PYP 2021 q4c

[muhdjabir]

Source

AY 20/21 q4c

Description

Hi everyone Im having issues with justifying question 4c. I know there 2 laws/criteria in being a functor. But the s.toUpperCase() throws me off abit. Does that affect it being a functor? and If not, is there anything that proves that it is not a functor

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[Zhenyubbx]

Are you able to show the question for part c?

[pohja1999]

You can try putting these tests into your shell and see if both results yield true; if yes, then its a functor, else no.
Identity: Box.of("TestString").equals(Box.of("TestString").map(x -> x ));
** x -> x is the identity func;
Assoc: Box.of(f.apply("TestString")).equals(Box.of("Some other String").map(f));

[muhdjabir]

@Zhenyubbx edited to include part c!

[calvinseptyanto]

I checked with my TA, I think the Box class doesn't seem to obey both Functor requirements

The easiest counter example is from the identity law

Box.of("cs2030").map(x -> x).equals(Box.of("cs2030"));

The code above yields false since map function will return new Box object, therefore it is not a functor

[muhdjabir]

@calvinseptyanto Thanks alot calvin! That clears up a bit for me

[pohja1999]

mmmhmmmm yea I still think it is a functor. we need to override the default equality method by objects to check for equality if not it'll just be checking for the address of the objects.... maybe someone more knowledgeable can clarify on this. :-( I cant think of a way for Optional map(Function<T,S> ) to return the same object as another Optional without overriding the equals method.

[weiming21]

Hi I think I can offer an alternative counterexample that might be more specific to this Box implementation as it has to do with the toUpperCase part

Function<String, String> f1= x -> "a"
Function<String, String> f2= x -> x == "a"?"1":"2"
Box.of("a").map(f2.compose(f1)).equals(Box.of("a").map(f1).map(f2))
//returns false -> hence does not obey functor associative law

CMIIW thanks!

[pohja1999]

Hi I think I can offer an alternative counterexample that might be more specific to this Box implementation as it has to do with the toUpperCase part

Box.of("a").map(f2.compose(f1)).equals(Box.of("a").map(f1).map(f2))
//returns false -> does not obey functor associative law

CMIIW thanks!

Sick hehe, may i know what are your f1 and f2 functions hehe

[sinteary]

Hi I think I can offer an alternative counterexample that might be more specific to this Box implementation as it has to do with the toUpperCase part

Box.of("a").map(f2.compose(f1)).equals(Box.of("a").map(f1).map(f2))
//returns false -> does not obey functor associative law

CMIIW thanks!

Sick hehe, may i know what are your f1 and f2 functions hehe

f1 and f2 are shown in the lines above the one you quoted. But I believe you can also test with any dummy functions that could modify the string, eg. substring, string concatenation, toUpper/LowerCase, just as a proof of concept.