I wrote on the original post that I'd be interested to see if there was a super-fast way to do this puzzle but in retrospect what of course I meant was that I hope no-one solves it quicker than the four hours or so it took me. Seems Jeff's got me down by half but I'm glad that no-one slipped in under the hour.
ding ding
Second round. Since everyone needs a bit of work avoidance I'm going to drop another couple of these puzzles into the feed next week and give y'all an excuse to get head scratching again.The next one will be a physics problem (that I was asked at my university interview and failed dismally and embarassingly to answer correctly) and then the one after that will be probability (this one also needed to be explained very slowly to me).
Don't fear the possibility of intellectual alienation for all conundrums will of course wrapped up in a nice everyday scenario like needing to find the odd-bearing (and you can always reassure yourself with the fact that I floundered like a fish in a puddle with the next two).
7 comments:
Jeff Koke
I really enjoyed that puzzle. I've done several puzzles like that one, but had never heard it before.
I looked for a way to email you, but don't see anything on your site, so I'll post this here. Here is a puzzle a friend of mine told me a few months ago, that stumped me for a long while. It's a little bit more involved than the ball bearings one, but I hope you'll forgive the long comment.
You and a mate are stranded on a deserted island. You have a single gold bar and an old saw. Your buddy has nothing. A rescue ship could show up at any moment.
Your mate agrees to build you a shelter if you'll pay him 1/7th of the gold bar for each day he spends working, up to a week. Since the rescue could happen at any time, he wants to be paid at the end of every day, exactly 1/7th of the bar.
Here's the trouble-- your saw can only cut through the bar 2 times before it breaks. How can you make sure that at the end of each day, he has 1/7th more of the gold bar than the previous day. For example, at day 1, he has 1/7th of the bar, day 2 he has 2/7ths, and so on.
No trickery like placing two pieces of the bar on top of each other and cutting through both (that would count as 2 cuts). You also can't break the bar.
Winner is whomever Peter decides it is...
Sarah
Sarah
1 - give him the 1/7th
2 - trade him the 2/7th
3 - give him the 1/7th
4 - trade him the 4/7th
5 - give him the 1/7th
6 - trade him the 2/7th for the 1/7th
7 - give him the 1/7th and start working for him =)
Peter Nixey
Doesn't seem much doubt who the winner here is and it certainly wasn't me. What an elegant solution.
Sarah, it really should be Jeff awarding this but since he's handed me the reigns I'd like to pass on my congratulations. This one goes to you.
Well ladies and gents lube your engines, rest your brains and enjoy the weekend. Competitions will resume again on Monday.
Atacrawl
I read Jeff Koke's solution to the ball bearing riddle (well done, I had trouble with the 3 x 3 vs. 4 x 4 option myself) and I noticed that this part of it...
#1 4 - 4, leave 4 out
if they balance, then the oddball is one of the four you left out, and the remaining weighings are 2 vs. 2 and 1 vs. 1
...is incorrect. Just as you state that measuring 6 x 6 to begin with is pointless, so too is measuring 2 x 2 in the second weighing of this scenario, because you've already established that one of the four is odd, and weighing them by themselves doesn't determine anything.
Instead, what you should do for the second weighing in this case is take three of the four you didn't measure (the odd group) and put it up against three of the ones you did measure (the control group). If they're even, the oddball is the one you haven't measured yet. If they're uneven, then you can measure 1 x 1 to determine which ball is odd, since you've already determined whether it's light or heavy.
Jeff Koke
That's why I should have shown my work more carefully. When I said the next two weighings are 2v2 and 1v1, I meant 2 from the 4 you set aside vs. 2 from the control group (the 8 you weighed in the first weighing).
If the scales balance, then you know the oddball is one of the 2 left & you can figure out which of those is the oddball by weighing one against a control.
If the scales are uneven, then you know it's one of the two that's not in the control group and you also know whether it's heavy or light. So you just weigh those 2 against each other.
Sorry I wasn't more clear. In solving the puzzle I always worked on the worst case scenario & once I figured out it was trivial to find the oddball in a group of 4 with two weigings, I didn't spend much time explaining that possibility.
Peter Nixey
Do the final 2x2, 1x1 in the following way:
swap 1 control into the 2x2 so you have 2-unknown v. 1-control + 1-unknown.
- If the scales balance the odd one was the one you swapped out
If the scales don't balance then take one from the 2-unknown and move it to the other side to do a 1-unknown v. 1-unknown comparison.
- If the scales balance you know it was the one you removed, if they change you know it was the one you swapped sides and if they stay the same you know it's the one you left on the same side.
It's not as simple as Jeff's solution but it does work. I'm going to have to go back to my original post and point out that the solution is not in fact unique - apologies.