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DATE 2014-10-01

HANGOUT

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Key: archive Value: 2014-10-01

Key: id Value: 542565

MESSAGE
DATE 2014-10-06
FROM mrbrklyn@panix.com
SUBJECT Subject: [NYLXS - HANGOUT] HW 1
From owner-hangout-outgoing-at-mrbrklyn.com Mon Oct 6 04:29:20 2014
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Subject: [NYLXS - HANGOUT] HW 1
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Q 1 :

Express the decimal number 101 in base 2 to 9 and hexadecimal
1100101
122
65

Q 2 :

Perform the following binary multiplication of two binary numbers: 10011011 and 11001101

Yo have to be kidding me:
10011011
11001101
________
10011011
000000000
1001101100
10011011000
000000000000
0000000000000
10011011000000
100110110000000
___________________
111110000011111



Q 3 :
Perform the following additions of the 3-digit binary numbers, knowing that supplementation 2 is used to represent negative numbers:

000 000 101 111 101 010

001 111 101 110 110 011
--- --- --- --- --- ---
001 111 010 over 101 011 over 101 over


Specify what additions are giving invalid results.

3 > results < -4

Check your results by performing the same operations after converting the numbers into the decimal system.

Q 4 :
Convert the decimal number 8.625 into floating point IEEE Standard 754 single precision:


OK - I feel like an idiot. The sides have an almost carbo copy problem. I was looking for an
understandable spec. Not until I finally gave up and tried to build a spec from the examples,
only then did I see that this problem is almost verbatum of an example. But I'm left with a perplexed
question. the slides say normalization happens by putting the decimal left of the left most integer.

This spec puts the \d.\d\d\d etc. The notes are pushing the student off point in this section.
the example is negitive, converting -12.625(10). The binary is -1100.101, plain binary. They
normalized it to -1.100101 x 2^3

Why? Why not -0.1100101 x2^4. Is this just for negatives? By examination of the other examples,
it seems that the deimal is always placed to the RIGHT of the left most digit, not left of it like
the author describes as most common usuage.?? Nuh? What can be more common that the single float?

Anyway:

8.625(10) =

8 0
4 0
2 0
1 1

1000

.625
1/250 1
500 0
1/000 1

1000.1010(2)

1.0001010(2) * 2^3

Exponent Bits: 00000011 (2)
01111111 (2) (excess 127 for single precision)
_________
10000010 (2)


Fractional Bits:
Loss the leading One (Not sure why, but it is repesented in signed bit
000 1010 0000 0000 0000 0000

Sig Bit is 0

0 1000 0010 000 1010 0000 0000 0000 0000




As this was not covered in tonight’s class, please use following web page to find steps to follow.

http://www.madirish.net/240

Q5

Give the corresponding translation to the following word of 4 bytes encoded in hexadecimal: 49 55 50 31

As:

- A signed integer,

49 55 50 31
0100 1001 0101 0101 0101 0000 0011 0001

1 0,4,5,12,14,16,18,20,22,24,27,30
16
32
4096
16384
65536
262144
1049576
16777216
67108864
134217728
1073741824
---------------
1230327857 -

- An integer represented in 2's complement,

0100 1001 0101 0101 0101 0000 0011 0001
1011 0110 1010 1010 1010 1111 1100 1110
1
-----------------------------------------
1011 0110 1010 1010 1010 1111 1100 1111



- A number represented by single precision floating point following the IEEE 754 standard,


Overflow -- but ...

01.001001010101010101000000110001 x2^30


Exp 11110
01111111 (+127) (127 + 30 = 157)
---------
0|10011101|0010010101010101010000
^thats 23





- A sequence of ASCII characters (each represented with 8 bits, the most significant bit is unused and coded 0)

49 55 50 31

I U P I




~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Q6



Consider the following two numbers, encoded according to the IEEE 754 single precision and represented in hexadecimal standard:

3E E0 00 00 and 3D 80 00 00
Hex : 3E E0 00 00 3D 80 00 00

_________________________________________________________________________________________________________________
|Binary : 0011 1110 1110 0000 0000 0000 0000 0000 | 0011 1101 1000 0000 0000 0000 0000 0000 |
_________________________________________________________________________________________________________________|
|Dec : 1,054,867,456 | 1,031,798,784 |
_________________________________________________________________________________________________________________|
|754 S :0 100,1110,0 111,1011,10 00,0000,0000,0000 | 0 100,1110,0 111,011 0,0000,0000,0000,0000 |
_________________________________________________________________________________________________________________|


Calculate the sum and give the result as IEEE 754 single precision and decimal form.

***Note, I'm not sure if you want to convert first and then do the math or take the sums and then convert. I'll do harder one
although adding 754S floats is guessing on my part:


Decimal Sum: 2,086,666,240
Hex Sum: 7C 60 00 00
SINCE they have both the same exponent bits you can add them and remember you have a hidden on in the fraction 1....

1.111,101,110
1.111,011,000
11.111,000,110

carry the one to the exponent
100,1110,1

the new 754s binary for the sum:

0 100,1110,1 111, 1000, 1100, 0000, 0000, 0000: Which translates to

Exp
+ 0011110 = 30(2)

Fract: 1.111100011(2)

111 1100 0110 0000 0000 0000 0000 0000

which translates 7C 60 00 00 in hex -- bingo


Just to do it the other way

Sum 2086666240

Binary 1111100011000000000000000000000

1.111100011 2^30

30 + 127 = 157 = 10011101
0 100, 1110, 1 111, 1000,11 00,0000,0000,0000



**NOTE Tracking all these decimal places is very hard on my dyslexia
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~





Same question with the numbers: C8 80 00 00 and C8 00 00 00.


C8 80 00 00 and C8 00 00 00
Hex : C8 80 00 00 C8 00 00 00

_________________________________________________________________________________________________________________
|Binary : 1100 1000 1000 0000 0000 0000 0000 0000 | 1100 1000 0000 0000 0000 0000 0000 0000 |
_________________________________________________________________________________________________________________|
|Dec : 3,363,831,808 | 3,355,443,200 |
_________________________________________________________________________________________________________________|
|754 S :0 100,1111,0 100, 1000,1000,0000,0000,0000 | 0 100,1111,0 100 1000,0000,0000,0000,0000 |
_________________________________________________________________________________________________________________|

exp = 10011110 (2) exp = 10011110
frac = 100,100,010,000,000,000,000,00 frac = 100 1000 0000 0000 0000 0000

Sum 01 90 80 00 00
0000 0001 1001 0000 1000 0000 0000 0000 0000 0000

754: 1.1001 0000 1000 x 2^32 => 10011111 Sig 0

Sum in 754 S => 0 100,1111,1 100, 1000,0100,0000,0000,0000


Q7

Convert the following number represented as a signed binary integer of 32 bits:

011001010 11100010 10101011 11000101 into real number represented by the IEEE 754 single precision.
*********No - this is not 32 bits. This is 33 bits *****************
sig = 0
1.1001010 11100010 10101011 11000101 x 2^31
exp = 10011110

0 100, 1111,0 100,1010,1110,0010,1010,1011 ==>Lost Precision?




Convert the real number obtained into a signed binary integer 32-bit, and compare the results with original number.

There is a loss of precision BUT problem is broken


What conclusions can you deduct?



I'm tired....



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