Domination and Independence on Square Chessboard

Transcription

1 Engineering nd Technology Journl Vol. 5, Prt, No. 1, 017 A.A. Omrn Deprtment of Mthemtics, College of Eduction for Pure Science, University of bylon, bylon, Irq lon.edu.iq Domintion nd Independence on Squre Chessbord Abstrct- In this pper, new ide for the problems of independence nd domintion on chessbord is introduced. Two clssicl chessbord problems of independence nd domintion on squre chessbord with squre cells of size n re determined for some cses when two different types of pieces re used together. For independence, fixed number of the first type of pieces is plced on the bord with mximum number of pieces of the second type together. For domintion, fixed number of pieces of first type is plced on the bord with minimum number of pieces of the second type. The pieces which re used together in this pper re: kings with rooks, kings with bishops, nd rooks with bishops. Received on: 0/06/015 Accepted on: /0/017 eywords- Domintion, Independence, Squre chessbord, ings, ishops nd Rooks. How to cite this rticle: A.A. Omrn, Domintion nd Independence on Squre Chessbord, Engineering nd Technology Journl, Vol. 5, Prt, No.1, pp , Introduction In the chessbord there re six kinds of pieces. Let "P" be piece of ny kind on the chessbord. There re two clssicl chessbord problems; one of them is by plcing mximum number of pieces of single kind, such tht ech piece does not ttck other pieces. This problem is clled independence problem, nd the number of pieces tht stisfies this criterion is the independence number. Independence number for "P" kind is denoted by β(p). The other problem is by plcing minimum number of single kind P, such tht ll unoccupied positions re under ttck by t lest one of the plced pieces. This problem is clled domintion problem nd the number of pieces tht stisfies this criterion is clled domintion number of "P" nd denoted by γ(p). Previous studies were concerned with domintion nd independence problems with single kind of pieces only, while our current study is concerned in the sme problems but with two kinds of pieces t time. This study cn be used in gme theory or ny similr life problems. For squre chessbord (n n) the independence nd domintion numbers re determined for Rook "R", ishop "" nd ing "". They proved tht γ(r) = n, () = n, γ() = n+, lso, β(r) = n, β() = n nd β() = n+ (see [1], [] nd []). In [], JoeMio nd Willim proved tht γ(r) = min m, n} nd β(r) = min m, n}for m n Toroidl chessbord. In [5], the minimum number of rooks tht cn dominte ll squres of the STC is determined. In [6], the tringulr hexgon bord, in which the cells re hexgons nd the bord is tringle is considered. ishops ttck in stright lines through the vertices of their cells, rooks ttck long stright lines through the centers of the edges of their cells, nd queens hve both ttcks. The only generl upper bound they re ble to give on the independence number of the queens grph is by the rooks bound, which is n+1 for ll n. For n =,, 6, 7, 1, 16, 19, 5, 1, they found tht β = 1, nd for the other n 1, β = n+1. In [7],[8],[9],[10] nd [11] the independence nd domintion in Rhombus chessbord, isosceles tringulr chessbord nd cubic chessbord with squre cells re determined. In this pper, we pply the mening of the two clssicl problems on squre chessbord for two different types of pieces tht hve been chosen. The pieces of first type re plced on the chessbord, nd their number is fixed nd then the domintion or independence number of the other type is determined. Let n P be the number of pieces of the sme kind (P) which re chosen to hve fixed number. N P refers to the number of the cells which re ttcked by one piece (P) in ddition to the cell tht is occupied by the piece (P). n+1 The Chessbord The chessbord in this work is squre chessbord of size n with squre cells. Three types of pieces, rooks R, bishops nd kings re used Copyright 017 by UOT, IRAQ 68

