Math Puzzle Game Spent Three Years Reporting Impossible Perfect Scores

For three years, Digit Party ended each daily puzzle by telling players how close they had come to perfection. The number looked authoritative: the highest score theoretically possible for that exact collection of digits. But on 55 occasions, the game’s creators had set a target that no human, no matter how cleverly they arranged the board, could reach.

It was not a software bug in the usual sense. Robert Brignall and Vincent Vatter, the mathematicians who created the game, had calculated a number that was genuinely an upper limit. The trouble was that their calculation let each group of matching digits occupy its own best possible shape, without requiring all those shapes to coexist on the same limited 5-by-5 board.

Now, writing in Math Horizons, they describe how to calculate the actual maximum. Their solution turns a small numerical embarrassment into a neat example of a broader optimization problem: the best arrangement for every part of a system may be impossible once all of those parts have to share the same limited space.

A Board Full of Competing Shapes

Digit Party deals 25 digits, each from 1 to 9, one at a time. A player sees the current digit and the next one, then must put the current digit into an empty square on a 5-by-5 grid. Matching digits score when they touch horizontally, vertically or diagonally. Two neighbouring 7s earn seven points, for example, and a compact group of several 7s can generate several such scoring connections.

The original maximum-score system asked a sensible first question: if a digit appears n times, how many adjacent same-digit pairs can its n cells make? The answer is a discrete version of the classical isoperimetric problem, which asks how to enclose the greatest area with a fixed boundary. Here, the puzzle is to arrange a fixed number of grid cells so that as many pairs touch as possible.

For a grid in which diagonal contact counts, the maximum number of touching pairs among n cells is 4n – ceiling(sqrt(28n – 12)). The formula was conjectured by Nicholas Taliceo and Julian Fleron, then proved by Andrew Vince in 2024. Vince’s paper frames the result as an extremal graph problem on an infinite grid and proves the formula by solving a discrete isoperimetric problem for polyominoes.

That result tells Digit Party’s makers the best possible shape for a group of, say, five 8s. But it does not ensure that the ideal shapes for five 8s, four 7s, four 6s and every other digit can tile one small board without interfering with one another. Sometimes one group must lose a valuable adjacency so another can keep one.

Day 88 Exposed the Error

The clearest case came on the game’s 88th daily puzzle. The old calculation said its maximum was 192 points. Yet the true best arrangement was worth 186. The discrepancy came from two groups of two digits: preserving both groups’ theoretically ideal placements could not be done alongside the optimal layouts for all the other groups.

More generally, Brignall and Vatter found that their old score was correct for 1,041 of the 1,096 daily games they examined. The median overstatement in the other 55 was only two points and the mean was 2.47 points, but the authors had still been displaying percentages against an unreachable benchmark. A player could be told they had scored 98 percent of perfection when perfection was not a possible outcome.

They call the solution to the full-board problem an integer linear program. This is a method for choosing among discrete options while obeying constraints. In this case, the program assigns every square one digit, preserves the number of copies of each digit and maximizes the value of equal-digit connections. Modern optimization software can solve one board quickly: Google’s OR-Tools documentation offers a practical example of how integer programs use variables, constraints and an objective function to identify the best allowable solution. But carrying such a solver around in every player’s browser would be overkill.

Millions of Boards Became 1,291 Cases

The useful insight was that the geometry depends on how often digits occur, rather than on which values they have. A board with groups of sizes 7, 6, 4, 4 and 4 has the same spatial problem whether the group of seven is made of 2s or 9s. The values still matter for the final score, because sacrificing a neighbouring pair of 9s costs more than sacrificing a pair of 2s, but the possible layouts are shared.

This allowed the team to replace 13,884,156 possible digit distributions with 1,291 integer partitions of 25 into no more than nine groups. Of those partitions, 891 permit every digit group to attain its own optimum at once. The other 400 require compromises.

For each nonideal partition, the researchers calculated a Pareto front: the small collection of layouts for which no other layout loses fewer matching connections in every digit group. In plain terms, each surviving arrangement makes a different unavoidable compromise. The browser can now look up the relevant partition, compare the few remaining trade-offs and select the one that gives the highest attainable score for the actual digits drawn.

The precomputed data set compresses to less than three kilobytes. The largest Pareto front contained 14 possible trade-off patterns, for the unusual partition (7, 6, 4, 4, 4). Rather than making phones solve hundreds of binary decisions after every game, the revised system needs only a lookup and a handful of arithmetic comparisons.

The Game Still Has a Harder Problem

Correcting the maximum score does not reveal how to play Digit Party well. The new calculation assumes complete foreknowledge of all 25 digits, while a player sees only the current digit and the next one. A move that seems prudent early on can block a much more valuable cluster that arrives later.

That makes actual play an online-algorithm problem, where choices are permanent and information arrives gradually. The authors leave open the question of the best strategy, and how close any strategy can reliably come to the offline optimum. After fixing the arithmetic, the game’s central human question remains: what should you do with the first 3 when you do not yet know what the board will become?

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