## Project Euler Problem #243

### January 21, 2012

Read the details of the problem here

Summary

Find the smallest denominator d, having a resilience R(d) < 15499 ⁄ 94744

Solution

Due to real-life intrusion I haven’t been doing much Project Euler for the last few months but a conversation with a colleague at work the other day brought the subject up and I thought I’d revisit the site.

They’ve changed their format quite a lot now and though it’s still fundamentally the same, the concept of “badges” has been introduced – “awards” above and beyond just the old numbers of puzzles solved (which is now more granular). I’m a sucker for things like that and one of them – “Trinary Triumph” – just needed one more question to complete so it was, as the saying goes, like a red rag to a bull!

The problem was actually worded quite straightforwardly, and a little thought showed that this was very similar to Problem #69 and Problem #70. The numerator in the resilience function is going to be the count of terms that are relatively prime to the denominator i.e. Euler’s Totient (φ) function. The solution to Problem #69 was found by maximising n/φ(n) whereas, in this problem we’re looking to minimize φ(d)/(d-1) – so it’s essentially the same thing. The answer is likely to be found by looking for a number consisting of the product of many small prime factors perhaps with some multiplier (also a product of small primes) i.e. some of those factors may be of a higher power.

Problem #108 and Problem #110 also tackled the domain of products of powers of primes and the solution to the latter seemed a good place to start as the one to the former ended up as being a bit of a lucky hack. As per that solution, taking the approach of assuming that the answer was to be found somewhere at n * product(prime factors) and driving off that meant that factorisation of large numbers wasn’t needed so Groovy would remain a viable language to do this in.

Again, when doing these problems with Groovy, it’s best to try to work out the potential scope ahead of runtime as it doesn’t perform that well at number crunching.

What I did need to how to calculate the totient value for numbers with large numbers of factors – I’d kind of worked it out from a bit of guess work for two factors when doing Problem #70 but this wouldn’t work for this. Luckily Wikipedia came to the rescue giving the formula quite succinctly as:

φ(n) = n(1-1/p1)(1-1/p2)…(1-1/pr).

Using this I found that the product of a prime sequence 2..23 gave a totient above what was being sought, whilst 2..29 was a potential solution itself. The answer probably fell somewhere between these two and this therefore formed the search space.

I then just reused the helper function I’d used in Problem #110 with a bit of an educated guess as to what the maximum power might be. If the multipler was 25 i.e. 32, this would exceed the solution of the longer prime sequence – so a maximum power of 4 seemed reasonable to use. I also seeded the generator function with the assumption that all of the primes between 2..23 were going to be factors – this seemed a reasonable assumption given what I knew of the totient function characteristics.

```def gen_powers(range, maxlen, yield) {
gen_powers(range, maxlen, [], yield)
}

def gen_powers(range, maxlen, pows, yield) {
if (pows.size() == maxlen)
yield(pows)
else if (pows.size() == 0)
range.each { gen_powers(range, maxlen, [it], yield) }
else
range.findAll { it >= pows[0] }
.each { gen_powers(range, maxlen, [it]+pows, yield) }
}

def ( answer, TARGET, POWER_RANGE ) = [ Long.MAX_VALUE, 15499/94744, (0..4) ]
def primes = [ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ]

gen_powers(POWER_RANGE, primes.size(), [1] * primes.size()- 1) { powers ->
def (d, factors) = [ 1G, [] ]
powers.eachWithIndex { v, i ->
d *= primes[i] ** v
if (v > 0) factors << primes[i]
}
if ((d > 1) &&
(((int)(factors.inject(d) { p, v -> p * (1 - 1/v) })) / (d-1) < TARGET)) {