 Instructor: Rob Nash  Readings: Berger, Chapter 1; Rosen, Chapter 3

 A set of steps in a specific order, or  A procedure for performing some (usually arithmetic) operation  These algorithms, adapted for use with binary, form the basis for computer arithmetic.

 In addition to arithmetic operations…  We’ll take a look at Euclid's Algorithm  An algorithm for finding the base b expansion of a positive integer for any base b  And, we’ll take some time to consider an algorithm’s performance

 We can represent numbers in other bases, however, often for convenience of manipulation or translation.  We’ll need to know base 2, since those are our machine’s symbols  We’ll also see that base 16 (hex) is useful for representing bit strings in a compact fashion

 Early tube & transistor technology functions much like a lightswitch  Today’s integrated circuits also offer “microswitches” as the basis for feedback loops (memory)  We could have had to manipulate numbers in base 3 if our early technology functioned accordingly  In fact, we’ll try out base 3 for fun

 965 = (9 * 10^2) + (6 * 10^1) + (5 * 10^0)  Just an expanded form that highlights the sum of products that compose this number ▪ Also highlights the base here, which is 10  In fact, we can use any int > 1 as a base when expressing integers, and we can convert between these bases  We’ll often use base 10 as the intermediary base when converting ▪ But, we can also convert from one base to another directly (16 -> 2)

 (101) base 2 is also a compact form of:  101 = (1 * 2^2) + (0 * 2^1) + (1 * 2^0 ) = 5  What about 1011?

 Let b be a positive integer greater than 1. If n is a positive integer, it can be uniquely expressed in the form:  n = a k b k + a k-1 b k-1 + … + a 1 b 1 + a 0 b 0

 Let b be a positive integer greater than 1. If n is a positive integer, it can be uniquely expressed in the form:  n = a k b k + a k-1 b k-1 + … + a 1 b 1 + a 0 b 0  Where k is a nonnegative integer  And a 0 - a k are nonnegative integers < b  This simply means our coefficients can’t exceed our base  And a k is not equal to 0  i.e. eliminate leading zeroes  The base b expansion of n is then (a k, a k-1,…, a 0 ) b  Base is indicated by subscript b

 The idea here: we can denote base b expansion of an integer n as (a k, a k-1,…, a 0 ) b  From this we can calculate the decimal expansion as n = a k b k + a k-1 b k-1 + … + a 1 b 1 + a 0 b 0  So, (245) 8 represents…  (2 * 8^2) + (4 * 8^1) + (5 * 8^0)  (2 * 64) + (4 * 8) + (5 * 1)  = 165

 Try to convert the following from the various bases to base 10  498 = (4 * 10^2) + (9 * 10^1) + (8 * 10^0)   (110) 2 = (? * 2^2) + (? * 2^1) + (? * 2^0 )  (201) 3 = (? * 3^?) + (0 * 3^?) + (1 * 3^? )

 (110) 2 = (1 * 2^2) + (1 * 2^1) + (0 * 2^0 )  (201) 3 = (2 * 3^2) + (0 * 3^1) + (1 * 3^0 )  (1111) 2 = ?  (0xCAFE) 16 = ?

 Lets use 2 as the base  The binary expansion of an integer, then, is just a string of bits  Lets consider first the decimal expansion of the integer that has “ ” as its binary expansion?  First, use the full form  Then, consider a quick counting trick

 Hex is quite useful in computer science, and we’ll soon see why  Hex symbols: {0,1,…,9,A,B,C,D,E,F}  Where A == 10, B == 11, …, F == 15  (0x01F) 16 = (0*16^2) + (1*16^1) + (15*16^0) = 31  (0xCAFE1) 16 = (12 * 16^4) + (10 * 16^3) + (15 * 16^2) + (14 * 16^1) + 1 = big base 10 int

 Consider (0xCAFE1) 16  Each hex symbol represents (2^4) unique symbols  Thus, a hex symbol neatly encodes 4 bits  Two hex symbols comprise a byte  So, if we can memorize all the bit patterns from (only 16 of them, to represent the integers 0-15), we can quickly convert from base 16 to base 2.

 0000 –  0001 –  0010 –  0011 –  0100 –  0101 –  0110 –  0111 –

 So far, this is our tool for converting some integer n from an arbitrary base b to base 10.  We know now how to translate from any base to decimal, but…  How do we translate from decimal to an arbitrary base b?  We can derive a second algorithm from Theorem 1, which accomplishes expansions in arbitrary bases  Binary expansions  Hexadecimal expansions  …

 Now lets consider this algorithm for constructing a base b expansion from a decimal integer.  With the two algorithms combined, we can go from base x to base 10, and then from base 10 to some other base y  And we’ll learn shortcuts to cut out the “middleman” along the way

 Lets see a Base b expansion of an integer n  Approach:  First, divide n by b to obtain a quotient q0 and a remainder r0. ▪ r0 is your least significant digit (rightmost digit)  Next, divide q0 by b to produce another quotient q1 and a remainder r1 ▪ r1 is your next least significant digit (second to rightmost)  Repeat until qx == 0

 Convert to n=31 to binary, b=2  31/2 = 15 r 1 //the rightmost bit  15/2 = 7 r 1  7/2 = 3 r 1  3/2 = 1 r 1  1/2 = 0 r 1 //the most significant bit

 Given n=165, convert to hexadecimal, b=16  165 / 16 = 10 r 5  10/ 16 = 0 r A  Answer: 0xA5

 Given n=241, find the binary expansion, b=2  241 / 2 = 120 r 1  120 / 2 = 60 r 0  60 / 2 = 30 r 0  30 / 2 = 15 r 0  15 / 2 = 7 r 1  7 / 2 = 3 r 1  3 / 2 = 1 r 1  1 / 2 = 0 r 1  Answer:  Now try converting this bit string to hex using our “shortcut”

 //invariant: b, n are positive ints  Procedure baseBExpansion(int n, int b) {  q = n  k = 0  while( q != 0 ) { ▪ a k = q mod b ▪ q = floor( q / b ) //if q & b are ints, no need for floor ▪ k = k + 1  }  return (a k-1 …a 1 a 0 ) b  }