wolfgang
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Posts posted by wolfgang
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i am very sorry.. the spoiler doesnt seem to work ...
from the hint.. "we love brainden"
Yes....thats right...We love brainden
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If I want to write(Wolfgang) in this GT method,it will be :
T+3T-5G+5G-1G0G-6T-6G0 -
If G,T are 0,0
Find:
T+3G-2 G+5 T-5T+2G-2 G-5T-2G-6G+2T-6 G-3G-2T-6
T+3G-2 G+5 T-5T+2G-2 G-5T-2G-6G+2T-6 G-3G-2T-6
T is Take? Negative? . G is Give? Positive?
T+3G-2 = Take+3 Give -2 = -5
G+5 T-5T+2G-2 = Give + 5 Take -5 Take + 2 Give -2 = +6
G-5T-2G-6G+2T-6 = Give -5 Take -2 Give -6 Give +2 Take - 6 = -1
G-3G-2T-6 = Give -3 Give -2 Take - 6 = +1
Start 0:
End 1:( -5 +6 -1 + 1)
No... thats far away from the answer I am seeking
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Or you can call them as " points to start from "
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What do you mean "G,t are 0,0"? G=0 and T=0?
They are like Y axis
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If G,T are 0,0
Find:
T+3G-2 G+5 T-5T+2G-2 G-5T-2G-6G+2T-6 G-3G-2T-6
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yes, it can be done in 2 moves with one yellow ball on a basket
Yes....thats right
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You have two baskets,each one contains 15 small balls( 5 Yellow,5 Red,and 5 Blue),you should exchange each two balls from one basket with one ball from the other basket,according to these roles:
1- Yellow+Yellow....... should be exchanged with a red ball.
2- Red +Red..........should be exchanged with a Blue ball.
3- Blue +Blue..........should be exchanged with a Yellow ball.
4- Yellow +Red..........should be exchanged with a Blue ball.
5- Red + Blue........should be exchanged with a Yellow ball.
6- Blue + Yellow.....should be exchanged with a Red ball.
7- each of the above exchanges will be considered as one step.
Your aim is to get only ONE Yellow ball in one of these two baskets in less than 14 steps.
Have a nice trial
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I'm taking a slightly stricter view, that NEWS coding is not available.
Wolfgang has allowed that prisoners can either face toward or away from the door.
That is, NS (or EW, as I drew it) coding is available, and that requires two prisoners to signal a color.
That presents a particular problem for P2 on initial entry, but I think it can be worked around.
If only a binary signaling scheme is allowed, CaptainEd has shown that only prisoner 1 needs to change his mind. Even in the stricter NS view, there are some latitude to define a quadnary signaling scheme due to an implied necessary condition
the entering prisoners *must* be able to see the signaler rotate before taking their position. Otherwise, the last prisoner has no way to receive information about his hat color, and the puzzle is not solvable. Therefore, rotation *within sight* of entering prisoners is a necessary condition of the puzzle.
Since the entering prisoners are able to see rotation, it is then trivial to combine rotation speed, direction, and amount with NS coding to allow quadnary signals. (Again, I don't see anything in the OP prohibiting this; well, except for the 'no communication of any kind allowed', but we all know how that went). Once quadnary signals are allowed, no one needs to change position.
You're right.
NS signaling is enough.
Four easily distinguishable rotations: None, 180ocw, 180occw, and 360occw.
And entering prisoners must be able to see the rotation.
I think we spent most of the time on this puzzle with a poor understanding of the conditions.
But it was fun to finally solve something possible.
It's just that the problem became teasing out the constraints that admitted a solution!
Bertrand was helpful in this case.
Thanks, Wolfgang.
I want to thank you all....I was thinking like this:
After the first three prisoners took their right places,they all should be facing the door, the possible combinations would be:
Yellow,Green,Red,........
Yellow,Green,........,Blue
Yellow,.........,Red, Blue
.........,Green, Red, Blue
so when the 4th one enters, Let him to have X color..the X one standing in the raw will turn his face to the wall,and the new comer will stand in that raw facing the door....and when nobody turns to the wall so he should stand at the empty place facing the door.
when the four raws are made, each new comer will know where he belongs when the man with the same color turns his face toward the wall.
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Conveying position to next arrival. Consider the sequence of prisoners in rows Y, G, R, B. The first two of those prisoners will convey the newcomer's color to him.
Also, To match Bonanova's picture, I should say that Prisoner A, facing the door, starts in location 0, with his LEFT hand pointing to Position 1 = Green, further LEFT is Position 2 = Red, even further LEFT is Position 3 = Blue. The negative values are to his RIGHT.
OK, I think we are agreeing that we have to relax our interpretation of Wolfgang's proscription against communication.
The first two prisoners can establish the locations of all 4 colored rows, before the third one enters the room.
Then, the prisoners can inspect each arriving prisoner, rotate their positions to indicate the newcomer's color, and the newcomer can take his place in the appropriate row.
* Let's consider that there are (invisible) row numbers, from -4 to +4.
