Solution of
Assignment 4
CSC2100B
4.1.1 Open Hash Table
0
1
2
3
4
5
6
7
8
9
4371
1323 6173
4344
4199 9679 1989
Advantage:
1. Space can be saved if the
records are large compared
with the pointers
2. Simple and efficient collision
handling
3. The size of the hash table
can be smaller than the
number of records
4. Deletion is easy and quick
Disadvantage:
1. All links require space
2. Space is wasted if the
records are small
4.1.2 Linear Probing
0
1
9679
4371
2
3
4
1989
1323
6173
5
6
7
8
4344
9
4199
Advantage:
1. Simple implementation
2. Insertion and search are efficient when
the table is less than half full
3. All space can be used up before
insertion failure happens (though
clustering may occur)
Disadvantage:
1.When the table is more than half full,
insertion and search may be slow as
expected number of probes is large
4.1.3 Quadratic Probing
0
1
2
3
4
5
6
7
8
9
9679
4371
1323
6173
4344
1989
4199
Advantage:
1. Clustering is reduced
(compared with linear
probing)
Disadvantage:
1.Insertion may fail if the table
is more than half full
2.Deletion is more
complicated (needs lazy
deletion)
4.1.4 Double Hashing
0
1
4371
2
3
4
1323
6173
5
6
7
8
9679
9
4199
1989 mod 10 = 9 but address 9 is already
occupied. On the other hand, 7 - (1989
mod 7) = 6 but (9 + i*6) mod 10 is odd for
all integer i and all odd addresses have
already been used. So insertion will fail
Advantage of double hashing:
1. Good for avoiding clustering
Disadvantage:
1. Maybe slower than the other methods
4344
2. Need to choose the second hash
function carefully so all cells can be probed.
4.2 Rehashing
0
1
10
11
1989
21
2
12
1323
22
3
13
4199
4
14
5
15
6
16
7
17
8
18
9
4371
20
6173
19
9679
4344
The size of the new hash table
should be chosen as the
smallest prime that is larger
than twice that of the original
table. The smallest prime
>2*10 is 23 so the new table
has size 23 and the new hash
function is x mod 23
5.1.1 Insertion to Min-Heap
1
6
14
1
5
12
10
3
8
15
6
15
5
12
14
6
15 14 12
5
9
10
3
8
6
15 14 12
8
A new element is inserted to
the rightmost position of the
bottommost level. If some node
is larger than its
parent, swap them.
1
Repeat until the
heap property is
restored.
1
3
10
5
7
10
9
8
5.1.1 Insertion to Min-Heap
1
1
3
6
4
7
15 14 12
5
9
3
8
10
6
4
7
15 14 12
5
9
1
15 14 12
4
7
5
9
10 11
1
3
6
8
3
8
10 11 13
6
15 14 12
2
7
5
9
4
10 11 13
8
5.1.2 Build Heap
10
Initially the keys
are inserted from
left to right, top to
12
bottom
14
15
3
10
1
6
9
5
7
4
12
8
11 13
3
2
15 14
1
6
9
4
7
10
15 14
1
6
9
4
7
2
11 13
8
1
3
12
5
After the first 4
Percolate_Down
operations, these
become 4 heaps
5
3
2
11 13
After 2 more Percolate_Down,
these become 2 heaps
12
8
15 14
2
6
9
4
7
5
8
11 13 10
The heap is built after the last
Percolate_Down
5.1.1 Deletion in Min-Heap
1
2
3
6
2
7
15 14 12
5
9
3
4
10 11 13
6
8
4
7
15 14 12
5
9
3
15 14 12
4
7
5
9
10 11 13
4
6
13
8
10 11
6
8
13
15 14 12
5
7
10
9
11
8
6.1 Inversions



6.1.1 An O(n log n) time algorithm (modified from
merge sort) for finding the number of inversions
can be found here
(you don’t need to know it)
6.1.2 #inversion is maximized when the
sequence is decreasing.
So Max#inversion = (n-1)n/2
6.1.3 No, it depends on the algorithm you use. If
you use selection sort, the times taken for
sorting A and B are the same.
