Proof. We first construct a planar computable graph that does not have a computable $4$ -coloring. Let $\{\varphi _{e}\}_{e\in {\mathbb {N}}}$ be the standard listing of the partial computable functions from ${\mathbb {N}}$ to $\{0,1,2,3\}$ . We construct a graph $G=\langle V,E\rangle $ satisfying the following requirement for each $e\in {\mathbb {N}}$ .

$$ \begin{align*}R_{e}:\text{If }\varphi_{e}\text{ is total, then }\varphi_{e}\text{ is not a coloring of }G. \end{align*} $$

We build G in stages, and say that a requirement $R_{e}$ is met when it has found some witness $v_{e}^{*}$ such that there is some v adjacent to $v_{e}^{*}$ and $\varphi _{e}(v)=\varphi _{e}(v_{e}^{*}){\downarrow }$ . In other words, $\varphi _{e}$ fails to be a $4$ -coloring of G.

• Stage $0$ : Enumerate a $K_{4}$ graph into G and let the vertices of this $K_{4}$ be denoted as $v_{0,i}$ for each $i<4$ . In addition to these vertices, also enumerate a single (temporarily) isolated vertex $v_{0,4}$ into G. The vertices $v_{0,j}$ for various j will be referred to as the $0^{th}$ component of G.

• Stage $s>0$ : Begin the $s^{th}$ component of G by enumerating five new vertices $v_{s,j}$ for each $j<5$ and connect $v_{s,i}$ for each $i<4$ as a $K_{4}$ graph.

  For each $e<s$ , if $\varphi _{e}(v_{e,j}){\downarrow }$ for all $j<5$ and $R_{e}$ has yet to act, then enumerate $v_{e,5}$ into the $e^{th}$ component and connect $v_{e,5}$ to $v_{e,4}$ and $v_{e,i}$ for each $i<4$ where $\varphi _{e}(v_{e,i})\neq \varphi _{e}(v_{e,4})$ . Then declare $v_{e,5}$ as the witness for $R_{e}$ and proceed with the construction.

Observe that if some requirement $R_{e}$ never acts, then there must be some $i<5$ for which $\varphi _{e}(v_{e,i})[s]{\uparrow }$ at every stage s. Then $\varphi _{e}$ cannot possibly be total and thus $R_{e}$ is satisfied. Suppose then that there is some stage at which $R_{e}$ is discovered to be ready. Since there is no injury in the construction, (each requirement acts only on the component reserved for it), then $R_{e}$ must act at some stage s. When it does, recall that $v_{e,5}$ is then enumerated into the $e^{th}$ component and connected with vertices $v_{e,4}$ and $v_{e,i}$ for each $i<4$ where $\varphi _{e}(v_{e,i})\neq \varphi _{e}(v_{e,4})$ . Then $\varphi _{e}$ must fail as a $4$ -coloring as the neighbours of $v_{e,5}$ are colored four different colors.

The main idea was simply to wait until a potential coloring has used up all of its allowable colors in a certain configuration and then enumerating a new vertex connected to all the vertices colored with the allowable colors. Since $K_{4}$ is planar and must be colored with four colors, we can ensure that any potential $4$ -coloring uses all four colors available to it. Evidently, we cannot simply enumerate a new vertex and connect it to all vertices currently in the $K_{4}$ otherwise the resulting graph is no longer planar. Thus we have an additional isolated vertex in each component to occupy the final color. It is not difficult to see that when the number of colors increases, the diagonalisation strategy will become more complex as it has to be able to get any potential coloring to commit to more colors. However, since every planar graph is $4$ -colorable, it is evident that each component should become ‘layered’ in some way.

The general gadget will be as follows. Start by enumerating sufficiently many $K_{4}$ graphs. By pigeonhole principle, there will be at least two, say $K_{4}^{0},K_{4}^{1}$ , that are colored the same four colors. Now connect a new vertex $\blacktriangle $ to two vertices each from $K_{4}^{0}$ and $K_{4}^{1}$ all of different colors. As shown in Figure 2, this is still a planar graph. Furthermore, $\blacktriangle $ cannot be colored any of the four colors used to color $K_{4}^{0},K_{4}^{1}$ . We shall refer to the configuration in Figure 2 as $W_{1}$ and $\blacktriangle $ as the special vertex of $W_{1}$ .

Figure 2

$W_{1}$ .

We refer the reader to Figure 3 for the illustration of the general gadget. To construct $W_{n+1}$ , start by enumerating a large number of $K_{4}$ graphs and wait for the coloring c, to converge on these graphs. Once it does, there must be sufficiently many $K_{4}$ graphs all colored the same four colors, which we may assume to be $0,1,2,3$ . Then extend the $K_{4}$ graphs (in the sense of Definition 2.5) into $W_{1}$ graphs by pairing them up and enumerating the special vertices.

Notation 2.8. Let $l\leq 4$ and $i\in {\mathbb {N}}$ be given. We write $K_{l}(W_{i})$ to denote l many $W_{i}$ graphs, where the special vertices (to be defined shortly for $i>1$ , also see Figure 3) of the $W_{i}$ graphs are connected as a $K_{l}$ graph. For convenience, we also use $W_{0}$ to denote a single vertex; $K_{4}(W_{0})$ is isomorphic to $K_{4}$ .

Figure 3

$W_{n+1}$ .

Figure 3. Long description

A horizontal sequence of six graph structures is arranged across the top. From left to right, the first two structures are simple tetrahedral graphs with four solid black nodes each. The next two structures are larger triangular clusters where each of the four nodes is labeled W sub 1. An ellipsis indicates a continuation of the pattern. The final two structures are triangular clusters where each of the four nodes is labeled W sub n. Each triangular cluster consists of three outer nodes forming a triangle with a fourth node at the center, all interconnected. From the bottom-most nodes of every cluster, curved and straight lines descend to converge at a single solid black triangular vertex located at the bottom center of the diagram.

In addition, extend the resulting $W_{1}$ graphs into $K_{4}(W_{1})$ graphs by connecting the special vertices of four $W_{1}$ graphs as a $K_{4}$ . The idea here is that since the special vertex of each $W_{1}$ cannot be colored any of $0,1,2,3$ , then the new $K_{4}(W_{1})$ configuration forces the coloring to use the next four colors.

In general, once the coloring has converged on the vertices of all the $K_{4}(W_{n})$ graphs, we extend it to $W_{n+1}$ by enumerating a special vertex $\blacktriangle $ and connecting it to two copies each of $K_{4},K_{4}(W_{1}),\dots ,K_{4}(W_{n})$ as shown in Figure 3. We may assume that each pair of $K_{4}(W_{i})$ is colored with $4i,4i+1,4i+2$ , and $4i+3$ (by pigeonhole principle). Then by connecting $\blacktriangle $ the special vertices of these various $K_{4}(W_{i})$ , each colored $4i,4i+1,4i+2,4i+3$ , the coloring cannot use any of the first $4n$ many colors to color $\blacktriangle $ . Now take four copies of $W_{n+1}$ which are colored the same $4n$ colors (except the special vertex) and connect the special vertices as a $K_{4}$ graph to form $K_{4}(W_{n+1})$ . Evidently, for a fixed computable k-coloring $\varphi $ , $W_{n}$ cannot be colored by $\varphi $ for any $n\geq \lceil k/4\rceil $ .

