In this paper, we present a simple and fast method for counting the number of nonnegative integer solutions to the equality a1x1+a2x2+: : :+arxr = n where a1; a2; :::; ar and n are positive integers. As an application, we use the method for finding the number of solutions of a Diophantine inequality

Farzaneh , A.

Hassani, Hossein

Howell, G.

Mahmoudvand , R.

English

Bournemouth University Research Online

The Exact Number of Nonnegative IntegerSolutions for a Linear Diophantine InequalityRahim Mahmoudvand1, Hossein Hassani2, Abbas Farzaneh3, Gareth Howell4 ∗Abstract—In this paper, we present a simple andfast method for counting the number of nonnegativeinteger solutions to the equality a1x1+a2x2+. . .+arxr =n where a1, a2, ..., ar and n are positive integers. As anapplication, we use the method for finding the numberof solutions of a Diophantine inequality.Keywords: Counting, Nonnegative integer solutions,Diophantine inequality.1 IntroductionCounting techniques play an important role in comput-ing probabilities in random experiments of throwing dice,or classical occupancy problems. As a result, they havecome to form a major part of the mathematics curriculumin many statistical publications. First we will considersome important applications of counting techniques.Ross [3] showed that the number of ways for placing nidentical objects into the r distinct cells is equivalent tothe number of nonnegative integer solutions to the equa-tionx1 + x2 + . . .+ xr = n (with xi ≥ 0, i = 1, . . . , r). (1)He also showed that the number of positive integers solu-tions of (1) is(n−1r−1). The number of nonnegative integersolutions of (1), subject to the constraint xi ≥ bi fori = 1, . . . , r is(n+r−(b1+b2+...+br)−1r−1). Letting xi = yi+ bifor each i yields the equationy1 + . . .+ yr = n− (b1 + b2 + . . .+ br), (2)to be solved in nonnegative integers. The number of suchsolutions where xi ≤ bi (i = 1, . . . , r) can be obtained us-ing the inclusion/exclusion principle (see, for example,Rosen et al. [1]). For the latter situation, Murty [4] ob-tained a simple method of counting the favoured number∗1Rahim Mahmoudvand: Group of Statistics, Universityof Payame Noor, Toyserkan, Iran, E-mail: r mahmod-vand@yahoo.com;2Hossein Hassani: Centre for Optimisation and Its Applications,School of Mathematics, Cardiff University, UK, CF24 4AG, Tele-phone: +44 (0) 7703367456 Fax: +44 (0) 29 2087 4199, E-mail:HassaniH@cf.ac.uk;3Abbas Farzaneh: Group of Computer, Islamic Azad University ,Toyserkan, Iran, E-mail: farzaneh.comp@gmail.com;4Gareth Howell: Statistics Group, Cardiff University , UK, CF244AG, E-mail: HowellGL@cf.ac.uk.of solutions. One generalization of (1) is the number ofnonnegative integer solutions of the following equation,a1x1 + a2x2 + . . .+ arxr = n. (3)Equation (3) is well-known as a Linear DiophantineEquation. As is discussed above for the simple case, it ispossible to obtain the number solutions of equation (1)with some bounds on xi’s from (1) without any boundson xi’s. It has been shown that the number solutions of(3) by some bounds on xi’s can be expressed as a func-tion of the number solutions of (3) without any boundson xi’s (Eisenbeis et al. [5]). Therefore, it is enough torestrict our effort to determine the number solutions of(3) without any bounds on xi’s. Given positive integersa1, a2, . . . , ar that are relatively prime, it is well-knownthat for all sufficiently large n the equation (3) has a so-lution with nonnegative integers xi (Tripathi [2]). Thegenerating function of equation (3) has the formϕ(t) = [(1− ta1)(1− ta2) . . . (1− tar )]−1,and the number of non-negative integer solutions J(n) ofequation (2) is given by the formula:J(n) =1n!