They should be between 1 and 1/4 inches
Answer:
The encoding algorithm looks for pairs of characters that appear in the string more than once and replaces each instance of that pair with a corresponding character that does not appear in the string. ... Byte pair encoding is an example of a lossy transformation because it discards some of the data in the original string.
Explanation:
hope it helps!!
<span>In order to perform maintenance on a router and need to temporarily reroute traffic through another office the best way to perform this action would be to configure a static route on the router.
</span>By doing this, the router will use the manually-<span>configured routing entry to send the packets.</span>
Answer:
In C++:
int PrintInBinary(int num){
if (num == 0)
return 0;
else
return (num % 2 + 10 * PrintInBinary(num / 2));
}
Explanation:
This defines the PrintInBinary function
int PrintInBinary(int num){
This returns 0 is num is 0 or num has been reduced to 0
<em> if (num == 0) </em>
<em> return 0; </em>
If otherwise, see below for further explanation
<em> else
</em>
<em> return (num % 2 + 10 * PrintInBinary(num / 2));
</em>
}
----------------------------------------------------------------------------------------
num % 2 + 10 * PrintInBinary(num / 2)
The above can be split into:
num % 2 and + 10 * PrintInBinary(num / 2)
Assume num is 35.
num % 2 = 1
10 * PrintInBinary(num / 2) => 10 * PrintInBinary(17)
17 will be passed to the function (recursively).
This process will continue until num is 0
For the view of Exercise 4.18, explain why the database system would not allow a tuple to be inserted into the database through this view.
For reference
For the database of Figure 4.12, write a query to find the ID of each employee with no manager. Note that an employee may simply have no manager listed or may have a null manager. Write your query using an outer join and then write it again using no outer join at all.
OUTER JOIN
select e.ID from employee e left outer join manages m on e.ID = m.ID
where m.manager_id is null;
NO OUTER
select e.ID from employee e where e.ID not in (select m.ID from manages m) or e.ID in (select m.ID from manages m where m.manager_id is null);
