An improved lower-bound on the counterfeit coins problem

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1 An mproved lower-bound on the counterfet cons problem L An-Png Bejng , P.R. Chna apl0001@sna.com Hagen von Etzen hagen@von-etzen.de Abstract In ths paper, we wll gve an mprovement on the lower bound for the counterfet cons problem n the case that the number of false cons s unknown n advance. Keywords: counterfet cons problem, combnatoral search, nformaton theoretc bound.

2 Searchng for counterfet cons n a number of cons wth same semblance by a balance s a well-known combnatoral search problem called the counterfet cons problem. The problem has a longer hstory and gotten a ntensve researches, for the detal refer to see the papers [1]~[11]. The problem has several versons, e.g. the number of the fakes s assumed known n advance, and/or the fakes are known lghter or heaver than the normals. Usually, t s assumed that the cons wll be permtted to be remarked by numbers n order to be dstngushed each other. In paper [5], we dscussed a general case that the number of the fakes s unknown beforehand. Suppose that S s a set of n cons wth same semblance, n whch possbly there are some counterfet cons, whch are heaver (or lghter) than the normals. Denoted by gnthe ( ) least number of weghngs need to fnd all the fakes n S by a balance, assumed that addtonal normal cons wll be avalable f needed. Our man result n [5] s as followng but wth a excepton that g (3) = 3. n log 2 g( n) 7 n/11, 3 (1) Clearly, the lower-bound,.e. the left-hand sde of (1) s just the nformaton theoretc bound of gn, ( ) n ths paper, we wll gve an mprovement over t, our man result s that Proposton 1 gn (2) n n 5 n 6 n 7 n 13 n 1 n 16 n 17 ( ) log 3( ) At frst, we ntroduce some notatons. Let A be the set of the three symbols <=>,,,.e, A = { <=>,, }, a vector = ( v1,, v k ) k A wll be called a drecton, and a subset X S wll be called a objectve n the drecton, f the state of the balance n the -th weghng for X s equal to v,1 k.denoted by S the set of all the objectves n the drecton. A As usually, for a set A, 2 A and wll represent the set of all subsets of A and the set of k all the k -subsets of set A respectvely. Smlar to Cartesan product, for any two subsets ΔΓ, 2 S, we defne Δ Γ= { A B A Δ, B Γ }. In ths paper, we wll employ a combnatoral dentty stated n the followng lemma 1.

3 Lemma 1. Suppose that stare, two non-negatve ntegers, and k s an nteger, then k t k Cs C t = C s t. (3) The formula (3) may be obtaned by a basc combnatoral calculus, whch has been omtted. For a subset Γ S, denoted by S ( Γ ) = { X X S, X Γ}. Suppose that L : R, = 1, 2,, k, are the frst k weghngs n a drecton, denoted by t s easy to know that ( ) 2 n For an nteger k, defne ζ = max{ S = k}. k Γ= ( L R), then 1 k Γ S = S Γ. () Lemma 2. n ζ /128. (5) Proof. Suppose that the frst weghng s that A : B, and the second weghng s that L : R. Denoted by Γ= A B, C1 = L\ Γ, C2 = R\ Γ, C1 s, denoted by = C2 = t. For an nteger, By Lemma 1, t has C C σ (,) st = σ ( st, ) 1 2 k k k+, (6) C C t Cs t k k k+ + (7) 1 2 = = We wll also smply wrte σ (,) st as σ n some apparent cases. Moreover, for a subset X Γ, denoted by δ ( X ) = X L X R. Let Λ =Γ C1 C2, v {, <=>,}, t s clear that So, where m= s+ t. It s easy to know that S ( Λ ) = ( X L) ( X R). (8) σ δ ( v, = ) ( X) X S( v )( Γ) S ( Λ ) = = C S ( Γ ) C, (9) σ δ t δ ( X) [ m/2] ( v, = ) ( X) s+ t ( v) m X S( v) ( Γ) X S( v) ( Γ)

