Cantr II Discussion Forum

yeah dawg...i got the answer! if you want it meet me in the unnamed building in Drojf with 3000g rubber and a dead cow at turn 8

Everytime I read through this my stomach feels sick. I really, really, really, hope it was just a hypothetical math problem and nothing more.

Chrissy

Hypothetical to the extreme. The staff of any game would not make desicions this way.

They would at least not publish it.

And I think the answer is 10, Thomas.

Thank you for an explicit and correct solution. I guess, now you could also answer the same question with 10,000 locked accounts and 10,000 runs?

Yes, it's 100.

Did you change the problem?

When I did 10 manually just now it came out to 5 open and 5 closed.

1 open
2 closed
3 closed
4 open
5 closed
6 open
7 closed
8 closed
9 open
10 open

0000000000
+0101010101
=0101010101
+0010010010
=0111010101
+0001000100
=0110010011
+0000100001
=0110110010
+0000010000
=0110100010
+0000001000
=0110100010
+0000000100
=0110101010
+0000000010
=0110101110
+0000000001
=110101100

I'm glad I didn't even try.

Yes, the algorithm has slightly changed.

1 = locked
0 = unlocked

D'oh

But the error was not due to the changed algorithm. It was only a slight change, that changes the outcome but not essentially the algorithm itself. Only a different start.

The answer for 100 accounts is 10. The answer for 10,000 accounts is 100. The perfect squares, which have an odd number of factors are flipped an odd number of times and therefore are open at the end. All the other numbers have an even number of factors and therefore stay closed. In general, for n accounts, the number open at the end is the greatest integer less than or equal to the square root of n.

Absolutely accurate! *applauds*