Software Testing · End Term · Jan 2025 · AN

**1. Identify the verb phrase:** The main verb phrase is will give. It uses will followed by the base form give. **2. Match the tense/aspect label:** In traditional exam terminology, will + base verb is future indefinite, also called simple future. There is no have + past participle construction to indicate future perfect, and no been + -ing construction to indicate perfect continuous. **Answer: C — Future indefinite.** **Convention note:** The question uses aspect loosely; future indefinite is the traditional tense-form label intended by these options. A: **Incorrect:** Future perfect would use will have given. B: **Incorrect:** Future perfect continuous would use will have been giving. C: **Correct:** Will give is the simple-future/future-indefinite construction. D: **Incorrect:** Simple present would use give or gives without will.

**1. Recognize the polite-request pattern:** The standard expression is Would you mind if ...? It asks politely whether the listener would object to a proposed action. **2. Match the following clause:** The past-form stayed expresses polite distance in this hypothetical request. The complete sentence is Would you mind if I stayed back after class for remedial classes? **Answer: A — Would.** A: **Correct:** Would you mind if I stayed ... is the standard polite-request construction. B: **Incorrect:** Could commonly begins Could I stay ...?, but Could you mind if ...? is not the idiomatic construction needed here.

**1. Explain the graph:** An application transition graph represents application pages or states as nodes and user/application navigation as edges. **2. Classify the trigger:** Here the user enters data into a form and submits it. The resulting navigation depends on submitting those form fields, making it a form-link transition in the terminology of this question. **Answer: C — Form Link Transition.** A: **Incorrect:** A simple-link transition follows an ordinary hyperlink without submitting form data. B: **Incorrect:** Operational transitions concern an operation/state change rather than the specific form-navigation category requested here. C: **Correct:** Submitting a form and moving to the resulting page is a form-link transition. D: **Incorrect:** A redirect is navigation initiated by redirect behavior; that is not specified merely by clicking Submit.

**1. Identify the constraint:** Check-out must occur after check-in. This is a relationship between two submitted parameters rather than a format/range constraint on one isolated field. **2. Explain bypass testing:** A tester can alter a request directly to bypass restrictions imposed by the browser interface. Server-side validation must still reject the invalid relationship. **3. Classify the case:** Because the test breaks the dependency between the check-in and check-out parameters, it is parameter-level bypass testing in this classification. **Answer: B — Parameter level bypass testing.** A: **Incorrect:** Control-flow bypass skips required steps or pages; this test changes submitted field relationships. B: **Correct:** The altered request violates an inter-parameter constraint. C: **Incorrect:** Value-level bypass tests a single field’s own validity, rather than the cross-field ordering emphasized here. D: **Incorrect:** This case does not involve modifying stored user-session data.

**1. Identify the load pattern:** The user count rises abruptly to a very large value. The distinctive feature is the sudden change in demand. **2. Match the performance test:** Spike testing examines how a system responds to a sharp load increase and, often, how it recovers afterward. Relevant observations include response time, errors and resource exhaustion. **Answer: A — Spike testing.** A: **Correct:** A sudden large increase in users is the defining spike-load pattern. B: **Incorrect:** Soak testing sustains load for a long duration to reveal issues such as resource leaks. C: **Incorrect:** Compatibility testing checks operation across environments, platforms or configurations. D: **Incorrect:** Regulatory testing checks compliance with applicable requirements rather than response to abrupt load.

**1. Explain maintenance backlog:** During maintenance, new problem reports arrive while existing problems are resolved. The backlog is the set still awaiting resolution. **2. Identify the maintenance measure:** The backlog management index compares problems closed during a period with problems arriving during that period, often as \[\mathrm{BMI}=\frac{\text{problems closed}}{\text{problems arriving}}\times100.\] A value above 100% indicates that more problems were closed than arrived, under this convention. **3. Match the category:** Tracking whether the backlog is being reduced is specifically a maintenance-quality concern. **Answer: B — Backlog Management Index (BMI).** A: **Incorrect:** Defect density relates defects to product size and is generally a product-quality measure. B: **Correct:** BMI monitors problem-resolution progress relative to incoming maintenance work. C: **Incorrect:** Code coverage measures how much program structure a test suite executes. D: **Incorrect:** Cyclomatic complexity measures control-flow complexity rather than maintenance-backlog progress.

