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Lecture notes

Lecture Notes on Lists (COMP11120)
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Master the fundamentals of lists with these comprehensive lecture notes for COMP11120. Covering essential topics such as list operations, properties, and various traversal techniques, these notes provide clear explanations and practical examples to help you understand and utilize lists effectively....

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  • May 30, 2024
  • 2
  • 2023/2024
  • Lecture notes
  • Andrea schalk
  • Lists

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Lists

Definition : A list over set S is defined as follows : ~
Listsy = Set of all lists over S
·
base case : The empty list [J is a list
·

Step case :
IfI is a list over S and SES S: l is a list over S
. > How are lists built ?
,


S =
IN

[J ,
1 : []
,
2 : (1 ; 53)
,
1 :
(2 (1 : :
[]))

[i] ta ,
is in aii)

Operations on recursive data structures



Idea : ·
Define operation for base case (s) Example : Want binary operation Listsgx Lists -
Listsy ,
which

takes 2 lists and concatenates them
.

Assume know how to define operation structures built
·
we on so far.
·
Define operation for new structures built using step case(s).

Let (i I' be I concatenated with

Example Define :
operation Sum : Lists
N
IN adding all the elements


of the list. base case [I H 11 = 1
given

Step case (5 : () + l = s : ((II))

base case sum[] = 0


Step case sum(5 () : =
s + Sum( [1 21 H 10
, ,
2
,
11 =
(1 : [2]) [3 ,
2
, 1] Sc concat
=
1 :
((2 [l) + :
[3 ,
2
, 11) Sc concat


Sum [1 ,
2
,
1] = sum (1 : [2, 1] sc Sum
= 1 : (2 ([I
:
+ [3 , 2, 11) s concat

= 1 + sum(2 : [1]) Sc Sum
=
1 : (2 : [3 , 2
, 17) bc concat

=
1 + 2 + sum(1 : []) Sc Sum
=
[1 ,
2 3
, ,
2
,
1 append
= 1 + 2 + 1 + sum[] sc sum



= 1 + 2 + 1 + 0 be sum
Example : ForI t given by base case [l + 11 = 1
=
4 Step case (5 : 1) H = 5 : ((11 ()

Show that for all Le Lists (+ [T =
,

[J
So is unit for it


Example Define : a operation rev : Lists >
Lists that reverses a list.

Proof by induction :
L
L


base rev[] = [] base case [J + []
=
[]
case



Step case rev(s : 1) = revL #ISI Step case (5 : (15) =
5 :
((it)) #
base Case !
=
S
: step case


Example : Show summer) = sum for all Le Lists
You may use sum (1H1') =
Suml + suml Example : Show forthat all 1, 11E Lists IN
,


sum((H(1) = sum) + Sum(

Proof by induction :




base case Sumrev[] =
SumI] be nev Proof by induction :




Step case sum rev (5 : 1) =
Sum(rov1H [s]) Sc rev base case Sum([]111) = Sum I' be concat
("
=
Sum(rev() + sum(s : []) ind . hyp
.
= O + Sum


= sumY + 5 + sumI]
Sc Sum = sum[I + sumI' be sum

Suml
=
+ 5 + 0 be Sum


=
S + sum Step case Sum(15 : 1) H() =
sum(s : (1 + 1)) Sc concert
= Sum(5 : () = S + Sum(( + 1) sc sum


= S + sum) + suml' ind .

hyp .




= Sum (s : 1) + suml" sc Sum

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