Field

Applied

A rule for choosing, stated exactly, and what it forces on whoever adopts it.
the majority tournament100 voters, every pair decidedevery profile of the space216 of them, one cell each+34+36+30ABCno Condorcet winner (12)a winner exists (204)the 3 arcs of the ring are the majority in each pair, and following them returns to A: A → B → C → A12 of the 216 profiles of 3 voters over 3 candidates have no Condorcet winner — 5.6% of the space, every oneof them built and tested

The majority that goes in a circle

Every voter hands in a ranking, and a ranking is transitive by construction. Compare the candidates two at a time and let the majority decide each pair, and the verdicts need not fit together into a ranking at all.

the profileone column per group, size abovethe rulesand what each returns7655221st2nd3rd4th5thDACEACEBDBBEBEBCDACDADCDACEAEBwinnerpluralityBordainstant runoffCondorcetCoombsABCDEthe deciding count8 first places63 points19 of 27 at the end4 of 4 pairs14 of 27 at the endthe candidate that rule returnsthe count it was decided on27 voters in 6 groups over 5 candidates; a majority is more than 13.5the five rules return 5 different winners: plurality A, Borda B, instant runoff C, Condorcet D, Coombs Einstant runoff eliminates B, E, D; Coombs eliminates A, C, D

Five rules and five winners

Twenty-seven ranked ballots, five entirely reasonable ways of counting them, and five different candidates declared the winner. Every count is correct, every rule is defensible, and the answer turns out to be a property of the rule rather than of the ballots.

first profile4 voters, one column eachsecond profilethe same voters on A and Bv1v2v3v41st2nd3rdAABBBCAACBCCv1v2v3v41st2nd3rdAABBBBACCCCAA 6 · B 5 pointsA 5 · B 6 pointsBorda: A ≻ BBorda: B ≻ AflipsABthe candidates that movedevery voter ranks A against B the same way in both profiles; only the third candidate moves — and Bordareverses its verdictno pair of profiles flips Borda at 3 voters; at 4 voters 3456 of the 104976 ordered pairs inside a class do

Four conditions, and no rule that has all of them

The rung below shows five reasonable rules returning five different winners, which invites the obvious question of which one is right. The answer is that the conditions anybody would write down cannot all hold at once — and here each named rule's own violation is found by search rather than quoted.

the electoratethe first column is the manipulating voterevery ballot that voter could submitand the winner it producestrue1121st2nd3rdAABCBBCBCCAAelectsfor the voterA ≻ B ≻ CA ≻ C ≻ BB ≻ A ≻ CB ≻ C ≻ AC ≻ A ≻ BC ≻ B ≻ AChonestCno gainBbetterBbetterCno gainCno gainthe honest ballota misreport that paysthe control: the same voters, A and B only0 of 2 ballots payelectsfor the voterA ≻ BB ≻ ABhonestBno gainthe voter's true ranking is A ≻ B ≻ C; the honest ballot elects C under instant runoff2 of the 6 ballots the voter could submit elect somebody the voter ranks higher: B ≻ A ≻ C; B ≻ C ≻ Athe control runs the identical search with only A and B left: 0 of the 2 ballots pay, which is what astrategy-proof contest looks like

A lie that pays

Three rungs of this ladder have read a ballot as a report of a preference. This one reads it as a move, and walks every move one voter has — all six rankings, the winner each produces, and the ones that beat honesty.

quota = population × 27 ÷ 10000populationexact quotaquotafloorremainderseatsABCDEsum571015417/100015.41715417/10001526707209/10007.2097209/10007650351/2001.7551151/20025301431/10001.4311431/10002440297/2501.188147/2501100002727.00025227rounded up from its floorHamilton: floors, then the largest remainders, over 27 seats and 5 regions of 10000 peoplethe exact quotas sum to 27 and so do the awarded seats; the floors account for 25, leaving 2the 2 spare seats went to the largest remainders: C and D

The seat that vanishes when the house grows

Twenty-seven whole seats have to be divided between five regions whose exact shares are 15.417, 7.209, 1.755, 1.431 and 1.188. Every rule for rounding those five numbers breaks something, and the instance drawn here breaks all three of the classical ways at once.

the cutter's measure, segment by segmentthe chooser's measure, segment by segment52530201553010552030cut at 4/9the left piecethe right piecethe cutterthe chooser5050130/3≈ 43.33170/3≈ 56.67the cutter's piecethe chooser's piecethe cutter's running total crosses 50 inside segment 3, 2/3 of the way through it, so the cut is at 4/9both pieces are worth exactly 50 to the cutter; the chooser takes the right one at 170/3 ≈ 56.67 andgains 20/3 ≈ 6.67 over half

