Here's a sample:
Thinking Entirely Too Much About Sudoku
Monday, December 3, 2018
Something new
As you can see, I ran out of steam. But my interest has been sparked again by a new video blog called Cracking the Cryptic over on youtube, and I highly recommend it to anyone who enjoys not only sudoku, but other puzzles as well.
Here's a sample:
Here's a sample:
Sunday, November 17, 2013
Zone defense
I don't usually spend a lot of time looking for patterns when I'm solving sudokus. Bifurcation works much faster and easier than x-wings or y-wings or any of the fishy patterns. But there's one kind of pattern that's worth looking for -- in order to avoid it!
You see, if the puzzlemaker did it right, a puzzle will have a single solution. That means that a pattern where four squares in the same zone that are the corners of a rectangle can have only two symbols between them if at least one of them is a given.
1|4
4|1 for example, would be easy to reverse and become
4|1
1|4 unless at least one corner can't be changed.
Of course, the corners aren't generally next to each other, since the symbols have to fall into different squares. But it's fairly easy to see that you can go from single symbols to twoshares and apply the same principle.
So, if you have two spaces such that 1/4 | | 1/4 in some triad, then there cannot be an identical parallel anywhere within the zone. You can either use this to decide you are going to use the 2 in a configuration where you've got a parallel that looks like 1/2/4 | | 1/4, or you can find a pair of 1s that parallel and know that there can't be a 4 in either space joining them.
That's short enough and easy enough that even I'll take a minute to check to see if I can use the pattern.
You see, if the puzzlemaker did it right, a puzzle will have a single solution. That means that a pattern where four squares in the same zone that are the corners of a rectangle can have only two symbols between them if at least one of them is a given.
1|4
4|1 for example, would be easy to reverse and become
4|1
1|4 unless at least one corner can't be changed.
Of course, the corners aren't generally next to each other, since the symbols have to fall into different squares. But it's fairly easy to see that you can go from single symbols to twoshares and apply the same principle.
So, if you have two spaces such that 1/4 | | 1/4 in some triad, then there cannot be an identical parallel anywhere within the zone. You can either use this to decide you are going to use the 2 in a configuration where you've got a parallel that looks like 1/2/4 | | 1/4, or you can find a pair of 1s that parallel and know that there can't be a 4 in either space joining them.
That's short enough and easy enough that even I'll take a minute to check to see if I can use the pattern.
Saturday, August 31, 2013
numbers numbers numbers
Okay, so I haven't been around for a while. But while I was gone I spent some time (a lot of time) working out the interactions by hand. And then inputting them into a google doc. And getting some of the stats. The first major result being this:
Now these are the numbers overall, rather than broken down by how crowded the squares are. That's why you see some hits and pairs in the "E"s. Those are the results for squares that have eight or seven knowns respectively.
(As a reminder, A=four influences, B=three influences, C=two influences, D=one influence, E=no influence. A "hit" is a deduction.)
Now the interesting thing is at the end, where I've collected the results for having two influences within the same zone affecting a square. You'll notice that it's a good bit more powerful for getting hits than the Cs are, where the influences can be coming from two different zones.
Overall, the stats look something like this:
This makes it seem like hits are easier to get than pairs, but since it includes squares that solve themselves, it probably doesn't represent the way we really solve the puzzles. I'll need to work out the stats for each level of crowdsquare to really see where the drop off is for results.
2376 | 7132 | 4692 | 2440 | 7208 | 2366 | 2954 | 3154 | 300 | 746 | 527 | 9 | 40 | 12274 | 8262 | 3626 |
100% | 100% | 66% | 34% | 100% | 33% | 41% | 100% | 10% | 24% | 100% | 2% | 8% | 100% | 67% | 30% |
A Valid/Hits | B Valid | B Hits | B Pairs | C Valid | C Hits | C Pairs | D Valid | D Hits | D Pairs | E Valid | E Hits | E Pairs | 2zone Valid | 2 Zone Hits | 2z pairs |
Now these are the numbers overall, rather than broken down by how crowded the squares are. That's why you see some hits and pairs in the "E"s. Those are the results for squares that have eight or seven knowns respectively.
(As a reminder, A=four influences, B=three influences, C=two influences, D=one influence, E=no influence. A "hit" is a deduction.)
Now the interesting thing is at the end, where I've collected the results for having two influences within the same zone affecting a square. You'll notice that it's a good bit more powerful for getting hits than the Cs are, where the influences can be coming from two different zones.
Overall, the stats look something like this:
25823 | 5462 | 21% | 20361 | 79% | 4416 | 22% | 9751 | 48% | 6172 | 30% | 15923 | 78% |
total | Invalid | percent invalid | Valid 1 | percent valid | None | none as percent of valid1 | Hits | hits as percent of valid1 | Pairs | Pairs as percent of valid1 | all (calc) | all as percent of all valid1 |
This makes it seem like hits are easier to get than pairs, but since it includes squares that solve themselves, it probably doesn't represent the way we really solve the puzzles. I'll need to work out the stats for each level of crowdsquare to really see where the drop off is for results.
