February 2026: Subtiles 2

This was a pretty standard Jane Street place-a-bunch-of-numbers-in-a-grid puzzle. Having recently become more convinced that these puzzles can always be solved by hand, I endeavoured to do this month’s puzzle without coding. Luckily, this was a relatively easy one, and a clear enough path to a solution was evident. That being said, I still managed to complicate things and take much longer than necessary, but I’m glad to have solved this puzzle without relying on code.

I collaborated on this month’s puzzle with Kelly Su.

Stats:

Difficulty: 5/10

Enjoyability: 6/10

Leaderboard Placing: 61/970

The first order of business was determining \(a\), \(b\), and \(c\). It’s tempting to fall into the trap of believing that these must be integers, but it’s important to stay flexible and make no unjustified assumptions, and indeed \(a\) and \(c\) end up not being integers after all.

I solved this part of the problem without Kelly, who came in later to help place the polyominoes. My first observation was that since the grid was of size \(13 \times 13\), the largest polyomino can have no more than \(17\) squares. This means that each algebraic expression must evalute to an integer in the set \({1,\cdots,17}\). In particular, we may deduce from the cells containing \(b^2\) and \((b-1)^2\) that both \(b^2\) and \((b-1)^2 = b^2 - 2b + 1\) are integers. It follows that, \(2b\) must also be an integer, hence \(b\) is rational. But for \(b\) to be a rational number whose square lies in the set \({1,\cdots,17}\), we must have \(b \in {\pm 1, \pm 2, \pm 3, \pm 4}\). Applying the same logic to \(b-1\) then shows that in fact \(b \in {-1, \pm 2, \pm 3, 4}\).

Now, the expression \(\frac{b+a}{\sqrt{c-a}}\) implies that \(c > a\), from which the expression \(\frac{b}{a-c}\) implies \(b < 0\), hence \(b \in {-1, -2, -3}\). From here, I’m sure there are many ways to proceed, some of which are likely better than the argument I found. I used the expression \(4a - 5b\) to conclude that \(a\) must be of the form \((n+k)/4\) for some \(n \in {1,\cdots,17}\) and some \(k \in {-5, -10, -15}\). Since the cell containing \(\frac{\sqrt{a+2}}{a}\) tells us that this expression must also be a positive integer, one can check all cases by hand to determine that \(a\) must be either \(1/4\) or \(2\). However, the cell containing \(\frac{b}{a-1}\) reveals that \(a < 1\), hence \(a = 1/4\).

Using what we now know about \(a\), one can use the cell containing \(a^b - \frac{12}{a}\) to deduce that \(4^{-b} \in {59, 50, \cdots 65}\), from which \(b = -3\) is easily determined.

Knowing \(a\) and \(b\), one can then use any of the cells whose expression features \(c\) to infer a finite set in which \(c\) must lie. If chosen correctly, taking the intersection of these sets reveals that \(c\) must be \(1/2\) (I used the expressions \(\frac{2^b+1}{ac}\), \(\frac{c+3}{a}\), and \(c^b\), though I’d be surprised if there weren’t some far better choices).

In summary, after various algebraic manipulations, we’re able to conclude that \(a = 1/4\), \(b = -3\), and \(c = 1/2\). Evaluating all expressions in the grid then reveals that polyominoes up to size \(16\) must be placed in the grid, and provides some strong direction on how to start making these placements.

It’s at this point that Kelly joined me in solving the puzzle, and immediately suggested starting with the \(12\)-omino given that the span of the \(12\)’s already placed in the grid constrains the \(12\)-omino placement to only a handful of possibilities. By a mixture of luck and intuition, we started with the correct placement, and started deducing the placements of some of the other polyominoes under the assumption that our \(12\)-omino placement was correct. We eventually hit a roadblock and had to backtrack, and ended up wasting lots of time on alternative \(12\)-omino placements. At the end of the day, we returned to our original choice, and after a more careful analysis, realised that we hadn’t tried a slightly different placement of the \(14\)-omino which quite easily led to the full solution. Essentially, the correct placement of the \(12\)-omino uniquely determines the placements of the \(6\)- \(10\)- and \(11\)-ominoes, following which a cascade of deductions reveals the placements of everything else.

Overall, I thought this puzzle was fairly fun and satisfying, though not particularly inspired. I was a little frustrated to have taken such a long detour from the correct polyomino placements, ultimately spending way more time on the puzzle than was necessary, but the process was still enjoyable overall. It was also nice to work on the puzzle with Kelly — our first joint effort in a while. She definitely took to the art of polyomino placement a little better than I did!