We prove that deciding if a diagram of the unknot can be untangled using at most Riedemeister moves (where is part of the input) is NP-hard. We also prove that several natural questions regarding links in the -sphere are NP-hard, including detecting whether a link contains a trivial sublink with components, computing the unlinking number of a link, and computing a variety of link invariants related to four-dimensional topology (such as the -ball Euler characteristic, the slicing number, and the -dimensional clasp number).


The Unbearable Hardness of Unknotting\thanksAdM is partially supported by the French ANR project ANR-16-CE40-0009-01 (GATO) and the CNRS PEPS project COMP3D. AdM and MT are partially supported by the Czech-French collaboration project EMBEDS II (CZ: 7AMB17FR029, FR: 38087RM). This work was partially supported by a grant from the Simons Foundation (grant number 283495 to Yo’av Rieck). MT is partially supported by the project CE-ITI (GAČR P202/12/G061) and by the Charles University projects PRIMUS/17/SCI/3 and UNCE/SCI/004.

1 Introduction

Unknot recognition via Reidemeister moves.

The unknot recognition problem asks whether a given knot is the unknot. Decidability of the unknot recognition problem was established by Haken [Hak61], and since then several other algorithms were constructed (see for example the survey of Lackenby [Lac17a]).

One can ask, naively, if one can decide if a given knot diagram represents the unknot simply by untangling the diagram: trying various Reidemeister moves until there are no more crossings. A first issue is that one might need to increase the number of crossings at some point in this untangling: examples of “hard unknots” witnessing this necessity can be found in Kaufman and Lambropoulou [KL14] . The problem then, obviously, is knowing when to stop: if we have not been able to untangle the diagram using so many moves, is the knot in question necessarily knotted or should we keep on trying?

In [HL01], Hass and Lagarias gave an explicit (albeit rather large) bound on the number of Reidemeister moves needed to untangle a diagram of the unknot. Lackenby [Lac15] improved the bound to polynomial thus showing that the unknot recognition problem is in NP (this was previously proved in [HLP99]). The unknot recognition problem is also in co-NP [Lac16] (assuming the Generalized Riemann Hypothesis, this was previously shown by Kuperberg [Kup14]). Thus if the unknot recognition problem were NP-complete (or co-NP-complete) we would have that NP and co-NP coincide which is commonly believed not to be the case. This suggests that the unknot recognition problem is not NP-hard.

It is therefore natural to ask if there is a way to use Reidemeister moves leading to a better solution than a generic brute-force search. Our main result suggests that there may be serious difficulties in such an approach: given a 3-SAT instance we construct an unknot diagram and a number , so that the diagram can be untangled using at most Reidemeister moves if and only if is satisfiable. Hence any algorithm that can calculate the minimal number of Reidemeister moves needed to untangle unknot diagrams will be robust enough to tackle any problem in NP.

The main result of this paper is:

Theorem 1.

Given an unknot diagram and an integer , deciding if can be untangled using at most Reidemeister moves is NP-complete.

Lackenby [Lac15] proved that the problem above is in NP, and therefore we only need to show NP-hardness.

For the reduction in the proof of Theorem 1 we have to construct arbitrarily large diagrams of the unknot. The difficulty in the proof is to establish tools powerful enough to provide useful lower bounds on the minimal number of Reidemeister moves needed to untangle these diagrams. For instance, the algebraic methods of Hass and Nowik [HN10] are not strong enough for our reduction. It is also quite easy to modify the construction and give more easily lower bounds on the number of Reidemeister moves needed to untangle unlinks if one allows the use of arbitrarily many components of diagrams with constant size, but those techniques too cannot be used for Theorem 1. We develop the necessary tools in Section 4.

Computational problems for links.

Our approach for proving Theorem 1 partially builds on techniques to encode satisfiability instances using Hopf links and Borromean rings, that we previously used in [dMRST18] (though the technical details are very different). With these techniques, we also show that a variety of link invariants are NP-hard to compute.

Precisely, we prove:

Theorem 2.

Given a link diagram and an integer , the following problems are NP-hard:

  1. deciding whether admits a trivial unlink with components as a sublink.

