element (L[k], L[2k], ...).– If K is greater, then go on.– If K is less, then use linear search on the k
elements.
• This is called Jump Search.
Analysis of Jump Search
If mk <= n < (m+1)k, then the total cost is at most m + k – 1 3-way comparisons.
What should k be?
11),(
kk
nkmknT
1min
1k
k
nnk
Jump Search Analysis (cont)
Take the derivative and solve for T'(x) = 0 to find the minimum.
This is a minimum when
What is the worst case cost?
Roughly
nk
n2
Lessons
We want to balance the work done while selecting a sublist with the work done while searching a sublist.
In general, make sub-problems of equal effort.
This is an example of divide and conquer.
What if we extend this to three levels?
Interpolation Search
(Also known as Dictionary Search)
Search L at a position that is appropriate to the value K.
Repeat as necessary to recalculate p for future searches.
]1[][
]1[
LL
L
n
Kp
Quadratic Binary Search
(This is easier to analyze.)
Compute p and examine
If then sequentially probe
Until we reach a value less than or equal to K.
Similar for
][L pn
][ pnK L
...3,2,1],[ inipnL
][ pnLK
Quadratic Binary Search (cont)
We are now within positions of K.
ASSUME (for now) that this takes a constant number of comparisons.
We now have a sublist of size .
Repeat the process recursively.
What is the cost?
n
n
QBS Probe Count
QBS cost is (log log n) if the number of probes on jump search is constant.
From Cebysev’s inequality, we can show that on uniformly distributed data, the average number of probes required will be about 2.4.
Is this better than binary search? Theoretically, yes (in the average case).
Comparison
n log n log log n Diff
16 4 2 2
256 8 3 2.7
64k 16 4 4
232 32 5 6.4
n log n-1 2.4 log log n Diff
16 3 4.8 worse
256 7 7.2 same
64k 15 9.6 1.6
232 31 12 2.6
Lists Ordered by Frequency
Order lists by (expected) frequency of occurrence.– Perform sequential search
Cost to access first record: 1Cost to access second record: 2
Expected search cost:
....21 21 nn npppC
Examples(1)
(1) All records have equal frequency.
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nniCn
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Examples(2)
(2) Geometric Frequency
ni
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if2/1
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1
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Zipf Distributions
Applications:– Distribution for frequency of word usage in
natural languages.– Distribution for populations of cities, etc.
80/20 rule:– 80% of accesses are to 20% of the records.– For distributions following 80/20 rule,
n
iennn nnniiC
1
.log/H/Η/
.122.0 nCn
Self-Organizing Lists
Self-organizing lists modify the order of records within the list based on the actual pattern of record accesses.
Self-organizing lists use a heuristic for deciding how to reorder the list. These heuristics are similar to the rules for managing buffer pools.
Heuristics
1. Order by actual historical frequency of access. (Similar to LFU buffer pool replacement strategy.)
2. When a record is found, swap it with the first record on list.
3. Move-to-Front: When a record is found, move it to the front of the list.
4. Transpose: When a record is found, swap it with the record ahead of it.
Text Compression Example
Application: Text Compression.
Keep a table of words already seen, organized via Move-to-Front heuristic.
• If a word not yet seen, send the word.• Otherwise, send (current) index in the table.
The car on the left hit the car I left.The car on 3 left hit 3 5 I 5.
This is similar in spirit to Ziv-Lempel coding.
Searching in Sets
For dense sets (small range, high percentage of elements in set).
Can use logical bit operators.
Example: To find all primes that are odd numbers, compute:0011010100010100 & 0101010101010101
Document processing: Signature files
277
Indexing
Goals:– Store large files– Support multiple search keys– Support efficient insert, delete, and range
queries
278
Files and Indexing
Entry sequenced file: Order records by time of insertion.– Search with sequential search
Index file: Organized, stores pointers to actual records.– Could be organized with a tree or other data
structure.
279
Keys and Indexing
Primary Key: A unique identifier for records. May be inconvenient for search.
Secondary Key: An alternate search key, often not unique for each record. Often used for search key.
280
Linear Indexing (1)Linear index: Index file organized as a
simple sequence of key/record pointer pairs with key values are in sorted order.
