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Archive for March, 2009

4024: Waterloo’s Symmetry (or Going for a Spin)

Posted by Listen With Others on 27 March 2009

The junior 8 X 8 team was faced with a new challenge this week. We’re nonplussed when faced with a carte blanche by Salamanca or Zebedee, in the Listener or in the Crossword Club’s ‘Crossword’. Perhaps it is the cold solving that separates the men from the boys (you experts from us learners!)

Waterloo’s clues seemed very fair to us and the wordplay slightly more decipherable than we are coming to expect in the Listener. We had the good fortune to cold solve 5ac. RUNG, 7 ac. LEAF, 30 ac. STOT, and 32 ac. EKES, as well as about half of the other clues, including a rather surprising ELIDES that reversed SEDILE, and REWARD that reversed DRAWER (didn’t think much of the ‘pant?’ part of that clue!) but the light didn’t dawn and we reached the limit of our solving skills.

Despair almost ensued with Crossword Compiler and a vast consumption of paper failing to produce any way that the solutions would fit into a grid. With the four-letter answers in the obvious place and ALPACAS going in the only direction it could, as well as eight intersecting five-letter words coming out from the centre, and four nine-letter words fitted in, we simply could not find anywhere for the final eight-letter words.

Those four six-letter down clues were the obvious culprits. Spotting that they were down clues that could reverse across clues and serve a dual purpose was an epiphany. Sheer joy! (After a few hours of frustration.) The good sense of the word play meant that it was pleasure all the way to the finish line with TENNER going in as our final word. We didn’t like the three unches there and still don’t know what RENNE has to do with a cathedral city (Is this an English spelling of RENNES?).

Just like last week’s colouring in, there was still a monkey trick to perform. (Is it a Listener rule that there has to be a final hoop to leap through?) Filling in the bars was no problem but we had a real squabble over the orientation of the numbers with Mr Math finally prevailing and pointing out how obvious and easy it was.

We were delighted with the ingenuity of ‘An Additional Symmetry’ and solving it was a wonderful tutorial in grid structure. Thank you, Waterloo for a superb and very, very challenging puzzle (for us!) We are only hoping that next week’s Listener does not need quite so much cold solving.

Shirley Curran.

Posted in Solving Blogs | 4 Comments »

