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1 Programming Interest Group Tutorial Four High-Precision Arithmetic.

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1 1 Programming Interest Group Tutorial Four High-Precision Arithmetic

2 2 Machine Arithmetic 32-bit machine An integer is roughly in the range ±2 31 = ±2,147,483, bit machine An integer is roughly in the range ±2 63 = ±9,223,372,036,854,775,808 Unsigned integer can double the upper limit Sometimes we need to operate on very large integers, especially in the area of cryptography High-precision integers, or multiple precision integers

3 3 Integer Libraries C:, C++: GNU C++: Integer class berlin.de/chemnet/use/info/libgpp/libgpp_20.html berlin.de/chemnet/use/info/libgpp/libgpp_20.html Java: BigInteger class in java.math teger.html teger.html

4 4 Java BigInteger Example Factorial n (usually written n!) is the product of all integers up to and including n (1x 2 x 3 x... x n). import java.math.BigInteger; public class Factorial { public static void main(String[] args) { //-- BigInteger solution. BigInteger n = BigInteger.ONE; for (int i=1; i<=20; i++) { n = n.multiply(BigInteger.valueOf(i)); System.out.println(i + "! = " + n); } //-- int solution (BAD IDEA BECAUSE ONLY WORKS TO 12). int fact = 1; for (int i=1; i<=20; i++) { fact = fact * i; System.out.println(i + "! = " + fact); } } }

5 5 Java BigInteger Example (Cont.) 1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = ! = ! = ! = ! = ! = ! = ! = ! = ! = ! = ! = ! = ! = ! = 1 2! = 2 3! = 6 4! = 24 5! = 120 6! = 720 7! = ! = ! = ! = ! = ! = ! = BAD 14! = BAD 15! = BAD 16! = BAD 17! = BAD 18! = BAD 19! = BAD 20! = BAD

6 6 What are we doing now? As a CS student, it’s not enough to just know how to use those high-precision integer libraries. It’s better to know how to design and implement a high-precision integer arithmetic library. One student from our department has done this as his Honours Project. Addition, subtraction, multiplication, and division, modular, exponentiation, etc

7 7 Review of Number Systems Positional notation using base b (or radix b) is defined by the rule: (…a 3 a 2 a 1 a 0.a -1 a -2 …) b = …+a 3 b 3 +a 2 b 2 +a 1 b 1 +a 0 +a -1 b -1 +a -2 b -2 +… Binary number system: b = 2 2 digits: 0 and 1 Decimal number system: b = digits: 0, 1, 2, 3, …, 9 Hexadecimal number system: b = digits: 0, 1, 2, …, 9, A, B, C, D, E, F Radix Point

8 8 High-precision Integers How to represent enormous integers? Arrays of digits The initial element of the array represents the least significant digits:  An integer is stored as an array: {0, 9, 8, 7, 6, 5, 4, 3, 2, 1} Maintain a counter for the number of digits Maintain the sign of the integer: positive or negative? Linked list of digits It can support real arbitrary precision arithmetic But:  Waste of memory (each node takes a pointer)  The performance is not as good as arrays

9 9 An Example: Using Array #define MAXDIGITS 100/*maximum length bignum */ #define PLUS 1/* positive sign bit */ #define MINUS -1/* negative sign bit */ typedef struct { char digits[MAXDIGITS];/* represent the number */ int signbit;/* PLUS or MINUS */ int lastdigit;/* index of high-order digit */ } bignum; Remarks: 1.Each digit (0-9) is represented using a single-byte character. 2.In fact, using higher numerical bases (e.g., 64, 128, 256) is more efficient.

10 10 Print a Bignum void print_bignum(bignum *nPtr) { int i; if (nPtr->signbit == MINUS) printf(“-”); for (i = nPtr->lastdigit; i >= 0; i--) printf(“%c”, ‘0’+nPtr->digits[i]); printf(“\n”); }

11 11 Convert an Integer to Bignum int_to_bignum(int s, bignum *nPtr) { int i;/* counter */ int t;/* int to work with */ if (s >= 0) nPtr->signbit = PLUS; else nPtr->signbit = MINUS; for (i=0; i digits[i] = (char) 0; nPtr->lastdigit = -1; t = abs(s); while (t > 0) { nPtr->lastdigit ++; nPtr->digits[ nPtr->lastdigit ] = (t % 10); t = t / 10; } if (s == 0) nPtr->lastdigit = 0; }

12 12 Initialize a Bignum initialize_bignum(bignum *nPtr) { int_to_bignum(0,nPtr); } int max(int a, int b) { if (a > b) return(a); else return(b); }