ϕn(0). (4)Calculation of J(n) is difficult in most situations. An-timirov and Matvejevs, in [6] have discussed several pos-sible methods methods for its calculation. Eisenbeis etal.(1992) [5] presented fast methods for computing theexact or approximate number of solutions. In summary,there are two main problem for finding the number ofnonnegative integer solution solutions of (3); the presentmethods, owing to the difficulty of the problem, are com-plicated, time consuming, and encounter difficulties whenone wishes to extract a list of such solutions. These is-sues motivated us to obtain a simple method for findingthe number of nonnegative integer solutions of (3) andprovide a list of the obtained solutions.2 New MethodAmong the two problems considered, i.e., computing thenumber solutions and generating the solutions, the firstone is by far the most complex. Therefore, it is vitalIAENG International Journal of Applied Mathematics, 40:1, IJAM_40_1_01______________________________________________________________________________________(Advance online publication: 1 February 2010)to simplify the problem as much as possible in order toobtain efficient computation. Let us first consider ai = 1for i = 2, ..., r in (3). In this case, we must find thenumber of nonnegative integer solutions fora1x1 + x2 + . . .+ xr = n. (5)For solving (5), we can give the possible values of x1 andreform (5) to form (1). Therefore,[n/a1]∑w1=0(n− a1w1 + r − 2r − 2)(6)is the number of nonnegative integer solutions for equa-tion (5), where [u] is the integer part of u and r is apositive integer and r > 2. If r = 2 we must use[n/a1]∑w1=0I(a2, w1) as the number of nonnegative integer so-lutions, whereI(a2, w1) ={1 a2|n− a1w10 otherwise (7)Now, let ai = 1 for i = 3, ..., r. In this case, we must findthe number of nonnegative integer solutions fora1x1 + a2x2 + x3 + . . .+ xr = n. (8)For solving (8), we can give the possible values of x1, x2and reform (8) to form (1). Therefore,[n/a1]∑w1=0[(n−a1w1)/a2]∑w2=0(n− a1w1 − a2w2 + r − 3r − 3)(9)is the number of nonnegative integer solutions for thisequation. It should be noted that, the formula is truewhen r is a positive integer and r > 3. However, if r = 3we use[n/a1]∑w1=0[(n−a1w1)/a2]∑w1=0I(a3, w1, w2) as the number ofnonnegative integer solutions, whereI(a3, w1, w2) ={1 a3|n− a1w1 − a2w20 otherwise (10)Continuing the procedure, we can get the following for-mula for the number of nonnegative integer solutions of(3).s(a1, . . . , ar;n) :=[n/a1]∑w1=0[(n−a1w1)/a2]∑w2=0. . .[(n−a1w1−...−ar−2wr−2)/ar−1]∑wr−1=0I(ar;w1, . . . , wr−1)(11)whereI(ar;w1, . . . , wr−1) ={1 ar|n− a1w1 − ...ar−1wr−10 otherwise.(12)Note also that if ai = 1 for all i, then s(a1, . . . , ar;n) isequal to(n+r−1r−1), sinces(a1, . . . , ar;n) =n∑w1=0n−w1∑w2=0. . .n−w1−...−wr−2∑wr−1=01=n∑w1=0n−w1∑w2=0. . .n−w1−...−wr−3∑wr−2=0(n− w1 − ...− wr−2 + 11)=n∑w1=0n−w1∑w2=0. . .n+1−w1−...−wr−3−1∑wr−2=0(1 + wr−21).(13)Now equality is obtained using the fact thatn−m∑k=0(m+ km)=(n+ 1m+ 1).3 An applicationThere are many problems which can be solved using theproposed algorithm. As a useful example, we use thealgorithm for solving the Diophantine inequalitya1x1 + . . .