A math problem for you...

[Alex Franke]

I first saw this problem maybe 20 years ago in some magazine -- I think Discover magazine if my memory serves me correctly. Anyway, I thought I had lost it forever. I was looking through some old computer files from the late 80's and lo and behold I had actually typed it in!

I kept it because I wanted to some day solve it. I did, but I don't think I could anymore.

How about you? Here it is...

Having disposed of the Nemean lion, Hercules traveled on to Lerna. There he found a herd of Hydra with various numbers of heads. As he took his club in hand, his uncle's messenger, Iole, stopped him and told him the rules. "The Hydra herd has regenerative powers," explained Iole, "which grow stronger with each clubbing. On the first clubbing a Hydra loses a head and then a duplicate Hydra appears. On the second clubbing a Hydra loses a head and then two more Hydras appear. On the tenth clubbing a Hydra looses a head and then ten extra copies of it appear, and this continues. Moreover, the clubbings are counted for the entire herd. Only when a one-headed Hydra is clubbed does it simply die."

To complicate the story, the jealous Juno, wife of Hercules' father Jupiter (remember mythology?), forces Hercules to club the beasts in the worst possible order, so that he takes as long as possible to destroy the Hydra herd. You're working for Jumo, and Hercules confronts two herds.

1) The herd consists of two singles and two doubles. How many clubbings can Juno force Hercules to use?

2)The herd consists of one four-headed Hydra. Hercules clubs one head every second. Aiding the vengeful Juno, for how long can you force Hercules to club the Hydra herd?

Good luck!

[cwithboat]

I'm outa here in my Oldsmobile Hydromatic.

[LCHIEN]

so late on a Friday nite...
OK, well, let me guess: ummmm... 42?

[cgallery]

I'm going to say that the answer involves some multiple of 1.

[bthorne]

*SPLAT*
I'm gonna guess 17 clubbings on the first one.

That second one will require a bit of coffee before I try to get my head around it.



Bryan

[Alex Franke]

So far so good Hercules did end up leaving the battle in a (slightly rusted) Oldsmobile Hydromatic, the answer to both is in fact a multiple of 1, and the answer to question one is 17 -- which of course added to Hercules age at the time equals 42.

Question #2 is still up for grabs. (...and the answer may surprise you.)

[crokett]

Hrrm...

I think the second scenario is a fibinacci sequence.

Something like:
(4(3(2*1)(2(2*1)(1))))

Not sure I have the parens in the right spots though.

[Russianwolf]

Without looking at anyone elses answers

1) 9
1) 1 minute 6 seconds

[Alex Franke]

The answer to Part 1 is 17.

Part 2 is still up for grabs...

Don't forget "jealous Juno" -- he needs to club them in "the worst possible order".

[Russianwolf]

I missed that part. I figured Herc was a smart guy (knowing the stable story) and would figure a way to whack them in the best order anyway.

[Alex Franke]

I missed that part. I figured Herc was a smart guy (knowing the stable story) and would figure a way to whack them in the best order anyway.

Yeah, that's what I did the first time through, too.

BTW, if anyone wants to know the answer -- and you promise not to post a spoiler -- then just PM me and I'll share it.

[dlminehart]

I don't understand the meaning of "the clubbings are counted for the entire herd". Could you give a simple example of a situation like one single + one double? If I hit the double, it becomes a single, that then duplicates into two singles. Does the original single then also lose a head and hence not duplicate? Or does it simply duplicate so I have 2 singles, plus the two singles from the one I hit? Or something else?

[Alex Franke]

No problem. Since the first part is already solved, I'll share the solution. Remember that is has to be done in the worst possible order. The number to the left of the colon is the club number (the number of hits that Hercules makes), and the numbers to the right each indicate one hydra with the specified number of heads. The number in brackets is the one that he is clubbing at the time, and the note explains what happens after he clubs it.

(Note: No hydra were harmed during this experiment. )

Another note -- this "write it out" approach is most defiantly not the right way to solve the problem.

1: 2 2 1 [1] - kills it
2: 2 2 [1] - kills it
3: 2 [2] - loses a head and 3 more appear
4: 2 1 1 1 [1] - kills it
5: 2 1 1 [1] - kills it
6: 2 1 [1] - kills it
7: 2 [1] - kills it
8: [2] - loses a head and 8 more appear
9: 1 1 1 1 1 1 1 1 [1] - kills it
10: 1 1 1 1 1 1 1 [1] - kills it
11: 1 1 1 1 1 1 [1] - kills it
12: 1 1 1 1 1 [1] - kills it
13: 1 1 1 1 [1] - kills it
14: 1 1 1 [1] - kills it
15: 1 1 [1] - kills it
16: 1 [1] - kills it
17: [1] - kills it

[bthorne]

Ahhh - coffee makes it all better. That and a few minutes of computer modeling

Just in case it's right, I hid the answer to not spoil the fun for others - Select the space from here to the to view my answer.
22,539,988,369,405 whacks, @ 1 whack / s = 714,738 years, 125 days, and 18 minutes.

That's a LONG time for Juno to pay attention and make sure Hercules doesn't start cheating.

[Alex Franke]

You got it! That must have been sooome coffee! You a math major or something?

...except my conversion math yielded a slightly different result (off by 474 on the most significant figure - remember one in four is just a hair longer), but the raw figure is exactly the same.

Congrats!

BTW -- clever way to hide the answer!