Linear indexing is good for searching variable-length records.
281
Linear Indexing (2)
If the index is too large to fit in main memory, a second-level index might be used.
282
Tree Indexing (1)
Linear index is poor for insertion/deletion.
Tree index can efficiently support all desired operations:– Insert/delete– Multiple search keys (multiple indices)– Key range search
283
Tree Indexing (2)
Difficulties when storing tree index on disk:– Tree must be balanced.– Each path from root to leaf
should cover few disk pages.
284
2-3 Tree
A 2-3 Tree has the following properties:1. A node contains one or two keys2. Every internal node has either two children
(if it contains one key) or three children (if it contains two keys).
3. All leaves are at the same level in the tree, so the tree is always height balanced.
The 2-3 Tree has a search tree property analogous to the BST.
285
2-3 Tree Example
The advantage of the 2-3 Tree over the BST is that it can be updated at low cost.
286
2-3 Tree Insertion (1)
287
2-3 Tree Insertion (2)
288
2-3 Tree Insertion (3)
289
B-Trees (1)
The B-Tree is an extension of the 2-3 Tree.
The B-Tree is now the standard file organization for applications requiring insertion, deletion, and key range searches.
290
B-Trees (2)
1. B-Trees are always balanced.2. B-Trees keep similar-valued records
together on a disk page, which takes advantage of locality of reference.
3. B-Trees guarantee that every node in the tree will be full at least to a certain minimum percentage. This improves space efficiency while reducing the typical number of disk fetches necessary during a search or update operation.
291
B-Tree Definition
A B-Tree of order m has these properties:– The root is either a leaf or has two children.– Each node, except for the root and the
leaves, has between m/2 and m children.– All leaves are at the same level in the tree,
so the tree is always height balanced.
A B-Tree node is usually selected to match the size of a disk block.
– A B-Tree node could have hundreds of children.
292
B-Tree SearchGeneralizes search in a 2-3 Tree.
1. Do binary search on keys in current node. If search key is found, then return record. If current node is a leaf node and key is not found, then report an unsuccessful search.
2. Otherwise, follow the proper branch and repeat the process.
293
B+-Trees
The most commonly implemented form of the B-Tree is the B+-Tree.
Internal nodes of the B+-Tree do not store record -- only key values to guild the search.
Leaf nodes store records or pointers to records.
A leaf node may store more or less records than an internal node stores keys.
294
B+-Tree Example
295
B+-Tree Insertion
296
B+-Tree Deletion (1)
297
B+-Tree Deletion (2)
298
B+-Tree Deletion (3)
299
B-Tree Space Analysis (1)
B+-Trees nodes are always at least half full.
The B*-Tree splits two pages for three, and combines three pages into two. In this way, nodes are always 2/3 full.
Asymptotic cost of search, insertion, and deletion of nodes from B-Trees is (log n).
– Base of the log is the (average) branching factor of the tree.
300
B-Tree Space Analysis (2)
Example: Consider a B+-Tree of order 100 with leaf nodes containing 100 records.
1 level B+-tree:2 level B+-tree:3 level B+-tree:4 level B+-tree:
Ways to reduce the number of disk fetches:– Keep the upper levels in memory.– Manage B+-Tree pages with a buffer pool.
Graphs
A graph G = (V, E) consists of a set of vertices V, and a set of edges E, such that each edge in E is a connection between a pair of vertices in V.
The number of vertices is written |V|, and the number edges is written |E|.
Graphs (2)
Paths and Cycles
Path: A sequence of vertices v1, v2, …, vn of length n-1 with an edge from vi to vi+1 for 1 <= i < n.
A path is simple if all vertices on the path are distinct.
A cycle is a path of length 3 or more that connects vi to itself.
A cycle is simple if the path is simple, except the first and last vertices are the same.
Connected Components
An undirected graph is connected if there is at least one path from any vertex to any other.
The maximum connected subgraphs of an undirected graph are called connected components.
