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The next step is to find the best line passing through the EI Salvador data point. We form the slopes (Yi - 40.2)/(x i - 11.5), put them in increasing order, and construct a table similar to Table 4.1 with a column for Ix i - 11.51 and a column for the cumulative sums. If you construct this table, you will find that the total cumulative sum is 265.5 and that the country whose cumulative sum first exceeds 265.5/2 = 132.75 is the United States. Therefore the best line passing through the EI Salvador data point also passes through the United States data point. The next step is to find the best line passing through the United States data point. Constructing a table similar to Table 4.1 and looking at the cumulative sums, we find that the best line passing through the United States data point also passes through the EI Salvador data point. But this is the same line we obtained at the previous step, and so the algorithm stops. The LAD regression line is the line passing through the data points of EI Salvador and the United States. Its slope is = (40.2 - 16.0)/(11.5 56.5) = - 0.5378 and its intercept is Ii = 40.2 - ( - 0.5378)(11.5) = 46.38. That is, the LAD regression line is Y = 46.38 - 0.5378X. See Figure 1.1.

if( ++x == theArraysize( x = 0;

da(v)

theArray( 1 1

1 / / Test if the queue is logically empty 2 / / Return true if empty, false, otherwise 3 template <class Object> 4 boo1 Queue<Object>:: isEmpty ( ) const 5 { 6 return currentsize == 0;

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stored in an array starting at location 0 Thus items must be copied carefully We discuss doubleQueue along with enqueue Many of the public methods resemble their stack counterparts, including the constructor shown in Figure 1612 and isErnpty, shown in Figure 1613 This constructor is not particularly special, except that we must be sure that we have the correct initial values for both f r o n t and back This is done by calling makeErnp t y The e n q u e u e routine is shown in Figure 1614 The basic strategy is simple enough, as illustrated by lines 7-9 in the e n q u e u e routine The doubleQueue routine, shown in Figure 1615 begins by resizing the array However, if f r o n t is not 0, we must move items because the implied wraparound of the original is no longer wrapped around once the array has become larger The simplest solution is to take the wrapped-around portion (those items in positions 0 to back) and move them to the new part of the array

= c,

The algorithm described in the preceding section will find the LAD regression line for most data sets, but occasionally problems arise with non uniqueness or degeneracy. Nonuniqueness means that there is more than one best

1 / / Insert x into the queue 2 template <class Object> 3 void Queue<Object>::enqueue( const Object 4 (

increment ( back theArray[ back ]

da(v) 2 dy = c(1 - y) ,

1 / / Internal method to double capacity 2 template <class Object> 3 void Queue~Object~::doubleQueue( ) 4 ( 5 theArrayresize( theArraysize( ) * 2 + 1 1 ; 6 if( front ! = 0 )

9 10 11 12 1

line passing through a data point. Degeneracy means that the best line through a data point also passes through two or more other data points. Recall that the algorithm proceeds in steps. At each step we find the best line passing through a given data point. The best line always passes through another data point, and this data point is used in the next step. But when there is non uniqueness, there is more than one best line. And when there is degeneracy, the best line passes through more than one other data point. In either case, there is more than one choice for the data point to be used in the next step. By making unlucky choices, the algorithm may go around in circles, or it may stop with a line that is not the LAD regression line. The possibility of such problems is indicated either when equality occurs in condition (4.2) or when the slope {3* = (Yk - yo)/(x k - xo) in (4.3) is equal to (Yk-l - Yo)/(X k - 1 - xo) or (Yk+l - Yo)/(X k + 1 - x o). In such instances we can resort to the algorithm in the following paragraph. Also see the subsection on nonuniqueness and degeneracy at the end of Section 4.6.

for( int i = 0; i < front; i++ theArray[ i + currentsize ] back += currentsize;

(12.79)

theArray[ i I ;

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