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Sacrifice Mice

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Seven prisoners are to be shot next week on Mon,Tue,wed,Thu,Fri,Sat & Sun.
Nobody knows the schedule of the others but anyone who can guess them all
correctly would be granted pardon.They had a plan.They expected the sentence.
When someone is caught communicating to other prisoners he/they will be
denied the chance to guess before the death penalty. No one must talk.
One of them has a pair of mice,trained to move forward keeping one of their
side on the wall. A good strategy to escape mazes.
The mice are their key to life. How is it possible ? The mice can go from
cell to cell through food tray slots.
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Posted · Report post

That should work. I had a different answer in mind.

Encode days of the week with numbers: Sun=1, Mon=2, etc.

Whoever starts off with the mice sends a mouse clockwise N times, where N is the day of his execution, and each of the other prisoners will let the mouse continue on its way so all prisoners will see a mouse going clockwise N times. The starting prisoner then sends a mouse counterclockwise, with everyone passing the mouse along, to let everyone know they've reached the end of the number.

To get the mice to the next prisoner, the prisoner who started off with the mice sends one mouse clockwise and one mouse counterclockwise. The next prisoner (going clockwise) will see a mouse going clockwise after the counterclockwise mouse signaled the end of a number, and that will be his cue to keep the mice and become the next prisoner to signal his execution day. The other prisoners will see the mouse going counterclockwise (which will be the second time in a row they see one going counterclockwise), and will know that it's not their turn to signal yet and will let the counterclockwise mouse continue along until it reaches the next person to signal their date.

Then just repeat that process for all the prisoners. It should work as long as the distances between cells aren't vastly different -- if they were, then on the last step where the mice are released at the same time going in opposite directions, it might cause the counterclockwise mouse to arrive at the next cell going clockwise before the clockwise mouse.

A slightly more efficient method:

Assuming the mice move at the same speed (since they are alike in every other way and move at fixed speed), there are four possible easily distinguishable configurations:

A)1 mouse clockwise

B)1 mouse counterclockwise

C)2 mice clockwise

S)2 mice counterclockwise

Denote one of the last two as the 'STOP' code, let's say 2 mice counterclockwise. Agree that the order of signaling (i.e. the way the mice will be sent after a prisoner is done signalling his day of execution) will be the same direction, i.e. here counterclockwise.

You should be able to code all 7 days with 2 trips of the mice around, since there are 3*3=9 different combinations, as long as each time the prisoner allows the mice to return to his cell before sending the next trip (the time between trips that the other prisoners see may be significantly different, but that should be okay, since it's the order that matters, not the time). For example:

AA - Mon

AB - Tue

AC - Wed

BA - Thur

BB - Fri

BC - Sat

CC - Sun

After coding his day, the first prisoner who has the mice will send them both counterclockwise (S-stop) to allow everyone to know he's done. The first time each prisoner sees the stop code he will let the mice pass. The second time a prisoner sees this, he will know it's his turn and pick up the mice and start his code, so that the prisoner adjacent counterclockwise to the previous prisoner will be the one to pick up the mice and start his coding.

Edit: Actually, to make it the tiniest bit more efficient, since we're using a STOP code, you could code 3 of the days with just one trip, then STOP. Eh.

...if the prisoners decide to code each day in 2 trips, then there's no need for every prisoner to see the STOP code, so the first time a prisoner sees two mice together counterclockwise he can pick them up. That way it will require a total of 2*7+1(for the mice to be passed all the way around) =15 trips around the prison.

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Posted (edited) · Report post

Can I ask some questions?

1- Can any of the prisoners catch the mice?

2- Is the speed of the mice fix?

3- Is it possible to hang any paper around the mice`s neck?

4- Do each one of the prisoners have a watch?

Edited by wolfgang
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Posted · Report post

Can I ask some questions?

1- Can any of the prisoners catch the mice?

Yes , they can hold the mice & can feed them rice.

2- Is the speed of the mice fix?

Yes, Left or Right on wall

3- Is it possible to hang any paper around the mice`s neck?

Paper or pen are prohibited.

4- Do each one of the prisoners have a watch?

No way of measuring the time.

Note: Cruelty to animals is not encouraged in this puzzle.

