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Molly Mae

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Posts posted by Molly Mae


  1. Some amount of primes.

     

    27

    2, 3, 5, 7, 23, 29, 31, 37, 53, 59, 71, 73, 79, 233, 239, 293, 311, 313, 317, 373, 379, 593, 599, 719, 733, 739, 797

    So I started with single digit prime numbers, then I just built a tree for two-digit primes that begin with one of those one-digit primes.  I did the same thing for the three-digit primes that begin with one of the two-digit primes.


  2. I think I've got it.  Nah, I definitely don't have it.  I can't account for one particular case.

    First, I'd like to say that any dealer who shorts a dime is a criminal.  It's just a dime!

    My method can't tell the difference between the first weighing having 2 light vs 2 light and then a second weighing having 1 vs 1 compared to no light dimes being present during any weighing.  I'll detail my answer anyway, though, just to get it out of my head.

    First weigh 16 vs 16.  If they balance, the least optimal possibility is 2 vs 2 or 0 vs 0, which will be indistinguishable* with only one weighing remaining.  1 vs 1 would be doable and any unbalanced amount would also be doable.

    *Note: If I can come up with a way to make this distinction, I would have a solution.

    I think 16 is a good number to begin, though.  It's 1/3 of the pile and allows itself to be split in half and be left with more than the required number of coins.  It's also a solid square number and a power of 2, which likely have no bearing on this puzzle, but it does please me.


  3.  

    I'm confident there can be at most 8 cases based on the way points would be related to each other.

    plot point 1
    plot point 2 at distance called A
    plot point 3, possibilities: AA, AB, BA, BB
    plot point 4, possibilities:  AAAA, AAAB, AABA, AABB, ABAA, ABAB, ABBA, ABBB

    Maybe I'm over-generalizing though.  Maybe some of these would be colinear.  I'll put pen to paper when I can and see if they're viable.

    EDIT: Now I'm doubting myself.  I think I missed a distance.


  4. 6 hours ago, bonanova said:

    Clue:

      Hide contents

    What property of the painted squares (collectively) remains constant as more squares are painted?

     

    Oh, wow.

    The total perimeter of painted area remains the same.  For an nxn board, you have a perimeter of 4n, so the same must be true of the initially painted squares.  Since each square is 4 units in perimeter, you need n squares initially painted to make 4n.


  5. 10 hours ago, rocdocmac said:
      Hide contents

    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.


  6.  


    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, one two 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.


  7. 10 hours ago, bonanova said:

    Sounds good. FTW!

      Hide contents

    What might you be able to conclude if you added a distinguishable third hand that moved 12 times as fast as the minute hand? i.e. so that the third hand and the minute hand moved like minute and hour hands?

     

     

    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 360and 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.


  8. 2 hours ago, bonanova said:

    Having second thoughts about your answer, or difficulty seeing it.

      Hide contents

    Do the numbers represent degrees for the two hands, and you counted the coincidences? They seem to be multiples of 30, which are the hours. But 12 is the only hour the hands coincide. What am I missing?

     

    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.


  9.  

    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).


  10.  

    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 360of one hand, the other does 30o. Reduce that down to 13:1, then iterate through all possibilities.  Then we'll look for x|y that has a matching y|x.  

