Having got some basics down in regard to addition and explaining it in terms of primitive operations (addition and multiplication), I am now again stuck on understanding the more complicated long multiplication.

I have read in my introductory book (Mehlhörn's 'Algorithms and Data Structures') that each partial product requires 2n + 1 primitive operations.

Now I understand, I think, where the +1 comes from for the first partial product at least, as in there being one extra digit in the result for the first partial product, but I don't understand why it is "2n" + 1, as it seems to me it would be more like n + 1, so I am defintely missing something important here. By 'partial product', I mean the result arising from the first intermediate calculation of the multiplication problem.

I am of course also totally confused about quantifying the rest of the primitive operations required to get the final result/product! But I would just like to understand where the 2n is coming from for now for the first partial product. Just to add my guess from what I have seen in addition: does the 2 in 2n refer to a maximum guiding number, rather than a definitive one?

Here is a link to the chapter in the book I am learning from: http://people.mpi-inf.mpg.de/~mehlhorn/ftp/Toolbox/Appetizer.pdf

I understood primitive operations to mean: addition and multiplication. The author does write at bottom of pg.1

"...we have two primitive operations at our disposal: the addition of three digits with a two-digit result (this is sometimes called a full adder), and the multiplication of two digits with a two-digit result"

The product of two n-digit numbers when they multiplied together needs then: 3n^2 + 2n primitive operations.

It is the algorithm as per the method taught at school for long multiplication problems.

(All of this is on pgs 1 and 2 of chapter)

  • 2
    $\begingroup$ Welcome to CS.SE! "I have read in my introductory book that each partial product requires 2n + 1 primitive operations." - I think we're missing some context here. This is probably going to depend on the specific algorithm. Can you summarize the algorithm, and what is meant by partial product in the context of that algorithm? Also, what counts as one primitive operation? $\endgroup$
    – D.W.
    Jul 7, 2016 at 5:42
  • 1
    $\begingroup$ You want to analyse this algorithm, but you don't give the algorithm. Can you see how that won't work? See how our reference question can help you along. $\endgroup$
    – Raphael
    Jul 7, 2016 at 6:44
  • $\begingroup$ Thanks for the edits. Don't just link to the chapter; please make your question self-contained, so people don't have to read an external link to understand your question. I suggest you include the algorithm and the definition of "partial product" and all other explanation needed to understand your question. Also, see meta.cs.stackexchange.com/q/657/755; don't use "EDIT:" -- instead, edit your question to be whta it should have been from the start, so it reads nicely to people who are seeing it for the first time. $\endgroup$
    – D.W.
    Jul 7, 2016 at 18:53
  • $\begingroup$ $2n+1$ still means multiply the number $n$ by two, then add one. That still doesn't seem to be what your question is, so your title still isn't very helpful. $\endgroup$ Jul 8, 2016 at 8:39

1 Answer 1


Perhaps the following example will be useful. Suppose that $n = 3$, and we want to multiply $a_2 a_1 a_0$ by $b$. We first compute $$ \begin{align*} a_0 \times b &= c_0 d_0 \\ a_1 \times b &= c_1 d_1 \\ a_2 \times b &= c_2 d_2 \end{align*} $$ (Here $c_0 d_0$ is a two-digit integer.) So far we have done 3 operations (for general $n$, this will be $n$ operations).

Now we wish to perform the addition $$ \begin{array}{cccc} 0&d_2&d_1&d_0\\ c_2&c_1&c_0&0 \\\hline r_3&r_2&r_1&r_0 \end{array} $$ The book states that you need 4 operations for this (and in general, $n+1$), though 2 are enough (in general, $n-1$): $$ \begin{align*} r_0 &= d_0 \\ s_1 r_1 &= d_1 + c_0 \\ s_2 r_2 &= d_2 + c_1 + s_1 \\ r_3 = s_2 \end{align*} $$ Here $s_1,s_2$ are carry bits, and the first and last operations are for free (though the author charges one operation for each).

Summarizing, if you only charge addition and multiplication, only $2n-1$ operations are needed. If you also charge moves, $2n+1$ operations are needed. (The moves can be eliminated, though. Exercise.)

  • $\begingroup$ Ah.. I get it. At least I think I do. So the 2n comes from 1*n for multiplication, and 1*n for the addition with the + 1 for the extra carry. Is that correct? $\endgroup$
    – hinterbu
    Jul 7, 2016 at 11:42
  • $\begingroup$ There are $n$ multiplications and $n+1$ additions. Here $n+1$ is the number of bits in the sum. $\endgroup$ Jul 7, 2016 at 11:43
  • $\begingroup$ thanks - I thought the 2n + 1 referred to the number of operations needed, not the number of bits in the sum? Is there overlap? $\endgroup$
    – hinterbu
    Jul 7, 2016 at 11:45
  • $\begingroup$ That's right. I meant digits rather than bits. $\endgroup$ Jul 7, 2016 at 11:53
  • $\begingroup$ Is it just a matter of coincidence that the number of operations needed = number of digits in the sum? $\endgroup$
    – hinterbu
    Jul 7, 2016 at 14:55

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