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Integer Representations & Bases

Integer Representations & Bases. Section 4.2 Wednesday 4 June. 1. Section Summary. Integer Representations Base b Expansions Binary Expansions Octal Expansions Hexadecimal Expansions Base Conversion Algorithm Algorithms for Integer Operations. Representations of Integers.

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Integer Representations & Bases

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  1. Integer Representations & Bases Section 4.2 Wednesday 4 June INTEGER REPRESENTATIONS & BASES 1

  2. Section Summary • Integer Representations • Base b Expansions • Binary Expansions • Octal Expansions • Hexadecimal Expansions • Base Conversion Algorithm • Algorithms for Integer Operations INTEGER REPRESENTATIONS & BASES

  3. Representations of Integers • Societies today use decimal notation, or base 10notation, to denote integers. • For example: 965 abbreviates 9∙102 + 6∙101 + 5∙100 . • But we could use any base b, where b is a positive integer greater than 1, to represent integers. • The bases b = 2 (binary), b = 8 (octal) , and b= 16 (hexadecimal) are important for computing and communications • The ancient Mayans used base 20 and the ancient Babylonians used base 60. INTEGER REPRESENTATIONS & BASES

  4. Why binary, octal, hex? • Physical representations for 0 and 1 • 0 volts v.s. 5 volts • lit v.s. dark • rough spot v.s. shiny spot (on CD surface) • etc. • Octal and hexidecimal are more compact and readable; easy to convert back and forth. INTEGER REPRESENTATIONS & BASES

  5. Base b Representations • Theorem 1: Let b be a positive integer greater than 1. If n is a positive integer, then n can be expressed uniquely in the form: n = akbk + ak-1bk-1 + …. + a1b + a0 where k is a nonnegative integer; a0, a1,…, ak are nonnegative integers less than b; and ak ≠ 0. • The integers aj in this theorem are called the base-b digits of the representation of n. • The representation of n in Theorem 1 is called the base b expansion of nand is denoted by (akak-1….a1a0)b. • We usually omit the subscript 10 for base 10 expansions. INTEGER REPRESENTATIONS & BASES

  6. Example: Binary Expansions • The binary, or base 2, expansion, uses the digits 0 and 1 only. (Why?) • Exercise: What decimal number equals (1 0101 1111)2? • (1 0101 1111)2 = 1∙28 + 0∙27 + 1∙26 + 0∙25 + 1∙24 + 1∙23 + 1∙22 + 1∙21 + 1∙20 =351. • Exercise: What decimal number equals (1 1011)2 ? • (11011)2 = 1 ∙24 + 1∙23 + 0∙22 + 1∙21 + 1∙20 = 27. INTEGER REPRESENTATIONS & BASES

  7. Example: Octal Expansions • The octal expansion (base 8) uses the digits {0,1,2,3,4,5,6,7}. • Exercise: What decimal number equals (7016)8? • 7∙83 + 0∙82 + 1∙81 + 6∙80 = 3598 • Exercise: What decimal number equals (111)8 ? • 1∙82 + 1∙81 + 1∙80 = 73 INTEGER REPRESENTATIONS & BASES

  8. Hexadecimal Expansions • The hexadecimal, or base 16, expansion uses {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F} for digits. • (A)16 through (F)16 represent (10)10 through (15)10. • Exercise: What decimal number equals (2AE0B)16 ? • 2∙164 + 10∙163 + 14∙162 + 0∙161 + 11∙160 =175627 • Exercise: What decimal number equals (E5)16 ? • 1∙162 + 14∙161 + 5∙160 = 256 + 224 + 5 = 485 INTEGER REPRESENTATIONS & BASES

