According to this German Wikipedia article, the time required to merge relations $R$ and $S$ is $\in \mathcal{O}(|R| + |S|)$ if both relations are already sorted. [Note: You don't really need to read the text and the link jumps right to where the time complexity is stated. But I added a translation of that section to this question for some clarity.]

Assume $R = S$ with $R$ having 2 columns. One is a random number and one is always 5 for every line. Join $R$ and $S$ on the column which is 5 for every line. The resulting output's space complexity is $\in \Theta(|R| \cdot |S|)$. Time complexity is always $\in \Omega(\text{spaceComplexity})$.

How can the time complexity stated by the Wikipedia article be true?

Here is the "Sort-Merge Join" section translated to English. Sorry, no markdown quote, just everything from here on is the quote. It's translated poorly (not how you would write it in English if you know what you're writing) in some parts to preserve the meaning as well as possible. I marked my own inline comments with ///.

Quote of https://de.wikipedia.org/wiki/Joinalgorithmen#Sort-Merge_Join

Sort-Merge Join

Both relations get sorted by their join attributes. The result can be determined via a single scan through both sorted relations.

The algorithm is only suited for natural join and equi-join.


Implementation of $R\bowtie_{R.a=S.a} S$ in pseudocode:

p_r := first tuple in R
p_s := first tuple in S
while(p_r != endof_r && p_s != endof_s)
    // Collect all tuples in S with the same join attributes.
    M_s := set with contents p_s /// Yes, it says "with", not "of".

    foreach(t_s in S > p_s)
        if(t_s.a = p_s.a)
            M_s += set with contents t_s
        elseif // I think they mean "else".
            p_s := t_s
            break foreach

    // Seach suitable start tuple in R. /// "passend" can also be translated "fitting" or "matching", not just "suitable".
    foreach(t_r in R > p_r)
        p_r = t_r
        if(t_r.a >= t_s.a)
            break foreach

    // Output tuples.
    foreach(t_r in R > p_r)
        if(t_r.a > t_s.a)
            break foreach

        foreach(t_s in M_s)
            Write output: (t_r, t_s)
        p_r = t_r


Sorting can be done with effort $\mathcal{O}(|R|\log|R|+|S|\log|S|)$. The number of block accesses to sort $S$ is $b_s\left(2\log\frac{b_s}{b_{free}}\right)+b_s$ in the worst case, analogous for $R$.

A merge of both relations after sorting them costs $\mathcal{O}(|R|+|S|)$. In the best case – i.e. the relations are already sorted –, the costs of merging are the only ones.

In the normal case, the total costs are $\mathcal{O}(n\log n)$.


[Not translated because it doesn't seem to be important for the question.]

  • $\begingroup$ It is probably referring to I/O complexity (number of I/O operations) rather than what we typically think as runtime complexity. The English version of the page makes this distinction. en.wikipedia.org/wiki/Sort-merge_join $\endgroup$ – mdxn Jan 2 '17 at 1:15
  • $\begingroup$ @mdxn Thanks. You have to write your output, though, for the "O" part of "I/O". $\endgroup$ – UTF-8 Jan 2 '17 at 8:30
  • $\begingroup$ I/O complexity is the number of total disk (or cache) transfer operations. Dumping the contents of RAM (regardless of size in bits) to disk at the very end would constitute a single I/O operation. The rough idea is to capture the number of back and forth queries necessary between your current machine and a potentially much larger external disk to solve a problem. This is not the same situation as with a regular or random access Turing Machine where you have immediate access to the entire input. $\endgroup$ – mdxn Jan 2 '17 at 13:55
  • $\begingroup$ @mdxn I thought about that but was able to defeat it in the same thought, so I throw it away immediately. For that to work, you need to be able to hold something $\in \theta(|R|^2)$ in your main memory. If you're able to do that, you can hold your entire input in your main memory, as well. Therefore, the I/O complexity of the entire merge operation has to be $\in \mathcal{O}(1)$. And the article wouldn't say it's $\in \mathcal{O}(|R| + |S|)$ if it's $\in \mathcal{O}(1)$, would it? $\endgroup$ – UTF-8 Jan 2 '17 at 19:42
  • $\begingroup$ @mdxn Is there something wrong with my line of argument? $\endgroup$ – UTF-8 Jan 4 '17 at 14:26

You are absolutely correct. Wikipedia has an error -- or perhaps, if we are feeling more charitable, we could call it an oversimplification.

It is not true that the running time is at most $O(|R|+|S|)$. For instance, if we consider the case where the value of attribute $a$ is 42 for all elements of $R$ and $S$, we output $|R| \times |S|$ tuples. It is also easy to see that the pseudocode does $|R| \times |S|$ iterations of the nested inner loop (i.e., that many iterations of the statement "Write output:"), so the running time is also $O(|R| \times |S|)$.

Here are some statements that are true:

  • If there are no repeated values of $a$ (i.e., each value for $a$ appears at most once in $R$ and at most once in $S$), then the running time for the merge is $O(|R|+|S|)$, and the size of the output is also $O(|R|+|S|)$.

  • If any value appears at most $c$ times in attribute $a$, where $c$ is a constant, then the running time for the merge is $O(|R|+|S|)$, and the size of the output is also $O(|R|+|S|)$.

  • If we count only the I/O complexity (the number of disk/cache transfer operations), and if memory is large enough to hold $|S|$ items, then the I/O complexity is $O(|R|+|S|)$... though the running time and the size of the output might be as large as $O(|R| \times |S|)$. More generally, the same statement is true if any value for attribute $a$ appears at most $k$ times in $S$, and main memory is large enough to hold at least $k$ items (here $k$ does not need to be a constant).

So Wikipedia's statement is misleading or wrong or (at best) over-simplified. Perhaps what they really meant was one of the bullet items above. Your understanding is absolutely correct.


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