2 Engineering nd Technology Journl Vol. 5, Prt, No. 1, 017 with their usul moving or ttcking. Let the number of cells (squres) in side be the length of tht side. To simplify the form of our results, the mtrix form is used where r i denote the i th row mesured from bove to down, i = 1,,, n nd let c i denote the j th column mesured from left to right, j = 1,,, n. Let the cell (squre) of i th row nd j th column is denoted by s i,j, i = 1,, n, nd j = 1,, n. Theorem.1 [8] In isosceles tringulr chessbord of size n the domintion number of king pieces (γ()) is given by γ() = (1) n 1 ( n 1 ( n 1 +1) (see Figure 1(); n = 6 ). + ), if n 1 is even, if n 1 is odd } Domintion nd Independence of pieces with fixed number of R pieces. In the squre chessbord in this work let ( n r ) be the number of rook pieces nd let γ(, n r ) nd β(, n r ) be the domintion number nd independence number of king pieces () with fixed number of Rook pieces (R) respectively. Theorem.1. The domintion number of king pieces with fixed pieces( n r ) of rooks in squre chessbord of size n is given by γ(, n r ) = n n r Proof. If the pieces re distributed in the chessbord by plcing the R pieces in the cells s i,i, i = 1,,, n 1 in order to keep the minimum number of domintion (ny other plcing mke prtition on the chessbord) then the mximum of N r of these pieces is gotten, nd squre chessbord of length n n r of cells which re not ttcked by R pieces. (see Figure (b); n = 9). Since, from [5] the domintion number of king in squre chessbord size n is γ() = n+ wich eequl to n. Hence, γ(, n r ) = n n r b Figure 1 This result is used for determining domintion in the isosceles semi tringulr chessbord of size n. In this chessbord, there re two equl sides nd every i th row contin i cells s shown in Figure 1(b), n = 6 Proposition.. In isosceles semi tringulr chessbord of size n the domintion number of king pieces (γ()) is given by R R R n n r γ() = n ( n + ), if n is even ( n +1), if n is odd } () Figure To prove ech of the following theorems we must refer to (1) The remining P pieces in ny step of the proof which we shll denoted by z equl to the difference between n r nd the mximum number n r of the previous step. 96

3 Engineering nd Technology Journl Vol. 5, Prt, No. 1, 017 Figure () The blck cells which pper in our figures represent the plces of pieces. Theorem.. For n 1 the independence number of pieces with fixed number n r of R pieces, where 1 n r n is given by β(, n r ) = n (n r + 1), if n 1, ( mod ) } ( n ) (n r ), if n 0, ( mod ) Proof. The independence number of the king piece on squre chessbord is β() = n ccording to [6], where the independence distribution of pieces is s shown in Figure (); where n = 1 (shded cells). The ide is to distribute the pieces of R such tht, it ttck minimum number of pieces to keep mximum number of pieces on the chessbord, nd no piece ttcks ny of R pieces. The column nd row for ny cell of the piece contin n pieces of where n 1, ( mod ), nd n 1 pieces of where n 0, (mod ). For this ide, we hve two cses tht depend on the length n of the squre chessbord nd s follows. (i) If n 1, ( mod ): the suitble cell to plce the first R piece is s,, since the neighborhood is the minimum. In this plce the first R piece does not ttck ny of the pieces, but there re four pieces djcent to it. So we must remove the djcent pieces nd we denote ech cell of these pieces by "x" s shown in Figure (b); for n = 1. To tke dvntge from the first R piece, we plce the second R piece in cell such tht N(R) is shred with the most number of pieces in the neighborhood of the first R piece. Now, if the neighborhood of the second R piece is shred with two pieces from the neighborhood of the first R piece, then the plce of the first nd second R pieces re in the sme column or the sme row nd this men tht the two R pieces re not independent. So the most shred pieces between the neighborhood of the first nd second R piece is one piece, thus suitble cell to plce the second R piece is s,. The second R piece is djcent to other three pieces, so we must remove these pieces nd we denote ech cell of the removble king by "x". Continue to plce other R pieces in the cells s i,i, i = 1,,.., n in order (see Figure (b); n = 1). Then we get β(, n r ) = n (n r + 1). (ii) If n 0, ( mod ): In the sme mnner in the previous cse, the suitble cells to plce the R pieces re s i+,i+1, i = 0,,.., n 1, in order. The first R piece does not ttck ny piece nd djcent to one piece, so this piece must be removed. The second R piece is djcent to other three pieces, so these pieces must be removed (see Figure ()). Continue to this procedure until n r = n. Therefore, β(, n r) = n (n r ). The following exmple illustrting the bove theorem for different vlues of n r. Exmple.. 1) n = 1, n r = 6, implies β(, 1) = 0, (see Figure ). ) n = 1, n r =, implies β(, 8) = 9 (see Figure ).