Let some $n\geq 4$ be given and let $\{\varphi _{e}\}_{e\in {\mathbb {N}}}$ be a listing of the computable functions $\varphi _{e}:{\mathbb {N}}\to \{0,1,\dots ,n-1\}$ . Once again we aim to construct a graph G satisfying the conditions below:

$$ \begin{align*} R_{e}:\,\text{If }\varphi_{e}\text{ is total, then }\varphi_{e}\text{ is not a coloring of }G. \end{align*} $$

The graph will be constructed in components where each component aims to satisfy a single $R_{e}$ . The action on the $e^{th}$ component will depend solely on the behaviour of $\varphi _{e}$ .

Fix some $e\in {\mathbb {N}}$ . In the $e^{th}$ component, we aim to construct a $W_{m}^{\varphi _{e}}$ where m is the least such that $m\geq \lceil n/4\rceil $ . Since we know exactly how many $K_{4},K_{4}(W_{1}),\dots ,K_{4}(W_{m-1})$ are needed, the required starting amount of $K_{4}$ graphs can be found recursively. Begin by enumerating a sufficiently large number of $K_{4}$ graphs into the $e^{th}$ component. Proceed to construct $W_{m}^{\varphi _{e}}$ as described previously; wait for $\varphi _{e}$ to converge on the vertices in $K_{4}(W_{i}^{\varphi _{e}})$ before extending them to $W_{i+1}^{\varphi _{e}}$ .

If $\varphi _{e}$ is total, it must converge on every vertex within the $e^{th}$ component. Then we must succeed in constructing $W_{m}^{\varphi _{e}}$ . However, in order for $\varphi _{e}$ to successfully color $W_{m}^{\varphi _{e}}$ , $4m+1$ colors are needed. By choice of m, we obtain that $4m+1\geq n+1>n$ ; $\varphi _{e}$ cannot be a computable n-coloring.

Proof. Let $T\subseteq 2^{<{\mathbb {N}}}$ be an infinite tree. We construct a countable planar graph, $G=\langle V,E\rangle $ in stages as follows. An illustration of G can be found in Figure 4. The vertex set V will contain the following type of vertices:

• • $W=\{w_{\tau },v_{\tau }\mid \tau \in 2^{<{\mathbb {N}}}\}.$

• • $F=\{f_{\tau }\mid \tau \in 2^{<{\mathbb {N}}}\}$ .

• • $N=\{n_{\tau }\mid \tau \in 2^{<{\mathbb {N}}}\}$ .

• • A, some infinite set of vertices to be used in the construction.

Figure 4

Reducing ${{\mathtt {WKL}}}$ to ${\mathtt {COL}}(4)$ .

Figure 4. Long description

The diagram consists of two panels.

Left panel titled The tau-th, tau 0-th and tau 1-th component.

At the bottom is node n sub tau. Above it is a central node f sub tau flanked by v sub tau and w sub tau, forming a small diamond shape connected to n sub tau. This structure is enclosed in a larger triangle with vertices n sub tau, n sub tau 0, and n sub tau 1.

Above n sub tau 0 is a second hierarchical level. It features a base node n sub tau 0 connected to a central node f sub tau 0, flanked by two smaller nodes v sub tau 0 and w sub tau 0. This is enclosed in a triangle with top vertices n sub tau 00 and n sub tau 01.

Above n sub tau 1 is a third hierarchical level. It features a base node n sub tau 1 connected to a central node f sub tau 1, flanked by two smaller nodes v sub tau 1 and w sub tau 1. This is enclosed in a triangle with top vertices n sub tau 10 and n sub tau 11.

Right panel titled Acting in the sigma super th component.

At the bottom is node n sub sigma. Lines radiate upward to a wide top edge defined by nodes n sub sigma 0 on the left and n sub sigma 1 on the right.

In the center of the top edge is node f sub sigma. Below f sub sigma are nodes v sub sigma and w sub sigma.

To the left of f sub sigma, two triangular markers are placed on the connecting lines. To the right of f sub sigma, one triangular marker is placed on a connecting line.

All nodes are connected by straight lines forming a complex mesh of nested triangles.

• Stage $0$ : Enumerate $f_{\epsilon },w_{\epsilon },v_{\epsilon },n_{\epsilon },n_{0}$ , and $n_{1}$ . Connect them as shown in Figure 4 (take $\tau =\epsilon $ ).

• Stage $s>0$ : For each $\tau \in 2^{<{\mathbb {N}}}$ of length s, enumerate the vertices $n_{\tau 0},n_{\tau 1},f_{\tau },v_{\tau }$ , and $w_{\tau }$ , and connect them as shown in Figure 4. We shall refer to this as the $\tau $ -th component of the graph.

  We say that the $\sigma $ -th component has acted when a vertex from A has been enumerated and connected to the vertices in the $\sigma $ -th component. We once again refer the reader to Figure 4 for an illustration, where the vertices from A are denoted by the triangular vertices. Furthermore, once the $\sigma $ -th component has acted, we also say that it has chosen $\sigma x$ if there is only one vertex from A between $w_{\sigma }$ or $v_{\sigma }$ and $n_{\sigma x}$ . As an example, the $\sigma $ -th component in Figure 4 has chosen $\sigma 1$ .

  For convenience, we assume that at each stage, we compute the value of $T(\tau )$ for the next $\tau $ . If at stage s, $\tau x$ is found to be in the complement, then pick the maximal prefix $\sigma \subseteq \tau $ such that the $\sigma $ -th component has yet to act. Let y be such that $\sigma y\subseteq \tau x$ and act for the $\sigma $ -th component, choosing $\sigma ^{\frown }(1-y)$ .

Since each (diagonalisation) component is clearly planar and they are connected as shown in Figure 4, the graph as defined above is connected and planar. Applying Proposition 2.2 allows us to conclude that whenever some $\tau x$ is discovered to be in the complement of T, the maximal prefix $\sigma \subseteq \tau $ that has yet to act can always be found. Furthermore, if the $\sigma $ -th component acts and chooses $\sigma ^{\frown }(1-y)$ , then the subtree above $\sigma y$ is finite.

By ${\mathtt {COL}}(4)$ , there is a coloring $c:V\to \{0,1,2,3\}$ of G. We extract a path h through T using c as follows. Let $\tau _{i}=h\restriction i$ .

• • If $c(v_{\tau _{i}})=c(n_{\tau _{i}0})$ , then let $h(i)=0$ .

• • Otherwise, let $h(i)=1$ .

We proceed by induction on i to verify that $\tau _{i}\in T$ for every i. Base case is trivial since $\tau _{0}=\epsilon $ . Suppose inductively that $\tau _{i}\in T$ and consider the following cases:

• Case 1: $h(i)=0$ . If the $\tau _{i}$ -th component never acts, since $\tau _{i}\in T$ , then both subtrees extending $\tau _{i}0$ and $\tau _{i}1$ must be infinite (by Proposition 2.3). We may thus assume for a contradiction that the $\tau _{i}$ -th component has acted and chooses $\tau _{i}1$ ; there are two vertices from A between $v_{\tau _{i}}$ and $n_{\tau _{i}0}$ . However, by definition of h, we also have that $c(v_{\tau _{i}})=c(n_{\tau _{i}0})$ . This leads to a contradiction since c has only one color to color the two adjacent vertices from A between $v_{\tau _{i}}$ and $n_{\tau _{i}0}$ .

• Case 2: $h(i)=1$ . Once again, we may assume for a contradiction that the $\tau _{i}$ -th component acts and chooses $\tau _{i}0$ . In other words, there is exactly one vertex from A between $v_{\tau _{i}}$ and $n_{\tau _{i}0}$ (the $\tau $ -th component chooses $\tau _{i}0$ ). Since $h(i)=1$ , then $c(v_{\tau _{i}})\neq c(n_{\tau _{i}0})$ ; the vertex from A between $v_{\tau _{i}}$ and $n_{\tau _{i}0}$ is connected to four vertices all colored different colors by c. Therefore, c cannot be a valid coloring of G.

