Asksia AI LOGO

Sia

Question
Computer Science
Posted 4 months ago
Given a set of N coins C 
=
 
{
c
1
,
 c
2
,
 c
3
,
 
.
.
.
,
 cN
}
 and a monetary value M 
>
 
0
,
 find the smallest subset of C that can pay value K 
(
without change
)
.

Note that in this problem, C can have duplicates, and a coin can be used only once.

Examples. C 
=
 
{
1
,
 
1
,
 
4
,
 
2
0
,
 
2
0
,
 
2
1
,
 
2
5
,
 
4
0
}
 

 M 
=
 
4
5
  Answer 
=
 
{
2
0
,
 
2
5
}
M 
=
 
1
3
2
 Answer 
=
 
{
1
,
 
1
,
 
4
,
 
2
0
,
 
2
0
,
 
2
1
,
 
2
5
,
 
4
0
}
 M 
=
 
3
  Answer 
=
 No subset meets the requirements

Design a backtracking algorithm to solve this problem. Follow the steps below:

A
.
 Draw a tree showing how a brute
-
force enumeration algorithm enumerates the solutions for the following instance: C 
=
 
{
2
,
 
2
,
 
4
}
 M 
=
 
4
.

Clearly show on the tree what each node and edge represents and which solutions are feasible and which are not.

B
.
 Suggest a backtracking strategy to improve the running time of the brute
-
force algorithm.

C
.
 Provide an example of values for C 
{
at least 
4
}
 and M that will lead to:

 
1
.
 No pruning in the tree.

 
2
.
 Maximum pruning in the tree.
Sign in to unlock the answer
Answer from Sia
Posted 4 months ago
 A. Draw a tree showing how a brute-force enumeration algorithm enumerates the solutions for the following instance: C = {2, 2, 4} M = 4.

Explanation
The tree will represent all possible combinations of coins from the set C to sum up to the monetary value M. Each node represents a state with a particular subset of coins, and each edge represents the decision to include or not include a coin in the subset. Feasible solutions are those subsets whose sum equals M, and infeasible solutions are those that do not.
Step-by-step Instruction
Draw branches to represent the decision to include or exclude the next coin
Mark nodes with subsets that sum up to M as feasible solutions
Mark nodes that exceed M or cannot reach M with the remaining coins as infeasible
Time Complexity
The brute-force algorithm has a time complexity of O(2^N), where N is the number of coins, as it explores every possible combination.
Space Complexity
The space complexity is O(N) due to the depth of the recursion tree.


 B. Suggest a backtracking strategy to improve the running time of the brute-force algorithm.

Explanation
Backtracking is an optimization over the brute-force approach that involves making decisions one at a time and undoing these decisions (backtracking) when we determine that they lead to infeasible solutions.
Step-by-step Instruction
Recursively try to include each coin in the subset
After including a coin, decrease the target sum by the value of the coin
If the target sum becomes negative, backtrack and try the next possibility
If the target sum reaches zero, record the current subset as a solution
Continue exploring other possibilities by including or excluding the next coin
Time Complexity
The time complexity is still exponential in the worst case but is generally better than O(2^N) due to pruning of the search space.
Space Complexity
The space complexity remains O(N) due to the depth of the recursion tree.


 C. Provide an example of values for C {at least 4} and M that will lead to:

# 1. No pruning in the tree.

Explanation
A scenario with no pruning would occur when all subsets need to be explored to determine the solution, typically when M is larger than the sum of all coins in C.
Step-by-step Instruction
Ensure that no early decisions lead to a sum greater than M
Example: C = {1, 2, 3, 4} and M = 11
Time Complexity
The time complexity is O(2^N) as all subsets are explored.
Space Complexity
The space complexity is O(N) due to the depth of the recursion tree.


# 2. Maximum pruning in the tree.

Explanation
Maximum pruning occurs when the backtracking algorithm can eliminate many branches early in the search process, typically when there are larger coins that quickly exceed M.
Step-by-step Instruction
Include smaller coins that can combine in limited ways to reach M
Example: C = {1, 2, 5, 10} and M = 7
Time Complexity
The time complexity is better than O(2^N) due to early pruning.
Space Complexity
The space complexity is O(N) due to the depth of the recursion tree.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Upgrade to Asksia Pro

Join a AskSia's Pro Plan, and get 24/7 AI tutoring for your reviews, assignments, quizzes and exam preps.

Unlimited chat query usages
Strong algorithms that better know you
Early access to new release features
Study Other Question