13 13 Compare Two Bignums compare_bignum(bignum *a, bignum *b) { int i;/* counter */ if ((a->signbit == MINUS) && (b->signbit == PLUS)) return(PLUS); if ((a->signbit == PLUS) && (b->signbit == MINUS)) return(MINUS); if (b->lastdigit > a->lastdigit) return (PLUS * a->signbit); if (a->lastdigit > b->lastdigit) return (MINUS * a->signbit); for (i = a->lastdigit; i>=0; i--) { if (a->digits[i] > b->digits[i]) return(MINUS * a->signbit); if (b->digits[i] > a->digits[i]) return(PLUS * a->signbit); } return(0); }

14 14 How to avoid leading zeros? zero_justify(bignum *n) { while ((n->lastdigit > 0) && (n->digits[ n->lastdigit ] == 0)) n->lastdigit --; if ((n->lastdigit == 0) && (n->digits[0] == 0)) n->signbit = PLUS;/* hack to avoid -0 */ }

15 15 Addition of Two Bignums add_bignum(bignum *a, bignum *b, bignum *c) { int carry;/* carry digit */ int i;/* counter */ initialize_bignum(c); if (a->signbit == b->signbit) c->signbit = a->signbit; else { if (a->signbit == MINUS) { a->signbit = PLUS; subtract_bignum(b,a,c); a->signbit = MINUS; } else { b->signbit = PLUS; subtract_bignum(a,b,c); b->signbit = MINUS; } return; }

16 16 Addition of Two Bignums (cont.) c->lastdigit = max(a->lastdigit,b->lastdigit)+1; carry = 0; for (i=0; i lastdigit); i++) { c->digits[i] = (char) (carry+a->digits[i]+b->digits[i]) % 10; carry = (carry + a->digits[i] + b->digits[i]) / 10; } zero_justify(c); }

17 17 Subtraction of Two Bignums subtract_bignum(bignum *a, bignum *b, bignum *c) { int borrow;/* has anything been borrowed? */ int v;/* placeholder digit */ int i;/* counter */ initialize_bignum(c); if ((a->signbit == MINUS) || (b->signbit == MINUS)) { b->signbit = -1 * b->signbit; add_bignum(a,b,c); b->signbit = -1 * b->signbit; return; } if (compare_bignum(a,b) == PLUS) { subtract_bignum(b,a,c); c->signbit = MINUS; return; }

18 18 Subtraction of Two Bignums (Cont.) c->lastdigit = max(a->lastdigit,b->lastdigit); borrow = 0; for (i=0; i lastdigit); i++) { v = (a->digits[i] - borrow - b->digits[i]); if (a->digits[i] > 0) borrow = 0; if (v < 0) { v = v + 10; borrow = 1; } c->digits[i] = (char) v % 10; } zero_justify(c); }

19 19 Digit Shift /* E.g., shift by 2 will get In memory: {6,5,4,3,2,1}  {0,0,6,5,4,3,2,1} */ digit_shift(bignum *nPtr, int d)/* multiply *nPtr by 10^d */ { int i;/* counter */ if ((nPtr->lastdigit == 0) && (nPtr->digits[0] == 0)) return; for (i=nPtr->lastdigit; i>=0; i--) nPtr->digits[i+d] = nPtr->digits[i]; for (i=0; i digits[i] = 0; nPtr->lastdigit = nPtr->lastdigit + d; }

20 20 Multiplication of Two Bignums multiply_bignum(bignum *a, bignum *b, bignum *c) { bignum row;/* represent shifted row */ bignum tmp;/* placeholder bignum */ int i, j;/* counters */ initialize_bignum(c); row = *a; for (i=0; i lastdigit; i++) { for (j=1; j digits[i]; j++) { add_bignum(c,&row,&tmp); *c = tmp; } digit_shift(&row,1); } c->signbit = a->signbit * b->signbit; zero_justify(c); }

21 21 Division of Two Bignums divide_bignum(bignum *a, bignum *b, bignum *c) { bignum row; /* represent shifted row */ bignum tmp; /* placeholder bignum */ int asign, bsign;/* temporary signs */ int i, j; /* counters */ initialize_bignum(c); c->signbit = a->signbit * b->signbit; asign = a->signbit; bsign = b->signbit; a->signbit = PLUS; b->signbit = PLUS;

22 22 Division of Two Bignums (Cont.) initialize_bignum(&row); initialize_bignum(&tmp); c->lastdigit = a->lastdigit; for (i=a->lastdigit; i >= 0; i--) { digit_shift(&row,1); row.digits[0] = a->digits[i]; c->digits[i] = 0; while (compare_bignum(&row,b) != PLUS) { c->digits[i] ++; subtract_bignum(&row,b,&tmp); row = tmp; } zero_justify(c); a->signbit = asign; b->signbit = bsign; }

23 23 For more information: Donald E. Knuth: The Art of Computer Programming, Volume 2. Seminumerical Algorithms, Chapter 4 Arithmetic  Positional number systems  Floating point arithmetic  Multiple precision arithmetic  Radix conversion  Rational arithmetic  Polynomial arithmetic  Manipulation of power series

