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Puzzles / Riddles: Star and STone Question -2-

Posted on 2008-06-19
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Last Modified: 2011-10-19
Please refer
http://www.experts-exchange.com/Other/Puzzles_Riddles/Q_23497738.html

One of the correct answer suggested by ozo is
Step 1. b->d->e
Step 2. i->j->b
Step 3. f->h->i
Step 4. c->d->f
Step 5. j->b->c
Step 6. g->h->j
Step 7. d->f->g

In practice, When I try to solve the problem, without seeing the answer, I am always stuck at step
7 or Step 8.

My question now is.
1) Is Starting Position is important, or is it important to place the 1st stone in a particular position. Can a rule be devised to locate starting positoin?

2) Can a rule be devised so that, after each step one can tell, whether the solution is possible or impossible in next steps.

Riddle By Mat

star-and-stone.doc
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Comment
Question by:Mehram
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13 Comments
 
LVL 84

Expert Comment

by:ozo
ID: 21828584
Starting position is not important.
Note that the above steps form a cycle, you can start anywhere along the cycle.
0
 
LVL 84

Expert Comment

by:ozo
ID: 21828597
If you redraw the graph to connect points that are 2 step apart,
and change the rule to say that stones make one step in the new graph rather than two steps in the original graph, the solution should become clearer
0
 

Author Comment

by:Mehram
ID: 21836336
Hi ozo,
I tried hard on your first comment and suddenly the word (cycle) makes me everything mirror clearl. Alternately I can say,

Cycle Rule: [ The position of the stone to be placed at any step is the starating
                            postion ot the previous step.]
                      ( With the above logic, I can place the stones easily)
The solutions provided by u and infinity08 are in accordance with the cycle rule.
Keeping the solutions in front of me, I can place the stones randomly ( avoiding cycle rule), but the problem is how to remember / grasp the random aspect.

I could not uderstand, your second comment. May be, If I understand your 2nd comment I can place the ston randomly ( breaking cycle rule).

Could u, please, make your 2nd statement more clear, probably giving example of two or three steps.
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LVL 84

Expert Comment

by:ozo
ID: 21836465
from any point, where can you get to or come from in 2 steps?
0
 

Author Comment

by:Mehram
ID: 21836586
<<from any point, where can you get to or come from in 2 steps?>>

b->d->e        Start Positon b, which is the end position of next step
i->j->b          Start Positon i, which is the end position of next step
f->h->i          Start Positon f, which is the end position of next step
c->d->f         Start Positon c, which is the end position of next step
j->b->c         Start Positon j, which is the end position of next step
g->h->j         Start Positon g, which is the end position of next step
d->f->g

This rule is easy for me. I am going through certain cycle.
Can I go through randomly and also remeber it by some rule / pattern
0
 
LVL 53

Accepted Solution

by:
Infinity08 earned 1000 total points
ID: 21836724
>> Can I go through randomly and also remeber it by some rule / pattern

Sure. Just choose a point ... From that point, you will have exactly two options to place the next stone (no matter which point you chose, there will always be exactly two). Choose one direction, and from then on, you simply follow the same logic. In that next position, you will have two options again. One of these options will be the previous stone though, so you have to take the other. Continue that until all stones are placed.

Note that there are only two variables : the original position, and the original direction you choose.
0
 
LVL 84

Assisted Solution

by:ozo
ozo earned 1000 total points
ID: 21836761
You can also start at th endiing position, and ask what does the previous position need to look like
in order to get to the desired ending position,
then what did the step before that need to liie like?

Another observation to make is that whether of not count 2 is occupied is irrelevant.
the only points that mater are counts 1 and 3
So it matters not how 1 connects to 2 or how 2 connects to 3
it only matters how 1 connects to 3

0
 
LVL 84

Expert Comment

by:ozo
ID: 21836885
> so you have to take the other
Actually you can add on to either end of the list,
going around the circle in either direction.
this may be easier to see if you redraw the graph as a circle, ignoring the irrelevant connections.
0
 
LVL 53

Expert Comment

by:Infinity08
ID: 21836899
>> going around the circle in either direction.

I meant, if you're actually placing the stones according to the original puzzle (one stone after the other following the rules), then you can only choose the first stone (anywhere), and for the second stone you can choose the direction (one of both). For all the next stones however, there's no choice, and only one possible position to place them in (following the chosen direction).
0
 

Author Closing Comment

by:Mehram
ID: 31469045
Thanks
0
 
LVL 84

Expert Comment

by:ozo
ID: 21836928
There is a choice after the first stone
after
b->d->e        Start Positon b, which is the end position of next step
you could have continued continued
a->j->h
d->b->a
g->f->d
j->h->g
c->b->j
and you can switch back to the original side
i->j->b          Start Positon i, which is the end position of next step
f->h->i          Start Positon f, which is the end position of next step
c->d->f         Start Positon c, which is the end position of next step




j->b->c         Start Positon j, which is the end position of next step
g->h->j         Start Positon g, which is the end position of next step
d->f->g
(note that this is the reverse of
g->f->d
j->h->g
c->b->j)
0
 
LVL 84

Expert Comment

by:ozo
ID: 21836985
> Can a rule be devised so that, after each step one can tell, whether the solution is possible or impossible in next steps.
A solution is possible if the occupied points are all adjacent in the circular sequence
...->e<->b<->i<->f<->c<->j<->g<->d<->a<->h<->e<-...
note that in the sequence
... -e-d-b-j-i-h-f-d-c-b-j-h-g-g-d-b-a-j-h-f-e- ...
it makes no difference whether the alternate points are occupied,
since you just jump over then just jump over them in any case,
so you can just ignore them fot the purposes of the problem
0
 

Author Comment

by:Mehram
ID: 21844713
Thanks ozo for more detail.
0

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