Thus, h must be a path through T. Since h can be computed solely from c, we obtain that ${{\mathtt {WKL}}}{\leq _{\mathsf {sW}}}{\mathtt {COL}}(4)$ as desired.

Proof. Fix some $f:{\mathbb {N}}\to {\mathbb {N}}$ , and let $\left \{\varphi _{e}^{f}\right \}_{e\in {\mathbb {N}}}$ be the standard listing of the f-computable functions. G will consist of components, each consisting of six $K_{3}$ graphs originally, labelled $K_{3}^{j}$ for each $j<6$ . At each stage s, check if $\varphi _{e}^{f}(e)[s]{\downarrow }$ . If $\varphi _{e}^{f}(e)[s]{\uparrow }$ , then do nothing. Otherwise, enumerate vertices into the $e^{th}$ component and connect them as illustrated in Figure 6, where $\circ $ represents the newly enumerated vertices, and the numbered $K_{3}$ graphs correspond to $K_{3}^{j}$ for $j<6$ .

Figure 6

The gadget for Theorem 2.14.

Figure 6. Long description

A four-panel diagram illustrating graph gadgets. Each panel contains six triangular components labeled 0 through 5, with 0, 1, and 2 on the bottom row and 3, 4, and 5 on the top row. Each triangle has three black nodes.

* Top-left panel: Condition is If phi sub e super f of e diverges or phi sub e super f of e converges to a value greater than 2. The six triangles are disconnected from each other.

* Top-right panel: Condition is If phi sub e super f of e converges to 0. The top three triangles are connected to a central horizontal structure of three white nodes, and the bottom three triangles are connected to a separate identical structure below it.

* Bottom-left panel: Condition is If phi sub e super f of e converges to 1. All top triangles are connected to a single white node, which is connected by a vertical edge to a second white node that connects to all bottom triangles.

* Bottom-right panel: Condition is If phi sub e super f of e converges to 2. All six triangles are connected to a single central white node, creating a unified star-like structure.

G as constructed above is clearly planar, and has an f-computable planar drawing, and thus by ${\mathtt {COL}}^{*}(5)$ has a $5$ -coloring c. For convenience, in the $e^{th}$ component, define $A_{e}=\{c(v)\mid v\in K_{3}^{j}\text { for }j<3\}$ and $B_{e}=\{c(v)\mid v\in K_{3}^{j}\text { for }3\leq j<6\}$ . Then define g as follows:

• • Define $g(e)=0$ if $|A_{e}|=3$ or $|B_{e}|=3$ .

  We now claim that if this is the case, $g(e)\neq \varphi _{e}^{f}(e)$ . Suppose otherwise, then $\varphi _{e}^{f}(e)=g(e)=0$ , and hence it must be that one of the sets $A_{e}$ or $B_{e}$ has size $3$ . Referring to Figure 6, observe that in either case, c cannot possibly extend to a $5$ -coloring of the $e^{th}$ component.

• • Define $g(e)=1$ , if $A_{e}=B_{e}$ and $|A_{e}|=|B_{e}|=4$ .

  Once again, we check that $g(e)\neq \varphi _{e}^{f}(e)$ . Suppose to the contrary that $\varphi _{e}^{f}(e){\downarrow }=g(e)=1$ . Then the two new vertices $v,w$ enumerated in this case cannot be colored by any of the four colors in $A_{e},B_{e}$ . However, as v is also adjacent to w, c cannot color both $v,w$ with the single remaining color.

• • Define $g(e)=2$ if both of the above cases do not hold.

  Assume for a contradiction that $\varphi _{e}^{f}(e)=g(e)=2$ , then a single vertex v must have been enumerated into the $e^{th}$ component. Furthermore, this vertex is connected to all other vertices in the $e^{th}$ component. But c is assumed to have already used up all five colors to color the original vertices in the $e^{th}$ component, leaving it no color for v.

Thus g as defined above satisfies ${\mathtt {DNR}}(3)$ and can be computed from c (a solution to the constructed instance G of ${\mathtt {COL}}^{*}(5)$ ). Therefore, ${\mathtt {DNR}}(3){\leq _{\mathsf {sW}}}{\mathtt {COL}}^{*}(5)$ .

Proof. Fix some $f:{\mathbb {N}}\to {\mathbb {N}}$ , and let $\left \{\varphi _{e}^{f}(e)\right \}_{e\in {\mathbb {N}}}$ be a standard listing of the f-computable functions. Once again, G will be made up of components, each of which now contains 24 $K_{3}$ graphs, denoted as $K_{3}^{j}$ for each $j<24$ .

At each stage s, do the following.

• Stage s: Enumerate the new $s^{th}$ component by enumerating 24 new $K_{3}$ graphs into G. For each $e<s$ , which is not yet declared satisfied, do the following.

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=0$ , then enumerate 24 new vertices, $v_{j}$ , for each $j<24$ , into the $e^{th}$ component. For each j, connect $v_{j}$ to all vertices in $K_{3}^{j}$ , and then for each $i<6$ , connect the vertices $v_{j}$ where $4i\leq j<4(i+1)$ in a $K_{4}$ graph. (See Figure 7 for an illustration; the labels of the $K_{3}$ graphs correspond to j and the newly enumerated vertices are represented by $\circ $ .)

    Figure 7

    The gadget for Theorem 2.16. We use ‘ $\cdots $ ’ to replace the edges to avoid cluttering the diagram.

    

    Figure 7. Long description

    The diagram consists of two panels, each with a logical statement at the top.

    Left Panel: Labeled ‘If phi super f sub e (e) down-arrow = 0, then for each i < 6’. It shows a graph with four triangular clusters labeled K sub 3 super 4i+2, K sub 3 super 4i+3, K sub 3 super 4i+1, and K sub 3 super 4i. Three clusters on the left (top, middle, and bottom) converge toward two central white nodes. The middle cluster, K sub 3 super 4i+3, is nested within the structure. The rightmost cluster, K sub 3 super 4i, is connected to the central nodes via a horizontal bridge of two white nodes. Black dots represent vertices within the clusters, while white circles represent connecting nodes.

    Right Panel: Labeled ‘If phi super f sub e (e) down-arrow = 1, then for each i = 0, 12’. This panel shows a more complex arrangement of rectangular blocks containing labels.

    - A top row of four blocks is labeled K sub 3 super i+4, K sub 3 super i+5, K sub 3 super i+6, and K sub 3 super i+7.

    - A bottom row of four blocks is labeled K sub 3 super i, K sub 3 super i+1, K sub 3 super i+2, and K sub 3 super i+3.

    - These two rows converge toward a central vertical pair of white nodes.

    - To the right, a large trapezoidal block contains a vertical stack of four labels: K sub 3 super i+11, K sub 3 super i+10, K sub 3 super i+9, and K sub 3 super i+8. This block is connected to the central nodes by a single white node. Ellipses (three dots) are used between the blocks and the central nodes to indicate repeated edges.

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=1$ , then enumerate six new vertices, $v_{i}$ for each $i<6$ , into the $e^{th}$ component and connect $v_{i}$ to all vertices of $K_{3}^{j}$ for each $4i\leq j<4(i+1)$ . Then connect $v_{0},v_{1},v_{2}$ and $v_{3},v_{4},v_{5}$ as two $K_{3}$ graphs, respectively. (See Figure 7 for an illustration.)

