Other examples: The Travelling
Salesperson Problem
Given a set of
locations V and a matrix of distances dij between each
pair of locations i and j in V, is it possible to visit
all locations exactly once and then return to your starting point while
covering a total distance less than k.
You may assume,
without loss of generality, that you can start from any one of the locations
(it actually makes no difference).
● Think about how to solve a yes instance
● Think about how to solve a no instance
To solve a
"yes" instance of the problem, where it is possible to visit all
locations exactly once and then return to the starting point while covering a
total distance less than k, we can use the following algorithm:
1.
Initialize
a list of visited locations and a variable total_distance to 0.
2.
Iterate
through the locations in V.
3.
At each
location, mark it as visited and add the distance from the previous location to
the total distance.
4.
If the
total distance is greater than or equal to k, return False.
5.
If all
locations have been visited, return True.
This algorithm has
a time complexity of O(n), where n is the number of locations in V, because it
requires one pass through the list of locations.
To solve a
"no" instance of the problem, where it is not possible to visit all
locations exactly once and then return to the starting point while covering a
total distance less than k, we can use the following algorithm:
1.
Initialize
a list of visited locations and a variable total_distance to 0.
2.
Iterate
through the locations in V.
3.
At each
location, mark it as visited and add the distance from the previous location to
the total distance.
4.
If the
total distance is greater than or equal to k, return False.
5.
If all
locations have been visited and the total distance is less than k, return True.
This algorithm has
a time complexity of O(n), where n is the number of locations in V, because it
requires one pass through the list of locations.
The Python code for the "yes" instance algorithm
with detailed line-by-line comments:
# Initialize list
of visited locations and variable total_distance to 0
visited = []
total_distance = 0
# Iterate through
locations in V
for i in V:
# Mark location as visited
visited.append(i)
# Add distance from previous location to
total distance
total_distance += dij[i][visited[-2]]
# If total distance is greater than or
equal to k, return False
if total_distance >= k:
return False
# If all locations
have been visited, return True
return True
the Python code for the "no" instance algorithm:
# Initialize list
of visited locations and variable total_distance to 0
visited = []
total_distance = 0
# Iterate through
locations in V
for i in V:
# Mark location as visited
visited.append(i)
# Add distance from previous location to
total distance
total_distance += dij[i][visited[-2]]
# If total distance is greater than or
equal to k, return False
if total_distance >= k:
return False
# If all locations
have been visited and total distance is less than k, return True
if len(visited) ==
len(V) and total_distance < k:
return True
else:
return False
Another example
Consider this
problem:
Find out if a
Boolean expression in disjunctive normal form (DNF) is a tautology.
● A tautology is an expression that evaluates
to true for all combinations of inputs
○ E.g. (a ∨ ¬a) ∧ (b
∨ ¬b)
● Disjunctive normal form is an or of ands
○ e.g. (a ∧ ¬b ∧ c)
∨ (a
∧ b ∧ ¬c) ∨ …
Think about yes and
no instances again - is it the same as the previous examples?
The problem of determining whether a Boolean expression in
disjunctive normal form (DNF) is a tautology is similar to the previous
examples of evaluating "yes" and "no" instances. A
"yes" instance would be a tautology that evaluates to true for all
combinations of inputs, while a "no" instance would be a Boolean
expression that does not evaluate to true for all combinations of inputs.
To solve a "yes" instance of the problem, we can
iterate through all possible combinations of inputs and check if the Boolean
expression evaluates to true for each combination. If the expression evaluates
to true for all combinations, we can conclude that it is a tautology.
To solve a "no" instance of the problem, we can
also iterate through all possible combinations of inputs and check if the
Boolean expression evaluates to true for each combination. If the expression
does not evaluate to true for at least one combination, we can conclude that it
is not a tautology.
Both solutions have a time complexity of O(2^n), where n is
the number of variables in the Boolean expression, because they require
checking all possible combinations of inputs.
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