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Posts posted by Molly Mae



My initial assumption, without diving too deep yet, is that it will happen once per hour. I'll have to verify that, though, since noon/midnight might actually be that time for the 00/12 hour. I can say almost certainly it's an even number, since any AM ambiguity will be the same for PM. That's not much of on observation, though.

9 hours ago, bonanova said:@Molly Mae Consider that the sheriffs are simply talking to each other, using statements like, "my pair of suspects has this property: (describes property.)" Would generating and using common key agreements be doable with such lowcomplexity statements?
If the list of suspects can be ordered publicly, they could use a key exchange to come up with a secret and use that to identify a particular number using a public modulus. I'll type up a demonstration when I get a moment.

9 hours ago, bonanova said:True, and that is a proof by construction, getting you a point.
There is an unexpected proof that comes from looking at the hexagon from a slightly different angle, literally. Can you find it for the win?
I think I might have it.
We can construct a stylised 3D cube out of the three diamonds in the shape of a 2D hexagon. In order to complete the cube/hexagon, we require one of each diamond orientation. For a cube of length n, there will be n^2 number of diamonds of each orientation.
I hope that makes sense. I wish I could scan and upload my drawing. Once I realised that it's a stylised cube for n=1, it took me a bit to actually draw out n=2. Once I did, though, it's easy to see that as one side of the cube grows, the others must (obviously) grow as well in order to maintain the shape of the regular hexagon (and cube).

9 hours ago, bonanova said:Can you prove that's minimal?
Can you think of a necessary property for your red squares?
I can try to put into words why I think it's minimal, but I don't know if it will qualify as a proof.
Some observations:
1. There must be a red square in a rank for that rank to become red. The same is true of files.
2. You could create a red square in a new rank only if two red squares already exist in different ranks along the same file, which is never more optimal. The same is true of files.
3. Since we can't add a new rank without giving up a file, the optimal placement is along a diagonal.
4. Any other diagonal other than the long diagonal doesn't cover every rank and file.
5. There are other configurations that still work by manipulating the diagonals, but all of them are n length and all of them cover the same colour square. 
If I understand correctly:
For n=1, yes.
For n=2, also yes: a triangle with a point in the centreFor n>2, should I assume that the lines are 1) redrawn for next possible colouring of each point, 2) evaluated for crossings after each redraw, 3) erased if if passes the evaluation, and 4) iterated back 14?

I can definitely do it in n by painting either diagonal that begins (and ends) in a corner.

First, we'll look at each "ring" of hexagons. The centre hexagon contains 6 diamonds in 2 sets of 3. In order to not overlap diamonds, choosing the first one forces the orientation of the other two to be different. For each ring around the centre hexagon, the same is true. Once a single orientation is chosen, the orientation of all of the diamonds in that ring are chosen. Additionally, two adjacent edges will share an orientation before the orientation is forced to change (if you want me to prove why the orientation is forced to change, I can't come up with anything other than "you've run out of room"). This pattern continues for all rings, regardless of how many you add.

From what I know of DiffieHellman, this shouldn't be too difficult!
Can I use the DiffieHellman key exchange as an answer? Or are we trying to come up with a way that doesn't involve common key agreement algorithms?

Ah
Is it beautiful like diamonds in the sky?
Shine bright like a diamond!

It makes me think of teeth, especially braces. While the standard set of teeth would have 16 on top and 16 on bottom, maybe braces don't connect to the last tooth. I don't know that much about them.

For 3 prisoners:
If any prisoner observes 2 hats of the same colour, he should guess the other colour.
GGG  they die
YGG  they live
GYG  they live
GGY  they live
GYY  they live
YGY  they live
YYG  they live
YYY  they dieThat also succeeds with 75%.
EDIT: Ninja'd by Izzy.

I'lI can do better than a standard coin toss, but the worst case scenario still comes down to a coin toss.
For two prisoners, let the first prisoner pass is he sees that his partner has a green hat. His partner is informed to always guess green. If his partner has a yellow hat, the first person guesses any colour.
50% of the time, they are guaranteed to live. 25% of the time, they live based on a correct guess from the first person. 25% of the time, they die.
75% success rate is better than a coin toss, but I'm convinced there's better.

Hahaha
1. Abominable.
2. Noble

While I've never seen this arrangement of words, I've seen a similar observation on a study of the same word.
Btw, I laughed when I reread it.

