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Inf2-IADS Sample Exam

创建日期：2024-04-11 17:18:23

所属分类：数学mathematics

标签： Inf2-IADS

Sample Exam

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Inf2-IADS Sample Exam

PART A:
1. (a) The diagram is essentially
[3,9,8,2,4,1,6,5]
[3,9,8,2] [4,1,6,5]
[3,9] [8,2] [4,1] [6,5]
[3] [9] [8] [2] [4] [1] [6] [5]
[3,9] [2,8] [1,4] [5,6]
[2,3,8,9] [1,4,5,6]
[1,2,3,4,5,6,8,9]
with the obvious arrows added.
[5 marks; roughly 2 for the splittings and 3 for the mergings.]
(b) Best, worst and average case times are all Θ(m ∗ 2m).
[1 mark each.]
(c) To mergesort a list of length 2m we only need an array of size 2 ∗ 2m (proof
is an easy induction). So space needed is Θ(2m).
[1 mark for the answer; 1 mark for any reasonable supporting point.]
2. This is the question about the different fi; we have f1(n) = (lg(n))
lg(n), f2(n) =
nlg(n), f3(n) = n
2 and f4(n) = n.
note: this is an example of where it’s important to know the “rules of logs”.
(a) We are asked to identify the i ∈ {1, 2, 3, 4} such that
fi(n) ∈ O(fj(n)) for all j = {1, 2, 3, 4}.
Basically we want the fi that is “asymptotically slowest”.
This function is f4 = n.
The (informal) justification for this . . .
• n certainly grows slower than f3 = n2.
• Also n certainly grows slower than f2 = nlgn (lg n ≥ 2 for every n ≥ 4)
• To consider relative growth of f4 against f1, we will take the lg of
both functions: then lg(f1(n)) = lg(lg(n)
lg(n)) which is equal to lg(n) ·
lg lg(n) by “rules of logs” and lg(f4(n)) = lg(n). Hence lg(f1(n)) =
lg(n) · lg(f4(n)) so lg(f1(n))) grows asymptotically faster than lg(f4(n)),
and f1(n) grows much much faster than f4(n). Hence f4 = O(f1).
[5 marks: 2 for identifying the correct i as 4, 1 mark each for the justification
wrt each j 6= 4].
Page 1 of 9
(b) We have to identify the k ∈ {1, 2, 3, 4} such that
fk(n) ∈ Ω(fj(n)) for all j = {1, 2, 3, 4},
and formally justify this.
The function fk is f2 = n
lg(n).
To prove the fk = Ω(fj) for another j we need to find a n0 ∈ N, c ∈ R>0
such that nlg(n) ≥ c · fj for all n ≥ n0.
• For j = 1, we can set c = 1, n0 = 2, then n ≥ lg(n) for n ≥ 2 and
hence nlg(n) ≥ lg(n)lg(n) (as the exponent lg(n) is a positive value > 1
for n ≥ 2)
• For j = 3, we set c = 1, n0 = 4, then for n ≥ 4 we have lg(n) ≥ 2 and
hence f3(n) = n
lg(n) ≥ n2.
• For j = 3, we set c = 1, n0 = 2, then for n ≥ 2 we have lg(n) ≥ 1 and
hence f3(n) = n
lg(n) ≥ n.
[5 marks: 2 for identifying the correct i as 4, 1 mark each for the formal
justification wrt each j 6= 4].
3. (a) The hash table will have better average case performance. If 10,000 integers
are stored, the average bucket size will be 5, so on average, 5 key comparisons
will be required for an insertion or unsuccessful lookup, and fewer for a
successful lookup. A binary tree storing 10,000 integers, even if perfectly
balanced, will have depth around lg 10, 000 ' 13, so a lookup will involve
12 comparisons on average.
However, the hash table has terrible worst case performance: if all 10,000
integers hash to the same value, a lookup might require 10,000 comparisons.
For red-black trees, the depth (hence the number of comparisons) will be
bounded by something around 2 lg 10, 000 ' 26.
[3 marks for discussion of average case, 3 marks for worst case. Ballpark
figures are acceptable, as in the above answer. Any other reasonable points
are acceptable.]
(b) The 1 is first added as the left child of 2, and initially coloured R [1 mark].
We then apply the ‘red uncle’ rule, turning 2 and 4 B and 3 R [2 marks].
Finally, since 3 is the root, we turn it B [1 mark]. So in the end, 2,3,4 are
B and 1 is R.
4. A decimal counter of length n consists of an array D indexed by 0, . . . , n− 1, in
which each cell contains one of the decimal digits 0, . . . , 9. The value v(D) of
such a counter is the sum
∑n−1
i=0 D[i] ∗ 10i.
The following operation increments the current value of the counter modulo 10n:
Page 2 of 9
Inc():
i = 0
while iD[i] = 0
i = i+1
if iD[i] = D[i]+1
(a) Best case time is Θ(1) [1 mark]. This is illustrated by any scenario where
D[0] 6= 9 [1 mark].
(b) Worst case time is Θ(n) [1 mark]. Illustrated by the case where D[i] = 9 for
all i [1 mark].
(c) Suppose we gauge runtime cost by the number of times the while-test is
evaluated (the total number of line executions will be within a factor of 6
of this). Each of the 10n Inc’s will do this test at least once; a tenth of
them will do it at least twice; a hundredth at least three times, etc. So total
runtime cost will be
10n(1 +
1
10
+
1
100
+ · · ·+ 1
10n
) < 10n ∗
∞∑
i=0
10−i = 10n ∗ 1.1111 · · ·
Thus total cost is between 10n and 10n ∗ 10/9, and so is Θ(n). Amortized
average cost per operation is thus between 1 and 10/9, hence Θ(1). [1 mark
for each of the answers; 4 marks for the justification.]
5. Trying to reduce Independent Set to 3-SAT.
(a) Given C = (`1 ∨ `2), we can replicate the exact conditions under which C is
true by introducing any logical variable xC and replacing C by the con-
junction of the following two 3-CNF clauses: C ′ = (`1 ∨ `2 ∨ xC) and
C ′′ = (`1 ∨ `2 ∨ ¬xC)
[2 marks for the correct construction]
(b) For a k-literal clause C = (`1 ∨ . . . ∨ `k) for k ≥ 4, we define k − 2 vari-
ables xC,2, . . . , xC,k−2 and replace C by the conjunction of the following
clauses:
C(2) = (`1 ∨ `2 ∨ ¬xC,2)
