Numbers in Olympic Rings
Rouben Rostamian
In article <3335d24e...@news.otago.ac.nz>,
Andrew Gray <ag...@commerce.otago.ac.nz> wrote:
>I've been trying to solve this for a while now and I'm willing to
>admit defeat. I could brute force the numerical solutions, but I'm
>sure that there is a general solution to this type of problem.
>
>Given the five olympic rings how can the digits one through nine be
>placed within the nine regions (five non-overlap ring regions and four
>overlap regions shared between two rings) so that each ring contains
>the same total.
>
>Or alternatively, using the labels A through I for the regions, how
>can the numbers one through nine be assigned to the variables such
>that:
>
>A+B = B+C+D = D+E+F = F+G+H = H+I
I gave this a try without much success. For whatever it's worth,
A, B, H and I can be solved in terms of the remaining variables:
A = C + D
B = E + F - C
H = D + E - G
I = G + F
Furthermore, from A+B+C+D+E+F+G+H+I=1+2+3+...+9 we obtain an additional
equation:
3*D + 3*E + 3*F + C + G = 45,
which implies, in particular, that C+G is a multiple of 3.
A brute-force search yields the following solutions:
A B C D E F G H I
solution 1: 4 9 1 3 8 2 5 6 7
solution 2: 9 4 1 8 3 2 5 6 7
solution 3: 9 2 5 4 6 1 7 3 8
--
Rouben Rostamian <rou...@math.umbc.edu>
Andrew Gray
I've been trying to solve this for a while now and I'm willing to
admit defeat. I could brute force the numerical solutions, but I'm
sure that there is a general solution to this type of problem.
Given the five olympic rings how can the digits one through nine be
placed within the nine regions (five non-overlap ring regions and four
overlap regions shared between two rings) so that each ring contains
the same total.
Or alternatively, using the labels A through I for the regions, how
can the numbers one through nine be assigned to the variables such
that:
A+B = B+C+D = D+E+F = F+G+H = H+I
Any help would be much appreciated.
Cheers,
Andrew
Software Metrics Research Laboratory, University of Otago
Phone: +64 3 479 5282 Fax: +64 3 479 8311
email: ag...@commerce.otago.ac.nz
http://divcom.otago.ac.nz:800/COM/INFOSCI/SMRL/home.htm
Mike Naylor
ag...@commerce.otago.ac.nz (Andrew Gray) wrote:
>I've been trying to solve this for a while now and I'm willing to
>admit defeat. I could brute force the numerical solutions, but I'm
>sure that there is a general solution to this type of problem.
>
>Given the five olympic rings how can the digits one through nine be
>placed within the nine regions (five non-overlap ring regions and four
>overlap regions shared between two rings) so that each ring contains
>the same total.
>
>Or alternatively, using the labels A through I for the regions, how
>can the numbers one through nine be assigned to the variables such
>that:
>
>A+B = B+C+D = D+E+F = F+G+H = H+I
>
>Any help would be much appreciated.
I don't know what you have against brute force. I think brute force is cool
if you have a computer to do it for you!
There are eight solutions. Four of them are:
Solution 1: A = 4, B = 9, C = 1, D = 3, E = 8, F = 2, G = 5, H = 6, I = 7
Solution 2: A = 5, B = 9, C = 2, D = 3, E = 4, F = 7, G = 1, H = 6, I = 8
Solution 3: A = 7, B = 6, C = 5, D = 2, E = 3, F = 8, G = 1, H = 4, I = 9
Solution 4: A = 7, B = 6, C = 5, D = 2, E = 8, F = 3, G = 1, H = 9, I = 4
The other four are the mirror images of these.
-
Mike Naylor - myfirstname...@mail.serve.com
Play Five by Five Poker at http://www.serve.com/games
Wei-Hwa Huang
ag...@commerce.otago.ac.nz (Andrew Gray) writes:
>Given the five olympic rings how can the digits one through nine be
>placed within the nine regions (five non-overlap ring regions and four
>overlap regions shared between two rings) so that each ring contains
>the same total.
