THE PIT AND THE PENTIUM LOGIC - not adequate for real life s

[mith]

unless we're talking about the death of bill gates or something...

[mole]

Well, we're certainly in trouble this time...

Spoilers Ahead! I've divided them into sections, so you can read my earlier thoughts if you want soemthing to start this discussion off...

I guess we need to find the number that would set off both traps at once. Since the rules say that can't happen, neither would be set off, and I can escape. Either that or the paradox would cause this place to collapse...

I'm willing to take the risk. So what's the rule then?

Well, I think it's odd that the output numbers all consist of 0, 1, 2 and 3... And what's so special about a ten-digit code? Let's try labelling the digits of the output code from zero to nine.

It would have been much easier if we could input some more logical numbers...

-shrugs- Yeah, I guess. Something like 0123456789 or 1111122222 would have shown you the answer quicker.

But there's enough information there for you to see the pattern, isn't there?

Each digit in the output is the number of corresponding digits in the input:
6626012798 has one 0, one 1, two 2s, no 3s, no 4s, no 5s, three 6s, one 7, one 8 and one 9, so the output is 1120003111.

Wait! I recognise the problem now!

Yeah, find the ten-digit number x such that the first digit is the number of 0s, the second is the number of 1s, the third is the number of 2s...

So what do we do now?

Well, since there are ten digits, the sum of the digits has to be 10. Because of this, we can't have more than one of the digits 5, 6, 7, 8 or 9 in the number, so the rest of them have to be zero. That's a lot of zeros, let's not try to work out how many we need until later, ok?

Let's say we have x digits that are zero. We then have to put a 1 in the place corresponding to x. Since we now have a 1, we need to put a 1 underneath the second place (the number of 1s).

But now we have two 1s! Don't panic! We'll just change the second place to a 2, and hope nobody notices that it's wrong. Now we have a single 2 around somewhere, so we'll put a 1 in the third spot.

We now have x 0s, two 1s, one 2, and one x. Hey, that works out after all! Our number now looks like this: x2100I0000 (the I could be in the place of any one of the 0s). Looks like there are six 0s there, and 6 + 2 + 1 + 1 = 10, so it fulfils the second part of the rule (the one I invented to check it quickly)!

Yay! And if you check it properly it also works out!

So I'm going to lock in 6210001000 for the game, thanks mith...

[El Blobbo revived]

But it can permutate! Inputting 0001000216 would presumably make Ned fire his programmer!

[This message has been edited by El Blobbo revived (edited 01-19-2002 10:07 PM).]

[El Blobbo revived]

Unless this way is offish.