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The cycle lemma turns a constraint at every prefix into a choice of where to start reading a circular sequence. Think of a lattice path whose vertical increments record votes for two candidates: $+1$ for one candidate and $-1$ for the other. Its partial sums are the running lead. Rotating the increments preserves the final lead, but changes whether that lead stays strictly positive throughout. This makes rotation useful for counting lattice paths subject to prefix constraints; the sum-one cycle lemma also supplies the cyclic prefix condition in the cyclic Turán covering construction.

The sum-one form. Let $b=(b_1,\dots ,b_n)$ be any nonempty sequence of integers with ${\sum }_{i=1}^n{b}_i=1$. A starting position $r\in \{0,\dots ,n-1\}$ means reading the rotated sequenceCall $r$ good when every nonempty partial sum of $b^{(r)}$ is strictly positive, including the full sum:The cycle lemma says there is exactly one good starting position. Negative entries, zero entries and positive entries larger than one are all allowed in this sum-one form. The empty partial sum is zero and is excluded from the positivity requirement. Douglas West's exercise 1.3.46(a) states a stronger result for arbitrary integer entries: take its case with all $n$ partial sums positive. ^[ref]

A worked sum-one example. For $b=(2,-2,0,1)$, the partial sums starting at zero are $S_0,S_1,S_2,S_3,S_4=0,2,0,0,1$. Starts $0,1,2$ fail: they have, respectively, a zero second partial sum, a negative first partial sum, and a zero first partial sum. Start $3$ reads $(1,2,-2,0)$ and has nonempty partial sums $1,3,1,1$, all positive. This example both allows an increment larger than one and shows why touching zero fails.

Why the last minimum works. Put $S_0=0$ and $S_j={\sum }_{i=1}^j{b}_i$. Choose the last index $r$ attaining the minimum of the proper partial sums $S_0,\dots ,S_{n-1}$: include $S_0$ and exclude $S_n$. After that index, later proper partial sums are strictly larger than $S_r$; also $S_n=1>S_r$, since $S_r\le 0$. Before wrapping, the new partial sums are therefore positive differences $S_j-S_r$. After wrapping, they are $1+S_j-S_r\ge 1$. In the worked example the last minimum is at $r=3$; choosing the first minimum at $r=0$ would fail. Uniqueness takes one further observation: two good starts would split the circle into two nonempty arcs, each having a positive integer sum, yet their sum would be one.

A standard positive-total form of the Dvoretzky-Motzkin cycle lemma uses a nonempty sequence $a=(a_1,\dots ,a_n)$ of integers satisfyingExactly $k$ starting positions have every nonempty partial sum strictly positive, using the same rotation convention as the sum-one cycle lemma. Negative entries can be arbitrarily large in magnitude, and zero entries are allowed. Addario-Berry's proposition 2.2.1 gives an equivalent form with increments at least $-1$ and negative total; reversing the sequence and changing all signs gives the form here. ^[ref]

Four starts, two successes. For $a=(1,1,1,-1)$, the sum is $k=2$. Starts $0$ and $1$ give partial sums $(1,2,3,2)$ and $(1,2,1,2)$, respectively. Start $2$ gives $(1,0,1,2)$ and touches zero; start $3$ gives $(-1,0,1,2)$ and begins below zero. Thus exactly two starts are good. The rotated sequence $(-1,1,1,1)$ has the same positive total, but its first start is bad: a positive final lead alone does not control earlier partial sums.

Positions can give identical words. The sequence $(1,1,-1,1,1,-1)$ repeats a block of length three and has sum two. Its good starts are $0$ and $3$, both giving the same word $(1,1,-1,1,1,-1)$ with partial sums $(1,2,1,2,3,2)$. Starts $1$ and $4$ touch zero; starts $2$ and $5$ begin negatively. The cycle lemma with increments at most one counts two starting positions even though there is only one distinct good rotated word. Counting distinct words would lose the multiplicity that makes the statement true.

The upper bound matters. The one-entry sequence $(2)$ has positive integer sum two, but only one starting position. Thus the sum-$k$ conclusion fails for arbitrary positive increments when $k>1$. The sum-one cycle lemma above needs no such upper bound. Strict positivity also matters: counting paths that merely stay at or above zero would include additional starts in both diagrams.

From rotation to enumeration. For a fixed admissible sequence, a start chosen from a uniform distribution on a finite set of its $n$ positions is good with probability $k/n$. More generally, let $\mathcal{F}$ be any finite set of length-$n$ sequences satisfying the increment bound and total $k$, closed under rotation. Double counting pairs consisting of a sequence and a good starting position givesFor each good output sequence and each of the $n$ positions, inverse rotation gives one such pair, even when several rotated words coincide. For example, the four sequences containing three $+1$ votes and one $-1$ vote have exactly $\frac{2}{4}\left(\genfrac{}{}{0ex}{}{4}{3}\right)=2$ vote orders with a strictly positive running lead: $(1,1,1,-1)$ and $(1,1,-1,1)$. This is how the cycle lemma with increments at most one converts many prefix inequalities into a simple counting factor.

With $r$ balanced cyclically ordered vertex classes and $2\le \ell \le r+1$, include an $\ell$-set if some start has at least $k+1$ vertices in its first $k$ classes for every $1\le k\le \ell -1$. The cycle lemma implies every $(r+1)$-set contains an edge. Its asymptotic density is $((\ell -1)/r{)}^{\ell -1}$; taking the hypergraph complement gives a lower bound for the Turán density of a complete uniform hypergraph.