>Or alternatively, using the labels A through I for the regions, how
>can the numbers one through nine be assigned to the variables such
>that:
>A+B = B+C+D = D+E+F = F+G+H = H+I
First, make the problem symmetrical by adding J=0 to the equation:
J+A+B = B+C+D = D+E+F = F+G+H = H+I+J
This turns the problem into putting the numbers 0..9 on the edges
and corners of a pentagon so that the three numbers on each edge
has the same sum (and the artificial restriction that 0 is in a
corner).
This problem has many solutions. First, note that if we add all
five edges and subtract all the numbers, we get the five corners.
This implies that the five corners is a multiple of five
and related to the sum on the edges. That is:
Magic sum Sum of corners Sum of edges
----------------------------------------
11 10 35
12 15 30
13 20 25
14 25 20
15 30 15
16 35 10
The sums stop at 11 and 16 because boundaries are met; i.e.,
10 = 0+1+2+3+4 and 35 = 5+6+7+8+9.
Now, note that if replace each digit k by (9-k), we get another
solution with a sum of (27-s). So we'll only have
to consider half of these. (s is the magic sum)
Another piece of helpful information is when we add two adjoining
edges and the edge opposite it, then subtract all the numbers.
We find that if we label one corner X and the two non-adjacent,
non-opposite edges Y and Z, we get
X + 12 = Y + Z when s = 11
X + 9 = Y + Z when s = 12
X + 6 = Y + Z when s = 13
or, more generally, X is congruent to Y+Z mod 3.
These allow you to pare down your search considerably.
--
Wei-Hwa Huang, whu...@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
God only knows why we're here -- not what the **** we're doing here.
Chris Lomont
Disregard my last post. I was wrong....
Chris Lomont
On Mon, 24 Mar 1997, Andrew Gray wrote:
> I've been trying to solve this for a while now and I'm willing to
> admit defeat. I could brute force the numerical solutions, but I'm
> sure that there is a general solution to this type of problem.
>
> Given the five olympic rings how can the digits one through nine be
> placed within the nine regions (five non-overlap ring regions and four
> overlap regions shared between two rings) so that each ring contains
> the same total.
>
> Or alternatively, using the labels A through I for the regions, how
> can the numbers one through nine be assigned to the variables such
> that:
>
> A+B = B+C+D = D+E+F = F+G+H = H+I
>
We have a total of 1+2+..+9 = 9*10/2 = 45 points to distribute. Divided
by 5 regions gives 9 per region. There are 5 odd numbers 1,3,5,7,9, so
one goes in each region since each region totals an odd number. Since
placing 9 in any region fills its capacity, any other number added to it
overflows the region, so the question as posed has no solution.
Chris Lomont
Mike Naylor
According to the time stamps, it took you one minute to realize that. But
if it makes you feel any better, you sounded very convincing!
I guess you didn't see my post from the day before, which was directed to
rec.puzzles but not to sci.math. I posted that there are eight solutions.
Four of them are:
Solution 1: A = 4, B = 9, C = 1, D = 3, E = 8, F = 2, G = 5, H = 6, I = 7
Solution 2: A = 5, B = 9, C = 2, D = 3, E = 4, F = 7, G = 1, H = 6, I = 8
Solution 3: A = 7, B = 6, C = 5, D = 2, E = 3, F = 8, G = 1, H = 4, I = 9
Solution 4: A = 7, B = 6, C = 5, D = 2, E = 8, F = 3, G = 1, H = 9, I = 4
and the other four are the mirror images of these.
Bill Daly
Chris Lomont wrote:
>
> On Mon, 24 Mar 1997, Andrew Gray wrote: ...
