Mba admission in india

MBA admission in India
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Binary numbers and arithmetic

addition

Addition (decimal)
1
+
4
5
5
+
14
19
15
+
5
10
16
+
5
11
1 1
+
12
99
111

Addition (binary)
0
+
0
0
1
+
0
1
0
+
1
1
11
+
1
10

Addition (binary)
1 1 1 1
+
01101
01011
11000

Addition (binary)
0
+
0
1
+
0
So can w0e count in binary?
1
0
+
1
1
11
+
1
10

Counting in binary (4 bits)
0
0000
1
0001
2
…
3
4
5
6
7
8
9
10
11
12
13
14
15

multiplication

Multiplication (decimal)
13
11
13
130
143
+
´

Multiplication (binary)
1101
1011
1101
´
11010
1101000
+
10001111

Multiplication (binary)
1101
1011
1101
´
11010
1101000
+
10001111
It’s interesting to note
that binary multiplication
is a sequence of shifts
and adds of the first
term (depending on the
bits in the second term.
110100 is missing here
because the
corresponding bit in the
second terms is 0.

Representing signed (positive and
negative) numbers

Representing numbers (ints)
Fixed, finite number of bits.
bits bytes C/C++ Intel
Sun
8 1 char [s]byte byte
16 2 short [s]word half
32 4 int or long [s]dword word
64 8 long long [s]qword xword

bits Intel signed unsigned
8 [s]byte -27..+27-1 0..+28-1
16 [s]word -215..+215-1 0..+216-1
32 [s]dword -231..+231-1 0..+232-1
64 [s]qword -263..+263-1 0..+264-1
In general, for k bits, the unsigned range is [0..+2k-1] and the signed
range is [-2k-1..+2k-1-1].

Methods for representing signed ints.
1. signed magnitude
2. 1’s complement (diminished radix complement)
3. 2’s complement (radix complement)
4. excess bD-1

Signed magnitude
Ex. 4-bit signed magnitude
1 bit for sign
3 bits for magnitude
+ N - N
0 0000 1000
1 0001 1001
2 0010 1010
3 0011 1011
4 0100 1100
5 0101 1101
6 0110 1110
7 0111 1111

1’s complement
(diminished radix complement)
Let x be a non-negative number.
Then –x is represented by bD-1+(-x) where
b = base
D = (total) # of bits (including the sign bit)
Ex. Let b=2 and D=4.
Then -1 is represented by 24-1-1 = 1410 or 11102.

1’s complement
b = base & D = (total) # of bits (including the sign bit)
Ex. What is the 9’s complement of 1238910?
Given b=10 and D=5. Then the 9’s complement of 12389
= 105 – 1 – 12389
= 100000 – 1 – 12389
= 99999 – 12389
= 87610

1’s complement
b = base
D = (total) # of bits (including the sign bit)
Shortcut for base 2?
All combinations used, but 2 zeros!
+ N - N
0 0000 1111
1 0001 1110
2 0010 1101
3 0011 1100
4 0100 1011
5 0101 1010
6 0110 1001
7 0111 1000

2’s complement
(radix complement)
Then –x is represented by bD+(-x).
Ex. Let b=2 and D=4. Then -1 is represented by 24-1 = 15 or
11112.
Ex. Let b=2 and D=4. Then -5 is represented by 24 – 5 = 11
or 10112.
Ex. Let b=10 and D=5. Then the 10’s complement of 12389
= 105 – 12389 = 100000 – 12389 = 87611.

