There are 10 tables in the Conversing Club and 15 members. Each day, 3 people sit together around each of 5 of 10 possible tables in the club talking to each other.

Every week (7 days) everyone is at the same table with everyone else exactly once. Also, nobody is at the same table twice in the course of a week to provide a change of scenery each time. The first day is as following:

ABC DEF GHI JKL MNO

(The second day A couldn't sit with B, or C; B couldn't sit with C; D couldn't sit with E or F, but could sit with A, B, or C.)

How could their schedule be configured?

(Based on Fifteen Schoolgirls)

(In reply to re: Solution (no computer program used) by Charlie)

I began by arbitrarily assigning meetings to A. I could do this in any way, since the names A through O are arbitrary. (I knew that if the end result did not have the day 1 meetings specified in the puzzle, I could transpose letters to change it to match the puzzle.)

Day1: ABC; Day2: ADE; Day3: AFG; Day4: AHI; Day5: AJK; Day6: ALM; Day7: ANO

Now it was a simple matter to assign everybody else to the remaining meetings in such a way that everyone meets every other person exactly once. I started with B. I just had B neeting with A and C, so I arranged for her to meet with the remaining girls D through O: BDF, BEG, BHJ, BIK, BLN and BMO. Now it was C's turn. I already had her meeting with A and B, so I set her up with D through O: CDG, CEF, CHK, CIJ, CLO, CMN. Now for D, whom I now had meeting with A, B, C, E, F and G, I set up meetings with H through O: DHL, DIM, DJN, and DKO. I now had E meeting with A, B, C, D, F and G, so I assigned her the following meetings: EHM, EIL, EJO and EKN. Now F was meeting with A, B, C, D, E and G, so I set her up with the remaining girls: FHN, FIO, FJL, and FKM. So now G was meeting with A, B, C, D, E, and F, so I filled out her schedule with the following meetings: GHO, GIN, GJM and GKL.

So now I had a set of 35 meetings with everbody meeting everybody else exactly once:

ABC, ADE, AFG, AHI, AJK, ALM, ANO, BDF, BEG, BHJ, BIK, BLN, BMO, CDG, CEF, CHK, CIJ, CLO, CMN, DHL, DIM, DJN, DKO, EHM, EIL, EJO, EKN, FHN, FIO, FJL, FKM, GHO, GIN, GJM, and GKL

The next step was to assign the groupings by day, from 1 to 7. This involved some minor trial and error, and I didn't verify that this is the only arrangement:

Day 1: ABC, DHL; EJO; FKM; GIN;
Day 2: ADE; BHJ; CMN; FIO; GKL;
Day 3: AFG; BIK; CLO; DJN; EHM;
Day 4: AHI; BLN; CEF; DKO; GJM;
Day 5: AJK; BMO; CDG; EIL; FHN;
Day 6: ALM; BDF; CIJ; EKN; GHO;
Day 7: ANO; BEG; CHK; DIM; FJL

Now I had to assign tables. I guessed correctly that I should use all ten tables on the first two days:

Day 1: ABC(1), DHL(2); EJO(3); FKM(4); GIN(5);
Day 2: ADE(6); BHJ(7); CMN(8); FIO(9); GKL(10);

Then it was not too hard to assign successive tables to successive groups of three people, based on avoiding all tables that any of the three had sat at previously.

Day 1: ABC(1), DHL(2); EJO(3); FKM(4); GIN(5);
Day 2: ADE(6); BHJ(7); CMN(8); FIO(9); GKL(10);
Day 3: DJN(1); AFG(2); BIK(3); CLO(4); EHM(5);
Day 4: AHI(4); DKO(5); BLN(6); CEF(7); GJM(9);
Day 5: EIL(1); BMO(2); CDG(3); AJK(8); FHN(10);
Day 6: GHO(1); EKN(2); ALM(3); CIJ(6); BDF(8);
Day 7: BEG(4); FJL(5); ANO(7); CHK(9); DIM(10);

Now all that remained was to transpose letters to arrive at the correct solution. By replacing H with E, L with F, E with G, J with H, O with I, F with J, M with L, G with M, I with N, and N with O, the result is the solution to the puzzle:

Day 1: ABC(1), DEF(2); GHI(3); JKL(4); MNO(5);
Day 2: ADG(6); BEH(7); CLO(8); JNI(9); MKF(10);
Day 3: DHO(1); AJM(2); BNK(3); CFI(4); GEL(5);
Day 4: AEN(4); DKI(5); BFO(6); CGJ(7); MHL(9);
Day 5: GNF(1); BLI(2); CDM(3); AHK(8); JEO(10);
Day 6: MEI(1); GKO(2); AFL(3); CNH(6); BDJ(8);
Day 7: BGM(4); JHF(5); AOI(7); CEK(9); DNL(10)

Edited on October 14, 2007, 12:01 pm

Posted by Penny on 2003-12-30 13:26:36