  2. deciding whether an intermediate invariant has value on ,

  3. deciding whether ,

  4. deciding whether admits a smoothly slice sublink with components.

We refer to Definition 12 for the definitions of , the -ball Euler characteristic, and of intermediate invariants. These are broadly related to the topology of the -ball, and include the unlinking number, the ribbon number, the slicing number, the concordance unlinking number, the concordance ribbon number, the concordance slicing number, and the -dimensional clasp number. See, for example, [Shi74] for a discussion of many intermediate invariants.

Related complexity results.

The complexity of computational problems pertaining to knots and links is quite poorly understood. In particular, only very few computational lower bounds are known, and as far as we know, almost none concern classical knots (i.e., knots embedded in ): apart from our Theorem 1, the only other such hardness proof we know of [KS, Sam18] concerns counting coloring invariants (i.e., representations of the fundamental group) of knots. More lower bounds are known for classical links. Lackenby [Lac17b] showed that determining if a link is a sublink of another one is NP-hard. Our results strengthen this by showing that even finding an -component unlink as a sublink is already NP-hard. Agol, Hass and Thurston [AHT06] showed that computing the genus of a knot in a -manifold is NP-hard, and Lackenby [Lac17b] showed that computing the Thurston complexity of a link in is also NP-hard. Our results complement this by showing that the -dimensional version of this problem is also NP-hard.

Regarding upper bounds, the current state of knowledge is only slightly better. While, as we mentioned before, it is now known that the unknot recognition problem is in co-NP, many natural link invariants are not even known to be decidable. In particular, this is the case for all the invariants for which we prove NP-hardness, except for the problem of finding the maximal number of components of a link that form an unlink, which is in NP (see Theorem 4).

Shortly before we finished our manuscript, Koenig and Tsvietkova posted a preprint [KT18] that also shows that certain computational problems on links are NP-hard, with some overlap with the results obtained in this paper (the trivial sublink problem and the unlinking number). They also show NP-hardness of computing the number of Reidemeister moves for getting between two diagrams of the unlink, but their construction does not untangle the diagram and requires arbitrarily many components. Theorem 1 of the current paper is stronger and answers Question 17 of [KT18].


This paper is organized as follows. After some preliminaries in Section 2, we start by proving the hardness of the trivial sublink problem in Part I because it is very simple and provides a good introduction for our other reductions. We then proceed to prove Theorem 1 in Part II and the hardness of the unlinking number and the other invariants in Part III. The three parts are independent and the reader can read any one part alone.

2 Preliminaries


Most of the notation we use is standard. By knot we mean a tame piecewise linear embedding of the circle into the -sphere . By link we mean a tame, piecewise linear embedding of the disjoint union of any finite number of copies of . We use interval notation for natural numbers, for example, means ; we use to indicate . We assume basic familiarity with computational complexity and knot theory, and refer to basic textbooks such as Arora and Barak [AB09] for the former and Rolfsen [Rol90] for the latter.

Diagram of a knot or a link.

All the computational problems that we study in this paper take as input the diagram of a knot or a link, which we define here.

A diagram of a knot is a piecewise linear map in general position; for such a map, every point in has at most two preimages, and there are finitely many points in with exactly two preimages (called crossing). Locally at crossing two arcs cross each other transversely, and the diagram contains the information of which arc passes ‘over’ and which ‘under’. This we usually depict by interrupting the arc that passes under. (A diagram usually arises as a composition of a (piecewise linear) knot and a generic projection which also induces ‘over’ and ‘under’ information.) We usually identify a diagram with its image in together with the information about underpasses/overpasses at crossings; see, for example, Figure 3, ignoring the notation on the picture. Diagrams are considered up-to isotopy.

Similarly, a diagram of a link is a piecewise linear map in general position, where denotes a disjoint union of a finite number of circles , and with the same additional information at the crossings.

By an arc in the diagram we mean a set where is an arc in (i.e., a subset of homeomorphic to the closed interval).

The size of a knot or a link diagram is its number of crossings plus number of components of the link. Up to a constant factor, this complexity exactly describes the complexity of encoding the combinatorial information contained in a knot or link diagram.