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=2$ , then enumerate two new vertices, $v_{0},v_{1}$ into the $e^{th}$ component. Connect $v_{0}$ to $v_{1}$ and all vertices in $K_{3}^{j}$ for each $j<12$ , and connect $v_{1}$ to all vertices in $K_{3}^{j}$ for each $j\geq 12$ .

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=3$ , then enumerate one new vertex $v_{0}$ into the $e^{th}$ component and connect $v_{0}$ to all vertices in each $K_{3}^{j}$ for every $j<24$ .

  In any of the cases, once the new vertices are enumerated, we declare e to be satisfied.

The graph G as described above is easily seen to be planar and having an f-computable plane diagram. Thus, a $6$ -coloring c may be obtained by applying ${\mathtt {COL}}^{*}(6)$ . Using c, define the following sets for each $e\in {\mathbb {N}}$ :

• • Let $A_{i}=\{c(v)\mid v\in K_{3}^{j}\text { for some }4i\leq j<4(i+1)\}$ for each $i<6$ .

• • Let $B_{i}=\{c(v)\mid v\in K_{3}^{j}\text { for some }12i\leq j<12(i+1)\}$ for each $i<2$ .

We are now ready to define $g(e)$ satisfying ${\mathtt {DNR}}(4)$ as follows:

• • If there is some i for which $|A_{i}|=3$ , then define $g(e)=0$ .Suppose now for a contradiction that $\varphi _{e}^{f}(e){\downarrow }=g(e)=0$ . By the construction, vertices $v_{j}$ for each $4i\leq j<4(i+1)$ must have been enumerated into the $e^{th}$ component and each $v_{j}$ is connected to all the vertices in $K_{3}^{j}$ . Since $|A_{i}|=3$ and c is a coloring, we have that $c(v_{j})\notin A_{i}$ for each $4i\leq j<4(i+1)$ . But the four vertices $v_{4i},v_{4i+1},v_{4i+2},v_{4i+3}$ are connected as a $K_{4}$ graph. Then c cannot possibly be a valid $6$ -coloring.

• • If the previous case does not hold and there is some i such that $|B_{i}|=4$ , then define $g(e)=1$ .Once again, suppose that $\varphi _{e}^{f}(e){\downarrow }=g(e)=1$ . Since the previous case does not hold, then it must be that for each $j<6$ , we have that $|A_{j}|\geq 4$ . An analysis of the second case of the construction would allow us to conclude that the six new vertices enumerated, $v_{j}$ for each $j<6$ is such that $c(v_{j})\notin A_{j}$ . In particular, since $A_{j}\subseteq B_{i}$ for each $3i\leq j<3(i+1)$ , and $|B_{i}|=4$ , then $c(v_{j})\notin B_{i}$ for each $3i\leq j<3(i+1)$ . Then c must fail as a $6$ -coloring as the vertices $v_{3i},v_{3i+1},v_{3i+2}$ are connected as a $K_{3}$ .

• • If neither of the previous cases hold and $B_{0}=B_{1}$ , then define $g(e)=2$ .Assume that $\varphi _{e}^{f}(e){\downarrow }=2$ . Then the two new vertices $v_{0},v_{1}$ enumerated into the $e^{th}$ component are, respectively, connected to all vertices in $K_{3}^{j}$ for each $j<12$ and $K_{3}^{l}$ for each $12\leq l<24$ . In particular, $c(v_{0})\notin B_{0}$ and $c(v_{1})\notin B_{1}$ . Since the previous cases all do not hold and $B_{0}=B_{1}$ , then $|B_{0}|=|B_{1}|\geq 5$ . c cannot possibly be a $6$ -coloring.

• • If none of the previous cases hold, then define $g(e)=3$ .If none of the previous cases hold, then $|B_{0}|,|B_{1}|\geq 5$ and $B_{0}\neq B_{1}$ . This means that $|B_{0}\cup B_{1}|=6$ . If $\varphi _{e}^{f}(e){\downarrow }=3$ , then in the $e^{th}$ component, a vertex $v_{0}$ must have been enumerated into the $e^{th}$ component and connected to all other vertices within the component. In particular, $c(v_{0})\notin B_{0}\cup B_{1}$ , and hence c fails to be a coloring.

Since the sets $A_{i}$ and $B_{j}$ can be computed using c, then g can similarly be obtained from c. Thus, ${\mathtt {DNR}}(4){\leq _{\mathsf {sW}}}{\mathtt {COL}}^{*}(6)$ .

Proof. Fix some $f:{\mathbb {N}}\to {\mathbb {N}}$ and let $\left \{\varphi _{e}^{f}(e)\right \}_{e\in {\mathbb {N}}}$ be a standard listing of the f-computable functions. Once again we construct the graph G via components. Each component will consist of 288 $K_{4}$ graphs indexed with $j<288$ . For each $K_{4}$ graph, we pick an arbitrary vertex and label it the special vertex.

We proceed with the construction as follows.

• Stage s: For each $e<s$ not yet declared satisfied, consider the following cases. (Refer to Figure 8 for an illustration. The nodes $\circ $ represent the newly enumerated vertices whilst $\blacktriangle $ and $\bullet $ represent the special and non-special vertices, respectively.)

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=0$ , then for each $j<48$ , enumerate vertices $v_{3j},v_{3j+1},v_{3j+2}$ connected as a $K_{3}$ graph. In addition, for each $i<3$ , connect each $v_{3j+i}$ to the special vertex of $K_{4}^{6j+2i}$ and the non-special vertices of $K_{4}^{6j+2i+1}$ .

    Figure 8

    The gadget for Theorem 2.17. To reduce clutter, we use $\bullet $ and $\blacktriangle $ to represent all the non-special and special vertices of the $K_{4}^{m}$ , respectively.

    

    Figure 8. Long description

    The diagram consists of four panels arranged in a two by two grid. Each panel begins with a conditional statement followed by a graph structure.

    * Top-left panel: Statement reads If phi sub e super f of e equals 0, then for each j less than 48. Below this, three white circular vertices are connected in a line with an arched edge connecting the outer two. Each white vertex is connected to two rectangular blocks below it. There are six blocks labeled K sub 4 super 6j through K sub 4 super 6j plus 5. Each block has a black dot and a black triangle on its top edge.

    * Top-right panel: Statement reads If phi sub e super f of e equals 1, then for each j less than 12. The structure features two central white vertices stacked vertically. The top white vertex connects to two upper rectangular blocks labeled K sub 4 super 24j plus 12 through 17 and K sub 4 super 24j plus 18 through 23. The bottom white vertex connects to two lower rectangular blocks labeled K sub 4 super 24j through 5 and K sub 4 super 24j plus 6 through 11. Black dots and triangles mark the connection points.

    * Bottom-left panel: Statement reads If phi sub e super f of e equals 2, then for each j less than 6. A single white vertex is positioned between an upper block labeled K sub 4 super 48j plus 24 through 47 and a lower block labeled K sub 4 super 48j through 23. An arrow points from the white vertex to the lower block.

    * Bottom-right panel: Statement reads If phi sub e super f of e equals 3, then for each j less than 2. A central white vertex connects to one upper block labeled K sub 4 super 144j plus 96 through 143 and two lower white vertices. These two lower white vertices are connected by a horizontal line and each connects to one of two large lower blocks labeled K sub 4 super 144j through 47 and K sub 4 super 144j plus 48 through 95.