Wednesday, June 12, 2013
There will be a slight delay...
Not that anyone is reading this, but it's going to take a bit longer to get to the next bit than I hoped.
So now we're getting a handle on this.
I've come up with a pattern that I can use to collect my data.
There are forty-nine interactions which I can show with this format. The first -- no influences -- is going to be expressed by the color of the blanks or Xs in the square where we'll show the crowd patterns. The single influences and double influences within a zone will be shown by the segments in the zones. I.E., if I'm looking at the effect of the influence "tm" I'll do it by coloring the lettering in the vertical segment that has a "t" and an "m" - in the horizontal zone.
In this particular example we're looking at the empty crowdsquare. There are no effects from the crowd, and none from any single influence or pair of influences (in black). Three influences always produce a pair (in green), and four influences always produce a hit (in blue). There are no invalid combinations (so nothing in red.)
The ooo's at the end of the zones will be how we'll identify crowdsquares. I'll explain that plan tomorrow.
_
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_
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_
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t
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-
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-
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t
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t
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-
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o
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_
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_
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_
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-
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m
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-
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m
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-
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m
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o
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_
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_
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_
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-
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-
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b
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-
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b
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b
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o
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l
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-
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-
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tl
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tc
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tr
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tml
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tmc
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tmr
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-
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c
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-
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ml
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mc
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mr
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tbl
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tbc
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tbr
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-
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-
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r
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bl
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bc
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br
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mbl
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mbc
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mbr
|
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l
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c
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-
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tmlc
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tmlr
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tmcr
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tlc
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mlc
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blc
|
|
l
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-
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r
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tblc
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tblr
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tbcr
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tlr
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mlr
|
blr
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-
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c
|
r
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mblc
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mblr
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mbcr
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tcr
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mcr
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bcr
|
|
o
|
o
|
o
|
There are forty-nine interactions which I can show with this format. The first -- no influences -- is going to be expressed by the color of the blanks or Xs in the square where we'll show the crowd patterns. The single influences and double influences within a zone will be shown by the segments in the zones. I.E., if I'm looking at the effect of the influence "tm" I'll do it by coloring the lettering in the vertical segment that has a "t" and an "m" - in the horizontal zone.
In this particular example we're looking at the empty crowdsquare. There are no effects from the crowd, and none from any single influence or pair of influences (in black). Three influences always produce a pair (in green), and four influences always produce a hit (in blue). There are no invalid combinations (so nothing in red.)
The ooo's at the end of the zones will be how we'll identify crowdsquares. I'll explain that plan tomorrow.
Tuesday, June 11, 2013
Some interactions between crowdedness and influences.
Let's start with one of the simplest sets of interactions. We'll look at squares with three spaces left and a single influence.
For each influence, I've color coded the result.
X
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X
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X
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t
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t
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|||||
X
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X
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X
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m
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m
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|||||
b
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b
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||||||||
l
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c
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r
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|||||||
l
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c
|
r
|
For each influence, I've color coded the result.
Black is no effect
Green is pair
Blue is deduction
Red is impossible/invalid
So in this first chart we can see that out of six occasions, three have a result and one can never happen in the game. So 3/5ths or 60% of the time we get at least a pair.
But of course, the "crowd" in the square mighte be in a different configuration.
How about something that looks like this:
X
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X
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X
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t
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t
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|||||
X
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X
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m
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m
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||||||
X
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b
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b
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|||||||
l
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c
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r
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|||||||
l
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c
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r
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Here, our results are different. There are no invalid interactions, and while two of the interactions have no result, two of them (b) will give us a hit, not just a pair. If 2/3rds of our interactions give us a result here, that's 66%.
We can create different charts with three spaces, but are they functionally equivalent to this one?
Nothing to do but test it.
X
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X
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t
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t
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||||||
X
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X
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m
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m
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||||||
X
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X
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b
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b
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||||||
l
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c
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r
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|||||||
l
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c
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r
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And clearly the answer is no. We're looking at one result with no influence, but we've raised our pairs to four and reduced our hits to one.
5/6ths into percentages is more than my brain is up to just now. But I've got one more chart for us to investigate.
This time, we'll break things up a bit.
X
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X
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t
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t
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||||||
X
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X
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m
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m
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||||||
X
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X
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b
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b
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||||||
l
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c
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r
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|||||||
l
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c
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r
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And we get a similar result. One dud, one hit, and four pairs.
What makes this arrangement functionally equivalent to the one above it?
Something to think about.
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