Some Results on the Counterfeit Coins Problem. Li An-Ping. Beijing , P.R.China Abstract

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1 Some Results on the Counterfet Cons Problem L An-Png Bejng , P.R.Chna apl0001@sna.om Abstrat We wll present some results on the ounterfet ons problem n the ase of mult-sets. Keywords: ombnatoral searh, ounterfet ons problem, nformaton theoret bound.

2 I. Introduton Searhng the ounterfet ons from the gven ons by a balane s a well-known problem of ombnatoral searh, whh there are several versons and a ntensve researhes, for the detal to see the papers [1]~[9]. From ths paper, we wll dsuss some the alled mult-sets ases. Let S,, 1 Sm be m sets of ons, n whh eah set S just ontans one false on, and the fakes are known wth same weght but heaver than the normals, where S S =, for all j. Suppose that S = n, 1 m, defne g 1 ( n 1, n 2,, n m ) the least number of weghngs to fnd all the m fakes n S, whh wll be smply wrtten as ( ) 1 m 1 m wll gve some estmatons for g ( ) 1 n m. j g n when n = n,1 m. The present paper In the rest of the seton, we ntrodue some symbols and notatons whh wll be used n ths paper. L : R : A weghng or a omparson of on set L aganst on set R. L> R( L< R, L= R) : L s heaver ( lghter, equal to) than R respetvely. X Y : Carteson produt of sets X and Y. : A partal order of the spae k X, αβ k α β α= 1 k = 0 k = 0, X, (,, ), l and β = (,,, ). 1 l X : {, j,,} l = X X j Xl, and smply wrtten as X, j,, lf no onfusablty. : The least nteger no less than the real number. Ct( X ): The number of fakes n the set X. k 1 Let A0 = {0}, A= { +,0, }, for a postve nteger k, we defne A = A A, k 0 A = Ak, k = 1 and W = { L: R L, R S, L = R }. m = 1 It s lear that an algorthm of searhng for the ounterfet ons s just a map F : A W.

3 Let m S = S, then a algorthm F wll ndues a map f : S A : X (0, a1, a2, ), = 1 where a1 = sgn( L1 X R1 X ), F(0) = L1: R1, a = sgn( L X R X ), F((0, a,, a )) = L : R, for > Hene, α A, α determnes a subset α S α as the objetve set on the dreton α. S of S : = { P P, α f( P) } S S, we all Defnton 1. For a algorthm F and a postve nteger k, f α A, S 1, then algorthm F s alled k - ompleted. α k α 2. Man results It s well-known that g1( n) = log3n. (2.1) So, there s a smple estmaton m m log 3n g1( n1,, nm) log 3n. = 1 (2.2) = 1 In ths paper, our man results are followng estmate Proposton 1 g1( n 2) log3n + log 3( n/5) + 1, (2.3) g1( n 3) log3n + log 3( n/ 4) + log 3( n/ 6) + 2, (2.4) g1( n 4) log3n + log 3( n/ 5) + log 3( n/) + log 3( n/ 20) + 5, (2.5) g1( n 5) log3n + log 3( n/ 7) + log 3( n/) + log 3( n/13) + log 3( n/17) + 7. To prove the result above need the followng ndvdual results (2.6) Lemma 1 g (2 ) = 2, (2.7) 1 3 g (2, 4) = 2, (2.8) 1 g (4,20) = 4, (2.9) 1