  9. Base Conversion To construct the base b expansion of an integer n: • Divide n by b to obtain a quotient q0 and remainder a0. n = b ∙ q0 + a0 0 ≤ a0 <b • The remainder, a0 , is the rightmost digit in the base b expansion of n. • Divide q0 by b to obtain a quotient q1 and remainder a1. q0 = b ∙ q1 + a1 0 ≤ a1 <b • The remainder, a1, is the second digit from the right in the base b expansion of n. • Continue by successively dividing the quotient by b, obtaining the next base-b digit as the remainder. • The process terminates when the quotient is 0. continued→ INTEGER REPRESENTATIONS & BASES

  10. Base Conversion: Why does this work? This process produces:n = b ∙ q0 + a0 0 ≤ a0 <b q0 = b ∙ q1 + a10 ≤ a1 <b q1 = b ∙ q2 + a20 ≤ a2 <b …… qk-2 = b ∙ qk-1 + ak-10 ≤ ak-1<b qk-1 = b ∙ 0 + ak0 ≤ ak<b • n = b ∙ q0 + a0 • = b ∙ (b ∙ q1 + a1) + a0 • = b2∙ q1 + b ∙ a1 + a0 • = b2 ∙ (b ∙ q2 + a2) + b ∙ a1 + a0 • = b3∙ q2 + b2∙a2 + b ∙ a1 + a0 … • = bk ∙ qk-1 + bk-1 ∙ak-1 + bk-2 ∙ak-2 + … … + b2∙ a2 + b ∙ a1 + a0 • = bk ∙ ak + bk-1 ∙ak-1 + bk-2 ∙ak-2 + … … + b2∙ a2 + b ∙ a1 + a0 • Thus, n = (ak ak-1 … a2 a1 a0)b INTEGER REPRESENTATIONS & BASES

  11. Algorithm: Constructing Base b Expansions • q represents the quotient obtained by successive divisions by b, starting with q = n. • The digits in the base b expansion are the remainders of the division given by qmodb. • The algorithm terminates when q = 0 is reached. procedurebase b expansion(n, b: positive integers with b > 1) q := n k := 0 while (q ≠ 0) ak := qmodb q := qdivb k := k + 1 return(ak-1 ,…, a1,a0) // {(ak-1 … a1a0)bis baseb expansion of n} INTEGER REPRESENTATIONS & BASES

  12. Base Conversion Exercise: Find the octal expansion of (12345)10 Solution: Successively dividing quotients by 8 gives: • 12345 = 8 ∙ 1543 + 1 (q1 = 1543, r1 = 1) • 1543 = 8 ∙ 192 + 7 (q2 = 192, r2 = 7) • 192 = 8 ∙ 24 + 0 (q3 = 24, r3 = 0) • 24 = 8 ∙ 3 + 0 (q4 = 3, r4 = 0) • 3 = 8 ∙ 0 + 3 (q5 = 0, r5 = 3) The remainders are the digits from right to left yielding (30071)8. INTEGER REPRESENTATIONS & BASES

  13. Base Conversion Exercise: Find the octal expansion of (12345)10 Solution: Successively dividing quotients by 8 gives: • q1 = 12345 div 8 = 1543, a1 = 12345 mod 8 = 1 • q2 = 1543 div 8 = 192, a2 = 1543 mod 8 = 7 • q3 = 192 div 8 = 24, a3 = 192 mod 8 = 0 • q4 = 24 div 8 = 3, a4 = 24 mod 8 = 0 • q5 = 3 div 8 = 0, a5 = 3 mod 8 = 3 The remainders are the digits from right to left yielding (30071)8. INTEGER REPRESENTATIONS & BASES

  14. Comparison of Hexadecimal, Octal, and Binary Representations Initial 0s are not shown When padded with initial 0’s: Each octal digit corresponds to a block of 3 binary digits. Each hexadecimal digit corresponds to a block of 4 binary digits. This makes conversion between binary, octal, and hexadecimal easy. INTEGER REPRESENTATIONS & BASES

