Tuesday, June 09, 2009

Off-Day Puzzle #5: Solution

Thank you Jose for a stellar puzzle! Great job of mixing up the puzzle styles - I for one would not have come up with something like this one.

Before I get to the solution, keep in mind the next one is in just 2 days - Thursday, June 11, 7am. And it will be SoSG's first ever "Grand Slam" Puzzle! What is a Grand Slam puzzle? Stay tuned, we will explain in about 1 hour (at 7:30am).

Anyhow, on to the solution, as written by the man himself, Jose:

Since Ethier attends both seatings (and we are given the two lefties Stults and Milton), there is at least 1 pitcher and fielder at each seating. Since pitchers and fielders can’t be alone at the table, we know there are at least two pitchers and two fielders at each seating.

Now, at each table, there are eight spots and four are taken up by the two pitchers and two fielders. There are only four attendees who are not pitchers or fielders: Martin, Ausmus, Torre and Mattingly. Any seating with at least two pitchers and two fielders needs at least two of those four Dodgers to form a buffer, otherwise a pitcher will be across from or next to a fielder, so all four must attend – two at each seating.

Let’s look at our complete list of attendees:

  • Pitchers (P): Brazoban, Milton, Mota, Ohman, Stults, Troncoso
  • Fielders (F): Blake, Ethier (twice), Hudson, Kemp, Pierre (because of Ausmus)
  • Catchers (C): Ausmus, Martin
  • Managers/Coaches (M): Mattingly, Torre

Now, each seating has six spots spoken for – two pitchers, two fielders, and two manager/coach/catchers – and two spots left. The remaining two fielders and two pitchers must be allocated to these four spots. Suppose the two pitchers go to the same seating. Then at that seating, we have four pitchers, two fielders, and two manager/coach/catchers. The only way to avoid having a pitcher next to or across from a fielder is to have the four pitchers together in a square on one end and the two fielders at the other end like this:

Then, the other table must look like this:

If we have this, then the catchers have nowhere to sit, since they must sit across from a pitcher. So, the two remaining pitchers must go to different seatings, and both seatings must look like this:

or this:

Pierre has to sit next to Ausmus. Since Pierre is a lefty and Ausmus is a righty, Pierre must sit on his left (row 1, column 3). So the seating that Ausmus and Pierre attend must look like the second of the two choices.

Ethier, a lefty, sits in the same seat both times. He can’t sit in Pierre’s seat, so he must sit in column 4. He can’t sit in row 2 because then, in both seatings, everyone sitting to his left would have to be a lefty too, and the last seating only has at most three lefties. So he must sit in row 1, column 4.

We know that Blake is a fielder and sits in the same row as Ethier. Therefore, Blake must also sit in row 1, column 3 (same seat as Pierre, but in the seating Pierre doesn't attend.) Since Blake and Pierre are both fielders, we know that both seatings look like this:

Mattingly is a lefty, and so the pitcher on his left must be Ohmans, the only LHP left. The last seating already has two lefties (Milton and Ethier), and can’t have more than 3, so Mattingly and Ohmans must attend the first seating. This means Torre attends the last seating, and therefore Brazoban attends the first. Now we have:

Both catchers are right-handed, and so, in the last seating, Troncoso’s chicken dance would disturb his neighbor if he sat in row 1, column 1. So Mota, the only pitcher left, must sit there while Troncoso takes the spot in row 2, column 2.

Since Blake sits on the same side (row) as Ethier, we know that one seating has Ausmus and Pierre in the two remaining spots in the top row and the other has Martin and Blake. Also, since Mota attends the last seating, Hudson must be at the first seating. Row 1, column 3 is taken up by either Pierre or Blake in both seatings, so Hudson must sit in the bottom right corner. Then, the only spot left for Kemp is the bottom right corner of the last seating and we have:

Pierre must attend the last seating to make three lefties, and now we have:

Now, for part 2: Who did the Dodgers at the first seating make a toast to? Let’s look at their first names:

Casey
Orlando
Russell
Yhency
Will
Andre
Don
Eric

Hmm, how about the last names of the Dodgers at the last seating?

Milton
Ausmus
Torre
Troncoso
Kemp
Ethier
Mota
Pierre

Congratulations to QuadSevens, Mr Customer, Fanerman, Loney Fan, UBragg, Keven C, Awesome, and Berkowit28 for solving it correctly. Updated PCS Ranking to be posted soon.

Next Off-day puzzle in 2 days - Thursday, June 11, 7am. See you there!

12 comments:

jose said...

Glad you enjoyed it, SoSGers. Hope it wasn't too frustrating.

I just realized I skipped a step in the explanation - the paragraph that starts with "Pierre" should read:


Pierre has to sit next to Ausmus. Since Pierre is a lefty and Ausmus is a righty, Pierre must sit on his left (row 1, column 3). So the seating that Ausmus and Pierre attend must look like the second of the two choices.

Ethier, a lefty, sits in the same seat both times. He can’t sit in Pierre’s seat, so he must sit in column 4. He can’t sit in row 2 because then, in both seatings, everyone sitting to his left would have to be a lefty too, and the last seating only has at most three lefties. So he must sit in row 1, column 4.

We know that Blake is a fielder and sits in the same row as Ethier. Therefore, Blake must also sit in row 1, column 3 (same seat as Pierre, but in the seating Pierre doesn't attend.) Since Blake and Pierre are both fielders, we know that both seatings look like this:

Eric Karros said...

corrected.

berkowit28 said...

It was a great puzzle - very intricate, very logical.

For the first time in two years, I actually got up in time to start right at 7:00 pm. Unfortunately, my mind was still fogged, or something, and I copied down "[Eric] Milton" from table 2 as "Martin". That set me on a wild goose chase that could not be solved. I never even noticed that one of the questions here (about whether the three lefties included Milton or not) specifically referred to "Milton".

When I saw my mistake in the evening, I managed to get the answer in under 3 hours (including my drive home). Yack. I might have been first (though Quad did it faster, starting later). Oh well - I'm glad I got it in time, anyway. Very satisfying puzzle. I'm no good at anagrams, so I'm sort of surprised I even got step 2.

rbnlaw said...

As soon as I saw this puzzle I knew I'd spend my day reading Dodger Thoughts and waiting for updates.

And EK says grammar makes his head hurt. Sheesh.

Steve Sax said...

rbnlaw: As soon as I saw this puzzle I knew I'd spend my day reading Dodger Thoughts and waiting for updates.

I don't get it--do they update DT with SoSG puzzle hint news? Or are these separate items?

Mr. LA Sports Fan said...

I'm getting tired of these brain-solving puzzles. I'm running out of paper.

QuadSevens said...

This puzzle was frustrating at first, then a lot of fun once I got it figured out. Great job on making this puzzle Jose!

Eric Karros said...

I declare SoSG's first reader-written puzzle a success! Now I'm curious to see what kind of puzzle other folks like Quad, Mr Customer, J Steve, etc, would come up with...

QuadSevens said...

This puzzle must have taken a good amount of time to come up with. I'd have to rack my brain to come up with something this good.

Neeebs said...

I'm much better at Limericks.


There once was a reader named JOSE

Who came up with a puzzle yesterday

The puzzle racked my brain to wit.

So finally I just had to quit.

And SOSG commended him with a big Horray!

rbnlaw said...

Funny Sax. Funny.

No they didn't. Wasted a whole day waiting.

jose said...

Thanks Quad, glad it turned out to be not impossible and not too easy - it was hard to tell when I came up with it.