TimeSpaceLightForce Posted October 26, 2014 Report Share Posted October 26, 2014 (edited) The warden will release the 7 prisoners to be hang on everyday next week if they can mark their hats with a number card from a playing card deck so that anyone may see and guess right the others prisoner's day and time of executions. The cards shall be erected on hat through a small straight cut on top. The schedule will be randomly drawn for everyone. No prisoner must tell his scheduled day (Sun,Mon..Sat) and hour of execution (1:00,2:00,3:00 ... 12:00). Since these prisoners had freed themselves a couple of times on this same prison it should not be hard for them to agree on a plan. What code strategy works for them? Edited October 26, 2014 by TimeSpaceLightForce Quote Link to comment Share on other sites More sharing options...
0 karthickgururaj Posted October 26, 2014 Report Share Posted October 26, 2014 Not sure if I understand the conditions of the puzzle well.. sticking a card on their head, that is (say) 1 day and 1 hour ahead of their scheduled time? Each prisoner knows their own scheduled time of execution right? Quote Link to comment Share on other sites More sharing options...
0 gavinksong Posted October 26, 2014 Report Share Posted October 26, 2014 Are all the executions on different days? Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted October 26, 2014 Author Report Share Posted October 26, 2014 (edited) . but what prevents them from.. sticking a card on their head, that is (say) 1 day and 1 hour ahead of their scheduled time? They have more time than that. After they know their own day and time of execution they can stick the card on their hat any day until the next Sunday (1st day of execution). They are not prevented to roam around the prison yard or to eat on same table wearing their hats. They are prevented to tell, hint or give any signal to each other. Right .Each prisoner knows their own scheduled day and time of execution. Edited October 26, 2014 by TimeSpaceLightForce Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted October 26, 2014 Author Report Share Posted October 26, 2014 Are all the executions on different days? Yes, everyday next week starting Sunday. Quote Link to comment Share on other sites More sharing options...
0 gavinksong Posted October 26, 2014 Report Share Posted October 26, 2014 . but what prevents them from.. sticking a card on their head, that is (say) 1 day and 1 hour ahead of their scheduled time? They have more time than that.After they know their own day and time of execution they can stick the card on their hat any day until the next Sunday (1st day of execution).They are not prevented to roam around the prison yard or to eat on same table wearing their hats.They are prevented to tell, hint or give any signal to each other.Right .Each prisoner knows their own scheduled day and time of execution. I guess the prisoners can wear the card that corresponds to the hour of their execution exactly a week before their execution. That way, they can broadcast the day and the hour. But that can't be right. It's too simple. I'm probably missing something...? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted October 27, 2014 Report Share Posted October 27, 2014 These questions probably have obvious answers, but just to clarify: Do the prisoners meet initially to decide on a strategy? Are the times all different? Does 1:00 - 12:00 signify a span of 12 hours or of 24 hours? That is, are all times a.m.? Or all p.m.? Or can they be either? The prisoners would need to know the number of possibilities. Finally, to escape, is it required that every prisoner inform the warden, correctly, of all seven execution schedules, prior to the first execution? It seems that the task is to encode a day and time with ace-10 in one of four suits. That's 40 possibilities. But 7x12 has 84 possibilities. More info is needed. Something akin to timing of the display of the cards must come into play. Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted October 27, 2014 Author Report Share Posted October 27, 2014 These questions probably have obvious answers, but just to clarify: Do the prisoners meet initially to decide on a strategy?Yes, after the warden's announcement and conditions. Are the times all different? Depends on random outcome (12 marks Clock roulette) Does 1:00 - 12:00 signify a span of 12 hours or of 24 hours? Not 24 hours That is, are all times a.m.? Or all p.m.? Either The prisoners would need to know the number of possibilities. Daylight time (dawn to dawn) 6am to 5pm Finally, to escape, is it required that every prisoner inform the warden, correctly, of all seven execution schedules, prior to the first execution? Prisoners write their guess collected before Sunday. A list of :Name___Time____Day___ It seems that the task is to encode a day and time with ace-10 in one of four suits. That's 40 possibilities. But 7x12 has 84 possibilities. More info is needed. Something akin to timing of the display of the cards must come into play. No timing here..the yard time and meal time has 1 hr fixed schedule while clocks and watches were confiscated by warden. Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted October 27, 2014 Author Report Share Posted October 27, 2014 Order of event: 1. the death sentence 2. warden's clemency announcement and conditions (OP) 3. plan agreement 4. random schedules of execution 5. card selecting 6. wearing of hats 7. writing the guess list 8. escaping or hanging (the warden would make the release look like an escape) Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted October 27, 2014 Author Report Share Posted October 27, 2014 I guess the prisoners can wear the card that corresponds to the hour of their execution exactly a week before their execution. That way, they can broadcast the day and the hour.Unless all the sentence,announcement, planning,agreement,scheduling,card selection and hat wearing did not happened on Sunday . But that can't be right. It's too simple. I'm probably missing something...? No, it is just unlikely solution..can you give it another go? . Quote Link to comment Share on other sites More sharing options...
