Another Prisoner Problem... ... ...

[Yoruichi-san]

"Gooood morning, inmates of Braingate Denitentiary!  And what a morning it is, my fine felonious fellows, for I have riveting news!  Now, I know you were expecting the warden to give his usual morning announcements, but, well, he's...otherwise occupied...hoohoohoo..."

 

"Umm...I tink we mighta used too much gas on the warden, I tink he's kinda...dead..."

 

"Not now Harley!  I'm. Doing. The. Announcements."

 

"Sow-wy, Mr. J."

 

"Anyhoo...where was I?  Oh yes.  Riveting news!  I hope you are enjoying your morning walk around the prison yard, it will be your last!  Yes, that's right, no more outrageous orange accouterments, mounds of mystery mush, or hoarding cigarettes for bartering after today!  Isn't that a blast?  Hoohoohoo..."

 

*Psss...*...*Psss...*...*Psss...*...

 

"Hello?  Am I back?  What happened?  Oh, nevermind.  As I was saying, in fifteen minutes the prison guards, who I believe you'll find to have developed a much more...cheerful demeanor..."

 

"Yeah, cuz we dosed 'em with Joker gas!  Keke..."

 

"Harley!"

 

"Sow-wy."

 

"The prison guards will escort you misunderstood miscreants back into your cells, where you'll each to assigned a number representing the order (within your cell block) in which you will be executed at sundown today.  Hoohoohoo...however, to make things fun, if every evildoer in your cell block can tell me the every other evildoers' numbers, I will let you go free, off to terrorize the streets of Gotham to your hoodlum heart's delight.  But don't be thinking of cheating now!  Hoohoo...no talking to each other...that would make it too easy...or doing anything out of the ordinary.  If the gayly grinning guards catch you, you'll all be executed on the spot!  Hoohoohoooo...that will be all!" 

 

*Psss...*...*Psss...*...*Psssssssssssss...*...

 

"Well...that's a development.  There are five of us in the cell block, and our cells are all in a row, so there goes any chance of seeing each other.  And if we try some knocking scheme, we'll probably get caught, seeing as how the walls are pretty thick and we'd have to be knocking pretty loud to hear each other, which would probably alert the guards.  What do you think, Bainanova?"

 

"Hmmm...well, Sbarcrow, I believe you are correct about the knocking or noise-making scheme, but perhaps there are some options we have that aren't 'out of the ordinary'?  Any ideas, Plainguin?"

 

"The way the plumbing is wired, I can hear when my neighbors on either side flush the toilet or turn on the faucet.  Maybe we can use that?"

 

"But if we start flushing our toilets or turning on the water every few seconds, don't you think they're going to notice something's off?"

 

"Good point, Plasin Ivy.  But we have until sundown, so maybe if we spread out and minimize the number of actions required, we can get away with it?"

 

"Sounds like as good a plan as any, TSLFace, now how to go about it..."

 

 

 

 

 

 

 

[Barcallica]

How about this one.

 

First prisoners will convey even/odd info (f=odd t=even). 

 

Phase 1.

 

a -> a -> a -> a ->

   <- b -> b -> b ->

   <- c <- c - > c ->

 

Now they all know each others even/odd. And should know their one of TFTFF, FFFTT, FTTFF ...

and can uniquely name them LF, MF, RF, LT, RT (L=left R=right)

 

Phase 2.

 

T - F=2

T - T=4

F - F=1

F - T= 3 or 5

 

first LT sends out f or t, which will determine 2 or 4. 

second LF will send out f or t.

third MF will send out f or t.

 

Phase 3.

Now left of the two who got 3 or 5 will send f if 3, T if 5.

 

Total 26 signals. And this is time independent as all actions will be determined by order of receiving and sending signals.

[Barcallica]

I am sure there is a better way but.

 

Prisoners A, B, C, D, E (in order of their cell) their given numbers a, b, c, d, e.  Faucet=f Toilet=t.

1=1f

2=2t

3=2f

4=2t

5=3f

 

first.

 

b: B->C->D

c: C->B->D

d: D->C->B

a: A->B->C

 

it interprets to 

A: 1) he hears b from B.

    2) hears c from B.

    3) hears d from B.