**1. Compare the expressions:** The original discriminant is b*b - 4*a*c. The mutant changes it to b*b - 4*a*b. **2. Identify exactly what changed:** The scalar variable c was replaced by the scalar variable b in one occurrence. The multiplication/subtraction operators and the surrounding square-root/division structure remain the same. **3. Name the mutation:** Replacing one scalar variable reference with another is scalar variable replacement. **Answer: A — Scalar variable replacement.** A: **Correct:** The mutation replaces c with b, another scalar variable. B: **Incorrect:** No new unary operator such as negation or increment was inserted. C: **Incorrect:** The arithmetic operators did not change. D: **Incorrect:** No absolute-value operation was added.

**1. Explain the TDD cycle:** Test-driven development starts with a failing test, adds enough production code to pass it, and then refactors while keeping tests passing. **2. Connect it to pairing:** The two developers can share this short feedback cycle: one writes a test and the other implements the behavior, then they switch roles. Both discuss requirements and review the code continuously. **3. Interpret the option:** The code-writing/test-writing division describes a useful pairing style, not permanently separate jobs or the removal of collaboration. **Answer: B — Because it allows one developer to write code and the other to test..** A: **Incorrect:** Pair programming depends on collaboration; TDD does not remove it. B: **Correct:** Alternating test and implementation responsibilities supports the TDD feedback cycle. C: **Incorrect:** Automated test frameworks are commonly used; avoiding them is not the reason TDD suits pairing. D: **Incorrect:** Writing and running tests during development is central to TDD.

**1. Identify the communication mechanism:** The User Module sends a message to a queue, and the Driver Module receives it through the messaging service. **2. Classify the interface:** The modules interact by exchanging messages rather than by directly passing arguments in an ordinary procedure call or accessing one common memory variable. The appropriate interface type is message passing. **3. Explain the coupling:** A queue can separate sender and receiver in time and execution, although both still depend on a shared message format and delivery contract. **Answer: D — Message-passing interfaces.** A: **Incorrect:** The described interaction is queued messaging rather than direct function-parameter transfer. B: **Incorrect:** The prompt does not describe both modules reading/writing a common data store as their communication mechanism. C: **Incorrect:** A shared external hardware device is not the stated interface. D: **Correct:** The queue explicitly carries messages between the modules.

**1. Read the productions:** The grammar has S → aSbb and S → abb. The second production terminates a derivation with one a followed by two b symbols. **2. Track a recursive expansion:** Applying S → aSbb adds one a before the existing string and two b symbols after it. It preserves the pattern of all a symbols preceding all b symbols. **3. Express all generated strings:** After k recursive expansions followed by the terminating rule, there are k + 1 a symbols and 2(k + 1) b symbols. Letting n = k + 1 gives \[L=\{a^n b^{2n}\mid n\ge1\}.\] The empty string cannot be generated. **Answer: A — \(\{a^n b^{2n}\mid n\ge1\}\).** A: **Correct:** The terminating string abb has n = 1, and recursion preserves twice as many b symbols as a symbols. B: **Incorrect:** This reverses the required count ratio; abb is a counterexample. C: **Incorrect:** Equal counts do not match even the shortest generated string. D: **Incorrect:** This includes the empty string at n = 0, but the grammar has no empty production.

**1. Interpret the states as parities:** The initial state S1 represents an even number of a symbols and an even number of b symbols. Reading a toggles the a parity, and reading b toggles the b parity. **2. Assign the remaining states:** S2 is (odd a, even b), S3 is (even a, odd b), and S4 is (odd a, odd b). Each drawn transition agrees with these toggles. **3. Use the accepting state:** Only S4 is double-circled. Therefore a string is accepted exactly when both symbol counts are odd; ab and ba are simple accepted examples. **Answer: C — Odd number of a symbols and odd number of b symbols.** A: **Incorrect:** Even a and odd b lead to S3, which is not accepting. B: **Incorrect:** Odd a and even b lead to S2, which is not accepting. C: **Correct:** Odd counts of both symbols lead to accepting state S4. D: **Incorrect:** Even counts of both symbols lead back to S1.

**1. Rewrite the predicate:** The first if sets r to the truth value of x > y. Thus the second predicate is \[p=a\land b,\qquad a:(x>y),\quad b:(x=z).\] **2. State RACC requirements:** To show that a independently controls p, hold b true and use both truth values of a. To show that b independently controls p, hold a true and use both truth values of b. The required clause combinations are TT, FT and TF. **3. Evaluate the selected tests:** (5,4,4) gives TF and p false. (5,4,5) gives TT and p true. (4,4,4) gives FT and p false. The latter two form the pair for a; the first two form the pair for b. **Answer: D — Tests (5,4,4), (5,4,5), (4,4,4).** A: **Incorrect:** The tests give FT and FF; no test makes the predicate true or makes a true. B: **Incorrect:** These tests give TT, FF and FT, so the TF combination needed for b is missing. C: **Incorrect:** The tests never make x > y true, so they cannot demonstrate both clauses independently controlling the predicate. D: **Correct:** The tests supply TF, TT and FT, meeting both required RACC pairs.

