1 to 8 Can wait!

[wolfgang]

Eight persons having numbers on their backs( 1 to 8) were standing randomely behind each other in a straight raw( so that the first person can see nothing,while the others,each one can see only the number of the person standing infront of him), each one has a mini cube in his hand containing one number,from 1 to 8 at random, no one of them know neither the number on his back nor the number inside his cube, but the number inside the cube for each person is either lager than or smaller than the number on his back by 1.

They have to enter another room,starting from the last one,,, in the other room there is a rectangular table drawn on it 8 squares in a straight raw, in each square there is a number, 1 to 8.

Each person when enters the room ,he should put his cube on the table( according to previous plan ),and get out through another door, and only the last person is allowed to put all the cubes inside the squares so that they should match each other,i.e. the 1st square should have number 1, 2nd square number 2...and so on.

Any kind of comunications between them is forbidden.

They are allowed to set a plan before starting the mission.

What is the PLAN ?

[TimeSpaceLightForce]

<1st<2nd<3rd<4th<5th<6th<7th<8th row order and the number in their backs are the same

[1] [2] [3] [4] [5] [6] [7] [8] numbered square order

8x table order the nearest to numbered square the earliest to put cube

7x

6x

5x

4x

3x

2x

The last or 8th person must place the cube as to hint his row order and the back number he see (+1/-1).

The next person do the same until the 2nd in row (7th to enter). Therefore the last to enter or 1st in row

know his back=1 and his cube=2 (no negative) ,then he knows that 8th has has no option but his cube=7 ,

so 7th has cube=6...2nd has cube=1 . Then arranges them accordingly . In case of random back number

the last to enter can know his back number from previous person and all cube numbers.

[TimeSpaceLightForce]

<1st<2nd<3rd<4th<5th<6th<7th<8th row order and the number in their backs are the same

[1] [2] [3] [4] [5] [6] [7] [8] numbered square order

8x table order the nearest to numbered square the earliest to put cube

7x

6x

5x

4x

3x

2x

The last or 8th person must place the cube as to hint his row order and the back number he see (+1/-1).

The next person do the same until the 2nd in row (7th to enter). Therefore the last to enter or 1st in row

know his back=1 and his cube=2 (no negative) ,then he knows that 8th has has no option but his cube=7 ,

so 7th has cube=8...2nd has cube=1 . Then arranges them accordingly . In case of random back number

the last to enter can know his back number from previous person and all cube numbers.

edit:

Note: back1 cube2,back2 cube1,back8cube7,back7cube8,back3cube4,back4cube3,back5cube6,back6cube5

[Yoruichi-san]

The way I'm reading the problem it seems that row order is random, i.e. the last person does not need to have 8 on their back, so...

Due to the constraints of the problem (each of 1 to 8 must be used once and exactly once on the cubes and each back number must have +/- 1 the cube number), there is only one possible one the cubes can be allocated to the numbers, as TSLF specified previously. So if you know the person's back number, you know the person's cube number.

So the plan is this:

Calling the dimension parallel to the numbered squares X, and the dimension perpendicular Y, the nth person who enters places

their cube at Y=n/8 the distance from the row of numbered squares to the edge of the table, at X=the number he sees on the person in front of him or the number he knows they must have on their cube (they agree on which before hand). This will 'graph' the numbers, and the last person will be able to figure out where to place each box fairly easily by 'reading' it, and figure out which number the last person had by process of elimination.

Edited February 18, 2013 by Yoruichi-san

[phaze]

If the grid form may not be used due to a lack of table space

If the orientation of the cube is known 24 states can be represented.

6 sides x 4 easily assigned orientations = 24.

This can be achieved with a dot in one corner of one of the faces on the cube or similar marker (such as noting were the hinge is).

Whilst this on it's own cannot represent the 56 states required to establish a complete map there are needs to still be enough room on the table for 7 cubes outside the 8 square receptacles.

These can be assigned the numbers 1-7 from left to right and top to bottom.

If there is less than 7 non receptacle spaces, the cubes need to be stacked.

This brings the total number of states that can be represented to 168 (much more for stacked version), more than enough without having to worry about grids with correct spacing.

Edited February 18, 2013 by phaze

[wolfgang]

If the grid form may not be used due to a lack of table space

If the orientation of the cube is known 24 states can be represented.

6 sides x 4 easily assigned orientations = 24.

This can be achieved with a dot in one corner of one of the faces on the cube or similar marker (such as noting were the hinge is).

Whilst this on it's own cannot represent the 56 states required to establish a complete map there are needs to still be enough room on the table for 7 cubes outside the 8 square receptacles.

These can be assigned the numbers 1-7 from left to right and top to bottom.

If there is less than 7 non receptacle spaces, the cubes need to be stacked.

This brings the total number of states that can be represented to 168 (much more for stacked version), more than enough without having to worry about grids with correct spacing.

The cubes are not allowed to be marked.

[wolfgang]

The way I'm reading the problem it seems that row order is random, i.e. the last person does not need to have 8 on their back, so...

Due to the constraints of the problem (each of 1 to 8 must be used once and exactly once on the cubes and each back number must have +/- 1 the cube number), there is only one possible one the cubes can be allocated to the numbers, as TSLF specified previously. So if you know the person's back number, you know the person's cube number.