    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
    85 145
    98 146
    111 147
    124 148
    137 149
    150 150
    163 151
    176 152
    189 153
    202 154
    215 155
    228 156
    241 157
    254 158
    267 159
    280 160
    293 161
    306 162
    319 163
    332 164
    345 165
    358 166
    11 167
    24 168
    37 169
    50 170
    63 171
    76 172
    89 173
    102 174
    115 175
    128 176
    141 177
    154 178
    167 179
    180 180
    193 181
    206 182
    219 183
    232 184
    245 185
    258 186
    271 187
    284 188
    297 189
    310 190
    323 191
    336 192
    349 193
    2 194
    15 195
    28 196
    41 197
    54 198
    67 199
    80 200
    93 201
    106 202
    119 203
    132 204
    145 205
    158 206
    171 207
    184 208
    197 209
    210 210
    223 211
    236 212
    249 213
    262 214
    275 215
    288 216
    301 217
    314 218
    327 219
    340 220
    353 221
    6 222
    19 223
    32 224
    45 225
    58 226
    71 227
    84 228
    97 229
    110 230
    123 231
    136 232
    149 233
    162 234
    175 235
    188 236
    201 237
    214 238
    227 239
    240 240
    253 241
    266 242
    279 243
    292 244
    305 245
    318 246
    331 247
    344 248
    357 249
    10 250
    23 251
    36 252
    49 253
    62 254
    75 255
    88 256
    101 257
    114 258
    127 259
    140 260
    153 261
    166 262
    179 263
    192 264
    205 265
    218 266
    231 267
    244 268
    257 269
    270 270
    283 271
    296 272
    309 273
    322 274
    335 275
    348 276
    1 277
    14 278
    27 279
    40 280
    53 281
    66 282
    79 283
    92 284
    105 285
    118 286
    131 287
    144 288
    157 289
    170 290
    183 291
    196 292
    209 293
    222 294
    235 295
    248 296
    261 297
    274 298
    287 299
    300 300
    313 301
    326 302
    339 303
    352 304
    5 305
    18 306
    31 307
    44 308
    57 309
    70 310
    83 311
    96 312
    109 313
    122 314
    135 315
    148 316
    161 317
    174 318
    187 319
    200 320
    213 321
    226 322
    239 323
    252 324
    265 325
    278 326
    291 327
    304 328
    317 329
    330 330
    343 331
    356 332
    9 333
    22 334
    35 335
    48 336
    61 337
    74 338
    87 339
    100 340
    113 341
    126 342
    139 343
    152 344
    165 345
    178 346
    191 347
    204 348
    217 349
    230 350
    243 351
    256 352
    269 353
    282 354
    295 355
    308 356
    321 357
    334 358
    347 359
    0 0
    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
    85 145
    98 146
    111 147
    124 148
    137 149
    150 150
    163 151
    176 152
    189 153
    202 154
    215 155
    228 156
    241 157
    254 158
    267 159
    280 160
    293 161
    306 162
    319 163
    332 164
    345 165
    358 166
    11 167
    24 168
    37 169
    50 170
    63 171
    76 172
    89 173
    102 174
    115 175
    128 176
    141 177
    154 178
    167 179
    180 180
    193 181
    206 182
    219 183
    232 184
    245 185
    258 186
    271 187
    284 188
    297 189
    310 190
    323 191
    336 192
    349 193
    2 194
    15 195
    28 196
    41 197
    54 198
    67 199
    80 200
    93 201
    106 202
    119 203
    132 204
    145 205
    158 206
    171 207
    184 208
    197 209
    210 210
    223 211
    236 212
    249 213
    262 214
    275 215
    288 216
    301 217
    314 218
    327 219
    340 220
    353 221
    6 222
    19 223
    32 224
    45 225
    58 226
    71 227
    84 228
    97 229
    110 230
    123 231
    136 232
    149 233
    162 234
    175 235
    188 236
    201 237
    214 238
    227 239
    240 240
    253 241
    266 242
    279 243
    292 244
    305 245
    318 246
    331 247
    344 248
    357 249
    10 250
    23 251
    36 252
    49 253
    62 254
    75 255
    88 256
    101 257
    114 258
    127 259
    140 260
    153 261
    166 262
    179 263
    192 264
    205 265
    218 266
    231 267
    244 268
    257 269
    270 270
    283 271
    296 272
    309 273
    322 274
    335 275
    348 276
    1 277
    14 278
    27 279
    40 280
    53 281
    66 282
    79 283
    92 284
    105 285
    118 286
    131 287
    144 288
    157 289
    170 290
    183 291
    196 292
    209 293
    222 294
    235 295
    248 296
    261 297
    274 298
    287 299
    300 300
    313 301
    326 302
    339 303
    352 304
    5 305
    18 306
    31 307
    44 308
    57 309
    70 310
    83 311
    96 312
    109 313
    122 314
    135 315
    148 316
    161 317
    174 318
    187 319
    200 320
    213 321
    226 322
    239 323
    252 324
    265 325
    278 326
    291 327
    304 328
    317 329
    330 330
    343 331
    356 332
    9 333
    22 334
    35 335
    48 336
    61 337
    74 338
    87 339
    100 340
    113 341
    126 342
    139 343
    152 344
    165 345
    178 346
    191 347
    204 348
    217 349
    230 350
    243 351
    256 352
    269 353
    282 354
    295 355
    308 356
    321 357
    334 358
    347 359
    0 0


  11. On ‎1‎/‎31‎/‎2018 at 6:17 AM, plainglazed said:

    Not a generalized formula but better than strategy for three?

      Hide contents

    Label prisoners A-G.  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 0

    These 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 sneaky-clever.


  12. 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 n-1xn-1 (and n-2xn-2, 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


  13. 16 minutes ago, plasmid said:

    Would that be equivalent to the following?

      Hide contents

    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.


  14. All right, here we go:

    1. Let the sheriffs publicly order the list from 1-8
    2. They publish two prime numbers, p1 and p2.
    3. Each sheriff chooses a secret number, a and b.
    4. The first sheriff calculates p1mod p2 and sends the result publicly to the second sheriff.  We'll call this result A.
    5. The second sheriff calculates p1mod pand sends this result to the first sheriff.  We'll call this result B.
    6. The first sheriff can take Band the second sheriff can take Ato 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 pre-shared key or well-known trusted certificate.


     

    An alternative:

    The sheriffs can take turns eliminating non-suspects 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.

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