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Binary numbers Primary memory • Memory = where programs and data are stored – Unit = bit • “BIT” is a contraction for what two words? • Either a 1 or a 0 (because computers are made primarily of transistors which act as switches). – Smallest addressable (i.e., can be read from or written to memory) unit is the byte. • byte = 8 bits (an 8-bit number) Binary numbers • How do computers represent numbers and do arithmetic? • How do we? What digits do we use? Why? – What is 903? • 9 is the MSD above; 3 is the LSD. Binary numbers • Can we represent numbers is other bases (radix)? – What is 11012 in base 10? – What is 110110 in base 10? – What is 11013 in base 10? – Note that they are different! Potential programming asssignment Write a program that reads in a string (not an int) representing a binary number and outputs the corresponding base 10 number. Do not use any methods from Java’s Integer class! Potential programming asssignment Write a program that reads in a string (not an int) (representing a # in base b), reads in the base b (an int), and outputs the corresponding base 10 number. (What Java String method can be used to obtain an individual char at a particular position in a string?) Do not use any methods from Java’s Integer class! Binary numbers • What are the range of digits for base 10 numbers? – For binary? – For base 5? – For some arbitrary base k? – Is 11022 a valid base 2 number? – Is 12644 a valid base 4 number? Numbers in mathematics • I = the integers – Finite or infinite? • R = the reals – Contains rational numbers like 2/7 and irrational numbers like – Finite or infinite? • C = the complex numbers – Finite or infinite? Numbers in computers • All represented by a finite sequence of bits (1’s and 0’s). • I = the integers – Finite/bounded. – char (8 bits), short (16 bits), int (32 bits), long long (64 bits) • R = the reals – Finite/bounded. (So what about ?) – float (32 bits), double (64 bits), long double (128 bits) Numbers in computers • R = the reals – Finite/bounded. (So what about ?) – Representation is called floating point. – float (32 bits), double (64 bits), long double (128 bits) – Can also be represented by fixed point and BCD (binary coded decimal). Common ranges of values (Java) • int -2 billion • double 4.9x10-324 • float 1.4x10-45 (closest to 0) (closest to 0) to +2 billion to 1.8x10+308 to 3.4x10+38 Converting from base 10 to base 2 • Convert 6110 to base 2. • Method 1: Subtract largest power of 2 repeatedly. – Also works for both ints and floats. – First, make a table of powers of 2. • • • • • • • • • 28 = 256 27 = 128 26 = 64 25 = 32 24 = 16 23 = 8 22 = 4 21 = 2 20 = 1 Converting from base 10 to base 2 • Convert 6110 to base 2. • Method 1: Subtract largest power of 2 repeatedly. 1. Make a table of powers of 2. • • • • • • • • • 28 = 256 27 = 128 26 = 64 25 = 32 24 = 16 23 = 8 22 = 4 21 = 2 20 = 1 2. Repeatedly subtract (largest power of 2) 61 32 29 16 13 8 5 4 1 1 0 1 25 msb 1 24 1 23 1 22 0 21 1 20 lsb Reading forward, the result is 1111012. Use this method to convert from base 10 to base 3. • Method 1: Subtract largest powers of 3 repeatedly. Step 1: Create table of powers of 3. • • • • • 34 = 81 33 = 27 32 = 9 31 = 3 30 = 1 Step 2: Repeatedly subtract (largest) multiples of largest power of 3. (Why? Recall digit range for a particular base.) Use this method to convert from base 10 to base 3. • Method 1: Subtract largest powers of 3 repeatedly. Step 1: Create table of powers of 3. • • • • • 34 = 81 33 = 27 32 = 9 31 = 3 30 = 1 Step 2: Repeatedly subtract (largest) multiples of largest power of 3. • Convert 6110 to base 3. 61 54 2 33 msd 7 0 32 6 2 31 1 1 1 30 lsd 0 Reading forward, the result is 20213. Potential programming assignment Write a program that takes as input a base 10 number (an int) and outputs the corresponding base 2 number. Do I need to remind you again? Do not use any methods from Java’s Integer class! Converting base 10 floating point numbers to base 2 using Method 1 (repeated subtraction) Step 1: Create table of powers of 2. 