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Inf2-IADS Sample Exam
Creation date：2024-04-11 17:18:23
Belonging category：数学mathematics
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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关键词一 关键词二 关键词三 关键词四 关键词五 和法规和的发更好的风格和 ECON 615 FINC 612 CSC148 COMP 639 ACCT 203 ACCT 211 PSYC 101 MTH2009 COMP52315 PSYCO 432 statistics MATH4063 11 xxx 1 MA1MSP MAST30028 MATH3888 CS341 EEE30002 ES2F5 MCD4160 CC0933 ECON 1101 MCHM3001 CHEM3118 COMMGMT 2511 COMP41280 STAT 631 24T1 IRE379 SENG365 ECON433 FIT3139 MATH4602 LING5621M FINS5516 ECON3208 MN2177 L2 SM EEET1415 Finance DS4023 COMP3004 7CCSMRTS K1S5B6 Stat230 CSCI4211 CSI2132 ECON30130 MAS6100 ENVS222 COM2107 INFO1112 STAT 4004 FIT2100 comp2022 COMP4691 Physics 307 Econ 659 probability CSCI316 fir29 CSCI203 COMP3160 COMP6714 INFO1113 Physics 7B Physics Photonics S203031 STAT2005 COMP90015 CSCI 2110 CSCC46 Information CSC477 COMP4650 Statistical COMP 346 COMP1017 ELEC340 MATH1007 Programming function STAT 244 CS 134 Java cse13s Math 108A FW20 ECON3389 CSE 130 Science 331 COMP9020 engineering COSC 445 CSE 5523 GAME2012 visualization MATH 235 Math 370 CSC343 COGS 108 EE524 COMP10200 STAT 416 3FK4 MSCI 311 ECON6113 ACC40700 EMESTER1 MAS8383 ECO 512 STAT 8310 CMN3102 MA577 CMPT 225 COMP0045 Methods Equations MATH 113 Math 3283W ICS-33 EN4302/ENT603 BIA B350F MA318 MTHM506 MECH 5315M ECONOMICS MAS8382 COM 2004 PX3142 XJEL 3662 STAT 231 MGT B399F PX4131 COM2108 30AM MSCI521 ECONM1015 ACFI304 GR5241 MATH10101 ECON 2070 COMS31900 EFIM20011 CSE 250B ACSE-5 STAT4207 SOFT2412 ELEC 331 AY2020 MSIN0209 FIT2094 CS 241L ST404 CS 411 3D CA3 INF4002 COMP3506 SIT718 CS 188 CS 170 MATH 138 CSE 251A CMT118 DATA5207 CMPT307 CMPT459 T1 PGEE11108 COSC 3P32 ME 163 MATH96007 7QQMM203 EE6227 BA 574 STA 4234 ECO206Y1Y ECO2008 MT2300 MATH 11158 CS 527 MACM 316 CSE474 2DI70 ISE 525 COMP3411 EMAT30007 7CCSMBDT MFIN704 MFIN704W21 MfIB ECE 438 CS 436 F36 CS5344 FE 621A FE621 TBA 612 ECM3151 ECMM422 MF1 MSIN0107 ESSMENT CS 5854 MFIT 5008 MIPS R4000 ENG5274 BU52018 ENG2079 BIOA02 STA2570 F0 INF553 STAT 440 STAT3008 COMP9417 COMP6451 6COM1042 BU52006 ST303 HW-1 CISC-235 CS 510 MATH3161 ECN 140 ECE391 Comp 3613 CVEN 9625 PRG455 UESTION 1 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MGMTMFE 405 25579 1ECE 273 M122 COMP4434 1ELEC ELEC5620M STAT0010 BFF2401 MXB361 MATH3811 CS330 CMT307 FIT1033 CptS 583 UP 431 MATH6005 AF5343 CS2435 ECOM30002 ES50061 CS 2506 ECMT6002 CODE 00099F MT4539 ESSAY ICM317 MATH1231 MFIT5010 BMA547 MATH334 CIVL2201 MATH39032 T-04 CIVE97119 EMESTER CAB302 CAB301 59PM STAT802 ISYS3412 RM3 COSC 2123 AcF 311 MATH2022 1FINAL 4CCS1PHC ICM 142 ENVS3023 Math 4023 I1 ACSE8 FIN B385F CS-275 STA 4109-01 A31021 EE140 MT1 EE124 ECO-6004B Inf2-IADS CS126 IS2184 UL18 MATH5092 ECOM009 COMP 6211 ELEC202 MKT3019 IAB230 FIT5046 584-S21 COMP3210 CO250 A29401 MATH 136 MGT6174 ECMT6007 DATA2001 MAT 443 MATH10222 PHDE0071 ELE00123M