There is a genuine error in the printed threshold. With $\ell =3$, its only nontrivial requirement is that some two consecutive classes contain at least one vertex of the triple. Every triple meets some class, so every triple qualifies. For example, when $r=4$, the printed construction is the complete 3-uniform hypergraph, of edge density $1$, whereas the claimed density is $(2/4{)}^2=1/4$. Thus the stated edge-count assertion is false even in a nondegenerate parameter range.

The intended cyclic Turán covering construction, which gives the claimed density and the stated lower bound, uses threshold $k+1$, in the natural range $2\le \ell \le r+1$. We solve this repaired version explicitly. Denote it by $G^{+}$ to keep it separate from the literal printed construction.

A cyclic prefix argument gives the covering property. We first prove the relevant cycle lemma. If integers $b_1,\dots ,b_m$ satisfy ${\sum }_i{b}_i=1$, exactly one cyclic starting position has every nonempty partial sum strictly positive. Existence follows by starting after the last minimum among the proper partial sums $0,b_1,\dots ,b_1+\cdots +b_{m-1}$. Subsequent nonwrapped sums are strictly above that minimum, and wrapped sums add the total $1$. For uniqueness, if two starts split the cycle into arcs with sums $x$ and $1-x$, both positive, their integrality would require $x\ge 1$ and $1-x\ge 1$, a contradiction.

Take any $r+1$ vertices and let $a_i$ count them in class $V_i$. The increments $b_i=a_i-1$ sum to one. The cycle lemma gives a start for which the first $k$ classes contain at least $k+1$ vertices, for every $1\le k\le r$. Take the first $\ell$ vertices encountered from that start in class order. The first $\ell -1$ classes already contain at least $\ell$ vertices; the chosen vertices therefore satisfy all the repaired prefix inequalities. HenceEvery repaired edge also satisfies the weaker printed inequalities. Thus this argument proves the covering assertion for the printed construction too, although it cannot repair its false edge count.

Count the repaired edges by labelled class assignments. Put $m=\ell -1$. For a fixed cyclic start, an edge satisfying the repaired inequalities has all $\ell$ vertices in its first $m$ classes. Assign $\ell$ labelled vertices to these $m$ positions. There are $m^{\ell }$ assignments in total. Their occupancy counts $a_1,\dots ,a_m$ satisfy ${\sum }_i(a_i-1)=1$; the cycle lemma says exactly one of their $m$ cyclic starts has the required prefixes. Rotational symmetry therefore gives exactly $m^{\ell -1}$ good assignments for a specified start.

We must check that summing over the $r$ starts does not count an edge twice. Suppose two starts cut the circle into arcs of lengths $h$ and $r-h$. If an arc has length at most $m$, goodness of its starting point forces that arc to contain at least its length plus one vertices. If its length is greater than $m$, it contains all $\ell$ vertices, while the other good start demands vertices in the disjoint complementary arc, impossible. If both lengths are at most $m$, their combined demand is at least $r+2>\ell$ vertices, again impossible. Thus a repaired edge has a unique good start.

There are consequently $r{m}^{\ell -1}$ good assignments of class labels to $\ell$ labelled vertices. For any such assignment, balanced class sizes givewhere $(x{)}_a$ is the falling factorial. Divide the total count of ordered distinct vertices by $\ell !$. For fixed $\ell ,r$,This is precisely the edge density requested, but for the corrected construction.

Pass from covering to a forbidden-hypergraph clique construction. Use $r-1$ balanced classes in $G^{+}$ and take its hypergraph complement. Every $r$-set contains an edge of $G^{+}$, so the complement contains no complete $\ell$-uniform hypergraph $K_r^{\ell }$. The preceding count givesHere Turán density means $\pi (F)={\lim }_{n\to \infty }\mathrm{ex}(n,F)/\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)$ for a fixed $\ell$-uniform forbidden hypergraph; the limit exists because these normalized extremal densities are nonincreasing under averaging over smaller vertex subsets.

A corresponding classical upper bound is the de Caen bound for hypergraph Turán density:Equivalently, an $\ell$-uniform covering hypergraph in which every $r$-set contains an edge has asymptotic edge density at least $1/\left(\genfrac{}{}{0ex}{}{r-1}{\ell -1}\right)$. Its proof recursively counts complete subhypergraphs by their one-vertex extensions. Double counting the extensions and applying the Cauchy-Schwarz inequality to common extension neighbourhoods gives a lower bound for the ratio of successive hypergraph clique counts. If the missing-edge density were smaller than $1/\left(\genfrac{}{}{0ex}{}{r-1}{\ell -1}\right)$, those bounds would force a positive number of $r$-vertex hypergraph cliques, a contradiction. Applying the resulting covering bound to the complement of a $K_r^{\ell }$-free hypergraph gives the displayed upper bound. This incidence argument supplies the extra factor beyond the simpler direct count of one covering edge in each $r$-set, which alone yields only $1-1/\left(\genfrac{}{}{0ex}{}{r}{\ell }\right)$.

The lower bound is therefore valid via $G^{+}$, while the printed enumeration cannot be proved with threshold $k-1$. Both the counterexample and the necessary repair are part of the solution, rather than a silent change of the definition.

The general upper bound is recorded, for example, in Turán densities of some hypergraphs related to complete uniform hypergraphs.