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Lacan once remarked on the Cartesian cogito that etymologically, “the French verb *penser* (to think)…means nothing other than *peser* (to weigh)” (1961-2: 14). Lacan’s mix of bad puns, abuse of notation, cryptic aphorisms & immense erudition has created a new style of writing, and even of thinking. A new language, in short, lacking any method for non-experts to weigh its words.

Of all things, Lacan’s earliest papers take the form of math puzzles—a lucid (albeit horrendously verbose) derivation, then reframing of the problem as metaphor. One such paper—“The Number Thirteen and the Logical Form of Suspicion”—has largely been forgotten. This post aims to recast the puzzle using discrete mathematics, and show how it bears upon Lacan’s later ideas.

What I like about this mathematical allegory is that even if one believes Lacan is a charlatan, here there’s no need to immerse oneself in psychoanalytic concepts, but only to *think*.

I hope that Lacanians will find working through my derivation a challenging exercise, and that mathematicians will be piqued at the idea of treating a math problem as a philosophical ‘text’. I for one would be glad to see more such texts.

#### 1. Lacan’s Algorithm

We are given 12 identical pieces, and told one of them is ‘bad’—either lighter or heavier than the rest, we’re not sure which. Having only a scale with two plates, and no way to gauge numerical weight, we must find the bad piece in 3 weighings.

If we knew whether the bad piece was lighter or heavier, the problem would be easy: just split the pieces into two groups of 6, then split the ‘bad’ half into two groups of 3, then simply weigh two of the bad three. But we don’t.

Here, we’ll overview Lacan’s account for 12 pieces, and then in the next section we’ll consider *n* pieces, and try to explain *why* Lacan’s algorithm works.

Lacan begins by placing on the scales two groups of 4. Suppose they balance. Then the bad piece is in the remaining 4, so we can just weigh any 2 of the 4. If those balance, the 2 left-out pieces are bad; if they don’t, the 2 pieces on the scale are bad. So weigh one of the bad 2 against a good piece: if they balance, the other piece is bad; if they don’t, then the piece on the scale is bad. Simple.

Note how this was equivalent to the sub-problem of finding a bad piece out of 4 pieces, in 2 weighings. The sub-problem is embedded in the larger problem.

If the two groups of 4 don’t balance, we use the method of *tripartite rotation*.

Tripartite rotation

The scales don’t balance, so one is heavier (H), one lighter (L). So, we select 3 pieces from H, L, and the remainder (R), and rotate them: H → L → R → H.^{[1]}

**Case 1**: Scales balance — the bad piece is in the 3 moved to R, and too light.

**Case 2**: Balance shifts — the bad piece is in the 3 moved to L, and too heavy.

**Case 3**: Unbalance doesn’t change — the bad piece is in the 2 unmoved pieces.

In cases 1 and 2, just weigh 2 of the bad pieces: if they’re equal, the remainder is bad; if not, we know the bad piece is the lighter (case 1) or heavier piece (case 2). For case 3, just pick one and weigh it against a good piece. And we’re done.

Lacan then considers the case of 13 pieces: 4 on each scale, 5 remainders. It’s clear that if the scales don’t balance, the problem is the same as with 12 pieces when the scale didn’t balance—the remainders are all good, whether 5 or 4.

Here, when the scales balance, we have a new problem. Recall how we could treat 4 pieces as a separate problem. So let’s examine the 5-piece sub-problem.

Start with 2 pieces on each scale and 1 remainder. If we’re lucky, the scales balance and the remainder is bad. If not, we have 4 pieces, but we know the 4-piece case takes two weighings, so the 5-piece case must take three weighings.

It’s the same even for 1 piece on each scale and 3 remainders. If we’re unlucky, the scales balance, giving a new sub-problem with 3 pieces—the smallest solvable version of Lacan’s problem. Weigh any 2. If they balance, the remainder is bad. If not, weigh a piece on the scale against the good piece. Total: three weighings.

So both the 3-piece and 4-piece cases take two weighings, 5 pieces takes three weighings, so it would seem that 13 pieces must take four weighings. Nope.

Actually, for 13 pieces, the 5 remainders aren’t truly a separate sub-problem. There’s a difference: we have 8 good pieces. For 3 or 4 pieces, this doesn’t matter, but for 5 pieces, Lacan can introduce a new trick: the *‘by-three-and-one’ position*.

The ‘by-three-and-one’ position

Here, we have 2 pieces in each plate, with one of the 4 a good piece, and 2 remainders. If the scales balance, just weigh one remainder against a good piece and we’re done. If they don’t balance, here’s the trick: we can do the smallest possible tripartite rotation, H → L → R → H, where R is a good piece.

**Case 1**: Scales balance — the bad piece is in R.

**Case 2**: Balance shifts — the bad piece is in L.

**Case 3**: No change — the unmoved piece is bad.

Thus, the 5 remainders take two weighings, and the 13-piece case takes three.

In this case, treating the 5 remainders as a sub-problem was the wrong way to go, making it seem impossible to solve in 3 weighings. More pieces means more ways to divide between scales and remainder, increasing the risk of such pitfalls.

Thus, Lacan’s task is to find a general algorithm for any number of pieces, including a uniform way to divide them. The algorithm must minimize the maximum amount of weighings—i.e. find the minimum, assuming we don’t get lucky.

The problem also raises some new questions. The main one is: for a given number of pieces, how many weighings are needed? As in the solutions outlined above, Lacan answers this question, but fails to explain *why* his solution works.

Hence, the next section will diverge from Lacan’s exposition, using discrete mathematics to give an algorithm for *n* pieces. This will help us see how Lacan’s problem relates to the logic of suspicion, which we will outline in the final section.

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