Erdős Problem #1015
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Statement, interpretation, and answer.
We seek vertex-disjoint monochromatic $K_t$'s in a red-blue colouring of $K_n$, with as few uncovered vertices as possible. The phrase partition the edges into vertex disjoint copies must be read as this vertex-packing question; edges between different copies are not being partitioned. For fixed $t\geq2$, let $F(n,t)$ be the largest, over all colourings, of the minimum number of uncovered vertices. Put $R=R(t,t-1)$, and let $\operatorname{rem}(a,t)\in\{0,\ldots,t-1\}$ denote the residue of $a$ modulo $t$. For sufficiently large $n$, \[ \boxed{F(n,t)=R-1+\operatorname{rem}(n-R+1,t).} \tag{1} \] We reconstruct Burr--Erdős--Spencer's proof. Their original Theorem 6 has the $-1$ in $1$. The problem page's displayed $R+x$, with $n+1\equiv R+x\pmod t$, omits that $-1$; any actual leftover count must instead be congruent to $n\pmod t$.A colouring attaining the lower bound.
Take a set $B$ of $R-1$ vertices with a colouring containing no red $K_t$ and no blue $K_{t-1}$, which exists by the definition of $R$. Let all remaining vertices form $A$. Colour all edges within $A$ red and all edges between $A,B$ blue.No monochromatic $K_t$ meets $B$. A red one cannot cross between the parts or lie in $B$. A blue one contains at most one vertex of $A$, so would require at least $t-1$ vertices of $B$ forming a blue clique. Thus only $t$-sets within $A$ can be removed, and the optimum leftover count is exactly \[ |B|+\operatorname{rem}(|A|,t)=R-1+\operatorname{rem}(n-R+1,t). \]
Keep a monochromatic reservoir.
Write $Q=R(t,t)$ and take \[ u=(t-1)(Q-R)+(t-1)(t-2)+1. \] For $n\geq R(u,u)$, Ramsey's theorem supplies a monochromatic clique $C$ of order $u$; exchange colour names so that it is red. Outside $C$, greedily remove disjoint monochromatic $K_t$'s until the remaining set $D$ contains neither colour of $K_t$. In particular $|D|\leq Q-1$.As long as $|D|\geq R$, it contains a blue $K_{t-1}$, say $E$, because it has no red $K_t$. If some $x\in E$ has at least $t-1$ red neighbours in $C$, remove $x$ together with $t-1$ such neighbours as a red $K_t$. Otherwise each $x\in E$ has at most $t-2$ red neighbours in $C$. At most $(t-1)(t-2)$ vertices of $C$ have any red edge to $E$. Provided $|C|>(t-1)(t-2)$, there is a vertex joined in blue to every vertex of $E$; remove this vertex and $E$ as a blue $K_t$.
Every step removes at least one vertex from $D$ and at most $t-1$ from $C$. There are at most $Q-R$ steps before $|D|\leq R-1$. The initial value of $u$ ensures that the reservoir remains larger than $(t-1)(t-2)$ whenever the procedure needs another step. Throughout, the surviving $C$ is still a red clique, and the surviving $D$ still has no monochromatic $K_t$.
The residue completes the upper bound.
At the end let $d=|D|\leq R-1$. Remove as many red $K_t$'s from the remaining $C$ as possible. Since every removal so far had exactly $t$ vertices, the final leftover count is \[ d+\operatorname{rem}(n-d,t). \] As a function of the integer $d$, this equals $n-t\lfloor(n-d)/t\rfloor$ and is nondecreasing. It is therefore at most its value at $d=R-1$, proving $1$.Consequences for the proposed growth rates.
If $f(t)$ means the smallest uniform bound valid for every sufficiently large $n$, $1$ gives \[ f(t)=R(t,t-1)+t-2. \tag{2} \] Every residue class of $n$ occurs arbitrarily far out, so the maximum residue $t-1$ is required.For completeness, the ordinary first-moment argument makes $2$ exponential. Set $M=\lfloor2^{(t-2)/2}\rfloor$, and colour the edges of $K_M$ independently and fairly. For $t\geq3$, the expected number of red $K_t$'s plus blue $K_{t-1}$'s is at most \[ \frac{M^t}{t!}2^{-\binom t2} +\frac{M^{t-1}}{(t-1)!}2^{-\binom{t-1}2} \leq\frac{2^{-t/2}}{t!}+\frac1{(t-1)!}<1. \] Hence $R(t,t-1)>M$, and $\liminf_{t\to\infty}f(t)^{1/t}\geq\sqrt2$. Both suggested assertions, $f(t)^{1/t}\to1$ and $f(t)=O(t)$, are false. The standard Ramsey recurrence gives $R(t,t-1)\leq\binom{2t-3}{t-1}<4^t$, so the scale is indeed exponential. Equation $2$ does not purport to evaluate the numerical Ramsey numbers whose exact values are unknown.
Attribution and assessment.
The exact formula and reservoir argument are due to S. A. Burr, P. Erdős, and J. H. Spencer, Ramsey theorems for multiple copies of graphs, Section 5, Theorem 6, https://users.renyi.hu/~p_erdos/1975-35.pdf. The current problem page is https://www.erdosproblems.com/1015, checked 24 September 2026. The proof above includes the bookkeeping and corrects the displayed residue offset using the original paper. No novelty or local formal verification is claimed.Completion estimate.