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=1$ , then for each $j<12$ , enumerate vertices $v_{2j},v_{2j+1}$ and connect $v_{2j}$ to $v_{2j+1}$ . We also connect $v_{2j}$ to all the special vertices of $K_{4}^{24j+i}$ and all the non-special vertices of $K_{4}^{24j+i+6}$ for each $i<6$ . Similarly, connect $v_{2j+1}$ to all the special vertices of $K_{4}^{24j+i+12}$ and all the non-special vertices of $K_{4}^{24j+i+18}$ for each $i<6$ .

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=2$ , then for each $j<6$ , enumerate a vertex $v_{j}$ and for each $i<24$ , connect $v_{j}$ to all the special vertices of $K_{4}^{48j+i}$ and to all the non-special vertices of $K_{4}^{48j+i+24}$ .

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=3$ , then for each $j<2$ , enumerate $v_{3j}$ , $v_{3j+1}$ , $v_{3j+2}$ connected as a $K_{3}$ graph. Also connect $v_{3j+i}$ to the non-special vertices of $K_{4}^{144j+48i+l}$ for each $l<48$ .

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=4$ , then enumerate $v_{0},v_{1}$ and connect them. Connect $v_{0}$ to all the non-special vertices of $K_{4}^{i}$ for each $i<144$ and connect $v_{1}$ to all the non-special vertices of $K_{4}^{j}$ for each $144\leq j<288$ .

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=5$ , then enumerate a single vertex $v_{0}$ and connect it to all the non-special vertices of $K_{4}^{i}$ for each $i<288$ .

  If one of the above cases holds, then declare e satisfied.

To maintain connectedness of the graph described above, we can arrange the various $K_{4}$ graphs in a ‘row’, and for each $K_{4}$ , enumerate some vertex and connect it to one of the non-special vertices. Then connect these newly enumerated vertices as a path. It is clear that planarity is preserved and the resultant graph is now connected. Hence by ${\mathtt {ConnCOL}}(6)$ , there is a $6$ -coloring c of G. It remains to prove that g satisfying ${\mathtt {DNR}}(6)$ may be computed from c.

Fix some e, and within the $e^{th}$ component, for each $i<288$ , let

• • $N_{i} =\{c(v)\mid v\in K_{4}^{i}\text { and }v\text { non-special}\}$ ;

• • $S_{i} =\{c(v)\mid v\in K_{4}^{i}\text { and }v\text { special}\}$ .

Now consider the following cases:

1. (1) If there exists some $j<6$ such that $|\bigcup _{48j\leq i<48(j+1)}N_{i}|=3$ , then we consider the following subcases:

   1. (a) If there exists some i such that $8j\leq i<8(j+1)$ and $|\bigcup _{l<6}S_{6i+l}|=1$ , then define $g(e)=0$ . In other words, for each $l<6,\,N_{6i}=N_{6i+l}$ and $S_{6i}=S_{6i+l}$ .

   2. (b) If the previous subcase does not hold and there exists some i such that $2j\leq i<2(j+1)$ and $|\bigcup _{l<24}S_{24i+l}|=2$ , then define $g(e)=1$ . Since the previous subcase does not hold, then $|\bigcup _{l<6}S_{24i+6m+l}|>1$ for each $m<4$ . In particular, it must be exactly $2$ . Therefore, we obtain that for each $m<4$ ,

      $$ \begin{align*}\bigcup_{l<6}S_{24i+l}=\bigcup_{l<6}S_{24i+6m+l}\text{ and }\bigcup_{l<6}N_{24i+l}=\bigcup_{l<6}N_{24i+6m+l}. \end{align*} $$

   3. (c) If neither of the previous subcases hold, then define $g(e)=2$ . In particular, it must be the case that for each i such that $2j\leq i<2(j+1),\,|\bigcup _{l<24}S_{24i+l}|=3$ since c is a $6$ -coloring. Therefore,

      $$ \begin{align*}\bigcup_{l<24}S_{48j+l}=\bigcup_{l<24}S_{48j+24+l}\text{ and }\bigcup_{l<24}N_{48j+l}=\bigcup_{l<24}N_{48j+24+l}. \end{align*} $$

   Combining the conclusions above with the corresponding actions taken during the construction if $\varphi _{e}^{f}(e)=k\leq 2$ allows us to conclude that c cannot possibly extend to a $6$ -coloring if $g(e)=\varphi _{e}^{f}(e)=k\leq 2$ .

2. (2) Suppose that the previous case does not hold; for every $j<6$ , $|\bigcup _{48j\leq i<48(j+1)}N_{i}|\geq 4$ . Now consider the following subcases:

   1. (a) If in addition, there is some $j<2$ for which $|\bigcup _{144j\leq i<144(j+1)}N_{i}|=4$ , then define $g(e)=3$ . This implies that for each $m<3$ ,

      $$ \begin{align*}\bigcup_{l<48}N_{144j+l}=\bigcup_{l<48}N_{144j+48m+l}. \end{align*} $$

   2. (b) If the previous subcase does not hold and $|\bigcup _{j<288}N_{j}|=5$ , then define $g(e)=4$ . Note that we have

      $$ \begin{align*}\bigcup_{l<144}N_{l}=\bigcup_{l<144}N_{144+l}. \end{align*} $$

   3. (c) Otherwise, define $g(e)=5$ . If neither of the previous subcases hold, we must have that $|\bigcup _{l<288}N_{l}|=6$ .

   A simple analysis of the actions taken during the construction if $\varphi _{e}^{f}(e)=k\geq 3$ allows one to conclude that it cannot be that $g(e)=\varphi _{e}^{f}(e)=k\geq 3$ .

Thus we have that ${\mathtt {DNR}}(6){\leq _{\mathsf {sW}}}{\mathtt {ConnCOL}}(6)$ .

Proof. Fix some $f:{\mathbb {N}}\to {\mathbb {N}}$ and let $\left \{\varphi _{e}^{f}\right \}$ be a standard listing of the f-computable functions. We adopt a similar gadget as before and construct the graph G in diagonalisation components. Each component will contain $9216\, K_{4}$ graphs (indexed in the superscript).

Now construct G as follows.

• Stage s: For each $e<s$ not yet declared satisfied, do the following:

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=0$ , then for each $j<1152$ , enumerate vertices $4j+i$ for each $i<4$ connected as a $K_{4}$ into the $e^{th}$ component. For each $i<4$ , connect each $v_{4j+i}$ to the special vertex of $K_{4}^{8j+2i}$ and the non-special vertices of $K_{4}^{8j+2i+1}$ .

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=1$ , then for each $j<192$ , enumerate vertices $3j+i$ for each $i<3$ connected as a $K_{3}$ . For each $i<3$ and $l<8$ , connect each $v_{3j+i}$ to all the special vertices of $K_{4}^{48j+16i+l}$ and the non-special vertices of $K_{4}^{48j+16i+8+l}$ .

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=2$ , then for each $j<48$ , enumerate vertices $v_{2j},v_{2j+1}$ and connect $v_{2j}$ to $v_{2j+1}$ . For each $i<2$ and $l<48$ , connect $v_{2j+i}$ to all the special vertices of $K_{4}^{192j+96i+l}$ and all the non-special vertices of $K_{4}^{192j+96i+48+l}$ .

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=3$ , then for each $j<24$ , enumerate a vertex $v_{j}$ into the $e^{th}$ component. In addition, for each $l<192$ , connect $v_{j}$ to all the special vertices of $K_{4}^{384j+l}$ and all the non-special vertices of $K_{4}^{384j+192+l}$ .

