One of those 6 possibilities is true:tintenklexs wrote: ↑08 September 2023, 16:59I am sorry to ask, but this is seriously melting my brain - how exactly would I know Yellow is not 1?
What if Yellow is 1 in the code; and I set it to 1; and the computer is testing for yellow?
The question I would pose by coding 1 for Yellow is: "Is Yellows NOT greater than 1?" correct? And the computer would then say "true"?
Arg, I feel like this is where I go wrong, but I can't really grasp it. Appreachiate any help!!
1) b=1 and the verificator test if b=1 [144/244]
2) b>1 and the verificator test if b>1 [144/244]
3) y=1 and the verificator test if y=1 [414/424]
4) y>1 and the verificator test if y>1 [414/424]
5) p=1 and the verificator test if p=1 [441/442]
6) p>1 and the verificator test if p>1 [441/442]
By testing the two possibilities under brackets you will know if the verificator is testing that part or not. By using 6 whole turns, you can know for sure which one it is (even if you can be much more effecient).
Now for your question: "What if Yellow is 1 in the code; and I set it to 1; and the computer is testing for yellow?" => you can't decide what the computer will test. He is already set to answer from one of the six possibilities above.
So if you test 212, and the verificator tell it's true, that mean the verificator test for either 2 (b>1), 3 (y=1), or 6 (p>1).
You can't know more than that at this point.
If then you test 111 and it's false, that mean the verificator test for 2 (b>1), 4 (y>1), or 6 (p>1)
And so it's either 2 (b>1) or 6 (p>1).