C(3) = (xC,2 ∨ `3 ∨ ¬xC,3)
. . . . . . . . . . . . . . .
C(k − 1) = (xC,k−2 ∨ `k−1 ∨ `k)
[2 marks for the correct construction]
Page 3 of 9
(c) The graph G = (V,E) (n = |V |,m = |E|) will have an Independent Set
of size ≥ 4 if and only if for every subset V ′ ⊆ V, |I| = n − 3, V ′ has at
least one vertex in I. Using the boolean variable xv to indicate “v is in I”
(xv = 1 in this case), we can represent “V
′ has at least one vertex in I” as
the length-(n− 3) clause ∨v∈V ′ xv.
We need to consider every subset of size (n − 3), so we have (n
3
)
of these
length-(n−3) clauses, joined by a ∧. By (b) we know any one of the length-
(n− 3) clauses can be re-cast as (n− 5) 3-CNF clauses. Doing this for each
of the big clauses gives a 3-CNF Φ which has (n− 5)(n
3
)
clauses in total.
To ensure adjacent vertices can’t belong to I, we add (¬xu ∨¬xv) for every
e = (u, v) ∈ E. By (a), these can be coded as 2 3-CNF clauses, giving 2m
extra clauses.
[4 marks - 1 for the length-(n−3) clause, 1 for the characterisation as 3-CNF,
1 for the counting of all big clauses, 1 for the “non-adjacent” clauses.]
(d) If we consider generalising the reduction in (c) to the case where we were
interested in an Independent Set of size ≥ h, for an arbitrary h, we would
find ourselves considering taking
(
n
h−1
)
different subsets of V , and adding a
“big clause” (eventually realised as the ∧ of a number of 3-CNF clauses) for
each of these. If h is variable, then the number of “big clauses” is no longer
polynomially-sized. So no, this approach doesn’t work.
[2 marks for a decent explanation.]
Page 4 of 9
FOR PART B:
1. (a) After the first Heap-Extract-Max, the array has 18, 11, 8, 1, 6, 5.
After calling Max-Heap-Insert(17), the array then has 18, 11, 17, 1, 6, 5, 8
After calling Heap-Extract-Max again, the array then has 17, 11, 8, 1, 6, 5
After calling Max-Heap-Insert(7), the array then has 17, 11, 8, 1, 6, 5, 7
After calling Max-Heap-Insert(19), the array then has 19, 17, 8, 11, 6, 5, 7, 1
[5 marks for the right answer]
(b) We are considering the implementation of ExtendedQueue.
i. There is not really any intelligent way to implement findElement on a
Heap. If we try to search from the root, then whenever we are exploring
a node v, if we find that v.key < k, we are allowed to ignore all nodes
below v. However, if v.key > k then we must search in both of the
subtrees below v, since the Heap structure does not tell us anything
more about where k might be.
We can however implement an algorithm which does a traversal of the
Heap, and which stops exploring when v.key < k. In the worst-case
this has a running time of Θ(n). Consider the case where k lies is the
lowest level of the Heap... In this case k is no more likely to be in any
spot (on this level) than another. However the lowest level of the Heap
may contain n/2 elements in total (if the total size of the Heap is n).
[3 marks for a correct findElement and 2 marks for justifying the Θ(n)
(1 for each of O(n) and Ω(n)).]
ii. To implement maxElement on a Red-Black tree, we observe that the
element with the Maximum key in a R-B tree is the rightmost “near-
leaf”, giving the following algorithm for maxElement:
I. Start at the root of the R-B tree and keep walking to the right-hand
child of the current node, until the next right child is a trivial leaf.
II. Return the element stored in this position.
This little algorithm has worst-case running-time Θ(h) = Θ(lg(n)),
given what we know about the height of R-B trees.
For removeMax, we initially perform the same steps as for maxElement.
Then we delete the node we arrived at - this will potentially require
the re-colouring of nodes up to the root of the tree (in the case that
we deleted a previously-black node). There are a number of cases to
patch-up the tree, but all can be executed in time proportional to the
height which is Θ(lg(n)).
[1 mark for removeMax and 1 for its running time, 2 marks for remove-
Max and 1 for its running time.]
iii. The operations removeMax and insertItem both have worst-case running
time Θ(lg(n)) on a Heap, and on a R-B tree. The operations findElement
and maxElement have different asymptotics on the different structures.
Page 5 of 9
If we are implementing an Extended Queue on a Heap, then the Θ(n)
running-time of findElement on a Heap might be prohibitive. If we
expect findElement to be used relatively frequently for a particular ap-
plication of ExtendedQueue, then we should use the Red-Black imple-
mentation.
However, the R-B implementation of maxElement is less efficient than
the Θ(1) implementation of the Heap. So if findElement calls are infre-
quent, we might choose to work with the Heap implementation.
[5 marks, awarded based on the quality of the arguments.]
(c) To implement UpdateKey(o, k′), we assume that o is a direct reference to
the object o = (e, k), so we don’t need to carry out findElement. We want
to update the key value of o to become k′, but doing this may violate the
Max-Heap property. There are two cases:
• If k′ > k, then k′ may be larger than the parent of o.
This scenario is very similar to the situation in Max-Heap-Insert after the
new key gets inserted at the available leaf - the solution is to “swap with
the parent” until o sits at a position where k′ is less than the parent’s
key. This is O(h) = O(lg(n)) time.

