Controlling statement execution

CS101 2012.1

CS101 2012.1

Statement block A block looks like{statement;statement;…statement;}

0 or 1 statement allowed for uniformity Walk down the list executing one statement

after another Effect of each

statement onmemory completesbefore next executed

Note on scope: Outside {…} cannot use variables declared inside

float x = 5, y = 11;float temporary = x;x = y;y = temporary;

CS101 2012.1

If-then-else Store in variable b the

absolute value of variable a

Store in variable c the smaller of the values of variables a and b

Else part is optional Cascades/nests allowed Statement blocks also

optional but best used

int a, b;cin >> a;if (a >= 0) { b = a;}else { b = -a;}

int a, b, c;cin >> a >> b;if (a < b) { c = a;}else { c = b;}

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Example: withdrawing money from bankcin >> deduct;if (deduct > balance) { cout << “Account overdrawn\n”;}else { cout << “Successfully withdrawing ” << deduct << endl;

balance -= deduct; // emit paper money}

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Curly brackets You can also write then or else parts without

curly brackets, but this could be dangerous Best to always use curly brackets even if not

needed

int m = 5;int n = 10;if (m >= n) ++m; ++n;cout << "m=" << m << "n=" << n << endl;

Increment of n happens outside the if-then action, which only

increments m

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Conditional expression Format: cond ? ifExpr : elseExpr Earlier examples rewritten

• b = (a > 0)? a : -a;• c = (a < b)? a : b;

If in doubt, use (parens) to make sure expression tree is correct

Nesting quickly gives unreadable code:(a > 0)? a : ((-a < b)? 100+c : c-100)

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Switch statement if (cond1) {…} else if (cond2) {…} else if (cond3) {…} can get tiring

Common use is to choose between different statements depending on the value of a variable

switch (dayOfWeek) {case 0: cout << “Sunday”; break;case 6: cout << “Saturday”; break;default: cout << “Weekday”;}

Can be char, int, long

intIf no break here,

control will “fall through” to the

next case

If none of the listed cases match dayOfWeek

CS101 2012.1

Switch mysteryint val = 5;switch (val) { case 5: cout << "five\n"; // suppose we forget this break; case 4: cout << "four\n"; break; default: cout << "default\n";}

This statement will be executed even though val

is not 4

CS101 2012.1

Switch exampleswitch (ch) {case 'a': case 'A':case 'e': case 'E':case 'i': case 'I':case 'o': case 'O':case 'u': case 'U': cout << ch << " is a vowel" << endl; break;default: cout << ch << " is not a vowel" << endl;

}

Can use fall-through as a feature to

combine many cases

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Another switch examplecout << "Enter simple expression: ";int left;int right;char operator;cin >> left >> operator >> right;cout << left << " " << operator << " " << right << " = ";switch (operator) {case '+' : cout << left + right << endl; break;case '-' : cout << left - right << endl; break;case '*' : cout << left * right << endl; break;case '/' : cout << left / right << endl; break;default: cout << "Illegal operation" << endl;

}

CS101 2012.1

The infinite loopwhile (true) { cout << “I will lecture clearly and slowly\n”;

cout << “I will remain quiet in class\n”;}

Insanity: doing the same thing over and over again and expecting different results.

Albert Einstein

CS101 2012.1

Are inputs in increasing order? Keep reading input number for ever, and check if it

is greater than the previous number read

#include <iostream>#include <limits>using namespace std;double prev = numeric_limits<double>::min();while (true) { double cur; cin >> cur; if (cur <= prev) { cout << “Not in increasing order\n”; } prev = cur;}

How do we get out of the infinite

loop?

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Printing an integer in binary Surprisingly, standard C++ does not offer a

library function But simple enough For brevity consider printing a byte num Take bit pattern mask = 10000000 Bitwise and: num & mask If result is nonzero, print ‘1’ else ‘0’ Shift mask right to 01000000, then

00100000, 00010000, etc.

