I’ve recently been working on automated testcase reduction tools for the MathJax project and thus I had the chance to study Jesse Ruderman’s Lithium tool, itself inspired from the ddmin algorithm. This paper contains good ideas, like for example the fact that the reduction could be improved if we rely on the testcase structure like XML nodes or grammar tokens instead of just characters/lines (that’s why I’ve started to write a version of Lithium to work with abstract data structure). However, the authors of the ddmin paper really don’t analyse precisely the complexity of the algorithm, except the best and worst case and there is a large gap between the two. Jesse's analysis is much better and in particular introduces the concepts of monotonic testcase and clustered reduction where the algorithm performs the best and which intuitively seems the usual testcases that we meet in practice. However, the monotonic+clustered case complexity is only “guessed” and the bound O(Mlog2(N))O(M\log _{2}(N)) for a monotonic testcase (of size NN with final reduction of size MM) is not optimal. For example if the final reduction is relatively small compared to NN, say M=Nlog2(N)=o(N)M=\frac{N}{\log _{2}(N)}=o(N) then Mlog2(N)=N=Ω(N)M\log _{2}(N)=N=\Omega(N) and we can’t say that the number of verifications is small compared to NN. In particular, Jesse can not deduce from his bound that Lithium’s algorithm is better than an approach based on MM binary search executions! In this blog post, I shall give the optimal bound for the monotonic case and formalize that in some sense the clustered reduction is near the best case. I’ll also compare Lithium’s algorithm with the binary search approach and with the ddmin algorithm. I shall explain that Lithium is the best in the monotonic case (or actually matches the ddmin in that case).

Thus suppose that we are given a large testcase exhibiting an unwanted behavior. We want to find a smaller test case exhibiting the same behavior and one way is to isolate subtestcases that can not be reduced any further. A testcase can be quite general so here are basic definitions to formalize a bit the problem:

  • A testcase SS is a nonempty finite sets of elements (lines, characters, tree nodes, user actions) exhibiting an “interesting” behavior (crash, hang and other bugs…)

  • A reduction TT of SS is a testcase TST\subseteq S with the same “interesting” behavior as SS.

  • A testcase SS is minimal if TS,T is not a reduction of S\forall T\subsetneq S,T\text{ is not a reduction of }S.

Note that by definition, SS is a reduction of itself and \emptyset is not a reduction of SS. Also the relation “is a reduction of” is transitive that is a reduction of a reduction of SS is a reduction of SS.

We assume that verifying one subset to check if it has the “interesting” behavior is what takes the most time (think e.g. testing a hang or user actions) so we want to optimize the number of testcases verified. Moreover, the original testcase SS is large and so a fast reduction algorithm would be to have a complexity in o(|S|)o(|S|). Of course, we also expect to find a small reduction TST\subseteq S that is |T|=o(|S|)|T|=o(|S|).

Without information on the structure on a given testcase SS or on the properties of the reduction TT, we must consider the 2|S|-22^{{|S|}}-2 subsets TS\emptyset\neq T\neq S, to find a minimal reduction. And we only know how to do that in O(2|S|)O\left(2^{{|S|}}\right) operations (or O(2|S|/2)O\left(2^{{|S|/2}}\right) with Grover’s algorithm ;-)). Similarly, even to determine whether TST\subseteq S is minimal would require testing 2|T|-22^{{|T|}}-2 subsets which is not necessarily o(|S|)o(|S|) (e.g. |T|=log2(|S|)=o(|S|)|T|=\log _{2}{(|S|)}=o(|S|)). Hence we consider the following definitions:

  • For any integer n1n\geq 1, SS is nn-minimal if TS,|S-T|nT is not a reduction of S\forall T\subsetneq S,|S-T|\leq n\implies T\text{ is not a reduction of }S.

  • In particular, SS is 11-minimal if xS,S{x} is not a reduction of S\forall x\in S,S\setminus\{ x\}\text{ is not a reduction of }S.

  • SS is monotonic if T1T2S,T1 is a reduction of ST2 is a reduction of S\forall T_{1}\subseteq T_{2}\subseteq S,\, T_{1}\text{ is a reduction of }S\implies T_{2}\text{ is a reduction of }S.

