I used this set of slides for the lecture on Complexity I gave at the University of Zurich for the 1st year students following the course of Formale Grundlagen der Informatik.

1. Complexity
www.tudorgirba.com

2. computation
information information
computer

3. memory
bandwidth
Computation requires resources
time
...

5. 1 public long factorial (int n) {
2 long factorial = 1;
3 int i = 1;
4 while ( i ≤ n ) {
5 factorial = factorial * i;
6 i = i + 1;
7 }
8 return factorial
9 }

6. 1 public long factorial (int n) {
2 long factorial = 1; 1
3 int i = 1; 1
4 while ( i ≤ n ) { 4
5 factorial = factorial * i; 4
6 i = i + 1; 3
7 }
8 return factorial 1
9 }

7. 1 public long factorial (int n) {
2 long factorial = 1; 1 1
3 int i = 1; 1 1
4 while ( i ≤ n ) { 4 n
5 factorial = factorial * i; 4 n
6 i = i + 1; 3 n
7 }
8 return factorial 1 1
9 }

8. 1 public long factorial (int n) {
2 long factorial = 1; 1 1
3 int i = 1; 1 1
4 while ( i ≤ n ) { 4 n
5 factorial = factorial * i; 4 n
6 i = i + 1; 3 n
7 }
8 return factorial 1 1
9 }
f(n) = 1+1+4*n+4*n+3*n+1 = 11*n+3

9. Big O Notation
f: N -> N
g: N -> N
f ∈ O(g)
∃ c, n0 :
(c,n0∈N) ∧ (c>0) :
(∀n : n∈N ∧ n>n0 : f(n) ≤ c*g(n))

10. Big O Notation
f: N -> N
g: N -> N
f ∈ O(g)
∃ c, n0 :
(c,n0∈N) ∧ (c>0) :
(∀n : n∈N ∧ n>n0 : f(n) ≤ c*g(n))
f: N % N, f(n)=11*n+3.
f ∈ O(n)

11. !"#$%$&'()("'*$+ ,'-.$/'--
O(1) Constant
O(log n) Logarithmic
O(n) Linear
O(n2) Quadratic
O(nk) Polynomial
O(kn) Exponential

12. O(1) ∈ O(log n) ∈ O(n) ∈ O(n2) ∈ O(nk) ∈ O(2n)
!"#$%$&'()("'*$+ ,'-.$/'--
O(1) Constant
O(log n) Logarithmic
O(n) Linear
O(n2) Quadratic
O(nk) Polynomial
O(kn) Exponential

13. N log N N log N N2 2N N!
10 3 33 100 1024 2*106
20 4 86 400 106 2*1018
100 7 664 10’000 1031 10157
1’000 10 10’000 106
10’000 13 130’000 108
100’000 17 106 1010
1’000’000 1’000 2*107 1012

14. N log N N log N N2 2N N!
10 3 33 100 1024 2*106
20 4 86 400 106 2*1018
100 7 664 10’000 1031 10157
1’000 10 10’000 106
10 2 s =
1.
10’000 13 130’000 108 10 4 s = 7 minutes
2.
10 5 s = 8 hours
1.1 day
100’000 17 106 1010
10 s =
6 s
1.6 we
10 s =
7 eks
3.8 mo
10 s =
8 nths
1’000’000 1’000 2*107 1012 3.1 yea
10 s =
9 rs
3.1 dec
10 10 s a
= 3.1 c des
10 11 s enturie
= neve s
r :)

15. O(c*f) = O(f)
O(f) + O(g) = O(f+g)
= max{O(f),O(g)}
O(f*g) = O(f)*O(g)

16. O(c*f) = O(f)
O(f) + O(g) = O(f+g)
= max{O(f),O(g)}
O(f*g) = O(f)*O(g)
O(f) ⊆ O(g) f ∈ O(g)
O(f) = O(g) (O(f) ⊆ O(g)) ∧ (O(g) ⊆ O(f))
O(f) ⊂ O(g) (O(f) ⊆ O(g)) ∧ (O(g) ≠ O(f))

17. a b c d e f g
a 0 2 3 0 0 0 0
a e b 0 0 1 0 0 0 0
3 5 3
2 c 0 0 0 2 0 0 0
2 c d g
d 0 0 0 0 5 4 0
1 4 3 e 0 0 0 0 0 0 3
b f
f 0 0 0 0 0 0 3
g 0 0 0 0 0 0 0

18. a b c d e f g
a 0 2 3 0 0 0 0
a e b 0 0 1 0 0 0 0
3 5 3
2 c 0 0 0 2 0 0 0
2 c d g
d 0 0 0 0 5 4 0
1 4 3 e 0 0 0 0 0 0 3
b f
f 0 0 0 0 0 0 3
g 0 0 0 0 0 0 0
procedure FloydWarshall ()
for k := 1 to n
for i := 1 to n
for j := 1 to n
path[i][j] = min ( path[i][j], path[i][k]+path[k][j] );