4 Engineering nd Technology Journl Vol. 5, Prt, No. 1, 017 Figure Domintion nd Independence of pieces with fixed number of pieces Let the domintion (Independence) number of pieces with fixed number n b of pieces re denoted by γ(, n b ) ( β(, n b )). Theorem.1. The domintion number of pieces with fixed pieces n b in squre chessbord is given s follows. For n >, let m = ( n n b ), if μ 1 = μ = m 1 μ = m ( m +1) ( m + ), if m is even },, if m is odd ( m 1 + ), if m 1 is even ( m 1 +1) m+ 1, if m 1 is odd } ( m ), if m + 1 is even ( m+ 1 +1) μ = nd μ 5 = m m 1 ( m 1 +1), if m + 1 is odd } ( m 1 + ), if m 1 is even }, if m 1 is odd ( m + ), if m is even ( m +1), if m is odd },,, then γ(, n b ) = μ μ 5 + μ + μ μ + μ μ + μ 1 + μ 5., if n nd n b re odd, if n is odd nd n b is even }, if n is even nd n b is odd, if n nd n b re even Proof. In ech cses below 1 n b n1. Cse 1) If n is odd: There re two cses tht depend on n b s follows. (i) If n b is odd, then n b pieces of re plced in the middle of the column c n, since these cells give the mximum of N b, so tht the minimum number of pieces is gotten. These pieces of re distributed on the cells which re not ttcked. These cells from four isosceles tringulr chessbord with sme size ( n n b ) in the squre chessbord s shown in Figure 8(); n = 11. Using equtions (1) nd () in section, the result is gotten. (ii) If n b is even, gin the n b pieces of re plced in the middle of the column c n, nd the cells which re not ttcked by these pieces form four shpes. Two of these shpes form two semi isosceles tringulr chessbord with the size ( n n b ), nd the other two form two isosceles tringulr chessbord with different sizes ( n n b ) nd ( n n b + 1), s shown in Figure 8(b); n = 11. y using equtions (1) nd (), the result is obtined.

5 Engineering nd Technology Journl Vol. 5, Prt, No. 1, 017 b Figure 8 b Figure 9 Cse ) If n is even: gin there re two cses with size ( n depend on the vlue of n b s follows. n b ), nd the other two re isosceles semi tringulr chessbord one of them (iii) If n b is odd, then n b pieces of in the middle of column c n +1, re plced. So, the mximl of N with size ( n b n b ) nd the other with size (n nd the minimum number of pieces re obtined. n b 1) s shown in Figure 9(b). Using the These pieces re distributed on the cells which equtions (1) nd () the result is obtined. re not ttcked by n b pieces of. The cells which The following exmple illustrting the bove re not ttcked by n b pieces of form four theorem for different vlues of n nd n b. isosceles tringulr chessbords, two of them re of size ( n n b + 1) nd the others re of size ( n n b ) s shown in Figure 9(). y using the equtions (1) nd (), we get the result. (iv) If n b is even, gin we plce the n b pieces of in the middle of the column c n +1, the cells not ttcked by these pieces form four shpes. Two of these shpes re isosceles tringulr chessbord Exmple.. 1) n = 11, n b =, then γ(, ) = 16, (see Figure 8()) ) n = 11, n b =, then γ(, ) = 10, (see Figure 8(b)) ) n = 10, n b =, then γ(, ) = 1, (see Figure 9()) ) n = 10, n b =, then γ(, ) = 8, (see Figure 9(b))

6 Engineering nd Technology Journl Vol. 5, Prt, No. 1, 017 Theorem. The independence of pieces with fixed number n b of pieces is given by the following: (i) If n is odd, then β(, n b ) = n (n b + 1), if 1 n r n n (n b + ), if n < n r n 1 } (ii) If n is even, then β(, n b ) = ( n ) n b, if 1 n r n ( n ) (n b + 1), if n < n r n 1 } Proof. There re two cses tht depend on the vlue of n s follows. (i) If n is odd: two successive cses tht depend on n b re obtined s follows. () If 1 n b n, the distribution of the pieces is s before in Section, nd cell is looked for plesing one piece such tht the minimum number of pieces is ttcked by it. The suitble cells of this ide re one of s n,j ; j = 1,,, n, where no pieces re ttcked. For the first piece there re two djcent of pieces to this piece, nd there is one piece djcent to the second piece. So, the djcent pieces must be removed. Continue with the sme mnner for the other pieces until reching cell s n, n s shown in Figure 10(); n = 11. According to [6], where β() = n, hence, (n b + 1). β(, n b ) = n (b) n < n r n 1, by plcing the remining z pieces in the cells s 1,j ; j = 1,,, n in order (see Figure 10; n = 11), s in (i), we get β(, n b ) = n (n b + ). (ii) If n is even: Agin we hve two successive cses depend on n b s follows: () If 1 n b n, the distribution of the pieces is s before in Section. Now, the next step is looking for cell to plce one piece such tht the minimum number of pieces is ttcked by it. The suitble cell of this ide is one of the cells s 1,j ; j = 1,,, n in order. There is one piece djcent to the first piece; therefore removing of this piece is necessrily. y continuing with other bishop pieces s the sme mnner (see in Figure 10(b); n = 1). Thus, β(, n b ) = n n b. (b) If n < n b n 1, by plcing the remining z pieces in the cells s n,+j ; j = 0,1,, n in order. There re two pieces djcent to the first piece, so we must remove these pieces. There is one djcent piece to the second piece, so we must remove this piece. We continue with the sme mnner for the other pieces until reching the cell s n,n 1 s shown in Figure 10(b); n = 1. Hence, we get β(, n b ) = n (n b + 1). b Figure 10 Note.. When n 1 < n r n, it is not esy to find generl formul for β(, n b ). The following exmple illustrting the bove theorem for different vlues of n nd n b.

