This is a first, as I don't think anyone has asked before for the "minumum movement" property. With 14 teams and 4 rooms you will need to have 23 rounds to go through the whole round-robin. I don't really have any algorithms set up specifically to solve this sort of problem, but managed to put together the schedule below.
(F G) (K C) (E M) (D B)
(G J) (I C) (H M) (D L)
(E L) (C A) (M B) (D H)
(G K) (C B) (M A) (J F)
(E G) (F C) (L M) (D A)
(A G) (C J) (M K) (E H)
(N K) (L C) (I M) (E J)
(L N) (C E) (F B) (K D)
(J A) (N C) (B L) (D E)
(A H) (C G) (K B) (J D)
(F N) (J L) (B H) (I D)
(A E) (J K) (N B) (L H)
(N G) (L F) (A B) (H K)
(N E) (L I) (B G) (F H)
(A F) (I N) (B E) (M D)
(H N) (I K) (B J) (F D)
(N A) (E I) (M F) (H J)
(K E) (A I) (J N) (G H)
(K A) (I J) (G M) (H C)
(G L) (I F) (M C) (D N)
(F K) (G I) (M J) (C D)
(L A) (B I) (N M) (D G)
(E F) (K L) (H I)
In particular there is a long run of C and I in room 2 (column 2) and runs of Bs and Ds in rooms 3 and 4 respectively. I am fairly sure you can do a lot better than this (any other takers?).
Ian.
Hi,
I took Ian's matrix and performed a convolution and found the following where every team plays in every room 3 times and one room 4 times. No team plays in the same room twice in a row.
(H E) (G A) (C J) (K M)
(I B) (N M) (D G) (A L)
(K A) (C H) (J I) (M G)
(B G) (I L) (N E) (H F)
(C L) (E J) (I M) (K N)
(E D) (L B) (C N) (J A)
(N A) (M F) (J H) (E I)
(E G) (A D) (L M) (C F)
(L N) (C E) (B F) (D K)
(D J) (B K) (H A) (G C)
(B C) (J F) (G K) (A M)
(F L) (N G) (A B) (H K)
(M D) (F A) (I N) (B E)
(I K) (J B) (F D) (N H)
(A C) (H D) (M B) (L E)
(G F) (M E) (K C) (B D)
(M H) (I C) (L D) (J G)
(J K) (L H) (E A) (N B)
(C M) (D N) (G L) (F I)
(G I) (F K) (M J) (C D)
(H B) (D I) (N F) (L J)
(N J) (H G) (K E) (I A)
(F E) (K L) (H I)
This should work for you as well.
Regards,
Jeff Bond