Hello guys:
I have been thinking half hr or so on this problem, so pls forgive me
for anything wrong but I think I have an idea that might help:
-Assuming:
1) Giving the nature of the number of colors I supposed that both the
inner and outer disks are divided into an even number of pieces (n)
Hello guys:
I have been thinking half hr or so on this problem, so pls forgive me
for anything wrong but I think I have an idea that might help:
-Assuming:
1) Giving the nature of the number of colors I supposed that both the
inner and outer disks are divided into an even number of pieces (n)
Hello guys:
I have been thinking half hr or so on this problem, so pls forgive me
for anything wrong but I think I have an idea that might help:
-Assuming:
1) Giving the nature of the number of colors I supposed that both the
inner and outer disks are divided into an even number of pieces (n)
On 3/25/07, Rajiv Mathews [EMAIL PROTECTED] wrote:
On 3/25/07, Prunthaban Kanthakumar [EMAIL PROTECTED] wrote:
If you see carefully his proof does not assume anything about sections
colored continuously. His proof assumes only one thing Half of them
are
red and half of them are white
Ouch I got the question completely wrong assuming the inner disc is
continuous.Sorry for the confusion.
On 3/25/07, Prunthaban Kanthakumar [EMAIL PROTECTED] wrote:
On 3/25/07, Rajiv Mathews [EMAIL PROTECTED] wrote:
On 3/25/07, Prunthaban Kanthakumar [EMAIL PROTECTED] wrote:
If you
I did assume that the outer disk is painted half (contiguous) red and half
white.
However the 'equivalence' should do the trick and the same proof applies.
As far as Stone's proof goes, I did not understand -
For each inner section,no matter white or black ,there is 100
color-matching
@Vishal
When you rotate the inner section through it's 200 configurations, each of
the inner section happens to come in tune with each of the outer sections,
so there will 100 'matchings' and 100 mismatches.
On 3/25/07, Vishal [EMAIL PROTECTED] wrote:
I did assume that the outer disk is painted
When both disks are not painted continuous 'equivalence' does not work.
Because in your proof when one half does not give the answer, you just take
take the other half and align it. But for arbitrary configuration, when one
configuration does not work, you cannot align the other half. It will not
@alfredo:
I dint get this part:
1) Lets say that we have the outer circle painted in the following
manner: 1 red(C1), 1 white(C2), 1 red(C3), etc. We call this sections
(RWR)
2) Then if we have that the inner circle is painted in the same way we
finish.
3) if not then lets say that we have a
That's a breadth first search. The only gotcha is that the usage of
push and pop maybe misleading (implying that we're using stack). But
actually it should use FIFO. You should view each push as enqueue and
pop as dequeue. It will traverse the tree layer by layer, left to
right.
On 3/24/07, vim
@Karthik
I think the confusion arises from my lack of explanation in the nature
of the pseudo-induction
Lets take 2 sections in the outer circle that are painted red (S1,
S2). My variable m messures the number of whites between S1 and S2
with no red sections.
What I wrote is the basis of the
Got it.
Stone's solution seems to be right.
On 3/25/07, Prunthaban Kanthakumar [EMAIL PROTECTED] wrote:
When both disks are not painted continuous 'equivalence' does not work.
Because in your proof when one half does not give the answer, you just take
take the other half and align it. But for
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