A formula in conjunctive normal form in variables is a Boolean formula of the form where each is a clause, that is, a formula of the form where each is a literal, that is, a variable or its negation . A formula is satisfiable if there is an assignment to the variables (each variable is assigned or ) such that evaluates to in the given assignment.

A 3-SAT problem is the well-known NP-hard problem. On input there is a formula in conjunctive normal form such that every clause contains exactly222Here we adopt a convention from [Pap94]. Some other authors define require only ‘at most three’ literals. three variables; see, e.g., [Pap94, Proposition 9.2].

Part I Trivial sublink

Informally, the trivial sublink problem asks, given a link and a positive integer , whether admits the -component unlink as a sublink. We define:

Definition 3 (The Trivial Sublink Problem).

An unlink, or a trivial link, is a link in whose components bound disjointly embedded disks. A trivial sublink of a link is an unlink formed by a subset of the components of . The trivial sublink problem asks, given a link and a positive integer , whether admits an component trivial sublink.

Theorem 4.

The trivial sublink problem is NP-complete.

Note that Theorem 4 is just a slight extension of Theorem 2(a), claiming also NP-membership. The essential part is NP-hardness.


It follows from Hass, Lagarias, and Pippenger [HLP99] that deciding if a link is trivial is in NP. By adding to their certificate a collection of components of we obtain a certificate for the trivial sublink problem, showing that it is in NP (NP-membership of the trivial sublink problem was also established, using completely different techniques, by Lachenby [Lac15]). Thus all we need to show is that the problem is NP-hard. We will show this by reducing 3-SAT to the trivial sublink problem.

Given a 3-SAT instance , with variables (say ) and clauses, we construct a diagram as follows (see Figure 1):

Figure 1: for )

we first mark disjoint disks in the plane. In each of the first disks we draw a diagram of the Hopf link, marking the components in the th disk as and . In the remaining disks we draw diagrams of the Borromean rings and label them according to the clauses of . We now band each component of the Borromean rings to the Hopf link component with the same label. Whenever two bands cross we have one move over the other (with no “weaving”); we assume, as we may, that no two bands cross twice. It is easy to see that this can be done in polynomial time. The diagram we obtain is and the link it represents is denoted (note that has exactly components). We complete the proof by showing that admits an -component trivial sublink exactly when is satisfiable.

Claim 4.1.

If is satisfiable then admits an -component trivial sublink.

Given a satisfying assignment we remove from the components that correspond to satisfied literals, that is, if we remove from and if we remove from . We claim that the remaining components form an unlink. To see this, first note that since the assignment is satisfying, from each copy of the Borromean rings at least one component was removed. Therefore the rings fall apart and (since we did not allow “weaving”) the diagram obtained retracts into the first disks. In each of these disks we had, originally, a copy of the Hopf link; by construction exactly one component was removed. This shows that the link obtained is indeed the -component unlink; Claim 4.1 follows.

Claim 4.2.

If admits an -component trivial sublink, then is satisfiable.

Suppose that admits an -component trivial sublink . Since itself does not admit the Hopf link as a sublink, for each , at most one of and is in . Since has components we see that exactly one of and is in . If is in we set and if is in we set . Now since does not admit the Borromean rings as a sublink, from each copy of the Borromean rings at least one component is not in . It follows that in each clause of at least one literal is satisfied, that is, the assignment satisfies ; Claim 4.2 follows.

This completes the proof of Theorem 4. ∎

Part II The number of Reidemeister moves for untangling

3 A restricted form of the satisfiability problem

For the proof of Theorem 1, we will need a slightly restricted form of the 3-SAT problem given by lemma below.

Lemma 5.

Deciding whether a formula in conjunctive normal form is satisfiable is NP-hard even if we assume the following conditions on .

  • Each clause contains exactly three literals.

  • No clause contains both and for some variable .

  • Each pair of literals occurs in at most one clause.


The first condition says that we consider the 3-SAT problem. Any clause violating the second condition can be removed from the formula without affecting satisfiability of the formula as such clause is always satisfied. Therefore, it is sufficient to provide a recipe to build in polynomial time a formula satisfying the three conditions above out of a formula satisfying only the first two conditions.

