The short side is not in the lemma

The short side only says how many gave a two-sided Maundy Cake its value in closed form — the running products of the long side’s prime factors, largest first, with as many terms as the difference of the two counts — and was careful about what it had. It had an identity checked on forty thousand cakes. It did not have a proof, and it said exactly where the gap was:

What that needs is the missing half of the rung below — that no other divisor beats the extreme prime — and that is the one statement on this ladder still resting entirely on a grid.

The gap closes, and it closes for a reason worth more than the closure. The missing statement, written out carefully, does not mention the short side at all.

What the induction needs #

The recursion is short enough to restate. A Maundy cut divides a side into equal pieces, so Left cutting an $m \times n$ cake at a divisor $j$ of $n$ leaves not one position but $j$ of them — a disjunctive sum of $j$ copies of the $m \times (n/j)$ cake, worth $j$ times what one copy is worth. Every Maundy cake is a whole number, so the simplicity rule reduces to arithmetic: when Left’s best option is worth $\lambda \geq 0$, the cake is worth $\lambda + 1$.

The step the induction wants is therefore

$$V(m, n) = 1 + P_1 \cdot V(m, n/P_1)$$

where $P_1$ is the largest prime factor of $n$, holding whenever the long side has strictly more prime factors than the short one. Applied over and over it peels one prime off the long side each time and adds one each time, and what comes out is the closed form: the running products $1, P_1, P_1P_2, \dots$, stopping when the two sides level.

The step has one assumption in it and everything else is the definition. The assumption is that $P_1$ is a best cut — that no other divisor of $n$, prime or composite, gives Left a larger option. That is the missing half, and it is the whole of the missing half: with it the step is a rewriting of the recursion, and without it the step is a guess about where a maximum sits.

Where the short side goes #

The rung below checked the assumption the only way a grid can: it took every cake in a square of sizes, computed the best cut, and found the largest prime among the best cuts on every one of them. That is a claim about a finite rectangle of $(m, n)$ pairs, and no amount of widening the rectangle makes it a claim about all of them.

Write the inequality down, though, and the rectangle turns out to be the wrong shape for it.

Let $S$ be the multiset of prime factors of $n$, written largest first, and let $d = \Omega(n) - \Omega(m)$ be the difference of the two counts. A cut at a divisor $j$ with $\Omega(j) = e$ leaves $j$ copies of an $m \times (n/j)$ cake, and by the closed form at a smaller total that cake is worth $F(S \setminus J, \, d - e)$, where $J$ is the multiset of primes of $j$ and

$$F(T, k) = 1 + T_1 + T_1T_2 + \cdots$$

is the running-product sum with $k$ terms. So the cut is worth $j \cdot F(S \setminus J, \, d - e)$, and what has to be shown is

$$j \cdot F(S \setminus J, \, d - e) \; \leq \; P_1 \cdot F(S \setminus P_1, \, d - 1)$$

for every divisor $j$ of $n$. The short side has vanished. It entered the recursion as a size and it leaves as the single number $d$, and $d$ is a term count rather than a cake.

That changes what a check is worth. A pair $(S, d)$ is a statement about every short side whatsoever that gives that difference — the sixty-two numbers below two hundred with two prime factors, and every other number with two prime factors there will ever be. Checking one pair settles infinitely many cakes, and the sweep over every long side to twenty thousand is 65,533 such pairs rather than 65,533 cakes.

Two boundary cases have to be swept out of the way first, and both are one line. A divisor $j$ with $\Omega(j) > d$ cuts the long side past the short one, leaving a cake whose value is negative, and a negative option cannot be the largest when $P_1$’s option is not negative — so over-cuts never win, and they never win without any reference to the short side either, because a sign is all that is needed. There are 277,997 of them in the sweep and not one is a candidate. And the cut $j = P_1$ is always available, since $d \geq 1$ means $n$ has a prime factor to spend.

Term by term #

Stated on multisets, the inequality is not a search. It is a comparison of two sums, and the two sums line up.