Directed Representation
Undirected Representation
Representation Costs
Adjacency Matrix:
Adjacency List:
Graph ADTinterface Graph { // Graph class ADT public void Init(int n); // Initialize public int n(); // # of vertices public int e(); // # of edges public int first(int v); // First neighbor public int next(int v, int w); // Neighbor public void setEdge(int i, int j, int wght); public void delEdge(int i, int j); public boolean isEdge(int i, int j); public int weight(int i, int j); public void setMark(int v, int val); public int getMark(int v); // Get v’s Mark}
Graph Traversals
Some applications require visiting every vertex in the graph exactly once.
The application may require that vertices be visited in some special order based on graph topology.
Examples:– Artificial Intelligence Search– Shortest paths problems
Graph Traversals (2)
To insure visiting all vertices:
void graphTraverse(Graph G) { int v; for (v=0; v<G.n(); v++) G.setMark(v, UNVISITED); // Initialize for (v=0; v<G.n(); v++) if (G.getMark(v) == UNVISITED) doTraverse(G, v);}
Depth First Search (1)// Depth first search void DFS(Graph G, int v) { PreVisit(G, v); // Take appropriate action G.setMark(v, VISITED); for (int w = G.first(v); w < G.n(); w = G.next(v, w)) if (G.getMark(w) == UNVISITED) DFS(G, w); PostVisit(G, v); // Take appropriate action}
Depth First Search (2)
Cost: (|V| + |E|).
Breadth First Search (1)
Like DFS, but replace stack with a queue.– Visit vertex’s neighbors before continuing
deeper in the tree.
Breadth First Search (2)
void BFS(Graph G, int start) { Queue<Integer> Q = new AQueue<Integer>(G.n()); Q.enqueue(start); G.setMark(start, VISITED); while (Q.length() > 0) { // For each vertex int v = Q.dequeue(); PreVisit(G, v); // Take appropriate action for (int w = G.first(v); w < G.n(); w = G.next(v, w)) if (G.getMark(w) == UNVISITED) { // Put neighbors on Q G.setMark(w, VISITED); Q.enqueue(w); } PostVisit(G, v); // Take appropriate action }}
Breadth First Search (3)
Topological Sort (1)
Problem: Given a set of jobs, courses, etc., with prerequisite constraints, output the jobs in an order that does not violate any of the prerequisites.
Topological Sort (2)void topsort(Graph G) { for (int i=0; i<G.n(); i++) G.setMark(i, UNVISITED); for (int i=0; i<G.n(); i++) if (G.getMark(i) == UNVISITED) tophelp(G, i);}
void tophelp(Graph G, int v) { G.setMark(v, VISITED); for (int w = G.first(v); w < G.n(); w = G.next(v, w)) if (G.getMark(w) == UNVISITED) tophelp(G, w); printout(v);}
Topological Sort (3)
Queue-Based Topsortvoid topsort(Graph G) { Queue<Integer> Q = new AQueue<Integer>(G.n()); int[] Count = new int[G.n()]; int v, w; for (v=0; v<G.n(); v++) Count[v] = 0; for (v=0; v<G.n(); v++) for (w=G.first(v); w<G.n(); w=G.next(v, w)) Count[w]++; for (v=0; v<G.n(); v++) if (Count[v] == 0) Q.enqueue(v); while (Q.length() > 0) { v = Q.dequeue().intValue(); printout(v); for (w=G.first(v); w<G.n(); w=G.next(v, w)) { Count[w]--; if (Count[w] == 0) Q.enqueue(w); } }}
Shortest Paths Problems
Input: A graph with weights or costs associated with each edge.
Output: The list of edges forming the shortest path.
Sample problems:– Find shortest path between two named vertices– Find shortest path from S to all other vertices– Find shortest path between all pairs of vertices
Will actually calculate only distances.
Shortest Paths Definitions
d(A, B) is the shortest distance from vertex A to B.
w(A, B) is the weight of the edge connecting A to B.– If there is no such edge, then w(A, B) = .
Single-Source Shortest Paths
Given start vertex s, find the shortest path from s to all other vertices.
Try 1: Visit vertices in some order, compute shortest paths for all vertices seen so far, then add shortest path to next vertex x.
Problem: Shortest path to a vertex already processed might go through x.
Solution: Process vertices in order of distance from s.
