CSE 1400 Applied Discrete Mathematics Conversions Between Number Systems

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1 CSE 400 Applied Discrete Mathematics Conversions Between Number Systems Department of Computer Sciences College of Engineering Florida Tech Fall 20 Conversion Algorithms: Decimal to Another Base Conversion of Natural Numbers Conversion of Integers 3 Biased Notation for Integers 4 Conversion of Rational Numbers 5 Conversion of Floating Point Numbers 7 Problems on Converting Decimal to Another Base 7 Conversion Algorithms: Another Base to Decimal 8 Conversion of Natural Numbers 9 Conversion of Integers 0 Conversion of Rational Numbers Conversion of Floating Point Numbers Floating Point Arithmetic 2 Machine Epsilon 3 Problems on Converting Another Base to Decimal 4 Abstract Decimal notation using Arabic numerals is the standard system for writing numbers in everyday life. However, in many computing applications, other notations are used. Chiefly: binary and headecimal. Learning algorithms to translate names between these numerical systems forms a foundation for developing translations between more comple languages. Conversion Algorithms: Decimal to Another Base Conversion of Natural Numbers The unsigned whole numbers, the natural numbers, written in decimal 0,, 2, 3, 4, 5,... can be converted to binary by repeated division

2 cse 400 applied discrete mathematics conversions between number systems 2 by 2. Dividing a number by 2 produces one of two remainders: r = 0 or r =. The natural numbers can be converted to headecimal by repeated division by 6. Dividing a number by 6 produces one of 6 remainder: r = 0 through r = 5 = F. Convert q = 4 to binary. Repeatedly divide 4 and following quotients by 2 pushing the remainder bits onto a stack. Repeated Remaindering Mod 2 Quotients Remainders 0 (4) 0 = (0) 2 Convert q = 4 to headecimal. Repeatedly divide 4 and following quotients by 6 pushing the remainder he-digits onto a stack. Repeated Remaindering Mod 6 Quotients 4 Remainders E (4) 0 = (E) 6 Convert q = 44 to binary. Repeated Remaindering Mod 2 Quotients Remainders (44) 0 = ( ) 2 Convert q = 44 to headecimal. Repeated Remaindering Mod 6 Quotients 44 9 Remainders 0 9 (44) 0 = (90) 6

3 cse 400 applied discrete mathematics conversions between number systems 3 Convert q = 43 to binary. Repeated Remaindering Mod 2 Quotients Remainders (43) 0 = (000 ) 2 Convert q = 43 to headecimal. Repeated Remaindering Mod 6 Quotients 43 8 Remainders F 8 (43) 0 = (8F) 6 Convert q = 255 to binary. Repeated Remaindering Mod 2 Quotients Remainders (255) 0 = ( ) 2 Convert 6 to binary. Repeated Remaindering Mod 2 Quotients Remainders (6) 0 = (00 000) 2 Convert 6 to ternary. Repeated Remaindering Mod 3 Quotients Remainders (6) 0 = (2222) 3. Notice that 6 =

4 cse 400 applied discrete mathematics conversions between number systems 4 Conversion of Integers The signed integers can be represented in two s complement notation. To write a decimal integer m in two s complement notation do the following Convert the natural number m to binary using repeated remaindering as described in the previous section. 2. Append a leading sign bit If m = +6, return else (if m = 6) return If m 0, return the result of step 2. Otherwise, if m < 0, return the two s complement of the result of step 2. The two s complement operation is performed as follows.. From right-to-left, copy each bit up to and including the first. 2. Flip (one s complement) the remaining bits to the left. The two s complement of a positive integer is a negative integer. The two s complement of a negative integer is a positive integer. The idea is: A number m and its negative m add to 0. Therefore, if m = is a signed integer written in two s complement notation, then m and m s complement (its negative) m add to 0.. This integer m = is positive: It s leading, left-most, bit is The value of m is decimal The two s complement of m is m = The sum of m and m is shown below with carry bits on the top row. c m = m = By construction, the sum of m and it s two s complement m is 2 s where s is the word length, size, or number of bits in m. m + m = 2 s = s 0 s To write 2 s in binary requires s + bits. On an s bit computer the overflow, last carry, bit is discarded. Therefore, on an s bit computer s 0 s m + m = 0 = To write 2 = (0) 2 requires 2 bits. To write 2 2 = (00) 2 requires 3 bits. To write 2 8 = (000) 2 requires 4 bits. Consider this eample. Let m = 227 which when written as an unsigned binary number, is To write m as a signed binary number, append a leading 0 so that +227 =