First, we consider an auxiliary formula

We observe that for any satisfying assignment of we get that is assigned . Indeed, if were assigned then translates as , that is all , , and are equivalent. However, then cannot be satisfied.

On the other hand, we also observe that there is a satisfying assignment for where is assigned and, for example, it is sufficient to assign with , with and arbitrarily.

Now we return to the formula discussed above. Suppose there exists a pair literals and contained in two clauses of , say and . We replace them with

where and are newly added variables, obtaining a new formula . We aim to show that is satisfiable if and only if is satisfiable.

Let us first assume that is satisfiable and fix a satisfying assignment. If is in this assignment, then we may extend it to variables of by setting to , and to and so that is satisfied. If is in the assignment for , then and must be assigned . Then we may extend by setting to and as before.

Now let us assume that is satisfiable. Then and are set to due to the properties of . If is , then both and must be true and the restriction of the assignment on to the variables of is therefore a satisfying assignment for . Similarly, if is , then must be true and the restriction is again a satisfying assignment for .

We also observe that in we have reduced the number of clauses containing the literals and simultaneously, and for any other pair of literals we do not increase the number of clauses containing that pair. Therefore, after a polynomial number of steps, when always adding new variables, we arrive at a desired formula satisfying all three conditions. ∎

4 The defect

Reidemeister moves.

Reidemeister moves are local modifications of a diagram depicted in Figure 2 (the labels at the crossings in a move will be used only later on). We distinguish the move (left), the move (middle) and the move (right). The first two moves affect the number of crossings, thus we further distinguish the and the moves which reduce the number of crossings from the and the moves which increase the number of crossings.

Figure 2: Reidemeister moves
The number of Reidemeister moves for untangling a knot.

A diagram of an unknot is untangled if it does not contain any crossings. The untangled diagram is denoted by . Given a diagram of an unknot, an untangling of is a sequence where , (recall that diagrams are only considered up to isotopy) and is obtained from by a single Reidemeister move. The number of Reidemeister moves in is denoted by , that is, . We also define where the minimum is taken over all untanglings of .

The defect.

Let us denote by the number of crossings in . Then the defect of an untangling is defined by the formula

The defect of a diagram is defined as . Equivalently, where the minimum is taken over all untanglings of .

The defect is a convenient way to reparametrize the number of Reidemeister moves due to the following observation.

Observation 6.

For any diagram of the unknot and any untangling of we have . Equality holds if and only if uses only moves.


Every Reidemeister move in removes at most two crossings and the move is the only move that removes exactly two crossings. Therefore, the number of crossings in is at most and equality holds if and only if every move is a move. ∎

Crossings contributing to the defect.

Let be an untangling of a diagram of an unknot.

Given a crossing in , for , it may vanish by the move transforming into if this is a or a move affecting the crossing. In all other cases it survives and we denote by the corresponding crossing in . Note that in the case of a move there are three crossings affected by the move and three crossings afterwards. Both before and after, each crossing is the unique intersection between a pair of the three arcs of the knot that appear in this portion of the diagram. So we may say that these three crossings survive the move though they change their actual geometric positions (they swap the order in which they occur along each of the three arcs); see Figure 2.

With a slight abuse of terminology, by a crossing in we mean a maximal sequence such that is the crossing in corresponding to in for any . By maximality we mean, that vanishes after the st move and either or is introduced by the th Reidemeister move (which must be a or move).

An initial crossing is a crossing in . Initial crossings in are in one-to-one correspondence with crossings in . For simplicity of notation, is also denoted (as a crossing in ).

A Reidemeister move in is economical, if both crossings removed by this move are initial crossings; otherwise, it is wasteful.

Let be the number of moves affecting a crossing . The weight of an initial crossing is defined in the following way.

For later purposes, we also define and for a subset of the set of all crossings in .

Lemma 7.

Let be an untangling of a diagram . Then

where the sum is over all initial crossings of .


In the proof we use the discharging technique, common in graph theory.