24 = 16 23 = 8 22 = 4 21 = 2 20 = 1 2-1 = 0.5 2-2 = 0.25 2-3 = 0.125 2-4 = 0.0625 2-5 = 0.03125 2-6 = 0.015625 Step 2: Subtract largest power of 2 repeatedly. Converting base 10 floating point numbers to base 2 using Method 1 (repeated subtraction) Step 1: Create table of powers of 2. 24 = 16 23 = 8 22 = 4 21 = 2 20 = 1 2-1 = 0.5 2-2 = 0.25 2-3 = 0.125 2-4 = 0.0625 2-5 = 0.03125 2-6 = 0.015625 Step 2: Subtract largest power of 2 repeatedly. Convert 11.07812510 to base 2. 11.078125 8 1 23 3.078125 0 22 2 1 21 1.078125 1 1 20 0.078125 0 2 1 0 2 2 0.0625 0 2 3 1 2 4 0.015625 0 2 5 0.015625 1 2 6 0 Reading forward, the result is 1011.0001012. Conversion methods Method 1. Using a table, repeatedly subtract largest multiple. - directly works for both ints and floats - requires a table Method 2: Repeated (integer) division (with remainder) by desired base. - works for ints but needs to be “modified” to work for floats - doesn’t require a table Method 2: Repeated (integer) division (with remainder) by desired base. Convert 6110 to base 2. 61/ 2 30 61%2 1 30 / 2 15 30%2 0 15 / 2 7 7/2 3 15%2 1 7% 2 1 3/ 2 1 1/ 2 0 3% 2 1 1% 2 1 Reading backwards, the result is 1111012. lsb msb Method 2: Repeated (integer) division (with remainder) by desired base. Convert 6110 to base 3. 61/ 3 20 61%3 1 lsd 20 / 3 6 20%3 2 6/3 2 2/3 0 6% 3 0 2%3 2 msd Reading backwards, the result is 20213. Method 2 for fp numbers • Convert 61.687510 to base 2. – We just saw how to convert 6110 to base 2 (by repeated division with desired base). – So all we need to do is convert 0.687510 to base 2. – This can (analogously) be accomplished by repeated multiplication by the desired base. Method 2 • Convert 0.687510 to base 2 by repeated multiplication by desired base. 0.6875 2 1.375 0.375 2 0.75 • Then read integer part forwards. 0.75 2 1.5 0.5 2 1.0 0.0 2 0.0 0.0 2 0.0 0.687510 0.10112 Method 2 • Convert 0.110 to base 2 by 0.1 2 0.2 repeated multiplication by 0 .2 2 0 .4 desired base. 0 .4 2 0 .8 • Then read integer part forwards. Do you see the repetition/loop? 0 .8 2 1.6 0.6 2 1.2 0 .2 2 0 .4 0.110 0.000110011 ...2 Method 2 • Convert 0.110 to base 2 by 0.1 2 0.2 repeated multiplication by 0 .2 2 0 .4 desired base. same 0 .4 2 0 .8 • Then read integer part forwards. Do you see the repetition/loop? 0 .8 2 1.6 0.6 2 1.2 0 .2 2 0 .4 same 0.110 0.000110011 ...2 Converting between arbitrary bases • In general, to convert from base u to base v, we typically: – Convert from base u to base 10. – Then convert from base 10 to base v. • When converting from base 2 to two other important bases, we can skip the intermediate conversion through base 10 and convert directly from base 2 to these two other bases (and vice versa). Other important bases related to base 2: base 8 • Octal, base 8 – Digits: 0..7 base 8 0 1 2 3 4 5 6 7 • Problems: base 2 000 001 010 011 100 101 110 111 – Convert 1008 to base 10. – Convert 1008 to base 2. – Convert 11012 to base 10. – Convert 11012 to base 8. – Convert 7028 to base 10. – Convert 7028 to base 2. – Convert 8028 to base 2. Other important bases related to base 2: base 16 • Hex (hexadecimal), base 16 – Digits: 16 of them (more than base 10) so we’ll will need 6 extra symbols: {0..9,a..f} base 16 0 1 2 3 4 5 6 7 8 9 a b c d e f base 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 base 2 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 base 8 00 01 02 03 04 05 06 07 10 11 12 13 14 15 16 17 Problems: • Convert 7258 to base 2 to base 16 to base 10. Problems: • Convert 7258 to base 2 to base 16 to base 10. – 111 010 1012 – 1 1101 01012 = 1d516 – 1x162 + 13x161 + 5x160 = 46910 Convert 153.51310 to base 8. Approach: convert 153 and 0.513 separately. 153/ 8 19 1 19 / 8 2 3 2/8 0 2 0.513 8 4.104 0.104 8 0.832 remainder So the result is 231.40651… base 8. 0.832 8 6.656 0.656 8 5.248 0.248 8 1.984 Problem: Convert 3115 to X10 to Y2. • 3x52+1x51x1x50 = 75+5+1 = 8110. 81/ 2 40 40 / 2 20 20 / 2 10 10 / 2 5 5/ 2 2 2/2 1 1/ 2 0 1 lsb 0 0 0 1 0 1 msb 10100012 64 16 1 81