Description
Requirements:
1. a. Write a C program to read two ordered integer input sequences and then a sequence of ranks (queries). For each rank
you should trace the binary search (one line per “probe”) and then indicate which element of which sequence has the
desired rank. The first line of the input file will give m, n, and p where m is the number of elements in the first sequence,
n is the number of elements in the second sequence, and p is the number of ranks (queries) for which binary searches will
be performed. The integer input sequence elements will be in the range 0 . . . 999,999, inclusive. The ranks will be in the
range 1 . . . m + n.
b. The output line for each time probe should include the values of low, high, i, and j. (Note that i serves the
role of “mid” for the binary searches.). You may also include other debugging information, such as the result of the probe.
2. Submit your C program on Canvas by 5:00 pm on September 21. Comments at the beginning of the source file should
include: your name, your ID number, and the command used to compile your code on Omega (5 point penalty for noncompliance).
Getting Started:
1. Your program should read the input files via Linux shell redirection (e.g. a.out < lab1.dat on omega.uta.edu,
https://en.wikipedia.org/wiki/Standard_streams). Do NOT prompt for file names! You should
dynamically allocate tables for storing the input. To simplify the binary search code, it is recommended that the first
sequence be stored in subscripts 1 . . . m of its table (a) and that subscripts 0 and m + 1 store “low” and “high” sentinel
values, respectively. The second sequence (b) would be stored similarly.
2. The rank of an element in one of the two sequences is the index of the position that it would occupy if the two sequences
were merged into one monotonically increasing sequence (starting at subscript 1 of the output sequence). Should equalvalued elements appear, those in the first sequence are copied to the output before those in the second sequence. The
following (linear-time) code will compute (and store) the ranks for all elements if stored as suggested in (1.):
void ranksByMerge()
{
int i,j,k;
i=j=k=1;
while (i<=m && j<=n)
if (a[i]<=b[j])
aRank[i++]=k++;
else
bRank[j++]=k++;
while (i<=m)
aRank[i++]=k++;
while (j<=n)
bRank[j++]=k++;
}
These tables of ranks will allow testing your code in an exhaustive fashion, e.g. for all ranks in the range 1 . . . m + n.
The aRank and bRank tables should NOT be computed in the version you submit.
Θ(1)
3. For a given rank, the corresponding element could be in either of the two input sequences. If stored as suggested in (1.),
the following (symmetric) observations allow a binary search to be used:
a. If the corresponding element is at a[i], then there must be an index j such that all of the following hold:
1. i + j == rank,
2. b[j] < a[i], and
3. a[i] <= b[j+1]
b. If the corresponding element is at b[j], then there must be an index i such that all of the following hold:
1. i + j == rank,
2. a[i] <= b[j], and
3. b[j] < a[i+1]
So, a probe consists of: a) compute i as the middle of the search range of the first table, b) determine whether 3.a.2 or
3.b.2 applies, and c) determine whether the desired element has been found or make an appropriate adjustment in the
search range. (Technically, there is also a search range for the second table that follows from the i + j == rank
notion.)
4. Be careful in initializing the search range of the first table (e.g. low and high). Otherwise, slots outside of the tables
could be referenced (leading to a segmentation fault).
5. Recursion should not be used.
i a[i] aRank[i]
0 -99999999 0
1 1 2
2 1 3
3 1 4
4 1 5
5 2 7
6 5 12
7 6 14
8 8 17
9 8 18
10 9 22
11 99999999 0
j b[j] bRank[j]
0 -99999999 0
1 0 1
2 1 6
3 2 8
4 3 9
5 4 10
6 4 11
7 5 13
8 6 15
9 7 16
10 8 19
11 8 20
12 8 21
13 9 23
14 9 24
15 9 25
16 99999999 0