Let us put charges on crossings in and on Reidemeister moves used in . The initial charge will be

  • on each Reidemeister move;

  • on each initial crossing; and

  • on each non-initial crossing.

We remark that the sum of the initial charges equals to by the definition of the defect.

Now we start redistributing the charge according to the rules described below. The aim is that, after the redistribution, the charge on each initial crossing will be at least and it will be at least on non-initial crossings and on Reidemeister moves. This will prove the lemma, as the sum of the charges after the redistribution is still equal to the defect, whereas it will be at least the sum of the weights of initial crossings.

We apply the following rules for the redistribution of the charge.

  • Every move sends charge to the (non-initial) crossing it creates.

  • Every move sends charge to the crossing it removes.

  • Every move sends charge to each of the two (non-initial) crossings it creates.

  • Every economical move sends charge to each of the two (initial) crossings it removes.

  • Every wasteful move that removes exactly one initial crossing sends charge to this initial crossing.

  • Every move sends to every crossing it affects.

  • Every non-initial crossing which is removed by a wasteful move sends charge to this move.

Now it is routine to check that the desired conditions are satisfied, which we now explain.

Every move has charge at least : The initial charge on moves is . The rules are set up so that every move distributes charge at most with exception of the rule (R5). However, in this case, the wasteful move that removes exactly one initial crossing from (R5) gets charge from rule (R7).

Every non-initial crossing has charge at least : The initial charge is . The only rule that depletes the charge is (R7); however in such case, the charge is replenished by (R1) or (R3).

Every initial crossing has charge equal at least : The initial charge is . First we observe that (R6) sends the charge to . If vanishes by an economical move, it gets additional charge by (R4). If vanishes by a move, it gets additional charge by (R2). Finally, if vanishes by a wasteful move then this move removes exactly one initial crossing, namely . Therefore gets an additional charge by (R5). ∎

We will also need a variant for a previous lemma where we get equality, if we use the and moves only.

Lemma 8.

Let be an untangling of a digram which uses the and moves only. Then

where the sum is over all initial crossings of .


Let be the number of initial crossings removed by a move and the number of initial crossings removed by a move. Then and from the definition of the defect we get . It follows directly the definition of the weight that . ∎

Twins and the preimage of a bigon.

Let be an initial crossing in an untangling removed by an economical move. The twin of , denoted by is the other crossing in removed by the same move. Note that is also an initial crossing (because of the economical move). We also get . If , then we also extend the definition of a twin to in such a way that is uniquely defined by . In particular, we will often use a twin of a crossing in (if it exists).

Furthermore, the crossings and in form a bigon that is removed by the forthcoming move. Let and be the two arcs of the bigon (with endpoints and ) so that is the arc that, after extending slightly, overpasses the crossings and whereas a slight extension of underpasses these crossings. (The reader may remember this as is ‘above’ and is ‘below’.)333Note that and are uniquely defined by as well as by . The choice of in the notation will be more useful later on. Now we can inductively define arcs and for so that and are the unique arcs between and which are transformed to (already defined) and by the th Reidemeister move. We also set and . Intuitively, and form a preimage of the bigon removed by the st move and they are called the preimage arcs between and .

Close neighbors.

Let be a subset of the set of crossings in . Let and be any two crossings in (not necessarily in ) and let be a non-negative integer. We say that and are -close neighbors with respect to if and can be connected by two arcs and such that

  • enters and as an overpass;

  • enters and as an underpass;

  • and may have self-crossings; however, neither nor is in the interior of or ; and

  • and together contain at most crossings from in their interiors. (If there is a crossing in the interior of both and , this crossing is counted only once.)

Lemma 9.

Let be a subset of the set of crossings in , let . Let be the crossing in which is the first of the crossings in removed by an economical move (we allow a draw). If , then and its twin are -close neighbors with respect to .


Let and be the preimage arcs between and . We want to verify that they satisfy the properties of the arcs from the definition of the close neighbors. The first two properties follow immediately from the definition of preimage arcs.

