TimeSpaceLightForce Posted September 8, 2014 Report Share Posted September 8, 2014 It is Sunday morning.. Loudspeaker announcement : "the 4 prisoners,locked in cells 1,2,3 & 4, will be executed next Sunday at exactly 1:00,2:00,3:00 & 4:00 pm .Schedule shall be random .." The 4 prisoner ,while having their final breakfast outside the cells, was informed that they all must guess ,before taken out of their cells : Who is 1st, 2nd, 3rd & 4th to hang ? in order to be pardoned by the warden. But any communication among them forfeits their chance of survival. One of the prisoners told the others that he have observed,every 12 noon ,the 4 white pigeons that always feeds on bread crumbs he places on his cell's window sill. And they can be caught if wanted to. That's how he finds out there are 2 females among those birds.. After a brief planning they pocketed some bread and returned to their cells where they were asked to draw stick for the time of their own execution.. What possibly is their scheme to get out alive and free? Quote Link to comment Share on other sites More sharing options...
0 plasmid Posted September 20, 2014 Report Share Posted September 20, 2014 (edited) Name the prisoners A, B, C, D. The first Sunday at noon, the first prisoner to be executed feeds the birds and captures 1 male and 1 female if he's A 1 male and 0 females if he's B 0 males and 1 females if he's C 0 males and 0 females if he's D Monday at noon, the second prisoner to be executed feeds the birds and sees how many are missing, so he knows who will be executed first. The second prisoner captures the number of birds to signal his name using the same table that's shown above (there will be at least one male and one female for him to capture). The first prisoner releases his pigeons after noon on Monday. Tuesday at noon, the third prisoner to be executed feeds the birds and sees how many are missing, so he knows who will be executed second. He then captures the number of birds to signal his name using the same table that's shown above. The second prisoner releases his birds after noon on Tuesday. Wednesday at noon, the first prisoner to be executed feeds the birds again and sees how many are missing, so he knows who will be executed third. He doesn't capture any pigeons and the third prisoner doesn't release any pigeons. Thursday at noon, the second prisoner to be executed feeds the birds again and sees how many are missing, so he knows who will be executed thrid, so now he knows when everyone will be executed. He then captures the number of birds needed to signal THE FIRST PRISONER'S NAME, and the third prisoner releases his birds. Friday at noon, the third prisoner to be executed feeds the birds and sees how many are missing, so now he knows who will be executed first and he knows when everyone will be executed. He captures the number of pigeons needed to signal THE SECOND PRISONER'S NAME, and the second prisoner releases his birds. Saturday at noon, the first prisoner feeds the birds and sees how many are missing, so he knows who will be executed second and he knows when everyone will be executed. At this point: the first, second, and third prisoners to be executed all know the execution order, but the fourth prisoner doesn't know anything yet. The third prisoner might still be holding some pigeons, and the rest of the birds are at the first prisoner's window. Now they change plans. Re-label the prisoners X, Y, and Z based on who the fourth prisoner is: If the fourth prisoner is A, then B=X, C=Y, D=Z. If the fourth prisoner is B, then A=X, C=Y, D=Z. If the fourth prisoner is C, then A=X, B=Y, D=Z. If the fourth prisoner is D, then A=X, B=Y, C=Z. Next, they plan to send a code to the fourth prisoner to tell him their execution order. XYZ = 0 males 0 females XZY = 0 males 1 female YXZ = 0 males 2 females YZX = 1 male 0 females ZXY = 1 male 1 female ZYX = 1 male 2 females The first prisoner and third prisoner are holding the pigeons right now, so the first prisoner releases however many pigeons he needs to in order to send that code. The third prisoner knows what code the first prisoner needs to send and how many pigeons the first prisoner has, so the third prisoner releases enough pigeons to finish sending the code if the first prisoner doesn't already have enough pigeons. Sunday at noon, the fourth prisoner feeds the birds and now everyone knows the execution order. Edited September 20, 2014 by plasmid 1 Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted September 12, 2014 Author Report Share Posted September 12, 2014 Possible Schedule: Cell 1 Cell 2 Cell 3 Cell 4 1pm 2pm 3pm 4pm 1pm 2pm 4pm 3pm 1pm 3pm 2pm 4pm 1pm 3pm 4pm 2pm 1pm 4pm 2pm 3pm 1pm 4pm 3pm 2pm 2pm 1pm 3pm 4pm 2pm 1pm 4pm 3pm 2pm 3pm 1pm 4pm 2pm 3pm 4pm 1pm 2pm 4pm 1pm 3pm 2pm 4pm 3pm 1pm 3pm 1pm 2pm 4pm 3pm 1pm 4pm 2pm 3pm 2pm 1pm 4pm 3pm 2pm 4pm 1pm 3pm 4pm 1pm 2pm 3pm 4pm 2pm 1pm 4pm 1pm 2pm 3pm 4pm 1pm 3pm 2pm 4pm 2pm 1pm 3pm 4pm 2pm 3pm 1pm 4pm 3pm 1pm 2pm 4pm 3pm 2pm 1pm Feed birds * 1) MFMF 2) FMF 3) MFM 4) MM 5) FF 6) MF 7) M 8) F 9) none *absent birds are held captive by other prisoner/s Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted September 22, 2014 Author Report Share Posted September 22, 2014 Yes, the finest plan.. Quote Link to comment Share on other sites More sharing options...
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