1.
REVISION
Martin Chapman
martin.chapman@kcl.ac.uk

2.
•By demand, from highest to lowest:
•Recursion
•Break
•Recursion
•Objects and Classes
•Break
•Arrays and Array Lists

3.
RECURSION
Martin Chapman
martin.chapman@kcl.ac.uk

4.
What do we need to make a loop run?
for (int i = 0; i<=10; i++) {
System.out.println(i);
}

5.
for (int i = 0; i<=10; i++) {
System.out.println(i);
}
What do we need to make a loop run?
the ‘movement’

6.
for (int i = 0; i<=10; i++) {
System.out.println(i);
}
What do we need to make a loop run?
the ‘change’

7.
for (int i = 0; i<=10; i++) {
System.out.println(i);
}
What do we need to make a loop run?
‘when to stop’

8.
for (int i = 0; i<=10; i++) {
System.out.println(i);
}
What do we need to make a loop run?
the content

9.
How do we loop without a loop?

10.
How do we loop without a loop?
the ‘movement’

11.
the ‘change’
How do we loop without a loop?

12.
‘when to stop’
How do we loop without a loop?

13.
the content
How do we loop without a loop?

14.
How do we loop without a loop?
base case
recursive case

15.
Reverse, reverse.
What happens here?

16.
Reverse, reverse.
methodA
3
3

17.
Reverse, reverse.
methodA
3
3
3
3
3

18.
Reverse, reverse.
methodA
3
3
3

19.
Reverse, reverse.
methodA
3
2
2
methodA
3

20.
Reverse, reverse.
methodA
3
2
!
!
2
3
methodA

21.
If we tell the compiler to execute non-
conditional code then at some point
it must be executed.

22.
Reverse, reverse.
methodA ! 3

23.
Reverse, reverse.
methodA
3
2
!
2
methodA

24.
Reverse, reverse.
methodA
3
2
!
2
methodA
2
2
2

25.
Reverse, reverse.
methodA
3
2
!
methodA
2
2

26.
Reverse, reverse.
methodA
3
2
!
methodA
1
1
methodA
2

27.
Reverse, reverse.
methodA
3
2
!
methodA
1
1
!
!
methodA
2

28.
Reverse, reverse.
methodA !
methodA !
3
2

29.
Reverse, reverse.
methodA
1
!
methodA
0
!
methodA !
-1
methodA
-1

30.
Reverse, reverse.
methodA
1
!
methodA
0
!
methodA !
-1
methodA
-1
-1

31.
Reverse, reverse.
methodA
1
!
methodA
0
!
methodA !
-1
methodA
End of the method, so
return

32.
Reverse, reverse.
methodA
1
!
methodA
0
!
methodA !
-1
methodAbase case

33.
Reverse, reverse.
methodA !
methodA !
methodA !
methodA ! 0
1
2
3

34.
Reverse, reverse.
methodA !
methodA !
methodA !
methodA ! 0
1
2
3

35.
Reverse, reverse.
methodA !
methodA !
methodA !
methodA ! 0
1
2
3

36.
Reverse engineering is fine, but
how do I build something from scratch?

37.
0 1 2 3 4 5 6 7 8 9 10
0 1 1 2 3 5 8 13 21 34 55

38.
0 1 2 3 4 5
0 1 1 2 3 5
Fibonacci Sequence
Every number is the sum of the
two previous, when possible.nth
fib

39.
base case
Ask, when do we not need to do any
computation? When can we simply return a number
rather than doing any addition?
Determining the
0 1 2 3 4 5
0 1 1 2 3 5

40.
Ask, when do we not need to do any
computation? When can we simply return a number
rather than doing any addition?
Determining the
0 1 2 3 4 5
0 1 1 2 3 5
base case

41.
0
0

42.
0 1 2 3 4 5
base case
Ask, when do we not need to do any
computation? When can we simply return a number
rather than doing any addition?
Determining the
0 1 1 2 3 5

43.
1
1

44.
0 1 1 2 3 5
Determining the
recursive case
0 1 2 3 4 5
What is the general case? Can this case be carried on
indefinitely?

45.
0 1 1 2 3 5
0 1 2 3 4 5
Determining the
+ =
recursive case
What is the general case? Can this case be carried on
indefinitely?

46.
0 1 1 2 3 5
0 1 2 3 4 5
Determining the
+ =
recursive case
+ =
What is the general case? Can this case be carried on
indefinitely?

