CS2012 Programming Techniques II
26 Sep 2014Lecture 31
Last lecture:Experimental observation & predictionCost
models: Counting the number of executions ofEvery single kind of
commandOnly some important commands (e.g., array accesses)We also
assumed:Each command executes in 1 time unitOnly the highest-order
term of calculations counts (T(n) ~ 5n2)Today:One more cost model:
Counting the number of executionsLines of Code *Best case / Worst
case * / Average caseA correctness argument Loop Invariants26 Sep
2014Lecture 32How can we sort a deck of cards?26 Sep 2014Lecture
33
wikipediaInsertionSortVideo:
http://www.youtube.com/watch?v=cFoLbjGUKWs
What is the algorithm?
26 Sep 2014Lecture 34
wikipediaSpecification of an AlgorithmWhat are the inputs?What
are assumed properties of the inputs?What is the output?What is the
important property of the output?26 Sep 2014Lecture 35InsertionSort
SpecificationFirst lets make sure we know what we want to do:The
specification of the algorithm AKA the problem we are trying to
solve26 Sep 2014Lecture 36Input: sequence of n numbers A=(a1,
an)Output: a permutation (reordering) of the input(a1, an)Such that
a1 a2 an
ExpressionWe can express the algorithm at different levels of
detail:In English: impreciseCan convey the main idea of the
algorithmBut hides the details some of them are important!We cannot
use it to analyse the algorithm
In a programming language: very preciseNecessary for
implementing the algorithmDoes express the detailsSometimes too
much detail can confuse the idea
In pseudocode: happy mediumResembles a programming languageHas a
good amount of detail26 Sep 2014Lecture 37Modern high-level
programming languages try to be more like pseudocode by hiding
details.InsertionSortAlgorithm (in English)Start from 1st element
of the array (optimisation: start from 2nd)Shift element back until
its right positionContinue to next elementRepeat (2) and (3) until
the end of the array26 Sep 2014Lecture 38
InsertionSortAlgorithm (in pseudocode)for (j = 1; j=0 and
A[i]>A[i+1] { swap A[i], A[i+1] i=i-1}}return A26 Sep
2014Lecture 39
InsertionSortAlgorithm (in Java)left as an exercise.26 Sep
2014Lecture 310AssumptionsWe use unspecified time units (tu)Each
command takes 1tuNumerical data are stored in binary formatSize of
an int is 1 memory word. Of an array A[0..n-1] is n words.Program
variables can store arbitrarily large numbers no overflowSimple
numerical operations takes 1tu (+,-,*,/,mod,exp,..)
26 Sep 2014Lecture 311for (j = 1; j=0 and A[i]>A[i+1] { swap
A[i], A[i+1] i=i-1}}return ABest Casecostno of timesfor (j = 1; j=0
and A[i]>A[i+1] {1 swap A[i], A[i+1]1 i=i-11}}return A1
26 Sep 2014Lecture 312Best Casecostno of timesfor (j = 1; j=0
and A[i]>A[i+1] {1 swap A[i], A[i+1]1 i=i-11}}return A1In the
best case the array is already sorted.26 Sep 2014Lecture 313Best
Casecostno of timesfor (j = 1; j=0 and A[i]>A[i+1] {1
(1+1++1)n-1 times swap A[i], A[i+1]10 i=i-110}}return A11In the
best case the array is already sorted.The time (as a function of
the input size n):
T(n) = n + n-1 + n-1 + 1 = 3n - 126 Sep 2014Lecture 314Worst
Casecostno of timesfor (j = 1; j=0 and A[i]>A[i+1] {1 swap A[i],
A[i+1]1 i=i-11}}return A126 Sep 2014Lecture 315Worst Casecostno of
timesfor (j = 1; j=0 and A[i]>A[i+1] {1 swap A[i], A[i+1]1
i=i-11}}return A1In the worst case the array is in reverse sorted
order.26 Sep 2014Lecture 31626 Sep 2014Lecture 317
Worst Case26 Sep 2014Lecture 318costno of timesfor (j = 1; j=0
and A[i]>A[i+1] {12++n swap A[i], A[i+1]11++(n-1)
i=i-111++(n-1)}}return A11In the worst case the array is in reverse
sorted order.
