

A160511


Number of weighings needed to find lighter coins among n coins.


0



1, 2, 3, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 16, 16, 17, 18, 18, 19
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OFFSET

1,2


COMMENTS

a(n) is the minimal worstcase number of weighings needed by an algorithm to sort a set of n coins of two possible weights into heavy vs. light coins by using a balance; additional known good (heavy, say) coins are available (esp. to distinguish "all heavy" from "all light").
It is known that ceiling(n*log_3(2) + log_3(135/128)) <= a(n) <= ceiling(7n/11) for all n except for n=3. This implies 19 <= a(30) <= 20 and in fact a(n) = ceiling(7n/11) for several n <= 187, esp. for all n with n <= 50 except n=3, n=30, n=41, n=49.


LINKS

Table of n, a(n) for n=1..29.
AnPing Li, A note on the counterfeit coins problem, arXiv:0902.0841v8 [math.CO]


EXAMPLE

For n=1, compare the given coin A with a known heavy coin X. If A < X, then A is a light coin; if A=X, then A is a heavy coin; the outcome A > X is not possible. Since one weighing was needed, we have a(1)=1.
For n=3, to sort coins A,B,C, one optimal algorithm is: First compare A:B. If A < B, we know that A is light and B is heavy and can find out about C by comparing, e.g., B:C in a second weighing. The case A > B is symmetric. If, however, A=B, compare A:C. If A < C, we know that A,B are light and C is heavy and vice versa for A > C. The worst case is A=C, which requires a third weighing, e.g., A:X, against a known heavy coin X. Since no algorithm exists that never uses more than 2 weighings, we have a(3) = 3.


CROSSREFS

Cf. A156301.  Jonathan Vos Post, May 18 2009
Sequence in context: A317596 A057355 A171975 * A079952 A055930 A090638
Adjacent sequences: A160508 A160509 A160510 * A160512 A160513 A160514


KEYWORD

nonn,hard,more


AUTHOR

Hagen von Eitzen, May 16 2009


STATUS

approved