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BCPM0008 ECN 620 QBUS6830 ECSE-415 ECON5102 MGMT20005 MAT 653 ECO 480 COMP 5/600 CSSE 5600 ME 571 7CCMMS61 CS 550 COMP2865 MATH3075 STATS 506 QTM 220 ISYS2120 FIT5106 STAT 153 MKTG3112 ECON2102 ELE2018 COMP3031 MSIN0010 SCC461 MATH237 CS246 CSC3065 CSE 101 ICTWEB411 COMP9313 ECOS 2025 FN620 MATH11111 MATH 3018 KE1068 FINA 6112 C++ MISM 6202 MSIN0105 CSC8631 MECO6935 COMP4121 ACCT2011 DPBS1150 EEL 3701C F19PL OPM 661 CSE 310 CS918 BIOL30001 ACCT3563 BANK3600 MET CS622 MATH367 ACCT5907 Continuing COMP5048 COMP61021 CS3334 GSOE9210 CSCI-570 COMP 5322 HW 2 ALY-6020 IEOR 4706 DATA 200 EBUS504 CGE12412 COMP4108 7SSMM800 ECMM171P MATE50002 CS225 RS 6003 BH2274 ELEC3111 MAT6669 ECON-UA 227 PHIL 1012 7SSMM603 TB028A P6 ECS708 CSC 427 CS6476 CSE 250A A1A2 ITS61504 EECS 376 model HW6 UV0010 ECON 6074 MG4F7 CMP-7030Y INFO3008 BFIP013 ECON 5068 MATH6002 CS3483 PSTAT 131 C31FM CSOR W4246 AAEC 5307 DATA ELEC362 COMSM0047 ECN6006 MA20224 BST351 MATH 323 MATH323 AA2021 COMP3811 COMP2013 IB9X60 INVP001 EMATM0051 COMP220 BLAW1002 E20 TCP8001 ECON61001 EART40005 ECONM1022 MATH 365 MATH 367 ELEC6218 COMP 5812M EDUC 5061M ELEC ENG 1101 equation MATH6174 ST334 C4 EC9570 A33648 EBUS603 ECON47815 stata WFMS0001 PEM102 INF6060 ionically ELECTRICAL MATH 375 MMW 12 MKTG860 ECON60401 ALY6010 SOST20131 Introduction COM3524 MME2 GGR305 JNB 538 EECS101 COMM5030 IEOR E4601 2B CSC336 EN2315 LAW 432 CSC263 MATH2001 OLET1143 CSCB63 SIT772 COMM1170 STAT 441 MS930 INFT3050 GLBH0030 Stat 120B MAST10007 MGEC61 Machine GGR 252H1S ECON7030 Financial Marketing MGMT30019 ECON5120 SDSC 8009 CSC165H1S DSCI 551 3081W ACM41000 ECA 5304 MATH 1064 ECSE444 DATA 604 COMP7105 MIE1615 ECON4004 ENV200H1S EC481 MKT 306 EC1B1 MATH08051 ALY6140 CS260C CSE30 MSIN0226 CS4103 CSC3064 AM16 COMP11212 CSE4101 PSTAT 174 Math 380W CNT 5106C STAT 310 FIN2028 CS5014 STAT 425 COMP 4102A ELEC2140 MANF9544 QRM 9 QRM 8 ECON 457 COMPSCI 4091 MATH 523 DNSC-4211 PEAS 6000 MSIN0180 COMP2119 A2A HT 2022 IB93F0 MAS316/414 ACCT2522 MSIN0181 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