2’s complement
(radix complement)
Then –x is represented by bD+(-x).
Ex. Let b=2 and D=4. Then -1 is
represented by 24-1 = 15 or 11112.
Ex. Let b=2 and D=4. Then -5 is
represented by 24 – 5 = 11 or 10112.
+ N - N
0 0000 0000
1 0001 1111
2 0010 1110
3 0011 1101
4 0100 1100
5 0101 1011
6 0110 1010
7 0111 1001

2’s complement
(radix complement)
Yes! Flip the bits and add 1.
+ N - N
0 0000 0000
1 0001 1111
2 0010 1110
3 0011 1101
4 0100 1100
5 0101 1011
6 0110 1010
7 0111 1001

2’s complement
(radix complement)
Are all combinations of 4 bits used?
No. (Now we only have one zero.)
1000 is missing!
What is 1000?
Is it positive or negative?
Does -8 + 1 = -7 work in 2’s complement?
+ N - N
0 0000 0000
1 0001 1111
2 0010 1110
3 0011 1101
4 0100 1100
5 0101 1011
6 0110 1010
7 0111 1001

excess bD-1 (biased representation)
For pos, neg, and 0, x is represented by
bD-1 + x
Ex. Let b=2 and D=4. Then the excess 8 (24-1)
representation for 0 is 8+0 = 8 or 10002.
Ex. Let b=2 and D=4. Then excess 8 for -1 is 8 – 1 = 7 or
01112.

excess bD-1
For pos, neg, and 0, x is represented by
bD-1 + x.
Ex. Let b=2 and D=4. Then the excess 8
(24-1) representation for 0 is 8+0 = 8 or
10002.
Ex. Let b=2 and D=4. Then excess 8 for -1
is 8 – 1 = 7 or 01112.
+ N - N
0 1000 1000
1 1001 0111
2 1010 0110
3 1011 0101
4 1100 0100
5 1101 0011
6 1110 0010
7 1111 0001

2’s complement vs. excess bD-1
In 2’s, positives start with 0; in excess,
positives start with 1.
Both have one zero (positive).
Remaining bits are the same.
+ N - N
0 1000 1000
1 1001 0111
2 1010 0110
3 1011 0101
4 1100 0100
5 1101 0011
6 1110 0010
7 1111 0001

Summary of methods for
representing signed ints.
signedMag sComp sComp excess
1 2 8
N + n - n - n - n - n +
n
0 0000 1000 1111 0000 1000 1000
1 0001 1001 1110 1111 0111 1001
2 0010 1010 1101 1110 0110 1010
3 0011 1011 1100 1101 0101 1011
4 0100 1100 1011 1100 0100 1100
5 0101 1101 1010 1011 0011 1101
6 0110 1110 1001 1010 0010 1110
7 0111 1111 1000 1001 0001 1111
1000=-8| 0000 unused

Binary arithmetic
Signed magnitude
1’s complement
2’s complement
Excess K (biased)

Binary Arithmetic
Signed magnitude

Addition w/ signed magnitude
algorithm
For A - B, change the sign of B and perform addition of A + (-B) (as
in the next step)
For A + B:
if (Asign==Bsign) then { R = |A| + |B|; Rsign = Asign; }
else if (|A|>|B|) then { R = |A| - |B|; Rsign = Asign; }
else if (|A|==|B|) then { R = 0; Rsign = 0; }
else { R = |B| - |A|; Rsign = Bsign; }
Complicated?

Binary Arithmetic
2’s complement

Representing numbers (ints) using
2’s complement
bits Intel signed
8 sbyte -27..+27-1
16 sword -215..+215-1
32 sdword -231..+231-1
64 sqword -263..+263-1
In general, for k bits, the signed range is [-2k-1..+2k-1-1].
So where does the extra negative value come from?

bits Intel signed
8 sbyte -27..+27-1
16 sword -215..+215-1
32 sdword -231..+231-1
64 sqword -263..+263-1
+ n - n
0 0000 0000
1 0001 1111
2 0010 1110
3 0011 1101
4 0100 1100
5 0101 1011
6 0110 1010
In general, for k bits, the signed range is
[-2k-1..+2k-1-1].
So where does the extra negative value come
from? 8 1000
7 0111 1001

