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A realm is a new security layer that provides containment and eliminates inadvertent access based on system ANY privileges System ANY privileges are required for many application and administration scenarios These privileges are heavily used in operating a database, but the privileges have no boundaries and therefore must be restricted The security gap exists between the database objects controlled by object privileges and the system controlled by system ANY privileges Realms were designed as a security mechanism that establishes boundaries around a set of objects that limit the use of system ANY privileges to authorized realm users or roles Therefore, any attempt to access a realm protected object by an unauthorized user with system ANY privileges will be blocked and raise a realm violation Access based on object-level privileges are not considered inadvertent and therefore are not enforced by the realm Objectlevel privileges are explicit permissions on individual objects that can be verified through database dictionary views Realms are a simple and critical security measure that address many of the issues we face today regarding compliance and separation of duties for different job responsibilities They also serve as an affective defensive measure for cyber security by limiting the extent to which system privileges can be exploited by an attacker Therefore, realms not only provide another layer of security, but they add another level of defense DBAs who are responsible for the health and well-being of production databases should consider the serious nature of realm violations, either legitimate or not For example, an application needing additional accesses is legitimate and a malicious attempt to access data within the database is not

xb = incsearch(func,xmin,xmax,ns) function incremental search root fo-aLor incsearch: xb - incsearch (func, xmin, xmax,ns) : % finds brackets of x that contain sign chattges % on an interval of a function inpuL: furLc - name of funct-iorr % Y xni -, xmax cndpoitr(s oI inrerva] (default = 50) ns = number of subintervals % outpuL: xb(k,1) is the lower bound of Lhe kth % 2 xb (k,2 ) is tire upper bound of t-he kth If no brackefs found, xb - Ll % if nargin < 4, ns = 50; encl %rf ns blank sigtn chatrge sign change

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search Z Increm ntal x = I i n s p a c e ( x m r - n ,x m a x , n s ) ; f - func (x) ; unless sign c|ranqe det ec t ed nb = 0r :{b - i I ; exb is: null for k = 1; length (x) -1 if sign(f (k)) ^-- sign(f (k+-L)) licheck for sign chanqe nb = nb + 1; x{kl; xbrnl,l) xb(t'tb,2) = x (k+1); end end jf isempt),,(xb) displzr-v that no brackets disp ( 'no brackets forind' ) or increase ns' ) disp ( 'check rnterval disp('nr:mber disp (nb) end ol brackets:')

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ProblemStotement UsetheM-flle i ncsearch (Fig5a)to identifybrackets within the interval[3, 6] fbr the function: J G ) : s i n (l 0 r ) * c o s ( 3 r ) (50) is Solution The MATLAB session usingthe detaultnumberof intervals

>> incscarch(@r of possiblc 5 afnS 3 2 1 , 49 33061 3 /34-t 46531 \{:i2 l t3061 336't3 3 t9\9 4'-t14j, a6939 sin(10*x) rooLs: +cos (3*x),3, 6) numbcr

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