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 The Conversing Club 2 (Posted on 2004-02-06)
Unfortunately the Conversing Club has lost members so now they are down to 12 members. Many of the members felt talking with 3 members about things ended up with 2-against-1 situations. To address this pain, the members decided to have only 2 people sit at a table each day. They also decided that with the decline in members, they only needed 6 tables and didn't need "move every time" rule. (In other words, there aren't any extra tables once two people are placed at each of 6 tables and a person can sit at a particular table more than once each "cycle") The first day they sit as following:

AB CD EF GH IJ KL

(So the second day A could sit with C or D (and others too), but not B. C could sit with A, B or E (and others too) but not D)

After 11 days, they wouldn't have anyone left to talk with, so would repeat their schedule. Their schedule was still constructed such that everyone would talk to everyone else exactly once. How was it constructed?

 See The Solution Submitted by Gamer Rating: 3.0000 (2 votes)

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 solution | Comment 2 of 5 |
This problem is solved similarly to the original Conversing Club problem, where three people met each day. In that one, the sets could be found by imagining one person's name in the center of a circle and the other names surrounding it along that circle. Triangles would be formed, most connecting three points (names) on the circle, but one connecting two points on the circle with the center. The edges that form chords would have to all span different lengths. Then each day, the set of triangles was rotated one position relative to the names, or the names relative to the triangles--same thing.

This is similar, except that instead of triangles we just have lines connecting two names. Again, one name is in the center and 11 names are on the circumference, and one line connects the center to a point (call it point 1) on the circle. The chords are among the remaining 10 positions and must each be a different length. One such way is to connect points 2 and 3 for length 1; 4 and 7 for length 3; 5 and 9 for length 5; 6 and 11 for length 5; and 8 and 10 for length 2, summarized as
2 3 ( 1); 4 7 ( 3); 5 9 ( 4); 6 11 ( 5); 8 10 ( 2)
Note that spans, n, larger than 5 are just formed by the same chord as 11 - n.

All the ways of doing this are:
```
2  3 ( 1);  4  7 ( 3);  5  9 ( 4);  6 11 ( 5);  8 10 ( 2)

2  3 ( 1);  4  9 ( 5);  5  7 ( 2);  6 10 ( 4);  8 11 ( 3)

2  4 ( 2);  3  6 ( 3);  5 10 ( 5);  7 11 ( 4);  8  9 ( 1)

2  4 ( 2);  3  7 ( 4);  5  8 ( 3);  6 11 ( 5);  9 10 ( 1)

2  4 ( 2);  3 10 ( 7);  5 11 ( 6);  6  9 ( 3);  7  8 ( 1)

2  5 ( 3);  3  7 ( 4);  4  9 ( 5);  6  8 ( 2); 10 11 ( 1)

2  5 ( 3);  3  8 ( 5);  4  6 ( 2);  7 11 ( 4);  9 10 ( 1)

2  5 ( 3);  3  9 ( 6);  4 11 ( 7);  6  7 ( 1);  8 10 ( 2)

2  6 ( 4);  3  4 ( 1);  5 10 ( 5);  7  9 ( 2);  8 11 ( 3)

2  6 ( 4);  3  8 ( 5);  4  5 ( 1);  7 10 ( 3);  9 11 ( 2)

2  6 ( 4);  3 11 ( 8);  4 10 ( 6);  5  7 ( 2);  8  9 ( 1)

2  7 ( 5);  3  4 ( 1);  5  8 ( 3);  6 10 ( 4);  9 11 ( 2)

2  7 ( 5);  3  5 ( 2);  4  8 ( 4);  6  9 ( 3); 10 11 ( 1)

2  8 ( 6);  3  6 ( 3);  4 11 ( 7);  5  7 ( 2);  9 10 ( 1)

2  8 ( 6);  3 10 ( 7);  4  7 ( 3);  5  6 ( 1);  9 11 ( 2)

2  8 ( 6);  3 11 ( 8);  4  5 ( 1);  6 10 ( 4);  7  9 ( 2)

2  9 ( 7);  3  4 ( 1);  5 11 ( 6);  6  8 ( 2);  7 10 ( 3)

2  9 ( 7);  3  5 ( 2);  4 10 ( 6);  6  7 ( 1);  8 11 ( 3)

2  9 ( 7);  3 11 ( 8);  4  6 ( 2);  5 10 ( 5);  7  8 ( 1)

2 10 ( 8);  3  7 ( 4);  4  6 ( 2);  5 11 ( 6);  8  9 ( 1)

2 10 ( 8);  3  8 ( 5);  4 11 ( 7);  5  6 ( 1);  7  9 ( 2)

2 10 ( 8);  3  9 ( 6);  4  5 ( 1);  6  8 ( 2);  7 11 ( 4)

2 11 ( 9);  3  6 ( 3);  4 10 ( 6);  5  9 ( 4);  7  8 ( 1)

2 11 ( 9);  3  9 ( 6);  4  8 ( 4);  5  6 ( 1);  7 10 ( 3)

2 11 ( 9);  3 10 ( 7);  4  9 ( 5);  5  8 ( 3);  6  7 ( 1)

