ASF Spoilers Thread

[bellenuit]

The answer is yes. Anna must either be married or unmarried either way the answer will become yes.

Correct. I was one of the 80% that got it wrong. I was too quick to jump to conclusions.

 

[cynic]

Correct. I was one of the 80% that got it wrong. I was too quick to jump to conclusions.

I am guessing that many would have opted for unable to determine. That was my first inclination before examining it a bit more thoroughly.

 

[pixel]

Try it this way:

A dozen, a gross, and a score
increased by three square roots of four,
divided by seven,
plus five times eleven
will give you nine squared, nothing more.

PS: It's true 182 / 7 = 26; add 55; result is 81

 

[bellenuit]

Try it this way:

A dozen, a gross, and a score
increased by three square roots of four,
divided by seven,
plus five times eleven
will give you nine squared, nothing more.

PS: It's true 182 / 7 = 26; add 55; result is 81

Yep, great effort and almost identical to the one I read in the paper that posed the problem (see below).



A dozen a gross and a score
Plus three times the square root of four
Divided by seven
Plus five times eleven
Is equal to nine squared, and no more.

 

[pixel]

Yep, great effort and almost identical to the one I read in the paper that posed the problem (see below).

View attachment 65638

A dozen a gross and a score
Plus three times the square root of four
Divided by seven
Plus five times eleven
Is equal to nine squared, and no more.

well, the similarity isn't surprising.
Once you have the "names" of the first 3 numbers lined up, the rest falls more or less in place.

... it probably helps that I've dabbled in composing and translating Limericks for fun.
couple examles:

An investor, betrayed by his broker,
cracked his skull with a brass fire poker.
His defense was "This clown
just kept averaging down.
Now I'm broke, thanks to that stupid Joker."

A goldie's share trading was halted
when continuous disclosure was faulted.
ASX asked "We're troubled
that your price more than doubled.
Could it be that your drill hole was salted?"

Our new Prime Minister, Ms Julia,
Implored indies "Please let me rule ya!
I'll swear by your lives,
There won't be no knives
Allowed in our Chambers to cool ya!"

 

[cynic]

Well, I am now glad that I didn't complete my answer to that one. What I had forming was somewhat creative but terribly clunky by comparison.

Well done pixel!

 

[keithj]

I'm sure the people here will solve this in an instant !

'No winning answers yet' to British spy agency's Christmas puzzle challenge

http://www.abc.net.au/news/2016-01-...o-british-spy-agency-christmas-puzzle/7127758

Complete puzzle here http://www.gchq.gov.uk/press_and_media/news_and_features/Pages/Directors-Christmas-puzzle-2015.aspx

 

[skc]

A shadowy puzzle

You have a light source that throws light out in all directions. You are to construct a closed room around that light source using straight perpendicular walls which meet each other at corners. The shape you end up with must be such that every wall is either partly or fully shaded from the light source (e.g. every wall of the room is either partly or fully shaded from the light source by other walls in the room).

There is no limit to the number of walls you require to do this, but those that can do it with the least number of walls are the best solutions.

There is no trick to the question. We are just looking for a two dimensional shape (as if looking down from above). You assume the position of the light source is fixed and design your shape from that, so you are NOT required to produce a shape that subsequently must have all walls partly/fully shaded if the light source is move to another position. Also, it is a continuous shape, where the walls join each other at their ends only with the first wall meeting up with the last to form a complete closed shape with the light source within it.

Something like this? Not drawn perfectly but the idea is to have a deep enough nook at each corner that will cast a shadow on each wall.

 

[bellenuit]

Something like this? Not drawn perfectly but the idea is to have a deep enough nook at each corner that will cast a shadow on each wall.

View attachment 65765

That is a correct answer, but can be done with less walls. I hope I haven't confused people by describing the walls as perpendicular. I meant perpendicular to the ground, not perpendicular to each other. They do not have to meet at 90 degree angles. The reason I specified perpendicular (to the ground) was that some people who were attempting to solve the problem when it was posed in the paper I read had asked if the walls could be at other than 90 degrees to the ground and such solutions were excluded by the problem poser. I will update the problem with this info.

 

[keithj]

That is a correct answer, but can be done with less walls.

I can't see it being done with less than 6 walls.

 

[bellenuit]

I can't see it being done with less than 6 walls.View attachment 65769

The one on the right is what I also came up with and many who replied to the newspaper column had the same or variations of it. That has 6 sides. The poser claims that it can be done with 5 walls, but hasn't produced a 5 walled solution yet. If he does I will post it here.

 

[skc]

That is a correct answer, but can be done with less walls. I hope I haven't confused people by describing the walls as perpendicular.

That was how I interpreted the question.. Doesn't really change much and still a fun exercise nonetheless.

