A note on the counterfeit coins problem

Transcription

1 A note on the counterfeit coins problem Li An-Ping Beijing , P.R.China Abstract In this paper, we will present an algorithm to resolve the counterfeit coins problem in the case that the number of false coins is unknown in advance. Keywords: counterfeit coins problem, combinatorial search, information theoretic bound.

2 The counterfeit coins problem is a well-known combinatorial search problem, namely, search for the counterfeit coins in a number of coins by a balance, where the fakes and the normals have same semblance but different weights. The problem has several versions, for instance, the fakes are known lighter or heavier than the normals, and the number of the fakes is assumed known. For the detail results and researches in these aspects refer to see the papers [1]~[6]. In this paper, we will consider a more general case that the number of the fakes is unknown beforehand. In general, it is assumed that the coins will be permitted to be remarked by numbers in order to be distinguished each other. Our main result is as following Proposition 1 Suppose that S is a set of n coins with same semblance, in which possibly there are some counterfeit coins, and the normals and the fakes have different weights. Denoted by gnthe ( ) least weighings need to sort S into the normals and fakes by a balance, then it has log (2 1) gn ( ) 7 n/11 n 3. (1) The proof of Proposition 1 will largely rely on the following lemma, which is one of our unpublished results obtained in Lemma 1 Let A be a set of 11 coins with two possible distinct weights, then A can be sorted by a balance with at most 7 weighings, and in the sixth weighing it will be found if A contains only one kind weight of coins. Proof. It will be suffice to provide a feasible algorithm for prove the lemma. We have succeeded such an algorithm, which sketch is provided as an appendix in the end of the paper, for to describe the whole algorithm require much more space. The readers are encouraged yet to design own algorithms for it as a puzzle. Proof of Proposition 1. The left-hand of (1) is trivial for the information theoretical bound of gn ( ) is equal to n log3 2, so we are provided to prove the right-hand of (1). It is easy to verified that the right-hand of (1) is true for n 11. Let n= 11 m+ r, S= A i, Ai = 11, 0 i< m, A m = r. 0 i m Applying Lemma 1 to each A i, i= 0,1,, m 1, and assume that there are t A i s that contains only one kind of coins, i.e. normals or fakes. Then by the induction, we have gn () 7( m t) + 6t+ 7/11 t + 7 r/11 7m t+ 7 t/ r/11 7m+ 7 r/11 7 n /11.

3 n Maybe, it should be mentioned that it is easy to know that log 3(2 1) = n log32 when a b n > 2, for the equation 2 1= 3 have just two integer solutions (1, 0) and (2,1). As the lower bound and upper bound of gnare ( ) much tighter, so we have following conjecture, Conjecture gn ( ) = n log 2. 3 (2) In addition, as an auxiliary result, there is a corollary of Lemma 1 Corollary 1 Let A i,1 i 11, be 11 sets of coins, each Ai consists of two coins, one is normal and one is fake, the normals and the fakes have distinct weights, then at most spend 7 weighings to find all the fakes in A i s, 1 i 11. Proof. From each A take a coin,1 i 11, to form a coin set A, applying Lemma 1 to A, i if there are two kinds of coins in A, then each A,1 i 11, will be sorted, otherwise, that is, i if A contains only one kind of coins, then a additional weighing will work well. Likely, there may be the algorithms independent of Lemma 1 for the result of Corollary 1, which will be left to the readers. Finally, we wish to acknowledge Hagen von Eitzen for his criticism and investigation. References [1] M. Aigner, Combinatorial Search. Wiley-Teubner 1988 [2] Anping Li, On the conjecture at two counterfeit coins, Disc. Math. 133 (1994), [3] Anping Li, Three counterfeit coins problem, J. Comb. Theory Ser. A, 66 (1994), [4] Anping Li, M. Aigner, Searching for counterfeit coins, Graphs Comb. 13 (1997), 9-20 [5] R. Bellman, B. Glass, On various versions of the defective coin problem. Inform. Control 4, (1961) [6] L. Pyber, How to find many counterfeit coins? Graphs Comb. 2, (1986)

