Yes...the proof is correct and this is what stone had suggested in his
earlier post.
Consider one red sector in the inner disk in each of the 200
different positions, it will match against exactly 100 sectors in the
outer disk since there are 100 of the red sectors in the outer disk.
Similarly
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
@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
Let R be the total reds on inner disk.
Consider a half with 'r' reds.
Half1 = r reds and 100-r whites
Half2 = R-r reds and 100-R+r whites.
If r = R-r, match half1 with Red half of outer disk.
Total matching = r + 100 - R + r = 100 - R + 2*r
Now r = R - r = 2*r - R = 0
Total matching = 100 - R +
The problem statement particularly mentions that inner disk has red and
white sections painted **arbitrarily**. It doesn't say any such thing about
outer disk.
On 3/24/07, Rajiv Mathews [EMAIL PROTECTED] wrote:
On 3/24/07, Vishal [EMAIL PROTECTED] wrote:
If r = R-r, match half1 with Red
yes...but that does not mean that you can assume that the 100 reds and
100 whites are contiguous blocks.It just says that the outer disk has
a sum of the 100 reds and 100 whites and does not say that they are
contiguous.
--~--~-~--~~~---~--~~
You received this
Yes, I don't think we can assume that the reds and whites are contiguous.
They might be arbitrarily distributed. The only information is that there
are 100 reds and 100 whites.
On 3/25/07, Karthik Singaram L [EMAIL PROTECTED] wrote:
yes...but that does not mean that you can assume that the 100
We rotate the inner disk at 200 different positions.
Now we count the total number of color-matching events.
For each inner section,no matter white or black ,there is 100 color-
matching events.
So total number of color-matching events is 100*200
The summation of color-matching events of 200
Hi,
The proof given by Vishal is correct irrespective of the arrangement (which
he himself did not realize).
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
Half does not
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
Half does not mean it should be continuous. So the proof
19 matches
Mail list logo
