Algorithm Correctness and Running Time, CNF and DNF, Derivations, and Exponential Time Hypothesis

Algorithm Correctness and Running Time

The correctness of an algorithm depends on the input.Therefore to produce desired results,we have to enter the input expected by algorithm else it will take more time to execute.Despite correctness,running tine of an algorithm typically increases or grows with respect to input size,though it may also vary.Examples of running time include;constant,linear,exponential,logarithmic runtime and quadratic runtime.

(i).Constant runtime-It means that regardless of the input size provided to a function,its time complexity will not change.

(ii)Linear-Runtime will increase proportionally with respect to input size.

(iii)Exponential-Runtime increases exponentially since it's behaviour is expressed as 2^n where n is the input size.

2^n.n^o(1)

n≥3....

(2^3)((3^o(1))

Since distance covered is the input and time taken is the output,then we are likely to plot a curve of distance against time.

In the above salesman scenario,we ought to know that this algorithm implements both exponential and constant runtime.Therefore,exponential runtime will dominate since time for the touring salesman cannot be constant.Running time as the output is not more effective,as a result of improper or reliable input.The number of cities he visited is unable to calibrate the time taken,thus results to an exponential curve.

Nevertheless,exponential graph implies the running time was in worst case because regardless of number of cities visited,time complexity will not be intefered readily.

a).Write a CNF computing  d(x,y,z)

In computing most problems are solved with boolean formula of conjunctive and disjunctive normal forms.

The normal form of conjunctive  expresses formulas as conjunctions of clauses with an AND or OR.It takes a conjunction(^) if negations of falsifying truth assignments.

It's negation which is the normal form disnjunction, takes a disjunction(v) of all satisfying truth assignments.

Let us express d(x,y,z)=xy+yz

Here,we are to find the DNF and later negate it...

DNF of -d=(x^-y^z)v(x^-y^-z)v(-x^-y^z)

ΔCNF of d= -(DNF)....

Lets find the CNF by negating DNF

CNF=(-x+y-z)^(-x+y+z)^(x+y-z)...then solve to find the conjunctive literal.

CNF=1

b).

First step,lets derive the function of 'f' where it's variables  are (y,z,x).

Then let 'x' be constant.

d(xy)/dx=y

d(xy)=ydx

_/¯d(xy)=_/¯ydx

Since x.y=y _/¯dx,but x.y is the final output of d(y,x);

Δy_/¯dx=-(y_/¯dx)

Derivation:   y_/¯dx=yx^x

c).

Here,we are required to find the satisfiability element then equate to it's variable

{d,0}

CNF=x^y....Vi=_/¯dxy

_/¯{d,0,3}

d3=C1^C3...

Δd3=(x1+x2+x3)^(x2+y1)

d3-SAT...

Derivation:   Var(d3)=(x1+x3+y1)^n/3.

3.Recall that the size of a CNF is it's number of literal occurences.A minimal CNF computing a boolean function f is one of smallest size

a).Show that for fixed boolean inputs,if VLj=0 for some j=1...m then f(x,y,z)=0

Find the derivation of f(y,z,x)

d(y,z,x)/dz=yx

d(z,y,z))=y.x.dz....This happens to be the first equation (i)

Having f(jklm),we will have to break and solve it's satisfiable values

Let f(jklm)=jk+lm...expand

=jk+lm=(j+l)^(j+m)^(k+l)^(k+m)

=(z1+z2)^(z1<->jk)^(z2<->lm)

=(z1+z2)^(z1->jk)^(jk->z1)^(z2->lm)^(lm->z2)

b).

Let (x.-(y+z)+u=f(x,y,z)

     But f(x,y,z)=y.x.dz

   DNF= -1(y.x.dz)

=-x(y+z)+z(x+y)

=-x(y+z)+zx+yz

Conjuction of (x,y,z)will be (zx+yz)-x(y+z)

solution = (z.x+y.z)-1(x.y+x.z)

In your words what do you understand about Exponential Time Hypothesis

Time complexity is an algorithm used to represent the amount of required by the algorithm to run to completion.Definately,exponential time is a computational assumption that applies when finding out  the boolean expression of literals in conjunctive and disjunctive normal forms where polynomial time problem cannot equal non-deterrministic problem.

The exponential time hypothesis  posits that  non-deterninistic polynomial complete problems cannot be solved in sub-exponential time in the worst case.Therefore,if the exponential time hypothesis is true,it implies that polynomial isn't equal to non_deterministic but it is refered as a  strong statement enough to solve and prove the 'three-satisfiability' problem.It can be used to show that many computational problems are equivalent in complexity,in the sense that if one of them has a sub-exponential time then they all function.

In the same matters,to illustrate Exponential Time Hypothesis,we apply the satisfiability problem in that the boolean formula ø is in conjunctive normal form if it is a conjunction of clause where a clause is a disjunction of literals.Each literals either a boolean variable (x) or it's negation (-x).

This algorithm is composed of 'K-satisfiability' problem.K-Satisfiability tests whether a boolean expression in conjunctive normal form with at most k variables per clause can be made true by some assignment of boolean values to it's variables.

Likewise,the K-Satisfiability is non deterministic polynomial hard for k is greater or equal to 3 and thus P≠PN tells that there is no polynomial time algorithm for k-satisfiability and for k≥3.A satisfying assignment for ø is an assignment of true or false values.