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CS103 Guest Lecture
Number Systems & ConversionBitwise Logic Operations
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Why 1’s and 0’s• Transistors are electronic devices used to build computer hardware
– Like a switch (2 positions)– Conducting / Non‐conducting– Output voltage of a transistor will either be high or low
• 1’s and 0’s are arbitrary symbols representing high and low voltage outputs.
• 2 states of the transistor lead to only 2 values in computer hardware
Low Voltage0V
-12V
High Voltage+5V
+12V1
0or
on
offControlling
Input(Gate )
Output (Drain )
Source
The voltage here determines if current
can flow between drain and source
Schematic Symbol of a Transistor
Functional View of a Transistor as a Switch
circuit is open (off) –no current can flow
circuit is closed (on) –current can flow
Circuit Diagram of a Switch Circuit Diagram of a Switch
---
-
- -- -
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POSITIONAL NUMBER SYSTEMS
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Interpreting Binary Strings
• Given a string of 1’s and 0’s, you need to know the representation system being used, before you can understand the value of those 1’s and 0’s.
• Information (value) = Bits + Context (System)
01000001 = ?
6510 ‘A’ASCII
Unsigned Binary system ASCII
system
Signed System
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Binary Number System
• Humans use the decimal number system– Based on number 10– 10 digits: [0‐9]
• Because computer hardware uses digital signals with 2 states, computers use the binary number system– Based on number 2– 2 binary digits (a.k.a bits): [0,1]
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Number Systems• Number systems consist of
1. A base (radix) r2. r coefficients [0 to r‐1]
• Human System: Decimal (Base 10): 0,1,2,3,4,5,6,7,8,9
• Computer System: Binary (Base 2): 0,1• Human systems for working with computer systems (shorthand for human to read/write binary)– Octal (Base 8): 0,1,2,3,4,5,6,7– Hexadecimal (Base 16): 0‐9,A,B,C,D,E,F (A thru F = 10 thru 15)
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Anatomy of a Decimal Number• A number consists of a string of explicit coefficients (digits).• Each coefficient has an implicit place value which is a power
of the base.• The value of a decimal number (a string of decimal
coefficients) is the sum of each coefficient times it place valueradix
(base)
(934)10 =
(3.52)10 =
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Anatomy of a Decimal Number• A number consists of a string of explicit coefficients (digits).• Each coefficient has an implicit place value which is a power
of the base.• The value of a decimal number (a string of decimal
coefficients) is the sum of each coefficient times it place value
Explicit coefficientsImplicit place values
radix (base)
(934)10 = 9*102 + 3*101 + 4*100 = 934
(3.52)10 = 3*100 + 5*10‐1 + 2*10‐2 = 3.52
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Positional Number Systems (Unsigned)
• A number in base r has place values/weights that are the powers of the base
• Denote the coefficients as: ai
r -1 r -2r 1 r 0
.r 2r 3
Left-most digit = Most Significant
Digit (MSD)
Right-most digit = Least Significant
Digit (LSD)
...... a -1 a -2a 1 a 0a 2a 3
Nr = Σi(ai*ri) = D10
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Examples
(746)8 =
(1A5)16 =
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Examples
(746)8 = 7*82 + 4*81 + 6*80= 448 + 32 + 16 = 48610
(1A5)16 = 1*162 + 10*161 + 5*160= 256 + 160 + 5 = 42110
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Anatomy of a Binary Number
• Same as decimal but now the coefficients are 1 and 0 and the place values are the powers of 2
(1011)2 = 1*23 + 0*22 + 1*21 + 1*20
Least Significant Bit (LSB)
Most Significant Digit (MSB)
coefficientsplace values= powers of 2
radix (base)
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Binary Examples
(1001.1)2 =
(10110001)2 =
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Binary Examples
(1001.1)2 = 8 + 1 + 0.5 = 9.510.51248
(10110001)2 = 128 + 32 + 16 + 1 = 177101632128 1
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Powers of 220 = 121 = 222 = 423 = 824 = 1625 = 3226 = 6427 = 12828 = 25629 = 512210 = 1024
512 256 128 64 32 16 8 4 2 11024
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Practice On Your Own
(11010)2 = 1*24 + 1*23 + 1*21
= 16 + 8 + 2 = (26)10
(6523)8 = 6*83 + 5*82 + 2*81 + 3*80= 3072 + 320 + 16 + 3 = (3411)10
(AD2)16 = 10*162 + 13*161 + 2*160
= 2560 + 208 + 2 = (2770)10
• Decimal equivalent is…… the sum of each coefficient multiplied by its
implicit place value (power of the base)
= Σi(ai * ri) [ai = coefficient, r = base]
