Daily puzzles/problems.

[prendrelemick]

Don't worry about that, it was up for 5 days but I reckon people only spend a few spare minutes here and there on it.

So, onwards-

Rorden Gamsey, a foul mouthed, scary, celebrity chef, barks an order to the timid Fish Fryer Freddy.

"I need three rare tuna steaks in exactly 3 minutes, or I'll have your guts for garters."

Fish Fryer Freddy knows rare tuna steaks must be fried for one minute on each side. He also knows that his frying pan will only hold 2 steaks at once.

"Yes Chef. No problem Chef" he shouts back.

How does he do it?

[kasie]

He puts two steaks in the pan; after one minute he turns one steak, removes the other steak and keeps it warm and puts in the third steak; after a further minute he removes the first steak, keeps it warm, returns the second steak to the pan to cook the other side and turns the third steak. At the end of the third minute he plates all three rare steaks and shouts 'Ready, Chef.'

[Scheherazade]

Excellent solution, Kasie!

Just going back to the previous puzzle (I am not arguing but trying to understand)... Because there were two last posts to the thread, I had assumed that one of those was a direct reply to the OP and the other one was a reply to one of the other posts in the thread. And since Mark was mentioned as having replied to another poster, I, again, assumed that his would be the very last post in the thread and IaN would be the last person to reply to the OP.

I am not sure how the solution given accommodates these. Can someone explain please?

• Six posts total.
• Only one post received more than one reply.
• No one replied to the 4th post.
• IamNobody was the last person to participate in the thread.
• Mark was the last person to participate in the thread.
• IamNobody replied to Scher twice.
• Scher replied to IamNobody and Mick.
• Mark replied to the reply to the OP.

1- Mick is op
2- Scher replies
3- IaN replies to Scher for first time.
4- Mark replies to Scher (2nd reply for 2nd post)
5- Sher replies to IaN
6- Ian replies to Scher for the second time.

[billl]

Scher, I'm not sure what you are saying when you say that 1) a reply to the OP, and 2) post number 6 being a reply to something in the middle four posts would mean two "last posts" or whatever--I think that the whole "reply" angle is being looked at in a way I hadn't forseen or something, maybe. And I'll admit that the wordplay and spaghetti-type nature of this means I might be missing something pretty interesting.

Here's how Mick's solution works for IamNobody and Mark being "last":
--IamNobody is last because he did the final post (number 6).
--Mark is last because his post comes after everyone else has posted (he was the last one to join the discussion).
In this interpretation/solution, the matter of "replies" doesn't affect who is "last".

AHHHHHHHHHH....
It just occurred to me what you might be getting at!
Are you saying that you had envisioned a pair of "strings of replies"? Like someone replies to someone who had replied to someone, and the last in that string can be considered "the last to participate" in that series of replies? With one string beginning with the OP, and the other string involving the other posts that are not part of the OP's string of replies?

Since that particular pair of clues is built on wordplay, I'd have to say such a solution would be permissable, yeah! (If there is one...)

[Scheherazade]

--Mark is last because his post comes after everyone else has posted (he was the last one to join the discussion).
In this interpretation/solution, the matter of "replies" doesn't affect who is "last".

Gotcha!

With one string beginning with the OP, and the other string involving the other posts that are not part of the OP's string of replies?

Yeah, something along those lines.

Since that particular pair of clues is built on wordplay, I'd have to say such a solution would be permissable, yeah! (If there is one...)

Well, obviously , I could not come with it if it exists! Not with 6 posts, at least (I think I managed it with 7, though).

Thanks for the reply, Bill

[billl]

No problem! Now I have to see if I can resist the urge to re-check all of the possibilities you already explored using the "strings of replies" interpretation, trying to match your seven.

[prendrelemick]

He puts two steaks in the pan; after one minute he turns one steak, removes the other steak and keeps it warm and puts in the third steak; after a further minute he removes the first steak, keeps it warm, returns the second steak to the pan to cook the other side and turns the third steak. At the end of the third minute he plates all three rare steaks and shouts 'Ready, Chef.'

Correct! Now Kasie your turn, I'm not sure we can accept any more IOU's

[kasie]

Oh dear, I knew I shouldn't have posted that answer - I am puzzled for a puzzle atm so I will have to think about it for a while.....

OK - I've thunk. I did a quick check and I don't think we've had this one before. If we have, then I apologise.

A farmer goes to market. He buys twenty sheep; he also buys twenty hurdles to make a small temporary fold in which to keep his new acquisitions. He calculates that twenty hurdles will just hold them comfortably. After a swift half at the Dog and Crook (Ramsbottom's Best, of course) he returns to the auction and buys a further twenty sheep, a bargain too good to miss. He then realises that he needs to buy more hurdles as he has doubled the size of his new flock. So - how many more hurdles does he need to buy to double the size of his enclosure? Remember he has already spent more on the sheep than he intended so he doesn't want to spend too much on the hurdles - what's the smallest number he can get away with?

[prendrelemick]

He goes back to the pub, and works out on a beer mat 20 Hurdles arranged 5x5 makes 25 sq Hurdles so twice the sheep will need 50sq Hurdles which is a pen 10x5. As he contemplates the 10 extra Hurdles he will need, he orders a pie and a pint.


As he stares at his round pint glass and nibbles on his Pie an Idea is slowly forming in his head, pie...pie...pi! And he realises he can get away with fewer hurdles by making a round pen. only 5 or 6 extra .

Sorry I can't do the maths (lost my calculator) but the square root of 50 over pi will be the necessary radius, then times 2 and times pi. and that is the number of hurdles (I think.)

[billl]

Imprecise, yes. But flashy!

[Scheherazade]

ESOL / NON-NATIVE ALERT!!!

What are hurdles???

[kasie]

Scher - hurdles are like fencing panels, portable frames with bars or brushwood in them: the farmer can move then around easily to make temporary enclosures. Sorry, it's a bit archaic, it's a very old puzzle, one I dragged back from my long-lost youth.

Mick, that is of course a shepherd's pie, isn't it? Which can be made rectangular, not necessarily round..... Have another pint and re-calculate.

[billl]

7.14 x 7.14

No way is that correct, but I might be just short of the correct answer. Unless Mick is on the right path. Orrrr any other number of reasons for me likely being wrong.

[kasie]

No, the farmer does not have to cut or bend any of the hurdles to make his sheep fold.

[billl]

Mick's math, approximated.

area = pi R*R
50 = pi R*R
R = square root of (50/pi)
R = square root of (15.9)
Circumference = 2 pi R = 2 pi (4.0) = 8 pi = 25.12 = 26 hurdles, so he'd need six more hurdles. And they'd have to be bendable, which Kasie rules out, OR the math gets more difficult and he needs maybe one or two (zero?) extra to make up for the relatively blocky nature of this "circular" pen.

************************************************** **********

Alternately, going back to the 5 x 5 = 25 sq feet being fine for 20 cows, then we can go with 8 x 7 rectangle having 56 sq feet being plenty enough room for 40 cows (and 7 x7 not being quite big enough).

So he'd need 10 more hurdles, if we do an 8 x 7 rectangular enclosure.

Maybe?