An objection one player makes to another

The objection nobody can make louder picks a split by sorting every coalition’s complaint and minimising the list in dictionary order. It works, it always produces exactly one answer — which a split nobody can walk away from does not, since its region can be empty — and the cascade of tie-breaks is bookkeeping rather than argument: nothing about a game suggests that its answer should be found by comparing sorted lists left to right.

There is a rule reaching the same split without any of that.

Narrowing what an objection is #

In the core’s reading, a coalition objects to a split by pointing out that it could earn more alone. The complaint is directed at everybody and its size is the excess $e(S,x) = v(S) - x(S)$.

The pairwise reading keeps the excesses and changes who is complaining to whom. Player $i$ objects to player $j$ by naming the coalition that contains $i$, excludes $j$, and is worst off — the idea being that $i$ can threaten to walk out with allies who would keep $j$ out. The size of that threat is

$$s_{ij}(x) \;=\; \max_{S \ni i,\; S \not\ni j} e(S, x),$$

the maximum surplus of $i$ against $j$. It is a maximum over coalitions rather than a single coalition’s excess, and computing it means scanning every coalition holding one player and not the other.

The condition, and why it needs no tie-breaks #

Player $i$ can out-argue $j$ when $s_{ij} > s_{ji}$: the worst-off group $i$ could leave with is worse off than the worst-off group $j$ could leave with, so $i$ has the louder grievance and $j$ has less to complain about. A split where that happens is one where $i$ has a case for a transfer.

The kernel is the set of splits where nobody can out-argue anybody: for every pair, $s_{ij} = s_{ji}$ — with one exception, which is not a technicality. If $j$ is already receiving exactly what it could earn alone, it cannot be paid less, so $i$’s louder complaint cannot be acted on and the pair is settled. So the condition is: for every ordered pair, either the two surpluses are equal, or the quieter player is at its own floor.

That is a system of equalities, and it produces an answer for the same reason the nucleolus does: solving it pins the split down. What it does not need is any order of precedence among the complaints. Six comparisons, each between two numbers, and a split satisfying all six.

The relationship between the two rules is the theorem worth knowing: the nucleolus is always in the kernel, and for games with few players the kernel is often the single point that the nucleolus is. So the dictionary order was producing a kernel point all along, and the pairwise condition explains why that point rather than another.

The glove game, where the answer looks harsh and the reason is visible #

Two left gloves and one right; a pair is worth one, two lefts nothing. Write $x = (x_1, x_2, x_R)$ and work the surpluses out.

For left glove one against left glove two: the coalitions holding the first and not the second are $\{L_1\}$, worth nothing, and $\{L_1 R\}$, worth one. Their excesses are $-x_1$ and $1 - x_1 - x_R = x_2$. So $s_{12} = \max(-x_1, x_2) = x_2$, since $x_2 \ge 0$. By the same argument $s_{21} = x_1$, so the pair is balanced exactly when $x_1 = x_2$.

For left glove one against the right: the coalitions are $\{L_1\}$ and $\{L_1 L_2\}$, with excesses $-x_1$ and $-x_1-x_2$, so $s_{1R} = -x_1$. Against that, the coalitions holding the right and not the first left are $\{R\}$ and $\{L_2 R\}$, with excesses $-x_R$ and $x_1$, so $s_{R1} = x_1$. Balanced when $-x_1 = x_1$, which is $x_1 = 0$.

So $x_1 = x_2 = 0$ and $x_R = 1$. Three lines of arithmetic, and the answer that needed a lexicographic cascade before. What the pairwise reading adds is the reason: a left glove’s loudest threat is to leave alone, which is worth nothing to it, while the right glove’s loudest threat against a left glove is to leave with the other left glove, which is worth a whole unit. The threats are not symmetric and the split reflects it.

Why a maximum and not a sum #

The surplus of $i$ against $j$ is a maximum over coalitions, and the choice of a maximum rather than an average or a sum is the one modelling decision in the definition. It is worth defending, because the alternatives give different rules.

A maximum says that $i$’s case against $j$ is as strong as its strongest available threat. That matches how a threat works: a player pointing out that some group it belongs to is badly treated does not need every such group to be badly treated, only one. An average would say the case is as strong as its threats are on average, which makes a player with one devastating threat and many weak ones look weak — and a player can only carry out one threat.

The maximum also makes the condition checkable. A maximum over coalitions is computed by scanning them, and it is achieved by a definite coalition that can be named — which the figures do, in the row under each cell. An average over coalitions would depend on which coalitions were counted and would have no witness to point at.