**1. Identify the two fields:** Device declares memorySizeGB, while Laptop declares another field with the same name. In Java, fields are hidden rather than dynamically overridden. The two fields occupy distinct state locations. **2. Follow the update:** Laptop.setMemorySize writes the Laptop field, setting it to 16. The inherited Computer.printSpecs method refers to the Device field visible in its defining class hierarchy; that field remains at its default integer value 0. **3. Identify the anomaly:** The operation that defines the intended memory state and the operation that reads it refer to different hidden fields. This is the stated state-definition inconsistency anomaly. **Answer: C — State definition inconsistency anomaly.** **Convention note:** The source describes the field as overridden. Java fields are hidden; dynamic overriding applies to instance methods. A: **Incorrect:** The problem is two distinct state fields, not use of an incompatible runtime type. B: **Incorrect:** A generic state-definition issue is less specific than the mismatch between the field written and the field subsequently read. C: **Correct:** The setter and printer disagree about which hidden state field represents memory size. D: **Incorrect:** The fields are accessible under their protected visibility; visibility denial is not causing the wrong value.

**1. Identify the deleted statement:** Line 8 is super.draw(), which explicitly invokes the parent Shape implementation before the subclass prints Drawing circle. **2. Explain the effect:** Removing this line removes the superclass call. The overriding Circle.draw method remains, so a call to it now prints only the circle-specific message. **3. Match the mutation family:** Among the choices, this is the superclass/super deletion mutation category labeled SKD. **Answer: D — Super keyword deletion (SKD).** **Convention note:** The statement, not just the literal keyword super, is deleted in the described edit. Deleting only super. while retaining draw() would create a recursive self-call, which is a different mutation effect. A: **Incorrect:** The overriding draw method is retained; only its superclass-call statement is deleted. B: **Incorrect:** The method is not moved to another class or location. C: **Incorrect:** No overloaded method with a different parameter list is deleted. D: **Correct:** The mutation removes the super-qualified parent-method call.

**1. Propagate symbolic inputs:** Initially a = a₀ and b = b₀. The helper calls produce \(t_1=a_0+5\) and \(t_2=b_0-3\). The outer condition is \[C:(a_0+5)^2<2(b_0-3).\] **2. Keep the path prefix:** Reaching LINE-1 already requires C to hold. Symbolic execution must retain that condition when splitting on the inner test t1 > t2. **3. Form the two child constraints:** The true branch has \[C\land(a_0+5>b_0-3),\] and the false branch has \[C\land(a_0+5\le b_0-3).\] Both include the same true outer condition. **Answer: D — Outer condition conjoined with each branch of the inner condition.** **Convention note:** This is the usual mathematical-integer symbolic interpretation; an implementation modeling finite-width overflow would need corresponding machine-arithmetic constraints. A: **Incorrect:** These split the outer condition, not the inner conditional reached at LINE-1. B: **Incorrect:** These omit the outer path constraint required to reach LINE-1. C: **Incorrect:** The second constraint negates the outer condition and retains the wrong inner branch. D: **Correct:** It retains the outer true condition and conjoins the inner condition or its negation.

**1. Explain propagation:** For a nonempty input, skipping index 0 changes the temporary reversed string. That infected state propagates to an observable failure only if it changes the final palindrome decision/message. **2. Use the length argument:** The mutant reverse has length n - 1, so it cannot equal a nonempty input of length n. The mutant therefore reports not a palindrome for every nonempty candidate. **3. Compare the original decisions:** river and little are already non-palindromes, so both versions print not a palindrome. refer and level are palindromes, so their printed decisions change. The non-propagating cases are river and little. **Answer: B, D — river; little.** A: **Incorrect:** refer is a palindrome; the mutant changes its reported result, so the infection propagates. B: **Correct:** river is not a palindrome, and both versions report that; the internal reverse difference does not change output. C: **Incorrect:** level is a palindrome and its message changes, so the infection propagates. D: **Correct:** little is already not a palindrome, so both versions produce the same classification.

