What is the minimum number of bits required to store each binary string of lengt
ID: 3841067 • Letter: W
Question
What is the minimum number of bits required to store each binary string of length 50? What is the minimum number of bits required to store each number with 9 base ten digits? What is the minimum number of bits required to store each length 10 fixed-density binary string with 4 ones? In terms of n, what is the minimum number of bits required to store each subset of a set with n elements? What is the minimum number of bits required to store each rearrangement of the numbers 1 through 8? What is the minimum number of bits required to store each three-letter string? (26 alphabetical letters, not case-sensitive)Explanation / Answer
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The number bits required to hold an integer is calculated as follows:
An integer m has b bits when 2 (b-1) <= m <= (2b) – 1
Therefore, they have given the length of the binary string as 50. Choose a number that satisfies the above equation.
Let b=6 then 2 (b-1) = 2 (6-1)
=2 (5)
=32
(2 b) – 1 = (2 6)-1
= 63
Thus, 32 <= 50 <= 63,
The minimum number of bits required to store each binary string of length 50 is 6 bits.
b.
A ten digit number with base 9 has the numbers ranges from 0 to 8.Therefore the highest possible ten digit number will be (8888888888)9 which is (3486784400)10.
Represent the decimal number in binary format, it will results a number which is having 32-bits as follows: 11001111110101000001101110010000.
Hence, the number of bits required to store the number of 10 digits with base 9 is 32-bits.
c.
Consider length of 10 a fixed-density binary string with 4 ones as 0000001111.
If we convert this value in decimal the value will be 15.
It is the minimum number therefore, the minimum number of bits required to store each length 10 fixed-density binary string with 4 ones is 4 bits.
d.
In terms of n, the minimum number of bits required to store each subset of a set with n elements would be 2n because in a binary system the minimum number of bits can be represented using the powers of 2.
e.
1-1
2-10
3-11
4-100
5-101
6-110
7-111
8-1000
Therefore, the number of bits = 1+2+2+3+3+3+3+4 =21
Thus, the minimum number of bits required to store each rearrangement of the numbers 1 through 8 is 21.
f.
"A" is the minimum value in as per ASCI in bits representation it's value is 01000001 and a character should need to have a minimum of 8 bits.
Thus, the minimum number of bits required to store each three-letter string is 3x8=24 bits
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