4 Appendices I The Sketch of Algorithm to Sort 11 Coins We at first introduce some notations. The expression L : R means a comparison or a weighing with L and R are the left-hand and the right-hand of the balance respectively, and for simplicity, the expression {1,2,3} means the total weight of the coins with marks 1,2 and 3, and so on. Fig.22 (<) Fig.21 (=) {3,4,7,10 } :{1,2,6,11} (<) {5,7,9 } :{6,8,10 } (<) {1,7,8 } :{2,9,10} (<) {1,2,3 } :{4,5,6 } (>) (=) Fig.20 {2,9,11 } :{ 3,4,5} (=) Fig.19 Fig.18 Fig.17 Fig.16 (<) Fig.15 (=) {2,3,6} :{5,8,9} (<) {1,5,7,9}:{2,6,8,10} (=) (>) (=) Fig.14 {1,2,4} :{ 3,7,10} Fig.13 Fig.12 Fig.11 Fig.10 (<) Fig.9 (=) {1,5,6}:{2,3,7} (<) {7,9}:{8,10} (<) {1,7.8}:{ 4,9,10} (>) Fig.8 (=) Fig.7 (<) {5,7,10}:{ 6,8,9} (=) (=) Fig..6 Fig.5 (<) Fig.4 (=) {1,3,4}:{8,9,11} (<) {1,2,7}:{ 3,4,8} (>) Fig.3 (=) Fig.2 (<) {1,3 } :{ 2,4} (=) Fig.1 Fig. 0

5 ( 7,8,9,10) ( 1,4) {1 } :{2 } (=) {10 } :{ 11} (<) {1,2 } :{10,11 } (>) {1 } :{ 2} (=) {10 } :{11 } ( 11 ) ( 2,3,5,6) {1 } :{ 2} {10 } :{11 } (~) {10 } :{11 } Fig.1 {6}:{11} {10}:{ 11} {9 }:{10 } (=) {3,5 } :{1,11 } (<) {1 } :{7 } (>) {5,11 } :{6,9 } (=) {5 } :{ 6} {9}:{10} {10}:{11} { 5,11} :{6,9} {10}:{11} {9}:{10} {10}:{11} Fig.2 {5,9}:{7,10} { 6,9} :{ 7,10} { 10} :{11} (=) {1,9}:{8,11} (<) {5}:{6} (>) {1,9}:{8,11} (=) {10}:{11} {1}:{ 7} {1}:{7} {2} :{ 7} { 10} :{11} {1}:{3} { 9} :{11} Fig.3

6 { 2} :{ 3} { 2} :{3} {1}:{2} (=) {9}:{ 10} (<) {5 } :{6 } (>) {9 } :{10 } (=) {1 } :{ 2} {2}:{3} { 2} :{ 3} {9 } :{ 11} {3}:{ 10} {9}:{10} {3}:{ 4} Fig.4 {9}:{11} {7}:{ 8} {10}:{11} (=) {3,6}:{7,9} (<) {3,5}:{6,9} (>) {9}:{ 11} (=) { 2} :{10} {1,4}:{10,11} { 2} :{3} {9 } :{10} {2,10}:{3,11} {1}:{2} {2}:{11} Fig.5 {3}:{11} {2}:{11} {1}:{7} (=) {3,11}:{9,10} (<) { 2} :{3} (>) {2,11}:{9,10} (=) {1}:{7} {1}:{7} {1}:{7} {1,11 } :{2,3 } {1}:{11} {6}:{ 9} {1}:{11} Fig.6

7 {1}:{11} {1,4}:{10,11} {6}:{8} (=) {1,11}:{2,8} (<) {2,5}:{3,7} (>) {1,6}:{5,11} (=) { 6} :{8} {4,5}:{10,11} {5}:{11} {1,10} :{8,11} {8}:{11} {6}:{ 9} {2}:{11} Fig.7 {2,3}:{9, 11} {2,3}:{9,11} {9 } :{11} (=) {5,9 } :{6,11} (<) {2} :{3} (>) {5,9 } :{6,11} (=) {9 } :{11} {2,3}:{9,11} {2,3}:{9,11} {5,6} :{9,11} {9}:{ 11} {9}:{10} {9}:{11} Fig.8 {2,3}:{9,11} {2,3}:{9,11} {9 } :{11} (=) {5,9} :{6,11} (<) {2} :{3} (>) {5,9 } :{6,11} (=) {9 } :{11} {2,3}:{9,11} {2,3}:{9,11} {5,6} :{9,11} {9}:{ 11} {9}:{10} {9}:{11} Fig.9