Example Graph
Dijkstra’s Algorithm Example
A B C D E
Initial 0
Process A 0 10 3 20
Process C 0 5 3 20 18
Process B 0 5 3 10 18
Process D 0 5 3 10 18
Process E 0 5 3 10 18
Dijkstra’s Implementation// Compute shortest path distances from s,// store them in Dvoid Dijkstra(Graph G, int s, int[] D) { for (int i=0; i<G.n(); i++) // Initialize D[i] = Integer.MAX_VALUE; D[s] = 0; for (int i=0; i<G.n(); i++) { int v = minVertex(G, D); G.setMark(v, VISITED); if (D[v] == Integer.MAX_VALUE) return; for (int w = G.first(v); w < G.n(); w = G.next(v, w)) if (D[w] > (D[v] + G.weight(v, w))) D[w] = D[v] + G.weight(v, w); }}
Implementing minVertex
Issue: How to determine the next-closest vertex? (I.e., implement minVertex)
Approach 1: Scan through the table of current distances.– Cost: (|V|2 + |E|) = (|V|2).
Approach 2: Store unprocessed vertices using a min-heap to implement a priority queue ordered by D value. Must update priority queue for each edge.– Cost: ((|V| + |E|)log|V|)
Approach 1int minVertex(Graph G, int[] D) { int v = 0; // Initialize to unvisited vertex; for (int i=0; i<G.n(); i++) if (G.getMark(i) == UNVISITED) { v = i; break; } for (int i=0; i<G.n(); i++) // Now find smallest value if ((G.getMark(i) == UNVISITED) && (D[i] < D[v])) v = i; return v;}
Approach 2void Dijkstra(Graph G, int s, int[] D) { int v, w; DijkElem[] E = new DijkElem[G.e()]; E[0] = new DijkElem(s, 0); MinHeap<DijkElem> H = new MinHeap<DijkElem>(E, 1, G.e()); for (int i=0; i<G.n(); i++) D[i] = Integer.MAX_VALUE; D[s] = 0; for (int i=0; i<G.n(); i++) { do { v = (H.removemin()).vertex(); } while (G.getMark(v) == VISITED); G.setMark(v, VISITED); if (D[v] == Integer.MAX_VALUE) return; for (w=G.first(v); w<G.n(); w=G.next(v, w)) if (D[w] > (D[v] + G.weight(v, w))) { D[w] = D[v] + G.weight(v, w); H.insert(new DijkElem(w, D[w])); } }}
Minimal Cost Spanning Trees
Minimal Cost Spanning Tree (MST) Problem:
Input: An undirected, connected graph G.Output: The subgraph of G that 1) has minimum total cost as measured by
summing the values of all the edges in the subset, and
2) keeps the vertices connected.
MST Example
Prim’s MST Algorithm// Compute a minimal-cost spanning treevoid Prim(Graph G, int s, int[] D, int[] V) { int v, w; for (int i=0; i<G.n(); i++) // Initialize D[i] = Integer.MAX_VALUE; D[s] = 0; for (int i=0; i<G.n(); i++) { v = minVertex(G, D); G.setMark(v, VISITED); if (v != s) AddEdgetoMST(V[v], v); if (D[v] == Integer.MAX_VALUE) return; for (w=G.first(v); w<G.n(); w=G.next(v, w)) if (D[w] > G.weight(v, w)) { D[w] = G.weight(v, w); V[w] = v; } }}
Alternate Implementation
As with Dijkstra’s algorithm, the key issue is determining which vertex is next closest.
As with Dijkstra’s algorithm, the alternative is to use a priority queue.
Running times for the two implementations are identical to the corresponding Dijkstra’s algorithm implementations.
Kruskal’s MST Algorithm (1)
Initially, each vertex is in its own MST.
Merge two MST’s that have the shortest edge between them.– Use a priority queue to order the unprocessed
edges. Grab next one at each step.
How to tell if an edge connects two vertices already in the same MST?– Use the UNION/FIND algorithm with parent-
pointer representation.
Kruskal’s MST Algorithm (2)
Kruskal’s MST Algorithm (3)
Cost is dominated by the time to remove edges from the heap.– Can stop processing edges once all vertices
are in the same MST
Total cost: (|V| + |E| log |E|).
