Answer:
The solution is as follows.
class LFilters implements Lock {
int[] lvl;
int[] vic;
public LFilters(int n, int l) {
lvl = new int[max(n-l+1,0)];
vic = new int[max(n-l+1,0)];
for (int i = 0; i < n-l+1; i++) {
lvl[i] = 0;
}
}
public void lock() {
int me = ThreadID.get();
for (int i = 1; i < n-l+1; i++) { // attempt level i
lvl[me] = i;
vic[i] = me;
// rotate while conflicts exist
int above = l+1;
while (above > l && vic[i] == me) {
above = 0;
for (int k = 0; k < n; k++) {
if (lvl[k] >= i) above++;
}
}
}
}
public void unlock() {
int me = ThreadID.get();
lvl[me] = 0;
}
}
Explanation:
The code is presented above in which the a class is formed which has two variables, lvl and vic. It performs the operation of lock as indicated above.
<span>Validation of electronic signatures was designed to encourage a paperless society.</span>
Answer:
Since the language isn’t stated, I’ll give answers in the two most-used (?) languages: Java and Python.
a) To print a’s value 3 times in the same line, in Java we would do:
System.out.print(a+a+a);
In Python, we would write:
print(a*3)
b) 2 times in different lines using one print statement
In Java, we would write
System.out.println(a+”\n”+a+”\n”+a);
In Python we would write:
print(a,a,a,sep=’/n’)
Hope this helps!
Let P(n) be "a postage of n cents can be formed using 5-cent and 17-cent stamps if n is greater than 63".Basis step: P(64) is true since 64 cents postage can be formed with one 5-cent and one 17-cent stamp.Inductive step: Assume that P(n) is true, that is, postage of n cents can be formed using 5-cent and 17-cent stamps. We will show how to form postage of n + 1 cents. By the inductive hypothesis postage of n cents can be formed using 5-cent and 17-cent stamps. If this included a 17-cent stamp, replace this 17-cent stamp with two 5-cent stamps to obtain n + 1 cents postage. Otherwise, only 5-cent stamps were used and n 65. Hence there are at least three 5-cent stamps forming n cents. Remove three of these 5-cent stamps and replace them with two 17-cent stamps to obtain n + 1 cents postage.Hence P(n + 1) is true.