Finding a nn-minimal reduction will give a minimal testcase that is no longer interesting if we remove any portion of size at most nn. Clearly, SS is minimal if it is nn-minimal for all nn. Moreover, SS is always nn-minimal for any n|S|n\geq|S|. We still need to test exponentially many subsets to find a nn-minimal reduction. To decide whether TST\subseteq S is nn-minimal, we need to consider subsets obtained by removing portions of size 1,2,,min(n,|T|-1)1,2,...,\min(n,|T|-1) that is k=1min(n,|T|-1)(|T|k)\sum _{{k=1}}^{{\min(n,|T|-1)}}\binom{|T|}{k} subsets. In particular whether TT is 11-minimal is O(|T|)O(|T|) and so o(|S|)o(|S|) if T=o(|S|)T=o(|S|). If SS is monotonic then so is any reduction TT of SS. Moreover, if TST\subsetneq S is a reduction of SS and xSTx\in S\setminus T, then TS{x}ST\subseteq S\setminus\{ x\}\subsetneq S and so S{x}S\setminus\{ x\} is a reduction of SS. Hence when SS is monotonic, SS is 11-minimal if and only if it is minimal. We will target 11-minimal reduction in what follows.

Let’s consider Lithium’s algorithm. We assume that SS is ordered and so can be identified with the interval [1,|S|][1,|S|] (think for example line numbers). For simplicity, let’s first assume that the size of the original testcase is a power of two, that is |S|=N=2n|S|=N=2^{n}. Lithium starts by n-1n-1 steps k=1,2,,n-1k=1,2,...,n-1. At step kk, we consider the chunks among the intervals [1+j2n-k,(j+1)2n-k](0j<2k)[1+j2^{{n-k}},(j+1)2^{{n-k}}]\ (0\leq j<2^{k}) of size 2n-k2^{{n-k}}. Lithium verifies if removing each chunk provides a reduction. If so, it permanently removes that chunk and tries another chunk. Because \emptyset is not a reduction of SS, we immediately increment kk if it remains only one chunk. The nn-th step is the same, with chunk of size 1 but we stop only when we are sure that the current testcase TT is 11-minimal that is when after |T||T| attempts, we have not reduced TT any further. If NN is not a power of 2 then 2n-1<N<2n2^{{n-1}}<N<2^{n} where n=log2(N)n=\lceil\log _{2}(N)\rceil. In that case, we apply the same algorithm as 2n2^{n} (i.e. as if there were 2n-N2^{n}-N dummy elements at the end) except that we don’t need to remove the chunks that are entirely in that additional part. This saves testing at most nn subtests (those that would be obtained by removing the dummy chunks at the end of sizes 2n-1,2n-2,,12^{{n-1}},2^{{n-2}},...,1). Hence in general if CNC_{N} is the number of subsets of SS tried by Lithium, we have C2n-nCNC2nC_{{2^{n}}}-n\leq C_{N}\leq C_{{2^{n}}}. Let MM be the size of the 11-minimal testcase found by Lithium and m=log2(M)m=\lceil\log _{2}(M)\rceil.

Lithium will always perform the n-1n-1 initial steps above and check at least one subset at each step. At the end, it needs to do MM operations to be sure that the testcase is 11-minimal. So CNlog2(N)+M-1=Ω(log2(N)+M)C_{N}\geq\lceil\log _{2}(N)\rceil+M-1=\Omega(\log _{2}(N)+M). Now, consider the case where SS monotonic and has one minimal reduction T=[1,M]T=[1,M]. Then TT is included in the chunk [1,2m][1,2^{{m}}] from step k=n-mk=n-m. Because SS is monotonic, this means that at step k=1k=1, we do two verifications and the second chunk is removed because it does not contain the TT (and the third one too if NN is not a power of two), at step k=2k=2 it remains two chunks, we do two verifications and the second chunk is removed etc until k=n-mk=n-m. For k>n-mk>n-m, the number of chunk can grow again: 2, 4, 8… that is we handle at most 21+k-(n-m)2^{{1+k-(n-m)}} chunks from step n-m+1n-m+1 to n-1n-1. At step k=nk=n, a first round of at most 2m2^{m} verifications ensure that the testcase is of size MM and a second round of MM verifications ensure that it is 11-minimal. So CN1+(k=1n-m2)+(k=n-m+1n21+k-(n-m))+2m+M=1+2(n-m)+2m-1+2m+MC_{N}\leq 1+\left(\sum _{{k=1}}^{{n-m}}2\right)+\left(\sum _{{k=n-m+1}}^{{n}}2^{{1+k-(n-m)}}\right)+2^{m}+M=1+2(n-m)+2^{m}-1+2^{m}+M and after simplification CN=O(log2(N)+M)C_{N}=O(\log _{2}(N)+M). Hence the lower bound Ω(log2(N)+M)\Omega(\log _{2}(N)+M) is optimal. The previous example suggests the following generalization: a testcase TT is CC-clustered if it can be written as the union of CC nonempty closed intervals T=I1I2ICT=I_{1}\cup I_{2}\cup...\cup I_{C}. If the minimal testcase found by Lithium is CC-clustered, each IjI_{j} is of length at most M2mM\leq 2^{m} and so IjI_{j} intersects at most 2 chunks of length 2m2^{m} from the step k=n-mk=n-m. So TT intersects at most 2C2C chunks from the step k=n-mk=n-m and a fortiori from all the steps kn-mk\leq n-m. Suppose that SS is monotonic. Then if cc is a chunk that does not contain any element of TT then TcT\setminus c is a reduction of TT and so Lithium will remove the chunk cc. Hence at each step kn-mk\leq n-m, at most 2C2C chunks survive and so there are at most 4C4C chunks at the next step. A computation similar to what we have done for T=[1,M]T=[1,M] shows that CN=O(C(log2(N)+M))C_{N}=O(C(\log _{2}(N)+M)) if the final testcase found by Lithium is CC-clustered. Note that we always have M=o(N)M=o(N) and log2(N)=o(N)\log _{2}(N)=o(N). So if C=O(1)C=O(1) then CN=O(log2(N)+M)=o(N)C_{N}=O(\log _{2}(N)+M)=o(N) is small as wanted. Also, the final testcase is always MM-clustered (union of intervals that are singletons!) so we found that the monotonic case is O(M(log2(N)+M))O(M(\log _{2}(N)+M)). We shall give a better bound below.