5 cse 400 applied discrete mathematics conversions between number systems 5 To write 227, perform the two s complement operation on to get 227 = Biased Notation for Integers Biased notation is an alternative method used to represent signed integers. Biased notation represents an integer n as n b = n + b where b 0 is called the bias or ecess. Adding a bias b to an integer n translates it to the right giving it a a new name. To compute the value n of a biased integer n b, solve the equation n b = n + b for n = n b b, that is, subtract b from n b. For illustration, we ll use 8 bit strings. (b 7 b 6 b 5 b 4 b 3 b 2 b b 0 ) b=27 Each b i is either a 0 or a for each i = 0,..., 7. Using a bias of b = 27 the 256 integers from 27 to 28 are named 0 to 255. For instance, consider the biased number ( 00) b=27 = n b = n + b = ( 00) Horner s rule converts the binary number ( 00) 2 to 252. For instance, pick a bias b = 32, and notice the integers from n = 32 to n = 3 map to biased integers from n 32 = 0 to n 32 = 63. Let n = 37 and let the bias be b = 27. Then the biased number ( 37) 27 = (n) b = n + b = = 90 is the biased name for the value 37. The subscript b = 27 is a hint to interpret the bit string as biased number with bias b = 27. Drop it when this is clear from contet. The diagram show the integers from 27 to 28 along the top line. The bottom line shows the biased numbers 0 to 255 which are used to name integers from 27 to Horner s Rule Subtract the bias b = 27 from n + b = 252 to get n = 25 as the value of the biased number ( 00) b=27 = 25 Biased notation is used to encode positive and negative eponents in floating point notation. That is, let t = ±. f 2 e b be a normalized floating point number. The eponent e b is a biased integer. As a small eample, the bias could be b = 4 and eponents from e 4 = 0 to e 4 = 7 represent values 4 to 3.

6 cse 400 applied discrete mathematics conversions between number systems 6 Conversion of Rational Numbers To convert a positive rational number a/b, written in decimal notation, to another base do the following.. Write the number in fied point notation: w. f 2. Convert the whole number part w as described in Conversion of Natural Numbers. 3. Convert the fractional number part f by repeated multiplication by 2 (or the appropriate base) inserting the overflow bit into a queue. Convert q = 39/4 to binary Often the fractional part will not terminate as a finite string.. Write 39/4 as Convert 9 to binary Convert 0.75 to binary. Repeatedly multiply 0.75 and following products by 2 inserting the overflow bits into a queue. Repeated Multiplying Mod 2 Fraction Overflow 0 (39/4) 0 = (00.) 2 Note that the binary fraction 0. represents the value 0.75 = Convert q = 4/3 to binary. Write 4/3 as Convert 4 to binary Convert 0.3 to binary. 0. = = 3 4 Repeatedly multiply 0.3 and following products by 2 inserting the overflow bits into a queue. Repeated Multiplying Mod 2 Fraction Overflow 0 0 (4/3) 0 = (00.00 ) 2 Note the identity 3 = = 4 ( 4 )

7 cse 400 applied discrete mathematics conversions between number systems 7 Convert q = /7 to binary. Write /7 as Convert 0 to binary Convert to binary. Repeatedly multiply and following products by 2 inserting the overflow bits into a queue. Repeated Multiplying Mod 2 Fraction Overflow (/7) 0 = ( ) 2 Convert q = 57/5 to headecimal. Write 57/5 as.4 2. Convert to headecimal B. 3. Convert 0.4 to headecimal. Repeatedly multiply 0.4 and following products by 6 inserting the overflow he-digits into a queue. Note the identity 7 = = 8 ( 8 ) Repeated Multiplying Mod 6 Fraction Overflow (57/5) 0 = (B.666 ) 2 Conversion of Floating Point Numbers Note the identity 2 5 = = 6 6 ( 6 ) A floating point number written in decimal can be converted to another base by writing the number in fied point notation and using the methods of the previous section. Convert to octal.. Write as Repeatedly multiply and following products by 8 inserting the overflow octal-digits into a queue. Repeated Multiplying Mod 8 Fraction Overflow = (0.3) 8

8 cse 400 applied discrete mathematics conversions between number systems 8 Convert to headecimal.. Write as Convert 299 to headecimal Repeated Remaindering Mod 6 Quotients Remainders B = (2B) 6 3. Repeatedly multiply 0.7 and following products by 6 inserting the overflow he-digits into a queue. Repeated Multiplying Mod 6 Fraction Overflow B = (2B.B3 ) 6 Problems on Converting Decimal to Another Base. Convert the following natural numbers (all named q) from decimal to binary. (a) q = 3. (b) q = 34. (c) q = 45. (d) q = Convert the following natural numbers from decimal to headecimal. (a) q = 3. (b) q = 34. (c) q = 45. (d) q = 257. Repeatedly divide q and following quotients by 2 pushing the remainder bits onto a stack. Instead of dividing by 6 and stacking remainders, it may be easier to convert to binary and group bits by fours. 3. Convert the following signed integers from decimal to two s complement binary. (a) q = 3. (b) q = +34. (c) q = 45. (d) q = Convert the following signed integers from decimal to biased decimal notation.

