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Posts posted by Molly Mae


10 hours ago, rocdocmac said:
What will happen if either Carroll or Kurt has 4 or 6 coins in their cell when Kleene's question on the "total = 15" is NO? One can still arrive at a total 9 or 21. Carroll and Kurt don't know that Kleene has 5.
One can possibly figure out a question to pose wrt the "total number of coins", but what if the OP stated "number of coins in each cell"? Then the answer is obviously 9 or 21 since no one would be able to figure out how to get to a total of 15 (four cases). So I think that, in such an instance, one should merely ask ... "Does the sum of the coins have an integer square root?" YES would mean a total of 9, NO would mean a total of 21.
Sorry for side tracking!
Since Kleene does have 5, though, if the answer to the question is no and the total isn't 15, Carroll and Kurt can't have 4 or 6 coins. They must have either 1 and 3 or 7 and 9.
Anyone who has 1 or 3 can deduce that they won't have enough coins for 21. Anyone who has 7 or 9 can deduce that they have too many for 9.
If the answer isn't 15, 4 and 6 coins aren't possible since Kleene has 5.

Kleene: Is the total value of the coins 15?There are currently 3 possibilities for their total: 9, 15, and 21. If the answer to the question is yes, everybody knows the total. If the answer to the question is no,
onetwo of the three (both Carroll and Kurt) should be able to deduce whether the total is 9 or 21.If anyone has 7, 8, or 9 coins, they can deduce that there must be 21 coins (they would have to total more than 9).
If anyone has 1, 2, or 3 coins, they can deduce that there must be 9 coins (they couldn't sum to 21 given the rules).Since Kleene has 5 coins, there is no way to make 21 coins without someone having 7 and the other having 9. There's no way to make 9 coins without someone having 1 and the other having 3.

A belt?

#3
Looking behind the facade, I'm going to have to say 5. Once I realised it wasn't a sequence, it was the only thing that made sense.

10 hours ago, bonanova said:
In that situation, the second:minute hand is the same as the minute:hour hand. There are many more cases to evaluate where the hands could be ambiguous, and you can't just compare them to the already ambiguous cases because it will create new ones (I'm fairly confident on this point). Knowing how long it took me to do just two hands, I don't think I'd like to run it against three. =P I wouldn't be surprised to find that there's an easier way to get to the answer, but I went with the way that made sense in my head. If I ever did do it again, I'd be certain to use a scheme that doesn't involve 360^{o }and go with something that uses whole numbers while being easier to calculate 1/12 of whatever unit I use.
Of course, now that I've typed this up and see that you used the word "distinguishable" instead of "indistinguishable" I'll have to reassess my whole last paragraph. Knowing how many seconds has passed
wouldn't help at all in that moment, though.would actually help, I believe. It breaks the minute hand down into yet finer movements to be tracked.Now I'm starting to think about the "paradox" of moving across a room (where you should never reach the end since you must cross half the room first, then half the remainder). There probably isn't actually a parallel between the two, but I can see how you could break the movements down smaller and smaller forever, but I don't think you'll ever come up with more ambiguous moments (you won't, actually). But yeah, next time I'd like to divide the clock into 24000 tocks and divide those into yet smaller ticks so that one tock is 12x ticks and come up with some reasonable x that doesn't make the working numbers too small or large (just for convenience).
Now I feel as though I've rambled.

I should also reiterate that I wasn't looking for xy where x=y. Those are the 22 results that I removed, since they weren't ambiguous. I looked for xy where there was a complimentary yx

2 hours ago, bonanova said:
Yeah, I ended up taking it a step further and moving the minute hand by 1 degree at a time instead of the hour hand. So the matching columns were a looooot longer. I couldn't include it in the post because it was too long.

So 286 times the hour and minute hand are interchangeable to make a realistic clock reading. 22 of those times aren't ambiguous, though, since the hands are both on the same position. So my answer is 264 (which is a whole lot higher than I expected).