Now, for each step k=1,2,,n-1k=1,2,...,n-1, Lithium splits the testcase in at most 2k2^{k} chunk and try to remove each chunk. Then it does at most NN steps before stopping or removing one chunk (so the testcase becomes of size at most N-1N-1), then it does at most N-1N-1 steps before stopping or removing one more chunk (so the testcase becomes of size at most N-1N-1), …, then it does at most M+1M+1 steps before stopping or removing one more chunk (so the testcase becomes of size at most MM). Then the testcase is exactly of size MM and Lithium does at most MM additional verifications. This gives CNk=1n-12k+k=MNk=2n-2+N(N+1)-M(M-1)2=O(N2)C_{N}\leq\sum _{{k=1}}^{{n-1}}2^{k}+\sum _{{k=M}}^{{N}}k=2^{n}-2+\frac{N(N+1)-M(M-1)}{2}=O(N^{2}) verifications. This bound is optimal if 1M2n-11\leq M\leq 2^{{n-1}} (this is asymptotically true since we assume M=o(N)M=o(N)): consider the cases where the proper reductions of SS are exactly the segments [1,k](2n-1+1k2n-1)[1,k]\ (2^{{n-1}}+1\leq k\leq 2^{n}-1) and [k,2n-1+1](2k2n-1-M+2)[k,2^{{n-1}}+1]\ (2\leq k\leq 2^{{n-1}}-M+2). The testcase will be preserved during the first phase. Then we will keep browsing at least the first half to remove elements at position 2n-1+2k2n-12^{{n-1}}+2\leq k\leq 2^{n}-1. So CNk=2n-1+22n-12n-1=2n-1(2n-1-2)=Ω(N2)C_{N}\geq\sum _{{k=2^{{n-1}}+2}}^{{2^{n}-1}}2^{{n-1}}=2^{{n-1}}\left(2^{{n-1}}-2\right)=\Omega(N^{2}).

