YaBASIC, score:15 Cracked by Dingus
The new and improved string to crack:
0 1 2 3 4 5 6 7 8 9 L S l s - Can you crack it? Yes they can!
My method relies on 1 variable,i, which defaults to 0. !i (NOT i) results in 1. Of course, with 0 and 1 we've got binary - the building block of all computing. However, in my case I just added as many !i as needed to get the result. Pretty ugly...
?i ?!i ?(!i)+(!i) ?(!i)+(!i)+(!i) ?(!i)+(!i)+(!i)+(!i) ?(!i)+(!i)+(!i)+(!i)+(!i) ?(!i)+(!i)+(!i)+(!i)+(!i)+(!i) ?(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i) ?(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i) ?(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i) ?chr$((!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)+(!i)) And so on with ?chr$ and as many !i as needed for the desired character. In hindsight maybe adding = to the string would have slowed Dingus down by a few more minutes... Their solution uses it whereas mine doesn't. 😁