An Algorithm to Solve the Games under Incomplete Information

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1 Annals of Pure and Applied Maematics Vol. 10, No.2, 2015, ISSN: X (P), (online) Published on 30 October Annals of An Algorim to Solve e Games under Incomplete Information S. C. Sharma 1 and Ganesh Kumar 2 Department of Maematics, Rajasan University Jaipur , India. 1 sharma-sc@uniraj.ernet.in 2 ganeshmandha1988@gmail.com Received 29 September 2015; accepted 16 October 2015 Abstract. Classical game eory assumes at e structure of e game is common knowledge among e players but in most of e real life game eoretic problems incomplete information available about payoffs of agents. But it is possible to determine lower and upper bounds on payoffs. In is paper an algorim to determine e value of game in fuzzy environment, using octagonal fuzzy numbers is proposed. Octagonal fuzzy numbers are converted into interval numbers using α -cut meod. The effectiveness of proposed algorim is illustrated by means of a numerical example. Keywords: Octagonal fuzzy numbers, interval numbers, saddle point, fuzzy payoffs AMS Maematics Subject Classification (2010): 91A80, 90C70 1. Introduction Fuzzy numbers are widely used in e research on fuzzy set eory. In order to identify e preference ranking of e fuzzy numbers, one such numbers need to evaluated and compared wi e oer fuzzy numbers. This is not easy in practical situations. Game eory had its beginnings in e 1920s, and significantly advanced at Princeton University rough e work of Nash [3,4]. The meod of finding e fuzzy game value using intervals as elements of matrix is explained in [5, 7]. By defining a ranking of octagonal fuzzy numbers, transportation problems solved in [1]. Also using e ranking of ese fuzzy numbers in [2] a two person zero sum game is solved. Definitions of intervals and interval arimetic are studied in [6]. By comparing e intervals and using dominance principle in [8] a fuzzy environment we find e fuzzy game value of e interval valued matrix. 2. Octagonal fuzzy numbers An octagonal fuzzy number denoted by A % w is defined to be e ordered quadruple, A % w = ( l1( r), s1( t), s2( t), l 2( r)), where r [0, k] and t [ k, w], where (1) l 1 ( r) Is a bounded left continuous non decreasing function over [0, w1],[0 w1 k] (2) s 1( t) Is a bounded left continuous non decreasing function over [ k, w2],[ k w2 w] 221

2 S.C. Sharma and Ganesh Kumar (3) s 2( t) Is a bounded left continuous non increasing function over [ k, w2],[ k w2 w] (4) l 2( r) Is a bounded left continuous non increasing function over [0, w1],[0 w1 k] If w=1, en e above defined number is called a normal octagon number. A fuzzy number A % is a normal octagon fuzzy number denoted by ( a1, a2, a3, a4, a5, a6, a7, a8 ) where a1 a2 a3 a4 a5 a6 a7 a8 are real numbers and its membership function µ % ( ) is given as following A x where 0 < k < 1. 0, x < a1 x a a a k, a2 x a3 x a a a ( x) = 1, x a x a a k, a6 x a7 a8 x k, a x a a8 a7 0, x a8 1 k, a1 x a k + (1 k), a3 x a4 4 3 µ A% a 4 a5 6 k + (1 k), a5 x The interval number system X, Y is e set of real numbers given by, Closed interval denoted by [ ] [ X, Y ] = { x : X x Y} Alough various oer types of intervals (open, half-open) appear roughout maematics, our work will center primarily on closed intervals. In is paper, e term interval will mean closed interval. We will adopt e convention of denoting intervals and eir endpoints by capital letters. The left and right endpoints of an interval X will be denoted by X and X, respectively. Thus, X = [ X, X ] Two intervals X and Y are said to be equal if ey are e same sets. Operationally, is happens if eir corresponding endpoints are equal: X = Y X = Y And X = Y We say at X is degenerate if X = X. Such an interval contains a single real number x. By convention, we agree to identify a degenerate interval [ X, X ] wi e real number x. Operations on intervals: Intersection of e intervals : X Y = [max{ X, Y}, min{ X, Y}] a 6 222

