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2008-11-03T05:20:18Z

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= Lab 7 =

== Task 1 ==
Task 1 was fun easy as hell... but I'm glad. Finally something I can handle HEHEHEHE! just jokes.
I've used every command and feel proficient in the program. ;)

== Task 2 ==
Ok so now I have to insert a-g in such an order that the tree is perfectly balanced.
Here's the characters: a b c d e f g
Here's how I chose to insert them:
a b c '''''[[D]]''''' e f g
a b c D e '''''[[F]]'''''
a '''''[[B]]''''' c D e F g
'''''[[A]]''''' B c D e F g
A '''''[[B]]''''' C D e F g
A B C D '''''[[E]]''''' F g
A B C D E F '''''[[G]]'''''

And this is the result:

> DISPLAY
a
b
c
d (*)
e
f
g

Basically my strategy was to choose the median value each time and split the array of values to be inserted about the median. The first time I chose [[D]] which was right in the middle a b c [[D]] e f g. This gave me 2 groups: '''''ABC''''' & '''''EFG'''''. Then I repeated this method with each group.

Just to see how the program would '''''[[BALANCE]]''''' these letters:
> SKEW
a (*)
b
c
d
e
f
g

> BALANCE
a
b
c
d (*)
e
f
g

Since the number of elements is odd, no matter in which order the elements are inserted using the '''''[[MEDIAN]]''''' method, they'll always give the same result for a perfectly balanced tree.

== Task 3 ==
After re-performing the SKEW:
> SKEW
a (*)
b
c
d
e
f
g

> r l
a (*)
b
c
d
e
f
g

> print
Tree[*a*,b,c,d,e,f,g]

Hence, the elements are still in the same order, but now the root is '''''[[b]]'''''.

b
c a
d
e
f
g

SE250:lab-7:rbha033 - Revision history

Mark: 14 revision(s)