COSC1295 MAT00045H MATH32032 1004GRC MAS223 COMP10001 DATA3404 CIVL2110 STAT2008 FNCE10002 EEE6431 COM6515 COMP9120 NBS8185 ECTE942 ECON 405 MATH314301 STAT 2011 COMPSCI 350 COMP20007 CMPSC-132 ECON5002 STA 35 MTH6108 FIT5147 MTH6150 100C FINA 5260 DESC9115 ECON4998 NBS8257 GMM ECON 6070 BSA2 ECE599 STATS 102B STA120 INFS601 BFC3241 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PHIL2617 PHYS3116 MIE250 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 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270M CIS 3207 Programs CSCI 2132 CS270 CSE-381 COMP3230A CIS 212 CSI 333 FINM 326 FINM-36702 Introduction to Software Analysis EEE102 GNG1106 ECS 60 CS 161 CSCI851 CSE1222 EDPS0249 TCHR5003 ANTA02 BBUS1SBY SOSC 1140 MDIA2091 FINS3641 INSY 661 ITEC3040 CPSC 481 computer DEE3519 COMP304 CSE536 Microtheory OFFICE 280 CIVE50003 COSC2673 ECS784U CMPSC/MATH 455 AMA 505 CSE 158 BUS1040 FIT3179 FIT5202 Math 104A Math 111A MBUS107 COMPSCI 761 COMP3710 COMPSCI 316 COMP559 COMP557 CP1 1902 COMP 310 ECED 3403 COMPX523 AGCM 3103 EMET 7001 ENGR3821 CISC 124 IPC comp20005 CS 323 CS105 STAT 326 MATH4007 CS4551 COMP1000 ACIT 4640 ENVX3002 CS2034 FC712 algorithms CSCI 4155 CSI4142 POL 850 MATH20811 2B03 CSCI803 Economies COMP 2016 Architect TRADE INFO1110 FIT1043 networks Robotics ICS 31 ECEC-301 FIT2099 CS 3640 CSE1010 Computational COMP1001 CSE 231 CS10 FIT5166 CSE 400 CSCI561 Math 350 CO-496 ENGG6400 CSCI 1133 COMP3055 MATH 819 CSE 6363 CIS 555 CS4420 COMP0037 ECS 140A Spanning Tree Protocol network 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EBU5376 STAT6118 ECON 20110 COMU7000 average acceleration BUSI4412 BFF3121 BIT233 ECMM462 ETC3430 DESN2348 INTE2584 ENME502 MA3XJ INTE2627 MA2815 MA3AM L0101 Math 3MB3 CPCCBC4004 EEEM061 0502E220 APPLIED 6SSMN310 ENV 3310 PM2PCOL3 CONS6201 MATH502 FIT5003 EEEN313 MATH3001 MECH3460 ECON30009 COSC1252 Assembler COMP 3023 ENGG7601 COMP20005 COMP SCI 1103 ICSI 402 INF2C CS320 COMPSCI7064 ECE 3220 PROGRAMMAMING MIPS Computer Science 2413 AM3611 MSOA TE2101 FFT CMPSC-121 CSE 109 COMP SCI 7064 HCI CMP-5013A CCN2042 COMP3400 EEEN30052 CO4215 EE101 KIT107 CSE 421 CSE521 MCD4720 EECS 280 MIT 403 GRPC VG101 APiE EECE 1080C FIT3143 CSE 325 ISOM 3029 CS3307 State SIT255 Engineering MQF Modelling ACX3150 SIA322 ACCT90004 COMU7201 MGMT90018 WRIT1000 COMP3702 BISM1201 COMP7505 IDEA9106 CSE3SMT BFC5926 BFC5280 BFC5935 QBUS3330 ECON3700 IFN521 ECON2121 COSC2626 MA413 DDES9903 ENGG1300 COMP9337 CO2201 ACCT2002 ICM289 MMM054 MSBA-204 DIGM707 CSC2250 Econ312 EVAT python CAN404 LUBS2840 CSCI-GA.2250 COMSM0093 COMP281 AFM 333 IBUS1001 CP2501 Math 105A VE320 TEC101 SOCY2166 COM240 EFIM20033 ENGI4467 EFIM20034 EC 979 BE432 7EJ502 CSSE6400 COMP90025 Econ 312 ENG2031 ECO0006 EXERSCI 207 CAN209 ECO 137W FT312 ECO 2199 MATH32052 BDW4213 APS 360 SAS 6 CHME0017 PO92Q Economics 01AV MATH39512 MATH 5486 COMP8620 ISE102 ECON235 ECON 206 FI 345 PHAY0020 MATH0099 MATH0085 EFB106 PHY 134 MAE115 