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Posted · Report post

I'm still not sure if I really understand the problem, and there's probably a really elegant way of doing it, but...

Since there's no limit on how many times the mice can be passed, you can gather information piece by piece until it's complete and then pass it on to everyone else. There are 7 clearly distinguishable ways to send the mice:

1 mouse on Right, 1 mouse on Left simultaneously

Both mice on Right simultaneously

Both mice on Left simultaneously

1st mouse on Right, 2nd mouse on Left staggered (i.e. count to 10 or sth)

1st mouse on Right, 2nd mouse on Right staggered

1st mouse on Left, 2nd mouse on Left staggered

1st mouse on Left, 2nd mouse on right staggered

There are also a bunch of ways not so clearly distinguishable, like sending just 1 mouse (might get confused with last 4 configurations) or sending the second mouse right after the first or waiting a longer interval to send the second mouse. Probably could design a more efficient way of doing it with those, but the simplest, albeit tedious way:

Assign each of the configurations a day of the week. The person who starts with the mice sends the mice to the person next to him in the configuration corresponding to his execution day, say both mice on left simultaneously (Wed). The second person sends them back in the configuration corresponding to his execution day, say both mice on right simultaneously (Tue). To indicate the string is done, the first person will send the mice back in the original configuration, both mice on left. Then the second person will know to send the mice on to the next person.

He can either send the string starting with the first person's day or his day, it doesn't really matter which as long as they agreed beforehand on the convention. Let's say they start the string with the first person's day, then the second person's day, etc. The second person would send the mice on the left simultaneously to the third person, the third person would send them back, the second person would send them on the right simultaneously, the third person sends them back, the second person sends them in the original configuration (both mice on left) to indicate string is done. Third person then adds his day to the string and sends to the fourth person, etc, until everyone knows everyone else's day.

There are definitely more efficient ways of doing it, like some kind of pendulum arrangement rather than a circular one, but this works, if I am understanding the OP correctly, and is fairly straightforward. And the extra exercise is not cruelty...exercise is good for you! ;)

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7 ways to send the mice w/Assigned day of the week:
Mon:RL
Tue:RR
Wed:LL
Thu:R then L
Fri:R then R
Sat:L then L
Sun:L then R
Direction:
1)The person who starts with the mice sends both mice on left simultaneously (Wed).
2)The second person sends both mice on right simultaneously (Tue).
3)the first person will send both mice on left.
4)The second person would send the mice on the left simultaneously to the third person,
5)the third person would send them back,
6)the second person would send them on the right simultaneously,
7)the third person sends them back,
8)the second person sends them in the original configuration (both mice on left) to indicate string is done.
9)Third person then adds his day to the string and sends to the fourth person, etc, until everyone knows everyone else's day.
1)P1-LL to P2 : Wed
2)P2-RR to P1 : Tue
3)P1-LL to P2 : reset
4)P2-LL to P3 : done
5)P3-RR to P2 : done
6)P2-RR t0 P1 : done
7)??
I guess you are on the right track for a solution, but i lost trail on step 7)

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Posted · Report post

If you were to attempt to use that strategy in a solution

Would the prisoner in cell 2 (just left of cell 1) be able to distinguish whether the prisoner in cell 1 sent one mouse left and then one mouse right, vs sending them both at the same time and having them arrive to cell 2 at different times simply due to different path lengths?

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Posted (edited) · Report post

If you were to attempt to use that strategy in a solution

Would the prisoner in cell 2 (just left of cell 1) be able to distinguish whether the prisoner in cell 1 sent one mouse left and then one mouse right, vs sending them both at the same time and having them arrive to cell 2 at different times simply due to different path lengths?

"

The mice can go from cell to cell through food tray slots", which (from the tv/movies I've watched :lol:) I would assume to be pretty short, and, like, be rectangular,prism shaped, so if sent through at the same time, I would imagine they would arrive at pretty much the same time...? I don't know, like I mentioned, I'm not sure if I'm actually understanding the OP correctly...