One cuts and the other chooses

The oldest rule in fair division promises each of two people at least half the cake by their own measure, and it keeps that promise exactly. It does not promise what the word "fair" is usually asked to carry, and the gap opens the moment the two measures disagree across the cut.

person 1 cuts three pieces of equal value to person 1piece 1piece 2piece 3person 2 trims the largest of them down to a tie with the secondthe trimming: 140/3 ≈ 46.67 to person 2person 3 chooses, then person 2, then person 1person 1person 3person 2person 3 cuts the trimming in three; person 2 chooses first, then person 1person 1person 2person 3the trimming, drawn at full widtheach person's value of each share — the diagonal is their ownperson 1's shareperson 2's shareperson 3's shareperson 1 valuesperson 2 valuesperson 3 values140/3≈ 46.6740/3≈ 13.33402040403540/3≈ 13.33155/3≈ 51.67own sharethe trimmingperson 1 cuts three pieces worth 100/3 each; person 2 trims 140/3 ≈ 46.67 off the largest, andperson 3 chooses firstthe verdict is the whole 3×3 matrix, not its diagonal: all nine comparisons hold, 1 of them as anexact tie

Three people and a trimmed piece

For two people, one cut and one choice deliver a division nobody would swap out of. For three, the same promise costs a trimming, a residue and a choosing order contrived so that an advantage once given cannot be taken back — and the verdict is not three numbers but a whole three-by-three matrix.

what each person would pay for each item, out of 100 for the setitem aitem bitem cperson 1person 2403525304525allocations: 8envy-free: 0up to one item: 4the control — the same search, on a matrix that has an answeritem aitem bitem cperson 1person 2503020203050allocations: 8envy-free: 2up to one item: 4round-robin pickingan envy-free allocationall 8 allocations of 3 items to 2 people were formed: 0 are envy-free, 4 are envy-free up to one itemround-robin picking, shaded, gives person 1 items a and c; person 2 item b — person 2 stopsenvying person 1 once item a is set asidethe control matrix underneath is searched by the same code and has 2 envy-free allocations, so anempty answer above is a finding rather than a broken search

Envy-free, up to one item

A cake can be cut anywhere, and every guarantee in this anchor was bought with that freedom. Take the knife away and the exhaustive search over every allocation of three objects returns nothing envy-free at all — so the subject weakened the word until taking turns was enough to reach it.

a matching that is not stableA1B2C3D4why that pair blocksD13would rather4has now23ADwould ratherChas nowBhow many of the 24 matchings have each number of blocking pairs40315283241516blocking pairsevery matching, sorted — the 4 stable ones first000011122222333333334456A1 B2 C3 D4 has 1 blocking pair: D and 3 would each rather have the otherall 24 matchings were formed and each put all 16 blocking questions; 4 of them came back with nonethe strip holds one cell per matching, sorted by that count with the 4 zeroes at the left

Nobody has a reason to run away

A matching is stable when no two people on opposite sides would both rather have each other than what they have — a condition that names nothing to build and everything to rule out. The surprise is that something always satisfies it, however perverse the rankings are made.

better for side one, worse for side twoM1 A4 B2 C1 D3side one's ranks 2, 2, 2, 2M2 A3 B2 C1 D4side one's ranks 3, 2, 2, 3M3 A4 B1 C2 D3side one's ranks 2, 4, 3, 2M4 A3 B1 C2 D4side one's ranks 3, 4, 3, 3join and meet, on every pairpairjoinmeetM1, M2M1M2M1, M3M1M3M1, M4M1M4M2, M3M1M4M2, M4M2M4M3, M4M3M44 of the 24 matchings are stable, drawn with the best for side one at the topevery one of the 16 ordered pairs was joined and met, and all 32 results were themselves stablethe top A4 B2 C1 D3 is what side one proposing returns; the bottom A3 B1 C2 D4 is what side two proposing returns

The side that proposes wins

An instance usually has several stable matchings, and the set of them is not a heap — it is a lattice, closed under taking the better partner and under taking the worse. The two ends of that lattice are exactly what deferred acceptance returns from the two sides, so whoever proposes decides which end the instance lands on.