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=4$ , then for each $j<6$ , enumerate vertices $v_{4j+i}$ for each $i<4$ connected as a $K_{4}$ . In addition, for each $i<4$ and $l<384$ , connect $v_{4j+i}$ to all the non-special vertices of $K_{4}^{1536j+384i+l}$ .

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=5$ , then for each $j<2$ , enumerate vertices $v_{3j+i}$ for each $i<3$ connected as a $K_{3}$ . Also connect for each $i<3$ and $l<1536$ , $v_{3j+i}$ to all the non-special vertices of $K_{4}^{4608j+1536i+l}$ .

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=6$ , then enumerate vertices $v_{0}$ and $v_{1}$ adjacent to each other. For each $l<4608$ , connect $v_{0},v_{1}$ to all the non-special vertices of $K_{4}^{l}$ and $K_{4}^{4608+l}$ , respectively.

  • • If $\varphi _{e}^{f}(e)[s]{\downarrow }=7$ , then enumerate a single vertex $v_{0}$ and connect it to all the non-special vertices of $K_{4}^{l}$ for each $l<9216$ .

We note here that if we were to make a similar modification to the gadgets above by attempting to connect them as before, the graph will no longer be planar. In fact there should be no easy modification of the gadget which would make the proof work for ${\mathtt {DNR}}(k){\leq _{\mathsf {W}}}{\mathtt {ConnCOL}}(7)$ for some k (as proved formally in Theorem 3.13). Similarly, this graph G also might not have an f-computable plane diagram since we are unable to f-compute if a special vertex is the ‘inner’ or ‘outer’ vertex in a standard drawing of $K_{4}$ .

It follows from the construction above that the graph is f-computable and planar. Thus, ${\mathtt {COL}}(7)$ is able to produce a $7$ -coloring c of the graph. In order to define $g(e)$ , we define the following sets using the behaviour of c in the $e^{th}$ component:

• • $N_{i}=\{c(v)\mid v\in K_{4}^{i}\text { and }v\text { non-special}\}$ ;

• • $S_{i}=\{c(v)\mid v\in K_{4}^{i}\text { and }v\text { special}\}$ .

Consider the following possibilities:

1. (1) First suppose that there exists $j<24$ such that $|\bigcup _{384j\leq i<384(j+1)}N_{i}|=3$ . We have the following subcases:

   1. (a) If there also exists i such that $48j\leq i<48(j+1)$ and $|\bigcup _{l<8}S_{8i+l}|=1$ , then define $g(e)=0$ . In this subcase, it is easy to see that for each $l<8$ ,

      $$ \begin{align*}S_{8i}=S_{8i+l}\text{ and }N_{8i}=N_{8i+l}. \end{align*} $$

   2. (b) If the previous subcase does not hold and there exists i such that $8j\leq i<8(j+1)$ and $|\bigcup _{l<48}S_{48i+l}|=2$ , then define $g(e)=1$ . Since the previous subcase does not hold, for each $m<6$ , $|\bigcup _{l<8}S_{48i+8m+l}|>1$ and thus must be exactly $2$ . Therefore, for each $m<6$ ,

      $$ \begin{align*}\bigcup_{l<8}S_{48i+l}=\bigcup_{l<8}S_{48i+8m+l}\text{ and }\bigcup_{l<8}N_{48i+l}=\bigcup_{l<8}N_{48i+8m+l}. \end{align*} $$

   3. (c) If both the previous subcases do not hold and there exists i such that $2j\leq i<2(j+1)$ and $|\bigcup _{l<192}S_{192i+l}|=3$ , then define $g(e)=2$ . Since Case (1b) does not hold, then for each $m<4$ , $|\bigcup _{l<48}S_{192i+48m+l}|>2$ . That is, it must be exactly $3$ . We thus obtain that for each $m<4$ ,

      $$ \begin{align*}\bigcup_{l<48}S_{192i+l}=\bigcup_{l<48}S_{192i+48m+l}\text{ and }\bigcup_{l<48}N_{192i+l}=\bigcup_{l<48}N_{192i+48m+l}. \end{align*} $$

   4. (d) If none of the previous subcases hold, then define $g(e)=3$ . Since we also have that c is a $7$ -coloring, then for each i such that $2j\leq i<2(j+1)$ , $|\bigcup _{l<192}S_{192i+l}|=4$ . Which is to say that

      $$ \begin{align*}\bigcup_{l<192}S_{384j+l}=\bigcup_{l<192}S_{384j+192+l}\text{ and }\bigcup_{l<192}N_{384j+l}=\bigcup_{l<192}N_{384j+192+l}. \end{align*} $$

   Once again, using the conclusions obtained above and some careful analysis of the cases in the construction, we can conclude that if $g(e)=\varphi _{e}^{f}(e)= k\leq 3$ , c fails to $7$ -color the $e^{th}$ component.

2. (2) Now suppose that for each $j<24$ , $|\bigcup _{384j\leq i<384(j+1)}N_{i}|\geq 4$ . We consider the following subcases:

   1. (a) If in addition, there exists $j<6$ such that $|\bigcup _{l<1536}N_{1536j+l}|=4$ , then define $g(e)=4$ . With the assumption made in Case (2), we obtain the following. For each $m<4,$

      $$ \begin{align*}\bigcup_{l<384}N_{1536j+l}=\bigcup_{l<384}N_{1536j+384m+l}. \end{align*} $$

   2. (b) If there exists $j<2$ such that $|\bigcup _{l<4608}N_{4608j+l}|=5$ and the previous subcase does not hold, then define $g(e)=5$ . Using a similar analysis as before, for each $m<3$ ,

      $$ \begin{align*}\bigcup_{l<1536}N_{4608j+l}=\bigcup_{l<1536}N_{4608j+1536m+l}. \end{align*} $$

   3. (c) If $|\bigcup _{l<9216}N_{l}|=6$ and neither of the previous subcases hold, then define $g(e)=6$ . Since the assumption in Case (2b) fails, then

      $$ \begin{align*}\bigcup_{l<4608}N_{l}=\bigcup_{4608\leq l<9216}N_{l}. \end{align*} $$

   4. (d) Otherwise, define $g(e)=7$ . It is evident that in this subcase, $|\bigcup _{l<9216}N_{l}|=7$ .

   A similar analysis as before can be done to conclude that if $g(e)=\varphi _{e}^{f}(e)=k\geq 4$ , c must fail to be a $7$ -coloring of the $e^{th}$ component.

Therefore, g as defined satisfies ${\mathtt {DNR}}(8)$ and is computable from c.

Proof. Fix some function f and a standard listing of the computable functions $\left \{\varphi _{e}\right \}_{e\in {\mathbb {N}}}$ . Let $\left \{T_{x}^{f}\right \}_{x\in {\mathbb {N}}}$ be a universal $2$ -bounded c.e. trace. Define p a computable function such that

$$ \begin{align*}T_{p(e,x)}^{f}= \begin{cases} \emptyset,&\text{if }\varphi_{x}^{f}(x){\uparrow}\\ \left\{\varphi_{x}^{f}(x)\right\},&\text{if }\varphi_{x}^{f}(x){\downarrow}\wedge\varphi_{e}^{f}(e){\uparrow}\\ \left\{\varphi_{x}^{f}(x),\varphi_{e}^{f}(e)\right\},&\text{if }\varphi_{x}^{f}(x){\downarrow}\wedge\varphi_{e}^{f}(e){\downarrow}. \end{cases} \end{align*} $$

By ${\mathtt {DNR}}(k,1)$ , there exists some $g:{\mathbb {N}}\to \{0,1,\dots ,k-1\}$ such that for all $e,x,\,g(p(e,x))\notin T_{p(e,x)}^{f}$ , and g has a computable $1$ -approximation $\tilde {g}$ . Define $g^{*}$ (non-uniformly) satisfying ${\mathtt {DNR}}(k)$ as follows:

• Case 1: If there is some e such that for every x, $\tilde {g}(p(e,x),0)\neq \varphi _{x}^{f}(x)$ , then define $g^{*}(x)=\tilde {g}(p(e,x),0)$ . $g^{*}$ clearly exists via $\Delta _{1}^{0}$ -comprehension and it is evident that $g^{*}(x){\downarrow }\neq \varphi _{x}^{f}(x)$ for every x.

