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55 prisoners are on the deathrow. The warden gives them a chance to live. He puts 100 empty jars into room A and randomly puts 10 balls under the 100 jars. Each jar is either empty or contains 1 ball. In the other room, call it room B, the warden puts 100 empty jars, and a stack of 10 balls.

The warden then divides the group of 55 into two groups of 54 and 1. The group of 54 he puts into room A. The last prisoner goes into room B.

Each prisoner from room A will take turn looking under the entire 100 jars, but can not move or rearrange the contents. He then can go to room B and must say one of two possible words to the prisoner there. The 2 possible words are Lakers and Rule. Assume he can not convey any other information besides that word (so no facial expression, tone, body language, hand gestures, etc. ). Any attempt to convey extra information like remaining silent, concatenating words, or walking a certain number of paces before stopping in room B will get all prisoners killed immediately. The prisoner in room B will then have to reconstruct the permutation of the balls in room A.

If the prisoner in room B can successfully reconstruct the permutation in room A after the 54 turns, all 55 will live. Otherwise they will die.

The night before, the warden tells the prisoners this scheme, so the prisoners know that there will be exactly 10 balls under the 100 jars. They have 1 night to discuss a strategy. They are not allowed to bring any mechanical computational aid to the game (yes, abacus are out too). Assume that each prisoner has the mental computational skills of a reasonable average person.

1) What strategy would give the prisoners the best chance to live? Describe the strategy.

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L=Lakers

R=Rule

Start each uncommitted sequence as 4 empty jars.

L=Commit sequence of four

RRR=Place ball in 1st jar

RRL=Place ball in 2nd jar

RLR=Place ball in 3rd jar

RLL=Place ball in 4th jar

To place 10 balls anywhere in the 100 will cost...

25 prisoners for 25 four-jar commits at one prisoner each

30 prisoners for 10 ball placements at 3 prisoners each

=====

55 prisoners

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9 is worst case

Sorts:

25 look at 25 groups of 4 ... ball = L ... no ball = R .... get 9/4

9 look at 9 groups of 4 ... multiple ball = L ... one = R .... locate multiple

8 look at first 2 of single ball groups ... ball = L ... no ball = R .... locate 8/2

8 look at first bottle of result ... ball = L ... no ball = R .... locate 8 balls

3 look in first three bottles of multiple ... ball = L ... no ball = R .... locate last 2 balls

Same sort can be used in every case with diminishing number of prisoners required. After 8 only the 25/4 are required you can look in each jar and say L or R. The last jar isn’t required but what the hell, the warden may try to pull a fast one.

10 -- 9 -- 8/2 -- 8 -- 7 -- 6 -- 5 -- 4 -- 3

25 - 25 - 25 - 25 - 25 - 25 - 25 - 25 - 25

0 --- 9 --- 8 --- 8 --- 0 --- 0--- 0 -- 0 -- 0

20 - 16 - 12 - 14 ---0-- - 0 ---0 ---0 -- 0

0 --- 3 -- 6 --- 3 -- 28 - 24 - 20 -16 - 12

45 - 53 -- 51 - 50 - 53 - 49 - 45 - 41 - 37

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L=Lakers

R=Rule

Start each uncommitted sequence as 4 empty jars.

L=Commit sequence of four

RRR=Place ball in 1st jar

RRL=Place ball in 2nd jar

RLR=Place ball in 3rd jar

RLL=Place ball in 4th jar

To place 10 balls anywhere in the 100 will cost...

25 prisoners for 25 four-jar commits at one prisoner each

30 prisoners for 10 ball placements at 3 prisoners each

=====

55 prisoners

:duh: ... which would fail the test and kill everyone because I've only got 54... except for one tiny modification.

On placement of the 10th ball, the prisoner interpreting the sequence auto-commits because he knows all other jars are empty.

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:duh: ... which would fail the test and kill everyone because I've only got 54... except for one tiny modification.

On placement of the 10th ball, the prisoner interpreting the sequence auto-commits because he knows all other jars are empty.

But what if the last ball is in the last 4-jar sequence?

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But what if the last ball is in the last 4-jar sequence?

If so, assuming the other nine balls were in prior sequences, I would have paid for 24 commits @1 prisoner each + 9 ball placements at 3 prisoners each, for a total of 51. Describing where the 10th ball goes would cost 3 more prisoners and no commits.

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If so, assuming the other nine balls were in prior sequences, I would have paid for 24 commits @1 prisoner each + 9 ball placements at 3 prisoners each, for a total of 51. Describing where the 10th ball goes would cost 3 more prisoners and no commits.

I see now, that would take advantage of the prior knowledge that there are only 10 balls. Well done.

Edited by bushindo
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