We now come back to the case where SS is monotonic. We will prove that the worst case is CN=Θ(Mlog2(NM))C_{N}=\Theta\left(M\log _{2}(\frac{N}{M})\right) and so our assumption M=o(N)M=o(N) gives Mlog2(NM)=(-MNlog2(MN))N=o(N)M\log _{2}(\frac{N}{M})=\left(-\frac{M}{N}\log _{2}(\frac{M}{N})\right)N=o(N) as we expected. During the steps 1km1\leq k\leq m, we test at most 2k2^{k} chunks. When k=mk=m, 2mM2^{{m}}\geq M chunks but at most MM distinct chunks contain an element from the final reduction. By monocity, at most MM chunks will survive and there are at most 2M2M chunks at step m+1m+1. Again, only MM chunks will survive at step m+2m+2 and so on until k=n-1k=n-1. A the final step, it remains at most 2M2M elements. Again by monocity a first round of 2M2M tests will make MM elements survive and we finally need MM additional tests to ensure that the test case is minimal. Hence CNk=1m2k+k=m+1n2M+M=2m+1-3+M(2(n-m)+1)=O(Mlog2(NM))C_{N}\leq{\sum _{{k=1}}^{{m}}2^{k}}+{\sum _{{k=m+1}}^{{n}}2M}+M=2^{{m+1}}-3+M(2(n-m)+1)=O(M\log _{2}(\frac{N}{M})). This bound is optimal: if M=2mM=2^{m}, consider the case where T={j2n-m+1:0j<2m}T=\{ j2^{{n-m}}+1:0\leq j<2^{{m}}\} is the only minimal testcase (and SS monotonic) ; if MM is not a power of two, consider the same TT with 2m-M2^{{m}}-M points removed at odd positions. Then for each step 1km-11\leq k\leq m-1, no chunks inside [1,2n][1,2^{n}] are removed. Then some chunks in [1,2n][1,2^{n}] are removed (none if MM is a power of two) at step mm and it remains MM chunks. Then for steps m+1kn-1m+1\leq k\leq n-1 there are always exactly 2M2M chunks to handle. So CNm+1kn-12M=2M(n-m-2)=2M(log2(NM)-2)=Ω(Mlog2(NM))C_{N}\geq\sum _{{m+1\leq k\leq n-1}}2M=2M(n-m-2)=2M(\log _{2}(\frac{N}{M})-2)=\Omega(M\log _{2}(\frac{N}{M})).

We note that we have used two different methods to bound the number of verifications in the general monotonic case, or when the testcase is CC-clustered. One naturally wonders what happens when we combine the two techniques. So let c=log2Cmc=\lceil\log _{2}C\rceil\leq m. From step 11 to cc, the best bound we found was O(2k)O(2^{k}) ; from step cc to mm, it was O(C)O(C) ; from step mm to n-mn-m it was O(C)O(C) again ; from step n-mn-m to n-cn-c, it was O(21-k-(n-m)C)O(2^{{1-k-(n-m)}}C) and finally from step n-cn-c to nn, including final verifications, it was O(M)O(M). Taking the sum, we get CN=O(2c+((n-m)-c)C+2(n-c-(n-m))C+(n-(n-c))M)=O(C(1+log2(NMC))+M(1+log2C))C_{N}=O(2^{c}+((n-m)-c)C+2^{{(n-c-(n-m))}}C+(n-(n-c))M)=O(C\left(1+\log _{2}{\left(\frac{N}{MC}\right)}\right)+M(1+\log _{2}{C})) Because C=O(M)C=O(M), this becomes CN=O(Clog2(NMC)+M(1+log2C))C_{N}=O(C\log _{2}{\left(\frac{N}{MC}\right)}+M(1+\log _{2}{C})). If C=O(1)C=O(1), then we get CN=O(log2(N)+M)C_{N}=O(\log _{2}(N)+M). At the opposite, if C=Ω(M)C=\Omega(M), we get CN=Ω(Mlog2(NM))C_{N}=\Omega(M\log _{2}{\left(\frac{N}{M}\right)}). If CC is not O(1)O(1) but C=o(M)C=o(M) then 1=o(log2C)1=o(\log _{2}{C}) and Clog2(C)=o(Mlog2C)C\log _{2}(C)=o(M\log _{2}{C}) and so the expression can be simplified to CN=O(Clog2(NM)+Mlog2C)C_{N}=O(C\log _{2}{\left(\frac{N}{M}\right)}+M\log _{2}{C}). Hence we have obtained an intermediate result between the worst and best monotonic cases and shown how the role played by the number of clusters: the less the final testcase is clustered, the faster Lithium finds it. The results are summarized in the following table:

Number of tests
Best case Θ(log2(N)+M)\Theta(\log _{2}(N)+M)
SS is monotonic ; TT is O(1)O(1)-clustered O(log2(N)+M)O(\log _{2}(N)+M)
SS is monotonic ; TT is CC-clustered (C=o(M)C=o(M) and unbounded) O(Clog2(NM)+Mlog2C)O(C\log _{2}{\left(\frac{N}{M}\right)}+M\log _{2}{C})
SS is monotonic O(Mlog2(NM));o(N)O\left(M\log _{2}\left(\frac{N}{M}\right)\right);o(N)
Worst case Θ(N2)\Theta(N^{2})
Figure 0.1: Performance of Lithium’s algorithm for some initial testcase SS of size NN and final reduction TT of size M=o(N)M=o(N). TT is CC-clustered if it is the union of CC intervals.