• Case 2: If for every e, there is some $x_{e}$ such that $\tilde {g}(p(e,x_{e}),0)=\varphi _{x_{e}}^{f}(x_{e})$ , then define $g^{*}(e)=\tilde {g}(p(e,x_{e}),s_{e})$ where $s_{e}$ is the first stage at which $\tilde {g}(p(e,x_{e}),s_{e})\neq \tilde {g}(p(e,x_{e}),0)$ .Since $\tilde {g}$ is a $1$ -approximation to g (satisfying ${\mathtt {DNR}}(k,1)$ ), then there must be some stage at which $\tilde {g}(p(e,x_{e}),s)\neq \tilde {g}(p(e,x_{e}),0)$ . Otherwise $\tilde {g}(p(e,x_{e}),s)=\tilde {g}(p(e,x_{e}),0)=\varphi _{x_{e}}^{f}(x_{e}){\downarrow }$ at every stage s which contradicts the assumption that $g(p(e,x_{e}))\notin T_{p(e,x_{e})}^{f}$ . Since s must exist, then there is always a least one and thus $g^{*}$ exists by $\Delta _{1}^{0}$ -comprehension.Suppose now for a contradiction that $\varphi _{e}^{f}(e){\downarrow }=g^{*}(e)=\tilde {g}(p(e,x_{e}),s_{e})$ . Since $\tilde {g}$ is a $1$ -approximation to g, that means that $g(p(e,x_{e}))=\tilde {g}(p(e,x_{e}),s_{e})$ . Then we have that $g(p(e,x_{e}))=\varphi _{e}^{f}(e){\downarrow }$ a contradiction. Thus $g^{*}$ satisfies ${\mathtt {DNR}}(k)$ .

In any case, we have that $g^{*}$ exists and satisfies ${\mathtt {DNR}}(k)$ .

Proof. Fix some function f and the standard listing of the computable functions $\{\varphi _{e}\}_{e\in {\mathbb {N}}}$ . Let $\left \{T_{x}^{f}\right \}_{x\in {\mathbb {N}}}$ and $\{U_{x}^{f}\}_{x\in {\mathbb {N}}}$ be an $l+1$ -bounded and $l+2$ -bounded universal c.e. trace, respectively. Define p a computable functions such that

$$ \begin{align*}U_{p(e,x)}^{f}=\begin{cases} \emptyset,&\text{if }\varphi_{x}^{f}(x){\uparrow}\\ \left\{\varphi_{x}^{f}(x)\right\}\cup T_{e}^{f},&\text{if }\varphi_{x}^{f}(x){\downarrow}. \end{cases} \end{align*} $$

By ${\mathtt {DNR}}(k,l+1)$ , there must exist g such that for all $e,x,\,g(p(e,x))\notin U_{p(e,x)}^{f}$ with a computable $l+1$ -approximation $\tilde {g}$ . We proceed non-uniformly and aim to define a function h which satisfies ${\mathtt {DNR}}(k)\vee {\mathtt {DNR}}(k,l)$ .

• Case 1: If there is some e such that for every x, $\tilde {g}(p(e,x),0)\neq \varphi _{x}^{f}(x)$ , then define $h(x)=\tilde {g}(p(e,x),0)$ . h clearly exists via $\Delta _{1}^{0}$ -comprehension and satisfies ${\mathtt {DNR}}(k)$ .

• Case 2: If for every e, there is some $x_{e}$ such that $\tilde {g}(p(e,x_{e}),0)=\varphi _{x_{e}}^{f}(x_{e})$ , then define $h(e)=\lim _{s\to \infty }\tilde {g}(p(e,x_{e}),s+s_{e})$ where $s_{e}$ is the first stage at which $\tilde {g}(p(e,x_{e}),s_{e})\neq \tilde {g}(p(e,x_{e}),0)$ . By assumption that g satisfies ${\mathtt {DNR}}(k,l+1)$ , then $g(p(e,x_{e}))\notin U_{p(e,x_{e})}^{f}$ . That is, $h(e)=g(p(e,x_{e}))\notin T_{e}^{f}$ . It remains thus to show that there is some computable $\tilde {h}$ which is an l-approximation of h.For similar reasons as before, $s_{e}$ exists and thus by defining $\tilde {h}(e,s)=\tilde {g}(p(e,x_{e}),s+s_{e})$ , $\tilde {h}$ exists via $\Delta _{1}^{0}$ -comprehension. Since we have that $\tilde {g}$ is an $l+1$ -approximation and $\tilde {g}(p(e,x_{e}),s_{e}-1)\neq \tilde {g}(p(e,x_{e}),s_{e})$ , then it is clear that $\tilde {h}$ is an l-approximation.

Thus we have that the h defined satisfies either ${\mathtt {DNR}}(k)$ or ${\mathtt {DNR}}(k,l)$ .

Proof. First, given n, we pick $k_{n},l_{n}$ as follows:

• • $l_{n}=\left \lfloor \frac {n}{2}\right \rfloor -1$ .

• • $q_{n,l_{n}+1}=2$ , the number of vertices in the $l_{n}+1$ -th layer of a component (both properly defined in the construction).

• • Proceed recursively for each $i<l_{n}+1$ and define $q_{n,i-1}$ as follows. Let $p_{l_{n}+1}=2$ , and let $p_{i-1}=\sum _{j=i}^{l_{n}+1}2p_{j}$ . Then define $q_{n,i-1}=(p_{i-1}-1)C(n,2)+1$ , where $C(x,y)$ is the binomial coefficient. Observe that by pigeonhole principle, $q_{n,i-1}$ is the number of pairs of vertices that when colored with some valid n-coloring, at least $p_{i-1}$ many pairs of vertices have the same two colors.

• • Define $k_{n}=\max \left \{\langle i,p_{i}^{*}\rangle \mid i<l_{n}+1\right \}+1$ , where $p_{i}^{*}$ is an encoding of a possible $p_{i}$ -tuple of pairs of vertices in the $i^{th}$ layer (to be defined later).

Fix some function $f:{\mathbb {N}}\to {\mathbb {N}}$ and the $l_{n}+1$ -bounded universal trace $\{T_{e}^{f}\}$ . We now construct an (f-computable) connected planar graph G with an f-computable plane diagram. The rough idea is that in order to obtain the ${\mathtt {DNR}}(k_{n},l_{n})$ function, we build the gadget in layers. Each time some new element is enumerated into the universal trace, we extend (in the sense of Definition 2.5) the gadget to the next layer, ensuring that any valid coloring encodes information regarding the element which just entered the universal trace. We now construct the graph in (diagonalisation) components as before.

