For the last problem, another possible deduction is that if there are no fours, since we have no duplicate numbers, then there can only be two as an even number. But this would make condition D redundant. Therefore, we can conclude the number contains exactly one 4.
Super specific question here: I can follow your logic through most of the B630EBF deduction but right at the beginning the logic you use on verifier 1 & 14 could be flipped to suggest that Triangle couldn't be 1 because it would give away the answer to verifier 14 that Triangle is the smallest? Obviously you've got the right answer but I can't tell why the logic breaks down depending on which way you look at it. Any insights?
Sorry for the delayed response! I guess “get back to you soon” is more like two weeks, though I did want to take some time to hopefully give an accurate response. I believe what you said isn’t possible because for a verifier to become useless/obsolete, it has to also provide no new information rather than just being certain of the verifier’s answer. Triangle being 1 doesn't by itself mean there couldn't be another 1 for square or circle, but the 'triangle is the smallest' criteria on verifier D would provide new information that it is the smallest (so the only 1). If triangle was 1 and that verifier D wasn’t there, then a code like 112 could be possible since there is could be multiple smallest number. The presence of verifier D provides information still that there is a single smallest number. Hopefully that kinda makes sense. I ran into a similar confusion with deductions a while back and posted about it on BGG. There are some helpful comments there if you want to check that out: boardgamegeek.com/thread/2962599/question-about-initial-deduction-confusion
That actually helps a lot, I followed that link as well and I can see why it would work one way and not the other. Thanks for taking the time to respond :)
What I find helpful in these cases, is to consider that test 14 actually has four outcomes: triangle is the minimum, square is the minimum, circle is the minimum, the code is invalid (there is no minimum). It is this fourth (hidden) case that "triangle = 1" cannot rule out. This principle can be applied to identify other tests where there is a "hidden" outcome not shown on the card. Hope this helps :)
I worked through B5WQUZ on my own and got close but a wrong result. I think I realized my mistake: I figured that Triangle = Purple was incorrect because it made Verifier D superfluous by forcing Square to be the greatest. However, I think this was untrue because the three choices of card D are mutually exclusive in the context of the problem, but not mutually exclusive for all possible numbers. So, while Triangle = Circle would decide Verifier D, D is still not redundant because it still gives necessary information by its very existence in the situation, even if not in an "active" capacity.
@@boardgameswiththomas the thing is: When you solve the problem it is all so fluid, when i try a problem, i can just eliminate the most obvious but i still need to query the machine 1 or 2
Not every question is but all of the ones in this video are. It’s not super common to be able to solve in 0 as you need to have just the right verifier cards
@@boardgameswiththomas thanks for your answer!just bought the game yesterday and thought it might loose some of its magic if it was possible to be solvable without asking the verification cards at all. cheers!
![Turing Machine [Solo Playthrough]](http://i.ytimg.com/vi/-tlfnuQXajE/mqdefault.jpg)
![Turing Machine [Solo Playthrough]](/img/tr.png)
Amazing video! Thank you!
Thanks for making this video, and walking through your process, and including the game links, so I can try (and fail) to play along on BGA.
Hey, thanks for the shout out, that was a nice surprise! Great video! :-)
No problem! :) Your posts were tremendously help in putting this video together
For the last problem, another possible deduction is that if there are no fours, since we have no duplicate numbers, then there can only be two as an even number. But this would make condition D redundant. Therefore, we can conclude the number contains exactly one 4.
Super specific question here: I can follow your logic through most of the B630EBF deduction but right at the beginning the logic you use on verifier 1 & 14 could be flipped to suggest that Triangle couldn't be 1 because it would give away the answer to verifier 14 that Triangle is the smallest? Obviously you've got the right answer but I can't tell why the logic breaks down depending on which way you look at it. Any insights?
Good question! I'll have to think on that a bit and get back to you here soon
Sorry for the delayed response! I guess “get back to you soon” is more like two weeks, though I did want to take some time to hopefully give an accurate response.
I believe what you said isn’t possible because for a verifier to become useless/obsolete, it has to also provide no new information rather than just being certain of the verifier’s answer. Triangle being 1 doesn't by itself mean there couldn't be another 1 for square or circle, but the 'triangle is the smallest' criteria on verifier D would provide new information that it is the smallest (so the only 1). If triangle was 1 and that verifier D wasn’t there, then a code like 112 could be possible since there is could be multiple smallest number. The presence of verifier D provides information still that there is a single smallest number.
Hopefully that kinda makes sense. I ran into a similar confusion with deductions a while back and posted about it on BGG. There are some helpful comments there if you want to check that out: boardgamegeek.com/thread/2962599/question-about-initial-deduction-confusion
That actually helps a lot, I followed that link as well and I can see why it would work one way and not the other. Thanks for taking the time to respond :)
What I find helpful in these cases, is to consider that test 14 actually has four outcomes: triangle is the minimum, square is the minimum, circle is the minimum, the code is invalid (there is no minimum). It is this fourth (hidden) case that "triangle = 1" cannot rule out. This principle can be applied to identify other tests where there is a "hidden" outcome not shown on the card. Hope this helps :)
I worked through B5WQUZ on my own and got close but a wrong result. I think I realized my mistake:
I figured that Triangle = Purple was incorrect because it made Verifier D superfluous by forcing Square to be the greatest. However, I think this was untrue because the three choices of card D are mutually exclusive in the context of the problem, but not mutually exclusive for all possible numbers. So, while Triangle = Circle would decide Verifier D, D is still not redundant because it still gives necessary information by its very existence in the situation, even if not in an "active" capacity.
Hey, can you solve this one, I don't understand how to deduce last step #C528O6E
Are cards guaranteed never to be redundant?
Nope! These were just specific problems where this occurs (which is not that common)
I feel stupid... I can't even do #A4 problems, but i can follow your stream of consciousness... :(
Don't worry! These types of problems take some time to get used to
@@boardgameswiththomas the thing is:
When you solve the problem it is all so fluid, when i try a problem, i can just eliminate the most obvious but i still need to query the machine 1 or 2
is every problem solvable with zero questions?
Not every question is but all of the ones in this video are. It’s not super common to be able to solve in 0 as you need to have just the right verifier cards
@@boardgameswiththomas thanks for your answer!just bought the game yesterday and thought it might loose some of its magic if it was possible to be solvable without asking the verification cards at all. cheers!
Try removing your picture and zooming in on the stuff people actually want to see.