Expand the candidate. Its $i$-th term is $j \cdot R_1 \cdots R_i$, where the $R$ are the primes of $n/j$ largest first — a product of $e + i$ prime factors of $n$, since $j$ contributes $e$ of them and the $R$ contribute $i$. Expand the target. Its $(e+i)$-th term is $P_1 \cdots P_{e+i}$: the product of the $e + i$ largest prime factors of $n$, which is the largest product of that many prime factors of $n$ there is.

So each term of the candidate is at most the term of the target sitting above it. The candidate has $d - e$ terms and the target has $d - 1$; for $e \geq 1$ the candidate’s terms map into the target’s, index by index, with none left over. Sum the inequalities and the lemma falls out.

There are no cases in that argument and nothing in it about $m$. It is the observation that a product of $k$ primes drawn from a multiset is largest when the $k$ largest are drawn, applied once per term.

The one place it could go wrong is the indexing, and it is worth being slow about it, because an off-by-one here would be an off-by-one in the theorem. The candidate’s terms are indexed $i = 0, \dots, d - e - 1$ and sit under the target’s terms $e, \dots, d - 1$. The target’s terms run $1, \dots, d - 1$. So the mapping lands inside the target exactly when $e \geq 1$, which every divisor above one satisfies, and it uses up the target’s last $d - e$ terms and leaves its first $e - 1$ unmatched. Those unmatched terms are the slack: a cut spending two primes at once is compared against a target that has already banked a term the candidate never gets, which is the arithmetic reason a composite cut loses even when its individual products are large. Cutting $360$ into nine is worth $9 + 45 = 54$ against $5 + 15 + 45 = 65$, and it draws on its larger term while giving up an entire smaller one.

The sweep is corroboration now rather than evidence, and it is worth having in that role: 839,006 term comparisons, 426,586 of them strict and 412,420 equalities, with nothing going the wrong way. The equalities are not noise. A term is equal exactly when the candidate has spent the same primes the target has, in the same order, which is the largest-prime cut agreeing with itself and the near misses agreeing with it as far as they go. The worked long side shows one: cutting $360$ into three gives $3 + 15 + 45$ against the target’s $5 + 15 + 45$, losing on the first term and drawing on the other two, for a value of 63 against 65.

The tie is where the rung below’s second case lives #

The anatomy of the equalities says something the rung below could not have seen, because the rung below was measuring a different population.

Above a term count of one, the largest prime is the only best cut — on all 45,534 pairs, with no tie anywhere. At a term count of one it ties with every other prime cut, on all 17,671 pairs where there is more than one prime to choose from. The reason is immediate once the sum is written out: at $d = 1$ every prime cut leaves a level cake, a level cake is worth nought, and the value is $1 + 0$ whatever was cut.

The rule for which prime to cut at settled the cut rule as two cases — the smallest prime when $\Omega(m) + 1 \geq \Omega(n)$, the largest otherwise — and that reads now as a rule whose second case is doing two quite different jobs. On the cakes where Left is level or behind it is doing real work, and those are the cakes this closed form says nothing about. On the cakes with $d = 1$ it is breaking a tie in which nothing whatever is at stake: every prime cut is a best cut, the smallest is a best cut, and so is the largest.

That is not a correction to the rung below, which scored its rules honestly on a population that included the level and behind cakes. It is a sharpening of what the score meant. A rule with two cases looks like a rule that has found two regimes, and on this ladder it has found one regime and one tie.

The level case is not a separate statement #

The induction bottoms out where the two sides have the same number of prime factors, and there it needs $V = 0$. That looked like a second statement wanting an argument of its own, and the obvious argument does not work: a mirroring strategy would have Right answer Left’s cut with a matching cut, but Left’s cut leaves $j$ copies and Right’s reply repairs one of them, which is not a strategy at all.

It is not a second statement. The $m \times n$ cake with the players swapped is the $n \times m$ cake — Left cuts columns, Right cuts rows, and transposing the cake exchanges the two — so $V(m, n) = -V(n, m)$ by the rules rather than by any measurement. On a level cake every one of Left’s cuts leaves the long side short, that transposed cake is a positive one at a smaller total, and so every Left option is negative; symmetrically every Right option is positive; and a position all of whose Left options are negative and Right options positive is worth nought.

So one induction on $\Omega(m) + \Omega(n)$ carries both cases, and it bottoms out at the empty cake with no cuts in it. What the whole ladder rests on is the transposition, which is a sentence about the rules.