+ arxr ≤ n. (14)Let us now briefly consider the characteristic of theDiophantine Inequality (for more information see, forexample, [16][17][18][19]). The main statement of theaforementioned theorem is in the language of lattices innumber theory. That is for any convex set in the r-dimensional Euclidean space Rr symmetric with respectto the origin, and with volume greater than 2r, must con-tain a lattice point other than that of the origin. In thelanguage of linear forms the problem is restated asr∑j=1aijxj = Li(X), 1 ≤ i ≤ r, (15)with real coefficients aij such that det(aij) 6= 0, suppos-ing that there exist r positive real numbers bi, i = 1 . . . rwith∏ri=1 bi ≥ det(aij). Then there exists an integerIAENG International Journal of Applied Mathematics, 40:1, IJAM_40_1_01______________________________________________________________________________________(Advance online publication: 1 February 2010)vector C such that Li(C) ≤ bi, 1 ≤ i ≤ r, thus im-plying that a solution exists for the above equations andindeed the implied inequality. The paper by Cheema,[13] suggests techniques similar to the programming ofthis research in its working, and indeed uses Minkowski’stheorem to state that, where || · || denotes the distanceof a number from its nearest integer, that there alwaysexists a nonzero integer-vector solution X = (x1, . . . , xr)to the inequalities:||Lj(X)|| ≤ C, (1 ≤ j ≤ r). (16)Another practical application of the discussed problem isthat of the “Knapsack” model, encountered in many areaswith a cleat explanation offered in [14]; “the question ofhow to fill a knapsack of limited weight capacity withdifferent items which best meet the needs of one’s trip”.Beged-Dov [14], first introduced bounds on the number,N , of solutions tor∑i=1aixi ≤ n with the ai’s all beingnatural-valued, asnrr!r∏i=1ai≤ N ≤ (n+ a1 + . . .+ ar)rr!r∏i=1ai. (17)These bounds were obtained in the following way. Denotethe rectangular box B(y1, . . . , yr) as the set of points Y =(y1, . . . , yr) such thataixi ≤ yi ≤ (xi + 1)ai for i = 1, . . . , r (18)which has r-dimensional volume∏ri=1 ai. Secondly, de-fine the pyramid P (n) with volume nrr! , which denotesthe set of points satisfying yi ≥ 0 for i = 1, . . . , r and∑ri=1 yi ≤ n. The bounds are obtained as a consequenceof the fact that each point xi as defined above belongsto a unique B, which is the one with xi =[yiai], andif that xi lies in the pyramid P (n), then it necessarilyobeys the linear diophantine inequality in question. Sothe union of the N boxes contains P (n). This somewhatsimple topological argument allows the derivation of theabove bounds. To add weight to Beged-Dov’s argumentin [14], some experimental results are calculated using analgorithm which could be considered to be an early pre-cursor to the results of this paper. The tendency of theupper and lower bounds of the number of solutions tothe linear Diophantine inequality to become close withincreased number of variables and right hand side is alsotouched upon.Padberg and Lambe sought to respectively improve uponBeged-Dov’s bounds. In the latter case an approximatenumber of solutions was eventually sought and found in[7]. Padberg [12] considered the following lower bound(n+ 1)rr!∑ri=1 ai≤ N (19)Very soon after [14] was submitted, Padberg took its re-sult in [12] and sharpened Beged-Dov’s result to the fol-lowing inequality:max((n+ 1)rr!∏ri=1 ai,(r + a∗r))≤ NandN ≤ min((n+∑rj=1 aj)rr!∏ri=1 ai,(r + a∗∗r)).(20)Here a∗ and a∗∗ are integers satisfying a∗ ≤ naj and a∗∗ ≥[naj]for all j = 1, . . . , r. The initial adjustment to theoriginal result is made by definition of the new pyramidP (n+ δ), whencer∑j=1xij ≤ n+δ, xij ≥ 0 for j = 1, . . . r, 0 ≤ δ < 1, (21)Then as above, taking a vector ξ ∈ P (n + δ), then sum-ming over each element of the vector we have (since[x] ≤ x, ∀x > 0):r∑j=1xj[ξjxj]≤r∑j=1xj(ξjxj)≤ n+ δ. (22)The lower bound(n+ δ)rr!