INTRODUCTION TO C – EXERCISE 7 HAND IN THE PRINTOUTS (CODE AND RESULTS): Dec. 17, 2014 (put them into the boxes provided after the lecture AND upload the C code to your FTP account!) DISCUSSION OF THE PROGRAMS: Jan. 14, 2015 Program 21 (4 points): Write a program that reads a floating point number from stdin to a double variable, and prints its IEEE 754 stanEnter floating point number : 3.141592654 dard byte representation in hexadecimal (see figure). Use an unsigned char HEX representation: 40 09 21 FB 54 52 45 50 pointer to individually access all eight bytes of the double variable; the byte with the highest address should be printed at leftmost position. Are you able to identify sign bit, biased exponent, and mantissa fraction? IEEE 754 DOUBLE PRECISION STANDARD ================================== Hint: The “%02X” format of printf prints a byte as two hexadecimal digits. Program 22 (5 points): There is a magical point between the Earth and the Moon, called the L1 Lagrange point, at which a satellite will orbit the Earth in perfect synchrony with the Moon, staying always in between the two. This works because the inward gravitational pull of the Earth and the outward pull of the Moon combine to create exactly the needed centripetal force that keeps the satellite in its orbit. Hopefully you are able to understand why the distance r from the centre of the Earth to the L1 point satisfies the equation (assuming circular orbits) GM E GM M − − ω 2r = 0 2 2 r (R − r ) where G = 6.674 × 10–11 m3 kg–1 s–-2 ME = 5.974 × 1024 kg MM = 7.348 × 1022 kg R = 3.844 × 108 m ω = 2.662 × 10–6 s–1 gravitational constant mass of the Earth mass of the Moon distance Earth-Moon 2π / orbital period of the Moon The above equation is a fifth-order polynomial equation in r that cannot be solved exactly in closed form. Write a program that calculates and prints its solution r using a binary search algorithm. First of all, we abbreviate the left hand side of the equation as f(r). Then choose two arbitrary values R > ra > 0 and R > rb > ra that satisfy f(ra) > 0 and f(rb) < 0 ; this defines your initial search interval [ra , rb] (hint: the L1 point neither lies within the Earth nor within the Moon). Calculate the interval midpoint rm = (ra + rb)/2 that splits the original interval into two halves of equal widths. If by some miracle f(rm) = 0 you are ready, but it is much more likely that the requested solution is located either in the left (if f(rm) < 0) or the right (if f(rm) > 0) subinterval, respectively. Now, set ra and rb to the boundaries of the correct sub-interval (the one that contains the solution of the equation) and repeat the whole process until the width of the search interval becomes less than 10 km. The midpoint of this final interval is your approximation to the L1 point distance. Remark: There are five Lagrange points in each gravitational two-body system. All but L4 and L5 are unstable, so no “space debris” will be caught in the L1 point of the Earth-Moon system. A satellite to be permanently placed there would need a propulsion system to correct the long-term drift due to orbital perturbations. . Don’t miss all the fun on the next page! Program 23 (6 points): The Sieve of Eratosthenes is an efficient method to calculate all prime numbers up to N: (1) Define a sufficiently long integer array A and initialise it with A[k] = 1 for k = 2,3,…,N. Set Index = 2. SIEVE OF ERATOSTHENES ( N=1000 ) ================================ 2 31 73 127 179 233 283 353 419 467 547 607 661 739 811 877 947 3 37 79 131 181 239 293 359 421 479 557 613 673 743 821 881 953 5 41 83 137 191 241 307 367 431 487 563 617 677 751 823 883 967 7 43 89 139 193 251 311 373 433 491 569 619 683 757 827 887 971 11 47 97 149 197 257 313 379 439 499 571 631 691 761 829 907 977 13 53 101 151 199 263 317 383 443 503 577 641 701 769 839 911 983 17 59 103 157 211 269 331 389 449 509 587 643 709 773 853 919 991 19 61 107 163 223 271 337 397 457 521 593 647 719 787 857 929 997 23 67 109 167 227 277 347 401 461 523 599 653 727 797 859 937 29 71 113 173 229 281 349 409 463 541 601 659 733 809 863 941 (2) Set A[n⋅Index] = 0 for n = 2,3,4,… (as long as n⋅Index ≤ N ). (3) Increment Index until A[Index] ≠ 0 or Index > N . If Index > N continue with (4), otherwise go back to (2). (4) The prime numbers are given by those n = 2,3,…,N with A[n] ≠ 0. Your program should print all prime numbers up to N = 1000 using this algorithm. Format the output to 10 numbers per line (cf. figure). Hint: The standard library function double sqrt(double x) returns the square root of a double argument x. You must include the header file math.h to get the correct function declaration.