number of all elements in the heap
n = 3k + 3k−1 + . . . + 9 + 3 + 1. (8)
In the last level of the heap, there does not have to be a complete number of
leaves. It may exist an extreme case, that in the last level we only have a few
items. Nevertheless we can write down that r = 3k is the number of elements in
the last level of the triple heap. As a result, equation (8) takes the form
n = r + 3k−1 + . . . + 9 + 3 + 1, (9)
where we consider geometric sequence. Therefore we calculate number of elements
in (9) as
n = r +
k−1
Xi=0
3i = r −
1
2
+
1
6
· 3k. (10)
Value determined in equation (10) we estimate by the maximum number of
elements in the heap as
n = r −
1
2
+
1
6
· 3k ≤
1
2
· 3k−1. (11)
On basis of (11) we can calculate number of levels defined as k. Height of the
heap we estimate using logarithmic function
log3 n ≤ log3
1
2
· 3k−1. (12)
Equation (12) is equal to
log3 n ≤ log3
1
2
+ (k − 1) · log3 3 ≤ k − 1. (13)
Therefore using formula (13) we can estimate number of levels in triple heap as
log3 n ≤ k − 1. (14)
Finally, height of triple heap is
k ≥ ⌊log3 n⌋ + 1. (15)
In discussed case of full triple heap, k is the maximum natural value of levels
comply with the condition (15). Therefore finally we can write down
k ∼=
⌊log3 n⌋ + 1. (16)
⊓⊔
In Fig.1 is shown triple heap structure made of k levels. Formula (15) helps to
calculate height of any triple heap. It is useful to calculate height of the entire
heap and heights of it’s components. There comes a question: how fast can we
create a triple heap?