20. O(2*n-1) = O(n)
O(n*(n+1)/2) = O(n2)
O(log n2) = O(log n)
O((3*n2 + 6*n+9)*log(1+2*n)) = O(n2log(n))

21. O(2*n-1) = O(n)
O(n*(n+1)/2) = O(n2)
O(log n2) = O(log n)
O((3*n2 + 6*n+9)*log(1+2*n)) = O(n2log(n))

22. 4 2 5 1 6 3
ch
: Line ar sear
E xample

23. 4 2 5 1 6 3
ch
: Line ar sear
E xample

24. 4 2 5 1 6 3
ch
: Line ar sear
E xample

25. Worst case: n steps
Best case: 1 step
Average case: n/2 steps
4 2 5 1 6 3
ch
: Line ar sear
E xample

26. Best case estimation
f: N -> N
g: N -> N
f ∈ Ω(g)
∃ c, n0 :
(c,n0∈N) ∧ (c>0) :
(∀n : n∈N ∧ n>n0 : f(n) ≥ c*g(n))

27. Best case estimation
f: N -> N
g: N -> N
f ∈ Ω(g)
∃ c, n0 :
(c,n0∈N) ∧ (c>0) :
(∀n : n∈N ∧ n>n0 : f(n) ≥ c*g(n))
(3n2+6n+9)*log(1+2n) ≤ 36 n2 * log n
(3n2+6n+9)*log(1+2n) ∈ O(n2 * log n)

28. Best case estimation
f: N -> N
g: N -> N
f ∈ Ω(g)
∃ c, n0 :
(c,n0∈N) ∧ (c>0) :
(∀n : n∈N ∧ n>n0 : f(n) ≥ c*g(n))
(3n2+6n+9)*log(1+2n) ≤ 36 n2 * log n
(3n2+6n+9)*log(1+2n) ∈ O(n2 * log n)
(3n2+6n+9)*log(1+2n) ≥ n2 * log n
(3n2+6n+9)*log(1+2n) ∈ Ω(n2 * log n)

29. Average case estimation
f: N -> N
g: N -> N
f ∈ θ(g)
∃ c 1, c 2, n 0 :
(c1,c2,n0∈N) ∧ (c1>0) ∧ (c2>0):
(∀n : n∈N ∧ n≥n0 : c1*g(n) ≤ f(n) ≤ c2*g(n))

30. Average case estimation
f: N -> N
g: N -> N
f ∈ θ(g)
∃ c 1, c 2, n 0 :
(c1,c2,n0∈N) ∧ (c1>0) ∧ (c2>0):
(∀n : n∈N ∧ n≥n0 : c1*g(n) ≤ f(n) ≤ c2*g(n))
(3n2+6n+9)*log(1+2n) ∈ O(n2 * log n)

31. Average case estimation
f: N -> N
g: N -> N
f ∈ θ(g)
∃ c 1, c 2, n 0 :
(c1,c2,n0∈N) ∧ (c1>0) ∧ (c2>0):
(∀n : n∈N ∧ n≥n0 : c1*g(n) ≤ f(n) ≤ c2*g(n))
(3n2+6n+9)*log(1+2n) ∈ O(n2 * log n)
(3n2+6n+9)*log(1+2n) ∈ Ω(n2 * log n)

32. Average case estimation
f: N -> N
g: N -> N
f ∈ θ(g)
∃ c 1, c 2, n 0 :
(c1,c2,n0∈N) ∧ (c1>0) ∧ (c2>0):
(∀n : n∈N ∧ n≥n0 : c1*g(n) ≤ f(n) ≤ c2*g(n))
(3n2+6n+9)*log(1+2n) ∈ O(n2 * log n)
(3n2+6n+9)*log(1+2n) ∈ Ω(n2 * log n)
(3n2+6n+9)*log(1+2n) ∈ θ(n2 * log n)

33. boolean f ( int[][] a , int n ) {
for ( int i = 0 ; i < n ; i++ ) {
for ( int j = i + 1 ; j < n ; j++ ) {
if ( a[i][j] == 0 ) {return false;}
}
return true;
}
}
E xample

34. boolean f ( int[][] a , int n ) {
for ( int i = 0 ; i < n ; i++ ) {
for ( int j = i + 1 ; j < n ; j++ ) {
if ( a[i][j] == 0 ) {return false;}
}
return true;
}
}
f ∈ O(n2)
f ∈ Ω(1)
f ∈ θ(n2)
E xample

35. 4
2 6
1 3 5
ch
: Bina ry sear
le
Examp

36. f ∈ O(log n)
f ∈ Ω(1)
f ∈ θ(log n)
4
2 6
1 3 5
ch
: Bina ry sear
le
Examp

37. solvable
in O(nk)

38. solvable
in O(nk) P

39. solvable
in O(nk) P NP

40. solvable
in O(nk) P NP verifiable
in O(nk)

41. solvable
in O(nk) P ⊂ NP verifiable
in O(nk)

42. solvable
in O(nk) P = NP
? verifiable
in O(nk)

43. NP-complete
NP, and one cannot do better

44. Hamiltonian path
a e
c d g
b f
:
p roblem
mplete path
NP-co onian
Hamilt

45. :
p roblem
NP-co mplete sman
le
Trave ling sa

46. :
p roblem
mplete ing
NP-co h color
Grap

47. S={x1, … , xn}
t ∈ N
∃ {y1, … , yk} ⊆ S : Σyi=t ?
:
p roblem
NP-co mplete m
u
S ubset s

48. S={x1, … , xn}
t ∈ N
∃ {y1, … , yk} ⊆ S : Σyi=t ?
S= {8, 11, 16, 29, 37}
t = 37
{11, 16}
{8, 29}
{37}
:
p roblem
NP-co mplete m
u
S ubset s

49. Will this program terminate?
g problem
Th e haltin dable
ci
is unde

50. Tudor Gîrba
www.tudorgirba.com
creativecommons.org/licenses/by/3.0/