9 cse 400 applied discrete mathematics conversions between number systems 9 (a) q = 3 with bias b = 32. (b) q = +34 with bias b = 256. (c) q = 45 with bias b = 256. (d) q = 257 with bias b = Convert the following rational numbers, written in decimal, to binary. (a) q = 27/8. (b) q = 5.. (c) q = (d) q = 0.. Conversion Algorithms: Another Base to Decimal Horner s rule is a key algorithm for converting a number, written in base b, to decimal. A number m written in base b can be epressed in polynomial form. m = a n b n + a n 2 b n a b + a 0 b 0 where the coefficients a n, a n 2,..., a, a 0 are numbers in the set Z b of integers mod b Z b = {0,, 2,..., (b )} Horner s rule can be comprehended by parenthesizing the operations required to evaluate the value of m. m = ( ((a n b + a n 2 )b + a n 3 )b + + a )b + a 0 There are at least two choices for storing the coefficients: Littleendian $[a_{0},\,a_{},\,a_{2},\ldots,\,a_{n-2},\,a_{n-}]$ and big-endian or $[a_{n-},\,a_{n-2},\,a_{n-3},\ldots,\,a_{},\,a_{0}]$ If the coefficients are stored in little-endian, then horner :: Float -> [Float] -> Float horner c [] = 0 horner c [] = horner c (:s) = + c * (horner c s) horner :: Float -> [Float] -> Float horner c [] = 0 horner c [] = horner c (:s) = horner c ([c * last s]++tail s)

10 cse 400 applied discrete mathematics conversions between number systems 0 horner :: Float -> [Float] -> Float horner c [] = 0 horner c [] = horner c (:s) = horner c ([c * + head s]++tail s) Conversion of Natural Numbers Consider the natural number forty-one. Written in binary, forty-one is (000 00) 2. Horner s rule will convert this string into its decimal form: (4) 0. Horner s Rule Table : Converting (00 00) 2 to (4) 0. In a like manner 23 written in binary is (0 0) 2. This can be checked by Horner s rule. Horner s Rule Horner s rule can convert octal (737) 8 to decimal using 8 as the multiplier. The computation is summarized in the table below. Horner s Rule Notice that (737) 8 = = 479 Choose a base 2 b < 0 and a string over that base. Convert the string to decimal using Horner s rule. Show that you understand how the string you wrote represents the decimal value computed by Horner s rule.

11 cse 400 applied discrete mathematics conversions between number systems Conversion of Integers Signed integers can be written in two s complement notation. By construction, an integer m and it s two s complement m sum to 2 s where s is the word size of m. To convert the two s complement binary number m = to decimal do the following.. Recognize that m is positive: It s leading, left-most, bit is Use Horner s rule to convert to decimal = Horner s Rule To convert the two s complement binary number m = to decimal do the following.. Recognize that m is negative: It s leading, left-most, bit is. 2. Method : Use Horner s rule to convert to decimal. Horner s Rule Subtract 2 9 = 52 to get = = Method 2: Compute the two s complement of to get Convert to decimal (in your head or by Horner s rule) to get 56. Negate the result = 56. Conversion of Rational Numbers The rational number two and nine-siteenth when written in binary is (000.00) 2. Horner s rule followed by division by 2 4 will convert this string into its decimal fraction form: 4/6. Notices the similarity of division by 2 to the fourth and writing 3.45 as 345/0 4. Horner s Rule Figure : Converting (00.00) 2 to 4/6.

12 cse 400 applied discrete mathematics conversions between number systems 2 Conversion of Floating Point Numbers Floating point numbers are commonly used to approimate the real numbers. It is beyond the scope of this course to provide a complete description of floating point numbers, but some essential ideas should be known before writing solutions to problems that involve real numbers. Floating point notation is built upon scientific notation. In general, rational numbers written in scientific notation have the form t = ±d. f 0 e where d is a non-zero digit, f is a decimal string, and e is a base 0 eponent. To gain understanding of floating point numbers let s consider how to write rational numbers in the form For instance, the scientific notation for is Restricting the leading digit to be nonzero is called This avoids multiple representations for identical values. For instance, is normalized as In computing practice, t could be stored as a concatenated string t = s e d f, where s is a sign bit. t = ±. f 2 e where 8 bits are used to encode the sign, the fraction f, and the eponent e. Denote this set of floating pointing numbers F. A binary floating point number t F is an 8-bit string that is parsed into three parts as shown below. Sign Bit Eponent (bias=3) Fraction s e 2 e e 0 f f 2 f 3 f 4 The sign bit s controls the whether the floating point number is positive or negative. The net three bits e = e 2 e e 0 encode a base 2 eponent written in biased notation, with bias b = 3. That means the biased eponents in the range 000,..., represent the natural numbers in the range 3,..., 4. The four fraction bits f = f f 2 f 3 f 4 encode the natural numbers from 0 to 5, but their value is interpreted as f 6 = ( f f 2 f 3 f 4 ) Finally, ecept for 0, which is the all zero string , we assume that all floating point numbers are normalized. Therefore, if The sign bit s conveniently encodes the sign of t by the function ( ) s. When s = 0 the floating point number is positive. When s = the floating point number is negative. This is similar to writing the decimal number 0.34 as 34/0 4. t = s e 2 e e 0 f f 2 f 3 f 4 = s e f is a floating point number, it represents the rational number ( t = ( ) s + f ) 2 e 6