What this does and does not settle #

It settles the positive half of the $\Omega$ rule. Maundy Cake established by census that a cake is worth nought exactly when the two sides have an equal count of prime factors, and reported it as a measured regularity over 1,296 cakes rather than a theorem — with the largest-odd-divisor claim it displaced refuted in the same sweep. The closed form is strictly positive whenever $d \geq 1$, so the sign rule’s two non-zero cases follow from it, and the zero case is the induction’s base. The census stands where it always did: it was right, and it is now a corollary.

It does not settle Cutcake. The game one clause away — a cut may fall anywhere, not only into equal pieces — has its own rule about binary length, from Cutcake, and nothing here touches it. That rule is in the same position the Maundy rule was in this morning: right on every board that has been tried and resting on a grid. Whether it has an induction of the same shape is a question this page cannot answer, and the two mechanisms have almost nothing in common — halving against factorising, a length against a multiset.

And it does not extend to a cake with more than two dimensions or to unequal pieces, which are not games this site has built. The lemma’s whole force comes from the option being $j$ copies of one cake, which is what makes its value a product and what makes the terms line up.

The convention is normal play throughout, as everywhere on this ladder. Under misère play none of the integer arithmetic survives, because the simplicity rule that turns the recursion into addition is a normal-play theorem.

And the tables cannot show the thing the page is about. Every figure here is a table of numbers, and a table of numbers is exactly the instrument the argument replaces: it can display 65,533 pairs holding and it cannot display why the sixty-five-thousand-and-thirty-fourth holds too. What a picture would have to draw is the term-by-term mapping between two sums — an alignment rather than a quantity — and the closest this page comes is the worked long side, where the two columns sit beside each other and the reader can see the candidate’s terms sliding down under the target’s. That figure is a proof for one $n$ and one $d$. The general statement is a sentence, and a sentence is not a figure; what the figures are honestly doing is showing that the sentence is not contradicted anywhere anyone has looked.

The shape this argument has #

It is worth naming what happened, because it is a shape that recurs and is easy to miss.

The obstacle was not the difficulty of the inequality. The obstacle was that the inequality had been written with a variable in it that it did not depend on. As long as the statement is no divisor of $n$ beats $P_1$ for a cake of short side $m$, the only instrument that fits is a sweep over $(m, n)$ pairs, and a sweep over pairs cannot be finished. Written as a statement about $(S, d)$ the same claim is finite in a useful sense — one statement per multiset per term count — and then it is short enough to see through.

A period is a proof is the other place on this site where a measured regularity became a theorem, and the mechanism there was the same in outline and different in every detail: a sweep long enough to cover a period is a proof for a game whose options reach back a bounded distance, because the sequence cannot do anything it has not already done. Here the sweep was never going to close, and what closed it was noticing which arguments the statement really has.

What solved means sets out the several things the word can mean in this subject, and this ladder has now moved along that list: from a value computed for one cake, to a rule fitted to a grid, to a formula checked on forty thousand cases, to a theorem. Each of the first three was worth writing and none of them was this.

Where the ladder goes next #

The cutcake anchor has six rungs: the game and its binary-length rule, the equal-cut variant and the $\Omega$ rule for its sign, the size on one row, which cut to make on two, the value itself, and now the proof of it.

The rung above is the other game. Cutcake’s own rule — a cake is worth nought when the two sides have the same binary length, and its size is read off the lengths — is exactly where Maundy Cake’s was before this page, and the interesting question is not whether it can be proved but whether it can be proved this way. The step would have to be cut in half, the analogue of the lemma would be that no other cut beats halving, and the object the short side collapses into would have to be a binary length rather than a multiset. If it collapses the same way, the two games share an argument as well as a shape; if it does not, the reason it does not is the sharpest available statement of what the equal-pieces clause is doing.

Two neighbours are worth the trip. The size of a cake is the one-row case, which is this formula with every term kept, and reading it beside the induction shows the walk that the running products are a record of. And a numeral in the empty squares is the other place on this site where a value turns out to be read off the position rather than searched for, and it is worth reading beside a lemma whose proof is a statement about which primes are largest.