Algorithms and Data Structures

N.Wirth. Algorithms and Data Structures. Oberon version 62
Fig. 2.7. Heap with 7 elements
06
42
12
55 94
18 44
Fig. 2.8. Key 44 sifting through the heap
Let us now assume that a heap with elements h L+1 ... h R is given for some values L and R, and that a
new element x has to be added to form the extended heap h L ... h R . Take, for example, the initial heap h 1
... h 7 shown in Fig. 2.7 and extend the heap to the left by an element h 0 = 44. A new heap is obtained by
first putting x on top of the tree structure and then by letting it sift down along the path of the smaller
comparands, which at the same time move up. In the given example the value 44 is first exchanged with 06,
then with 12, and thus forming the tree shown in Fig. 2.8. We now formulate this sifting algorithm as
follows: i, j are the pair of indices denoting the items to be exchanged during each sift step. The reader is
urged to convince himself that the proposed method of sifting actually preserves the heap invariants that
define a heap.
A neat way to construct a heap in situ was suggested by R. W. Floyd. It uses the sifting procedure
shown below. Given is an array h 0 ... h n-1 ; clearly, the elements h m ... h n-1 (with m = n DIV 2) form a
heap already, since no two indices i, j are such that j = 2i+1 or j = 2i+2. These elements form what may
be considered as the bottom row of the associated binary tree (see Fig. 2.6) among which no ordering
relationship is required. The heap is now extended to the left, whereby in each step a new element is
included and properly positioned by a sift. This process is illustrated in Table 2.6 and yields the heap
shown in Fig. 2.6.
PROCEDURE sift (L, R: INTEGER);
VAR i, j: INTEGER; x: Item;
BEGIN
i := L; j := 2*i+1; x := a[i];
IF (j < R) & (a[j+1] < a[j]) THEN j := j+1 END;
WHILE (j 0 DO DEC(L); sift(L, n-1) END
In order to obtain not only a partial, but a full ordering among the elements, n sift steps have to follow,