8 {10}:{11} {10}:{11} {5}:{6} (=) {1,3,11}:{ 2,5,6} (<) {1,10} :{8,9} (>) {1,3,11} :{ 2,5,6} (=) {5 } :{6 } {5}:{6} {5}:{6} {1,3,11} :{ 2,5,6} {5}:{6} {5}:{6} {10}:{11} Fig.10 {1,11}:{3,4} {2}:{11} {2}:{11} (=) { 1,5} :{7,11} (<) {1,2}:{8,9} (>) { 1,10} :{2,11} (=) { 2} :{ 11} {2}:{4} {7}:{11} {1,2} :{10,11 } { 4} :{5} {3,4}:{ 6,7} {10}:{11} Fig.11 {8}:{11} {8}:{ 11} { 6} :{8} (=) {3,4}:{ 9,11} (<) {1}:{2} (>) {3,4}:{9,11} (=) {10}:{11} {8}:{11} { 8} :{11} {3,4} :{ 9,11} {8}:{11} {6}:{8} {8}:{11} Fig.12

9 {3,4}:{9,11} {9}:{11} {3}:{4} (=) {3,8}:{4,11} (<) {3,7,10}:{4,5,6} (>) {3,5}:{9,11} (=) { 3} :{4} {3,4}:{9,11} {9}:{11} {3,4} :{9,11} {9}:{11} { 3} :{5} {9}:{11} Fig.13 {7}:{11} {10}:{11} {2,4}:{5,6} (=) {2}:{10} (<) {3,5,10}:{7,9,11} (>) {2,8}:{ 10,11} (=) {1,3}:{7,8} {4}:{6} { 7} :{8} {2,4} :{10,11 } {1}:{3} {10}:{ 11} {3,4}:{7,8} Fig.14 {3}:{11} {1,11}:{4, 7} {10}:{ 11} (=) {5,11}:{7,9} (<) {2,5,9}:{3,4,7} (>) {1,4}:{7,11} (=) { 2} :{ 4} {5,9}:{7,11} {7}:{11} {2,9} :{3,11 } {3}:{11} { 10} :{11} {10}:{ 11} Fig.15

10 {3,4}:{5,6} {4}:{5} {2}:{ 3} (=) {1,5}:{ 7,11} (<) {1,6,11}:{3,7,10} (>) {5,10}:{2,11} (=) {1,4}:{5,6} {4}:{ 6} {4}:{6} {5} :{ 6} {4}:{6} {4}:{ 6} {4}:{ 11} Fig.16 {8}:{ 10} {1}:{8} {2}:{7} (=) {2,9}:{ 10,11} (<) {1,4,11}:{3,5,10} (>) {5,6}:{7,11} (=) {2,4}:{5,6} {7}:{ 8} {2,6}:{4,5} {7,9} :{8,11} {6}:{8} { 1} :{ 7} {2}:{11} Fig.17 {5}:{ 7} {1}:{7} {3}:{4} (=) {1,6}:{5,8} (<) {2,6,8}:{3,5,11} (>) {1,6}:{ 8,9} (=) {2}:{7} {2}:{8} {6}:{10} {4} :{ 9} {1}:{8} {10}:{11} {2}:{ 7} Fig.18

11 {3,11}:{4,10} { 1,2} :{3,11} {2}:{11} (=) {3,6}:{7,10} (<) {1,5,9} :{2,4,11} (>) {1,4} :{8,9 } (=) {2,5 } :{3,4 } {6}:{ 11} { 3} :{ 4} {1,4} :{2,6 } {8}:{10} {5}:{7} {2,4}:{3,8} Fig.19 {2}:{ 11} {2,3}:{4,8} {2}:{3} (=) { 8,9} :{3,11} (<) {1,5,6} :{4,7,11} (>) {2,4} :{3,9 } (=) {3 } :{9 } {3,6}:{4,9} {3,4}:{8,9} {8,9} :{6,11} {2,4}:{3,5} {2,4}:{ 3,5} {2}:{3} Fig.20 {5,6}:{7,8} { 8} :{10} {5}:{8} (=) {2,4}:{5,9} (<) {1,2} :{7,11} (>) {5} :{9 } (=) {6,7} :{8,9 } {3}:{8} {1}:{ 8} {2,8} :{7,10 } { 5} :{9} {5}:{9} {3,6}:{5,9} Fig.21

