1d:T7b3,# Support of LR fuzzy number To determining the support of a LR fuzzy number one can use from this function. In other words, the `support` function is able to compute the smallest and biggest values $x$ for which $\mu(x)>0$. ```r support(M, Left.fun = NULL, Right.fun = NULL) ``` ## Arguments - `M`: A LR, RL or L fuzzy number - `Left.fun`: The left-shape function which usually defined before using LRFN.plot (see examples in bellow) - `Right.fun`: The right-shape function which usually defined before using LRFN.plot (see examples in bellow) ## Returns The "support" function return a interval-valued vector in which the membership function value of LR fuzzy number is bigger than zero. ## Author(s) Abbas Parchami ## Examples ```r Left.fun = function(x) { (1-x)*(x>=0)} Right.fun = function(x) { (exp(-x))*(x>=0)} T = LR(1, 0.6, 0.2) support(T) LRFN.plot( T, xlim=c(-5,20), lwd=2, lty=3, col=4) N = RL(3, 0.5, 2) support(N) Left.fun = function(x) { (1-x)*(x>=0)} M = L(2,4,3) support(M) Left.fun = function(x) { (1-x^2)*(x>=0)} Right.fun = function(x) { (exp(-x))*(x>=0)} support( LR(17,5,3)) ## The function is currently defined as function (M, Left.fun = NULL, Right.fun = NULL) { range1 = M[1] - M[2] - M[3] - 100 range2 = M[1] + M[2] + M[3] + 100 x = seq(range1, range2, len = 2e+05) if (M[4] == 0) { y = Left.fun((M[1] - x)/M[2]) * (x <= M[1]) + Right.fun((x - M[1])/M[3]) * (M[1] < x) } else if (M[4] == 1) { y = Right.fun((M[1] - x)/M[2]) * (x <= M[1]) + Left.fun((x - M[1])/M[3]) * (M[1] < x) } else if (M[4] == 0.5) { y = Left.fun((M[1] - x)/M[2]) * (x <= M[1]) + Left.fun((x - M[1])/M[3]) * (M[1] < x) } supp = c() supp[1] = min(x[0 < y & y < 1]) supp[2] = max(x[0 < y & y < 1]) if (supp[1] == min(x)) { supp[1] = -Inf } if (supp[2] == max(x)) { supp[2] = +Inf } return(supp) } ```

support function