CS101 2012.1

Shortcut Instead of shifting mask right, shift num left Use fixed mask of 10000000 Say num = 01101011 = 107 (decimal) 10000000 & 01101011 print ‘0’ Shift num left to 11010110 10000000 & 11010110 print ‘1’ Shift num left to 10101100 10000000 & 10101100 print ‘1’ And so on

CS101 2012.1

Raising a number to an integer power Given a double base and a non-negative integer

power int pow Find basepow

A simple and a more efficient version

double ans = 1;int todo = pow;while (todo > 0) { ans = ans * base; // or ans *= base; --todo;}

CS101 2012.1

Loop invariant Initially, ans=1 and todo=pow After each iteration, ans * basetodo = basepow

• ans multiplied by base, todo reduced by 1

Finally, todo=0, therefore ans = basepow

Number of multiplications: pow

double ans = 1;int todo = pow;while (todo > 0) { ans = ans * base; // or ans *= base; --todo;}

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The squaring method Suppose pow = 32 Instead of base*base*…*base

(31 multiplications), run the following

double b2 = base * base;double b4 = b2 * b2;double b8 = b4 * b4;double b16 = b8 * b8;double ans = b16 * b16;// only 5 multiplications

CS101 2012.1

General powers Suppose pow = 29 is a byte Express as 1 + 4 + 8 + 16 Answer is base * b4 * b8 * b16

Let bx take values 0 through 7 Keep squaring base as before If the bxth bit of pow is 1, include the power of

base in the product, else not Already know how to get bit by (val & 1)

0 0 0 1 1 1 0 1

bx

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Faster code

unsigned int todo = pow;double ans = 1, term = base;while (todo > 0) { if (todo & 1 == 1) { ans *= term; } term = (term * term); // squaring

todo = (todo >> 1);}

Exercise: Work out the loop invariant

What if pow is a real number?

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Factorial

long long int double fac = 1, n;cin >> n;while (n > 0.5) { fac *= (n--);}cout << fac << endl;

Can easily overflow even for moderate n Even if we use double

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float x;cin >> x;int numDivs = 0;while (x > 1) { x = x / 10; numDivs = numDivs + 1;}cout << numDivs;

Iteration: Approximating the logarithm How many times must we divide a number x by 10

until the result goes below 1? while (condition) statementOrBlock

Another shorthand:x /= 10;

More shorthands:numDivs += 1;

or++numDivs;

Prolog

Loopbody

Epilog

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Quick review Why do we care about how data types are

represented in RAM? To understand the limits of what can be

represented within hardware limitations Thus write more reliable and robust code Can give insights for better algorithms E.g., bit shifts squaring trick fast power

algorithm

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Perimeter of regular polygons Polygons inscribed in circle of

radius ½ First, a polygon with sides x Next double the number of

sides, each side is y

Let’s start with x = ½ (six sides)

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Code

double sides = 6, x = .5;while (true) { cout << (sides * x) << endl; double a = sqrt(1-x*x) / 2; double b = .5 – a; double y = sqrt(x*x/4 + b*b); x = y; sides *= 2;}

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Numerical integration Divide x-axis into

intervals of width h Approximate area under

function f(x) between x and x+h as h f(x)

As h becomes smaller, estimate should become more accurate

But remember finite precision limitations

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One loop nested in anotherdouble h = .01;while (true) { double area = 0, x = -1.; while (x + h <= 1.) { area += h * sqrt(1. – x*x); x += h; } cout << 2.*area << endl; h /= 2.; // halve interval width}

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Moral of the story On an ideal computer, increasing

number of polygon sides, or decreasing interval width h without bound should get you better and better estimates

Finite precision prevents this (This is why we need to know our

hardware representation!) In general limits are troublesome to

evaluate reliably Luckily there are other methods,

e.g., convergent series

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ex

Well-known series

double ans=1, base=x, fac=1, ix=1;bool keepGoing = true;while (keepGoing) { ans += base/fac; base *= x; fac *= (++ix); if (base/fac < epsilon) { keepGoing = false; }}

Tedious to have to exit from a loop

like this

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breakwhile (true) { ans += base/fac; base *= x; fac *= (++ix); if (base/fac < epsilon) { break; } cout << (base/fac) << endl;}

Terminates immediately enclosing while loop

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Pep Why are we developing algorithms to

compute powers of numbers or or ex if we already have library functions?