> > Or alternatively, using the labels A through I for the regions, how
> > can the numbers one through nine be assigned to the variables such
> > that:
> >
> > A+B = B+C+D = D+E+F = F+G+H = H+I
> >
>
> We have a total of 1+2+..+9 = 9*10/2 = 45 points to distribute. Divided
> by 5 regions gives 9 per region. There are 5 odd numbers 1,3,5,7,9, so
> one goes in each region since each region totals an odd number. Since
> placing 9 in any region fills its capacity, any other number added to it
> overflows the region, so the question as posed has no solution.
>
> Chris Lomont
Well, that's as may be, but the values ABCDEFGHI=837164529 are such that
A+B = B+C+D = D+E+F = F+G+H = H+I.
Regards,
Bill
Wei-Hwa Huang
0x49F...@pgp.key (Mike Naylor) writes:
>Chris Lomont <clo...@omni.cc.purdue.edu> wrote:
>>Disregard my last post. I was wrong....
>According to the time stamps, it took you one minute to realize that. But
>if it makes you feel any better, you sounded very convincing!
>I guess you didn't see my post from the day before, which was directed to
>rec.puzzles but not to sci.math. I posted that there are eight solutions.
>Four of them are:
>Solution 1: A = 4, B = 9, C = 1, D = 3, E = 8, F = 2, G = 5, H = 6, I = 7
>Solution 2: A = 5, B = 9, C = 2, D = 3, E = 4, F = 7, G = 1, H = 6, I = 8
>Solution 3: A = 7, B = 6, C = 5, D = 2, E = 3, F = 8, G = 1, H = 4, I = 9
>Solution 4: A = 7, B = 6, C = 5, D = 2, E = 8, F = 3, G = 1, H = 9, I = 4
>and the other four are the mirror images of these.
Forgive me, Mike, but your first solution *is* a mirror image of the fourth.
Even then, there are more than eight solutions: have you considered
8 3 7 1 6 4 5 2 9 ?
Mike Naylor
I wrote (concerning solutions to A+B = B+C+D = D+E+F = F+G+H = H+I, where
the digits 1 through 9 are assigned to the letters A through I):
>... there are eight solutions.
>Four of them are:
>Solution 1: A = 4, B = 9, C = 1, D = 3, E = 8, F = 2, G = 5, H = 6, I = 7
>Solution 2: A = 5, B = 9, C = 2, D = 3, E = 4, F = 7, G = 1, H = 6, I = 8
>Solution 3: A = 7, B = 6, C = 5, D = 2, E = 3, F = 8, G = 1, H = 4, I = 9
>Solution 4: A = 7, B = 6, C = 5, D = 2, E = 8, F = 3, G = 1, H = 9, I = 4
>and the other four are the mirror images of these.
whu...@ugcs.caltech.edu (Wei-Hwa Huang) wrote:
>Forgive me, Mike, but your first solution *is* a mirror image of the fourth.
>Even then, there are more than eight solutions: have you considered
>8 3 7 1 6 4 5 2 9 ?
Oops! I'd better have another look at this. I stand by my statement that
there are eight solutions, but I was careless when I picked out the four
that needed to be thrown out as mirror images. Here are all eight
solutions:
Solution 1: 4, 9, 1, 3, 8, 2, 5, 6, 7
Solution 2: 5, 9, 2, 3, 4, 7, 1, 6, 8
Solution 3: 7, 6, 5, 2, 3, 8, 1, 4, 9
Solution 4: 7, 6, 5, 2, 8, 3, 1, 9, 4 (mirror image of #1)
Solution 5: 8, 3, 7, 1, 6, 4, 5, 2, 9 (the one you mentioned)
Solution 6: 8, 6, 1, 7, 4, 3, 2, 9, 5 (mirror image of #2)
Solution 7: 9, 2, 5, 4, 6, 1, 7, 3, 8 (mirror image of #5)
Solution 8: 9, 4, 1, 8, 3, 2, 5, 6, 7 (mirror image of #3)