* Prisoner A enters and stands at 0, asserting his own color to be Yellow (regardless of his actual color, which he doesn't know). So at his right hand (position 1) should be Green, next (position 2) Red, further right (position 3) is Blue.
* Prisoner B enters, takes his position according to A's actual color. That is, B goes to the row A SHOULD be in = Row(Color(A)). Now A knows his own color and that of B.
* Prisoner A now changes the color framework by moving himself so that B is in the row B should be in. That is, A moves to Row(Color(A) + Row(B) - Color( B ) ). At this point, both prisoners know their own colors, and any observer can tell the new color framework, which may or may not be 0 = Yellow.
1) A is Green, B is Blue. A enters, establishes location of 0. B enters, sees that A is Green, so he moves to (Row(Color(A)) = 1. Now A knows A is Green and B is Blue. Now A moves to (Row(Color(A) + Row(B) - Color( B ) ) = (1+1-3) = -1. At this point, in absolute terms, A is in position -1, B is in position 1. However, to the remaining prisoners, A (visibly Green) is therefore at position 1 and B (visibly Blue) is at position 3. Anybody knows where Yellow and Red go.
2) A is Green, B is Green. A enters. B enters, moves to 1. A now knows that both A and B are Green, so A moves to (Row(Color(A) + Row(B) - Color( B ) )= (1+1-1)= 1 (ie. stands in the same row, directly behind or ahead of B). Now remaining prisoners can tell where Green is, therefore can infer where Red, Yellow and Blue are.
3) A is Green, B is Yellow. A enters. B enters, moves to 1. A now knows that A is Green and B is Yellow, moves to (Row(Color(A) + Row(B) - Color( B ) ) = (1 + 1 -0) = 2. Now remaining prisoners see B is Yellow and A is Green; they can easily infer where Red and Blue belong.
When prisoner K enters, two chosen prisoners tell him his color by facing him or facing away from him, giving two bits of information, therefore a number from 0 to 3. Which two chosen prisoners? We have a number of ways to establish a sequence; here are two ways:
* If the prisoners were permitted to choose their order of arrival, then prisoner K looks at prisoners I and J, who just entered before him.
* Alternatively, consider the sequence of all prisoners in the Yellow row, followed by Red row, followed by Green row, followed by Blue row. Take the first two prisoners in this sequence. They are the ones who will tell prisoner K his color.
your method is successful with the first two prisoners, but you should find a very simple,uncomplicated way for the 3rd one,if you find it, the rest will be very easy and straight forward job. -
A breath
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This is a real one....on the inner wall of a W.C. someone wrote this with red paint" long live the communist party of russia "
and underneath,another one replays with this " YES....but only in W.C. " !!
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Hi...
If
.........f= 3
........ h= 4
........n= 9
........y= 14
.......
Find T?
Shouldn't the OP be F=1, H=2,... etc.?
In the alphabet of (f, h, n, y) k=1!
Why should F=1??
Please see my above explanation
How is "f=3" in that counting system? The letter "f" has no straight lines.
Perhaps, you meant "F=3".
Yes...I numbered them according to capital letters,but I asked the question using small letters to make it hard
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I can add something which may help you... The raws may face the door or the opposit direction!
Continuing with the community solve approach, can we clarify:
- Must all the persons in a particular row face the same direction?
- If prisoners in say the Yellow row can face in differing directions does that constitute [illegal] communication?
- Can the first two prisoners choose the time that they [legally] alter their position, at any time up to the last prisoner takes his place?
During making raws...some of them can be..back to back (in any raw)...but by reaching the last prisoner,they all shoud be at the same direction
But only the first two prisoners can move.
This means the seventh prisoner, say can come into a row back to back with someone else in that row, then at a later time switch his direction so that at the end all in that row have the same direction. So as long as a prisoner stays in the same position, he can flip his direction back and forth during the formation of rows.
That sounds like communication.
Another piece of jello fell off the nail...
Yes....in my OP I said....each prisoner should choose his raw and once he did,he is not allowed to change his mind,but turning around himself was not mentioned
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Hi...
If
.........f= 3
........ h= 4
........n= 9
........y= 14
.......
Find T?
Shouldn't the OP be F=1, H=2,... etc.?
In the alphabet of (f, h, n, y) k=1!
Why should F=1??
Please see my above explanation
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Another point to clarify.
Must the rows be parallel, and in particular Y-G-R-B order?
OP seems to suggest this condition.
+-------------------------+
| |
| Y Y Y Y Y . . . |
| G G G . . . . . |
Door R R R R R R . . |
| B B B B B . . . |
| |
| |
+-------------------------+
Or could the rows e.g. all start at the middle of the room, and grow outward toward the four walls, in any sequence?
Say Yellow to the north, Green to the south, Red to the east and Blue to the west?
Yes...the raws should be as you mentioned in your diagram,but all should face one direction
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I can add something which may help you... The raws may face the door or the opposit direction!
Continuing with the community solve approach, can we clarify:
- Must all the persons in a particular row face the same direction?
- If prisoners in say the Yellow row can face in differing directions does that constitute [illegal] communication?