Techniques like limits, numerical integration, series summing are universal

Familiarizes us with loop and break constructs

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A counting problem Input is int len How many sequences of len bits are there?

• 2len

Of these, how many do not have two consecutive zeros?

If len = 1, answer is 2 (0, 1) If len = 2, answer is 3 (01, 10, 11) Suppose a(len) sequences end with 0 And b(len) sequences end with 1 Required answer is a(len) + b(len)

CS101 2012.1

Counting problem, continued If a sequence ends with 0, second last bit

must be 1 Therefore a(len) = b(len-1) If a sequence ends with 1, second last bit

could be either 0 or 1 b(len) = a(len-1)+b(len-1) Therefore b(len) = b(len-2)+b(len-1) First focus only on b(len) Hemachandra numbers, Fibonacci series

CS101 2012.1

Base cases for b() b(1) = 1 (the bit sequence 1) b(2) = 2 (sequences 01, 11)

int len;cin >> len;if (len == 1 || len == 2) { cout << “b(" << len << ")=" << len;

}else { … }

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After base cases Note that b(len) depends only on b(len-1) and

b(len-2) Enough to keep around last two values Rather than all of b(1), b(2), …, b(len)

1 2 3 5

lx

bLxbLxM1bLxM2

1 2 3 5 8

lx

bLxbLxM1bLxM2

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Else… len 3int bLxM2=1, bLxM1=2;int lx = 3;while (lx < len) { int bLx = bLxM2 + bLxM1; // shift forward one step ++lx; bLxM2 = bLxM1; bLxM1 = bLx;}cout << len << " " << (bLxM2+bLxM1) <<endl;

Invariant:bLxM2+bLxM1 = b(lx)

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Running average The user inputs a sequence of numbers After the third number is input, and for every

subsequent input, print the three-point running average

double a1, a2, a3;cin >> a1 >> a2 >> a3;while (true) { cout << (a1 + a2 + a3) / 3.; a1 = a2; a2 = a3; cin >> a3;}

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Migrations between Mumbai and Pune Initial populations m and p Every year, 1/4th of Mumbai migrates to Pune And ½ of Pune migrates to Mumbai Simulate for k years Do the populations stabilize?

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Migrations codemain() { double m, p; int k; cin >> m >> p >> k; for (int t=0; t<k; ++t) { p = p – p/2. + m/4.; m = m – m/4. + p/2.; // print }}

Synchronous vs. asynchronous update

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Greatest common divisor (gcd) Given integers m n 1, find their gcd gcd(m,n) = gcd(n, m%n)

int m, n;cin >> m >> n;// assume m >= n

while (n > 0) { const int tmp = n; n = m % n; m = tmp;}cout << m << endl;

Let h = gcd(m,n) h divides m, nLet m = n*q + rGiven h divides n …… h must divide r h divides gcd(n,r)(so h gcd(n,r))Also, gcd(n,r) dividesn and r, n and m h gcd(n,r)

CS101 2012.1

“String theory” Iterative computations are demonstrated well

on arrays We have already introduced strings, which are

arrays of characters Luckily the system manages the array space

for us Can assign and append to strings Can read a position: cout << message[px] Can write a position: message[px] = ‘q’ That’s all we need for now

CS101 2012.1

Printing a string in reverse

string message;getline(cin, message);int mx = message.size()-1;while (mx >= 0) { cout << message[mx]; --mx;} mx updated in a completely predictable way Ideal candidate to write as for loop

Character at position mx in

string message

CS101 2012.1

While and for

cond

stmt

t

f

while (cond) { stmt }

cond

stmt

stepper

init

f

t

for (init; cond; stepper) { stmt }

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Printing reversed string, for loop General syntax

for (prelude; condition; stepper) { body } Equivalent to

prelude; while (condition) { body; stepper; }

for (int mx = message.size()-1;mx >= 0; --mx) {

cout << message[mx];}

Look, a declaration

In ISO standard C++, mx is no longer available here

CS101 2012.1

Nested for loops Calculate (again!) by… Generate grid of x, y points in the planefor (double x=-radius; x<=radius; ++x) { for (double y=-radius; y<=radius; ++y) { // detect if they are within a circular disk