• Stage s: Begin the $s^{th}$ component by enumerating $2q_{n,0}$ many vertices and connecting them as a path. These vertices shall be referred to as the $0^{th}$ layer of the $s^{th}$ component.For each $e<s$ , let i be the greatest such that the $i^{th}$ layer has been defined in the $e^{th}$ component and do the following.For each $t\in T_{e}^{f}[s]$ , check if $t=\langle i,p_{i}^{*}\rangle $ for some $p_{i}^{*}$ , an encoding of a $p_{i}$ -tuple consisting of entries $<q_{n,i}$ . If there are no $t\in T_{e}^{f}[s]$ which satisfy the desired condition, then do nothing and attend to the next $e<s$ or proceed to the next stage if there are no more $e<s$ to attend to.If there is some $t_{i}\in T_{e}^{f}[s]$ such that $t_{i}=\langle i,p_{i}^{*}\rangle $ for some $p_{i}^{*}$ , then enumerate new vertices $v_{j}$ for each $j<2q_{n,i+1}$ into the $e^{th}$ component. We refer to these vertices as the $i+1$ -th layer of the $e^{th}$ component. For each $j<2q_{n,i+1}$ , and for each $r<i+1$ , connect $v_{j}$ to some pair of vertices encoded by $p_{r}^{*}$ in the $r^{th}$ layer, where $\langle r,p_{r}^{*}\rangle \in T_{e}^{f}[s]$ (see Figure 9 for an illustrationFootnote ⁵ ). Finally, within the $i+1$ -th layer, connect $v_{2j}$ to $v_{2j+1}$ for each $j<q_{n,i+1}$ . Note that this procedure is well-defined as the $i+1$ -th layer can only exist at the $s^{th}$ stage if for each $r\leq i$ , some element in $T_{e}^{f}[s]$ is of the form $\langle r,p_{r}^{*}\rangle $ . By some careful arrangement of the connections as illustrated in Figure 9, each component can be made planar.Once we are done acting for each $e<s$ , proceed to stage $s+1$ .

  Figure 9

  Reversal of ${\mathtt {COL}}(n)$ . The dashed lines represent some number of vertices and edges; the $0^{th}$ layer is a path.

  

It is easy to see that the graph can be made connected with an f-computable plane diagram. By the choice of $q_{n,i}$ , there is always ‘sufficient space’ to connect each new layer to the previous ones.

Applying ${\mathtt {ConnCOL}}(n)$ , there is an n-coloring c of the graph G as described in the construction (we could also use ${\mathtt {COL}}(n)$ or ${\mathtt {COL}}^{*}(n)$ in place of ${\mathtt {ConnCOL}}(n)$ ). Now we attempt to define an $l_{n}$ -approximation $\tilde {g}$ satisfying ${\mathtt {DNR}}(k_{n},l_{n})$ .

To define an $l_{n}$ -approximation, the intuitive idea is that each layer aims to ‘block’ off two colors and corresponds roughly to each change in the approximation. In the case that all the colors are used up, we then show that the final value must avoid the trace. We provide the details as follows.

Fix some $e\in {\mathbb {N}}$ and consider the coloring c within the $e^{th}$ component. Define $\tilde {g}(e,0)=\langle 0,p_{0}^{*}\rangle $ , where $p_{0}^{*}$ is the encoding of the first $q_{n,0}$ many pairs in the $0^{th}$ layer of the $e^{th}$ component that is colored the same two colors by c. Without loss of generality, we may assume that these two colors are $0$ and $1$ . Now suppose inductively that $\tilde {g}(e,s)$ is some pair $\langle i,p_{i}^{*}\rangle $ where $p_{i}^{*}$ is an encoding of a $p_{i}$ -tuple with entries $<q_{n,i}$ . Furthermore, this $p_{i}$ -tuple encoded by $\tilde {g}(e,s)$ is colored $2i,2i+1$ by c and the $i^{th}$ layer exists in the $e^{th}$ gadget. We now aim to define $\tilde {g}(e,s+1)$ .

If $\tilde {g}(e,s)\notin T_{e}^{f}[s+1]$ , then let $\tilde {g}(e,s+1)=\tilde {g}(e,s)$ . In particular, observe that the desired property is satisfied. If we instead have that $\tilde {g}(e,s)\in T_{e}^{f}[s+1]$ , then some element t must have entered $T_{e}^{f}$ at stage $s+1$ and must be equal to $\langle i,p_{i}^{*}\rangle $ (the value of $\tilde {g}(e,s)$ ). When this happens in the construction, the $i+1$ -th layer consisting of $q_{n,i+1}$ many pairs of vertices must have been enumerated into the $e^{th}$ component. Furthermore, each vertex in the $i+1$ -th layer is connected to some pair of vertices in each layer $\leq i$ , which is encoded by some $t_{r}=\langle r,p_{r}^{*}\rangle \in T_{e}^{f}[s+1]$ for each $r\leq i$ as described in the construction. By the inductive hypothesis, this means that each vertex in the $i+1$ -th layer is adjacent to vertices colored $0,1,\dots ,2i,2i+1$ . In order for c to be a valid coloring, none of the vertices in the $i+1$ -th layer can be colored with the first $2i+2$ colors. In addition, the vertices in the $i+1$ -th layer are also connected as pairs, and by pigeonhole principle, there are at least $p_{i+1}$ many pairs colored the same two colors. Without loss of generality, we may assume that these colors are $2i+2,2i+3$ as desired. Observe also that $\tilde {g}$ is computable from $f\oplus c$ .

Since we are defining $\tilde {g}$ , in order to make it an $l_{n}$ -approximation, we simply stop changing its value once it has used up all $l_{n}$ many changes. Now suppose for a contradiction that $\lim _{s\to \infty }\tilde {g}(e,s)\in T_{e}^{f}$ . Let $s_{0}=0$ and $s_{i}$ be the least stage such that $\tilde {g}(e,s_{i-1})\neq \tilde {g}(e,s_{i})$ . Since $\tilde {g}$ is an $l_{n}$ -approximation, we obtain that

$$ \begin{align*}\{s\mid\tilde{g}(e,s-1)\neq\tilde{g}(e,s)\}=\{s_{i}\mid 1\leq i\leq l_{n}\}. \end{align*} $$

By choice of $s_{i}$ , we also have that $\tilde {g}(e,s_{i})=\langle i,p_{i}^{*}\rangle $ for each i. This implies that

$$ \begin{align*}T_{e}^{f}=\{\tilde{g}(e,s_{i})\mid 0\leq i\leq n\}, \end{align*} $$

because $\tilde {g}(e,s)\neq \tilde {g}(e,s-1)$ only if $\tilde {g}(e,s-1)\in T_{e}^{f}[s]$ . We also have that $\tilde {g}(e,s_{i})=\langle i,p_{i}^{*}\rangle $ encodes $p_{i}$ many pairs of vertices in the $i^{th}$ layer colored with $2i,2i+1$ . Consider the $l_{n}+1$ -th layer (consisting of a pair of vertices) in the $e^{th}$ component. Each vertex in this layer is connected to a pair of vertices from each previous layer; in order for c to be a valid coloring, the vertices in the $l_{n}+1$ -th layer cannot be colored with colors $0,1,\dots ,2l_{n}+1$ . However, by choice of $l_{n}$ , we obtain that $2l_{n}+1\geq n-2$ . Then c has only one remaining color left to color the two adjacent vertices in the $l_{n}+1$ -th layer. Thus, it must be that $\lim _{s\to \infty }\tilde {g}(e,s)\notin T_{e}^{f}$ . Furthermore, by choice of $k_{n}$ , it is also evident that for each s, $\tilde {g}(e,s)<k_{n}$ . Therefore, $\tilde {g}$ satisfies ${\mathtt {DNR}}(k_{n},l_{n})$ .