Divine Discs?
Replies
Yellow should be in the first slot according to the 4th try. Once you had all 4 colors figured out (in your 4th attempt), it's easier to try different combinations of those 4 until you get it right instead of doing other colors.
The 4 circles beside each combination tells you how many designs you got right and if they're in the right location. 4 yellows means you have all 4 colors correct but they are all in the wrong spot. Green/blue in the 4 circles tell you how many of those designs are in the right pattern location. It is not in any order (1st green does not mean your first slot color is correct)
Yeah, what AUXIN said. You can run it through process of elimination since you found the colours pretty fast, you just got stuck on using red for the first slot continuously.
I love logic puzzles, so spoilers for solution:
With info from 1- 6th try
In 6th try, all positions are 2, 3, 4 for y are shown to be wrong, therefore y can only be y = [1].
In the 4th try, r, y, p, g is shown to be all wrong, therefore r = [2,3,4], y = [1], p = [2,4], g = [2,3].
In the first try, we know r = [1], y = [2], are the wrong possible positions by now, but it still shows 1 is correct, it can only be: g = [3].
In the second try, r, g, y, p. By now, we know all except p are in wrong possible positions, but since it shows 1 correct, therefore p = [4].
Final combo: y, r, g, p.
With info from just 1- 4th try.
Without 6th try.
In the 4th try, r, y, p, g are all wrong, therefore r = [2, 3, 4], y = [1,3,4], p = [1,2,4], g = [1,2,3].
Based on the 1st try, we know r = [1], y = [2] from 4th try are the wrong possible positions. It still shows 1 correct, so it can only be: g = [3]. r = [2, 4], y = [1,4]. p = [1,2,4], g = [3].
In third, we now know g =[3], so g is the correct dot. r, y, p must be in the wrong position then.
Based on that third try:
r = [2, 4], y = [1,4]. p = [1,2,4], g = [3].
--yellow cannot be 4, purple cannot be 2
r = [2, 4]. y = [1], p = [1,4], g = [3].
--remove 1 from p since that is occupied by y
r = [2, 4]. y = [1], p = [4], g = [3].
--remove 4 from r since that is occupied by p
r = [2], y = [1], p = [4], g = [3].
Final combo: y, r, g, p
Bonus:
With info from just 1- 3rd try.
This one needs more use of logic.
Using all 3, the most systematic process of elimination that happened here is red and yellow. Red is always at 1, while yellow covered 2, 3, 4. If r = [1], then at least one of the tries should have 2 correct. From that, we can tell that red cannot be in 1 ever. Because if it was in 1, then at least 1 of the tries would have 2 dots from having r and y in the right position.
From 1st try, where r =[1], y = [2], g = [3], one of these options are in the right position. r = [1] is not possible.
Then, either one of these must be true:
Opt 1: y is in 2, g cannot be 3. y = [2], g =[1,4].
Opt 2: g is in 3, y cannot be in 2. y = [1, 4], g = [3].
In 2nd try, r, y, and g are in positions from none of the two options, so it makes sense that there are 3 wrong positions. But somehow, there is 1 correct. Therefore the only possibility is that, p is in 4. Thus: r = [2,3], p = [4],... Opt 1: y = [2], g = [1]. Opt 2: y = [1], g = [3].
In 3rd try, we already know r and p are going to be wrong. But it still shows 1 correct. So either, g = [3] or y = [4] is correct. If Opt 1 is correct, all would be wrong. If Opt 2 is correct, then it is possible to show 1 correct. Therefore, only the 2nd option is possible. y = [1], g = [3], p = [4], therefore r can only be 2. Final solution: y, r, g, p Hopefully, maybe it'll help with understanding how people logic their way through these kind of puzzles