And the maximum is what makes the connection to the excesses work. The nucleolus minimises the sorted vector of excesses, whose first entry is the largest excess over all coalitions; the kernel balances maxima over restricted families of coalitions. Those two are about the same quantity looked at two ways, which is why they agree, and an average would have broken the connection entirely — there would be no reason for the balanced split to be the one minimising the worst complaint.

The one place the choice bites is the floor exception. A maximum over coalitions holding $j$ and avoiding $i$ always includes the singleton $\{j\}$, whose excess is $v(\{j\}) - x_j$ — so a player receiving exactly what it earns alone has that excess at nought and may have nothing louder available. That is the case the exception covers, and it exists because the maximum picks up the singleton whether or not the singleton is a serious threat.

Where the kernel sits when the core is not empty #

When the core is non-empty the kernel is inside it, and the reason is worth stating because it is the same kind of argument as the emptiness certificates.

Suppose a split is balanced pairwise and some coalition $S$ has a positive excess — it could do better alone. Then any player in $S$ has $s_{ij} > 0$ against any player outside $S$, since $S$ itself is a witness. Balance forces $s_{ji} > 0$ too, so some coalition holding $j$ and avoiding $i$ also has a positive excess. Iterating that argument across the players produces a family of coalitions all with positive excess and covering everybody in a balanced way, which sums — exactly as a balanced family does — to the statement that the game’s total is exceeded by what the family can earn. That is the failing certificate, so the core was empty after all.

The partnership game’s answer is $(\tfrac{19}{4}, \tfrac{11}{4}, \tfrac{3}{2})$, which the sweep finds when the lattice is fine enough to carry quarters and misses otherwise. That sensitivity is worth noticing rather than hiding: a kernel point generally has a denominator built out of the game’s numbers, and a search on the wrong lattice reports the nearest point and calls it the answer. The figures compute the imbalance at every split and mark the minimisers, so a lattice too coarse reports a positive smallest imbalance rather than a false answer.

The weaker test the kernel sits inside #

There is a third reading, older than the kernel, and it changes what counts as a successful objection rather than who makes it.

An objection to a split $x$ is a coalition $S$ together with a payment $y$ to its members, worth no more than $S$ can earn alone, giving every member of $S$ strictly more than $x$ did. The core asks that no objection exist. The bargaining set asks something weaker: that every objection can be answered.

A counter-objection to $(S, y)$ is a coalition $T$ holding at least one member of $S$ and at least one player outside it, together with a payment that gives $T$’s outsiders what $x$ gave them and $T$’s insiders what $y$ promised them, worth no more than $T$ can earn. The reading is that $T$ can peel a member off $S$ by matching the offer, while keeping its own outsiders no worse off than before.

The majority game’s even split is the example the whole apparatus was built for. The pair $AB$ objects, offering each of its members a half. The pair $AC$ answers: it needs to pay $A$ the half it was promised and $C$ the third it already had, which is five sixths, and $AC$ can earn a whole pound. So the objection is dismissed, and every other objection is dismissed the same way. The split is outside the core and inside the bargaining set, which is the gap the concept exists to occupy.

The bargaining set contains the kernel, which contains the nucleolus, and all three are non-empty for every game with three players. Where they differ is in what they promise: the bargaining set is usually a region, the kernel usually a small set, the nucleolus always a point. A subject that wants one answer takes the last; a subject that wants to know which answers are defensible takes the first.

Three rules on one game, and which disagree #

Setting the three single-valued rules these essays have produced against each other on one game is the quickest way to see what each is measuring, and the glove game separates all three.

The averaging rule gives the two left gloves a sixth each and the right glove two thirds, because in some orders of arrival a left glove walks in to find a right one waiting and adds a whole unit. The kernel and the nucleolus give the left gloves nothing. Neither answer is wrong; they measure contribution and threat, and the glove game is built so that the two come apart completely.

The three-partner game is the opposite case: its core is a region, its kernel is a single interior point, and the averaging rule lands somewhere else inside the same region. So on that game the two rules disagree about which stable split to choose and agree that a stable split exists — a much milder disagreement, and one no reader would notice without computing both.

The pattern is that the rules agree where the game is unambiguous and diverge where it is not, which is what one would want and is not automatic. A pair of rules could easily have disagreed on easy games and agreed on hard ones, and the fact that they do not is a property of these particular definitions rather than a law.

What separates them in every case is the same question: does a player’s share reflect what it adds or what it can threaten? The shop game is one where the two nearly coincide because every player both adds and threatens the same amount, and the glove game is one where a player adds a great deal in some orders and can threaten nothing at all.