Next, we want to check that neither nor is the interior of or . For contradiction, let us assume that this not the case. For example, suppose that also lies in the interior of . Let be the initial crossing corresponding to . The st Reidemeister move in removes . Any preceding Reidemeister move either does not affect at all, or it is a move swapping the crossing with other crossings. In any case, it preserves the self-crossing of at . However, this contradicts the fact that is an arc of a bigon removed by the st move.

In order to check the last property, let us assume that, for contradiction, and together contain at least crossings from in their interiors. These crossings have to be removed from the arcs and until we reach and . They cannot be removed by an economical move as is the first crossing from removed by such a move. Thus they have to be removed from the arcs either by a move, a wasteful move or a move (by swapping with or ). This contradicts . Indeed, if only and moves are used, we get a total weight at least on the crossings; if at least one move is used, we get a weight at least on the crossings and an additional on or . This is in total more than as . ∎

5 The reduction

Let be a formula in conjunctive normal form satisfying the conditions stated in Lemma 5 and let be the number of variables. Our aim is to build a diagram by a polynomial-time algorithm such that if and only if is satisfiable.

The variable gadget.

First we describe the variable gadget. For every variable we consider the diagram depicted at Figure 3 and we denote it .

Figure 3: The variable gadget .

The gadget contains crossings , and for . The variable gadget also contains six distinguished arcs and for , and and six distinguished auxiliary points which will be useful later on in order to describe how the variable gadget is used in the diagram .

We also call the arc between and which contains and the tentacle, and similarly, the arc between and which contains and is the tentacle. Informally, a satisfying assignment to will correspond to the choice whether we will decide to remove first the loop at by a move and simplify the tentacle or whether we remove first the loop at and remove the tentacle in the final construction of .

We also remark that in the notation, we use square brackets for objects that come in pairs and will correspond to a choice of literal . This regards , , and whereas we use parentheses for the remaining objects.

The clause gadget.

Given a clause in , the clause gadget is depicted at Figure 4. The construction is based on the Borromean rings. It contains three pairs of arcs (distinguished by color) and with a slight abuse of notation, we refer to each of the three pairs of arcs as a “ring”. Note that each ring has four pendent endpoints (or leaves) as in the picture. Each ring corresponds to one of the literals , , and .

Figure 4: The clause gadget for clause .
A blueprint for the construction.

Now we build a blueprint for the construction of . Let be the variables of and let be the clauses of .

For each clause we take a copy of the graph (also known as the star with three leaves). We label the vertices of degree of such a by the literals , , and . Now we draw these stars into the plane sorted along a horizontal line; see Figure 5.

Figure 5: A blueprint for the construction of .

Next for each literal we draw a piecewise linear segment containing all vertices labelled with that literal according to the following rules (follow Figure 5).

  • The segments start on the right of the graphs in the top down order , , .

  • They continue to the left while we permute them to the order , . We also require that occur above the graphs and occur below these graphs (everything is still on the right of the graphs).

  • Next, for each literal the segment for continues to the left while it makes a ‘detour’ to each vertex labelled . If is not the leftmost vertex labelled , then the detour is performed by a ‘finger’ of two parallel lines. We require that the finger avoids the graphs except of the vertex . If is the leftmost vertex labelled , then we perform only a half of the finger so that becomes the endpoint of the segment.

Note that the segments often intersect each other; however, for any the segments for and do not intersect (using the assumption that no clause contains both and ).

The final diagram.

Finally, we explain how to build the diagram from the blueprint above.

Step I (four parallel segments): We replace each segment for a literal with four parallel segments; see Figure 6. The outer two will correspond to the arc from the variable gadget and the inner two will correspond to ; compare with Figure 3.

Figure 6: Step I: Replacing segments.

Step II (clause gadgets): We replace each copy of by a clause gadget for the corresponding clause ; see Figure 7. Now we aim to describe how is the clause gadget connected to the quadruples of parallel segments obtained in Step I. Let be a degree vertex of the we are just replacing. Let be the literal which is the label of this vertex. Then may or may not be the leftmost clause containing a vertex labelled .

Figure 7: Step II: Replacing the .