3 An Algorim to Solve e Games Under Incomplete Information Union of intervals : X Y = [min{ X, Y},max{ X, Y}] Interval hull of intervals : X Y = [min{ X, Y},max{ X, Y}] Intersection plays a key role in interval analysis. If we have two intervals containing a result of interest, regardless of how ey were obtained, en e intersection, which may be narrower, also contains e result. In general, e union of two intervals is not an interval. However, e interval hull of two intervals is always an interval and can be used in interval computations. We have: X Y X Y The wid of an interval X is defined and denoted by: ω ( X ) = X X The absolute value of X, denoted by X, and is e maximum of e absolute values of its endpoints: X = max{ X, X } note at x X, x X The midpoint of X is given by: 1 m( X ) = ( X + X ) 2 Order Relations for Intervals: X < Y means at X < Y Transitive order relation for intervals is set inclusion: X Y Y X and X Y The sum of two intervals : X + Y = { x + y : x X, y Y} Subtraction of two intervals : X Y = { x y : x X, y Y} Product of two intervals : X Y = { xy : x X, y Y} Quotient of two intervals : X / Y = { x / y : x X, y Y}. α -cut of an octagonal fuzzy number The α -cut of an normal octagonal fuzzy number A % = ( a1, a2, a3, a4, a, a6, a, a ) α [0,1] is given by 4. Algorim [ A% ] α α α [ a1 + ( a2 a1),[ a8 ( a8 a7)], α [0, k] k k = α k α k [ a3 + ( a 4 a3), a6 ( a6 a5)], α ( k,1] 1 k 1 k Step 1: formulate e given problem in e form of octagonal fuzzy number for Step 2: convert e octagonal fuzzy numbers into interval numbers using α -cut meod to get e interval matrix. Step 3: For each row { 1,2,...,m } find e entry entries in e i row. g at is less an or equal to all oer ij 223

4 S.C. Sharma and Ganesh Kumar Step 4: For each column { 1,2,...,n } find e entry oer entries in e j column. Step 5: Determine if ere is an entry g at is greater an or equal to all ij g at is simultaneously e minimum of e ij i row and e maximum of e j column which is known as saddle point and value of saddle interval is known as value of game, if any of e above values cannot be found en saddle point does not exist. Then we go for dominance principle. Step 6: if all e elements of e elements of j row en Step 7: if all e elements of elements of r column en i row are less an or equal to e corresponding j row will be dominated row. k k column are greater an or equal to e corresponding column will be dominated column. Step 8: dominated row or column can be deleted to reduce e size of payoff matrix as optimal strategies will remain unaffected. A given strategy can also said to be dominated if it is inferior to an average of two or more oer pure strategies. More generally if some convex linear combination of some rows dominates e i row en i row will be deleted. Similar arguments follow for columns. Step 9: using e above arguments matrix can be reduced to a simple matrix for game value can be evaluated easily. Step 10: if ere is no saddle point and dominance principle also failed en we define fuzzy membership of an interval being a minimum and a maximum of an interval vector and en we define e notion of a least and greatest interval in as defined in next step. Step 11: The binary fuzzy operator of two intervals X and Y returns a real number between 0 and 1 as follows: 1, X = Y; X Y, X Y; X < Y < X < Y 0, Y X, X Y; Y < X < Y < X X Y = Y X, Y X X Y, ω( Y ) > 0, X Y ω( Y ) ω( X ) Y X, X Y Y X, ω( X ) > 0, X Y ω( X ) ω( Y ) The binary fuzzy operator ± of two intervals X and Y is defined as, X ± Y =1If X = Y and X ± Y = 1 ( X Y ) oerwise, i.e. 224

