Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 3 docx
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 3 docx
... sort on 3 elements:
≤ >
≤ >
1:2
2 :3 1 :3
〈1,2 ,3
1 :3
〈2,1 ,3
2 :3
〈1 ,3, 2〉 3, 1,2〉 3, 2,1〉
≤ >
≤ >
≤ >
〈2 ,3, 1〉
A[1] ≤ A[2] A[1] > A[2] (swap in array)
A[1] ≤ A[2]
A[2] > A [3]
A[1] ... Exercise B.5 -3) , # of internal nodes =
# of leaves − 1.
Lecture Notes for Chapter 7: Quicksort 7 -3
8164 039 5
p,j ri
8164 039 5
prj
1864 039 5
p,i rj
1864 039 5
p,...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 1 pptx
... in the
“cost” and “times” columns later.]
Example:
1 234 56
5246 13
1 234 56
2546 13
1 234 56
2456 13
1 234 56
2456 13
1 234 56
2456 13
1 234 56
2 4 5 61 3
j jj
jj
[Read this Þgure row by row. Each part shows ... 20 03. Corrected an error in the solution to Exercise 11 .3- 3. Affected
chapters: Chapter 11.
•
3 April 20 03. Reversed the order of Exercises 14.2 -3 and 14 .3- 3. Affe...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 2 pps
... n.
Substitution:
T (n) ≤ T (n /3) + T (2n /3) + cn
≤ d(n /3) lg(n /3) +d(2n /3) lg(2n /3) + cn
= (d(n /3) lg n − d(n /3) lg 3)
+ (d(2n /3) lg n −d(2n
/3) lg (3/ 2)) + cn
= dn lg n − d((n /3) lg 3 +(2n /3) lg (3/ 2)) + cn
= ... 2, 3 4/27
1, 3, 2 5/27
2, 1, 3 5/27
2, 3, 1 5/27
3, 1, 2 4/27
3, 2, 1 4/27
6-6 Lecture Notes for Chapter 6: Heapsort
(a) (b)
(c) (d)
(e)
1 234 7
2
13...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 4 pot
... pointer) to the new slot, and updating the pointer in the slot
that pointed to j to point to the new slot. Then insert the new element in
the now-empty slot as usual.
To update the pointer to j , ... lists. If we want to delete elements, it’s better to
use doubly linked lists.]
•
Slot j contains a pointer to the head of the list of all stored elements that hash
to j
[or t...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 5 potx
... hypothesis)
=
1
n
n−1
i=0
i + 3
3
=
1
n
n + 3
4
(lemma)
=
1
n
·
(n + 3) !
4! (n − 1)!
=
1
4
·
(n + 3) !
3! n!
=
1
4
n + 3
3
.
Thus, we’ve proven that E
[
Y
n
]
≤
1
4
n + 3
3
.
Bounding E
[
X
n
]
With ... keys are not distinct, because in order to Þnd the
path to the node to delete—a particular node with a given key—we have to
make some changes to how we...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 6 pps
... Solutions for Chapter 15: Dynamic Programming
= n
3
− n
2
−
2n
3
− 3n
2
+ n
3
=
n
3
− n
3
.
Solution to Exercise 15 .3- 1
Running RECURSIVE-MATRIX-CHAIN is asymptotically more efÞcient than enu-
merating ... Programming
a
1,1
e
1
enter
e
2
a
2,1
a
1,2
a
2,2
t
1,1
t
2,1
t
1,2
t
2,2
a
1 ,3
a
2 ,3
t
1 ,3
t
2 ,3
a
1,4
a
2,4
t
1,4
t
2,4
a
1,5
a
2,5
x
1
x
2
exit
S
1,1
S
2,1
S
1,2
S...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 7 ppsx
... (size
i−1
/2 − num
i−1
)
= 3 ·num
i−1
−
3
2
· size
i−1
+3
= 3 ·α
i−1
size
i−1
−
3
2
· size
i−1
+3
<
3
2
· size
i−1
−
3
2
· size
i−1
+3
= 3 .
Therefore, amortized cost of insert is < 3.
Delete:
•
If ... to go to get to 1 or 1/4, starting from 1/2, rate
of increase of differs.
•
For α to go from 1/ 2to1 ,num increases from size /2tosize, for a total
increas...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 8 potx
... point to
the second element of A.
Solution to Exercise 21 .3- 3
You need to Þnd a sequence of m operations on n elements that takes (m lg n)
time. Start with n MAKE-SETs to create singleton sets
{
x
1
}
,
{
x
2
}
, ... edge it visits to the cycle C, which is then added to the Euler
tour T , when we return from a call to VISIT(u), all edges entering or leaving
vertex u have be...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 9 pdf
... spanning tree of weight 7
and two second- best minimum spanning trees of weight 8:
1
24
35
minimum
spanning tree
1
24
35
second- best
minimum
spanning tree
1
24
35
second- best
minimum
spanning tree
b. ... for Chapter 23:
Minimum Spanning Trees
Solution to Exercise 23. 1-1
Theorem 23. 1 shows this.
Let A be the empty set and S be any set containing u but not v.
Solution to Exercis...
Introduction to Algorithms Second Edition Instructor’s Manual 2nd phần 10 potx
... 23. 1-1, 23- 8
Exercise 23. 1-4, 23- 8
Exercise 23. 1-6, 23- 8
Exercise 23. 1-10, 23- 9
Exercise 23. 2-4, 23- 9
Exercise 23. 2-5, 23- 9
Exercise 23. 2-7, 23- 10
Exercise 24.1 -3, 24- 13
Exercise 24.2 -3, 24- 13
Exercise ... additional unit along p:
[Continue from
G
left on board from
before.]
3
1
1
3
2
3
1
1
1 4
2
3
G
f
3/ 3
1/2
3/ 3
2/2
1
2 /3
1/1
2/2
3
2 /3
2
G
s t
w
y
x
z
s...
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