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Shifting in Binary
• Move bits to the left or right with 0's shoved in one side and dropping bits on the other
• Useful for multiplying and dividing by power of 2.– Right shift by n‐bits = Dividing by 2n
– Left shift by n‐bits = Multiplying by 2n
0 0 ... 0 0 1 1
Right Shift by 2 bits:
... 0 1 1 0 0 0 0 0
Left Shift by 3 bits:
0’s shifted in… 0’s shifted in…
0 ... 0 1 1 0 0 = +12
= +3 = +96
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Bottom‐Up Conversion & Shifting
• X = 0112 =
• 01102
• 011012
• 0110102
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Unique Combinations
• Given n digits of base r, how many unique numbers can be formed? rn
– What is the range? [0 to rn‐1]
Main Point: Given n digits of base r, rn unique numbers can be made with the range [0 - (rn-1)]
2-digit, decimal numbers (r=10, n=2)
3-digit, decimal numbers (r=10, n=3)
4-bit, binary numbers (r=2, n=4)
6-bit, binary numbers (r=2, n=6)
0-90-9
100 combinations:00-99
0-10-10-10-1
1000 combinations:000-999
16 combinations:0000-1111
64 combinations:000000-111111
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CONVERSION FROM DECIMAL TO OTHER BASE
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Conversion: Base 10 to Base r
• X10 = (?)r• General Method (base 10 to arbitrary base r)
– Division Method for integer portion or number
• Alternate Method– Left‐to‐right (greedy) approach
(45)10= (?)r
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Division Method Explanation
24 23 22 21 20
a4 a3 a2 a1 a04510=
4510= a424 + a323 + a222 + a121 + a020
0.
4510= a424 + a323 + a222 + a121 + a020
2 2
22.510= a423 + a322 + a221 + a120 + a02-1
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Binary Division Method Example
4510 = (??)2
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Binary Division Method Example
4510 = (??)2
4522
22
11252
0Keep dividing until you reach 0
rem. = 1rem. = 0rem. = 1rem. = 1
MSB
LSB
4510 = (101101)2
Remainders form the number in
base r (order from bottom up)
2212 rem. = 0
rem. = 1
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Division Method• Converts integer portion of a decimal number to base r
• Informal Algorithm– Repeatedly divide number by r until equal to 0
– Remainders form coefficients of the number base r
– Remainder from last division = MSD (most significant digit)
19310 = (??)5
19338
55
75150
Keep dividing until you reach 0
rem. = 3rem. = 3rem. = 2rem. = 1
MSD
LSD
19310 = (1233)5
Remainders form the number in
base r (order from bottom up)
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How number conversion works
• Each time we divide by r, another coefficient “falls out” and all the other place values are reduced by a factor of r.
4510 = a4 a3 a2 a1 a0
24 23 22 21 20
4510 = a424 + a323 + a222 + a121 + a020
4510 = a424 + a323 + a222 + a121 + a020 = a423 + a322 + a221 + a120 + a0
2 2 2
Quotient Rem.
a423 + a322 + a221 + a120 = a422 + a321 + a220 + a1
2 2
Quotient Rem.
More bits may be required for this actual example, but we'll use 5 to illustrate…
This is just explanation for what you've learned…Focus on the conversion process outlined earlier
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How number conversion works
• Each time we divide by r, another coefficient “falls out” and all the other place values are reduced by a factor of r.
raarara
rr
rararara
rD n
n
nn
nn 0
11
22
10
11
22
1110 )(
a0 is the remainder
D10 written as coefficients * place values
Factor r out of numerator terms with an-1 – a1
This is just explanation for what you've learned…Focus on the conversion process outlined earlier
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Left‐To‐Right Method
• An alternative to the division method• To convert a decimal number, x, to binary:
– Only coefficients of 1 or 0. So simply find place values that add up to the desired values, starting with larger place values and proceeding to smaller values and place a 1 in those place values and 0 in all others
16 8 4 2 12510 = 1 1 1
32
For 2510 the place value 32 is too large to include so we include 16. Including 16 means we have to make 9 left over. Include 8
and 1.
0 00
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Left‐To‐Right Binary Examples
7310=128 64 32 16 8 4 2 1
.5 .25 .125 .0625 .03125
0 1 0 0 1 0 0 1
0 1 0 1 0 1 1 1
1 0 0 1 0 0 0 1
1 0 1 0 0
8710=
14510=
0.62510=
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Left‐To‐Right In Other Bases
• Can use the left‐to‐right method to convert a decimal number, x, to any base r:– Use the place values of base r (powers of r). Starting
with largest place values, fill in coefficients that sum up to desired decimal value without going over.
