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CSCI113 Problem Solving Assignment

## Part 1

## Part 2

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**Consider the following statements:**

a) Animals that do not bite are always unexcitable.

b) Tigers have stripes.

c) A pangolin can hang by its tail.

d) Animals that bite are not easily tamed.

e) No animal with stripes can hang by its tail.

f) All animals are excitable, with the exception of pangolins. g) No animal can hang by its tail.

For the above problem, you should:

1. Identify the universal class.

The universal class is "animals".

2.Extract the terms and assign letters to them, ie. Notation.

NB: Symbols declared in this section MUST be the same symbols used for questions 3 and 4 below.

a) Animals that do not bite -> A

Unexcitable animals -> U

b) Tigers -> T

Striped animals -> S

c) Pangolin -> P

Hanging by its tail -> H

d) Animals that bite -> B

Not easily tamed animals -> N

e) Striped animals -> S

Hanging by its tail -> H

f) Animals -> A

Excitable animals -> E

Pangolins -> P

3.For each statement a through f, in order: write the statement in symbolic form using the appropriate terms from step 2 and write its contrapositive.

a) Statement: A → U

Contrapositive: ¬U → ¬A

b) Statement: T → S

Contrapositive: ¬S → ¬T

c) Statement: P → H

Contrapositive: ¬H → ¬P

d) Statement: B → ¬N

Contrapositive: N → ¬B

e) Statement: S → ¬H

Contrapositive: H → ¬S

f) Statement: A ∧ ¬P → E

Contrapositive: ¬E → ¬(A ∧ ¬P)

4.Chain the symbolic representations together to reach a conclusion. You should already have everything you need from step 3.

a) A → U

b) T → S

c) P → H

d) B → ¬N

e) S → ¬H

f) A ∧ ¬P → E

We can chain them together using logical inference rules to derive a conclusion. Here's one possible way:

• Combining a and f using modus ponens, we get: ¬P → ¬U

• Combining b and e using modus tollens, we get: ¬S → H

• Combining c and e using modus ponens, we get: ¬H → ¬P

• Combining d and 3 using contraposition, we get: N → B

• Combining 1 and 4 using modus ponens, we get: N → ¬U

• Combining 2 and 5 using disjunctive syllogism, we get: ¬T ∨ ¬U

• Combining f and 6 using contraposition, we get: ¬E → ¬(¬T ∨ ¬U)

• Simplifying 7 using De Morgan's law, we get: ¬E → (T ∧ U)

Therefore, the conclusion is: "If an animal is not excitable, then it is either a tiger with stripes or an animal that does not bite."

The symbolic conclusion is: "¬E → (T ∧ U)"

5.Then translate the symbolic conclusion back into English

Translated back into English, it means: "If an animal is not excitable, then it is either a tiger with stripes or an animal that does not bite."

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After winning the bid, Neo has several assignments to complete, unfortunately they are all due at the same time later tonight. Neo has 10 hours available to work on his assignments.

If he completes the full assignment, he will get all the marks for that assignment. If he completes a fraction of the assignment, then he will get that fraction of the marks.

The following shows his assignments, along with how many marks each assignment is worth:

Assignment Problem Solving Java C++ Database security Project management Network design Web development

Marks 60 20 40 8 10 30 15

Hours 3 6.5 5 0.5 5 6 1

**1.Explain the basic idea of how to use greedy strategy to decide which assignments to do. **

The basic idea of using the greedy strategy to decide which assignments to do is to prioritize them based on their mark value per unit time (i.e., marks per hour), and complete the assignments with the highest mark value per unit time first, until either they are completed or the available time runs out.

**2.Describe the metric or measure Neo should use to decide which assignment is the “best”.**

The metric or measure Neo should use to decide which assignment is the "best" is the mark value per unit time, which is the number of marks he can earn per hour for each assignment. The higher the mark value per unit time of an assignment, the better it is, as it means that Neo can earn more marks for each hour of work.

**3.Show how your method works on this problem. Ensure that you show:**

• The value of your measure for each assignment,

• Which assignments Neo should choose in order,

• How many hours Neo should spend on each assignment, and

• The final expected mark.

Using the greedy strategy, we will calculate the value of the measure for each assignment and choose the one with the highest value, then repeat the process until we have no more time left.

Assignment A: 6 hours, 80 marks, value of measure: 13.33

Assignment B: 3 hours, 60 marks, value of measure: 20

Assignment C: 2 hours, 30 marks, value of measure: 15

Assignment D: 4 hours, 50 marks, value of measure: 11.25

Assignment E: 5 hours, 40 marks, value of measure: 8

Based on these values, Neo should start with Assignment B since it has the highest measure value. Then, he should move on to Assignment A, which has the next highest value. After that, he should work on Assignment C, and then divide his remaining time between Assignments D and E.

So, Neo should spend 3 hours on Assignment B, 6 hours on Assignment A, 1 hour on Assignment C, 3 hours on Assignment D, and 0 hours on Assignment E (since there is not enough time left).

The final expected mark would be:

Assignment B: 60 (completed)

Assignment A: 64 (80 marks * 6/10 = 48 + 16 marks for the remaining 4 hours)

Assignment C: 15 (completed)

Assignment D: 22.5 (50 marks * 3/4 = 37.5, but only 3 hours are available, so 37.5 * 3/4 = 22.5)

Assignment E: 0 (not completed)

Therefore, the final expected mark would be 161.5 (60 + 64 + 15 + 22.5 + 0).

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