4 g (5 ) = 3, (2.10) 1 2 g ( ) = 9, (2.) 1 4 g (7 ) = 9, (2.12) g ( ) =, (2.13) g (13 ) = 12, (2.14) g (17 ) = 13. (2.15) Proof. The denttes (2.7) and (2.8) are easer and left to the readers. The algorthms for the rest are put n the end as an append. The results above have the followng orollary Corollary 1 g (4 ) = 4, (2.16) 1 3 g (20 ) =. (2.17) 1 4 Proof. (2.12) s followed by applyng (2.7) twe, and (2.13) s from (2.9) and (2.10). Proof of Proposton 1. Let k n = λ3, 1 λ 5, Where λ s a real number and k s a non-negatve nteger. We wll take nduton on k. By (2.6) of Lemma 1, (2.3) s stand for k = 0. For k > 0, suppose that A and B are two on sets, A = B = n, Ct( A) = Ct( B) = 1. Take subsets A1, A2 A, and B1, B2 B, suh that A1 = A2 = B1 = B2 = n/3, and make two weghngs A1: A 2 and B 1 : B 2, then t wll be found the subsets A' A and B ' B wth A' n/3, B' n/3, Ct ( A ') = 1, Ct( B ') = 1. Wthout loss generalty, we may assume that A' = B ', then by the nduton, g1( n 2) 2+ log 3 n/3 + log 3 n/5 + 1 log n + log ( n/ 5) The proofs for (2.4), (2.5) and (2.6) are smlar but nstead apply (2.7), (2.12) and (2.10), (2.), (2.17) and (2.13), (2.14), (2.15), (2.16) respetvely, so whh are omtted. Note. The upper bounds n Proposton 1 may be wrtten as followng form

5 g1( n 2) 2 log3n , (2.3') g1( n 3) 3 log3n , (2.4') g1( n 4) 4 log3n (2.5') g1( n 5) 5 log3n (2.6') Referenes [1] M. Agner, Combnatoral Searh. Wley-Teubner 1988 [2] R. Bellman, B. Glass, On varous versons of the defetve on problem. Inform. Control 4, (1961) [3] A.P. L, On the onjeture at two ounterfet ons, Ds. Math. 133 (1994), [4] A.P. L, Three ounterfet ons problem, J. Comb. Theory Ser. A, 66 (1994), [5] A.P. L, M. Agner, Searhng for ounterfet ons, Graphs Comb. 13 (1997), 9-20 [6] A.P. L, A note on ounterfet ons problem, arxv, e-prnt arhve, [7] B. Manvel, Counterfet on problem, Math. Mag. 50, (1977) [8] L. Pyber, How to fnd many ounterfet ons? Graphs Comb. 2, (1986) [9] R. Tos, Two ounterfet ons, Ds. Math. 46 (1983),

6 Append Algorthms The sketh of algorthm g (4,20) = 4 1 Let X = { 1 4}, Y = { y 1 20}, Ct( X ) = Ct( Y) = 1. The followng s a 4 - ompleted searh algorthm (, y ) (, y ) ,16 (, y ) 7,19,20 {, y }:{, y } {, y }:{, y } 5, , (, y ) (, y ) ,18 {, y }:{, y } 1 5,6,7 2 8,9,10 (, y ) (, y ) 2 5,6,7 (, y ) (, y ) (, y ) 1 8,9,10 (, y ) Fg.1 The rest two weghngs are a routne work, so omtted.

7 The sketh of algorthm g 1(5 2) = 3 = =, Ct( A) = Ct( B) = 1. A 3 - ompleted algorthm Let A { a 1 5 }, B { b 1 5} s as followng ( a, b ) ( a, b ) ( a, b ) { b }:{ b } 1 3 ( a, b ) 1 4 ( a, b ) { a }:{ a } { a, b }:{ a, b } { a, b}:{ a, b } 1,2 1 3,4 2 ( a, b ) 2 3 { b}:{ b } 1 4 ( a, b ) ( a, b ) ( a, b ) { b}:{ b } ( a, b ) 3 1 ( ) { a }:{ a } { b }:{ b } ( a, b ) ( a, b ) ( a, b ) ( a, b ) Fg.2