If is the leftmost clause containing a vertex labelled , then there are four parallel segments for with pendent endpoints (close to the original position of ) obtained in Step I. We connect them to the pendent endpoints of the clause gadget (on the ring for ); see and at Figure 7. Note also that at this moment the two arcs introduced in Step I merge as well as the two arcs merge.

If is not the leftmost clause labelled then there are four parallel segments passing close to (forming a tip of a finger from the blueprint). We disconnect the two segments closest to the tip of the finger and connect them to the pendent endpoints of the clause gadget (on the ring for ); see at Figure 7.

Step III (resolving crossings): If two segments in the blueprint, corresponding to literals and have a crossing, Step I blows up such a crossing into crossing of corresponding quadruples. We resolve overpasses/underpasses at all these crossings in the same way. That is, one quadruple overpasses the second quadruple at all crossings; see Figure 8.

Figure 8: Step III: Resolving crossings.

However, we require one additional condition on the choice of overpasses/underpasses. If and appear simultaneously in some clause we have crossings on the rings for and in the clause gadget for . We can assume that the ring of passes over the ring of at all these crossings (otherwise we swap and ). Then for the crossings on segments for and we pick the other option, that is we want that the and arcs underpass the and arcs at these crossings. This is a globally consistent choice because we assume that there is at most one clause containing both and , this is the third condition in the statement of Lemma 5.

STEP IV (the variable gadgets): Now, for every variable , the segments and do not intersect each other. We extend them to a variable gadget as on Figure 9. Namely, to the bottom left endpoints of and we glue the parts of the variable gadget containing the crossings and and to the top left endpoints of and we glue the remainder of the variable gadget. At this moment, we obtain a diagram of a link, where each link component has a diagram isotopic to the diagram on Figure 3.

Figure 9: Step IV: Adding the variable gadgets.

STEP V (interconnecting the variable gadgets): Finally, we form a connected sum of individual components. Namely, for every we perform the knot sum along the arcs and by removing them and identifying with and with as on Figure 10. The arcs and remain untouched. This way we obtain the desired diagram ; see Figure 11.

Figure 10: Step V: Interconnecting the variable gadgets.
Figure 11: The final construction for the formula . For simplicity of the picture, we do not visualize how the crossings are resolved in Step III. (Unfortunately, we cannot avoid tiny pictures of gadgets.)

The core of the NP-hardness reduction is the following theorem.

Theorem 10.

Let be a formula in conjunctive normal form satisfying the conditions in the statement of Lemma 5. Then if and only if is satisfiable.

Theorem 1 immediately follows from Theorem 10 and Lemma 5:

Proof of Theorem 1 modulo Theorem 10.

Due to the definition of the defect, the minimum number of Reidemeister moves required to untangle equals . Therefore, setting , Theorem 10 gives that can be untangled with at most moves if and only if is satisfiable. This gives the required NP-hardness via Lemma 5. (Note also, that and can be constructed in polynomial time in the size of .) ∎

The remainder of this section is devoted to the proof of Theorem 10.

5.1 Satisfiable implies small defect

The purpose of this subsection is to prove the ‘if’ implication of Theorem 10. That is we are given a satisfiable and we aim to show that .

Let us consider a satisfying assignment. For any literal assigned we first remove the loop at the vertex in the variable gadget (see Figure 3) by a move. This way, we use one move on each variable gadget, that is such moves. Next we aim to show that it is possible to finish the untangling of the diagram by moves only. As soon as we do this, we get an untangling with defect by Lemma 8 which will finish the proof.

Thus it remains to finish the untangling with moves only. We again pick assigned and we start shrinking the tentacle by moves. This way we completely shrink and as due to the construction as all arcs that meet simultaneously meet and vice versa. See Figure 12 for the initial move and a few initial moves. Furthermore we can continue shrinking the tentacle until we get a loop next to the vertex; see Figure 13.

Figure 12: Initial simplifications.
Figure 13: The tentacle was shrunk to a loop next to . In this example we have .

We continue the same process for every literal assigned . In the intermediate steps, some of the other arcs meeting and might have already been removed. However, it is still possible to simplify the tentacle as before. See Figure 14 for the result after shrinking all tentacles assigned .

