/prog/ challenge

To what will this SKI calculus expression reduce? Post the reduction steps and the result.

(((S S K) (K (S K S) K) ((S K) K S K) ((S (K K) S K) K (K S)) K S K) (((K K) (S K S) (S K) ((K K S (S K K) K S) K S K) (K K) S) ((((K S) S K (K (S K (K (K (K S S (K (S (K S) K) S) K) K K) S) S) (K S)) K) (S K K)) K S S K S (K S) S) K S) K S K K S (S K K) (S (K S) K) S K K (K (K (K (S K)) S)) S)

[[['S', 'S', 'K'], ['K', ['S', 'K', 'S'], 'K'], [['S', 'K'], 'K', 'S', 'K'], [['S', ['K', 'K'], 'S', 'K'], 'K', ['K', 'S']], 'K', 'S', 'K'], [[['K', 'K'], ['S', 'K', 'S'], ['S', 'K'], [['K', 'K', 'S', ['S', 'K', 'K'], 'K', 'S'], 'K', 'S', 'K'], ['K', 'K'], 'S'], [[[['K', 'S'], 'S', 'K', ['K', ['S', 'K', ['K', ['K', ['K', 'S', 'S', ['K', ['S', ['K', 'S'], 'K'], 'S'], 'K'], 'K', 'K'], 'S'], 'S'], ['K', 'S']], 'K'], ['S', 'K', 'K']], 'K', 'S', 'S', 'K', 'S', ['K', 'S'], 'S'], 'K', 'S'], 'K', 'S', 'K', 'K', 'S', ['S', 'K', 'K'], ['S', ['K', 'S'], 'K'], 'S', 'K', 'K', ['K', ['K', ['K', ['S', 'K']], 'S']], 'S']

You are allowed to reduce it programmatically. If so, post the code of the evaluator. You can also do it by hand.

Someone already did your homework: http://esoteric.voxelperfect.net/wiki/Lazy_K

>>2
I already did my ``homeworks''. It's just a challenge.

great post OP i'm almost tempted to read it twice!

( ͡° ͜ʖ ͡°)