**1. Define strong killing:** A test strongly kills a mutant when the original and mutated programs produce different observable results for that input. **2. Determine the mutant output:** The changed loop uses i > 0, omitting the first character of every nonempty input. Its reverse string has one fewer character and therefore never equals the full input. It reports not a palindrome. **3. Select distinguishing inputs:** The original recognizes refer and level as palindromes, while the mutant rejects them. For river and little both versions reject, so those tests do not strongly kill the mutant. **Answer: A, C — refer; level.** A: **Correct:** The original accepts refer as a palindrome while the mutant rejects it. B: **Incorrect:** Both versions report that river is not a palindrome. C: **Correct:** The original accepts level as a palindrome while the mutant rejects it. D: **Incorrect:** Both versions report that little is not a palindrome.

**1. Define a definition-clear path:** Relative to a variable’s definition at the start, a def-clear path reaches the intended later use without redefining that variable along the way. **2. Locate assignments to y:** Node 1 defines y = j. Node 3 assigns y = x, and node 5 assigns y = 0. Node 2 only assigns x = y; node 6 only assigns x = 0. Node 7 prints y. **3. Choose the clear route:** The route [1,2,4,6,7] avoids nodes 3 and 5. Hence the definition from node 1 reaches the print at node 7 without a new assignment to y. **Answer: B — [1, 2, 4, 6, 7].** A: **Incorrect:** Node 5 redefines y as 0. B: **Correct:** Nodes 2, 4 and 6 do not redefine y, so this path is def-clear from node 1 to node 7. C: **Incorrect:** Nodes 3 and 5 both redefine y. D: **Incorrect:** Node 3 redefines y before the later print.

**1. Identify all definitions:** The graph defines y at nodes 1, 3 and 5. All-definitions coverage requires a test to exercise a definition-clear path from each applicable definition to at least one use. **2. Inspect the supplied paths:** Both [1,2,4,6,7] and [1,2,4,5,7] take the node-2 branch. Neither visits node 3, so its definition y = x is not covered by any definition-use tour. **3. Compare the criteria:** Because all-definitions coverage already fails, the stronger requirements all-uses and all-du-paths cannot be claimed either. Thus none of the first three descriptions holds. **Answer: D — None of these.** A: **Incorrect:** The definition at node 3 is missed entirely. B: **Incorrect:** All-definitions already fails, so both criteria cannot hold. C: **Incorrect:** The stronger combined claim fails because the node-3 definition is missed. D: **Correct:** None of the listed coverage claims is satisfied.

**1. Isolate the major clause:** The predicate is \(p=a\land(\neg b\lor\neg c)\). Let \(q=\neg b\lor\neg c\), so p = a AND q. **2. Test determination:** Clause a determines p when changing only a changes the predicate. With a true, p = q; with a false, p = false. These results differ exactly when q is true. **3. Write the condition:** \[p_a=p(a=T)\oplus p(a=F)=q\oplus F=\neg b\lor\neg c.\] **Answer: B — \(\neg b\lor\neg c\).** A: **Incorrect:** The c term has the wrong polarity; for b = c = true this expression is true even though a cannot affect p. B: **Correct:** This is precisely the condition that makes the other conjunct true, allowing a to control p. C: **Incorrect:** This is sufficient but not necessary; either b false or c false already allows a to determine p. D: **Incorrect:** When b and c are both true, the other conjunct is false, so a does not determine p.

**1. State restricted inactive coverage:** The major clause a must take both truth values while the minor clauses b and c stay fixed and a does not change p. Feasible pairs are sought for each predicate result. **2. Find when a is inactive:** For \(p=a\land(\neg b\lor\neg c)\), a is inactive only when \(\neg b\lor\neg c=F\). By De Morgan’s law, this requires \(b=T,c=T\). **3. Construct the feasible pair:** With those minor values, both assignments \((a,b,c)=(T,T,T)\) and \((F,T,T)\) give p = F. No inactive pair can give p = T, since making a false always makes this conjunction false. **Answer: B — No pair for p = T; for p = F use (T,T,T) and (F,T,T).** A: **Incorrect:** With c true and b false, p follows a, so the proposed pair changes the predicate rather than keeping a inactive. B: **Correct:** It correctly states no true-predicate pair and the false-predicate pair (T,T,T), (F,T,T). C: **Incorrect:** The proposed pairs make the other conjunct true, so changing a changes p. D: **Incorrect:** The purported true-predicate pair cannot stay true when a is false, and the other pairs fail the inactive/minor-fixed requirements.