    4) makes a.

B: 1) makes b.

    2) hears b from C.

    3) hears c from C

    4) makes c.

    5) hears d from C

    6) makes d.

    7) hears a from A

    8) makes a.

C: 1) hears b from B,

    2) makes b,

    3) hears b from D.

    4) makes c,

    5) hears c from B

    6) hears c from D

    7) hears d from D,

    8) makes d,

    9) hears d from B

   10) hears a from B

   11) makes a

D: 1) hears b from C,

     2) make b.

     3) hears c from C

     4) makes c.

     5) makes d.

     6) hears d from C.

     7) hears a from C

E:  1) hears b from D,

     2) hears c from D,

     3) hears d from D.

 

Worst case scenario 24 actions (total f and t), best case 18.

[plasmid]

Is it obvious which of the two neighbors is using the plumbing, or are they indistinguishable?

Edit:

And can you tell if a neighbor is both flushing and running a faucet simultaneously?

Edited August 20, 2014 by plasmid

[bonanova]

Borrowing Barcallica's notation:

Identify the prisoners by their cells ABCDE

Assign them execution orders abcde where {a, b, c, d, e} is some permutation of {1, 2, 3, 4, 5.}

 

Encode

Let F, T denote Faucet, Toilet, and prefer F to T with the assumption that T makes more noise and takes more time.

{1, 2, 3, 4, 5} <=> {F, T, FF, TF, FT} so communicating the five numbers requires 5 faucets and 3 toilets.

 

Communicate

Assume:

Prisoners know which cell they occupy

Prisoners BCD can tell which wall sounds are coming through

Prisoners signal multiple numbers from {a, b, c, d, e} in alphabetical order and with a slight pause between numbers.

 

Initially:

Aa, Bb, Cc, Dd, Ee, which means

Prisoner A knows the value of a, B knows b, ... E knows e.

 

Round 1. A sends a; E sends e.

B and D each hear one send and know it is from their respective outer neighbor.

Aa, Bab, Cc, Dde, Ee

 

Round 2. B sends ab; D sends de (ab and de are alphabetical with a slight pause between numbers.)

C knows the order of numbers in each send, and through which wall each send came.
Aab Bab Cabcde Dde Ede

 

Round 3. C sends abcde

B and D hear 5 sends and append the last three and prepend the first three sends, respectively.
Aab Babcde Cabcde Dabcde Ede

 

Round 4. B sends cde; D sends abc

A and E each hear three sounds and append and prepend them respectively.
Aabcde Babcde Cabcde Dabcde Eabcde

Total sends: 17.
Round 1: ae  (2)
Round 2: abde  (4)
Round 3: abcde (5)
Round 4: abccde (6)

Best case: a=T and e=F (or v.v.): TTTT  FFFFFF  FTFTFT  TTTTTT  FFFF = 13T and 13F
Worst case: neither a nor e are T or F: TTTTTTTT  FFF  FFFFFF  TTT  TFTFTFTF = 15T and 13F

[Yoruichi-san]

 

Is it obvious which of the two neighbors is using the plumbing, or are they indistinguishable?

Edit:

And can you tell if a neighbor is both flushing and running a faucet simultaneously?

 

 

Yes.

 

Hmm...good question.  Technically from experience, lol, I find the toilet flushing tends to drain out the sound of the faucet, but for the sake of the problem, let's say yes.

 

 

 

I am sure there is a better way but.

 

Prisoners A, B, C, D, E (in order of their cell) their given numbers a, b, c, d, e.  Faucet=f Toilet=t.

1=1f

2=2t

3=2f

4=2t

5=3f

 

first.

 

b: B->C->D

c: C->B->D

d: D->C->B

a: A->B->C

 

it interprets to 

A: 1) he hears b from B.

    2) hears c from B.

    3) hears d from B.

    4) makes a.

B: 1) makes b.

    2) hears b from C.

    3) hears c from C

    4) makes c.

    5) hears d from C

    6) makes d.

    7) hears a from A

    8) makes a.

C: 1) hears b from B,

    2) makes b,

    3) hears b from D.

    4) makes c,

    5) hears c from B

    6) hears c from D

    7) hears d from D,

    8) makes d,

    9) hears d from B

   10) hears a from B

   11) makes a

D: 1) hears b from C,

     2) make b.

