Outline

• infix, prefix, and postfix expressions
• queues
• queue interface
• queue applications
• queue implementation: array queue
• data structure traversal

infix expressions

• 2 + 3 * 4 has the value?
• some operators (*, /, %) have higher precedence than other operators (+, -)
• operators with the same precedence are evaluated in left-to-right order
• these expression can be evaluated using two stacks:
  • when an operand is read, it is pushed onto the operand stack
  • operators are:
    1. pushed onto the operator stack if the new operator has higher precedence than the old operator
    2. otherwise, the top of the operator stack is popped and evaluated with the top two elements of the operand stack, the result is pushed onto the operand stack, and the new operator is pushed onto the operator stack
• at the end of the expression, operators are popped off and evaluated (popping the operands and pushing the results) until the operator stack is empty
• at this point, the operand stack should have exactly one number in it
• more interesting with more precedence levels, e.g. ^ (exponentiation)
• the Java compiler must recognize (parse) Java expressions and either:
  • evaluate any constant-valued (sub-) expressions, or
  • emit code to evaluate the expression at run-time

parenthesized infix expressions

• (2 + 3) * 4 has the value?
• when reading a left parenthesis, push it onto the operator stack
• when reading a right parenthesis, behave as at the end of the expression, until the matching left parenthesis is popped from the operator stack
• everything else is the same as the previous algorithm

integer operators

• addition (+), subtraction (-), multiplication (*) work as expected
• division (/) always rounds down: 3 / 2 gives 1, 101 / 100 gives 1
• remainder (%), also known as modulo, returns the remainder from the division: 3 % 2 gives 1, 127 % 100 gives 27
• multiplication, division, and modulo have higher precedence than addition and subtraction, and so are evaluated first: 3 + 54 * 17 is 3 + (54 * 17)
• with equal-precedence operators, the expression is evaluated from left to right: 99 - 3 - 33 / 11 / 3 is ((99 - 3) - ((33 / 11) / 3))

other kinds of expressions

• in a prefix expression, the operator comes before the operands: / + 3 * 7 4 2 means (3 + 7 * 4) / 2
• in a postfix expression, the operator comes after the operands: 4 2 / 3 + 1 * means (4 / 2 + 3) * 1
• in an infix expression, the operator comes in-between the operands
• only infix expressions need:
  • precedence
  • parentheses (to override precedence)
  in prefix and postfix expression, the position indicates which operands are used with which operators
• converting from one notation to the other can benefit from using a stack or recursion
• computing in prefix or postfix is easy when using a stack

algorithm for postfix computation

• read the next input (next character) of the string
• if the character is an operand, push it on the stack
• if the character is an operator, pop the top two elements off the stack, apply the corresponding operation (the operands must be in the correct order!), and push the result back on the stack
• if the string is empty:
  • if the stack has one element, that element is the result
  • if the stack has 0 or multiple elements, the expression is malformed
• in-class exercise: use the above algorithm to evaluate the following expressions:

  9 7 /
  1 2 * 3 * 4 *
  3 4 * 1 2 + -
  

StringBuilder

• it is easy to concatenate strings
• however, it is not particularly efficient: it involves copying all the characters of the original and the new string
• a string builder is like an array list, but for strings: it is a data structure that efficiently supports growable (extensible) strings
• a StringBuffer is similar, but will work correctly in multi-threaded programs

queues

• a stack is a Last-In, First-Out (LIFO) data structure
• a First-In, First-Out (FIFO) data structure is known as a queue
• the word "queue" (pronounced the same as the letter "Q") is used in the UK for what in the US is known as a "line", e.g. at a supermarket or a bank or a movie theater
• the first person in the queue will be the first one served
• queues are commonly used with computers:
  • documents to be printed are queued
  • packets to be sent on a network are queued
• priority queues allow high-priority items to move to the head of a line, so are not strictly FIFO, whereas
• regular queues are strictly FIFO

queue interface

• a queue is a collection: Interface Queue<E>
• boolean isEmpty()
• boolean offer(E value) attempts to insert the value at the end of the queue, returning whether the insertion was completed
• E poll() removes and returns the object at the head of the queue, or null if the queue is empty
• void remove() is similar to poll, but throws an exception if the queue is empty
• E peek() is similar to poll, but does not remove the element
• Iterator<E> iterator() returns a new iterator over the elements of the queue
• isEmpty() and iterator() are inherited from Interface Collection<E>