5 An Algorim to Solve e Games Under Incomplete Information 1, X = Y; Y X, X Y 0, X Y, X Y; X < Y < X < Y ; Y < X < Y < X X Y X ± Y =, Y X X Y, ω( Y ) > 0, X Y ω( Y ) ω( X ) X Y, X Y Y X, ω( X ) > 0, X Y ω( X ) ω( Y ) 5. Numerical example If pay off matrix of a game is given as following (0,1,2,3,4, 5,6, 7) ( 8, 9,10,11,12,13,14,1 5) (4, 5,6, 7, 8, 9,10,11) (2,3,4, 5,6, 7, 8, 9) (11,12,14,1 5,16,17,18,21) ( 5,6, 8, 9,10,11,12,1 5) (3,6, 7, 8, 9,10,12,13) ( 5,6, 8,10,12,13,15,1 7) (1,3, 5,6, 7, 8,10,12) Now we will use α -cut meod to convert given octagonal fuzzy numbers into intervals. Using, α α [ a1 + ( a2 a1),[ a8 ( a8 a7)], α [0, k] k k [ A% ] α = α k α k [ a3 + ( a 4 a3), a6 ( a6 a5)], α ( k,1] 1 k 1 k For is problem taking k=0.5 and α =1, we get following, Octagonal fuzzy number Corresponding interval number (0, 1, 2, 3, 4, 5, 6, 7) [3, 4] (8, 9, 10, 11, 12, 13, 14, 15) [11, 12] (4, 5, 6, 7, 8, 9, 10, 11) [7, 8] (2, 3, 4, 5, 6, 7, 8, 9) [5,6] (11, 12, 14, 15, 16, 17, 18, 21) [15, 16] (5, 6, 8, 9, 10, 11, 12, 15) [9, 10] (3, 6, 7, 8, 9, 10, 12, 13) [8, 9] (5, 6, 8, 10, 12, 13, 15, 17) [10, 12] (1, 3, 5, 6, 7, 8, 10, 12) [6,7] Hence we get e following interval matrix, 225

6 S.C. Sharma and Ganesh Kumar [3,4] [11,12] [ 7, 8] [ 5,6] [15,16] [ 9,10] [ 8, 9] [10,12] [6, 7] First of all we check weaer saddle interval exist or not for is, min{[3,4],[11,12], 7, 8 ]} = [3,4],min{[ 5,6],[1 5,16],[ 9,10]} = [ 5,6],min{[ 8,9],[10,12],[6,7]}=[6,7] max{[3,4],[ 5,6],[ 8, 9 ]} = [ 8, 9],max{[11,12],[1 5,16],[10,12]} = [15,16],max{[ 7, 8],[ 9,10],[6, 7 ]} = [ 9,10] From above discussion saddle interval does not exist. Since all e interval numbers of first row are less an or equal to e interval numbers of second row hence first row is dominated by second row. So we can delete first row. Now all e interval numbers of second column are greater an or equal to e interval numbers of first column hence second column is dominated by first column. So we can delete second column. Remaining matrix will be, [ 5,6] [ 9,10] [ 8, 9] [6, 7] Now firstly we are looking for least intervals, min{[ 5,6] p [ 9,10],[ 5,6] p [ 8, 9],[ 5,6] p [6, 7 ]} = min{1,1,1} = 1 min{[ 9,10] p [ 5,6],[ 9,10] p [ 8, 9],[ 9,10] p [6, 7 ]} = min{0,0,0} = 0 min{[ 8, 9] p [ 5,6],[ 8, 9] p [ 9,10],[ 8, 9] p [6, 7 ]} = min{0,1,0} = 0 min{[6, 7] p [ 5,6],[6, 7] p [ 9,10],[6, 7] p [ 8, 9 ]} = min{0,1,1} = 0 Hence max{1, 0, 0, 0} = 1, is corresponding to interval [5, 6]. For greatest intervals, max{[ 5,6] f [ 9,10],[ 5,6] f [ 8, 9],[ 5,6] f [6, 7 ]} = max{0,0,0} = 0 max{[ 9,10] f [ 5,6],[ 9,10] f [ 8, 9],[ 9,10] f [6, 7 ]} = max{1,1,1} = 1 max{[ 8, 9] f [ 5,6],[ 8, 9] f [ 9,10],[ 8, 9] f [6, 7 ]} = max{1,0,1} = 1 max{[6, 7] f [ 5,6],[6, 7] f [ 9,10],[6, 7] f [ 8, 9 ]} = max{1,0,0} = 1 Hence min{0,1,1,1} = 0, is corresponding to interval [5, 6]. So from minimax eorem [5, 6] is saddle interval. 6. Conclusion In e above discussion an algorim for solving e games having payoffs in form of octagonal fuzzy numbers is suggested. A numerical example is given to clarify e proposed algorim. Many problems in e present scenarios having competitive situations require best strategy for e agents. These competitive situations may be seen in business, military battles, sports, elections, advertising, marketing and different cases of conflict. 226

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