7 R Engineering nd Technology Journl Vol. 5, Prt, No. 1, 017 Exmple.5. 1) n = 11, n b = 10, then β(, 10) =, (see Figure 10()) ) n = 1, n b = 11, then β(, 11) =, (see Figure 10(b)) R n n r b Figure 11 5 Domintion nd Independence of pieces with fixed number of R pieces. We denote the domintion (Independence) number of pieces with fixed number n r of R pieces by γ(, n r ) ( β(, n r )). Theorem 5.1. The domintion (independence) of pieces γ(, n r ) (β(, n r )) with fixed number n r of R pieces is given by the following: (i) γ(, n r ) = n n r (ii) β(, n r ) = (n n r ) Proof. The ide is to plce n r pieces of R nd then distributing the pieces to get the domintion (independence) number of the pieces together with fixed number n r. (i) We look for cell to plce the first R piece such tht we obtin the mximum of N R. Therefore we cn distribute minimum number of pieces such tht they re not ttcked by the first R piece. The suitble cells for (n 1) R pieces distribution re the min digonl of the chessbord s i,i, i = 1,.., r in order. The vcuum cells which re not ttcked by R pieces form squre chessbord of length n n r s shown in Figure 11(); n =10. We know tht γ() = n (see [6]), where n is the size of squre chessbord. Therefore by our distribution we get γ(, n r ) = n n r. (ii) Similrly s in (i) we plce the R pieces in the min digonl of squre chessbord s i,i, i = 1,.., r in order. The cells which re not ttcked by these pieces form s squre chessbord of length n n r. We know tht β() = n n n r (see [6]) where n is the size of the squre chessbord. So we distribute (n n r ) - in the chessbord of size n n r, but we must remove the piece from the min digonl, since it is ttcked by the R pieces (see Figure 11 (b), 11(c); n = 10). Thus we get β(, n r ) = (n n r ). The following exmple illustrting the bove theorem for different vlues of n r. Exmple 5.. 1) n = 10, n r = 1, then γ(,1 ) = 9, (see Figure 11()). ) n = 10, n r = 1, then β(, 1) = 15, (see Figure 11(b)). Open problems for two different types of pieces Find the generl formul of ech of the following numbers γ(r, n k ), γ(, n k ), γ(, n r ), β(r, n k ), β(, n k ) nd β(, n r ). ReferencesNo.6, [1] J. J. Wtkins, C. Ricci," ings Domintion on Torus ", Colrdo College, Colrdo Spring, Co [] O. Fvron," From Irredundnce to Annihiltion : A rief Overview of Some Domintion Prmeters of Grphs", Sber, Universidd de Oriente, Venezuel,1(), 6-69, 000. [] R.Lskr, C.Wllis,"Chessbord Grphs, Relted Designs, nd Domintion Prmeters", Journl of Sttisticl Plnning nd Inference, (76), 85-9,1999. [] J. DeMio nd W. Fust, "Domintion nd Independence on the Rectngulr Torus by Rooks nd ishops ", Deprtment of Mthemtics nd Sttistics ennesw Stte University, ennesw, Georgi, 01, USA [5] H. Chen, T. Hob,"The Rook Problem on Swtoothed Chessbords", Applied Mthemtics Letters,( 1),1 17,008. [6] H.Hrborth, V. ultn,. Nyrdyov, Z. Spendelov, "Independence on tringulr

8 Engineering nd Technology Journl Vol. 5, Prt, No. 1, 017 hexgon ords", in: Proceedings of the Thirty-Fourt Southestern Interntionl Conference on Combintorics, Grph Theory nd Computing,160, 15-, 00. [7] A. A.Omrn,et l., " Independence in Isosceles Tringle Chessbord ", Applied Mthemticl Sciences, 6 (11),651-65, (01). [8] A. A. Omrn et l., "Domintion in Isosceles Tringle Chessbord", Mthemticl Theory nd Modeling Journl, 7(),71-80, 01. [9] A. A. Omrn, et l., "Independence in Rhombus Chessbord ", Archive Des Sciences Journl,(66), 8-58, 01. [10] A. A. Omrn, et l., "Domintion in Rhombus Chessbord ", Journl of Asin Scientific Reserch, (5), 8-59, 01. [11] A. A. Omrn, Domintion nd Independence in Cubic Chessbord, E Eng. nd Tech. Journl, (1),Prt (), 6-59, 016. Author biogrphy Ahmed Abed Ali Omrn Deprtment of Mthemtics, Collegeof Eduction for Pure Science, University of bylon, bylon, Irq Prgrph He published previously in the field of combintorics. He is member of hwrizmi legue.