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As Figure 161 shows a stack can be implemented with an array and an integer The integer tos (top of stack) provides the array index of the top element of the stack Thus when tos is -1, the stack is empty To push,we increment tos and place the new element in the array position tos Accessing the top element is thus trivial, and we can perform the pop by decrementing tosIn Figure 161, we begin with an empty stack Then we show the stack after three push(b), pop and operations: push (a), Figure 162 shows the interface for the array-based Stack class It specifies two data members: theArray,which is expanded as needed, stores the items in the stack; and topof stack gives the index of the current top of the stack For an empty stack this index is -1 The constructor is shown in Figure 163 Because the data members are first-class objects, the Big-Three is automatically defined correctly Thus we do not need to provide a destructor, copy constructor, or copy assignment operator The public methods are listed in lines 21-27 of the interface Most of these routines have simple implementations The i sEmpty and makeEmpty routines are one-liners, as shown in Figure 164 The push method is shown in Figure 165 If it were not for the array doubling, the push routine would be only the single line of code shown at line 7 Recall that the use of the prefix + + operator means that topof s tack is incremented and that its new value is used to index theArrayThe remaining routines are equally short, as shown in Figures 166 and 167 The postfix - - operator used in Figure 167 indicates that, although topof Stack is decremented, its prior value is used to index theArray If there is no array doubling, every operation takes constant time A push that involves array doubling will take O ( N ) time If this were a frequent occurrence we would need to worry However, it is infrequent because an array doubling that involves N elements must be preceded by at least N / 2 pushes that do not involve an array doubling Consequently, we.

tos (0)

(12.70)

( 4.3)

tos(-])

How the stack routines work: (a) empty stack: (b) push (a) ; (c) push (b)and (d) pop ;

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

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where () is defined as in Fig. 12.13. From the figure, it is immediately apparent that the backward P"u -+ /L + d scattering (() = '17) is forbidden_by helicity considerati~ns. The cross sections for scattering from antiquarks, p"d -+ /L + U and v"u -+ /L - d, are given by (12.69) and (12.:70), respectively. We see that, for instance, vI' does not interact with either u or d quarks. To compare these results with experiment, we have to embed the constituent cross sections, (12.69) and (12.70), in the overall vN inclusive cross section. The procedure is familiar from 9. We obtain

/ / Stack class - - array implementation // / / CONSTRUCTION: with no parameters // / / ******************puBLIC OPERATIONS*********************

Justification That the Line Passes Through Two Data Points A key fact used in devising the algorithm is that the LAD regression line passes through two of the data points To see why this is so, imagine a line Y = a + bx drawn through a plot of the data points The absolute deviation of a data point from the line is the length of the vertical line segment from the point to the line The value of (41) is the sum of these absolute deviations Suppose the line does not pass through any data points If the line is moved upward a small amount, say e, then each absolute deviation either decreases by e or increases bYe, depending on whether the data point is above or below the line The value of (4.

// // // // // // //

/LX) =

void push( x ) void pop ( ) Object top( Object topAndPop( boo1 isEmpty( ) void makeEmpty ( )

--> *hxh**************ERRORS*********W*************X**

x most recently inserted item most recently inserted item and remove most recent item true if empty; else false all items

(12.71)

1) can be decreased (or at least not increased) by moving the line either upward or downward, depending on whether there are more data points above or below the line Move the line until it meets a data point If the line passes through exactly one data point, the line can be pivoted on its data point, either clockwise or counterclockwise, until it meets a second data point Of course the absolute deviation for the one data point remains zero, and each of the other absolute deviations either decreases or increases (by varying amounts) With respect to decreasing or increasing the sum of absolute deviations, pivoting clockwise and pivoting counterclockwise have opposite effects So by pivoting one way or the other, we can decrease (or at least not increase) the sum of absolute deviations.

public: Stack(

boo1 isEmpty ( ) const; const Object & top( ) const; void makeEmpty( ) ; void pop ( ) ; void push ( const Object Object topAndPop ( ) ;

NIp)

private: vector<Object> theArray; int topofstack;

1 2 3 4 5 6

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