Answer:
I will code in Javascript.
Preconditions:
The variables k, incompletes has been declared and initialized.
The variable nIncompletes contains the number of elements of incompletes.
The variable stududentID has been declared and initialized.
The variable numberOfIncompletes has been declared.
Script:
numberOfIncompletes = 0; <em>//initialize numberOfIncompletes</em>
for( k = 0; k < nIncompletes; k++) {
if(incompletes[k] == studentID){ <em>//if the element of array is equal to studentID</em>
numberOfIncompletes++; <em>//adds 1 to numberOfIncompletes</em>
}
}
Explanation:
The script uses a for loop to go through the entire array incompletes, comparing the k variable with nIncompletes and adding 1 on each pass.
Then inside the for, uses an if statement to compare if the actual element is equal to studentID, if equal adds 1 numberOfIncompletes.
Answer:
3. (9 points) The goal of this problem is to show that our formal definitions of P and NP imply that P = NP cannot have a non-constructive proof. Throughout this problem we assume a consistent, binary, encoding of both Turing machines and 3-SAT instances. Note that the complexities are intentionally not tight to address potential overheads from scan- ning the tape linearly in order to look for certain symbols. (a) (3 points) Show that for any constant cı, there is a constant c2 such that if Mi is a Turing machine encoded by at most bits that correctly decides 3-SAT in O(n10) time (aka. 3-SAT E TIME(n10), then there is a Turing machine M, encoded by at most most c2 bits that runs in O(n100) time such that if the 3-SAT instance has a satisfying assignment, M, accepts and writes a satisfying assignment at the end of the tape. (b) (3 points) Let M ...Mbe Turing machines show that the language {(k, M... Mk,t,w) | w is an instance of 3-SAT and one of M, writes a satisfying assignment to w in at most t steps can be decided by a Turing machine in (k2107 10) time. (c) (3 points) Under the assumption that there exists an unknown Turing machine encodable in c bits that decides 3-SAT is in O(n10) time, give implementation details for a Turing Machine M that decides 3-SAT in O(n10000) time. Hint: the constant factor in the big-O is able to hide any function related ci and c2, which independent of input sizes, and thus constants.
Explanation:
3. (9 points) The goal of this problem is to show that our formal definitions of P and NP imply that P = NP cannot have a non-constructive proof. Throughout this problem we assume a consistent, binary, encoding of both Turing machines and 3-SAT instances. Note that the complexities are intentionally not tight to address potential overheads from scan- ning the tape linearly in order to look for certain symbols. (a) (3 points) Show that for any constant cı, there is a constant c2 such that if Mi is a Turing machine encoded by at most bits that correctly decides 3-SAT in O(n10) time (aka. 3-SAT E TIME(n10), then there is a Turing machine M, encoded by at most most c2 bits that runs in O(n100) time such that if the 3-SAT instance has a satisfying assignment, M, accepts and writes a satisfying assignment at the end of the tape. (b) (3 points) Let M ...Mbe Turing machines show that the language {(k, M... Mk,t,w) | w is an instance of 3-SAT and one of M, writes a satisfying assignment to w in at most t steps can be decided by a Turing machine in (k2107 10) time. (c) (3 points) Under the assumption that there exists an unknown Turing machine encodable in c bits that decides 3-SAT is in O(n10) time, give implementation details for a Turing Machine M that decides 3-SAT in O(n10000) time. Hint: the constant factor in the big-O is able to hide any function related ci and c2, which independent of input sizes, and thus constants.
Answer: responsive
Explanation:
its called mobile responsivness and most good websites have it
Answer:
Click DATA, Look for "Sort & Filter", Click Filter Box (right above "Sort & Filter"). Done.