  15. Conversions between binary and octal • Binary to octal: replace 3-bit blocks with octal digit • Example: (11111010111100)2 (011 111 010 111 100)2 ( 3 7 2 7 4 )8 • Octal to binary: replace octal digit with 3-bit binary equivalent • Example: (37274)8 ( 3 7 2 7 4 )8 (011 111 010 111 100)2 INTEGER REPRESENTATIONS & BASES

  16. Conversions between binary and hex • Binary to octal: hexidecimal: replace 4-bit blocks with hexidecimal digit • Example: (11111010111100)2 (0011 1110 1011 1100)2 ( 3 E B C )16 • Hexidecimal to binary: replace hexidecimal digit with 4-bit blocks with binary block • Example: (3EBC)16 ( 3 E B C )16 (0011 1110 1011 1100)2 INTEGER REPRESENTATIONS & BASES

  17. Addition of Base-b Integers • Addition of decimal numbers generalizes for any base • For base-b digits, n1, n2,… nk ∈ { 0, 1, 2, …, b − 1 }, convert the decimal sum n1 + n2 + … + nkto base-b • Example: (0)2 + (1)2 = (1)2 since 0 + 1 = 1 = (1) 2 • Example: (1)2 + (1)2 = (10)2 since 1 + 1 = 2 = (10) 2 • Example: (1)2 + (1)2 + (1)2 = (11)2 since 1 + 1 + 1 = 3 = (11) 2 • Calculate the sum of the base-b digits from right to left, potentially carrying a bit. • Example: 1 1 carry bits (1 1 1 0 )2 + (1 0 1 1 )2 (1 1 0 0 1 )2 17 MSU/CSE 260 Fall 2009

  18. Addition of Base-b Integers • Addition of decimal numbers generalizes for any base • For base-b digits, n1, n2,… nk ∈ { 0, 1, 2, …, b − 1 }, convert the decimal sum n1 + n2 + … + nkto base-b • Example: (0)2 + (1)2 = (1)2 since 0 + 1 = 1 = (1) 2 • Example: (1)2 + (1)2 = (10)2 since 1 + 1 = 2 = (10) 2 • Example: (1)2 + (1)2 + (1)2 = (11)2 since 1 + 1 + 1 = 3 = (11) 2 • Calculate the sum of the base-b digits from right to left, potentially carrying a bit. • Example: 1 1 carry bits (1 1 1 0 )2 (14)10 + (1 0 1 1 )2 (11)10 (1 1 0 0 1 )2 (25)10 18 MSU/CSE 260 Fall 2009

  19. Addition of Base-b Integers • Addition of decimal numbers generalizes for any base • For base-b digits, n1, n2,… nk ∈ { 0, 1, 2, …, b − 1 }, convert the decimal sum n1 + n2 + … + nkto base-b • Example: (3)8 + (7)8 + (1)8 = (13)8 since 3 + 7 + 1 = 11 = (13) 8 • Calculate the sum of the base-b digits from right to left, potentially carrying a bit. • Example: 1 1 carry bits (1 3 5 1 )8 + (1 7 3 4 )8 (3 3 0 5 )8 19 MSU/CSE 260 Fall 2009

  20. Addition of Base-b Integers • Addition of decimal numbers generalizes for any base • For base-b digits, n1, n2,… nk ∈ { 0, 1, 2, …, b − 1 }, convert the decimal sum n1 + n2 + … + nkto base-b • Example: (3)8 + (7)8 + (1)8 = (13)8 since 3 + 7 + 1 = 11 = (13) 8 • Calculate the sum of the base-b digits from right to left, potentially carrying a bit. • Example: 1 1 carry bits (1 3 5 1 )8 (745)10 + (1 7 3 4 )8 (988)10 (3 3 0 5 )8 (1733)10 20 MSU/CSE 260 Fall 2009

  21. Exercise Calculate the following sums: (1100110)2 (60352)8 (A750C)16 +( 101111)2 +(32277)8 +(B9553)16 (10010101)2 (112651)8 (160A5F)16 INTEGER REPRESENTATIONS & BASES

