Cryptography Reference
In-Depth Information
10.4. Name three methodologies for a malicious hacker to break into a host
from the Internet. (See Section 10.2.)
10.5. Compare and contrast worms and viruses, and the mechanisms to protect
against them. (See Section 10.3.)
10.6. Review the Clipper Chip enciphering, deciphering, and law enforcement.
Decide on advantages and disadvantages of the Skipjack scheme, discussed
on pages 420-424.
G.11 Chapter 11 Exercises
11.1. Establish identity (11.4) on page 432.
11.2. Verify Inequality (11.5) on page 432.
11.3. Prove the equivalence of items 1-4 in the “Role of Independence” at the
bottom of page 432.
In Exercises 11.4-11.7, calculate the entropy for the set
,
where the probabilities for each s j is given in the exercises via p j for j =
1 , 2 , 3 , 4 . See pages 433-434.
S
=
{
s 1 ,s 2 ,s 3 ,s 4 }
11.4. ( p 1 ,p 2 ,p 3 ,p 4 )=(0 . 5 , 0 . 25 , 0 . 125 , 0 . 125).
11.5. ( p 1 ,p 2 ,p 3 ,p 4 )=(0 . 4 , 0 . 4 , 0 . 1 , 0 . 1).
11.6. ( p 1 ,p 2 ,p 3 ,p 4 )=(0 . 3 , 0 , 3 , 0 . 1 , 0 . 1).
11.7. ( p 1 ,p 2 ,p 3 ,p 4 )=(0 . 7 , 0 . 1 , 0 . 1 , 0 . 1).
11.8. Calculate the Huffman codes for each of the situations in Exercises 11.1-
11.4.
In Exercises 11.9-11.12, use the data given to calculate each of H (
K
) ,
H (
M
) , H (
C
) , and H (
K | C
) . Then compare the latter with the former three.
Assume that
M
=
{
s 1 ,s 2 ,s 3 ,s 4 }
and
C
=
{
c 1 ,c 2 ,c 3 ,c 4 }
. See pages 435-
436.
11.9. p s 1 =0 . 2, p s 2 =0 . 3, p S 3 =0 . 4, p s 4 =0 . 1,
with p k 1 =0 . 4,
p k 2 =0 . 6, E k 1 ( s 1 )= c 1 , E k 1 ( s 2 )= c 2 , E k 1 ( s 3 )= c 3 , E k 1 ( s 4 )= c 4 ,
E k 2 ( s 1 )= c 3 , E k 2 ( s 2 )= c 2 , E k 2 ( s 3 )= c 1 , and E k 2 ( s 4 )= c 4 .
K
=
{
k 1 ,k 2 }
11.10. p s 1 =0 . 1, p s 2 =0 . 1, p s 3 =0 . 5, p s 4 =0 . 3,
with
p k 1 =0 . 5, p k 2 =0 . 2, p k 3 =0 . 3, E k 1 ( s 1 )= c 1 , E k 1 ( s 2 )= c 2 , E k 1 ( s 3 )= c 3 ,
E k 1 ( s 4 )= c 4 , E k 2 ( s 1 )= c 2 , E k 2 ( s 2 )= c 3 , E k 2 ( s 3 )= c 4 , E k 2 ( s 4 )= c 1 ,
E k 3 ( s 1 )= c 3 , E k 3 ( s 2 )= c 4 , E k 3 ( s 3 )= c 2 , and E k 3 ( s 4 )= c 1 .
K
=
{
k 1 ,k 2 ,k 3 }
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