∏rj=1 aj≤ N is obtained with thesubstitution of P (n+ δ) with P (n) in the previous proof,which is then sharpened by taking the limit of this boundas δ → 1. The result is improved further by making theabove substitution for a∗ and a∗∗ above, noting thatr∑j=1xj ≤r∑j=1xjajamin≤ namin, (23)to obtain the bounds stated above.The paper of Padberg also introduces the formula forthe number of possible partitions explored in this paper,and quotes that another proof is mentioned in the book[15]. Lambe in his paper [11] of 1974 introduced boundswhich in most cases were better still, than what had beenpreviously discussed:(n+ rr) r∏i=11ai≤ N ≤(n+ rar) r∏i=11ai, (24)IAENG International Journal of Applied Mathematics, 40:1, IJAM_40_1_01______________________________________________________________________________________(Advance online publication: 1 February 2010)Table 1: Comparison between current methods and the new algorithm.{ai} n New (Exact) (17) (20) (24) (19)lower upper lower upper lower upper lower2, 3, 5 10 20 6 44 10 56 10 38 82, 3, 5 50 947 695 1200 737 1200 781 1140 7372, 3, 5 100 6518 5556 7394 5724 73946 5896 7194 57242, 3, 5 200 48202 44444 51450 45115 51450 45791 50717 451151, 1, 10 12 97 29 230 37 230 46 202 372, 3, 4, 4, 5 11 53 3 356 21 252 9 247 212, 4, 4, 4, 5 11 41 2 316 21 252 7 223 212, 3, 4, 4, 5, 7 15 162 5 1693 28 1693 16 1142 72, 3, 4, 5, 6, 7 20 364 18 2970 28 2970 46 2130 242, 3, 5, 7, 9, 11, 13 50 8872 574 73412 659 73412 978 59228 659where a =∑ri=1air . His new bounds were also able toshow that the ratio of upper to lower bounds tends tounity as r and n grow large. To attain the lower bound,the inequalitiesg−1∑i=1aiyi +r∑i=gyi ≤ n, (25)with the yi’s all integers and g ∈ {1, . . . r+1} are consid-ered. The proof requires - where Pi denotes the numberof feasible (that is, nonnegative) solutions to (25) - theproving ofPg ≤ agPg+1, for g = 1, . . . , r. (26)The proof of the upper bound is achieved using the in-equalitiesg−1∑i=1aiyi +r∑i=gyi ≤ n+r∑i=g(xi − 1), (27)and requires the assertion - where Qi denotes the numberof feasible solutions to (25) and (27) - thatQg ≥ agQg+1, for g = 1, . . . , r. (28)Both are achieved in similar fashion.As mentioned above, Lambe in [7], discovered upper andlower bounds for this number. However, the algorithmproposed here is able to compute the exact number ofsolutions. To do this, we convert (14) to (3) by addingan extra nonnegative integer variable xr to (14). Thenwe need to solve a1x1 + . . . + ar−1xr−1 + xr = n andusing the algorithm the number of nonnegative integersolutions to (14) is:s(a1, . . . , ar−1, 1;n) =[n/a1]∑w1=0[(n−a1x1)/a2]∑w2=0. . .[(n−a1x1−...−ar−2xr−2)/ar−1]∑wr−1=01.(29)It should be noted that in the reduced form of inequalitywe have ar = 1. Therefore I(ar;w1, . . . , wr−1) = 1 forall w1, . . . , wr−1. Let us first consider an simple exam-ple. Suppose we are interested in finding the number ofnonnegative solutions to10x1 + x2 + x3 ≤ 12. (30)The lower and upper bounds on the number of solutionsto this inequality, 4 and 455 respectively, are obtainedfrom the algorithm of (20), whilst we know the exactnumber of solution is 97. It can be seen easily that thesebounds represent a wide deviation from the actual num-ber of solutions. Let us now use the proposed algorithmfor solving (30). As we mentioned above, first we needto reform (30) to 10x1 + x2 + x3 + x4 = 12. Thus, thesolution is as follows[12/10]∑w1=0[(12−10w1)/1]∑w2=0[(12−10w1−w2)/1]∑w3=01 ==12∑w2=012−w2∑w3=01 +2∑w2=02−w2∑w3=01 = 97. (31)Table 1 shows the resulting lower and upper bounds givenfor the number of solutions to the inequality with coef-ficients ai and relevant n. The third column shows theexact number of solutions given by the method of thisnote.References[1] H. K. Rosen, G. J. Michaels, L. J. Gross, W. J.Grossman, R. D. Shier, Handbook of Discrete andCombinatorial Mathematics, CRC Press, New York,2000.