4's true ranking of side one, best firstCDAtruthfulBtruthfully: A4 B2 C1 D3one cell per ranking 4 could submit, labelled with the partner it returnsAAAAAABBBBBBAABBADAADDADthe true rankinga ranking that paysthe 4 profitable misreports, submitted ranking and partner obtainedsubmittedpartnertrue rankCDBAD2 of 4DBACD2 of 4DBCAD2 of 4DCBAD2 of 4the control: the same search, every member of both sides, 24 rankings eachside onepaysside twopaysA010B020C030D0444 truly ranks side one CDAB and gets A, its 3rd choiceof the 24 rankings it could submit instead, 4 return a partner it strictly prefersthe same sweep over every member of the proposing side searched 96 rankings and found none, which is whatmakes the 4 a finding

No stable rule is safe from a lie

A stable matching always exists, and the side that proposes gets the best one it could hope for. This essay closes the ladder with the result that spoils it — one participant's whole strategy space searched, four submissions found that beat the truth, and a theorem saying no rule anywhere escapes.

primal: max 3x₁ + 4x₂dual: min 12y₁ + 10y₂3x₁ + 2x₂ ≤ 12x₁ + 4x₂ ≤ 103y₁ + y₂ ≥ 32y₁ + 4y₂ ≥ 401234501234x₁x₂78/512100(14/5, 9/5)00.511.522.5301234y₁y₂78/53024(4/5, 3/5)05101520253078/51210078/53024they meet at 78/5primal vertex values, from belowdual vertex values, from aboveweak duality checked on all 12 pairs — each of the 4 feasible primal vertices against each of the 3 feasible dual vertices — andcᵀx ≤ bᵀy held every timestrong duality is the equality of the two optima: max 3x₁ + 4x₂ = 78/5 = min 12y₁ + 10y₂, one number reached from below andfrom abovethe dual region is unbounded upward and the drawing cuts it at the box; the enumeration ignores the cut

Two numbers that have to meet

Every linear program has a shadow — a second program built from the same numbers read the other way, whose minimum can never fall below the first's maximum. That much is a one-line calculation; the theorem is that the two numbers are always exactly equal.

51015202530051015202530b₁max 3x₁ + 4x₂b₁ = 5b₁ = 30slope 2 = y₁slope 4/5 = y₁slope 0 = y₁one row per linear piecethe optimum sits onslope of the optimumdual variable y₁b₁ from 2 to 5b₁ from 5 to 30b₁ from 30 to 32row 1 and x₁ = 0(0, 5/4)22row 1 and row 2(1/5, 49/20)4/54/5row 2 and x₂ = 0(10, 0)00the optimum is piecewise linear in b₁ with 3 pieces, breaking at b₁ = 5 and 30on each piece the slope, taken from the optima at the two ends, equals the dual variable y₁ at the middleof that piece — 2, then 4/5, then 0 — and it changes exactly where a different vertex becomes optimal

What a constraint is worth

The rung below settled that a linear program and its dual reach the same number. This one asks what the dual's variables are, and the answer converts a solution into a rate for every constraint — piecewise constant, zero on the constraints that are not doing any work.

the row chooser mixes00.20.40.60.81-4-20246weight on row Apayoff to the row choosercol Xcol Ycol Z8/15 → 19/15the most the row chooser can guarantee: 19/15the column chooser mixes00.20.40.60.81-4-20246weight on col Xpayoff to the row chooserrow Arow B7/15 → 19/15the least the column chooser can concede: 19/15both sides name 19/15 = 1.267the value is 19/15 = 1.267, reached by the row chooser mixing 8/15 on A and by the column chooser mixing 7/15 on Xthe lower envelope of 3 lines peaks where two cross; the upper envelope of 2 bottoms out at that heightchecked against every pure reply on both sides, and over 61 mixtures on one side and 91 on the other

The value from both sides

Two choosers move at the same instant, and each asks the cautious question — how much can be guaranteed, whatever the other does. With pure choices the two answers are usually different numbers; allow a probability and they are forced to be the same one.

before the link A→B existsSABTcost 3flow 3cost 7flow 3cost 7flow 3cost 3flow 33 units each way, every traveller takes 10after the link A→B is addedSABTcost 6flow 6cost 7flow 0cost 7flow 0cost 6flow 6cost 0flow 6all 6 units one way, every traveller takes 12S→A and B→T cost 1 for each unit on them; A→T and S→B cost 7 whatever the trafficaverage travel time per traveller02468101214before the link10after the link12least possible119/12every traveller takes 10 before the zero-cost link exists and 12 after it doesthe least total travel time on the larger network is 119/2, an average of 119/12 = 9.917, so theequilibrium costs 144/119 = 1.210 times the least possiblechecked over every route at both flows, and against all 703 splits of the traffic on a lattice of sixths

The road that makes everyone later

An equilibrium is a state nobody can improve alone, which is a much weaker thing than a state anybody would choose. Adding a link that costs nothing to use makes every traveller in this network strictly slower, and the arithmetic says by exactly how much.

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