4023: Pentomino Factory by Oyler

Posted by erwinch on 20 March 2009

I found this quite the most enjoyable Listener of 2009 to date and felt disappointed when it came to an end.  It also happened to be one of the most straightforward of puzzles, which just shows that difficulty is not everything. 
I first encountered pentominoes in Martin Gardner’s Mathematical Puzzles and Diversions (1959), which was a Christmas present circa 1966, but it lay forgotten on my shelves until Listener No.3685, Pentad by Tangent (Aug 2002), where we had to arrange the twelve shapes into an 8×8 square that had a 2×2 blocked centre (not a 2×3 hole as it says on the Listener site) with the cross already in place.   This puzzle totally enthralled me and it took five days to solve – I even made my own set of pentominoes.  With Gardner’s book, I went on to look at the rectangles including the most difficult to solve, the 3×20, which has just two distinct solutions ignoring rotations and reflections.  Those took me an absolute age to find but now you can see them in an instant on the Net.  You can also see in an instant that there are no less than 1010 distinct solutions for our 5×12 rectangle here so no point in looking for a shortcut there!
As the title hints, the key to this was factorisation and especially prime factorisation to start.  The 12th shape (B=11) could be placed immediately by finding those products that had eleven as a factor:
Row 1: 1497375000 = 11 × 11 × 11 × 1125000
Row 2: 14701500 = 11 × 11 × 121500
Col 1: 4356 = 11 × 11 × 36
Col 2: 12474 = 11 × 1134
Col 3: 15246 = 11 × 11 × 126
A product with zero at the end indicated multiples of five or ten but the single zero at Col 4 (3150) was too far from most of the zeros, Col 12 (36000), to be connected so must indicate multiples of five with the tens over towards Col 12:
Row 1: 1125000 = 10 × 10 × 10 × 5 × 5 × 5 × 9
Row 2: 121500 = 10 × 5 × 5 × 486
Col 4: 3150 = 5 × 5 × 126
Col 10: 640 = 10 × 64
Col 11: 8640 = 10 × 864
Col 12: 36000 = 10 × 10 × 10 × 36
The 3×3 square where the 1st shape (0) would appear was also clearly indicated so the grid now looked like this: 
Continuing in the SW corner, it was soon apparent that two, seven and nine would appear here in some order with Cols 2, 3 & 4 each containing one seven.
Col 9: 32 = 2 × 2 × 2 × 2 × 2 or 8 × 4 × 1 × 1 × 1 but it had to be the latter since the twos were accounted for elsewhere.
The remaining factorisation of Row 1: 9 = 3 × 3 × 1 and Col 12: 36 = 6 × 6
The initial completion of the grid was then far from a formality but no major problems were encountered.
The final stage was the colouring.  The four colour theorem excludes meetings at a point but the preamble told us that these had to be considered here as is also usual with maps.  The states of Utah, Colorado, New Mexico and Arizona all meet at a point but I know of nowhere on earth where five states do so although it is not inconceivable.  We have no four shapes meeting in our grid so a good starting point is to colour any one shape and then go around it alternately using two other colours.  A fourth colour is needed where a state has no coastline but an odd number of neighbours such as West Virginia with five.  However, our shapes each have a ‘coastline’ so the three colours sufficed:
Usually I have no idea how these numerical puzzles are set but the construction here is beautifully simple and I am able to extend the fun:
Here I looked at a different position for the numbers thinking that having the five-, ten- and two-shapes adjacent might make things more difficult.  In fact it turned out to be easier than Oyler’s grid.  The 7-shape could be entered immediately since we had one seven in row 2 and there is only one shape with that 1,2,1,1 arrangement.  With that as a base I was able to creep slowly to the right.
Would the puzzle have been too easy had twelve been used in place of zero?  Not really, the 11-shape could also be placed immediately but I noticed little difference in difficulty.  With the original puzzle, I had thought that placing the 1-shape could be a problem but you just had to look at the low products such as the 27 for Col 12 here.
Here I used Oyler’s numbers for the shapes except that zero and seven were swapped.  Note that we have to use four colours for the two reasons stated above: four shapes meeting at a point and the 10-shape having no ‘coastline’ with an odd number of neighbours (5), as indeed does the 3-shape. Note also that there are no solutions with the straight pentomino in any column other than 1 or 12.  This was a different sort of puzzle requiring a lot of trial and error using my home-made pentominoes but it did not prove to be too difficult.  I wonder if this is a unique solution?
Skimming through all 1010 solutions on the Net, I got the impression that those with one instance of four shapes meeting at a point occurred with the greatest frequency followed by those with none then two and finally three.  Three turned out to be very rare and the above may be the only one but I did not examine them all too closely.
Final Word
I wonder if we will get to see pentominoes again in the Listener – I certainly hope so. If considered one-sided, the six asymmetrical shapes become two so we have 18 shapes or 90 squares to play with. The five tetrominoes, seven if considered one-sided, could form a puzzle such as the 16-piece jigsaw of a chessboard we had in the sixties – we could never solve it.
Puzzles of this quality are a great recruiter to the Listener cause so many thanks to Oyler and I look forward to his next.
Footnote (added Dec 09)
Here are the two 3×20 solutions:
For me it is rather sad to find these readily available on the Net since I can’t see anyone ever again making their own set of pentominoes to then spend hours searching for the two solutions.
When it came to finding them, I knew that they couldn’t be split into two smaller rectangles so that any straight edge of three squares (1-, 3-, 8- & 12-shapes) would only appear in columns 1 or 20 and therefore the 12-shape must appear at one end. As for the other end, apart from the remaining straight edges, there were some very seductive couplings such as the 1- and 10-shapes – I wasted so much time on these.
Note that the six asymmetric shapes (3, 5, 6, 7, 10 & 11) all show opposite sides for each solution so that it would be possible to make a double-sided jigsaw of the twelve shapes with a different arrangement for each side.  We could have long, thin things such as a train on one side and snake on the other.  I would have just loved to have had such a toy as a child.