- Can the first two prisoners choose the time that they [legally] alter their position, at any time up to the last prisoner takes his place?
During making raws...some of them can be..back to back (in any raw)...but by reaching the last prisoner,they all shoud be at the same direction
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I can add something which may help you... The raws may face the door or the opposit direction!
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Wolfgang, I think the point that asks for clarification is this:
Must each prisoner wait, outside the room and blindfolded, until previous prisoners have taken their positions?
Yes...thats right....he should wait until the previous prisoner is standing in his raw and took his position.
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To me it seems that 'Entering singly and not changing one's position once taken" are conditions with no effect, if "hanging out" is permitted. What relevance to the solution can that guidance have, other than to prohibit that option? Yet that is what adds the difficulty that I see here.
If language issues have made the point debatable, Wolfgang will have to clarify.
I am at loss about the purpose of "Entering singly". "Not changing one's position once taken" seems to be a reasonable, good, and severe limitation upon those criminals, indeed.
Conversely, if OP requires each conman to take his place in the formation before the next inmate enters the hall, then I cannot imagine what method could possibly exist for the condemned men? After all, the rows would start forming without any knowledge of the color on the forehead of the next culprit to enter...
Yes, each one should choose his place where to belong,once he did, he is not allowed to change his mind.(Except the first two ...see OP)
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Plan: NEWS
North=Y East=G West=R South=B
Color Rows
Y _ _ _ _ _ _ _
G _ _ _ _ _ _ _
R _ _ _ _ _ _ _
B _ _ _ _ _ _ _
1 enters stand in front of door "stationary"
2 enters, 1 face (NEWS) according to color of 2,
2 knew his own color, stands on his color row facing
(NEWS) according to color of 1,1knew his own color.
3 enters, 1 face (NEWS) according to color of 3,
3 knew his own color, stands on his color row
4 enters, 1 face (NEWS) according to color of 4,
4 knew his own color, stands on his color row
..goes on until last prisoner then 1 stands on his color rows
Note: 1 reset initial position as the prisoners walks their rows
hey WG this is a nice puzzle for my friends
According to your method(NEWS),seems as if the first prisoner is giving a hint to the others,which is actually not allowed...see the roles in OP
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Color Rows
Y _ _ _ _ _ _ _
G _ _ _ _ _ _ _
R _ _ _ _ _ _ _
B _ _ _ _ _ _ _
position
2
1
3
door
4
5
6
7
8
9
10
11
12
13
14
.
.
L
Plan: using 4-way path
Y G R B
|o o| o/ \o
|o o| /o o\
set-up stage:
1 enter and face the door
2 enter, walk behind 1 and face the door
3 enter, pass 1& 2 ( 1 of 4 ways) according to color of 1,
then stands behind 2 and face the door ..so 1 knew his own color
1 pass 2&3 (1 of 4 ways) according to color of 2,
then stands behind 3 and face the door.. so 2 knew his own color
2 pass 3&1 (1 of 4 ways) according to color of 3,
then stands behind 1 and face the door ..so 3 knew his own color
sitting down stage:
4 enter , time for 3 to sit down in color row ,
pass 1 & 2 (1 of 4 ways) according to color of 4
Now 4 knew his own color, stand in front of 1 facing door and waits for 5..
5 enter , time for 4 to sit down in color row ,
pass 1 & 2 (1 of 4 ways) according to color of 5
Now 5 knew his own color, stand in front of 1 facing door and waits for 6..
..goes on until "Last" prisoner has nothing to wait on, but also now knew his color,
just sit down. Finally, 1 & 2 sit down in color row .
I said in the OP",each one should himslf choose where to stay, once he did,he can not change his place".and with this method each one stays infront of 1 making a raw,so can not change his place.
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The suggested solution assumes that their are equal number of each color slip and the each person, when he enters can see every other forehead and count the color papers glued there. We need a more innovative solution, I guess.
The colors are not equal
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I think the problem's set up in such a way that you can't use that strategy. The people are blindfolded until they enter the room. So the first person won't see anyone's color, the second person will only see the first person's color, etc.They assign values to each color: yellow = 0, green = 1, red = 2, blue = 3. The first prisoner adds the values of all the other prisoners' hats (modulo 4), then finds the corresponding color to that number and stands in that row. By doing this, everyone else will know the color of their own hat. Now the second prisoner does the same thing as the first one did. This tells the first prisoner the color of his/her hat. Then the first two prisoners change their minds if necessary, moving to the right row. And everyone else can move to the right row as well.
Just for clarification, are there set positions where each row should form? In other words, should we treat this as if there are four rows of seats and all of the yellows must sit in the first row and all the greens in the second row etc.? Or instead can they form rows wherever they want, so if the first person to enter is yellow he can he stand anywhere in the room as long as all the other colors stand somewhere behind him?
They can form raws whenever they want
How can you weigh a living fish?
in New Logic/Math Puzzles
Posted
How can you weigh a living fish( which is more than 500 gm),found in an Aquarium?
You are not allowed to:
1- take the fish out of water or out of the aquarium.
2- move the aquarium or weigh it.
3- replace the water with new one.