Cheating on a string theory exam
daemonology.net
Cheating on a string theory exam
1–10 of 76 posts
Re: Cheating on a string theory exam
#2- need 23 bit string to fill 23 yes/no answers
- the test is 90min * 60sec/min = 5400 sec
- log2 5400 time-indexes = ~13 bit time-address capability
23 bits required - 13 bits given = 10 bits that are unavailable. You can answer 13 questions.
Re: Cheating on a string theory exam
#3My attempt: - need 23 bit string to fill 23 yes/no answers - the test is 90min * 60sec/min = 5400 sec - log2 5400 time-indexes = ~13 bit time-address capability 23 bits required - 13 bits given = 10 bits that are unavailable. You can answer 13 questions.
(You can do better.)
Re: Cheating on a string theory exam
#4Re: Cheating on a string theory exam
#5This uses only 6 bits of information, and you should be able to pack some extra info rather easily into the remaining 7.
Thinking about alternative approaches leads me to think that the only cases that you should focus on to solve this are those where the number of true/false is almost equal. Figure out how to pack them together, layer a few solutions together, and you should be able to answer the exam to within some N guaranteed.
EDIT: 12 correct answers. not 13. And you can easily pack this into 1 bit. Leaving my wrong answer for posterity.
Re: Cheating on a string theory exam
#6Re: Cheating on a string theory exam
#7A simple approach that gives 13 correct answers minimum is to have him walk out after X seconds, where X is the number of true answers. Answer true for everything if X is greater than 12. Otherwise answer false for everything. This uses only 6 bits of information, and you should be able to pack some extra info rather easily into the remaining 7. Thinking about alternative approaches leads me to think that the only ca…
That can be reduced to a single bit:
If you should mark 'true' on all the answers, he leaves during the first half of the exam; if you should mark 'false' on all the answers, he leaves during the second part of the exam. This gets you at least 12 correct answers in 1 bit.
This leaves you with 11 bits to modify the pattern with to improve your score.
Re: Cheating on a string theory exam
#8A simple approach that gives 13 correct answers minimum is to have him walk out after X seconds, where X is the number of true answers. Answer true for everything if X is greater than 12. Otherwise answer false for everything. This uses only 6 bits of information, and you should be able to pack some extra info rather easily into the remaining 7. Thinking about alternative approaches leads me to think that the only ca…
Doesn't your method only guarantee 12 answers -- half of 23 (rounded up)? That can be reduced to a single bit: If you should mark 'true' on all the answers, he leaves during the first half of the exam; if you should mark 'false' on all the answers, he leaves during the second part of the exam. This gets you at least 12 correct answers in 1 bit. This leaves you with 11 bits to modify the pattern with to improve your s…
000,001,010,100,000 -> buddy signals 0, I write 000
111,110,101,011,111 -> buddy signals 1, I write 111
with 7 bit you definitely win 14 questions from 21
Re: Cheating on a string theory exam
#9Re: Cheating on a string theory exam
#10Earlier quoted context omitted.
Doesn't your method only guarantee 12 answers -- half of 23 (rounded up)? That can be reduced to a single bit: If you should mark 'true' on all the answers, he leaves during the first half of the exam; if you should mark 'false' on all the answers, he leaves during the second part of the exam. This gets you at least 12 correct answers in 1 bit. This leaves you with 11 bits to modify the pattern with to improve your s…
Using this on only 3 questions, you get 2/3 success rate: 000,001,010,100,000 -> buddy signals 0, I write 000 111,110,101,011,111 -> buddy signals 1, I write 111 with 7 bit you definitely win 14 questions from 21
I've been trying to think if there's a clever scheme to spread the knowledge around to beat 17 with statistical tricks.
(As a fun aside: 17/23 is about 73%, which typically gets you a C -- so you'd pass the exam.)