CS6515 EXAM 2 ACTUAL EXAMINATION
2026 QUESTIONS WITH ANSWERS GRADED
A+
⩥ master theorem. Answer: -if f(n) = O n^(logba -∈) for some const ∈>0
then T(n) =θn^(logb a)
-if f(n) =θn^(log b a ) then T(n)=θn^(logba) logn
-if f(n) =Ω(n^logba + ε) for some constant ε>0 and if af(n/b) ≥ cf(n) for
some constant c<1 and all sufficiently large n, then T(n) = θf(n)
f(n) dominates g(n) iff grows polynomially faster than g(n). this is a
stronger condition than f(n) = ωg(n)
does not apply for the gap between the tree cases (little o and little
omega)
⩥ when master theorem does not apply. Answer: a must be >= 1, b must
be >1
(they don't necessarily need to be integers)
,can check this by the limit test and leaving the + ∈ and if there is not an
∈ you can pick to make it polynomially faster, then the rule does not
apply
f(n) is negative
ex. nlogn / n^1+e
cuz both have n^1 in it
if a = 2^n (cuz not constant)
if constant work is subtracted
⩥ polynomially larger? n^2 vs. n. Answer: yes
⩥ polynomially larger? n^2logn vs. n^2. Answer: no
⩥ polynomially larger? n^3 log n vs. n^2. Answer: yes
⩥ polynomially larger? n^2.001 vs. n^2. Answer: yes
,⩥ polynomially larger? nlogn vs. n^log4 3. Answer: yes
⩥ applying the master theorem given recurrences. Answer: coefficient is
a, division is b and function is constant work
compare n^logb a vs. f(n)
⩥ special rules for master theorem. Answer: -case 2 only:
this happens when log_b a is equal to the degree of n in f(n). if there is
no log then master method does not apply. if there is a log, look at the
power.
if the power of the log is > -1, then T(n) = theta log _ b a * log^k+1 n
where k is the power of the log
if the power of the log = -1 then T(n) = theta log_ b a log log n
if the power of the log < -1 then T(n) = theta log_ b a
⩥ proof by induction on the binary search. Answer: base case: on the 0
iteration, x exists in A or x never existed in A at all
, inductive hypothesis: x exists in A iff x lies in the subarray A[L...H]
inductive step:
if(A[mid] <= x)
L<- mid
else
H<- mid-1
if A[L] = x
return L
⩥ heap sort. Answer: insert n inputs into the heap. call extractmin n
times. values sorted least to greatest
⩥ merge sort. Answer: based on linear merge of 2 sorted arrays. divide
and conquer algorithm
⩥ selection sort. Answer: goes thru and finds the smallest element and
brings it to the front of the array
⩥ running time of bubble sort. Answer: n^2
2026 QUESTIONS WITH ANSWERS GRADED
A+
⩥ master theorem. Answer: -if f(n) = O n^(logba -∈) for some const ∈>0
then T(n) =θn^(logb a)
-if f(n) =θn^(log b a ) then T(n)=θn^(logba) logn
-if f(n) =Ω(n^logba + ε) for some constant ε>0 and if af(n/b) ≥ cf(n) for
some constant c<1 and all sufficiently large n, then T(n) = θf(n)
f(n) dominates g(n) iff grows polynomially faster than g(n). this is a
stronger condition than f(n) = ωg(n)
does not apply for the gap between the tree cases (little o and little
omega)
⩥ when master theorem does not apply. Answer: a must be >= 1, b must
be >1
(they don't necessarily need to be integers)
,can check this by the limit test and leaving the + ∈ and if there is not an
∈ you can pick to make it polynomially faster, then the rule does not
apply
f(n) is negative
ex. nlogn / n^1+e
cuz both have n^1 in it
if a = 2^n (cuz not constant)
if constant work is subtracted
⩥ polynomially larger? n^2 vs. n. Answer: yes
⩥ polynomially larger? n^2logn vs. n^2. Answer: no
⩥ polynomially larger? n^3 log n vs. n^2. Answer: yes
⩥ polynomially larger? n^2.001 vs. n^2. Answer: yes
,⩥ polynomially larger? nlogn vs. n^log4 3. Answer: yes
⩥ applying the master theorem given recurrences. Answer: coefficient is
a, division is b and function is constant work
compare n^logb a vs. f(n)
⩥ special rules for master theorem. Answer: -case 2 only:
this happens when log_b a is equal to the degree of n in f(n). if there is
no log then master method does not apply. if there is a log, look at the
power.
if the power of the log is > -1, then T(n) = theta log _ b a * log^k+1 n
where k is the power of the log
if the power of the log = -1 then T(n) = theta log_ b a log log n
if the power of the log < -1 then T(n) = theta log_ b a
⩥ proof by induction on the binary search. Answer: base case: on the 0
iteration, x exists in A or x never existed in A at all
, inductive hypothesis: x exists in A iff x lies in the subarray A[L...H]
inductive step:
if(A[mid] <= x)
L<- mid
else
H<- mid-1
if A[L] = x
return L
⩥ heap sort. Answer: insert n inputs into the heap. call extractmin n
times. values sorted least to greatest
⩥ merge sort. Answer: based on linear merge of 2 sorted arrays. divide
and conquer algorithm
⩥ selection sort. Answer: goes thru and finds the smallest element and
brings it to the front of the array
⩥ running time of bubble sort. Answer: n^2
