Smart City
ADA midterm - 3
(a) : normal
DP[n][m] =
min(
DP[n-1][m]+W(n-1,m,(+1,0),DP[n-1][m]),
DP[n][m-1]+W(n,m-1,(0,+1),DP[n][m-1])
)
DP[n][m] =
1+min(
DP[n-1][m]+W(n-1,m,(+1,0),DP[n-1][m]),
DP[n][m-1]+W(n,m-1,(0,+1),DP[n][m-1])
)
(if n > 1)
(if m > 1)
Base Case
DP[0][0] = 0
Answer
DP[n][m]
(b) : violate no more than k tickets
DP[n][m][k] =
min(
DP[n-1][m][k]+W(n-1,m,(+1,0),DP[n-1][m]),
DP[n][m-1][k]+W(n,m-1,(0,+1),DP[n][m-1]),
DP[n][m-1][k-1], DP[n-1][m][k-1],
)
DP[n][m][k] =
1+min(
DP[n-1][m][k]+W(n-1,m,(+1,0),DP[n-1][m]),
DP[n][m-1][k]+W(n,m-1,(0,+1),DP[n][m-1]),
DP[n][m-1][k-1], DP[n-1][m][k-1],
)
(if n > 1)
(if m > 1)
(if m > 1 and k > 1)
(if n > 1 and k > 1)
Base Case
DP[0][0][*] = 0
Answer
DP[n][m][k]
Grading Criteria
(c) : synchronized traffic lights with greedy choice
According to random d if all green ...
Counter-example
| Time | ||
|---|---|---|
| 0 | 0 | 0 |
| 1 | 100 | 0 |
| 2 | 99 | 0 |
| 3 | 98 | 0 |
| 4 | 97 | 0 |
| 5 | 96 | 0 |
| ... | ... | ... |
Grading Criteria
Misunderstanding
Common flaws
2018 ada miterm 3
By Arvin Liu
2018 ada miterm 3
2018 ADA Midterm Prob 3's Grading Criteria
