Question

task b: how would you change the loop if the number of characters was chosen by the user instead of fixed at 5?

part 4: logic gate puzzle
look at this logic circuit: (a and b) or c
task a: fill in the truth table.
a | b | c | (a and b) | output ((a and b) or c)
0 | 0 | 0 |
0 | 0 | 1 |
0 | 1 | 0 |
0 | 1 | 1 |
1 | 0 | 0 |
1 | 0 | 1 |
1 | 1 | 0 |
1 | 1 | 1 |

task b: create a real-world analogy for this circuit.

part 5: solvability & performance reflection
an algorithm is designed to check every password in a list until it finds a match.
task a: is this algorithm solvable? why or why not?
task b: how does performance change as the number of passwords grows?

Explanation:

Answer:

Part 4: Logic Gate Puzzle - Task A (Truth Table)

A	B	C	(A AND B)	Output ((A AND B) OR C)
0	0	1	0	1
0	1	0	0	0
0	1	1	0	1
1	0	0	0	0
1	0	1	0	1
1	1	0	1	1
1	1	1	1	1

Part 4: Logic Gate Puzzle - Task B (Real - World Analogy)

A real - world analogy could be: Consider a party entrance system. Let A be "has an invitation", B be "is on the guest list", and C be "is a VIP". The output is "allowed entry". So, (A AND B) means a person has an invitation and is on the guest list, and (A AND B) OR C means a person is allowed entry if they have an invitation and are on the guest list OR they are a VIP.

Part 5: Solvability & Performance Reflection - Task A

Yes, this algorithm is solvable. Because it will eventually find a match (if it exists in the list) by checking each password one by one. If the password is in the list, the algorithm will stop when it reaches that password. If the password is not in the list, it will check all passwords (but in theory, it still completes the process of checking the entire list).

Part 5: Solvability & Performance Reflection - Task B

As the number of passwords in the list grows, the performance of the algorithm (in terms of the time it takes to find a match or determine there is no match) generally degrades. In the worst - case scenario (when the password is the last one in the list or not in the list at all), the algorithm has to check more passwords, so it will take longer to execute. The time complexity of this algorithm is linear with respect to the number of passwords in the list (O(n), where n is the number of passwords).