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4. Make sure that your trees (both general and binary) are instances of Show, Eq, and Ord. The Ord and Eq instantiation should implement a custom ordering, as follows: For some trees t and u we have t≤u iff traces t C traces u t == u iff traces t C traces u and traces u C traces t Pay attention to the fact that you are thus defining a partial order over trees (that is, two trees may not be comparable with each other). 2 Implementation note If you have a type with a parameter (such the type Tree a for trees holding values of type a) it is often the case that the "big" type (Tree a) can be an instance of some class (say, Eq) only if the "inner" type (a) is an instance of a certain type class (say, Ord). This kind of constraints must be specified at instantiation time. In the example mentioned above the instance Tree a of Eq will thus go as follows: instance Ord a => Eq (Tree a) where Note that I used Eq and Ord merely as examples, I am not claiming that the particular example above makes actual sense.

4. Make sure that your trees (both general and binary) are instances of Show, Eq, and Ord. The Ord and Eq instantiation should implement a custom ordering, as follows: For some trees t and u we have t≤u iff traces t C traces u t == u iff traces t C traces u and traces u C traces t Pay attention to the fact that you are thus defining a partial order over trees (that is, two trees may not be comparable with each other). 2 Implementation note If you have a type with a parameter (such the type Tree a for trees holding values of type a) it is often the case that the "big" type (Tree a) can be an instance of some class (say, Eq) only if the "inner" type (a) is an instance of a certain type class (say, Ord). This kind of constraints must be specified at instantiation time. In the example mentioned above the instance Tree a of Eq will thus go as follows: instance Ord a => Eq (Tree a) where Note that I used Eq and Ord merely as examples, I am not claiming that the particular example above makes actual sense.

C++ Programming: From Problem Analysis to Program Design
C++ Programming: From Problem Analysis to Program Design
8th Edition
ISBN:9781337102087
Author:D. S. Malik
Publisher:D. S. Malik
Chapter17: Linked Lists
Section: Chapter Questions
Problem 9PE
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in Haskell, question 4

4. Make sure that your trees (both general and binary) are instances of Show, Eq, and Ord. The Ord and Eq instantiation should implement a custom ordering, as follows: For some trees t and u we have t≤u iff traces t C traces u t == u iff traces t C traces u and traces u C traces t Pay attention to the fact that you are thus defining a partial order over trees (that is, two trees may not be comparable with each other). 2 Implementation note If you have a type with a parameter (such the type Tree a for trees holding values of type a) it is often the case that the "big" type (Tree a) can be an instance of some class (say, Eq) only if the "inner" type (a) is an instance of a certain type class (say, Ord). This kind of constraints must be specified at instantiation time. In the example mentioned above the instance Tree a of Eq will thus go as follows: instance Ord a => Eq (Tree a) where Note that I used Eq and Ord merely as examples, I am not claiming that the particular example above makes actual sense.
Transcribed Image Text:4. Make sure that your trees (both general and binary) are instances of Show, Eq, and Ord. The Ord and Eq instantiation should implement a custom ordering, as follows: For some trees t and u we have t≤u iff traces t C traces u t == u iff traces t C traces u and traces u C traces t Pay attention to the fact that you are thus defining a partial order over trees (that is, two trees may not be comparable with each other). 2 Implementation note If you have a type with a parameter (such the type Tree a for trees holding values of type a) it is often the case that the "big" type (Tree a) can be an instance of some class (say, Eq) only if the "inner" type (a) is an instance of a certain type class (say, Ord). This kind of constraints must be specified at instantiation time. In the example mentioned above the instance Tree a of Eq will thus go as follows: instance Ord a => Eq (Tree a) where Note that I used Eq and Ord merely as examples, I am not claiming that the particular example above makes actual sense.
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