Informatik I (D-ITET) Übungsstunde 6, Hossein Shafagh

Transcription

1 Informatik I (D-ITET) Übungsstunde 6, Hossein Shafagh shafagh@inf.ethz.ch

2 Self-Assessment

3 Self-Assessment 1 Typen und Werte int i = 11; unsigned int u = 12; double d = 2.0; int j = 0; double d1 = 1; double d2 = 2; double d3 = 3; (a) i u > 0 bool, true (b) i / 2 / d double, 2.5 (c) ++i * j++ int, 0 (d) d3 > d2 > d1 bool, false

4 Self-Assessment 2 Programmausgaben Ausgabe: #include <iostream> int main(){ unsigned int x = 8; do { std::cout << x << " ; x = (2 * x + 1) % 11; while (x!= 0); return 0;

5 Self-Assessment 3 Programm Schreiben #include <iostream> int main(){ int res = 0; for (int i = 0; i<4; ++i){ int digit; std::cin >> digit; res = res*2 + digit; std::cout << res; #include <iostream> int main(){ int res = 0; for (int i = 8; i>=1; i/=2){ int digit; std::cin >> digit; res += i*digit; std::cout << res;

6 Self-Assessment 4 Zahlendarstellungen 96 hexadezimal: 0x60 0x400 dezimal: b11001 dezimal: 25 0xfe binär: 0b

7 Self-Assessment 5 Fliesskommazahlen: F * (2, 6, -3, 4) eine exakte Darstellung /16 1.5/32

8 Self-Assessment 6 Programmausgaben A) Post: return number of digits of n B) Post: returns if n is a prime number

9 Übungsblatt 5 Problem 5.1. Floating-Point Number Representation a) F * (2,4,-3,3) b) Conversion: i) = * 2-3 ii) = iii) = iv) 7 10 = c) reverse: i) = Rounding error: ii) = 3.25 Rounding error: iii) = 2.75 Rounding error: iv) = 7 Rounding error: 0

10 Übungsblatt 5 Problem 5.1. Floating-Point Number Representation d) Addition Denormalize: = Renormalize and round Observation: ( ) hat keinen Einfluss auf das Endergebnis

11 Übungsblatt 5 Problem 5.2. Point on Line? double abs(double value) { if (value < 0.0) return -value; else return value; int main() { double abs_err = 10e-4; double a = 0.9; double b = 1.3; double c = 0.7; double res = a * x * x + b * x + c; // is point within absolute error tolerance? if (abs(res - y) < abs_err){ std::cout << "yes\n"; else { std::cout << "no\n";

12 Übungsblatt 5 Problem 5.3. Rounding int round(double x){ //The result at this point is the number cut off at the decimal point int trunc= x; double difference = (x - trunc); //Now we look at the digits after the decimal point to decide how/if to round if (difference >= 0.50){ trunc++; else if (difference <= -0.50){ trunc--; return trunc;

13 Übungsblatt 5 Problem 5.3. Rounding // PRE: x is roundable to a number in the value range of type int // POST: return value is the integer nearest to x, or the one further // away from 0 if x lies right in between two integers. int round (double x) { int trunc = x; // rounds towards 0 by standard double difference = x - trunc; // note: result is exact! if (difference >= 0.5) { // x was positive, and trunc + 1 is closer return trunc + 1; else if (difference <= -0.5) { // x was negative, and trunc - 1 is closer return trunc - 1; else { // difference < 0.5, trunc is closest integer return trunc;

14 Übungsblatt 5 Problem 5.4. Binary Expansion //PRE: input a decimal number 0<= x <2, and the number of digits to be output //POST: Prints to console (std::cout) the specified amount of digits of the binary expansion void binexp(double x, unsigned int digits){ int bi; for(int i=0; i<digits; i++){ bi = x; // truncated, e.g., 1.4 à 1 and 0.4 à 0 std::cout << bi; if (i==0) std::cout << "."; x = 2 * (x - bi);

15 Übungsblatt 5 Problem 5.5. Fixing Functions A) bool is_even (unsigned int i) { if (i % 2 == 0) return true; // POST: returns true if and only if i is even bool is_even (unsigned int i) { return i%2==0;

16 Übungsblatt 5 Problem 5.5. Fixing Functions B) double invert (double x) { double result; if (x!= 0) result = 1 / x; return result; // POST: returns either the inverse of // x, or zero if x is zero double invert (double x) { double result = 0; if (x!= 0) result = 1 / x; return result; // PRE: x is not zero // POST: returns the inverse of x double invert (double x) { assert (x!= 0); return 1 / x;

17 1 Standard Library #include <cmath> std::pow(3.3, 6.5); // computes 3.3 ^ 6.5 std::sqrt(9.1); // computes the square root of 9.1, the argument has to be >0 std::abs(-3.0); // returns 3.0, the absolute value of the argument Errors: std::sqrt(x) std::sqrt(x) x = err Costs sqrt expensive to compute std::sqrt(x*x + y*y) < radius; // expensive, because of sqrt x*x + y*y < radius*radius; // cheap, because no sqrt

18 1 Standard Library #include <algorithm> std::min (3.5, 5.1); // outputs 3.5 std::max (4.3, 7.9); // outputs 7.9 Advantages of libraries Written and published once, the code can be used by everyone (Usually) Programmed by an expert of the field. Easily maintainable (you have to change it only in one location rather than for every possible user) Less time wasted maintaining code implies more time to write especially fast and efficient code

19 2 References Part I void increment (int m) { m++; int main () { int n = 3; increment (n); std::cout << n << \n ; //? return 0;

20 2 References Part I void increment (int &m) { m++; int main () { int n = 3; increment (n); std::cout << n << \n ; // 4 return 0;

21 2 References Part I int i = 1; int& j = i; // int& j; Address Value i 1.. j i++; // i = 2 j++; // i = 3

22 2 References Part I int i = 1; int j = &i; Address Value i 1.. j i

23 2 References Part I void increment (int m) { m++; Call-by-value: Local copies! int main () { int n = 3; increment (n); std::cout << n << \n ; //? return 0;

24 2 References Part I void increment (int &m) { m++; Call-by-Reference: int main () { int n = 3; increment (n); std::cout << n << \n ; // 4 return 0;

25 2 References Part I // POST: return value is the number of distinct real solutions of the quadratic // equation ax^2 + bx + c = 0. If there are infinitely many solutions // (a = b = c = 0), the return value is -1. Otherwise, the return value // is a number n from {0,1,2 and the solutions are written to s1,..., sn int solve_quadratic_equation (const double a, const double b, const double x, double& s1, double& s2)

26 Exercise: Swap

27 Übungsblatt 6 Problem 6.1. Perpetual Calendar Given any date, return the weekday! Input: day, month, year Pay attention to leap years (February) Reference date: Monday, 1th January, 1900 Not allowed to used ready libraries! Problem 6.2. Run-length Encoding (Stepwise Refinement) 0 à encode, 1 à decode, -1 à end of sequence Not allowed to used ready libraries!

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