6.2.1 Bubble Sort
Original
3
1
4
1
5
9
2
6
5
P=1
1
3
1
4
5
2
6
5
9
P=2
1
1
3
4
2
5
5
6
9
P=3
1
1
3
2
4
5
5
6
9
P=4
1
1
2
3
4
5
5
6
9
6.2.1 Insertion Sort
Original
P=1
P=2
P=3
P=4
P=5
P=6
P=7
P=8
3
1
1
1
1
1
1
1
1
1
3
3
1
1
1
1
1
1
4
4
4
3
3
3
2
2
2
1
1
1
4
4
4
3
3
3
5
5
5
5
5
5
4
4
4
9
9
9
9
9
9
5
5
5
2
2
2
2
2
2
9
6
5
6
6
6
6
6
6
6
9
6
5
5
5
5
5
5
5
5
9
6.6.3 Quick Sort
1
2
3
4
5
6
7
8
9
10
11
m
n
3
3
1
1
1
1
1
1
1
1
1
1
1
1
4
3
3
3
3
3
3
1
1
3
3
3
3
3
5
3
3
3
3
3
3
9
9
9
9
5
5
4
3
5
5
5
5
5
5
6
6
6
6
6
4
5
5
5
5
5
5
5
5
3
4
4
4
4
6
6
5
5
5
5
9
9
9
1
11
1
4
1
3
6
11
6
10
6
8
10
10
6.6.3 Quick Sort


If different implementation is used, the
intermediate results may be different.
As long as the pivot is chosen according o the
requirement and all elements < pivot are on its
left and all elements > pivot are on its right
(equal elements can be on either side), the
answer will be accepted.
7.1 Topological Sort
A
s
B
D
C
E
A
F
t
B
D
C
E
G
H
I
G
H
I
A
B
C
A
B
C
D
E
H
F
I
t
E
H
Colored vertices have zero indegrees and are inserted into a queue.
At the beginning of each iteration, one vertex is dequeued. This vertex and all its
incident edges are removed from the graph.
I
F
t
F
t
7.1 Topological Sort
A
B
C
E
B
F
t
C
E
F
I
C
I
t
t
E
I
C
F
C
F
I
C
F
t
t
t
t
7.3.1 Single Source Shortest Path
Initial State
A known
C known
B known
v
Known
dv
pv
Known
dv
pv
Known
dv
pv
Known
dv
pv
A
0
0
0
1
0
0
1
0
0
1
0
0
B
0
∞
0
0
5
A
0
5
A
1
5
A
C
0
∞
0
0
3
A
1
3
A
1
3
A
D
0
∞
0
0
∞
0
0
10
C
0
10
C
E
0
∞
0
0
∞
0
0
10
C
0
8
B
F
0
∞
0
0
∞
0
0
∞
0
0
∞
0
G
0
∞
0
0
∞
0
0
∞
0
0
6
B
7.3.1 Single Source Shortest Path (Dijkstra’s
Algorithm
G known
E known
F known
D known
v
Known
dv
pv
Known
dv
pv
Known
dv
pv
Known
dv
pv
A
1
0
0
1
0
0
1
0
0
1
0
0
B
1
5
A
1
5
A
1
5
A
1
5
A
C
1
3
A
1
3
A
1
3
A
1
3
A
D
0
10
C
0
∞
E
0
10
E
1
10
E
E
0
7
G
1
∞
G
1
10
G
1
8
G
F
0
∞
0
0
∞
E
1
∞
E
1
∞
E
G
1
6
B
1
∞
B
1
∞
B
1
6
B
7.3.1 Single Source Shortest Path
∞
1
B
2
0
3
2
7
2
E
6
1
B
G
2
0
3
2
7
10 D
∞
∞
7
2
6
2
E
1
F ∞
∞
1
∞
3
C
7
2
E
1
6
F ∞
1
B
G
6
1
3
5
2
0
A
G
7
D
5
10
3
C
3
A
1
B
2
0
F ∞
1
3
5
1
3
5
A
∞
C
∞ D
5
∞
7
5
1
3
5
A
G
∞
3
2
7
10 D
8
3
C
7
2
6
E
1
F ∞
7.3.1 Single Source Shortest Path
1
5
B
2
0
3
A
2
7
10 D
7
2
B
G
2
0
3
2
9 D
1
F 8
7
2
E
1
6