Addition of 2’s complement binary
numbers
Consider 8-bit 2’s complement binary numbers.
Then the msb (bit 7) is the sign bit. If this bit is 0, then this is a
positive number; if this bit is 1, then this is a negative number.
Addition of 2 positive numbers.
Ex. 40 + 58 = 98
1 1 1
00101000
00111010
01100010
+

numbers
Consider 8-bit 2’s complement binary
numbers.
Addition of a negative to a positive.
What are the values of these 2 terms?
-88 and 122
-88 + 122 = 34
1 1 1 1
+
10101000
01111010
1 00100010

So how can we perform subtraction?

numbers
Consider 8-bit 2’s complement binary numbers.
Subtraction is nothing but addition of the 2’s complement.
Ex. 58 – 40 = 58 + (-40) = 18
1 1 1 1
+
00111010
11011000
1 00010010
discard carry

numbers
Carry vs. overflow when adding A + B
If A and B are of opposite sign, then overflow cannot occur.
If A and B are of the same sign but the result is of the opposite
sign, then overflow has occurred (and the answer is therefore
incorrect).
Overflow occurs iff the carry into the sign bit differs from the carry out
of the sign bit.

numbers
class test {
public static void main ( String args[] )
{
byte A = 127;
byte B = 127;
byte result = (byte)(A + B);
System.out.println( "A + B = "
+ result );
}
}
#include <stdio.h>
int main ( int argc, char* argv[] )
{
char A = 127;
char B = 127;
char result = (char)(A + B);
printf( "A + B = %d n", result );
return 0;
} Result = -2 in both
Java (left) and C++
(right). Why?

numbers
class test {
public static void main ( String args[] )
{
byte A = 127;
byte B = 127;
byte result = (byte)(A + B);
System.out.println( "A + B = "
+ result );
}
}
Result = -2 in both Java and C++.
Why?
What’s 127 as a 2’s complement
binary number?
01111111
+
01111111
11111110
What is 111111102?
Flip the bits: 00000001.
Then add 1: 00000010.
This is -2.

Binary Arithmetic
1’s complement

Addition with 1’s complement
Note: 1’s complement has two 0’s!
1’s complement addition is tricky (end-around-
carry).
+ N - N
0 0000 1111
1 0001 1110
2 0010 1101
3 0011 1100
4 0100 1011
5 0101 1010
6 0110 1001
7 0111 1000

8-bit 1’s complement addition
Ex. Let X = A816 and Y = 8616.
Calculate Y - X using 1’s complement.

Ex. Let X = A816 and Y = 8616.
Calculate Y - X using 1’s complement.
Y = 1000 01102 = -12110
X = 1010 10002 = -8710
~X = 0101 01112
1000 0110
+
0101 0111
(Note: C=0 out of msb.) 1101 1101
Y - X = -121 + 87 = -34 (base 10)

Ex. Let X = A816 and Y = 8616.
Calculate X - Y using 1’s complement.

Ex. Let X = A816 and Y = 8616.
Calculate X - Y using 1’s complement.
X = 1010 10002 = -8710
Y = 1000 01102 = -12110
~Y = 0111 10012
(Note: C=1 out of msb.)
1010 1000
0111 1001
+
1 0010 0001
1
+
end around
carry
0010 0010
X - Y = -87 + 121 = 34 (base 10)

Binary Arithmetic
Excess K (biased)

Binary arithmetic and Excess K
(biased)
Method: Simply add and then flip the sign bit.
-1 0111
+5 1101
-- ----
+4 0100 -> flip sign -> 1100
+1 1001
-5 0011
-- ----
-4 1100 -> flip sign -> 0100
+1 1001
+5 1101
-- ----
+6 0110 -> flip sign -> 1110
-1 0111
-5 0011
-- ----
-6 1010 -> toggle sign -> 0010
+ N - N
0 1000 1000
1 1001 0111
2 1010 0110
3 1011 0101
4 1100 0100
5 1101 0011
6 1110 0010
7 1111 0001
(Not used for integer arithmetic but employed
in IEEE 754 floating point standard.)

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