4 ms·

The puzzle is easy enough to solve by brute force, but it seems like rather a lot to solve by hand for a very junior-targeted game. Heuristics could reduce tri

The puzzle is easy enough to solve by brute force, but it seems like rather a lot to solve by hand for a very junior-targeted game. Heuristics could reduce trial-and-error to a reasonable search space, but is there a purely deductive way to solve this by hand?

Jtsummers 2y ago
With the 4 common answers (all 4 people answered them the same) we know that no more than 3 of them can be correct (because Lisa scored 3 points). We also know at least one must be correct (otherwise Mary can't score 7). Examine what Mary and Dan have in common besides those 4... Nothing. They disagreed on each of the remaining 6 answers. We can break this down into three cases, either 1, 2, or 3 of the common answers are correct: 1 - Impossible. Mary gets 6 of the remaining correct, Dan gets 4 of the 6 remaining correct, but they have none in common so this doesn't work. 2 - Impossible. Mary gets 5 of the remaining correct, Dan gets 3 of the remaining correct, but again their answers disagree so they cannot share a common answer out of the 6 remaining. 3. Mary needs 4 of the remaining 6 and Dan needs 2, this is feasible. At this point you can ignore the 4 common ones other than to say Colin has at least a score of 3. Since we now know that everything outside those common ones are incorrect for Lisa, cross everything off that Lisa and Colin share of the 6 non-common answers. There are only 3 that disagree with Lisa so they must be correct (because everything Lisa answered here is wrong). This gives Colin a final score of 6, three in common with everyone and three from disagreeing with Lisa. The reason there are 4 keys in Hillel's solution is actually kind of funny, they don't matter. Those 4 keys only disagree on which of the 4 common answers are incorrect. They agree everywhere else. EDIT: Presentation and some typos
- flanbiscuit 2y ago
  I'm more of a visual person when it comes to puzzles like these so here's my take: 1 2 3 4 5 6 7 8 9 10 Mary B B A B A B B A B B 7 Dan B A A A B A B A A A 5 Lisa B A A A B B B A B A 3 Colin B B A A A B B A A A ? First I remove the answers that everyone agreed on. We know that one of these 4 answers is wrong because of Lisa. 1 2 3 4 5 6 7 8 9 10 Mary _ B _ B A B _ _ B B 7 Dan _ A _ A B A _ _ A A 5 Lisa _ A _ A B B _ _ B A 3 Colin _ B _ A A B _ _ A A ? Next we can assume anything matching Lisa's remaining answers are wrong, so I will put an x for those. These we 100% know are wrong answers. 1 2 3 4 5 6 7 8 9 10 Mary _ B _ B A x _ _ x B 7 Dan _ x _ x x A _ _ A x 5 Lisa _ x _ x x x _ _ x x 3 Colin _ B _ x A x _ _ A x ? If you count both the remaining answers and the blanks, you'll see they equal to their score + 1. That 1 being that unknown of the 4 original blank columns where everyone agreed. So the answer for Colin would by 6 because he has 7 remaining answers.
  - thaumasiotes 2y ago
    > Next we can assume anything matching Lisa's remaining answers are wrong This is incorrect. The assumption is true, but that's just a coincidence - you haven't justified it.
supernewton 2y ago
Dan and Lisa's answers agree on all but problems 6 and 9, and Dan got two more correct than Lisa. So Dan is correct on both 6 and 9. Therefore, Mary is incorrect on 6 and 9. So she got seven of the other eight correct, and Dan got three of the other eight correct. But on those eight problems, Mary and Dan disagree on only problems 2, 3, 5, and 10, therefore Mary got all four of those correct. Then Mary got three of problems 1, 4, 7, 8 correct, but everyone answered the same way on those, so Colin got three right on those problems as well. Then read off the deduced answer key for the other six problems to get Colin's final score.
saagarjha 2y ago
To be fair, this is one of the hardest puzzles in the game.

Jtsummers 2y ago

With the 4 common answers (all 4 people answered them the same) we know that no more than 3 of them can be correct (because Lisa scored 3 points). We also know at least one must be correct (otherwise Mary can't score 7). Examine what Mary and Dan have in common besides those 4... Nothing. They disagreed on each of the remaining 6 answers. We can break this down into three cases, either 1, 2, or 3 of the common answers are correct: 1 - Impossible. Mary gets 6 of the remaining correct, Dan gets 4 of the 6 remaining correct, but they have none in common so this doesn't work. 2 - Impossible. Mary gets 5 of the remaining correct, Dan gets 3 of the remaining correct, but again their answers disagree so they cannot share a common answer out of the 6 remaining. 3. Mary needs 4 of the remaining 6 and Dan needs 2, this is feasible. At this point you can ignore the 4 common ones other than to say Colin has at least a score of 3. Since we now know that everything outside those common ones are incorrect for Lisa, cross everything off that Lisa and Colin share of the 6 non-common answers. There are only 3 that disagree with Lisa so they must be correct (because everything Lisa answered here is wrong). This gives Colin a final score of 6, three in common with everyone and three from disagreeing with Lisa. The reason there are 4 keys in Hillel's solution is actually kind of funny, they don't matter. Those 4 keys only disagree on which of the 4 common answers are incorrect. They agree everywhere else. EDIT: Presentation and some typos

flanbiscuit 2y ago
I'm more of a visual person when it comes to puzzles like these so here's my take: 1 2 3 4 5 6 7 8 9 10 Mary B B A B A B B A B B 7 Dan B A A A B A B A A A 5 Lisa B A A A B B B A B A 3 Colin B B A A A B B A A A ? First I remove the answers that everyone agreed on. We know that one of these 4 answers is wrong because of Lisa. 1 2 3 4 5 6 7 8 9 10 Mary _ B _ B A B _ _ B B 7 Dan _ A _ A B A _ _ A A 5 Lisa _ A _ A B B _ _ B A 3 Colin _ B _ A A B _ _ A A ? Next we can assume anything matching Lisa's remaining answers are wrong, so I will put an x for those. These we 100% know are wrong answers. 1 2 3 4 5 6 7 8 9 10 Mary _ B _ B A x _ _ x B 7 Dan _ x _ x x A _ _ A x 5 Lisa _ x _ x x x _ _ x x 3 Colin _ B _ x A x _ _ A x ? If you count both the remaining answers and the blanks, you'll see they equal to their score + 1. That 1 being that unknown of the 4 original blank columns where everyone agreed. So the answer for Colin would by 6 because he has 7 remaining answers.
- thaumasiotes 2y ago
  > Next we can assume anything matching Lisa's remaining answers are wrong This is incorrect. The assumption is true, but that's just a coincidence - you haven't justified it.

supernewton 2y ago

Dan and Lisa's answers agree on all but problems 6 and 9, and Dan got two more correct than Lisa. So Dan is correct on both 6 and 9. Therefore, Mary is incorrect on 6 and 9. So she got seven of the other eight correct, and Dan got three of the other eight correct. But on those eight problems, Mary and Dan disagree on only problems 2, 3, 5, and 10, therefore Mary got all four of those correct. Then Mary got three of problems 1, 4, 7, 8 correct, but everyone answered the same way on those, so Colin got three right on those problems as well. Then read off the deduced answer key for the other six problems to get Colin's final score.

saagarjha 2y ago

To be fair, this is one of the hardest puzzles in the game.