The one on the right is what I also came up with and many who replied to the newspaper column had the same or variations of it. That has 6 sides. The poser claims that it can be done with 5 walls, but hasn't produced a 5 walled solution yet. If he does I will post it here.

Might give this a shot when I have time.

 

[keithj]

The poser claims that it can be done with 5 walls, but hasn't produced a 5 walled solution yet.

Logically, the best possible solution.....

...can only have walls of 2 types -

1. A wall must either shade another wall AND be in complete shade itself
2. or a wall must be partially shaded by another wall.

In an optimal solution, a wall can only shade a max of 1 other wall. (Otherwise the 2 walls it shades can be straightened to make a single partially shaded wall)
In an optimal solution, 2 adjacent completely shaded walls can be straightened to into 1 wall.

Therefore the wall types must alternate....

...therefore there must be an even number of walls.
So I'd be impressed to see a 5 walled solution.

 

[bellenuit]

Logically, the best possible solution.....

...can only have walls of 2 types -

1. A wall must either shade another wall AND be in complete shade itself
2. or a wall must be partially shaded by another wall.

In an optimal solution, a wall can only shade a max of 1 other wall. (Otherwise the 2 walls it shades can be straightened to make a single partially shaded wall)
In an optimal solution, 2 adjacent completely shaded walls can be straightened to into 1 wall.

Therefore the wall types must alternate....

...therefore there must be an even number of walls.
So I'd be impressed to see a 5 walled solution.

You are correct. The minimum is 6. The original poser tweeted "A prize for whoever can do it with 5 walls". Since prizes are never offered for his puzzle answers, it seems that was made jokingly, but obviously p***ed off those who took it seriously (I being one, as I have wasted a lot of time on a 5 solution subsequent to working out the 6 wall solution). He has some interesting puzzles but is terrible in formulating them, usually only getting it right after several readers point out obvious mistakes in his puzzles.

 

[cynic]

 

[bellenuit]

View attachment 66065

Well done! That's what I got too.

Cynic, do you do Kakuro? I do Sudoku daily and can usually get the most difficult within an hour (not the above one, which took me about 4 hours). However, with Kakuro, I struggle on the simplest categories, rarely finishing what The West Australian categorise as Gentle or Moderate. They also have Tough and Diabolical, which I don't even attempt.

I have read all the tips and techniques that I have found on the internet, but still have huge difficulty solving them.

How do you fare on Kakuro if you do them, particularly in comparison to Sudoku?

 

[cynic]

Well done! That's what I got too.

Cynic, do you do Kakuro? I do Sudoku daily and can usually get the most difficult within an hour (not the above one, which took me about 4 hours). However, with Kakuro, I struggle on the simplest categories, rarely finishing what The West Australian categorise as Gentle or Moderate. They also have Tough and Diabolical, which I don't even attempt.

I have read all the tips and techniques that I have found on the internet, but still have huge difficulty solving them.

How do you fare on Kakuro if you do them, particularly in comparison to Sudoku?

This particular puzzle took me a similar amount of time to yourself. I somehow kept making the same mistake part way into it and having to start afresh. I was almost comvinced that there was an error with the actual puzzle. Luckily I persevered with it and finally got past the glitch in my workings.

I have only done a small handful of kakuro over the years, but, the one's I've encountered weren't particularly difficult and would probably fall into the gentle category.

Edit: forgot to mention that by comparison I do find kakuro easier than the harder classes of sudoko problems.

 

[cynic]

The larger quarter circle has double the radius of each of the two half circles. So the total surface are of the quarter circle works out to (pi x (2r)^2)/4) = pi x (r^2). Each of the two identical half circles works out to (pi x (r^2))/2.

So the two half circles would add up to the same area as the quarter if only they didn't overlap. The area taken up by these semicircles is equal to 2 x pi x (r^2) - (area of overlap). Hence the residual of the quarter circle has an area equal to (pi x (r^2)) - (((2 x pi x (r^2)/2) - (area of overlap)) = (area of overlap). Ie. The blue area must equal the red area.

 

[bellenuit]

The larger quarter circle has double the radius of each of the two half circles. So the total surface are of the quarter circle works out to (pi x (2r)^2)/4) = pi x (r^2). Each of the two identical half circles works out to (pi x (r^2))/2.

So the two half circles would add up to the same area as the quarter if only they didn't overlap. The area taken up by these semicircles is equal to 2 x pi x (r^2) - (area of overlap). Hence the residual of the quarter circle has an area equal to (pi x (r^2)) - (((2 x pi x (r^2)/2) - (area of overlap)) = (area of overlap). Ie. The blue area must equal the red area.

Yes, that turned out to be quite easy. My method was pretty much the same as yours.

 

[cynic]

1 to D1
T to B4
2 to A1
3 to B1
T to D4