7 ways to send the mice w/Assigned day of the week:
Mon:RL
Tue:RR
Wed:LL
Thu:R then L
Fri:R then R
Sat:L then L
Sun:L then R
Direction:
1)The person who starts with the mice sends both mice on left simultaneously (Wed).
2)The second person sends both mice on right simultaneously (Tue).
3)the first person will send both mice on left.
4)The second person would send the mice on the left simultaneously to the third person,
5)the third person would send them back,
6)the second person would send them on the right simultaneously,
7)the third person sends them back,
8)the second person sends them in the original configuration (both mice on left) to indicate string is done.
9)Third person then adds his day to the string and sends to the fourth person, etc, until everyone knows everyone else's day.
1)P1-LL to P2 : Wed
2)P2-RR to P1 : Tue
3)P1-LL to P2 : reset
4)P2-LL to P3 : done
5)P3-RR to P2 : done
6)P2-RR t0 P1 : done
7)??
I guess you are on the right track for a solution, but i lost trail on step 7)

Let's say P3 is executed on Mon,

1)P1-LL to P2 : Wed
2)P2-RR to P1 : Tue
3)P1-LL to P2 : reset
4)P2-LL to P3 : tells P3 string starts on Wed
5)P3-RL to P2 :
6)P2-RR t0 P3 : tells P3 second term in string is Tue
7)P3-RL to P2:
8)P2-LL (or RR or RL, anything that has already been used): tells P3 string is finished, which also indicated length of string
9)P3-LL to P4: tells P4 string starts on Wed
...etc...until everyone knows the full string
Like I said, this is definitely not the most efficient method, some kind of pendulum arrangement (i.e. P1->P2->P1->P7->P1->P2->P3->P2->P1->P7->P6->P7->P1->P2->P3->P4->P3->P2->P1->P7->P6->P5) would probably save the mice some travel time, but I'll leave that to someone else while I work on stuff like CW and MxM ;)

Edited by Yoruichi-san
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Posted · Report post

Oh, I see what you're thinking. In the solution you're proposing, one prisoner can send the mice to any other specific prisoner he wants, with the mice arriving from either the left or the right.

In my mind, the food slots face out through the main door of each cell and into a common area like in the picture. I was thinking that if, for example, prisoner 4 were to send one mouse right and one mouse left, then the mouse he sends right would visit prisoners 3, 2, 1, 7, 6 etc in that order, and the mouse he sends left would visit prisoners 5, 6, 7, 1, etc in that order. The mouse wouldn't have any indication that it's intended to visit one particular prisoner along that route and would visit everyone along the way, and might be picked up by any prisoner along the way and either have its direction switched or be held onto while waiting for the other mouse to arrive.

TSLF might be able to clarify what's intended in this problem.

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Posted (edited) · Report post

Actually I was thinking he could send them to either of the prisoners next to him ("cell to cell"? :huh: ), but I see what you mean, so yeah, it'd be good to have TSLF clarify ^_^.

Edited by Yoruichi-san
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Posted · Report post

Plasmid's notion is correct.. if the 1st person's day is Thursday sending a mouse left & sending a mouse right..both the adjacent cells would hold the mouse he catches and waits for further info indefinetly.

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Posted · Report post

The mice have different color or have spots and marks and everyone knew which cell they will start as they knew who own them. Therefore less mice work.

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Actually the mice are like clones, one is indistinguishable with the other and the mice owner and other prisoners are shuffled into different cells nos.. But the cells are numbered inside and are in clockwise order.

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Posted · Report post

If they worked out the code beforehand...

The first prisoner converts their execution day to a number, 1-7, and sends 1 mouse to the prisoner on the left, who returns it, until the number is reached at which point both mice are sent. If the number was 1, then both are sent the first time. if the number was 7, then there are 6 trips with 1 mouse and then the 7th trip has them both.



The second prisoner now knows the date for the 1st one. They convert their own day to a number, 1-7, but multiply that by 7, and add the number from the 1st prisoner. The new number will be between 8 and 50. Send 1 mouse to the 3rd prisoner with the same process until the new number is reached. That prisoner now knows the dates of 1 and 2. Just divide the number by 7 and drop the remainder to get date for prisoner 2. Use the remainder for the date of prisoner 3.

Prisoner 3's date is multipled by 7^2. Next prisoner's date is times 7^4, and so on until they have a number no larger than 823,542 encoding all their dates. Prisoner 7 is the first to get the full list, and can then pass it along to the others.
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Posted · Report post

That should work. I had a different answer in mind.

