🔁 Subtraction Through Addition — Method of Complements

📜 Example of how this works

Goal: Make Subtractions be in terms of Addition
==> A - B, where A >= B

(100 - 47) is the 10's complement of 73
(-100) is the remover of the leading digit, 

If we extend this idea further, removing the leading digit is the same as caring about grouping numbers within a fixed range

• mod is a math sign invented to mean “keep only the digits within this range”

This concept is then extended for calculations involving fixed ranges, called modulo arithmetic

• In this case, removing 100 from the answer is equivalent to keeping the digits within 100
• So instead of -100, we could generally do ... mod 100

And since we group numbers according to a fixed range, when numbers exceed the range, it will appear to loop back to the start of this fixed range

Thus, circular properties are present in systems using fixed ranges to represent numbers.

For example, the 2’s complement system for representing binary numbers in computers

In the following sections, we will extend the explanation more generally

🔧 Successor to the Most-Significant Value

Let S(x) be the next significant place, where the S stands for Successor of place-value i.e.

If x is some number where r is the radix, and n is highest power of that radix

Then S(x) = r^(n + 1)

Where

S(13)   = 100
S(99)   = 100
S(123)  = 1000
S(1234) = 10000

🧮 Subtraction Setup Using S(A)

We set up MSV Redundancy by doing this

A − B
= A − B + S(A) − S(A)

Which allows us to convert - S(A) to mod S(A)

A − B
= A − B + S(A) − S(A)
--------------------------
= [A − B + S(A)] mod S(A)

Then we group the subtractive components together, that will be used for pre-calculation

A − B
= A − B + S(A) − S(A)
= [A − B + S(A)] mod S(A)
------------------------------
= {A + [ S(A) − B ]} mod S(A)

Let S(A) − B be the r’s complement of B, denoted B'.

A − B
= A − B + S(A) − S(A)
= A − B + S(A) mod S(A)
= {A + [ S(A) − B ]} mod S(A)
------------------------------
= (A + B') mod S(A)

We have thus converted it to just addition and the removal of the MSV

✅ Subtraction through Radix Complement

A − B
= (A + B') mod S(A)

Where B' , which is S(A) - B, is the Radix Complement—— $\boxed{ \phantom{\big(} r^{n+1} - B \phantom{\big)}}$

⚙️ Optimizing for Per-Place Operation

Since, we still want the convenience of removing the MSV after calculation, we will keep using the MSV redundancy technique

However, instead of taking subtraction directly from the MSV, we can instead subtract from MSV - 1 for its convenient place-value calculation——as we simply subtract from radix - 1 at every place-value.

• The extra 1 that is taken out of MSV must be accounted for

Here’s how it goes:

We set up MSV Redundancy by doing this

A − B
= A − B + S(A) − S(A)

Which allows us to convert - S(A) to mod S(A)

A − B
= A − B + S(A) − S(A)
--------------------------
= [A − B + S(A)] mod S(A)

Then we set up for subtraction from MSV - 1 by balancing the 1s like so

A − B
= A − B + S(A) − S(A)
= [A − B + S(A)] mod S(A)
--------------------------------------
= {[ A − B + S(A) - 1 ] mod S(A)} + 1

A − B
= A − B + S(A) − S(A)
= [A − B + S(A)] mod S(A)
= {[ A − B + S(A) - 1 ] mod S(A)} + 1
----------------------------------------
= [{ A + [S(A) - 1 - B] } mod S(A)] + 1

Where now B' = S(A) − 1 − B, which we will call the r - 1’s complement of B

A − B
= A − B + S(A) − S(A)
= [A − B + S(A)] mod S(A)
= {[ A − B + S(A) - 1 ] mod S(A)} + 1
= [{ A + [S(A) - 1 - B] } mod S(A)] + 1
----------------------------------------
= [ (A + B') mod S(A) ] + 1

Thus, we reached the optimal setup for per-place operation that maximises addition-only operations, by pre-calculating one subtraction. And adding 1 at the end.

✅ Subtraction through Diminished Radix Complement

A − B
= [ (A + B') mod S(A) ] + 1

Where B', which is S(A) - 1 - B, is the Diminished Radix Complement—— $\boxed{ \phantom{\big(} r^{n+1} - 1 - B \phantom{\big)}}$

Conclusion

Diminished Radix Complement = Radix Complement - 1

• Or similarly:

Radix Complement = Diminished Radix Complement + 1