CMPS 10 Winter 2011- Homework Assignment 3 Chapter 2 (p.76 ...

6.) code_dist = dist + k
7.) if ( code_dist < 26 )
8.) Codei = ‘a’ + code_dist
9.) else
10.) Codei = ‘a’ + code_dist – 26
11.) set i to i+1
12.) print Code
13.) STOP
Explanation: This algorithm assumes that alphabets ‘a’ –‘z’ have consecutive values
assigned to them, for example, as in ASCII code. (see page 141 in Text, 5 th ed). Thus ‘b’
=’a’+1, ‘c’ = ‘a’ + 2 so on. Each character has a fixed distance from ‘a’. In step 5, we
compute the distance of ith letter in text from ‘a’. In step 6, we add k to this value to get
distance of coded letter. If this code_dist is less then 26, we can get the character by adding it
to ‘a’ as in step 8. If code_dist is greater than 26, we wrap around. A code_dist of 27 would
actually be distance 1.
22. Design and implement an algorithm that is given as input an integer value k 0
and a list of k numbers N1, N2, … , Nk. Your algorithm should reverse the order of the
numbers in the list. That is, if the original list contained:
N1 = 5, N2 = 13, N3 = 8, N4 = 27, N5 = 10 (k = 5)
Then when your algorithm has completed, the values stored in the list will be:
N1 = 10, N2 = 27, N3 = 8, N4 = 13, N5 = 5.
Solution:
There are several possible solutions to this problem. Here is one possible algorithm:
Step Operation
1.) Input k and N1 … Nk
2.) Create and set i to 1, and create Temp
3.) Repeat steps 4, 5, 6, and 7 while (i k/2) (See note in explanation)
4.) Set Temp to Ni
5.) Set Ni to N(k + 1 – i)
6.) Set N(k + 1 – i) to Temp
7.) Set i to i + 1
8.) STOP
Explanation: This algorithm reverses the list in place by utilizing a temporary storage
variable. One could replace lines 4-6 with a line saying “Swap Ni and N(k + 1 – i)” if an
appropriate “swap” algorithm were available. In step 3, we assume integer division for
“k/2,” which truncates the answer. For example, if k = 5, we get 5/2 = 2. Therefore, again
using k = 5, we swap N1 with N(5 + 1 – 1) (or N5) and we swap N2 with N(5 + 1 – 2) (or N4). Due
to integer division, we stop at this point and leave N3 alone, which is correct. If we are using
real number division, the algorithm will attempt to swap N3 with N3. This does not change