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4023: Oyler’s Pentomino Factory (or Painting by Numbers)

Posted by Listen With Others on 20 March 2009

This was the first numerical Listener puzzle the junior coffee break team had attempted. We had heard how dreadfully difficult these numerical puzzles are and were predisposed to settle down with a good book and not even look at No 4023. But we did!

Surprise, surprise! It looked do-able. We heaved a huge sigh of relief – unlike you regulars who felt that this was not a challenge. Thank you, Oyler for sparing a thought for the relative newcomers.

We had to suss out pentominoes first and Googled up the twelve that are different (not counting the rotation and reflection effects that clearly had to be considered). With a little 5 X 12 board, the less mathematical part of the team played with these and soon established that, spatially, fitting them together is no cakewalk. The Internet confirmed that there are 1200 or so possibilities! Meanwhile, Mr Math was working out the factors and sorting out possible positions.

It was obviously clear which nine squares the 0 pentomino had to occupy and which six potential pentominoes could fill that role. The 10 seemed a candidate for the top right hand corner because of all those high numbers in the row and column products and the 11 could be uniquely placed on the other side for the same reason – so now there were five 0 candidates.

The rest took time with the 5 and 3 pentominos required to lurk centrally in the top 2 or 3 rows putting 9 on the lower left, 2 and 7 causing us some concern, and some dithering about the way to fit in the multiples of 2 on the right, but we worked from that side and made it – with astonishment.

Our final challenge was that question of the colouring in. Clearly it makes judging the solutions easier if there are all those pretty colours BUT when we had put in two, could we consider the white spaces to be the third? Is the Listener so tricky? Will the winner have just two colours and white? No – that would be too silly: we decided that ‘colours’ were required and white is not, strictly speaking, a colour or, if it is as Chambers claims ‘the colour of snow’, it counts as a third one. (Incidentally, Mr Chambers hasn’t seen the ‘colour’ of the snow that has been lying in our garden since November!)

Thank you Oyler! After Salamanca’s Mazy, where we got into the maze but never even met the Minotaur and certainly couldn’t put together enough thread to get out, this was very refreshing for the less able gang.

Shirley Curran.

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Opening Time

Posted by Listen With Others on 20 March 2009

Welcome to the official relaunch of Listen With Others, and what could be more appropriate for the first day of Spring (here in the northern hemisphere)?! We have a band of bloggers ready to keep you entertained with their successes, and perhaps even failures, in solving the Listener crossword each week. I hope that we will have at least one blog per puzzle to amuse you.

Blogs will be posted at about 5pm on the Friday three weeks after publication of the puzzle. This is the day before its solution is printed in The Times, and about the same time as it appears on The Times web site. It also coincides with discussion of the puzzle on the Crossword Centre message board.

Comments are welcome from anyone … there is no need to set up a WordPress account or log in.

Moreover, if you would like to join the team of bloggers, please feel free to contact me by adding a comment to this post (to be found under the Sundries category on the left).

Best wishes,
Dave Hennings.
20 March 2009.

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Update, March 2009

Posted by Listen With Others on 9 March 2009

Unfortunately, due to other commitments, Chris Lancaster is stepping down as administrator for Listen With Others. In order that this excellent site continues, I have agreed to take it on … after all, the Crossword Database (see link on the right) just doesn’t take up enough of my time!

I hope to be up to speed on managing LWO by the middle of this month, and will shortly be contacting all those who have told Chris that they would be willing to contribute blogs, either frequent or occasional. To keep the ball rolling, I have posted a couple of submissions which Chris has forwarded to me from Shirley Curran, for which I thank her.

Before going any further, I would like to thank Chris for his outstanding contribution to the crossword community in general, and The Listener in particular. His posts here and on the Crossword Centre message board have been a source of great entertainment to me and many others, especially his sense of humour, which is very similar to my own … so beware! I know Chris hopes to be able to submit the occasional blog, but his setting of new puzzles is likely to be on hold for a time. Hopefully that will not be too long.

I hope to be able to devote as much time as necessary to ensuring that the site runs smoothly (famous last words). I am open for business pretty much 24/7 … better make that 15/6 … and will do my best to ensure that the site continues to provide pleasure to all those for whom The Listener is more than just a crossword.

Watch this space.

Dave Hennings.

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