queue applications

• simulating waiting lines:
  • random arrivals
  • random time to serve a client
  • single queue? multiple queues?
  • compute the average waiting time over many simulations
  • compute the worst case waiting time over many simulations
  • compute the average worst case waiting time for each simulation
• recognizing palindromes:
  • queue is FIFO, stack is LIFO
  • push each character onto the stack, offer each character to the queue
  • then remove one character at a time from both data structures
  • if it is a palindrome, the characters will be the same
• print queue: print jobs in the order submitted
• traversing data structures

queue implementation strategies

• similar to stack
• linked list implementation, or
• array implementation
• as for other collections, we don't need to know the type of the elements, only that they are objects, so a generic implementation is fine

array implementation of queues

• like stack: store all the elements in an array
• when inserting a new element (offer), just like stack, add the element at the end of the queue
• two choices when removing an new element from the front of the list (poll, remove):
  1. copy all subsequent elements down by one index, so the first element is still at the beginning of the array, or
  2. keep track of where the head of the queue is, with another index variable
• for example, removing the front element (a) from this queue:
• can be done by copying:
• or by keeping track of where the first element is:
• in-class exercise: what are the advantages and disadvantages of these two strategies?

in-place array implementation

• two integer variables, one the index of the head of the queue (front), the other the index of the tail of the queue (end)
• so valid elements are found from array[front] through array[end - 1]
• queue contents get ever higher in the array, while lower-numbered indices will no longer contain valid data
• eventually, even if the queue only has a few values, we run out of indices to put new data into
• solution: once we reach the end of the array, put the data beginning at index 0 again
• the modulo operation can be used to make this simple:

  index = (index + 1) % QUEUE_SIZE;
  

• for example, if queue size is 15 and index = 14,
  14 + 1 = 15,
  15 % 15 = 0
  so the new value of the index is 0
• the number of elements in the array is
  (end - front + QUEUE_SIZE) % QUEUE_SIZE,
  but it is easier to simply maintain a size field
• this is an example of adding elements 's' and 't' to a queue:

implementation of the method offer

• if there is room,
  • insert at end
  • increment end modulo MAX_SIZE
• offer calls the private method full to make sure room is available
  • if no room, simply returns false
  • implementation in textbook (p. 323) doubles the size of the array: what is the runtime of this method?

amortized runtime analysis

• so far, we have only considered worst-case runtime in our big-O analysis
• consider the offer method:
  • if there is room, it takes O(1) (constant time)
  • if more room is needed, it takes O(n) (linear time)
• how long does it take on average?
• assume we just doubled the size of the array, to size 2n
• so for the next n calls to offer, the calls will take constant time
• then, on the following calls to offer, it will take time n
• the total time for n calls is O(n)
• so the average time per calls must be O(1)
• this is known as the amortized runtime: sometimes the call is expensive, but this cost is amortized across a large (enough) number of inexpensive calls, so the average is low
• the hard part is guaranteeing that there will be all those inexpensive calls
• for example, assume that instead of doubling the array size, the array size increases by a constant, say 10 new elements
• then, O(n) time is spent for every 10 calls
• on average, the time is O(n/10), which is still O(n) or linear
• no constant is large enough to amortize the linear cost
• so the array size has to at least double to give O(1) amortized time

implementation of the method poll

• if there is at least one element,
  • element is taken from front
  • increment front modulo MAX_SIZE
• if there is no element, simply returns null
• peek is even simpler: no increment, no size change
• what is the runtime of this method?
• what is the runtime of the empty method?

data structure traversal

• in a linked list, each node has a link to at most one other node
• in a doubly-linked list, each node has a link to at most two other nodes
• there are more general data structures in which nodes can have links to multiple other nodes
• these data structures go by different names, including trees and graphs
• in some cases, we need to have a program start at one node and visit all the other nodes in the data structure: tree traversal or graph traversal
• in general, I can start from a given node and put all the nodes it is connected to into a queue
• then, I repeatedly remove one node from the queue, visit it, and put into the queue all the nodes it is connected to
• if I keep doing this, staying away from nodes that have already been visited, I will eventually visit all connected nodes
• this is called breadth-first traversal:
  • visually arranging the first node at the top
  • the nodes it is connected to right below it,
  • the nodes they are connected to right below them,
  • nodes are visited in left-to-right and top-to-bottom order

depth-first traversal

• for a traversal, I could push the nodes in a stack instead of a queue
• then, the node I will visit next is the node I put on the stack most recently
• that means visiting every node connected to the most recently visited node, before visiting any node that was pushed onto the stack earlier
• this is called depth-first traversal because the traversal tends to go top-to-bottom, then climb back up and explore the side branches