Question at position 11 Given the following Haskell expression Which of the following expressions is equivalent to the above, using the $ operator to reduce parentheses? Clear my selection

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BlackTom AI · Accepted Answer

We start by restating the question and the available answer forms to compare them clearly.
Question: Given the Haskell expression result = map (*2) $ filter (>3) [1, 4, 3, 6], which of the following expressions is equivalent to the above, using the $ operator to reduce parentheses?
Answer options:
- Option A: result = map *2 $ filter (>3) [1, 4, 3, 6]
- Option B: result = map (*2) $ filter (>3) [1, 4, 3, 6]
- Option C: result = map $ (*2) (filter (>3) [1, 4, 3, 6])
- Option D: result = map (*2) (filter >3) [1, 4, 3, 6]

Option A analysis:
This option uses map *2, which is missing the parentheses around *2. In Haskell, (*2) denotes a function that multiplies its argument by 2; without the parentheses, *2 is not a valid function value here, so this expression is syntactically invalid or would not type-check as intended. Moreover, the intended use of $ is to avoid parentheses around the second argument; this form does not correctly place the operator and the function as a single function-application chain.

Option B analysis:
This expression writes result = map (*2) $ filter (>3) [1, 4, 3, 6]. Here, (*2) is the function to be mapped, and $ is used to apply the map to the result of filter (>3) [1, 4, 3, 6] without extra parentheses. This matches the structure of the original expression: map (*2) applied to the result of filter (>3) [1, 4, 3, 6], with the $ making the grouping explicit and eliminating the need for parentheses around the second argument. Logically and syntactically, this is a correct reformulation using ($).

Option C analysis:
In this form, result = map $ (*2) (filter (>3) [1, 4, 3, 6]). The ($) operator is applied between map and the result of (*2) (filter (>3) [1, 4, 3, 6]). Because ($) has low precedence, this ends up supplying a single argument to map: the value (*2) (filter (>3) [1, 4, 3, 6]), which is the result of applying the function (*2) to the list (filter (>3) [1, 4, 3, 6]). This changes the meaning from mapping over the list to multiplying the entire list result by 2, which is not equivalent to the original expression. Hence, incorrect.

Option D analysis:
This expression uses filter >3 without parentheses, which is invalid syntax for the comparison function in the list context. The correct predicate should be (>3) to produce a list of elements greater than 3. As written, this option would not type-check or would be interpreted incorrectly, so it is incorrect.

Summarizing the reasoning across options, Option B is the one that preserves the intended structure, uses the $ operator properly to avoid extra parentheses, and matches the semantics of applying map (*2) to the filtered list.

Question at position 11 Given the following Haskell expression Which of the following expressions is equivalent to the above, using the $ operator to reduce parentheses? Clear my selection

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