8 The sketh of algorthm g 1( 4) = 9 Let A= { a 1 }, B= { b 1 }, C = { 1 }, D= { d 1 }, Ct( A) = Ct( B) = Ct( C) = Ct( D) = 1. For X = ABCorD,,,, let X = X, X = 3, = 0 3 1, 2, 3, and X 0 = 2. The sketh of a feasble algorthms s as followng A B C D 2 2,3 0,1 2,3 : 2 3 A B C D 0 2,3 0,1 0,1 A A A C : B D 0 0,1 0 0,1 A B C D : A B C D 2,3 2,3 0,1 0,1 0,1 0,1 2,3 2,3 A B C D 3 2,3 0,1 2,3 A : A 2 3 A B C D 2 1,2,3 0,1 0,1 A B C D A B C D 0 2,3 0,1 0,1 3 1,2,3 0,1 0,1 A : B 2,3 2,3 A B C D A B C D 2 0,1 0, ,1 3 0,1 A B C D 2 0,1 2 0,1 C : C 2 3 A B C D 2 0,1 0,1 3 A : A A C : A D A : A A B C D 3 0,1 2 0,1 A B C D A B C D 0,1 0,1 0,1 0,1 3 2,3 2,3 2,3 D : D A B C D ,3 2,3 2,3 A B C D 3 0,1 0,1 2 A B C D A B C D 3 0,1 3 0,1 3 0,1 0,1 3 Fg.3

9 The sketh of algorthm g 1( 5) = obj.2 obj.1 obj.3 abde : e abde : e abde : abd e 4 1,2,3,7,8 4 1,2,3,7, obj.4 ab : ab Fg.3 obj obj.5 ae : bd : d be : a d obj. 1 4, ,2 1,2 1 4, obj.6 obj.7 obj.8 obj.9 obj.10 a e : b e Fg.2 1,2 7,8 1,2 9,10 obj.16 b d e : a Fg.1 1, ,3 obj.15 obj.14 : d a b : e 1,2 1,2 1,2 1 obj.17 6 obj.13 obj.18 Fg.A obj.1 = a b d e, obj.4 = a b d, ,2, o a b d a b d b a d b a d bj.2 = e, obj.3 e 7,8 = 1,3 a b d a b d d a b d a b a b e d obj.5 = a e b d, obj.6 = a d e b d b e a d b d e a

10 obj.7 = e a b d, obj.9 = d e a b d e a b a e b d obj.8 =, obj.10 = d e a b a d e b a d obj. = b e a d, obj.12 = b e a d, d a a d e b a e b d obj.13 = b d e a, obj.14 = a d e b d e a b a b e d 2 1 9, , , , a e b 2 1 9, obj.15 = b e a, obj.16 = e a b d, 2 1 7, d e a b obj.17 = e a b d, obj.18 = d e a b 1 4,5, ,5, d e a b e : a b e a b d 1,2,3,9,10 1,2 1, a b e 2 3 d 2 3 d 2 3 e : e 1,2,3 4,5,6 a b e 2 3 d 2 3 d 2 3 a b e d 1 3 d 1 3 : 2 2 a b e : e a : b 7 8 a b e d 1 3 d 1 3 a b e d 2 3 d 2 3 a b e d 2 3 d 2 3 b e a d d ae : d b e a d d 2 2 a b e d a d b e a d a d a d 2 b e a d a d 2

11 Fg.1 a b d e a b d e : a 1,2 1,2 1,2 1,2 7 3 a b d e a e b d b d : a e a e b d 1,2 1,2 1, a b e d Fg.2a ab : e ,9,10 ab : e,2 2 7,9,10, a e b d a e b d d e a b a d e b a e b d d : e 1,2 2 7,9,10 a b d e e 1 4 a d b e e 2 1 a b e d Fg. 2 ae : be 1 2 a d e b a b d e a d e b a b d e a d e b a b d e Fg.2a

12 Fg.4 ab : d a b d e ,10, a : b b a d e ,10, Fg.3a e : e. 1,2,3,7 4,5,6,8 a b d e d : e 1,2 1,2 1 4 a b d e 3 3 1,2,3 a b e d 1,2 1,2,3,7 3 : d a b d e 1,2 1,2 3 1,2 1,2,3,7 a b d e 1 2 1,2,3,7 : d a b d e 2,2,3,7 a b d e Fg.3 d : d a b d e 1 3 9,10, 2 3 9,10, a b d e : 1 2 9,10, 1 2 a b d e d a b d e 3 3 9,10, : d 2 1 a b d e 2 1 9,10, a b d e 2 3 9,10, 1 3 9,10, a b d e Fg. 3a

