December 2016 - Page 2

tgan · 2017-01-24, 12:27

Hi this is our solution

As can be seen there are some papers on it

We reduced this months challenge to the generic coin weighting problem which is defined as the following with k=2:
Given a set of m coins out of a collection of coins of k unknown distinct weights, we wish to
decide if all the m given coins have the same weight or not using the minimum possible number of
weighings in a regular balance beam.

This led us to various papers including: "COINS AND CONES" by Dmitry N. Kozlov and Van H. Vu that proves that the limit for 4 weighings is 30 coins. We also found two additional papers with further cooperation of Kozlov with Noga Alon.

Apparently their "trivial" examples has proven to be hard to comprehend by many people but further research brought up the following solution for 30 people using groups of 4, 5, 6, 7, 8 people.

G4 = ABCD
G5 = EFGHI
G6 = JKLMNO
G7 = PQRSTUV
G8 = WXYZ0123

Contests:
ABCD EFGHI JKLMNO ; PQRSTUV WXYZ0123
ABCD PQRSTUV ; EFGHI JKLMNO
JKLMNO PQRSTUV ; EFGHI WXYZ0123
ABCD WXYZ0123 ; EFGHI PQRSTUV

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2017-01-24, 12:27	#12
tgan Jul 2015 3³ Posts	Hi this is our solution As can be seen there are some papers on it We reduced this months challenge to the generic coin weighting problem which is defined as the following with k=2: Given a set of m coins out of a collection of coins of k unknown distinct weights, we wish to decide if all the m given coins have the same weight or not using the minimum possible number of weighings in a regular balance beam. This led us to various papers including: "COINS AND CONES" by Dmitry N. Kozlov and Van H. Vu that proves that the limit for 4 weighings is 30 coins. We also found two additional papers with further cooperation of Kozlov with Noga Alon. Apparently their "trivial" examples has proven to be hard to comprehend by many people but further research brought up the following solution for 30 people using groups of 4, 5, 6, 7, 8 people. G4 = ABCD G5 = EFGHI G6 = JKLMNO G7 = PQRSTUV G8 = WXYZ0123 Contests: ABCD EFGHI JKLMNO ; PQRSTUV WXYZ0123 ABCD PQRSTUV ; EFGHI JKLMNO JKLMNO PQRSTUV ; EFGHI WXYZ0123 ABCD WXYZ0123 ; EFGHI PQRSTUV