```

------
Then we need to assign names to positions and then start rotating them. We have to assign the positions so that we start out with AB, CD, EF, GH, IJ and KL as the matchups. Here's a way:
```
2  3 ( 1);  4  7 ( 3);  5  9 ( 4);  6 11 ( 5);  8 10 ( 2);

A BCDEGIFKHLJ  AB CD EF GH IJ KL

A JBCDEGIFKHL  AJ BC DI EK FH GL

A LJBCDEGIFKH  AL BJ CG DF EH IK

A HLJBCDEGIFK  AH BE CI DK FG JL

A KHLJBCDEGIF  AK BG CF DJ EI HL

A FKHLJBCDEGI  AF BI CL DG EJ HK

A IFKHLJBCDEG  AI BH CE DL FK GJ

A GIFKHLJBCDE  AG BD CH EL FI JK

A EGIFKHLJBCD  AE BK CJ DH FL GI

A DEGIFKHLJBC  AD BL CK EG FJ HI

A CDEGIFKHLJB  AC BF DE GK HJ IL

```

----
where A is placed at the center, where he remains. B starts off at position 1 so in the first set he is paired with A. C and D are in positions 2 and 3 of the circumference as they are connected, while E and F are at positions 4 and 7, as that is the next connected pair, etc.

So each line above shows the center person, the order of names around that center, and then the pairings, shown alphabetically, based upon the positions listed at the top.

Other sets that work come from the other sets of chords, such as:

```
2  3 ( 1);  4  9 ( 5);  5  7 ( 2);  6 10 ( 4);  8 11 ( 3);

A BCDEGIHKFJL  AB CD EF GH IJ KL

A LBCDEGIHKFJ  AL BC DK EI FG HJ

A JLBCDEGIHKF  AJ BL CH DG EK FI

A FJLBCDEGIHK  AF BI CE DH GK JL

A KFJLBCDEGIH  AK BD CI EH FJ GL

A HKFJLBCDEGI  AH BG CL DI EJ FK

A IHKFJLBCDEG  AI BJ CG DF EL HK

A GIHKFJLBCDE  AG BE CK DJ FL HI

A EGIHKFJLBCD  AE BH CF DL GI JK

A DEGIHKFJLBC  AD BK CJ EG FH IL

A CDEGIHKFJLB  AC BF DE GJ HL IK

2  4 ( 2);  3  6 ( 3);  5 10 ( 5);  7 11 ( 4);  8  9 ( 1);

A BCEDGFIKLHJ  AB CD EF GH IJ KL

A JBCEDGFIKLH  AJ BE CG DL FH IK

A HJBCEDGFIKL  AH BD CJ EK FI GL

A LHJBCEDGFIK  AL BH CI DK EJ FG

A KLHJBCEDGFI  AK BF CH DG EI JL

A IKLHJBCEDGF  AI BL CF DE GJ HK

A FIKLHJBCEDG  AF BG CE DH IL JK

A GFIKLHJBCED  AG BC DJ EL FK HI

A DGFIKLHJBCE  AD BJ CK EH FL GI

A EDGFIKLHJBC  AE BI CL DF GK HJ

A CEDGFIKLHJB  AC BK DI EG FJ HL

2  4 ( 2);  3  7 ( 4);  5  8 ( 3);  6 11 ( 5);  9 10 ( 1);

A BCEDGIFHKLJ  AB CD EF GH IJ KL

A JBCEDGIFHKL  AJ BE CI DF GL HK

A LJBCEDGIFHK  AL BG CJ DK EI FH

A KLJBCEDGIFH  AK BL CG DJ EH FI

A HKLJBCEDGIF  AH BD CF EL GI JK

A FHKLJBCEDGI  AF BI CK DG EJ HL

A IFHKLJBCEDG  AI BH CL DE FK GJ

A GIFHKLJBCED  AG BK CE DL FJ HI

A DGIFHKLJBCE  AD BC EK FG HJ IL

A EDGIFHKLJBC  AE BJ CH DI FL GK

A CEDGIFHKLJB  AC BF DH EG IK JL

or by putting others than A in the center:

2  3 ( 1);  4  7 ( 3);  5  9 ( 4);  6 11 ( 5);  8 10 ( 2);

B ACDEGIFKHLJ  AB CD EF GH IJ KL

B JACDEGIFKHL  AC BJ DI EK FH GL

B LJACDEGIFKH  AJ BL CG DF EH IK

B HLJACDEGIFK  AE BH CI DK FG JL

B KHLJACDEGIF  AG BK CF DJ EI HL

B FKHLJACDEGI  AI BF CL DG EJ HK

B IFKHLJACDEG  AH BI CE DL FK GJ

B GIFKHLJACDE  AD BG CH EL FI JK

B EGIFKHLJACD  AK BE CJ DH FL GI

B DEGIFKHLJAC  AL BD CK EG FJ HI

B CDEGIFKHLJA  AF BC DE GK HJ IL

or

2  3 ( 1);  4  7 ( 3);  5  9 ( 4);  6 11 ( 5);  8 10 ( 2);

C DABEGIFKHLJ  AB CD EF GH IJ KL

C JDABEGIFKHL  AD BI CJ EK FH GL

C LJDABEGIFKH  AG BF CL DJ EH IK

C HLJDABEGIFK  AI BK CH DE FG JL

C KHLJDABEGIF  AF BJ CK DG EI HL

C FKHLJDABEGI  AL BG CF DI EJ HK

C IFKHLJDABEG  AE BL CI DH FK GJ

C GIFKHLJDABE  AH BD CG EL FI JK

C EGIFKHLJDAB  AJ BH CE DK FL GI

C BEGIFKHLJDA  AK BC DL EG FJ HI

C ABEGIFKHLJD  AC BE DF GK HJ IL