circuits Chemistry CSE12 MATH-UA 233 PHAS0038 COMP2140 CONMGNT 7049 COMP151101 ECON 485 ACX5903 ENGN6536 ARCH1162 Law PHAS0042 CCF1C03 MATH 351 Phys 139 AG942 MGT001371 APH106 ECON1051 MA4AM MA3Z7 NUMBER THEORY REST0006 SESS0026 STAT0023 TABL 2741 CMP4008Y EFIM10015 Acct 560 PSYC735 GEOG 5 ACCT3104 CSEC5616 PHIL1012 ITLS6111 ECON8022 5CCM226A compX123 MA4XJ R001 EWB LLP120 COMP643 ENG 103 ECON 23950 ACCT 207 FINC 616 Bio99 Math 137A Psychology BCPM0054 MATH20212 EC3450 CENV6141 PSY 205 CE335 CS365 Calculus ACSD INFT6800 MCIT INFO6030 PHYS 111LV H63EEJ ECON2151 ENGI 2181 MTRX1701 PLIT08009 CS240 Math 270 CMP7000A CMP-7038B Math 263 LAW3016 CS371 STAT 334 FIN0008 ACC3004 STAT 371 ​COMP1921 MTH201 ACCT2000 UN3412 CIV6000 MATHS 3021 MATE7014 ECON1 CIBC6035 LAW121G COMM 321 CSC8204R TTM2033 CSCI 2600 MAE 115 LNG302 CENV6175W1 ECON30019 CPS2015 BST969 DDES9015 AMATH242 CS 1151 N1592 ACCT600 EG503G SMM519 EC318 Ecstatistics EEE 471 MARK2051 Topology COM00037H ACCT3000 MTHM501 EG501V COM00055H BUSN9085 GGR273 MATH 2B ITD102 MA214 ECON0060 ECE3114 ECMM447 Mechanics CCT361H5S BSAN3210 WELF2001 Fin3001 RESP5001 STA 137 Linguistics 101 CSCI368 CMPSC497 LUBS3655 OMBA 5355 BUSM2562 ​BFF5902 CYBR7001 CULT2005 ECON 7720 MTRX3760 FSE598 ​LUE1001 MGSC 7100 CS3240 COSC7500 ACCT3091 INFOGR TRAN 2080 DESIGN CIVE215001 CIVE2360 CPSC 8430 FNCE2003 Stoichiometry INFS6023 Quantitative Method Math 21D ECE5179 Design ​ECON8022 Chemical DMS 213 ADM1370 BFF5902 BFF2701 AGRI10051 AMN400 ECON7322 ​FINC5090 ​MATH1052 ​SciComp Elec4622 ​FINS5516 ENG 101 MMM143 CS431CA2 BSAN2201 EBSC7070 ES9v9-10 PLIN0009 DATA 8 MATH 101 BCPM0097 ​MAT 237 BISM2201 EEE8099 Physics 11 ECE2238B Biostat 234 ​ECON30009 INF10025 CHEM252 CCMA4008 CIVE 3007 STAT0031 ELEC ENG 3108 MGMT90140 CHEM 252 STAT3016 MMM095 ​MMGT6016 Legal STAT 311 ELEC9764 COMP 370 TCS 3053 EECS 1021 FIT1051 C/C++ EE3C COMP2003J COM398 COMP51915 Simulator FIN42330 8PRO102 PGD ELEC5160 XJCO1921 COMP5123M INFS 2042 CHE1148H ECA5313 IN3329 Microsoft Modeling B15 2TT COMP3687 COMP2432 DMS2030 KAN4TSF CFKG CSE 5322 CSE 445 STATS 3DA3 CHC5223 COMP1048 EL3105 EECS 3221 CS231 ME 4434 ITP4510 ITP4501 CSCI 5708 Controller EECE 5699 CPEN 231L CPN programs CSC8016 MATU9D2 ENGN4213 2. SVM. (10 points) Consider an SVM classifier using RBF kernel, and finish the following python code to search for the optimum kernel parameter γ. You are required to calculate the RBF function and search yourself, instead of using functions in Scikit-Learn (sklearn). def parameter search(gamma list ,Xtr,ytr,Xts,yts ,...): """ Parameters gamma list: A list of RBF kernel parameter gamma to search for Xtr: Training data, shape (ntr,nrow,ncol) ytr: Training labels, shape (ntr,) Xts: Test data, shape (nts,nrow,ncol) yts: Test labels, shape (nts,) ... Add other parameters as needed Returns: gamma optimum: The gamma with minimum error on Xts, yts """ ... return gamma optimum
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