Binary is the basis of digital systems and it is important to be fluent when it comes to obtain the decimal value of a binary number. In this post we will start dealing with fast binary to decimal conversion for numbers up to 16 bits. This will be handy if you work in the domain of embedded systems and digital design.
In a previous post, we introduced the basics of binary number systems. I will reuse some paragraphs here:
” The binary system is a base2 positional numeral system. A baseB positional numeral system represents numbers with digits that can have values from 0 to B1 (from 0 to 1 for binary; from 0 to 9 for decimal, etc.), and they are written one after the other. A digit is in a position i which in turn determines its weight. Assuming that the digit located to the right of the number has position 0 and that the digit located to the left of the number has position N1 (because there are N digits), the weight of each digit is B^i. Given that an Ndigit number with values b_i (b_i\in [0\ldots B1] and i\in [0\ldots N1]), its value can be computed as:
value = \sum_{i=0}^{N1} b_i\cdot 2^i = b_{N1}2^{N1}+b_{N2}2^{N2}+\ldots +b_{1}2^{1}+b_{0}2^{0}.
For instance, the binary number 10011_2 has a decimal value of 1\cdot2^4+0\cdot2^3+0\cdot2^2+1\cdot2^1+1\cdot2^0=16+2+1=19_{10}. “
Learning by heart a few powers of two is useful in many ways. Firstly, it is necessary to do any mental conversion from binary to decimal. So do your homework and learn the powers of two from 2^0 to 2^{16}; that is: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768 and 65536. Once you know these powers of two, it is easy to find out the minimum number of bits that we need to represent an integer number. For instance, if we know the upper closest power of two of a number x, this will lead us to the number of bits required to represent that number. Let’s say that we want to convert to binary number 240. The upper closest power of two is 256, which is 2^8, which tells us that 8 bits are required. Note, that if the number is exactly a power of two (i.e. x=256) then the number of bits has to be increased by 1. And knowing these powers of two will be also essential in performing fast binary to decimal conversions.
So now, let’s start with some “shortcuts”
Consecutive ones
Let’s talk now about something quite trivial. If we add 1 to number 999, we obtain 1000. So 999=10001, that is 999=10^31, so, the exponent of the power of ten matches the number of consecutive nines. We can generalized this so that a number in base B composed of a sequence of N consecutive digits is equal to (B^N1)_B, and particularizing for binary (B=2) a sequence of N consecutive ones is equal to (2^N1)_{10}. So now, assuming that we master the powers of two, we can easily calculate the decimal value of numbers such as:

 11111_2=2^51=321=31

 1111111111_2=2^{10}1=10241=1023

 Etc.
Zeros on the right
Adding a zero to the right of a number is equivalent to multiplying the number by the base. For instance, 560_{10}=56\times10, 110_2=11_2\times 2, etc. Thus, adding N zeros to the right of a binary number is equivalent to multiplying the number by 2^N . This could be useful sometimes:
10100_2=101_2\cdot 2^2=5\cdot4=20 (also, 16+4=20)
1111111111000_2=1111111111_2\cdot 2^3=(2^{10}1)\cdot 8=1023\cdot8=8184
Etc.
More ones than zeros
If a binary number has just a few zeros you might as well consider the number as a sequence of consecutive ones that has a few ones missing. Thus, number 111111110111_2 can be seen as 111111111111_21000_2=(2^{12}1)2^3=4095_{10}8_{10}=4087_{10}. Let’s see some examples:
10111101_2=(2^81)2^72^1=25566=189
111111111100_2=(2^{12}1)3=4092 (or, 1023\cdot4=4092)
Putting all together
Let’s try to find the best way to perform fast conversion to several numbers. I am assuming here that you master powers of two and that you easily recognize short binary numbers.
10100000101=1024+256+5=1285
10100000101=5\cdot 2^8 + 5=5\cdot 257 = 1285
11110000=15\cdot16=240
11110111000=(2558)\cdot8=247\cdot8=1976
11110111000=2047647=204071=1976
1000000000001=4096+1=4097
1000000001110=4096+(7\cdot2)=4096+14=4110
111101000=8\cdot(632)=8\cdot61=488
Etc.
Well, it is not rocket science, but surely it is useful. You will find many occasions when these tricks cannot be applied and you have to add all the weights corresponding to the ones to obtain the decimal values. Also, you can use a calculator 🙂 .
In this post we have covered unsigned numbers. The next post we’ll be about fast conversion of signed binary numbers: two’s complement numbers. Stay tune and soon you will master the binary language of moisture vaporators (Star Wars rules!) . See you soon!