F 8
B
G
6
1
3
5
2
0
A
1
1
5
E
6
7
3
C
7
9 D
1
7
2
6
2
7
3
C
7
3
6
3
5
2
A
F ∞
1
5
E
G
3
0
1
6
B
5
7
3
C
1
5
1
3
5
A
G
6
3
2
7
9 D
7
3
C
7
2
6
E
1
F 8
7.3.2 Unweighted Shortest Path
Initial State
v
Known
A
0
B
0
C
dv
B dequeued
pv
Known
0
0
0
1
0
0
0
D
0
0
0
E
0
0
0
F
0
0
0
G
0
0
0
Q
B
0
C,E,G
dv
C dequeued
pv
Known
0
0
0
0
1
1
B
1
1
pv
Known
0
0
0
0
1
0
0
1
1
B
1
1
B
0
0
2
C
0
2
C
B
0
1
B
1
1
B
0
0
0
0
2
E
B
0
B
0
1
B
E,G,D
dv
E dequeued
1
G,D,F
dv
pv
0
7.3.2 Unweighted Shortest Path
G dequeued
v
Known
A
0
B
1
C
dv
D dequeued
F dequeued
A dequeued
pv
Known
dv
pv
Known
dv
pv
Known
dv
pv
0
0
3
D
0
3
D
1
3
D
0
0
1
0
0
1
0
0
1
0
0
1
1
B
1
1
B
1
1
B
1
1
B
D
0
2
C
1
2
C
1
2
C
1
2
C
E
1
1
B
1
1
B
1
1
B
1
1
B
F
0
2
E
0
2
E
1
2
E
1
2
E
G
1
1
B
1
1
B
1
1
B
1
1
B
Q
D,F
F,A
A
7.3.2 Unweighted Shortest Path
Initial
state
∞
∞
A
∞
C
After C is
dequeued
E
D
F ∞
0 B
G 1
∞
∞
1
A
D
∞
G 1
1
A
1
C
After E is
dequeued
E
F ∞
0 B
E
D
F ∞
0 B
G 1
∞
1
C
2
After B is
dequeued
G ∞
0 B
∞
1
A
1
C
2
D
E
F 2
7.3.2 Unweighted Shortest Path
After G is
dequeued
0 B
∞
1
A
1
C
After F is
dequeued
E
D
F 2
0 B
G 1
2
3
1
A
D
3
G 1
1
A
1
C
After A is
dequeued
E
F 2
0 B
E
D
F 2
0 B
G 2
2
1
C
2
After D is
dequeued
G 1
3
1
A
1
C
2
D
E
F 2
7.5 Prim’s
3
A
4
4
5
D
6
2
11
2
1
4
4
4
5
6
H
10
3
11
1
4
2
7
2
2
4
5
D
8
6
H
C
A
2
11
G
8
J
2
11
1
4
5
6
H
4
2
11
11
8
C
6
1
2
F
3
I
G
J
10
3
1
1
7
2
2
6
3
B
E
C
F
I
3
4
D
3
E
4
10
B
2
J
1
F
7
4
G
6
3
I
1
11
3
A
6
3
B
E
C
F
I
3
A
D
3
E
H
10
B
11
7
G
8
J
7.5 Prim’s
3
A
4
4
5
D
6
2
11
2
1
4
4
4
5
6
H
10
3
11
1
4
2
7
2
2
4
5
D
8
6
H
C
A
2
11
G
8
J
2
11
1
4
5
6
H
4
2
11
11
8
C
6
1
2
F
3
I
G
J
10
3
1
1
7
2
2
6
3
B
E
C
F
I
3
4
D
3
E
4
10
B
2
J
1
F
7
4
G
6
3
I
1
11
3
A
6
3
B
E
C
F
I
3
A
D
3
E
H
10
B
11
7
G
8
J
7.5 Kruskal’s
3
A
4
4
5
D
6
2
11
2
1
4
4
4
5
6
H
10
3
11
1
4
2
7
2
2
4
5
D
8
6
H
C
A
2
11
G
8
J
2
11
1
4
5
6
H
4
2
11
11
8
C
6
1
2
F
3
I
G
J
10
3
1
1
7
2
2
6
3
B
E
C
F
I
3
4
D
3
E
4
10
B
2
J
1
F
7
4
G
6
3
I
1
11
3
A
6
3
B
E
C
F
I
3
A
D
3
E
H
10
B
11
7
G
8
J
7.5 Kruskal’s
3
A
4
4
5
D
6
2
11
2
1
4
4
4
5
6
H
10
3
11
1
4
2
7
2
2
4
5
D
8
6
H
C
A
2
11
G
8
J
2
11
1
4
5
6
H
4
2
11
11
8
C
6
1
2
F
3
I
G
J
10
3
1
1
7
2
2
6
3
B
E
C
F
I
3
4
D
3
E
4
10
B
2
J
1
F
7
4
G
6
3
I
1
11
3
A
6
3
B
E
C
F
I
3
A
D
3
E
H
10
B
11
7
G
8
J