More observations:
We don't care how fast the hands actually move, only their relationship and how many times they move in a 24 hour period. For each 1 full rotation one hand does, the other does 1/12 of a rotation. We'll divide the clock into degrees instead of minutes, so for each 360^{o }of one hand, the other does 30^{o}. Reduce that down to 13:1, then iterate through all possibilities. Then we'll look for xy that has a matching yx.
Position in degrees:
13  1
26  2
39  3
and so on.We can continue to do this until the second column reaches 360 and then we can take the first column mod 360 to actually start comparing against itself.
I'm working on that now, in between my busy schedule. =P
EDIT: Forgot to add that we actually need to double the final result to get a complete 24 hours.
Just ran this up in excel: Probably too large for a post. Now I just have to find ones with a duplicate inverse.
13 1 26 2 39 3 52 4 65 5 78 6 91 7 104 8 117 9 130 10 143 11 156 12 169 13 182 14 195 15 208 16 221 17 234 18 247 19 260 20 273 21 286 22 299 23 312 24 325 25 338 26 351 27 4 28 17 29 30 30 43 31 56 32 69 33 82 34 95 35 108 36 121 37 134 38 147 39 160 40 173 41 186 42 199 43 212 44 225 45 238 46 251 47 264 48 277 49 290 50 303 51 316 52 329 53 342 54 355 55 8 56 21 57 34 58 47 59 60 60 73 61 86 62 99 63 112 64 125 65 138 66 151 67 164 68 177 69 190 70 203 71 216 72 229 73 242 74 255 75 268 76 281 77 294 78 307 79 320 80 333 81 346 82 359 83 12 84 25 85 38 86 51 87 64 88 77 89 90 90 103 91 116 92 129 93 142 94 155 95 168 96 181 97 194 98 207 99 220 100 233 101 246 102 259 103 272 104 285 105 298 106 311 107 324 108 337 109 350 110 3 111 16 112 29 113 42 114 55 115 68 116 81 117 94 118 107 119 120 120 133 121 146 122 159 123 172 124 185 125 198 126 211 127 224 128 237 129 250 130 263 131 276 132 289 133 302 134 315 135 328 136 341 137 354 138 7 139 20 140 33 141 46 142 59 143 72 144 85 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On 1/31/2018 at 6:17 AM, plainglazed said:
Not a generalized formula but better than strategy for three?
Label prisoners AG. Label Yellow hats 0 and Green hats 1.
Each prisoner must memorize these eight strings and their inverses as well as their relative position within the group:
A B C D E F G
S1  0 0 0 0 0 0 0
S2  0 0 0 1 1 1 1
S3  0 0 1 0 1 1 0
S4  0 0 1 1 0 0 1
S5  0 1 0 0 1 0 1
S6  0 1 0 1 0 1 0
S7  0 1 1 0 0 1 1
S8  0 1 1 1 1 0 0These sixteen strings and all strings created by changing just one bit of each of the sixteen comprise all 128 possible two
color hat combinations with seven prisoners.
So now the strategy: Each prisoner sees six bits (hats) and can create two seven bit strings by inserting a 0 and a 1 at
their respective position. If neither of those two strings match any of the sixteen memorized strings, abstain. If one of
those two strings matches one of the memorized strings, choose the other bit (hat color) that did not create a match.
This effectively packs seven wrong answers in each of the sixteen memorized strings and gives one correct response in each
of the other 112 cases. So with 7 prisoners (or more) freedom is possible 7/8ths of the time.
These are the kinds of things I feel like I'm normally good at, but I stared at it for a long time and I eventually unpacked it. Since each string is different from the other 15 by at least three bits, you can guarantee that two people won't choose a colour. That's sneakyclever.

I again think I've got it
The Andromeda Galaxy

A few observations that might lead to a proof:
1. There must be a red square on each extreme rank and file (this can be, but isn't required to be, the same red square for a given rank/file combination) for any nxn.
2. This must also be true for n1xn1 (and n2xn2, and so on)
3. Since we can always reduce n to 1, there should be no optimal strategy that uses fewer than n red squares for a board that's nxn.Probably

Yeah, now I think I've got it.
Glass/Silicon (which is in the Carbon group!)

My wife's guess:
The carbon group

16 minutes ago, plasmid said:
Would that be equivalent to the following?
I’ll be the first sheriff to speak. I’ll tell the other sheriff to write down the eight suspects’ names in a circle in the order that I specify, going clockwise with 45 degrees between each name. Unbenknownst to the other sheriff or the eavesdroppers, I’ll arrange the names in the list such that my two suspects are at opposite ends of a diameter of the circle. I’ll ask the other sheriff to name who’s halfway between his two suspects on the list along the shortest arc that joins them (and if his suspects are say #1 and #4 then he should say the midpoint is between names #2 and #3). I’ll know that the midpoint that he specifies – between the guilty party and an innocent that’s less than 180 degrees away – is closer to the guilty than to my innocent suspect so I’ll know who the guilty suspect is. Then I can tell the other sheriff whether to go clockwise or counterclockwise from his midpoint to find the guilty suspect. The eavesdroppers shouldn’t be able to tell who’s guilty with that approach.
Does that allow the other sheriff to make the arrest, though? That was the stumbling block I was having with my second approach.

All right, here we go:
1. Let the sheriffs publicly order the list from 18
2. They publish two prime numbers, p_{1} and p_{2}.
3. Each sheriff chooses a secret number, a and b.
4. The first sheriff calculates p_{1}^{a }mod p_{2} and sends the result publicly to the second sheriff. We'll call this result A.
5. The second sheriff calculates p_{1}^{b }mod p_{2 }and sends this result to the first sheriff. We'll call this result B.
6. The first sheriff can take B^{a }and the second sheriff can take A^{b }to arrive at the same secret result, even though the first sheriff never knows b and the second sheriff never knows a (and the public never knows either).We can then use any public arithmetic to arrive at a result without giving away our secret number, and hence our accusation.
This is the basis for any kind of secure internet connection that doesn't involve a preshared key or wellknown trusted certificate.
An alternative:
The sheriffs can take turns eliminating nonsuspects from the list or passing. They can easily get down to a list of 2. If, at any time, one sheriffs list has been eliminated, he can announce it. The sheriffs don't agree. If they ever reach a list of 3 where both sheriffs have both of their suspects, the sheriff will pass. They must agree on at least one suspect. The sheriffs can restart as many times as they like, without ever publicly revealing who is actually on their list.
I'm trying to determine how many times they could do that before the public can deduce the identities of the suspects.
It kind of turns into reverse Mastermind.
My second method actually doesn't work. It lets the sheriffs know that they share a suspect (or don't), but doesn't tell them who the shared suspect is.

Perhaps spectacles/glasses? Or more specifically, eyes that need glasses?


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.
What dus the poor man have
in New Logic/Math Puzzles
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