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== Cheprasov Algorithm ==
The Cheprasov <ref>Cheprasov, Artem. ''On a New Method of Multiplication and Shortcuts''. Scotts Valley: Createspace, 2010'. Print.</ref> <ref>Craver, Kevin. ''On the Record With… Artem Cheprasov''. Northwest Herald, 1/10/11'.</ref> method of multiplication can be applied for any n digit number multiplied by any other n digit number. The main uniqueness of this algorithm lies in its ability to:
1. Utilize negative numbers, even when multiplying only positive numbers, to reach the correct positive result.
2. Attach, or link, numbers to one another in order to avoid excessive addition.
3. As the n digit multiplication problem grows to ever larger numbers, the number of possible combinations one can use to reach the same answer grows as well; meaning the user can pick and choose the easiest and fastest route to reach the answer to each multiplication problem according to their own specific needs at the time. For example, using the Cheprasov algorithm, a 3 digit • 4 digit multiplication problem yields a total of 6 different combinations one can use to reach the correct answer using the same exact algorithm but through different permutations.
The basic algorithm, as best exemplified in a ''2'' • ''2'' multiplication problem, is as follows (where t represent a tens digit, u represent a units digit, and T symbolizes the tens digit multiplied by its respective power):
: Step 1 = (t <sub>1</sub> - u <sub>1</sub>) • u <sub>2</sub> = A
: Step 2 = u <sub>1</sub> • t <sub>2</sub> = B
: Step 3 = (u <sub>1</sub> • u <sub>2</sub>) + t <sub>(u <sub>1</sub> • u <sub>2</sub>) </sub> = C
Step 3 first multiplies both units digits together to reach an ''initial answer'' ((u 1 • u 2)). That ''initial answer’s'' tens digits (t <sub>(u <sub>1</sub> • u <sub>2</sub>)</sub>) is added back to step 3’s ''initial answer''.
A+B+C are then added together to reach a sum (D).
Step 4 is where the units digit to step 3’s ''initial answer'' to: (u <sub>1</sub> • u <sub>2</sub>) is attached (symbolized by: @) to the end of the sum of steps 1-3.
: Step 4 = D @ u <sub>(u <sub>1</sub> • u <sub>2</sub>)</sub> = E
Finally this number is taken and the following is added to it:
: Step 5 = E + (T <sub>1</sub> • T <sub>2</sub>) = Final Answer
For example, in the following problem: 79 • 26, by assigning subscripts of 1 to 79 and subscripts of 2 to 26, we would reach the answer as follows:
: Step 1 = (7 - 9) • 6 = -12
: Step 2 = 9 • 2 = 18
: Step 3 = (9 • 6) = 54 + 5 = 59
Steps 1-3 are added together to reach a sum of 65.
: Step 4 = 65@4 = 654
: Step 5 = 70 • 20 = 1400 + 654 = 2054
The reverse, where subscripts of 1 are assigned to 26 and subscripts of 2 are assigned to 79 would yield the same answer but with different results to each intermediary step.
== Fast multiplication algorithms for large inputs ==
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