47.
+
1
1
0
0
fib(0) fib(1)
fib(2) =

48.
static int fib(int n) {
if (n == 0) { return 0; }
if (n == 1) { return 1; }
else {
return fib(n-1) + fib(n-2); }
}
2
1 0

49.
Determining the
recursive case
0 1 1 2 3 5
0 1 2 3 4 5
+ =
+ =
fib
( )
fib
( )
fib
( )

50.
0 1 2 3 4 5
fib
( )
fib
( )
fib
( )
Determining the
recursive case
0 1 1 2 3 5+ =
+ =

51.
0 1 2 3 4 5
fib
( )
fib
( )
fib
( )
Determining the
recursive case
0 1 1 2 3 5+ =
+ =

52.
0 1 5
fib
( n )
fib
(n-2)
fib
(n-1)
Determining the
recursive case
0 1 1 2 3 5+ =
+ =
This operation is consistent, so we can generalise.

53.
0 1 5
fib
( n )
fib
(n-2)
fib
(n-1)
Determining the
recursive case
0 1 1 2 3 5+ =
+ =

54.
static int fib(int n) {
if (n == 0) { return 0; }
if (n == 1) { return 1; }
else {
return fib(n-1) + fib(n-2); }
}

55.
static int fib(int n) {
if (n == 0) { return 0; }
if (n == 1) { return 1; }
else {
return fib(n-1) + fib(n-2); }
}
5

56.
Recursion creates a tree

57.
1 = actual value
1 0
1 1 0 01
1

58.
1 = actual value
1 0
1 1 0 01
+
1

59.
1 = actual value
1 0
1 1 0 01
+
+
1

60.
1 = actual value
1 0
1 1 0 01
+
+
+
!
!
1

61.
static int fib(int n) {
if (n == 0) { return 0; }
if (n == 1) { return 1; }
else {
return fib(n-1) + fib(n-2); }
}
!

62.
1 = actual value
1 0
1 1 0 01
+
+
+
!
!
1

63.
1 = actual value
1 0
1 1 0 01
+
+
+
!
!!
1
+

64.
1 = actual value
1 0
1 1 0 01
1
+
+
+
!
!!
1
+

65.
0 1 2 3 4 5
0 1 1 2 3 5
Fibonacci Sequence
Every number is the sum of the
two previous, when possible.

66.
1 = actual value
1 0
1 1 0 01
1
+
+
+
!
!!
1
+

67.
1 = actual value
1 0
1 1 0 01
1
+
+
+
!
!!
1
+

68.
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
!
!!
1
+

69.
0 1 2 3 4 5
0 1 1 2 3 5
Fibonacci Sequence
Every number is the sum of the
two previous, when possible.

70.
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
!
!!
1
+

71.
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
!
!!
+
1
+

72.
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
!
!
1
+
1
+

73.
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
!
!
1
+
1
+

74.
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
!
!
1
3
+
1
+

75.
+
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
!
!
1
3
+
+
1

76.
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
!
!
1
3
+
++
1
+

77.
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
!
1
3
+
++
1
1
+

78.
+
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
!
1
3
+
++
1
1

79.
+
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
1
3
+
++
1
1
2

80.
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
1
3
+
++
1
1
2+

81.
1 = actual value
1 0
1 1 0 01
1
+
+
+
2
1
3
+
++
1
1
2
5
+

82.
static int fib(int n) {
if (n == 0) { return 0; }
if (n == 1) { return 1; }
else {
return fib(n-1) + fib(n-2); }
}
Think, when do I not have to do any computation?
This will also be when my code stops.
base case
SUMMARY
base case

83.
static int fib(int n) {
if (n == 0) { return 0; }
if (n == 1) { return 1; }
else {
return fib(n-1) + fib(n-2); }
}
When I do have to compute something, what is the
pattern? If I can prove it for one case, it should
work for them all.
recursive case
SUMMARY
base case
base case

84.
During the break...

85.
SUMMARY
• Recursion is just another way of looping, with information
changing each time.
• Recursion consists of a recursive call and a base case.
• If there is code after a recursive call it must be executed
and will be executed in reverse order.
• Recursion is a natural solution if you can visualise the call in
any instance i.e. Fibonacci numbers.

86.
WILL I SEE RECURSION
AGAIN?
• Yes, sorry.
• But it will give you more of a chance to understand it.
• Modules which touch on recursion (at least): DST (Semester
2, 1st year) and Algorithm Modules.