Figure 14: Simplifying the tentacles according to a satisfying assignment , .

Because we assume that we started with a satisfying assignment, in each clause gadget at least one ring among the three Borromean rings disappears. Consequently, if there are two remaining rings in some clause gadget, then they can be pulled apart from each other by moves as on Figure 15.

Figure 15: Untangling two rings in the clause gadget via moves.

After this step, for each assigned , the and form ‘fingers’ of four parallel curves. These fingers can be further simplified by moves so that any crossings among different fingers are removed; see Figure 16. For each variable gadget we get one of the two possible pictures at Figure 17 left. Both of them simplify to the picture on the right by three further moves.

Figure 16: Simplifying and via moves. First, we untangle the inner (horizontal) ‘finger’ and then we untangle the outer (horizontal) ‘finger’.
Figure 17: Results of the simplifications on the previous picture on the level of variable gadgets.

Finally, we recall how are the variable gadgets interconnected (compare the right picture at Figure 17 with Figure 10). Then it is easy to remove all remaining crossings by moves gradually from top to bottom. This finishes the proof of the ‘if’ part of Theorem 10.

5.2 Small defect implies satisfiable

The purpose of this subsection is to prove the ‘only if’ part of the statement of Theorem 10. Recall that this means that we assume and we want to deduce that is satisfiable. (Along the way we will actually also deduce that .) In this subsection, we heavily use the terminology introduced in Section 4.

Let be an untangling of with . For a variable let be the set of out of the self-crossings in the variable gadget (we leave out ) and let the weight of , denoted by , be the sum of weights of the crossings in .

Our first aim is to analyze the first economical move that removes some of the crossings in , using Lemma 9.

Claim 10.1.

Let be a variable with . Let be the crossing in which is the first of the crossings in removed by an economical move (we allow a draw). Then one of the following cases holds

  1. , and is removed by a move prior to removing and .

  2. , and is removed by a move prior to removing and .

Before we start the proof, we remark that the condition implies that there are (at least ) crossings in removed by economical moves. In particular, in the statement always exists.


First, we need to identify the possible pairs where . Such pairs are found by a case analysis, using Lemma 9 with and .

The general strategy is the following. For each element of we consider whether it may be . We analyze possible arcs and from the definition of -close neighbors.

There are two directions in which may emanate from . For each direction we allow up to one internal crossing from on , getting a candidate position for (even if passes through other variable gadgets we count only the crossings from ). We immediately disregard the cases when passes through again (this is not allowed by the third item of the definition of close neighbors). We also emphasize that we are interested only in the cases when enters the candidate as an overpass.

Next, we refocus on ; again there are two possible directions and we again identify possible the possible position of (this time entered as an underpass).

Finally, we compare the lists of candidate positions for obtained for and for ; it must be possible to obtain in both ways.

The candidate positions of as an endpoint of and as an endpoint of are summarized in Table 1; and they can be easily found with the aid of Figure 3. Note that it follows from the construction of that and are (usually) not in . For considerations in the table, we denote by the arc in between the points and which avoids (equivalently, any other crossing in ). Similarly, we let denote the arc in between the points and which avoids . The table also misses values for which requires a separate analysis.

In order to avoid any ambiguity, we explain how the first row of the table is obtained, considering the case (follow Figure 3):

Choice of in in Overlap
separate analysis
Table 1: Possible positions for depending on in Claim 10.1.

We know that may emanate to the left or to the right. Emanating to the left is immediately ruled out as we reach again in the next crossing. Emanating to the right allows to be some crossing on or seemingly it may be or ; however and are entered by as an underpass, so they cannot be . Therefore the only option, from the point of view of , is that belongs to , as marked in Table 1. Similarly when focusing on , emanating to the left is immediately ruled out whereas emanating to the right allows to belong or to be or (as in the table). We conclude that for there is no suitable both for and , using the fact that and do not intersect (in the whole ). (This is marked by the in the overlap column.) We deduce that cannot be .

In general, for identifying the overlaps for other options of , we use that no two of the arcs intersect.

Next, we want to rule out the case because this is not covered by Table 1. Considering the possible arcs , we get the following options for :