0 Barcallica Posted October 30, 2014 Report Share Posted October 30, 2014 Order of event: 1. the death sentence 2. warden's clemency announcement and conditions (OP) 3. plan agreement 4. random schedules of execution - what day is it? Monday? 5. card selecting 6. wearing of hats 7. writing the guess list 8. escaping or hanging (the warden would make the release look like an escape) Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted October 31, 2014 Author Report Share Posted October 31, 2014 I dont see what the days of event has to do with the strategy as long as they occur within the week before execution but not on just one day. Say they happened : 1. the death sentence -Sunday 2. warden's clemency announcement and conditions (OP)-Monday 3. plan agreement -Tuesday 4. random schedules of execution - Wednesday 5. card selecting-Thursday 6. wearing of hats-Friday 7. writing the guess list-Saturday 8. escaping or hanging-Sunday (sunrise to sunset) Quote Link to comment Share on other sites More sharing options...
0 Barcallica Posted October 31, 2014 Report Share Posted October 31, 2014 Monday prisoner goes first put his card and then tuesday etc... They need to identify only 12 hours with 40 cards. Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted October 31, 2014 Author Report Share Posted October 31, 2014 Monday prisoner goes first put his card and then tuesday etc... They need to identify only 12 hours with 40 cards. Correct, they must wear the hat with marker stucked already when going outside their cells. When they are on the yard or mess hall, the act of wearing or taking off the hats or sticking the card on hat can be mistaken as trying to communicate. Lots of suspicious eyes and ears around. It looks like the information they needed must be just on the marked hat alone. There is a clue in the OP. Quote Link to comment Share on other sites More sharing options...
0 Barcallica Posted November 4, 2014 Report Share Posted November 4, 2014 It seems that the task is to encode a day and time with ace-10 in one of four suits. That's 40 possibilities. But 7x12 has 84 possibilities. More info is needed. Something akin to timing of the display of the cards must come into play. We could place the card facing front or back, so we now have 80 possibilities. 4 to go. (3 if we can choose not to place any card) Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted November 4, 2014 Author Report Share Posted November 4, 2014 what if 2 did not place any card Quote Link to comment Share on other sites More sharing options...
0 plasmid Posted November 6, 2014 Report Share Posted November 6, 2014 Hearts, spades, and clubs are all have asymmetrically shaped symbols such that they can be oriented vertically upright or upside-down, and those two orientations are clearly distinguishable. Any playing cards of odd number will have a symbol for their suit in the center of the card that will break symmetry. So in addition to having 40 cards x 2 orientations (facing frontward or backward), there will be 15 cards (odd numbered cards with a heart, spade, or club) that can be either upright or inverted, so there are more than enough orientations to create a table linking each possible card orientation to a specific execution time. 1 Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted November 6, 2014 Author Report Share Posted November 6, 2014 Hearts, spades, and clubs are all have asymmetrically shaped symbols such that they can be oriented vertically upright or upside-down, and those two orientations are clearly distinguishable. Any playing cards of odd number will have a symbol for their suit in the center of the card that will break symmetry. So in addition to having 40 cards x 2 orientations (facing frontward or backward), there will be 15 cards (odd numbered cards with a heart, spade, or club) that can be either upright or inverted, so there are more than enough orientations to create a table linking each possible card orientation to a specific execution time. Affirmative!They might just do that, but the plan should not be hard to agree upon . Maybe discarding Ranks and suit ? Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted November 9, 2014 Author Report Share Posted November 9, 2014 1 week=7 ranks(21 cards) 1 card = 4 hours (3 suits) Ace,three,five,six, seven,eight,nine of heart Ace,three,five,six,seven,eight,nine of spade Ace,three,five,six,seven,eight,nine of spade Quote Link to comment Share on other sites More sharing options...
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