12 {6,7}:{ 8,9} {6,7}:{9,11} {6,7}:{8,9} (=) { 1,2} :{ 4,11} (<) {4,9}:{5,8} (>) {2,9}:{10,11} (=) {5,6}:{8,9} {2}:{9} {1}:{ 2} {1,6} :{ 8,11} {2}:{9} {1,2}:{6,7} {8}:{ 11} Fig.22 Where L> R, or L< R, or L= Rmeans L heavier than, or lighter than, or equal to R respectively, and the outputs (x), (a,b,c) and (~) means null, the coins with marks abc,, are heavier ones, and all coins have same weights respectively. The symmetric part of the algorithm has been omitted.

13 Appendices II An Alternative Algorithm to sort 11 coins The difference between the new and the previous one is mainly the Fig.1 to Fig.5 and the third weighing on the direction (=, =) in Fig.0, and the other keep unchanged. Fig.22 (<) Fig.21 (=) {3,4,7,10 } :{1,2,6,11} (<) {5,7,9 } :{6,8,10 } (<) {1,7,8 } :{2,9,10} (<) {1,2,3 } :{4,5,6 } (>) (=) Fig.20 {2,9,11 } :{ 3,4,5} (=) Fig.19 Fig.18 Fig.17 Fig.16 (<) Fig.15 (=) {2,3,6} :{5,8,9} (<) {1,5,7,9}:{2,6,8,10} (=) (>) (=) Fig.14 {1,2,4} :{ 3,7,10} Fig.13 Fig.12 Fig.11 Fig.10 (<) Fig.9 (=) {1,5,6}:{2,3,7} (<) {7,9}:{8,10} (<) {1,7.8}:{ 4,9,10} (>) Fig.8 (=) Fig.7 (<) {5,7,10}:{ 6,8,9} (=) (=) Fig..6 Fig.5 (<) Fig.4 (=) {1,9}:{4,10} (<) {2, 7 }:{ 3, 8} (>) Fig.3 (=) Fig.2 (<) {5} :{ 6} (=) Fig.1 Fig. 0

14 {2}:{3} {4}:{11} {1}:{ 11} (=) {4}:{5} (<) {1,11}:{7,8} (>) {4}:{5} (=) {1}:{ 11} {2}:{11} {2}:{3} {6} :{11 } { 2} :{ 3} (~) {2}:{ 3} Fig.1 {1}:{3} {1}:{ 2} {1}:{ 3} (=) {2,8}:{10,11} (<) {7,9}:{8,10} (>) {3,7}:{9,11} (=) {1}:{ 2} {10}:{11} {9}:{11} {7,10} :{4,11 } {1}:{11} { 1} :{2} {1}:{11} Fig.2 {5}:{ 6} {5}:{6} {5}:{6} (=) {3,5,6}:{2,8,11} (<) {1,4} :{2,8 } (>) {4,8} :{9,11 } (=) {5 } :{6} {9}:{11} {9}:{11} {2,11} :{ 7,10} { 5} :{6} {5}:{6} {9}:{11} Fig.3

15 {6}:{ 11} {6}:{7} {6}:{11} (=) {1,2}:{7,8} (<) {5,10} :{6,11 } (>) {1,4} :{ 7,11} (=) {1 } :{3 } {4}:{6} {10}:{11} {5} :{8} {4}:{ 6} (x) {4}:{6} Fig.4 {1}:{ 11} {4}:{ 11} {1}:{8} (=) {3,6}:{ 4,11} (<) {3,5}:{ 6,11} (>) {5,10}:{ 4,11} (=) { 1} :{2} {1}:{11} {4,10}:{5,11} {5,10} :{6,11 } { 6} :{11} {2}:{4} {2}:{4} Fig.5