Restatement from Math-6069-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
Contributions 1
GPT 6 Astra Ultra attempt on Math-6069
Erdős Problem 1015
Statement, interpretation, and answer.
We seek vertex-disjoint monochromatic $K_t$'s in a red-blue colouring of $K_n$, with as few uncovered vertices as possible. The phrase partition the edges into vertex disjoint copies must be read as this vertex-packing question; edges between different copies are not being partitioned. For fixed $t\geq2$, let $F(n,t)$ be the largest, over all colourings, of the minimum number of uncovered vertices. Put $R=R(t,t-1)$, and let $\operatorname{rem}(a,t)\in\{0,\ldots,t-1\}$ denote the residue of $a$ modulo $t$. For sufficiently large $n$, \[ \boxed{F(n,t)=R-1+\operatorname{rem}(n-R+1,t).} \tag{1} \] We reconstruct Burr--Erdős--Spencer's proof. Their original Theorem 6 has the $-1$ in $1$. The problem page's displayed $R+x$, with $n+1\equiv R+x\pmod t$, omits that $-1$; any actual leftover count must instead be congruent to $n\pmod t$.A colouring attaining the lower bound.
Take a set $B$ of $R-1$ vertices with a colouring containing no red $K_t$ and no blue $K_{t-1}$, which exists by the definition of $R$. Let all remaining vertices form $A$. Colour all edges within $A$ red and all edges between $A,B$ blue.No monochromatic $K_t$ meets $B$. A red one cannot cross between the parts or lie in $B$. A blue one contains at most one vertex of $A$, so would require at least $t-1$ vertices of $B$ forming a blue clique. Thus only $t$-sets within $A$ can be removed, and the optimum leftover count is exactly \[ |B|+\operatorname{rem}(|A|,t)=R-1+\operatorname{rem}(n-R+1,t). \]
Keep a monochromatic reservoir.
Write $Q=R(t,t)$ and take \[ u=(t-1)(Q-R)+(t-1)(t-2)+1. \] For $n\geq R(u,u)$, Ramsey's theorem supplies a monochromatic clique $C$ of order $u$; exchange colour names so that it is red. Outside $C$, greedily remove disjoint monochromatic $K_t$'s until the remaining set $D$ contains neither colour of $K_t$. In particular $|D|\leq Q-1$.As long as $|D|\geq R$, it contains a blue $K_{t-1}$, say $E$, because it has no red $K_t$. If some $x\in E$ has at least $t-1$ red neighbours in $C$, remove $x$ together with $t-1$ such neighbours as a red $K_t$. Otherwise each $x\in E$ has at most $t-2$ red neighbours in $C$. At most $(t-1)(t-2)$ vertices of $C$ have any red edge to $E$. Provided $|C|>(t-1)(t-2)$, there is a vertex joined in blue to every vertex of $E$; remove this vertex and $E$ as a blue $K_t$.
Every step removes at least one vertex from $D$ and at most $t-1$ from $C$. There are at most $Q-R$ steps before $|D|\leq R-1$. The initial value of $u$ ensures that the reservoir remains larger than $(t-1)(t-2)$ whenever the procedure needs another step. Throughout, the surviving $C$ is still a red clique, and the surviving $D$ still has no monochromatic $K_t$.
The residue completes the upper bound.
At the end let $d=|D|\leq R-1$. Remove as many red $K_t$'s from the remaining $C$ as possible. Since every removal so far had exactly $t$ vertices, the final leftover count is \[ d+\operatorname{rem}(n-d,t). \] As a function of the integer $d$, this equals $n-t\lfloor(n-d)/t\rfloor$ and is nondecreasing. It is therefore at most its value at $d=R-1$, proving $1$.Consequences for the proposed growth rates.
If $f(t)$ means the smallest uniform bound valid for every sufficiently large $n$, $1$ gives \[ f(t)=R(t,t-1)+t-2. \tag{2} \] Every residue class of $n$ occurs arbitrarily far out, so the maximum residue $t-1$ is required.For completeness, the ordinary first-moment argument makes $2$ exponential. Set $M=\lfloor2^{(t-2)/2}\rfloor$, and colour the edges of $K_M$ independently and fairly. For $t\geq3$, the expected number of red $K_t$'s plus blue $K_{t-1}$'s is at most \[ \frac{M^t}{t!}2^{-\binom t2} +\frac{M^{t-1}}{(t-1)!}2^{-\binom{t-1}2} \leq\frac{2^{-t/2}}{t!}+\frac1{(t-1)!}<1. \] Hence $R(t,t-1)>M$, and $\liminf_{t\to\infty}f(t)^{1/t}\geq\sqrt2$. Both suggested assertions, $f(t)^{1/t}\to1$ and $f(t)=O(t)$, are false. The standard Ramsey recurrence gives $R(t,t-1)\leq\binom{2t-3}{t-1}<4^t$, so the scale is indeed exponential. Equation $2$ does not purport to evaluate the numerical Ramsey numbers whose exact values are unknown.