13 Fg.4a ae : e e : e Fg.5 1,2,3,7 4,5,6,8 Fg.4b a : bd a b d e ,10, a b d e 2 3 9,10, a b d e 3 3 9,10, 2 2 b : d a d : b a : b d a e 1,2 2 1, ,10, a d b e b d a e 1 3 9,10, ,10, a d b e 3 3 9,10, a : b b a d e 3 3 9,10, b a d e 9,10, 2 3 b 2 3 () a b e 9,10, d d 2 3 a 2 3 Fg.4 e : e 4 8 a e b d a b d e a d e b a : d b : a d e b a d e b b d e a a d e b Fg.4a

14 a : d a e b d b d e a a e b d ,6 2 2 e : b d e : a b d e a 3, b e a d a d e b b a e 8 d d 2 3 b 2 3 a b d e a 2 3 Fg.4b Fg.5a a d e b , ,2 ab : abd ae : e 1 2 Fg.5b Fg.5d a : b d 1,2,3 2 1,2,3 2 e : e Fg.5e 3 7 Fg.5 a d : b 1,2,3 2 1,2,3 2 a d e b a b d e 2,3 2 1 a : b 2,3 2,3 b e a d a d e b ,3 b e a d 1 2,3 3 a d e b b d e a a d e b 2 2,3 3 Fg.5

15 a : bd a b d e a b d e 1 2 2,3 b d a e 1 d 2 3 b 2 3 a e b d Fg.5a a : bd b d e a a b d e a 2 3 a b d e Fg.5b a : b 2,3 2,3 a e b d 3 2,3 3 a d e b b d e a b d e a 3 2,3 3 Fg.5 b : b 4,5,6 7,8,9 a e b d b : b a e b d 3 4,5, ,8,9 3 a b e d a b d e a b d e a b e d Fg.5d

16 a : b d 1,2 2 1,2 2 a e b d a : b b d e a a e 7 b d 2 3 d b 2 3 () b d e 7 a 2 3 a 2 3 Fg.5e

17 The sketh of algorthm g 1(17 5) = 13 a e : bd e Fg. 9,2 1 6, ae : be Fg ,8 1 9,10 a b d e e : Fg , d a b e abde : e Fg. 6 d a b e Fg. B abd : e,2,3 d e a b ,8 9,10 e : e d e a b b e a d,2 7, a b e d a b d e 1, a d e b,2 9, Fg. 6

18 abe : e 1, a b e ,2,3,17 abe : e a b e e a b d e d a b a b d e 1,2 1, , e a b d Fg. 7 e a b d ab d e : a e a b d 1,2 1, a e 7 d b b d 1,2 1, e : e a b d e 1 4, a d b e,2 1, ab : 1,2 d e a b 1,2 1,2 7, a d e b a b e d 7, ,2 7,8, Fg. 8

19 e a b d ab d e ,2 1, : a e a b d : Fg. 9a 1 2 a b d e ,2 6 7 abd e : e a b d e e a b d e : abd e , e 6 b a d d a b 1, a b d e b d e,2 1 6 b : e a b d e d b e 1, a e b d Fg. 9 bd e : e 1,2 6 7 a b d e ,2 : b d a b d e a b d e a e b d a d b e , Fg. 9a

20 Sub-algorthm g (5,16) = 4 1 = 1 5, = 1 16, Ct( X ) = Ct( Y ) = 1. Let X { } Y { y } y 1 13 y y : y ,2,5,2 2 3,4 y : y y : y 1, , y y y : y 1 3,4 y ,2 y y y 5 2 y : y 1,3 4 2,4 5 y 3 1,2 y 1 6 y y : y y : y 1,2,6 3,4, y 3 3,4 y 1 7 y 5 13 y 3 5 y 4 4 y Fg. 10 The sub-algorthm g (5,16) = 4 1 wll be used n the algorthm g 1(17 5) = 13. The outputs n the skethes are the objetve sets settled easly by the known sub-algorthms. The skethes of algorthms for g 1(7 5) = 9 and g 1 (13 5 ) = 12 have been omtted to save the spae.