What the pairwise reading gives up #

It is not additive either. The complaint against the nucleolus — that playing two games at once and adding the answers is not the answer to the combined game — applies here too, for the same reason: the maximum surplus of a sum is not the sum of the maximum surpluses, since a maximum does not distribute over addition.

Its condition is a system of equations with cases in it. The floor exception is not cosmetic: for games where some player is worth a great deal alone, the exception is what makes the kernel non-empty, and a statement of the condition without it is simply false. So the rule is slightly awkward to state and much easier to check than the cascade it replaces.

And it needs a maximum over exponentially many coalitions. Each of the $n(n-1)$ surpluses is a maximum over up to $2^{n-2}$ coalitions, so computing the condition at one split is already expensive, and finding the split that satisfies it is a system of equations whose coefficients depend on which coalition achieves each maximum — which changes across the region. Solving it in general is done by a sequence of linear programs, and the practical algorithms exploit structure exactly as the nucleolus’s do.

A rule that always answers, and what that costs #

All three of the concepts here are non-empty for every three-player game, which is the property the core lacks and the reason any of them was invented. It is worth asking what that guarantee costs, because a rule that always answers has to be answering something weaker.

The core asks a hard question — is there a split no group can beat? — and reports failure honestly. The kernel asks a softer one: is there a split at which no player can out-argue another? The softening is precisely that a group’s complaint is replaced by the complaint of each of its members against each non-member, which is less demanding because a group with a case need not contain any individual who can win a two-sided argument.

The majority game shows the softening at work. Every pair has a genuine case at the equal split — any two of them could take the whole pound and are receiving two thirds of it — and no individual can out-argue any other, because the game is symmetric and every player’s best threat is the same size. So the equal split satisfies the weaker test and fails the stronger one, and the weaker test’s answer is not a repair of the stronger test’s failure but a different question with a different answer.

That is the honest way to read every always-answers rule in this subject. The nucleolus always answers because it minimises rather than demands; the kernel because it compares pairs rather than groups; the bargaining set because it lets an objection be dismissed. None of them makes an unstable game stable, and a reader who takes the answer as a guarantee of stability has read a weaker claim as a stronger one — which is the standing hazard with any solution concept whose definition ends in a minimisation.

What the pictures cannot show #

Every game here has three players and six ordered pairs, so the whole condition fits in one row of a table. Four players give twelve pairs and a tetrahedron, and the visible part of the argument ends there.

The surpluses are maxima over coalitions and the figures name the coalition achieving each one. Which coalition that is changes as the split moves — that is what makes the kernel’s defining equations piecewise rather than linear — and a static table at three splits shows three of the pieces without showing where the boundaries between them lie. The boundaries are where the argument is hardest and the pictures are least informative.

And the containment claims are checked at the drawn games rather than proved. That the kernel lies inside a non-empty core, and that the nucleolus lies inside the kernel, are theorems about every game; what the figures check is that nothing on their lattices contradicts them, which is the strongest thing a sweep can do.

Still open: what a defensible objection is #

The three concepts on this page — core, kernel, bargaining set — differ in what an objection has to survive, and the list does not stop at three. Variants differ in whether a counter-objection must keep its own outsiders at $x$ or merely no worse than some alternative, in whether an objection may be made by a coalition that is itself unstable, and in whether the objecting coalition must be able to enforce its proposal. Each variant is non-empty for some class of games and empty for another, and there is no agreed answer as to which captures what a reader would call a defensible objection.

The question is not mathematical and that is the honest thing to say about it. Each definition is precise, each has theorems, and choosing among them is a judgement about what a threat means — the same kind of judgement four conditions and no rule shows cannot be settled by listing desirable properties. What the mathematics supplies is the consequences of each choice, and for this family the consequences form a nested chain, which is at least a tidy state of affairs.

What the narrowing bought #

The core asks a question with a yes-or-no answer and frequently answers no. The nucleolus always answers, and its mechanism — sort, minimise, break ties — is a procedure rather than a reason. The kernel answers with the same split and its mechanism is a balance condition between pairs.

Narrowing who may object made the condition symmetric, and symmetry is what removed the need for an order. A coalition’s complaint against everybody has no natural partner to be compared with; one player’s complaint against another has exactly one, and asking the two to match is a condition rather than a ranking.

The lesson is about how a definition earns its answer. A rule that produces one output by breaking ties has to justify the tie-breaking, and usually cannot. A rule that produces one output because a system of equations has one solution needs no such justification — and the two rules agreeing, here, is the strongest available evidence that the cascade was not arbitrary after all.