So the plan is this:

Calling the dimension parallel to the numbered squares X, and the dimension perpendicular Y, the nth person who enters places

their cube at Y=n/8 the distance from the row of numbered squares to the edge of the table, at X=the number he sees on the person in front of him or the number he knows they must have on their cube (they agree on which before hand). This will 'graph' the numbers, and the last person will be able to figure out where to place each box fairly easily by 'reading' it, and figure out which number the last person had by process of elimination.

The OP should begin with the last one,who can see the number infront of him,but still don`t know his own number.

[TimeSpaceLightForce]

<1st7 <2nd4 <3rd2 <4th8 <5th1 <6th6 <7th5 <8th3 row order and the number on their backs

Table____________________________________________________

[1] [2] [3] [4] [5] [6] [7] [8] numbered square order

r8b3c4 row order the nearest to numbered square the earliest to put mini cube

r7b5c6

r6b6c5

r5b1c2

r4b8c7

r3b2c1

r2b4c3

r1b7c8

___________________________________________________________

The 8th in row must place the cube as to hint his row order and the back number he see (nearest to number squared in front of square 5).

The 7th in row will know his back number and cube number from previous cube position...6th, 5th, 4th,3rd & 2nd in row will know too.

The 1st in row will know too and since square [3] has nothing in front, it is the 8th in row's back number and his cube number must be 4..

All info complete..he place those mini cubes on numbered squares.

Note: back1 cube2, back2 cube1, back8cube7, back7cube8, back3cube4, back4cube3, back5cube6, back6cube5

[wolfgang]

<1st7 <2nd4 <3rd2 <4th8 <5th1 <6th6 <7th5 <8th3 row order and the number on their backs

Table____________________________________________________

[1] [2] [3] [4] [5] [6] [7] [8] numbered square order

r8b3c4 row order the nearest to numbered square the earliest to put mini cube

r7b5c6

r6b6c5

r5b1c2

r4b8c7

r3b2c1

r2b4c3

r1b7c8

___________________________________________________________

The 8th in row must place the cube as to hint his row order and the back number he see (nearest to number squared in front of square 5).

The 7th in row will know his back number and cube number from previous cube position...6th, 5th, 4th,3rd & 2nd in row will know too.

The 1st in row will know too and since square [3] has nothing in front, it is the 8th in row's back number and his cube number must be 4..

All info complete..he place those mini cubes on numbered squares.

Note: back1 cube2, back2 cube1, back8cube7, back7cube8, back3cube4, back4cube3, back5cube6, back6cube5

Very nice trial....I had another approach, but yours is ok and clever

[Yoruichi-san]

Lol...is my terminology really that hard to understand?

The exactness of the spacing isn't really important, as long as the 2nd person who enters places their cube noticeably behind the 1st in the Y direction, and the 3rd places theirs noticeably behind the 2nd's, etc. I specified that spacing in my solution to ensure the last few ppl who entered had room on the table to put their cubes,

in the case of limited space. 8 is actually not hard to space fairly accurately, since it's dividing in 2 three times, but if the number had been harder, as long as they make sure the last ppl have space, it doesn't need to be exact, since they're going in sequential order on Y.

But I do like you solution too...lol...just need to smuggle in some invisible ink and a couple matches ;P

[CaptainEd]

There's one row of numbered squares, and the OP requires that only the last player can put the cubes in the squares.

But it doesn't explicitly prevent other players from moving the cubes around. They just can't put them into the row of squares.

So, instead of establishing 8 rows above the row of squares (per Y-san), just establish two--the "home row" and the "indicator row".

First player (last in line), who knows the second player's number, puts his/her own cube in the indicator row, putting it in the column where the 2nd player's cube should go.

Remaining players (knowing the next player's number) come in, place their cube in the home row below the indicator cube, then move the indicator cube to the column where the next player's cube should go.

(In other words, the first player's cube is used as the indicator cube all the way through).

Last player places their cube in home row below the indicator cube, moves indicator cube to the empty column, then moves each cube into the row of squares.

Saves a little real estate in principle (though you'd think Y-san's approach would work in practice), but would start to look real good if there were 100 players rather than 8.

[Yoruichi-san]

Fair enough, mon capitan, but that would be one loooonnng table ;P

Nice catch tho...I wasn't sure it you were allowed to move the boxes due to language differences and translation issues, etc .

[CaptainEd]

yes long, but looooooong is better than looooooooong squared. :-).

I didn't know whether the OP intended to accept my proposal, either. But I was offended by the large amount of wasted real estate, so I thought I'd point out the elegance of the solution that you folks arrived at--each person (but the first) can immediately know the proper location of their own piece. And the last knows what to do with the first. The additional dimension is not needed for the logic of the solution, aside from preserving the arrival order.

Nice job, all!

[Yoruichi-san]

Yes, it is indeed an elegant solution, but unfortunately out of those 100 ppl the first 99 are germaphobes and refuse to touch the box that someone else has touched without sterilizing it first

Just kidding...I guess we'll have to wait to hear from wolfgang whether moving the boxes are allowed .