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August 2016: <a href="https://en.m.wikipedia.org/wiki/Hat_puzzle" rel="nofollow">https://en.m.wikipedia.org/wiki/Hat_puzzle</a>
The solution to the 4 person problem is actually pretty simple.
Let's assume that we know the answer to the 2 person problem.
We have 4 persons; A, B, C and D.
At first the persons guess their colour (without revealing it of course) based on their direct neighbor.
So for instance:<p><pre><code> A <-----> B
C <-----> D
</code></pre>
A and B being neighbors, they guess their colours based on the solution of the 2 persons problem. Same goes for C and D.
Without loss of generality, let's assume C guesses correctly his card colour.
Since he can see the cards of A and B, he knows who will guess their colour correctly.
For instance, if they have the same colour, A will guess correctly, if not, then B will guess correctly.
Without loss of generality, let's assume A guesses correctly his card colour. He then sees that C will guess correctly.<p>The participants agreed beforehand that clubs and hearts are associated with 0 while diamonds and spades are associated with 1.
A and C now apply the solution to the 2 person problem to guess their shape group (either (diamonds or spades) or (clubs or hearts)):<p><pre><code> A B
|
|
C D
</code></pre>
It is guaranteed that at least one has guessed the shape group correctly. Since they both guessed correctly the color, it is guaranteed that one of them will guess the shape correctly.
Some of these are poorly worded. Why does the answer to the May one keep talking about 30 seconds passing when the only time mentioned in the problem is 30 minutes? 30 seconds gives Nick at least 10 seconds per possible solution, he should have tried them all by then.<p>It should talk about the number of tries so far (one each), the length of time it took is irrelevant. But why are they even taking turns when they have separate padlocks and could easily brute-force it? I get the concept they're going for but the premise doesn't fit and just confuses things.
The NSA also has (had?) an internal puzzle in there cryptolog internal magazine. Some have been declassfied at <a href="https://www.nsa.gov/news-features/declassified-documents/cryptologs/" rel="nofollow">https://www.nsa.gov/news-features/declassified-documents/cry...</a>. An example of their puzzles: <a href="https://david.newgas.net/nsa-puzzle/" rel="nofollow">https://david.newgas.net/nsa-puzzle/</a>
It said no communication during but you could devise a strategy before play. What I would do on the first challenge is say I would write down the opposite color of what I see and tell the other person the write down the same color they see so you will definitely at least win $50 "or 100$ if you have the same color". For the suite challenge do the same strategy but split 2 people write down the same suite and have a plan for the other 2 to pick what they don't see.
The answer is pretty simple.<p>1. The strategy is that they should agree to write down the same colour(either red or black), irrespective of what they see on the opposite persons forehead<p>2. Same as above. This time all of them should agree to write any single suit.<p>By probability, at least one of them will get it right.<p>[Edit]: Didn't put it correctly before
Reminds me of "ponder this" from IBM research: <a href="https://researchweb.watson.ibm.com/haifa/ponderthis/index.shtml" rel="nofollow">https://researchweb.watson.ibm.com/haifa/ponderthis/index.sh...</a>
Make Bruce guess opposite of what he sees on Ava and Ava guess same color as what she sees on Bruce. This covers atleast one correct guess for all for Cases BB, RR, BR, RB.
<a href="https://github.com/zazaalaza/napp" rel="nofollow">https://github.com/zazaalaza/napp</a><p>see my solution here
For the July one:<p>Always go first.<p>Assuming the table is symmetrical around only one point, the centre of the table, place the first coin on that point.<p>Every move after that place your coin in the location directly opposite the last coin played, the mirrored location around the point of symmetry.<p>Since you are always playing a symmetrical position, you know that there will always be a space available for you after the previous coin was played.<p>[edit] looks like the answer on the site is almost the exact wording as mine :)