Week 1 Puzzle: Cicada Survival

Depends on what "survival is compromised" means.

Does it mean "has a lower chance of surviving" ? (because if it does, there's still a chance that brood D could survive with ANY life cycle)

Or does it mean " has zero chance of surviving", so that if two broods appear in the same year, they both die out completely ? i.e when broods A and B coincide in year 12, both are wiped out and never appear again ? If so, why not say so instead of using ambiguous language ?

Can you confirm if you can start a brood on a different year to it's emerging year, for example: "Brood D emerges every 2 years starting on year 1"? Thanks

I'm just taking it to mean that you have to avoid any other broods, and assume that the pattern is the same ie the start year is the same as the cycle length. Can anyone confirm these rules? Cheers.

Actually, now I look at this, I can see that A and B will coincide in year 12 so would therefore both be 'compromised'. If we then knock out A and B thereafter (assuming them dead), it gives us an answer that is NOT one of the options! Why is this? Are we to assume that we can't let nature take its course for Broods A-C?

If "compromised" is taken to mean dies out then the implied "counting from now" means all four broods have emerged in year 0 so none would be left.

So the question is really asking "what is the shortest life cycle Brood D can have and still survive the next 40 years without being compromised?"

This would also mean that broods A and B survive year 12, albeit in a compromised way.

Killerbadger's question is relevant - otherwise brood D could emerge in year 39 with a 2 year cycle and survive to year 41...

I think the setup of the question makes it clear that the answer to Killerbadger's question is "no". In particular, if brood D has a 2-year cycle then it shouldn't be absent for the first 39 years.

Also Hypericus must be right, since as happeecow points out otherwise the correct answer would be one which does not appear.

"otherwise brood D could emerge in year 39 with a 2 year cycle and survive to year 41..."

But if it had a 2 year cycle, it would also have appeared in years 37, 35, 33, 31, etc

There are 2 different answers to this problem though - we need an answer to Killerbadgers question above.

"otherwise the correct answer would be one which does not appear."

I disagree.

With repect to the largest dover sole caught, could Marcus du Sautoy (sorry, not sure of correct title) use the same (normal) distribution to predict how many negative weight dover sole have been caught?

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Do we know if the meeting and mating of broods A and B in year 12 won't "cause the cicadas to appear at a variety of intervals"? If it does, the problem has a different solution.

This is very simple and no need to complicate things, the answer is 25.

Oh dear, pattayalob, did you have to? Most of the words I'd use to describe you would be moderated out.

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I agree with happeecow @4. A puzzle can be difficult but still be a good puzzle, but I really dislike puzzles that have vague or ambiguous statements or inconsistencies. I too identified an answer not on the list subject to A and B disappearing after year 12. But if and B can appear in the same year and still survive why should D not equally survive even if comprimised?

think you guys are just over complicating things

The methodology for working out this solution is clearly illustrated in episode 1. I think some people just try and look for added complications that just aren't there.

Whilst I appreciate the "you're over-thinking/complicating" the problem comments, surely finding the non-obvious answer (for example the Monty Hall problem) is often part of the fun in logic/maths puzzles?

Even ignoring the possibility that A&B kill each other off in year 12, there is still a perfectly valid non-prime answer based on the wording of the question. If the question required us to use the methodology illustrated in the episode to arrive at a prime answer (by treating year '0' as year dot, as it were, before which there were no cicadas) then it could have been rephrased to say so.

Wouldn't the easiest way to find the solution be to gather all the numbers available in the 'Codebreaker' and choose the most fitting one?... ie 3 6 7 8 9 11 12 17 23 25 28 36 42 47 56..etc
Looks like 11 would be the best as it doesn't hit any multiples of 4 6 or 7 before 40... although I'm tempted to go with the Hitchhikers Guide and choose 42. Its the ultimate answer, afterall.

@witcheryk ... 9 isn't one of the answers is it? ... think you're looking at 6 upside down ... unless I'm blind ... and I wondered last night why a programme about real world maths highlighted 42 ... now I can see why!

"This is very simple and no need to complicate things, the answer is 25"

Its 11! :P

Considering the Cicada aspect of the show was about prime numbers, I have a suspicion the answer is prime. I was working on it for 30 minutes - but then it hit me. It's obvious when you just look at the numbers. Perhaps try what Marcus du Sautoy did in the video with the grid ;)

Would the answer be 11 since this is less than 25 suggested above and it is suggested that we look for "primality of these life cycles" ie prime numbers.

agree with 11 as every other number before it would have coincided with a.n.other broods cycle at some point before 40, thus putting it at risk.

The real answer is 8: the brood emerges in years 1, 9, 17, 25, 33, 39 and never encounters another. The rules didn't specify that it has to emerge in year 8 for the first time.

Obviously that is not the answer you are looking for, but you should have worded the puzzle better

I think the answer is 10, where the brood emerges first on year 9. Later broods will be on years 19, 29, 39. None of these years will be compromised by any other cicada broods.

elpaw has given a better answer. How did I miss that?

Re-watching the episode is to be recommended.

I think it makes clear that two cohorts of cicadas hatching simultaneously results in unwanted interbreeding that changes the lifecycles of their offspring. It doesn't result in all the cicadas for that year being wiped out as all cicadas from a cohort will die after mating (and the females after laying their eggs) to produce the next generation in any case.

This is what I understand by the use of the word 'compromised' in this context.

Read it carefully and I think this is all covered in Adrian's post.

I still think 11 is the answer. One has to adopt a baseline in the absence of one being stipulated, meaning all broods start from the same baseline (and not in say year 1 as others have posted). Brood A then will emerge with other broods (except D) 4 times within the 40yr period. Brood B 3 times and Brood C once. Given then survival is compromised by a single cross breed (Brood C once) one has to conclude no cross breeding can take place if survival were to be guaranteed. Therefore the SHORTEST brood cycle to survive past year 40 is 11 (prime number and whole point of piece on this). 11 could not survive past year 44 because in it's 4th brood cycle it would clash with Brood A (if they were not totally wiped out by then).

I think it is safe to assume that the puzzle is designed to test your understanding of the maths involved and not the laws of Mendelian genetic inheritance.

Nothing ambiguous about it... :)

11 :) 44th year will not be the good one but respects the "past 40" rule.

the solution is to find a number whose least common multiple with any the others 3 numbers is bigger that 40.