24 Practice I: Primary Arithmetic Children are taught to add multi-digit numbers from right-to- left one digit at a time. Many find the "carry" operation - in which a 1 is carried from one digit position to be added to the next - to be a significant challenge. Your job is to count the number of carry operations for each of a set of addition problems so that educators may assess their difficulty. Input Each line of input contains two unsigned integers less than 10 digits. The last line of input contains 0 0. Output For each line of input except the last you should compute and print the number of carry operations that would result from adding the two numbers, in the format shown below. 24

25 Practice I: Primary Arithmetic Sample Input Sample Output No carry operation. 3 carry operations. 1 carry operation. 25 Hint: Use an array of characters to store the digits of an integer

26 Practice II: Reverse and Add The "reverse and add" method is simple: choose a number, reverse its digits and add it to the original. If the sum is not a palindrome (which means, it is not the same number from left to right and right to left), repeat this procedure. For example: 195 Initial number Resulting palindrome 26 In this particular case the palindrome 9339 appeared after the 4th addition. This method leads to palindromes in a few step for almost all of the integers. But there are interesting exceptions. 196 is the first number for which no palindrome has been found. It is not proven though, that there is no such a palindrome.

27 Practice II: Reverse and Add Task : You must write a program that gives the resulting palindrome and the number of iterations (additions) to compute the palindrome. You might assume that all tests data on this problem: - will have an answer, - will be computable with less than 1000 iterations (additions), - will yield a palindrome that is not greater than 4,294,967,295. The Input The first line will have a number N with the number of test cases, the next N lines will have a number P to compute its palindrome. The Output For each of the N tests you will have to write a line with the following data : minimum number of iterations (additions) to get to the palindrome and the resulting palindrome itself separated by one space. 27

28 Practice II: Reverse and Add Sample Input Sample Output

29 29 Practice III: Archeologist’s Dilemma An archeologist seeking proof of the presence of extraterrestrials in the Earth's past, stumbles upon a partially destroyed wall containing strange chains of numbers. The left-hand part of these lines of digits is always intact, but unfortunately the right-hand one is often lost by erosion of the stone. However, she notices that all the numbers with all its digits intact are powers of 2, so that the hypothesis that all of them are powers of 2 is obvious. To reinforce her belief, she selects a list of numbers on which it is apparent that the number of legible digits is strictly smaller than the number of lost ones, and asks you to find the smallest power of 2 (if any) whose first digits coincide with those of the list. Thus you must write a program such that given an integer, it determines (if it exists) the smallest exponent E such that the first digits of 2 E coincide with the integer (remember that more than half of the digits are missing).

30 30 Practice III: Archeologist’s Dilemma Input It is a set of lines with a positive integer N not bigger than in each of them. Output For every one of these integers a line containing the smallest positive integer E such that the first digits of 2 E are precisely the digits of N, or, if there is no one, the sentence ``no power of 2". Sample Input: Sample Output: 7(2, 4, 8, 16, 32, 64, 128=2 7 ) 8(2, 4, 8, 16, 32, 64, 128, 256=2 8 ) 20(2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, , , , =2 20 )

31 31 Analysis of the problem The legible digits form a number N Assume there are k missing digits The we have the following inequalities: Nx10 k <= 2 E < (N+1)x10 k We perform the lg() operation: lg(Nx10 k ) <= lg(2 E ) < lg ((N+1)x10 k ), which leads to lgN + klg10 <= E < lg(N+1) + klg10 We can search for the k such that, there is an integer in the middle

32 32 Sample Solution #include void process(int); int length(int); int main() { int in; while( scanf("%u", &in) != EOF) process(in); return 0; } void process(int n) { int k, e; double d1, d2, d3, d4; double left, right; d4 = log(2); d3 = log(10) / d4; d1 = log(n) / d4; d2 = log(n+1) / d4; k = length(n)+1; left = d1+k*d3; right = d2+k*d3; e = (int)ceil(left); while(e >= right) { k++; left = d1+k*d3; right = d2+k*d3; e = (int)ceil(left); } printf("%d\n", e); } int length(int n) { int len = 0; if(n < 10) return 1; while(n > 0) { n = n/10; len++; } return len; }

33 Practice IV: How Many Fibs? Recall the definition of the Fibonacci numbers: f 1 = 1, f 2 = 2, f n = f n-1 +f n-2 (n >= 3) Given two numbers a and b, calculate how many Fibonacci numbers are in the range [a,b]. 33

34 Practice IV: How Many Fibs? Input Specification The input contains several test cases. Each test case consists of two non-negative integer numbers a and b. Input is terminated by a=b=0. Otherwise, a<=b<= The numbers a and b are given with no superfluous leading zeros. Output Specification For each test case output on a single line the number of Fibonacci numbers f i with a<=f i <=b. 34

35 Practice IV: How Many Fibs? Sample Input Sample Output Hint: Use the bignum functions!

36 36 More practice on basic arithmetic


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