Encode days of the week with numbers: Sun=1, Mon=2, etc.

Whoever starts off with the mice sends a mouse clockwise N times, where N is the day of his execution, and each of the other prisoners will let the mouse continue on its way so all prisoners will see a mouse going clockwise N times. The starting prisoner then sends a mouse counterclockwise, with everyone passing the mouse along, to let everyone know they've reached the end of the number.

To get the mice to the next prisoner, the prisoner who started off with the mice sends one mouse clockwise and one mouse counterclockwise. The next prisoner (going clockwise) will see a mouse going clockwise after the counterclockwise mouse signaled the end of a number, and that will be his cue to keep the mice and become the next prisoner to signal his execution day. The other prisoners will see the mouse going counterclockwise (which will be the second time in a row they see one going counterclockwise), and will know that it's not their turn to signal yet and will let the counterclockwise mouse continue along until it reaches the next person to signal their date.

Then just repeat that process for all the prisoners. It should work as long as the distances between cells aren't vastly different -- if they were, then on the last step where the mice are released at the same time going in opposite directions, it might cause the counterclockwise mouse to arrive at the next cell going clockwise before the clockwise mouse.

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Posted · Report post

Ah, your way is much more efficient. Mine would take weeks even if the mice could be sent every second.

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Posted (edited) · Report post

That should work. I had a different answer in mind.

Encode days of the week with numbers: Sun=1, Mon=2, etc.

Whoever starts off with the mice sends a mouse clockwise N times, where N is the day of his execution, and each of the other prisoners will let the mouse continue on its way so all prisoners will see a mouse going clockwise N times. The starting prisoner then sends a mouse counterclockwise, with everyone passing the mouse along, to let everyone know they've reached the end of the number.

To get the mice to the next prisoner, the prisoner who started off with the mice sends one mouse clockwise and one mouse counterclockwise. The next prisoner (going clockwise) will see a mouse going clockwise after the counterclockwise mouse signaled the end of a number, and that will be his cue to keep the mice and become the next prisoner to signal his execution day. The other prisoners will see the mouse going counterclockwise (which will be the second time in a row they see one going counterclockwise), and will know that it's not their turn to signal yet and will let the counterclockwise mouse continue along until it reaches the next person to signal their date.

Then just repeat that process for all the prisoners. It should work as long as the distances between cells aren't vastly different -- if they were, then on the last step where the mice are released at the same time going in opposite directions, it might cause the counterclockwise mouse to arrive at the next cell going clockwise before the clockwise mouse.

A slightly more efficient method:

Assuming the mice move at the same speed (since they are alike in every other way and move at fixed speed), there are four possible easily distinguishable configurations:

A)1 mouse clockwise

B)1 mouse counterclockwise

C)2 mice clockwise

S)2 mice counterclockwise

Denote one of the last two as the 'STOP' code, let's say 2 mice counterclockwise. Agree that the order of signaling (i.e. the way the mice will be sent after a prisoner is done signalling his day of execution) will be the same direction, i.e. here counterclockwise.

You should be able to code all 7 days with 2 trips of the mice around, since there are 3*3=9 different combinations, as long as each time the prisoner allows the mice to return to his cell before sending the next trip (the time between trips that the other prisoners see may be significantly different, but that should be okay, since it's the order that matters, not the time). For example:

AA - Mon

AB - Tue

AC - Wed

BA - Thur

BB - Fri

BC - Sat

CC - Sun

After coding his day, the first prisoner who has the mice will send them both counterclockwise (S-stop) to allow everyone to know he's done. The first time each prisoner sees the stop code he will let the mice pass. The second time a prisoner sees this, he will know it's his turn and pick up the mice and start his code, so that the prisoner adjacent counterclockwise to the previous prisoner will be the one to pick up the mice and start his coding.

Edit: Actually, to make it the tiniest bit more efficient, since we're using a STOP code, you could code 3 of the days with just one trip, then STOP. Eh.

Edited by Yoruichi-san
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Posted · Report post

That might be the most efficient way to do it; I can't think of a better one.

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That should work. I had a different answer in mind.

