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Julli [10]
3 years ago
15

If a schema is not given, you may assume a table structure that makes sense in the context of the question. (using sql queries)

Computers and Technology
1 answer:
adell [148]3 years ago
6 0

Answer:

The correct query is;

select * from EMPLOYEE where Employee_Name = 'JOE' or Employee_Name = 'Joe' or Employee_Name = 'joe';

where EMPLOYEE refer to the table name and type attribute name is Employee_Name

Explanation:

Here, the first thing we will do is to assume the name of the table.

Let’s assume the table name is EMPLOYEE, where the column i.e attribute from which we will be extracting our information is Employee_Name

The correct query to get the piece of information we are looking for will be;

select * from EMPLOYEE where Employee_Name = 'JOE' or Employee_Name = 'Joe' or Employee_Name = 'joe';

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5 0
3 years ago
find_cow(name, cows) Given a name and a Python list of Cow objects, return the Cow object with the specified name. If no such Co
nadya68 [22]

Using the computational knowledge in python it is possible to write a code that Given a name and a Python list of Cow objects

<h3>Writting the code in python:</h3>

<em>def load_cows(filename):</em>

<em>      cow_dict = dict()</em>

<em>    f = open(filename, 'r')</em>

<em> </em>

<em>    for line in f:</em>

<em>        line_data = line.split(',')</em>

<em>        cow_dict[line_data[0]] = int(line_data[1])</em>

<em>    return cow_dict</em>

<em>def greedy_cow_transport(cows, limit=10):</em>

<em>       trips = []</em>

<em>    cowsCopy = cows.copy()</em>

<em>    sortedCows = sorted(cowsCopy.items(), key=lambda x: x[1], reverse = True)</em>

<em>    while sum(cowsCopy.values()) > 0:</em>

<em>        ship = []</em>

<em>        total = 0</em>

<em>        for cow, value in sortedCows:</em>

<em>            if cowsCopy[cow] != 0 and value + total <= limit:</em>

<em>                ship.append(cow)</em>

<em>                total += value</em>

<em>                cowsCopy[cow] = 0</em>

<em>        trips.append(ship)</em>

<em>    return trips</em>

<em>def brute_force_cow_transport(cows,limit=10):</em>

<em>     trips = []</em>

<em>     possibilities = []</em>

<em>    for i in power_list:</em>

<em>        ship = []</em>

<em>        for j in i:</em>

<em>            ship_weights = []</em>

<em>            for k in j:</em>

<em>                ship_weights.append(cows[k])</em>

<em>                #print(ship_weights)</em>

<em>            ship.append(sum(ship_weights))</em>

<em>            #print(ship)</em>

<em>        if all(d <= limit for d in ship):</em>

<em>            possibilities.append(i)</em>

<em>      pruned_possibilities = []</em>

<em>    for k in possibilities:</em>

<em>        if k not in pruned_possibilities:</em>

<em>            pruned_possibilities.append(k)</em>

<em>    # now find the minimum list length:</em>

<em>    min_list_len = min(map(len, pruned_possibilities))</em>

<em>    for l in pruned_possibilities:</em>

<em>        if len(l) == min_list_len:</em>

<em>            return l</em>

<em>       </em>

<em>def compare_cow_transport_algorithms():</em>

<em>       greedy_start = time.time()</em>

<em>    greedy_results = greedy_cow_transport(cows, limit = 10)</em>

<em>    greedy_end = time.time()</em>

<em>    print('Greedy Algorithm time:', greedy_end -greedy_start)</em>

<em>    brute_force_start = time.time()</em>

<em>    brute_force_results = brute_force_cow_transport(cows, limit = 10)</em>

<em>    brute_force_end = time.time()</em>

<em>    print('Brute force time:', brute_force_end - brute_force_start)</em>

<em>    print('Greedy Algorithm results:', greedy_results)</em>

<em>    print('Number of trips returned by Greedy Algorithm:', len(greedy_results))</em>

<em>    print('Brute Force Algorithm results:', brute_force_results)</em>

<em>    print('Number of trips returned by Brute Force Algorithm:', len(brute_force_results))</em>

   

See more about python at brainly.com/question/13437928

#SPJ1

4 0
2 years ago
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