16 1
7510 = 4 B256
0 hex
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SHORTHAND FOR BINARYHexadecimal and Octal
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Binary, Octal, and Hexadecimal
• Octal (base 8 = 23)• 1 Octal digit ( _ )8 can represent: 0 – 7
• 3 bits of binary (_ _ _)2can represent: 000‐111 = 0 – 7
• Conclusion…1 Octal digit = 3 bits
• Hex (base 16=24)• 1 Hex digit ( _ )16 can represent: 0‐F (0‐15)
• 4 bits of binary (_ _ _ _)2 can represent: 0000‐1111= 0‐15
• Conclusion…1 Hex digit = 4 bits
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Binary to Octal or Hex
• Make groups of 3 bits starting from radix point and working outward
• Add 0’s where necessary• Convert each group of 3 to an octal digit
101001110.11101001110.11 0 00 000
• Make groups of 4 bits starting from radix point and working outward
• Add 0’s where necessary
• Convert each group of 4 to an octal digit
516.68 14E.C16
5 1 6 6 1 4 E C
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Octal or Hex to Binary
• Expand each octal digit to a group of 3 bits
• Expand each hex digit to a group of 4 bits
317.28 D93.816
011001111.0102 110110010011.10002
11001111.012 110110010011.12
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LOGIC OPERATIONS
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Bitwise Logical Operations
B1 B2 F0 0 0
0 1 0
1 0 0
1 1 1
ANDB1
B2
Pas
sFo
rce
'0'
ZX
YXOR
B1
B2F
B1 B2 F0 0 0
0 1 1
1 0 1
1 1 0
Pas
sIn
vert
B1 B2 F0 0 0
0 1 1
1 0 1
1 1 1
Pas
sFo
rce
'1'
B1
B2OR
0 | x = x1 | x = 1x | x = x
0 & x = 01 & x = xx & x = x
0 ^ x = x1 ^ x = ~ xx ^ x = 0
X F1 0
0 1
F = X or ~X
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Logical Operations• Logic operations on numbers means performing the operation on each pair of bits
0xF0AND 0x3C
0x30
1111 0000AND 0011 1100
0011 0000
0xF0OR 0x3C
0xFC
1111 0000OR 0011 1100
1111 1100
0xF0XOR 0x3C
0xCC
1111 0000XOR 0011 1100
1100 1100
NOT 0xAC0x53
NOT 1010 11000101 0011
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Logical Operations• The C language has two types of logic operations
– Logical and Bitwise• Logical Operators (&&, ||, !)
– Operate on the logical value of a FULL variable (char, int, etc.) interpreting that value as either True (non‐zero) or False (zero)char x = 1, y = 2, z;z = x && y;
– Result is z = 1; Why?• Bitwise Logical Operators (&, |, ^, ~)
– Operate on the logical value of INDIVIDUAL bits in a variablechar x = 1, y = 2, z;z = x & y;
– Result is z = 0; Why?
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Logical Operations• Logic operations on numbers means performing the operation on each pair of bits
0x7AAND 0xEC
0x30
0x36OR 0x91
0xFC
0x3C0x78
XOR 0x3C0x78
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Look Toward LFSR PA
• One of your next PA's will utilize the bitwise XOR operator and leverage the fact that:– a ^ b ^ a = b
• Proof:
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Logical Operations• Bitwise logic operations are often used for "bit fiddling"– Change the value of a bit in a register w/o affecting other bits
– C operators: & = AND, | = OR, ^ = XOR, ~ = NOT
• Examples (Assume an 8‐bit variable, v)– Clear the LSB to '0' w/o affecting other bits
• v = v & 0xfe;– Set the MSB to '1' w/o affecting other bits
• v = v | 0x80;– Flip the LS 4‐bits w/o affecting other bits
• v = v ^ 0x0f;
?v ? ? ? ? ? ? ?
7
& _________________
?v ? ? ? ? ? ? 0
?v ? ? ? ? ? ? ?
| _________________
?v ? ? ? ? ? ?1
?v ? ? ? ? ? ? ?
^ 0 0 0 0 1 1 1 1
?v ? ? ? ? ? ??
Bit #
6 5 4 3 2 1 0
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Exercises for Practice
• Q1‐Q15 on the posted worksheet – http://bits.usc.edu/files/cs103/coursework/NumberSys.pdf