13 cse 400 applied discrete mathematics conversions between number systems 3 where s B, 0 f 5, and 3 e 4. The eight line graphs that follow show the distribution of these floating point numbers at each eponential scale Floating Point Arithmetic Floating point number systems are rife with arithmetic errors. Even the most basic rules do not apply. Floating point arithmetic is not closed under addition. For instance t = 6 = = (0 0000) f p and s =.0 2 = 0.5 = ( ) f p are floating point number, but their t + s = 6.5 sum is not a number in this system. Floating point arithmetic is not associative. For instance ( ) = 6 but 6 + ( ) = 7 There are several measures of floating point errors that occur when approimating arithmetic over the real numbers. Let t() be the floating point number closest to real number. Then e() = t() is the rounding error in approimating by floating point number t(). As the graphs above show, the rounding error increases as the magnitude of increases. The relative error in approimating = 0 by t() is defined to be Addition is computed by (a) aligning eponents to the higher power: ; (b) and adding fractional parts , and truncating to the accuracy of the system, in our system, 4 places to the right of the binary point: = = 6. E() = e() = t() Look at the first line graph above, where the magnitude of scale is the smallest.

14 cse 400 applied discrete mathematics conversions between number systems The floating point number closest to = 8/28 = /6 is t = 0. Therefore /6 rounds to 0. The rounding error is e(/6) = /6 and the relative error is E(/6) =. The floating point number closest to = 6/28 = /8 is t = 7/28. Therefore /8 rounds to 7/28. The rounding error is e(/8) = /28 and the relative error is E(/8) = /6. The floating point number closest to = /256 is t = 7/28. Therefore /256 rounds to 7/28. The rounding error is e(/256) = /256 and the relative error is E(/256) = /. Real FloatingPoint RoundingError RelativeError t() e() E() Machine Epsilon In floating point arithmetic, n = 5 is the power of 2 such that + 2 n =. The value ɛ = 2 5 = /32 is called the machine s epsilon and measures accuracy or inaccuracy to floating point arithmetic. = = 65/64 F = = /32 F < = 7/6 F Problems on Converting Another Base to Decimal. Use Horner s rule to convert the following unsigned numbers into decimal notation. (a) ( ) 2 (b) ( ) 2 (c) (00 00) 2 (d) (00 00) 2 (e) (747) 8 (f) (23) 4 (g) (A47) 6 2. Convert the signed decimal integers below into two s complement notation.

15 cse 400 applied discrete mathematics conversions between number systems 5 (a) +76. (b) 76. (c) 37. (d) Negate the two s complement integers below. (a) (b) (c) (d) Convert the following two s complement integers into decimal notation. (a) (b) (c) (d) (e) 62A3. (f) C0B. 5. Using two s complement notation what range of integers from most negative to most positive can be represented using (a) 2 bits? (b) 8 bits? (c) 32 bits? (d) 64 bits? 6. Consider ten s complement notation. An integer and its complement should sum to zero. Writing in 4 digits the integer 234 has complement 8766 since = The smaller numbers from 0000 to 4999 are positive and have their normal value. The larger values 5000 to 9999 are negative and defined by the complement rule illustrated in the eample. (a) What are the complements of (a) 34? (b) 736? (c) 9999? (d) 5000? (b) What are the decimal numbers for the ten s complements numbers (a) 34? (b) 736? (c) 9999? (d) 5000? 7. The following binary strings are floating point numbers where the first (leftmost) bit is a sign bit, the net three bits are eponent bits written in biased notation with bias b = 3, and the last four bits are fraction bits. These floating point are normalized. What are decimal names for these floating point numbers? (a) (b) 00 0 (c) (d) Using the 8 bit floating point numbers from the course, eplain why 7/28 is the smallest positive floating point number and why /8 is not.

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