In the ddmin algorithm, at each step we add a preliminary round where we try to immediately reduce to a single chunk (or equivalently to remove the complement of a chunk). Actually, the ddmin algorithm only does this preliminary round at steps where there are more than 2 chunks for otherwise it would do twice the same work. For each step k>1k>1, if one chunk c1c_{1} is a reduction of SS then c1c2c_{1}\subseteq c_{2} for some chunk c2c_{2} at the previous step k-1k-1. Now if SS is monotonic then, at level k-1k-1, removing all but the chunk c2c_{2} gives a subset that contains c1c_{1} and so a reduction of SS by monocity. Hence on chunk survive at level kk and there are exactly 2 chunks at level kk and so the ddmin algorithm is exactly Lithium’s algorithm when SS is monotonic. The ddmin algorithm keeps in memory the subsets that we didn’t find interesting in order to avoid repeating them. However, if we only reduce to the complement of a chunk, then we can never repeat the same subset and so this additional work is useless. That’s the case if SS is monotonic.

Finally, if SS is monotonic Jesse proposes a simpler approach based on a binary search. Suppose first that there is only one minimal testcase TT. If kminTk\geq\min T then [1,k][1,k] intersects TT and so Uk=S[1,k]TU_{k}=S\setminus[1,k]\neq T. Then UkU_{k} is not a reduction of SS for otherwise a minimal reduction of UkU_{k} would be a minimal reduction of SS distinct from TT which we exclude by hypothesis. If instead k<minTk<\min T then [1,k][1,k] does not intersect TT and Uk=S[1,k]T[1,k]TU_{k}=S\setminus[1,k]\supseteq T\setminus[1,k]\supseteq T is a reduction of SS because SS is monotonic. So we can use a binary search to find minT\min T by testing at most log2(N)\log _{2}(N) testcases (modulo some constant). Then we try with intervals [1+minT,k][1+\min T,k] to find the second least element of TT in at most log2(N)\log _{2}(N). We continue until we find the MM-th element of TT. Clearly, this gives Mlog2(N)M\log _{2}(N) verifications which sounds equivalent to Jesse’s bound with even a better constant factor. Note that the algorithm still works if we remove the assumption that there is only one minimal testcase TT. We start by S=S1S=S_{1} and find x1=max{minT:T is a minimal reduction of S1}x_{1}=\max\{\min T:T\text{ is a minimal reduction of }S_{1}\}: if k<x1k<x_{1} then S[1;k]S\setminus[1;k] contains at least one mininal reduction with least element x1x_{1} and so is a reduction because SS is monotonic. If kx1k\geq x_{1} then S[1;k]S\setminus[1;k] is not a reduction of SS or a minimal reduction of S[1;k]S\setminus[1;k] would be a minimal reduction of SS whose least element is greater than x1x_{1}. So S2=S1[1;x1-1]S_{2}=S_{1}\setminus[1;x_{1}-1] is a reduction of S1S_{1}. The algorithm continues to find x2=max{min(T{x1}):T is a minimal reduction of S2,min(T)=x1}x_{2}=\max\{\min(T\setminus\{ x_{1}\}):T\text{ is a minimal reduction of }S_{2},\min(T)=x_{1}\} etc and finally returns a minimal reduction of SS. However, it is not clear that this approach can work if SS is not monotonic while we can hope that Lithium is still efficient if SS is “almost” monotonic. We remark that when there is only one minimal testcase T=[1,M]T=[1,M], the binary search approach would require something like k=0M-1log2(N-k)=Mlog2(N)+k=0M-1log2(1-kN)Mlog2(N)+Mlog2(1-MN)=Mlog2(N)+o(M)\sum _{{k=0}}^{{M-1}}\log _{2}(N-k)=M\log _{2}(N)+\sum _{{k=0}}^{{M-1}}\log _{2}\left(1-\frac{k}{N}\right)\geq M\log _{2}(N)+M\log _{2}\left(1-\frac{M}{N}\right)=M\log _{2}(N)+o(M). So that would be the worst case of the binary search approach whereas Lithium handles this case very nicely in Θ(log2(N)+M)\Theta(\log _{2}(N)+M)! In general, if there is only one minimal testcase TT of size MN2M\leq\frac{N}{2} then max(T)\max(T) can be anywhere between [M,N][M,N] and if TT is placed at random, max(T)34N\max(T)\leq\frac{3}{4}N with probability at least 12\frac{1}{2}. So the average complexity of the binary search approach in that case will be at least 12k=0M-1log2(14N)=12Mlog2(N)+o(M)\frac{1}{2}\sum _{{k=0}}^{{M-1}}\log _{2}(\frac{1}{4}N)=\frac{1}{2}M\log _{2}(N)+o(M) which is still not as good as Lithium’s optimal worst case of O(Mlog2(NM))O(M\log _{2}(\frac{N}{M}))