JavaScript Joel

Posted on Oct 8, 2018

The easiest problem you cannot solve.

#programming #challenge #javascript

Given the following code:

const K = a => b => a
const cat = 'cat'
const dog = 'dog'

Calling K like this will output cat

K(cat)(dog)
//=> "cat"

How can you call K to output dog without swapping the order of cat and dog?

cat and dog must appear exactly once.

The Given cannot be modified.

// INVALID: Cannot swap order!
K(dog)(cat)

// NO CHEATING: May only appear once.
K.bind(null, dog)(cat)(dog)

I know of two ways to solve this problem. Surprise me with another!

I should have also specified no bind, call, or apply, but the solutions are just too interesting :)

Hints: K in the problem is the K combinator, part of the SKI combinator calculus.

Spoilers are in the comments! BEWARE!

Many great solutions have been posted below! A lot I never considered. Also some very creative loopholes around the rules I created ;)

Here's one solution that went undiscovered. It's base64 encoded to hide the "spoiler". But if you are curious you can decode it using your console's atob function.

Syh4PT54KShjYXQpKGRvZyk=

Latest comments (40)

Jon • Aug 23 '19

Well, I looked up SKI combinators. If I am understanding it correctly, and if I've implemented S correctly here...I think the following would work:

const S = x => y => z => x(z)(y(z));
S(K)(K(cat))(dog);

I am going to have to add "Learn SKI Combinators" to my list of things to do. :D

JavaScript Joel • Aug 23 '19

After you research SKI Combinators, look into Church Encoding.

JavaScript Joel 🍻

@joelnet

Church Numerals use functions to represent numbers.
ie: `one = f => x => f (x)` and `two = f => x => f (f (x))`

Math can also be simulated with only functions and without + - / * operators.

#javascript #es6 #functional

22:45 PM - 01 Aug 2019

100 332

Jochem Stoel • Oct 14 '18

Hey Joel, I enjoy your articles and I think you deliver a good contribution overall to this website but I would like to see you use better use cases in your examples than cats and dogs meowing. Not for me but for others less experienced. :)

JavaScript Joel • Oct 14 '18

But cats and dogs are the best ;)

JavaScript Joel • Oct 10 '18 • Edited

Exactly! That is definitely the I combinator. It works beautifuly with the K combinator in the problem to solve this problem.

Two parts of SKI combinator calculus!

JavaScript Joel • Oct 10 '18 • Edited

eval(K.toString().replace('b', 'a'))(cat)(dog)

This one is pretty creative. Modifying the original K function to swap a and b. I like it.

Ian Laird • Oct 10 '18 • Edited

I didn't see this one in the comments

K(cat && dog)();
K()(cat)||dog;

And arrays are fun

K.call(null, [cat, dog].reverse()[0])();
K.apply(null, [cat, dog].reverse())();
K([ ...cat, ...dog].slice(3).join(""))();

Technically i think this follows the rules.

K({ cat: dog })()

And if the does then this should work too

K(`${cat} ${dog}`)()

JavaScript Joel • Oct 10 '18

Beautiful!

metalbrain28 • Oct 9 '18

Well...

K((function *() { yield(arguments); }))(cat)(dog).next().value[0];

JavaScript Joel • Oct 9 '18

Generators and iterators. quite a unique solution. I love it!

kip • Oct 9 '18 • Edited

> K.call(null, (cat, dog))()
'dog'

> K.apply(null, [cat, dog].reverse())()
'dog'

> (cat, K(dog))()
'dog'

> K(((...args) => args[1])(cat, dog))()
'dog'

JavaScript Joel • Oct 9 '18

So many unique solutions. Fantastic!

Lajos Koszti • Oct 9 '18

K(cat && dog)()

JavaScript Joel • Oct 9 '18

Technically within the rules and as valid as any other solution!

Md Abu Taher • Oct 9 '18

Brilliant.

Rong Sen Ng • Oct 9 '18

Do you really need the second dog since you already .bind?

JavaScript Joel • Oct 9 '18

Maybe I do not understand, what is the second dog?

Zhou Qi • Oct 9 '18

Comma operator:

K((cat, dog))()

JavaScript Joel • Oct 9 '18

Extra hearts for the comma operator. Kudos!

JavaScript Joel • Oct 9 '18

Awesome work!

You found the clue I left, the K combinator! And it does come from the SKI calculus.

So many interesting solutions!

Nick Karnik • Oct 9 '18

This call makes no sense in reality, but it will yield the desired output:

K.call(1,2)()

JavaScript Joel • Oct 10 '18

Pretty interesting solution. You are binding the first value to this so that the a gets set as your second argument. Very creative!

Tiago Romero Garcia • Oct 9 '18 • Edited

These are less clever yet might be valid solutions:

K('')(cat)+K(dog)()
K()(cat);K(dog)()

JavaScript Joel • Oct 9 '18

lol. Clearly I have to get better at writing rules! Technically these qualify!

Tiago Romero Garcia • Oct 9 '18

One more:

K([cat,dog])()[1];

Heiker • Oct 8 '18

I give up. This is still within the rules, right?

K.bind(null, [cat, dog].pop())()()

JavaScript Joel • Oct 8 '18

lol. not what I was looking for, but I would say it technically meets the rules.

Very creative!

Kasey Speakman • Oct 8 '18 • Edited

In FP, there is a fairly common operation for this called flip. Here is a solution I tested in Chrome console.

const flip = f => a => b => f(b)(a)
flip(K)(cat)(dog)
// "dog"

JavaScript Joel • Oct 8 '18

I did not consider this as a possible solution!

View full discussion (43 comments)

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The easiest problem you cannot solve.

Latest comments (40)

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