Encode days of the week with numbers: Sun=1, Mon=2, etc.

Whoever starts off with the mice sends a mouse clockwise N times, where N is the day of his execution, and each of the other prisoners will let the mouse continue on its way so all prisoners will see a mouse going clockwise N times. The starting prisoner then sends a mouse counterclockwise, with everyone passing the mouse along, to let everyone know they've reached the end of the number.

To get the mice to the next prisoner, the prisoner who started off with the mice sends one mouse clockwise and one mouse counterclockwise. The next prisoner (going clockwise) will see a mouse going clockwise after the counterclockwise mouse signaled the end of a number, and that will be his cue to keep the mice and become the next prisoner to signal his execution day. The other prisoners will see the mouse going counterclockwise (which will be the second time in a row they see one going counterclockwise), and will know that it's not their turn to signal yet and will let the counterclockwise mouse continue along until it reaches the next person to signal their date.

Then just repeat that process for all the prisoners. It should work as long as the distances between cells aren't vastly different -- if they were, then on the last step where the mice are released at the same time going in opposite directions, it might cause the counterclockwise mouse to arrive at the next cell going clockwise before the clockwise mouse.

A slightly more efficient method:

Assuming the mice move at the same speed (since they are alike in every other way and move at fixed speed), there are four possible easily distinguishable configurations:

A)1 mouse clockwise

B)1 mouse counterclockwise

C)2 mice clockwise

S)2 mice counterclockwise

Denote one of the last two as the 'STOP' code, let's say 2 mice counterclockwise. Agree that the order of signaling (i.e. the way the mice will be sent after a prisoner is done signalling his day of execution) will be the same direction, i.e. here counterclockwise.

You should be able to code all 7 days with 2 trips of the mice around, since there are 3*3=9 different combinations, as long as each time the prisoner allows the mice to return to his cell before sending the next trip (the time between trips that the other prisoners see may be significantly different, but that should be okay, since it's the order that matters, not the time). For example:

AA - Mon

AB - Tue

AC - Wed

BA - Thur

BB - Fri

BC - Sat

CC - Sun

After coding his day, the first prisoner who has the mice will send them both counterclockwise (S-stop) to allow everyone to know he's done. The first time each prisoner sees the stop code he will let the mice pass. The second time a prisoner sees this, he will know it's his turn and pick up the mice and start his code, so that the prisoner adjacent counterclockwise to the previous prisoner will be the one to pick up the mice and start his coding.

Edit: Actually, to make it the tiniest bit more efficient, since we're using a STOP code, you could code 3 of the days with just one trip, then STOP. Eh.

...if the prisoners decide to code each day in 2 trips, then there's no need for every prisoner to see the STOP code, so the first time a prisoner sees two mice together counterclockwise he can pick them up. That way it will require a total of 2*7+1(for the mice to be passed all the way around) =15 trips around the prison.

That is a good enough solution!

Using: Two passing solution
AA:1 mouse clockwise+1 mouse clockwise - Mon
AB:1 mouse clockwise+1 mouse counterclockwise - Tue
AC:1 mouse clockwise+2 mice clockwise - Wed
BA:1 mouse counterclockwise+1 mouse clockwise - Thu
BB:1 mouse counterclockwise+1 mouse counterclockwise - Fri
BC:1 mouse counterclockwise+2 mice clockwise - Sat
CC:2 mice clockwise+2 mice clockwise - Sun
S : 2 Pass to next
We have only 20 mice cycles..or say 140 trips cell to cell..
========================================================
Using:Clockwork solution
Mon-1 mouse
Tue-both mice head to tail
Wed-2 mice staggered or spaced
Thu-1 mouse twice
Fri-1 mouse then both mice
Sat-both mice then 1 mouse
Sun-1 mouse trice
Pass-2
Example :(mice starts from)
Sat Thu Fri (Sun) Mon Wed Tue : 1 2 3 4 5 6 7 respective cells
4 trips:Sun pass both mice clockwise to Mon (Mon informs Sun)+1 return
6 trips:Mon pass both mice clockwise to Wed (Wed informs Mon)+2 returns
6 trips:Wed pass both mice clockwise to Tue (Tue informs Wed)+2 returns
8 trips:Tue pass both mice clockwise to Sat (Sat informs Tue)+3 returns
6 trips:Sat pass both mice clockwise to Thu (Thu informs Sat)+2 returns
8 trips:Thu pass both mice clockwise to Fri (Fri informs Thu)+3 returns
8 trips:Fri pass both mice clockwise to Sun (Sun informs Fri)+3 returns
46 Trips (cell to cell)
1st round- informs everyone the day of the next person
2nd round- informs everyone the day of the next to next person
after 5 rounds everyone has a 5 day strings to end with his day and last persons day
so it will be 230 trips