     3) hears c from C

     4) makes c.

     5) makes d.

     6) hears d from C.

     7) hears a from C

E:  1) hears b from D,

     2) hears c from D,

     3) hears d from D.

 

Worst case scenario 24 actions (total f and t), best case 18.

 

Don't sell yourself short out of fear. ;P

 

That's a good solution, but you run into a problem of...

 

Conspicuousness vs. distinguishability.  

 

I.e. if the guard hears *flush**flush**flush*...*flush**flush**flush*...*flush**flush**flush*  it will probably draw his attention.  The idea to "spread out" as mentioned in the OP was to make the actions seem more natural, i.e *flush*... (wait a reasonable amount of time) *flush*...etc, but then you run into the problem of being able to distinguish 1f from 2f from 3f.  

 

Borrowing Barcallica's notation:

Identify the prisoners by their cells ABCDE

Assign them execution orders abcde where {a, b, c, d, e} is some permutation of {1, 2, 3, 4, 5.}

 

Encode

Let F, T denote Faucet, Toilet, and prefer F to T with the assumption that T makes more noise and takes more time.

{1, 2, 3, 4, 5} <=> {F, T, FF, TF, FT} so communicating the five numbers requires 5 faucets and 3 toilets.

 

Communicate

Assume:

Prisoners know which cell they occupy

Prisoners BCD can tell which wall sounds are coming through

Prisoners signal multiple numbers from {a, b, c, d, e} in alphabetical order and with a slight pause between numbers.

 

Initially:

Aa, Bb, Cc, Dd, Ee, which means

Prisoner A knows the value of a, B knows b, ... E knows e.

 

Round 1. A sends a; E sends e.

B and D each hear one send and know it is from their respective outer neighbor.

Aa, Bab, Cc, Dde, Ee

 

Round 2. B sends ab; D sends de (ab and de are alphabetical with a slight pause between numbers.)

C knows the order of numbers in each send, and through which wall each send came.
Aab Bab Cabcde Dde Ede

 

Round 3. C sends abcde

B and D hear 5 sends and append the last three and prepend the first three sends, respectively.
Aab Babcde Cabcde Dabcde Ede

 

Round 4. B sends cde; D sends abc

A and E each hear three sounds and append and prepend them respectively.
Aabcde Babcde Cabcde Dabcde Eabcde

Total sends: 17.
Round 1: ae  (2)
Round 2: abde  (4)
Round 3: abcde (5)
Round 4: abccde (6)

Best case: a=T and e=F (or v.v.): TTTT  FFFFFF  FTFTFT  TTTTTT  FFFF = 13T and 13F
Worst case: neither a nor e are T or F: TTTTTTTT  FFF  FFFFFF  TTT  TFTFTFTF = 15T and 13F

 

Good idea, but same issue here of distinguishability. 

 

I was having a little too much fun in the OP *whistling*, so let me rephrase the goal: you are trying to minimize the number actions while making sure each string is clearly distinguishable, regardless of time.

 

 

 

An aside:  You know you want to sign up for Justice League UNLEASHED ;P

[Barcallica]

With bainanova's {1, 2, 3, 4, 5} <=> {F, T, FF, TF, FT}  my solution yeilds best scenario=12 worst=21 signals (T+F). 

And it isn't C, for example, have to make b right after hearing b. Because all actions have order => A will make a only after and right after hearing 3 signal from B, and B knows he will hear a and do nothing until then etc . 

 

Actually there will be 12 signals overall and it's matter of distributing it evenly without suspicion and I think it is possible given enough time (but this is subjective).

Anyway I get your idea that we should look for solution that requires minimal action.

Edited August 21, 2014 by Barcallica

[Yoruichi-san]

With bainanova's {1, 2, 3, 4, 5} <=> {F, T, FF, TF, FT}  my solution yeilds best scenario=12 worst=21 signals (T+F). 

And it isn't C, for example, have to make b right after hearing b. Because all actions have order => A will make a only after and right after hearing 3 signal from B, and B knows he will hear a and do nothing until then etc . 