```

---
Each choice of set of chords and identity of the center person seems to produce a unique set.

The program, that finds the appropriate set of chords and then assigns letters is:
DECLARE SUB matchup ()
DIM SHARED cnct(11), dist(5), mcount
CLS
OPEN "conclub2.txt" FOR OUTPUT AS #2
matchup
CLOSE

SUB matchup
DIM ans\$(6)
FOR i = 2 TO 10
IF cnct(i) = 0 THEN
FOR j = i + 1 TO 11
d = j - i: IF d > 5 THEN d = 11 - d
IF dist(d) = 0 AND cnct(j) = 0 THEN
cnct(i) = j: cnct(j) = i
dist(d) = 1
mcount = mcount + 1
IF mcount = 5 THEN
'****** Report the Result
FOR p = 2 TO 11
IF p < cnct(p) THEN
PRINT USING "### ## (##);"; p; cnct(p); cnct(p) - p;
PRINT #2, USING "### ## (##);"; p; cnct(p); cnct(p) - p;
END IF
NEXT
PRINT
PRINT #2,
' Assign names A - L
c\$ = "A"
round\$ = "B" + SPACE\$(10)
strPsn = 1
FOR p = 2 TO 11
IF p < cnct(p) THEN
MID\$(round\$, p, 1) = MID\$("CDEFGHIJKL", strPsn, 1)
MID\$(round\$, cnct(p), 1) = MID\$("CDEFGHIJKL", strPsn + 1, 1)
strPsn = strPsn + 2
END IF
NEXT
FOR day = 1 TO 11
ans\$(1) = c\$ + LEFT\$(round\$, 1)
a = 2
FOR p = 2 TO 11
IF p < cnct(p) THEN
ans\$(a) = MID\$(round\$, p, 1) + MID\$(round\$, cnct(p), 1)
a = a + 1
END IF
NEXT
FOR a = 1 TO 6
IF RIGHT\$(ans\$(a), 1) < LEFT\$(ans\$(a), 1) THEN
ans\$(a) = RIGHT\$(ans\$(a), 1) + LEFT\$(ans\$(a), 1)
END IF
NEXT
' sort the pairs:
DO
fl = 0
FOR a = 2 TO 6
IF ans\$(a) < ans\$(a - 1) THEN
SWAP ans\$(a), ans\$(a - 1): fl = 1
END IF
NEXT
LOOP UNTIL fl = 0

PRINT c\$; " "; round\$; " ";
PRINT #2, c\$; " "; round\$; " ";
FOR a = 1 TO 6
PRINT ans\$(a); " ";
PRINT #2, ans\$(a); " ";
NEXT
PRINT
PRINT #2,
' rotate the people:
round\$ = RIGHT\$(round\$, 1) + LEFT\$(round\$, LEN(round\$) - 1)
NEXT
'*************************
ELSE
matchup
END IF
cnct(i) = 0: cnct(j) = 0
dist(d) = 0
mcount = mcount - 1
END IF
NEXT
EXIT FOR
END IF
NEXT
END SUB

 Posted by Charlie on 2004-02-06 13:09:23

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