Thanks for solving ! :)
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Posted · Report post

That might be the most efficient way to do it; I can't think of a better one.

I just thought if it could be better if the Sunday 4 mice code can be reduced to 2 mice only
and considering some opportunities to lessen the trips
Using: Pendulum solution
Mon-1 mouse once
Tue-2 mice w/gap
Wed-1 mouse twice
Thu-1 mouse once then 2 mice w/gap (or 1 mouse twice if Wed is known)*
Fri-2 mice w/gap then 1 mouse once (or 2 mice w/gap if Tue is known)
Sat-1 mouse trice (or 1 mouse once if Mon is known)**
Sun-Receives both mice counterclockwise
Pass-Both mice clockwise
Example 1 :(mice starts from)
Tue Wed (Mon) Sun Thu Sat Fri : 1 2 3 4 5 6 7 respective cells
12 trips:Mon pass both mice counterclockwise to seek Sun
23 trips:Sun pass both mice to Thu clockwise (1st signal is clockwise)
23 trips:Thu pass both mice to Sat clockwise (2nd signal is Counterclockwise)
23 trips:Sat pass both mice to Fri clockwise (3rd signal is clockwise)
Now everyone knows its Sun-Thu-Sat-Fri
16 trips:Fri pass both mice to Tue clockwise (4th signal is Counterclockwise)
16 trips:Tue pass both mice to Wed clockwise (5th signal is clockwise)
Now everyone knows its Sun-Thu-Sat-Fri-Tue-Wed+Mon
113 trips (cell to cell) max
Example 2 :(mice starts from)
Sat Thu Fri Sun (Mon) Wed Tue : 1 2 3 4 5 6 7 respective cells
2 trips:Mon pass both mice counterclockwise to seek Sun
9 trips:Sun pass both mice to Mon clockwise (1st signal is clockwise)
16 trips:Mon pass both mice to Wed clockwise (2nd signal is Counterclockwise)
16 trips:Wed pass both mice to Tue clockwise (3rd signal is clockwise)
Now everyone knows its Sun-Mon-Wed-Tue
9 trips:Tue pass both mice to Sat clockwise (4th signal is Counterclockwise)**
16 trips:Sat pass both mice to Thu clockwise (5th signal is clockwise)*
Now everyone knows its Sun-Mon-Wed-Tue-Sat-Thu+Fri
68 trips (cell to cell) min

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Posted · Report post

Oh, if we're allowed to send two mice in a sort of staggered fashion like that - letting one mouse go in one direction and then letting the next mouse go in the same direction several seconds or a minute later to send a signal that's clearly distinct from, say, one mouse making two trips around the prison - then it can be done even faster.

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Posted · Report post

Hmmm...interesting. I was only considering the obviously distinguishable cases since they don't a way of telling time and I wasn't sure how fast the mice went, how big the prison was, etc., but as plasmid points out, depending on the level of distinguishability, there are interesting ways of being even more efficient ^_^.

Thanks for the puzzles TSLF :).

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Posted · Report post

Oh, if we're allowed to send two mice in a sort of staggered fashion like that - letting one mouse go in one direction and then letting the next mouse go in the same direction several seconds or a minute later to send a signal that's clearly distinct from, say, one mouse making two trips around the prison - then it can be done even faster.

Glad to hope it can be done faster with mice of same speed by distinguishing the ff:

head to tail- <: )~<: )~

Foot gap- <: )~ <: )~

One round- <: )~ <: )~

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Posted · Report post

There is no sense to this puzzle if there is no "sacrifice" done like cutting a mouse tail or tying knot to a tail. In that case to have more distinct codes to lessen the distance of travel.

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