# Worksheet400 Assignment 1

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School of Engineering and Information Technology ASSESSMENT COVER SHEET Student Name Mrinalini Padmanbhan Student ID S252757 Assessment Title Assignment 2 Unit Number and Title HIT400 Lecturer/Tutor Peter Shaw Date Submitted 5/10/2015 Date Received 23/9/2015 Office use only KEEP A COPY Please be sure to make a copy of your work. If you have submitted assessment work electronically make sure you have a backup copy. PLAGIARISM Plagiarism is the presentation of the work of another without acknowledgement. Students may use a limited amount of information and ideas expressed by others but this use must be identified by appropriate referencing. CONSEQUENCES OF PLAGIARISM Plagiarism is misconduct as defined under the Student Conduct By-Laws.…show more content…
Assuming T(0)=1 n/3^k=1 n= 3^k log_3⁡n =k T(n)= 4^log⁡n Thus the solution to the recurrence relation equation is T(n)= 4^log⁡n Question 2 Solve T^0(k ) = T^0 (k-1)+T^1 (k) ………………(1) T^1 (k)= T^0 (k-1)+(d-2)T^1 (k-1) ………………(2) Considering T^0(k ) = T^0 (k-1)+T^1 (k) When k= k-1, then T^0(k -1) = T^0 (k-2)+T^1 (k) ………………(3) Substituting (3) in (1) T^0(k ) = T^0 (k-2)+〖T^1 (k) +T〗^1 (k) ……………..(4) When k= k-2, then T^0(k -2) = T^0 (k-3)+T^1 (k) …………….(5) Substituting (3) in (1) T^0(k ) = T^0 (k-3)+T^1 (k)+〖T^1 (k) +T〗^1 (k) ……………..(4) Generalising the equation T^0(k ) = T^0 (k-m)+〖m*T〗^1 (k) Assuming T^0(0)= 0 k-m =0 k=m T^0(m ) = 〖m*T〗^1 (k) Considering T^1 (k)= T^0 (k-1)+(d-2)T^1 (k-1) When k= k-1, then T^1 (k-1)= T^0 (k-1)+(d-2)T^1 (k-2) ……………….(5) Substituting (5 ) in (3) T^1 (k)= T^0 (k-1)+(d-2)〖{T〗^0 (k-1)+(d-2)T^1 (k-2)} …………(6) When k= k-2, then T^1 (k-2)= T^0 (k-1)+(d-2)T^1 (k-3) ……………….(7) Substituting(7) in(6) T^1 (k)= T^0 (k-1)+(d-2)〖{T〗^0 (k-1)+(d-2)〖{T〗^0 (k-1)+(d-2)T^1 (k-3)}}………..(8) T^1 (k)= T^0 (k-1)+(d-2)T^0 (k-1)+(d-2)^2 T^0 (k-1)+(d-2)^3 T^1 (k-3) ………………..(9) Consider d-2 as a constant ‘a’ then T^1 (k)= T^0*a^0+a^1*T^0 (k-1)+a^2*T^0 (k-1)+a^3*T^1…show more content…
Example:SAT,Graph colouring, Hamilton Cycle NP – complete is a complexity class in which an instance of a NP- complete problem can be transformed into another instance of solved NP- complete problem in polynomial times.NP complete is a subset of NP and NP hard Example: Euler and Rudrata, walking salesman problem, integer linear programming Co-NP is a complexity class where an instance of a decision problem has a solution no, can be verified in polynomial times. Co-Np is the exact opposite of NP as it verifies no as the solution and NP verifies yes as the solution. Example: Prime and factorisation b) consider P≠ NP which is usually proved under the assumption that PH would not collapse as it P≠NP has a constraint of ∑_(i+1)▒ . Polynomial hierarchal collapse occurs when PH = PSPACE, since the PSPACE- complete language of true quantified Boolean formulas falls in the constraint of∑_i▒ .