What is the Knowledge Horizon Contest and how can you participate?

In summary, a Knowledge Horizon Contest was proposed in which participants could enter by posting a single question on a given thread, with the restriction that the participant must know the correct answer to their question. After all questions were submitted, a solution period began where participants could send their solutions to the corresponding author. Each solved problem earned the author one point, and the participant with the most points would be declared the winner. A specific problem was given as an example, and a solution was provided. The next contest was planned to begin on Friday, APR 29 0:00 GMT, with a restriction on the subject area to only include physics and math questions. One participant suggested a chemistry question, but it was decided to stick to physics and math.
  • #1
hemmul
Well... let me offer a small contest - I called it Knowledge Horizon Contest, and you'll find out why in a couple of minutes:

1. Everyone who wishes to enter the contest has to post a single question in this thread till Friday 22 APR 0:00 GMT. The poster is recognized as an author of the question, till the end of the contest.
2. It can be any question - regarding physics, astronomy, biology, math, computer science, etc... The only restrictment is that You should know the correct answer to the question yourself. Also i suggest not posting history/literature and other location/nation-specific questions for it is not fair with respect to others ;) however Earth Science/Georgaphy - is welcomed! i hope you got what i mean :)
3. After all questions are in, the solution period starts! Once a player solves a problem he/she sends the solution as a private message to the corresponding author. You have only one chance to send the solution in, regardless if it is true or false... as soon as the author receives and checks the solution - he/she has to PM back to the sender and notify him whether he succeded.
Solutions are considered during a week from the start of the contest, e.i. the contest is over at 0:00 GMT, Friday 29th April.
4. After that each author has to post a valid solution to his problem, and a list of contestants who solved it. (maybe he has to post all of the solutions as well, to make sure he hadn't miss something)

1 solved problem = 1 point. (however it is better to create a corresponding weight for the problem, depending on the number of contestants who solved it)
One with most points is the WINNER!

what do you think? ready to go? :)

P.S. yes, the amount of fun depends on the honesty of contestants - no cheating - fair play!
 
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  • #2
Well, I may be the only participant here. I foresee this not leading to many correct answers as it is in each participant's best interest to post very difficult questions.

I saw this type of problem while reading my discrete math book and I think the way to solve it is pretty hard to figure out.

Here's the problem:
How many subsets of size 10 are there from the set {1, 2, ... , 500} so that no subset contains any consecutive integers?

I think that's sufficiently large so you wouldn't have much success if you try to write a brute force program. Just the expression is acceptable as an answer, you don't have to calculate it out.
 
  • #3
Also no "cheap" answers like writing your brute-force program as the expression and saying the answer is "the eventual output of this program."
 
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  • #4
Looks like I might win by default. (Now someone's going to post an impossible problem, no doubt)
 
  • #5
Okay, I declare myself winner by default! Here is how to solve the problem:
Consider the number of whole integers between the nearest members of the subset. For example, if the subset contains 1 and 3 then there would be 1 whole integer between them, namely 2. Also, to find the location of the first integer, count the number of whole integers between 0 and the first integer, and to find the location of the last, count the number of whole integers between the last and 501. The sum of all of these gaps must be the number of whole integers that are not selected, or 500-10 = 490, so the answer is the number of ways to have the gaps so they sum to 490.

Using the generating function method, the generating function for the number of ways to have the gaps sum to a given number is
(1 + x + x^2 + ...)((x + x^2 + x^3 + ...)^9)(1 + x + x^2 + x^3 + ...)
= x^9 * (1 + x + x^2 + ...)^11
= x^9 * (1/(1-x))^11
= x^9 * (1-x)^-11
The answer is the coefficient of x^490 in this expansion, which is the coefficient of x^481 in the expansion of (1-x)^-11. By the binomial theorem, this is
C(-11, 481) * (-1)^481
= (-1)^481 * C(11 + 481 - 1, 481) * (-1)^481
= 204639619731902815383

You could also do this without generating functions.
 
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  • #6
BicycleTree,

you're breaking the rules :)
no solutions in this thread till the end of the contest...
for a violation of the rules you should be eliminated from the contest, but taking into the account the whole number of participants - i declare YOU the winner! Congratulations! You're the number 1!
actually i did like your problem - but you really should not have posted the solution that early...

so... what about the second tournament? ;) point a date?
 
  • #7
Damn, I would have posted, but I didn't see this. How's a second tournament?
 
  • #8
Icebreaker said:
Damn, I would have posted, but I didn't see this. How's a second tournament?

very nice of you! thanks for being intereted :)
well, i like fridays :)
unfortunately i was too busy this week to come up with a problem and participate myself...
i think the next contest start time should be Friday, APR 29 0:00 GMT
the problems should be in by that time.
Also, you're surely welcomed to share your ideas concerning the contest till friday - maybe you have something that will make it far more better ;)
so, 5 days to go! make a right decision! participate! share your cool ideas! WIN!
 
  • #9
hemmul said:
i think the next contest start time should be Friday, APR 29 0:00 GMT
I'm in!

i think you should restrick the subject area to physics and math only, bcs it's not fair to ask say chemistry question in here...
however IMO the idea of such a contest is COOL :approved:
 
  • #10
K I learned this in science:


what are the five elements that combine to make Nicotinamide adenine dinucleotide phosphate? Slso, how many atoms are there in each one? :smile:




No cheating, I HATE CHEATERS!
 

Related to What is the Knowledge Horizon Contest and how can you participate?

1. What is the Knowledge Horizon Contest?

The Knowledge Horizon Contest is an annual competition that aims to recognize and reward innovative research and discoveries in various fields of science and technology. It provides a platform for scientists and researchers to showcase their work and compete for prestigious awards and prizes.

2. Who can participate in the Knowledge Horizon Contest?

The Knowledge Horizon Contest is open to all scientists, researchers, and individuals who have made significant contributions in their chosen field of study. Both individuals and teams can participate, and there are different categories for students, professionals, and organizations.

3. What are the benefits of participating in the Knowledge Horizon Contest?

Participating in the Knowledge Horizon Contest can bring various benefits, such as recognition for your work, networking opportunities with fellow scientists, and access to funding and resources to further your research. It can also enhance your professional profile and credibility in the scientific community.

4. How are winners chosen in the Knowledge Horizon Contest?

The winners of the Knowledge Horizon Contest are chosen through a rigorous evaluation process by a panel of expert judges. The judging criteria include the originality, impact, and feasibility of the research, as well as its potential for real-world applications and contributions to the advancement of knowledge.

5. How can I apply for the Knowledge Horizon Contest?

To apply for the Knowledge Horizon Contest, you will need to submit your research or project proposal through the official website or designated application portal. The submission requirements and deadlines may vary each year, so be sure to check the guidelines for the current edition of the contest.

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