# Froebenius coin problem

November 28, 2019
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A challenge from The Riddler last weekend came out as the classical Frobenius coin problem, namely to find the largest amount that cannot be obtained using only n coins of specified coprime denominations (i.e., with gcd equal to one). There is always such a largest value. For the units a=19 and b=538, I ran a basic R code that returned 9665 as the largest impossible value, which happens to be 19×538-538-19, the Sylvester solution to the problem when n=2. A recent paper by Tripathi (2017) manages the case n=3, for “almost all triples”, which decomposes into a myriad of sub-cases. (As an aside, Tripathi (2017) thanks a PhD student, Prof. Thomas W. Cusick, for contributing to the proof, which constitutes a part of his dissertation, but does not explain why he did not join as co-author.) The specific case when a=19, b=101, and c=538 suggested by The Riddler happens to fall in one of the simplest categories since, as ⌊cb⁻¹⌋ and ⌊cb⁻¹⌋ (a) are equal and gcd(a,b)=1 (Lemma 2), the solution is then the same as for the pair (a,b), namely 1799. As this was quite a light puzzle, I went looking for a codegolf challenge that addressed this problem and lo and behold! found one. And proposed the condensed R function

function(a)max((1:(b<-prod(a)))[-apply(combn(outer(a,0:b,"*"),sum(!!a))),2,sum)])

that assumes no duplicate and ordering in the input a. (And learned about combn from Robin.) It is of course very inefficient—to the point of crashing R—to look at the upper bound

$\prod_{i=1}^n a_i \ \ \ \ \ \ \ (1)$

for the Frobenius number since

$\min_{(i,j);\text{gcd}(a_i,a_j)=1} (a_i-1)(a_j-1)\ \ \ \ \ \ \ (2)$

is already an upper bound, by Sylvester’s formula. But coding (2) would alas take much more space…

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