T(n) = n + n-1 + Sumx=2..n(x) + 2Sumx=1..n-1(x-1) + 1 =n + n-1 +
(n(n+1)/2 - 1) + 2n(n-1)/2 + 1 = (3/2)n2 + (3/2)n - 1Average Case26
Sep 2014Lecture 319costno of timesfor (j = 1; j=0 and
A[i]>A[i+1] {1(2++n)/2 swap A[i], A[i+1]1(1++(n-1))/2
i=i-11(1++(n-1))/2}}return A11In the average case we shift each
A[j] about j/2 positions to the left
T(n) = n + n-1 + Sumx=2..n(x)/2 + Sumx=1..n-1(x-1)/2 + 1 =
ExercisesEstimate the worst case running time cost of 2SUM by
counting the number of times each line of code is executed.Assume
each line takes 1 time unit to executeNow give this estimate using
the tilde notation
Estimate the worst case running time cost of InsertSort by
counting the number of swap operationsNow give this estimate using
the tilde notation
26 Sep 2014Lecture 320Correctness26 Sep 2014Lecture 321Why is
our algorithm correct?We will make an argument of its correctness
using a loop invariant.
A loop invariant is a property which is true:At the beginning of
the algorithmAt the end of the algorithmBefore each iteration of
the algorithm26 Sep 2014Lecture 322Loop exit: InsertionSort loop
invariantfor (j = 1; j=0 and A[i]>A[i+1] { swap A[i], A[i+1]
i=i-1}}return A26 Sep 2014Lecture 323
Loop exit: j==nInsertionSort loop invariantfor (j = 1; j=0 and
A[i]>A[i+1] { swap A[i], A[i+1] i=i-1}}return A26 Sep
2014Lecture 324
n=A.lengthLoop exit: j==n sorted A[0..n-1]
InsertionSort loop invariantfor (j = 1; j=0 and A[i]>A[i+1] {
swap A[i], A[i+1] i=i-1}}return A26 Sep 2014Lecture 325
n=A.lengthLoop exit: j==n sorted A[0..n-1]Loop
entry:InsertionSort loop invariantfor (j = 1; j=0 and
A[i]>A[i+1] { swap A[i], A[i+1] i=i-1}}return A26 Sep
2014Lecture 326
n=A.lengthLoop exit: j==n sorted A[0..n-1]Loop entry:
j==1InsertionSort loop invariantfor (j = 1; j=0 and A[i]>A[i+1]
{ swap A[i], A[i+1] i=i-1}}return A26 Sep 2014Lecture 327
n=A.lengthLoop exit: j==n sorted A[0..n-1]Loop entry: j==1
sorted A[0..0]InsertionSort loop invariantfor (j = 1; j=0 and
A[i]>A[i+1] { swap A[i], A[i+1] i=i-1}}return A26 Sep
2014Lecture 328
n=A.lengthLoop exit: j==n sorted A[0..n-1]Loop entry: j==1
sorted A[0..0]unsorted A[1..n-1]
InsertionSort loop invariantfor (j = 1; j=0 and A[i]>A[i+1] {
swap A[i], A[i+1] i=i-1}}return A26 Sep 2014Lecture 329
n=A.lengthLoop exit: j==n sorted A[0..n-1]Loop entry: j==1
sorted A[0..0]unsorted A[1..n-1]After line 1:InsertionSort loop
invariantfor (j = 1; j=0 and A[i]>A[i+1] { swap A[i], A[i+1]
i=i-1}}return A26 Sep 2014Lecture 330
n=A.lengthLoop exit: j==n sorted A[0..n-1]Loop entry: j==1
sorted A[0..0]unsorted A[1..n-1]After line 1:1jA[i+1] { swap A[i],
A[i+1] i=i-1}}return A26 Sep 2014Lecture 331
n=A.lengthLoop exit: j==n sorted A[0..n-1]Loop entry: j==1
sorted A[0..0]unsorted A[1..n-1]After line 1: 1jA[i+1] { swap A[i],
A[i+1] i=i-1}}return A26 Sep 2014Lecture 332
n=A.lengthLoop exit: j==n sorted A[0..n-1]Loop entry: j==1
sorted A[0..0]unsorted A[1..n-1]After line 1: 1jA[i+1] { swap A[i],
A[i+1] i=i-1}}return A26 Sep 2014Lecture 333
n=A.lengthLoop exit: j==n sorted A[0..n-1]Loop entry: j==1
sorted A[0..0]unsorted A[1..n-1]After line 1: 1jA[i+1] { swap A[i],
A[i+1] i=i-1}}return A26 Sep 2014Lecture 334
n=A.lengthExercise: Loop invariant of 1SUM?26 Sep 2014Lecture
335int count = 0;for (int i = 0; i < N; i++) if (a[i] == 0)
count++;Properties:When we exit the loop (and the code)?At the
start of the code?Immediately after line 1 is executed?