41 minutes ago, rocdocmac said:Thanks Thalia,
Interesting twist in the tale, for sure!
Yes, surely 26 cubelets removed leaves a single cubelet, but where was the remaining one positionally located/"fixed" before removal of the other 26? Should be one specific remnant shape for each combination and, therefore, four variants/labels/possibilties.
"Nothing" or "no shape" may be regarded as a "blank" stacked cube.
(1) What's your suggestion for the end of the sequence?
(2) What do our other BrainDenners think?
(3) How many distinct shapes can one expect for a 4cubelet removal? Just a ballpark figure, perhaps!
2) I wouldn't use "what shapes are left" but instead keeping with "after removing 27 cubelets, how many possibilities are there?" Which is 1.
3) ~950

On 1/22/2018 at 5:40 PM, bonanova said:This is another puzzle where precise wording is important  I'll try to get it right, but if anything is unclear, please ask ... I'll start out by saying that all the circles in this puzzle have the same radius, the aspect ratio of the rectangle is not specified and does not matter, and its size, relative to the size of the circles is only indirectly implied. Only the constraints stated in the puzzle should be assumed.
 I've drawn 17 circles that at least partially overlap a rectangle.
 Their centers all lie within the rectangle.
 None of the circles overlap or even touch any of the other circles.
 There is no room for an 18th circle to be added to the group.
That is, the circles are drawn in such a way that even though there is space between them, it is impossible to draw another circle whose center lies within the rectangle that does not at least partially overlap one of the first 17 circles. That is all you know about the relative sizes of things. And it is enough information to answer the following question:
First, let's erase the circles that I drew. Then I will paint the rectangle red and give you a large supply of opaque white circles. What is the smallest number of circles you will need to completely cover the rectangle? (so that no red will be showing.) The centers of the circles, again, must lie within the rectangle, but now, of course, the circles can overlap each other.
Clarifying question: Is the rectangle as large as possible meeting those constraints? Or as small as possible meeting those constraints? I feel like the number of circles required depends on that, but maybe it doesn't.

Let's think more generally. Draw point A, the diameter containing A, point B, and its diameter (thanks, Iz!). This is certainly the right path.
What is the expected area of the quadrant opposite these two points? It can be as high as 1/2 (A, center, and B are colinear and found on the line in that order) or 0 (A, B, and center are colinear and found on the line in that order). Those two (and all possibilities in between) occur randomly with equal probability. The average of all them is 1/4.
If there's another (and better!) way to arrive at it, I'm sure Izzy can demonstrate it when she isn't busy.



The word buried here has but one letter. 
Did you find a jellyroll in Gearhardt's Bakery? 
It's the best oneI've ever seen. 
The rug at her stairway was made in India. 
He's an old friend. 
Amos sold his bicycle to an old friend.
A rolling stone gathers no moss.
That's a really neat puzzle!


1 hour ago, flamebirde said:A thought:
The first two points (call them A and B) don't matter, since if you are able to choose the third one, it's always possible to choose one such that the center is within the triangle. (draw a line from the center of the line segment AB to the center, and then place the third point on the line at any point after it intersects the center.)
Hence, the question is only really about the third point.
More thoughts:
The farther you go on that line, the more tolerance you have for moving off it (i.e. when you've only gone one unit on the line, you might only be able to deviate .1 units off it, whereas if you go ten units you might be able to deviate by a full unit). Hence, the region where the third point needs to be placed is always within a certain sector of the circle.
Oh, nice start.
My answer is 1/4.
Draw point A. Then draw the diameter of the circle that passes through A and the centre. This divides the circle into two distinct halves. We can then draw a line perpendicular to this line to divide the circle into 4 equal parts.
Choose point B at random. If point C is in the quadrant directly opposite point B, the cenre is inside the triangle. Otherwise, it isn't.
Note: I disregard all subsets of points that are colinear. I also include the centre as being "in" the triangle if it lies on an edge of the triangle.

46 minutes ago, Izzy said:So, the above answer can be generalized a bit. Let everyone pick a partner, and number the partners 0 and 1. Everyone agrees to say "green", unless they count the exact number of greens hats as their number, The possible colorings are:
green, green (guy on left lives)
green, yellow (guy on right lives)
yellow, green (guy on right lives)
yellow, yellow (guy on left lives)
Since we can have 50 such groups, 50 people are guaranteed to live (and die!).
Wouldn't green, green result in both saying yellow? [/spoier]
Edit nvm. I misread.

48 minutes ago, Izzy said:Okay, what about this. (Long shot. Does the warden assume the prisoners to be truthful? Probably not.)
The prisoners decide aloud, as MM mentioned, to each say the hat color that they see most frequently. (So, if they see 4950, they say the color of the 50.)
"Drats!" thinks the warden. "A majority of them will survive, which reflects poorly on my wardening ability. Unless! Hah. I will make the hats 5050 and then they will all die and I'll get a raise." The warden of course believes the prisoners, because why would they lie to each other during their only opportunity to plan?
Of course, the perfectly logical prisoners knew the warden would think of this solution. The hats are distributed 5050, and everyone lives by saying the opposite color of the majority that they see.
The only problem is that it doesn't guarantee that anybody lives.
Too much WiFoM

I can save many for all cases other than the 5050 and 4951 (likewise 5149, obviously). In either of these cases, I believe the warden has a direct riposte to the any strategy that is discussed.
I’m not a gardener
in New Word Riddles
Posted · Edited by Molly Mae · Report reply
Perhaps spectacles/glasses? Or more specifically, eyes that need glasses?