 

Actually there will be 12 signals overall and it's matter of distributing it evenly without suspicion and I think it is possible given enough time (but this is subjective).

Anyway I get your idea that we should look for solution that requires minimal action.

 

As you may have deduced , the inmates are not your run of the mill prisoners.  They are half logical genius and half psycho/product of a freak accident/experiment gone awry, and so their perception of time (and the world in general *whistling*) may be skewed in different directions ;P

 

Right, I get you get that part lol, it's the distinguishability part that is not guaranteed in your solution.  Like, if you hear F, how long do you wait to be sure it's not FF of FT?  It

is subjective, so I added the rephrasing to take the subjectivity out of it.  There should be no time dependence in the solution.

Edited August 21, 2014 by Yoruichi-san

[plasmid]

Code

1 = flush with no faucet

2 = faucet with no flush

3 = faucet on, flush starts, faucet off, flush ends

4 = faucet on, flush starts and ends, faucet off

5 = flush starts, faucet on, flush ends, faucet off

[Yoruichi-san]

Code

1 = flush with no faucet

2 = faucet with no flush

3 = faucet on, flush starts, faucet off, flush ends

4 = faucet on, flush starts and ends, faucet off

5 = flush starts, faucet on, flush ends, faucet off

 

Cute, I said you could hear when the faucet turns on, but that's not necessarily the case for when it turns off (from experience there's usually a gurgle at the beginning and then smoothness that kinda fades)...and I never said that all the toilets flush at the same speed all the time...it's an old prison .  

Edited August 21, 2014 by Yoruichi-san

[Yoruichi-san]

Sorry, technically what I should say is that we're looking for a time interval independent solution, i.e. it should not matter whether you wait a minute or an hour between actions, or whether or not you wait the same amount of time between actions, the information encoded should be distinguishable .  

[bonanova]

I did some of these schemes to see if they work and my neighbors started answering back. .

So I'm turn in spectator for the nonce, to see how it turns out.

[DeGe]

 

Lets say they are in rooms A to E in that order.

All signals are sent after the previous signal from the adjoining room has been heard

 

First all of they signal their own positions one by one from A to E

 

Now after hearing the signal from E,

D signals position of C

C signals position of D

B signals position of A

 

Then C signals position of B

B signals position of D and D signals position of B

 

Finally, B signals position of C

 

Each of them knows the position of 4 persons and can deduce the position of the 5th.

 

Code for sending the signals can be any for example 1 = F, 2 = T....

 

[Yoruichi-san]

How about this one.

 

First prisoners will convey even/odd info (f=odd t=even). 

 

Phase 1.

 

a -> a -> a -> a ->

   <- b -> b -> b ->

   <- c <- c - > c ->

 

Now they all know each others even/odd. And should know their one of TFTFF, FFFTT, FTTFF ...

and can uniquely name them LF, MF, RF, LT, RT (L=left R=right)

 

Phase 2.

 

T - F=2

T - T=4

F - F=1

F - T= 3 or 5

 

first LT sends out f or t, which will determine 2 or 4. 

second LF will send out f or t.

third MF will send out f or t.

 

Phase 3.

Now left of the two who got 3 or 5 will send f if 3, T if 5.

 

Total 26 signals. And this is time independent as all actions will be determined by order of receiving and sending signals.

 

Actually the flushing of toilets is not directional, lol, so you can save a few actions by doing B,C, and D rather than A, B, and C, but otherwise, great job.  

 

I did some of these schemes to see if they work and my neighbors started answering back.  .

So I'm turn in spectator for the nonce, to see how it turns out.

 

 You know, if your neighbors are anything like these inmates, you might want to call the Justice League...speaking of which...;P

 

 

Lets say they are in rooms A to E in that order.

All signals are sent after the previous signal from the adjoining room has been heard

 

First all of they signal their own positions one by one from A to E

 

Now after hearing the signal from E,

D signals position of C

C signals position of D

B signals position of A

 

Then C signals position of B

B signals position of D and D signals position of B

 

Finally, B signals position of C

 

Each of them knows the position of 4 persons and can deduce the position of the 5th.

 

Code for sending the signals can be any for example 1 = F, 2 = T....

 

 

Nice idea, but I think you're going to run into the distingishability issue discussed previously.