  22. Multiplying Base-b Integers • Multiplication of base-b numbers is similarly based on the usual method for multiplying decimal numbers. • For base-b digits, n1, n2,… nk ∈ { 0, 1, 2, …, b − 1 }, convert the decimal product n1∙n2∙ … ∙nkto base-b Example: (0)2∙ (0)2 = (0)2∙ (1)2 = (1)2∙ (0)2 = (0)2 (1)2∙ (1)2 = (1)2 • Example (1 1 0)2  (1 0 1)2 (1 1 0)2 (0 0 0)2 + (1 1 0)2___ (1 1 1 1 0)2 • Up to 2n bits may be needed to represent the product of two n-bit numbers. 22 MSU/CSE 260 Fall 2009

  23. Multiplying Base-b Integers • Multiplication of base-b numbers is similarly based on the usual method for multiplying decimal numbers. • For base-b digits, n1, n2,… nk ∈ { 0, 1, 2, …, b − 1 }, convert the decimal product n1∙n2∙ … ∙nkto base-b Example: (0)2∙ (0)2 = (0)2∙ (1)2 = (1)2∙ (0)2 = (0)2 (1)2∙ (1)2 = (1)2 • Example (1 1 0)2 (6)10  (1 0 1)2 (5)10 (1 1 0)2 + (0 0 0)2 (1 1 0)2___ (1 1 1 1 0)2 (30)10 • Up to 2n bits may be needed to represent the product of two n-bit numbers. 23 MSU/CSE 260 Fall 2009

  24. Multiplying Base-b Integers • Multiplication of base-b numbers is similarly based on the usual method for multiplying decimal numbers. • For base-b digits, n1, n2,… nk ∈ { 0, 1, 2, …, b − 1 }, convert the decimal product n1∙n2∙ … ∙nkto base-b Example: (3)8∙ (6)8 = (22)8 since 3∙6 = 18 = (22)8 • Example (7 6 1)8  ( 2 3)8 (2 7 2 3)8 + (1 7 4 2)8 (2 2 3 4 3)8 • Up to 2n bits may be needed to represent the product of two n-bit numbers. 24 MSU/CSE 260 Fall 2009

  25. Multiplying Base-b Integers • Multiplication of base-b numbers is similarly based on the usual method for multiplying decimal numbers. • For base-b digits, n1, n2,… nk ∈ { 0, 1, 2, …, b − 1 }, convert the decimal product n1∙n2∙ … ∙nkto base-b Example: (3)8∙ (6)8 = (22)8 since 3∙6 = 18 = (22)8 • Example (7 6 1)8 (497)10  ( 2 3)8 (19)10 (2 7 2 3)8 + (1 7 4 2)8 (2 2 3 4 3)8 (9443)10 • Up to 2n bits may be needed to represent the product of two n-bit numbers. 25 MSU/CSE 260 Fall 2009

  26. Exercise Calculate the following products: (10110)2 (705)8 (78)16 × (110)2 × (52)8 × (2B)16 INTEGER REPRESENTATIONS & BASES

  27. Exercise Calculate the following products: (10110)2 (705)8 (78)16 × (110)2 × (52)8 × (2B)16 (00000)2 (10110)2 (10110)2 (10000100)2 INTEGER REPRESENTATIONS & BASES

  28. Exercise Calculate the following products: (10110)2 (705)8 (78)16 × (110)2 × (52)8 × (2B)16 (00000)2 (10110)2 (10110)2 (10000100)2 (1612)8 (4331)8 (45122)8 INTEGER REPRESENTATIONS & BASES

  29. Exercise Calculate the following products: (10110)2 (705)8 (78)16 × (110)2 × (52)8 × (2B)16 (00000)2 (10110)2 (10110)2 (10000100)2 (1612)8 (4331)8 (45122)8 (528)16 (F0)16 (1428)16 INTEGER REPRESENTATIONS & BASES

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