}}

Just to demonstrate nested for loops• radius2 grid points unacceptably slow• Numerical integration much faster, at least as

accurate

CS101 2012.1

Finding needles in a haystack Given two strings, needles and haystack needles has no repeated characters haystack may repeat characters How many characters in needles appear in

haystack at least once? needles = “bat”, haystack = “tabla” 3 needles = “tab”, haystack = “bottle” 2

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One needle in a haystack Subproblem: given one character ch and a

string find if ch appears in string at least once

char ch; // suitably initializedstring haystack; // suitably initializedint ans = 0; // will change to 1 if foundfor (int hx = 0; hx < haystack.size(); ++hx) { if (ch == haystack[hx]) { ++ans; break; // quit on first match }}

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Many needles: nested loopmain() { string needles, haystack; getline(cin, needles); getline(cin, haystack); int ans = 0; for (int nx=0; nx < needles.size(); ++nx) { char ch = needles[nx]; for (int hx = 0; hx < haystack.size(); ++hx) {

if (ch == haystack[hx]) { ++ans; break; // quit on first match } } // ends haystack loop } // ends needles loop}

Generalize to work in case needles can also

have repeated characters

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Duplicate needles needles = “bat”, haystack = “tabla” 3 needles = “tab”, haystack = “bottle” 2 needles = “bata”, haystack = “tabla” 3 Two approaches

• Dedup needles before executing earlier code (reducing to known problem)

• Dedup needles “on the fly” (inside the nx loop)

Exercise: If the input strings have n and h characters, at most how much time does

the needle-in-haystack search code take?

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Parallel steppers Print the first character of a string, then the

last, then second, then second last … until you meet in the middle

for (int lx=0, rx=message.size()-1;lx < rx; ++lx, --rx) {

cout << message[lx] << message[rx];

}

Parallel steppers

Middle character of odd-length

string will not get printed

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Reversing a string

for (int lx=0, rx=msg.size()–1;lx < rx; ++lx, --rx) {

const char tmp = msg[lx]; msg[lx] = msg[rx]; msg[rx] = tmp;} H e l l o

lx rx

o e l l H

lx rx

o l l e H

lx,rxUsed in a rhs expression,

msg[lx] is a character value

Used as lhs, msg[rx] is a cell where a character is

to be written

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Comparing two strings of equal length For simplicity assume only lowercase chars Remember ‘a’ < ‘b’ < … < ‘y’ < ‘z’ Return difference of first differing character, 0 if none

(identical strings)

string as, bs; // suitably initializedint ans = 0;for (int ix=0, in=as.size(); ix<in; ++ix) { if ( (ans = as[ix] – bs[ix]) != 0) break;}cout << ans;

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Generalize to arbitrary lengths “Hello” < “Help” but “Hello” > “Hell” Scan both strings from the beginning If differing character found, same as before If a string ends, it is “less” than the otherint ans=0, ax=0, bx=0, an=as.size(), bn=bs.size();

for (; ans==0 && ax < an && bx < bn; ++ax, ++bx) {

if ( (ans = as[ax] – bs[bx]) != 0) break;}if (ans == 0) { ans = an – bn;}

This results in an arbitrary integer as the return value in case of

unequal input string lengths, but the sign of ans is what is important

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break and continue Sometimes, we need to exit from a loop in an

unexpected way Or, in the middle of the loop body, we may

want to abandon that iteration and start on the next iteration

The break and continue statements provide these capabilities

Use sparingly, only in a clean, well-lighted area

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Summary If then else, switch, while, for Printing numbers in any radix Raising number to power, squaring trick Factorial, summing convergent series Taking limits, calculating pi Numerical definite integration Computing discrete series recurrences Greatest common divisor Nested loops, string manipulations