4 1, f Γ = 1, max{ S( < )( Γ), S( = )( Γ), S ( > )( Γ) } 2, f Γ = 2, (10), f Γ = 3. Moreover, f Γ = 5, then max{ S ( Γ), S ( Γ) } 16. (11) ( < ) ( > ) On the other hand, f Γ 6, then Hence, the reman cases to be checked are ) Γ =, A = B = 2, m, ) Γ = 6, A = B = 3, m. For the case ), t has that For the case ), t has that max{ S ( Γ), S ( Γ) } (2 C ) / 2. (12) Γ [ Γ /2] ( < ) ( > ) Γ Λ max{ ( =<, )( Λ), ( ==, )( Λ), ( =>, )( Λ) } 2 /8, S S S (13) Λ max{ ( <<, )( Λ), ( <=, )( Λ), ( <>, )( Λ) } 15 2 /128. S S S (1) The estmaton (5) s from (10) to (1). Lemma 3. ζ. n Proof. Follow wth the proof of Lemma 2, suppose that the thrd weghng and the fourth are U : V and X : Y respectvely. From the proof above, we know that A = B = 3 and ( L R)\( A B). Smlarly, there are ( U V)\( A B L R) and ( X Y)\( A B L R U V). Moreover, by Lemma 2, t has And so ζ ζ. n n 1 a a 2 ( ) / 3, 0 a, ε = 1, or 2, as a s odd, or even. It follows ε ζ. (16) n

5 Proof of Proposton 1. Let k 1 n k <, t s known that the nformaton theoretc bound of gns ( ) equal to log3 2 n, hence gn ( ) k. Denoted by ω n 5 n 6 n 7 n 13 n 1 n 16 n 17 = , k n and suppose that 3 = 2 + x, f ω x, then 2 n + ω 2 n + x = 3 k. Namely, gn ( ) k= log (2 + ω) k n Thereby, we assume that x < ω, then 3 < 2 + ω, and n 3. (17) By Lemma 3 and (18), t has Ths means that ζ n k 2 > n 1707 k k > = (18) gn ( ) k+ 1= log (2 + ω). n 3 (19) The proof of Proposton 1 has been fnshed. Remark. Wth computer ad, t has shown that for the second weghng follow the drecton ( < ), there are followng sx types of ones whch gve 1) {1,2,}:{3,7,8} 2) {1,,5}:{6,7,8} 3) {1,,7}:{2,5,8} ) {1,7,8}:{2,9,10} 5) {,7,8}:{5,9,10} 6) {7,8,9}:{1,,10} ζ = up to the equvalency, n Wth Lemma2 and a short program, t may be known that the second weghng of the frst fve types all gve the estmaton ζ =, (20) n

6 and ζ = + =. (21) n 1 n 13 2 (60 2) / For the sxth type of the second weghng, follow the drecton (<,<), wth the thrd weghng {1,2,3,10,11,13}: {5,6,7,8,9,12}gve the estmaton n 10 ζ = (= 2 0 n ), and then ζ = =. (22) n 17 n Overall, t follows ζ = 2 5, ζ = (23) n 7 n 13 3 So, gn + ζ n +, 13 ( ) log3 log32 log 3(107 81/ 2 ) or, gn whch s a lttle better than (2) , (2) n n 5 n 6 n 7 n 9 n 10 n 12 n 13 ( ) log 3( ) References [1] M. Agner, Combnatoral Search. Wley-Teubner 1988 [2] R. Bellman, B. Glass, On varous versons of the defectve con problem. Inform. Control, 11-17(1961) [3] Rchard K. Guy; Rchard J. Nowakowsk, Con-Weghng Problems, The Amercan Mathematcal Monthly, Vol. 102, No. 2. (Feb., 1995), pp [] A.P. L, On the conjecture at two counterfet cons, Dsc. Math. 133 (199), [5] A.P. L, M. Agner, Searchng for counterfet cons, Graphs Comb. 13 (1997), 9-20 [6] A.P. L, A note on counterfet cons problem, arxv, e-prnt archve, [7] A.P. L, Some results on the counterfet cons problem, arxv, e-prnt archve, [8] A.P. L, Some results on the counterfet cons problem II, arxv, e-prnt archve, [9] B. Manvel, Counterfet con problem, Math. Mag. 50, (1977) [10] L. Pyber, How to fnd many counterfet cons? Graphs Comb. 2, (1986) [11] R. Tosc, Two counterfet cons, Dsc. Math. 6 (1983),