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Choose an option to check your answer.
A.
{A, B, C}
B.
{A} only
C.
{B, C} only
D.
{A, A, B, C}
Show Answer
Correct Answer: A. {A, B, C}
Explanation:
The two itemsets share the ordered prefix A and differ in the final item.
Their union forms the candidate {A, B, C}.
Choose an option to check your answer.
A.
Fewer candidates are generated
B.
All confidence values become zero
C.
The number of transactions decreases
D.
More itemsets may qualify as frequent and the search space can expand sharply
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Correct Answer: D. More itemsets may qualify as frequent and the search space can expand sharply
Explanation:
A lower threshold admits patterns with smaller counts.
This often produces many more candidates and frequent itemsets.
Choose an option to check your answer.
A.
It follows one itemset to maximum depth first
B.
It searches only rule consequents
C.
It explores itemsets level by level according to their size
D.
It orders transactions by time
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Correct Answer: C. It explores itemsets level by level according to their size
Explanation:
Apriori completes frequent one-itemsets before two-itemsets, then three-itemsets, and so on.
This level-wise strategy enables subset-based pruning.
Choose an option to check your answer.
A.
If an antecedent is frequent, every rule is valid
B.
If a consequent fails minimum confidence, its supersets can be pruned in standard rule generation
C.
Confidence rises whenever the consequent grows
D.
All consequents have equal confidence
Show Answer
Correct Answer: B. If a consequent fails minimum confidence, its supersets can be pruned in standard rule generation
Explanation:
For rules from a fixed frequent itemset, enlarging the consequent shrinks the antecedent and can reduce confidence.
This supports structured pruning of consequent candidates.
Choose an option to check your answer.
A.
The number of nonempty subset splits grows exponentially
B.
Support must be negative
C.
Long itemsets contain no subsets
D.
Confidence cannot be computed for them
Show Answer
Correct Answer: A. The number of nonempty subset splits grows exponentially
Explanation:
An itemset with k items has many possible antecedent subsets.
Efficient pruning is needed to avoid evaluating every possible rule.
Choose an option to check your answer.
A.
Use only L as both sides
B.
Use infrequent supersets as consequents
C.
Assign random rule directions
D.
Use nonempty proper subsets as antecedents and the remaining items as consequents
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Correct Answer: D. Use nonempty proper subsets as antecedents and the remaining items as consequents
Explanation:
Each split of L into disjoint nonempty parts defines a candidate rule.
Confidence filtering then retains sufficiently strong directions.
Choose an option to check your answer.
A.
Yes, adding items always lowers confidence
B.
Yes, every subset has identical confidence
C.
No, confidence is not anti-monotone in the same way as support
D.
No, because confidence cannot be calculated
Show Answer
Correct Answer: C. No, confidence is not anti-monotone in the same way as support
Explanation:
Support has a subset monotonicity property, but confidence depends on a changing denominator.
Adding or removing antecedent items can raise or lower confidence.
Choose an option to check your answer.
A.
{A, C} is infrequent
B.
{A, C} is frequent
C.
{A, C} has the same support necessarily
D.
{A, C} has confidence one
Show Answer
Correct Answer: B. {A, C} is frequent
Explanation:
Every subset of a frequent itemset must meet minimum support.
The subset may have greater support than the full itemset.
Choose an option to check your answer.
A.
{A, B, C} must be infrequent
B.
{A, B, C} must be frequent
C.
Its support must equal support of {A, B}
D.
Its confidence must be one
Show Answer
Correct Answer: A. {A, B, C} must be infrequent
Explanation:
A superset cannot have more support than its subset.
Because {A, B} is below threshold, any superset containing it is also below threshold.
Choose an option to check your answer.
A.
Changing item names during mining
B.
Recalculating only confidence
C.
Counting one-itemsets after long itemsets
D.
Introducing and counting new candidates before completing a full database pass
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Correct Answer: D. Introducing and counting new candidates before completing a full database pass
Explanation:
Dynamic counting reduces the rigid level-by-level waiting of standard Apriori.
Candidates can enter the process at checkpoints within a scan.
Choose an option to check your answer.
A.
Replacing support with confidence
B.
Using only the last transaction
C.
Mining a representative subset to obtain candidate patterns more cheaply
D.
Generating random item names
Show Answer
Correct Answer: C. Mining a representative subset to obtain candidate patterns more cheaply
Explanation:
Sampling reduces computation by working on fewer transactions.
A verification step may be needed because some true patterns can be missed.
Choose an option to check your answer.
A.
Splitting items into classes for prediction
B.
Mining local frequent itemsets in partitions and verifying their union globally
C.
Partitioning features for normalization
D.
Clustering transactions without support
Show Answer
Correct Answer: B. Mining local frequent itemsets in partitions and verifying their union globally
Explanation:
Any globally frequent itemset must be frequent in at least one partition.
This property limits global candidates and can reduce database scans.