[2] A. Tripathi, On a Linear Diophantine Problem ofFrobenius, Integers: Electronic Journal of Combi-natorial Number Theory, 6, no. A14, pp. 1-6, 2006.[3] S. M. Ross, A First Course in Probability, CRCPress, New York, 1976.IAENG International Journal of Applied Mathematics, 40:1, IJAM_40_1_01______________________________________________________________________________________(Advance online publication: 1 February 2010)[4] V. N. Murty, Counting the Integer Solutions of aLinear Equation with Unit Coefficients. Mathemat-ics Magazine, 54, no. 2, pp. 79-81, 1981.[5] C. Eisenbeis, O. Temam, and H. Wijshoff, On ef-ficiently characterizing solutions of linear diophan-tine equations and its application to data depen-dence analysis. In Seventh International Symposiumon Computer and Information Sciences, Antalya,Turquie, 1992. Note: Appeared also as INRIA Re-search Report No 1616, Fvrier 1992.[6] M. Ya. Antimirov, A. Matvejevs, Evaluation of theNumber of Non-Negative Solutions of DiophantineEquations, 5th Latvian Mathematical Conference,Daugavpils, Latvia, 2004.[7] T. A. Lambe, Upper bound on the Number ofNonnegative integer Solutions to a linear equation.SIAM, Journal of Applied Mathematics, 32, no. 1,pp. 215-219, 1977.[8] K. Aardal, and C. A. J. Hurkens and A. K. Lenstra,Solving a Linear Diophantine Equation with Lowerand Upper Bounds on the Variables, Proceedings ofthe 6th International IPCO Conference on IntegerProgramming and Combinatorial Optimization, pp.229-242, 1998.[9] S. Mertens, The Easiest Hard Problem: NumberPartitioning. ArXiv Condensed Matter e-prints,http://adsabs.harvard.edu/abs/2003cond.mat.10317M[10] D. H. Lehmer, A note on the Linear DiophantineEquation, The American Mathematical Monthly, 48(4), pp. 240-246, 1941.[11] T. A. Lambe, Bounds on the Number of FeasibleSolutions to a Knapsack Problem. SIAM Journal onApplied Mathematics, 26(2), pp. 302-305, 1974.[12] M. W. Padberg, A Remark on “An Inequality forthe Number of Lattice Points in a Simplex”, SIAMJournal on Applied Mathematics, 20 (4), pp. 638-641, 1971.[13] M. S. Cheema, Integral Solutions of a Systemof Linear Equations, The American MathematicalMonthly,73 (5), pp. 487-490, 1966.[14] A. G. Beged-Dov, Lower and Upper Bounds for theNumber of Lattice Points in a Simplex, SIAM Jour-nal on Applied Mathematics, 22 (1), pp. 106-108,1972.[15] W. Feller, An Introduction to Probability Theory andits Applications, Volume 1, John Wiley, New York,1968.[16] S. Lang, Survey on Diophantine Geometry. Springer-Vlarge, New York, 1997.[17] V. G. Sprindzuk, Classical Diophantine Equation,Springer-Vlarge, New York, 1993.[18] B. M. M. de Weger, Algorithms for Diophantineequations. Centrum Voor Wiskunde en Informatica,1989.[19] H. Davenport, Analytic Methods for DiophantineEquations and Diophantine Inequalities. CambridgeUniversity Press, 2005.IAENG International Journal of Applied Mathematics, 40:1, IJAM_40_1_01______________________________________________________________________________________(Advance online publication: 1 February 2010)

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On the Number of Nonnegative Solutions to the Inequality a1 +....ar < n

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