nicow wrote:

The problem must be the power in a 4 cell cage. An overflow occurs at bad programming.

I'm thinking the program exits without finding a solution.

The values submiited are actually MAXINT (so probably some initial "impossible" value)

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pnm wrote:

oldmathtchr wrote:

I've been seeing this for several weeks. Every day "chip" solves all the puzzles in about 5 seconds. I thought it was something you had put in to check puzzles.

Today his automatic solver failed on the difficult 7x7 though, submitting a solution with only the (given) 5 and 3 filled in,

and a number over 2 milliard (billion for those in the US) as a solution for every other cell..

Ha ha ha!

The problem must be the power in a 4 cell cage. An overflow occurs at bad programming.

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rafaelhoukes wrote:

There's one thing I still wonder about these rankings by year:

because they seem to be only counting the points for the regular puzzles, wouldn't it be more logical to move them to the 'regular puzzle' rankings?

Then we could maybe make another ranking counting all points, and we would still have the regular rankings for the die-hard solvers

The regular rankings include bonus puzzles but don't include extra puzzles. The opposite applies under the year to date rankings.

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Then we could maybe make another ranking counting all points, and we would still have the regular rankings for the die-hard solvers

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pnm wrote:

oldmathtchr wrote:

I've been seeing this for several weeks. Every day "chip" solves all the puzzles in about 5 seconds. I thought it was something you had put in to check puzzles.

Today his automatic solver failed on the difficult 7x7 though, submitting a solution with only the (given) 5 and 3 filled in,

and a number over 2 milliard (billion for those in the US) as a solution for every other cell..

April Fools!

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eclipsegirl wrote:

Today, I lost my shared place at the top of the rankings. (I had in the past because I didnt do the EXTRA or finish a 12 x 12 on the day it was published, and get caught up a day or two later. Through March 30, I was able to compete

Today, I couldnt get all the dailies done because I did not realize this was when Europe “sprung ahead” and I thought I had one more hour to compete the puzzles and I did not have access to the internet in the morning when I susally do many of the puzzles andnternet service in the developing world.

:sad:

That's very unlucky (and very frustrating). Having said that, it kind of takes the pressure off and I hope you enjoy the rest of your vacation without having to worry about trying to maintain your streak.

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oldmathtchr wrote:

I've been seeing this for several weeks. Every day "chip" solves all the puzzles in about 5 seconds. I thought it was something you had put in to check puzzles.

Today his automatic solver failed on the difficult 7x7 though, submitting a solution with only the (given) 5 and 3 filled in,

and a number over 2 milliard (billion for those in the US) as a solution for every other cell..

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Today, I couldnt get all the dailies done because I did not realize this was when Europe “sprung ahead” and I thought I had one more hour to compete the puzzles and I did not have access to the internet in the morning when I usally do many of the puzzles (bad internet service in the developing world).

:sad:

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michaele wrote:

Quote:

very surprised and still wondering why (this will require me some time in the future, ... ). The michaele's considerations on cell h7 (which is the only cell with four possibilities, [5,6,7,8], are very interesting) (we could also start by f7 where the 5 is possible only once while 7 and 8 have both three possibilities, ... ), but my confusion is mainly related to the global concept of permutations in these situations.

I think the quickest way to find the number of solutions is to start with the cell with the most pencil marks, if you start by looking at f7 (5 or 6) it does not help very much trying either 5 or 6 leaves several cells unsolved.

My previous post was a bit vague, so below is a step by step method for finding all solutions.

The steps to determine the number of possible solutions are:

1. Find the cell with the most values, cell h7 has 4 values (5,6,7,8) this indicates that there are at least 4 possible solutions.

2. Try each value:

h7=5 gives a solution

h7=6 gives a solution

h7=7 does not provide a single solution

h7=8 does not provide a single solution

3. There is no solution for 7 & 8, so do the same as steps 1&2 to find the number of possible solutions:

h7=7 cells h4 and d7 have 3 values (this indicates a minimum of 3 solutions if h7=7), choose one of these cells (it does not matter which cell you use), and try each value, this will give 3 solutions

h7=8 several cells have 2 values (this indicates a minmum of 2 solutions if h7=8), choose one of them and try each value, this will give 2 solutions.

This give a total of 7 solutions.

This is a little different to what I posted before, previously I had made testing h7=7 more complicated than it needed to be.

This is quite quick to do, it seems like a lot, but there are only 14 cells with pencil marks so each possible value is quick and easy to test.

Thank you, michaele, your steps look very well. BTW, I assume that when you say "

Best, clm.

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eclipsegirl wrote:

ddarthez wrote:

Just a question for the time pressed users among us (like me at the moment): is there any time limit to submit solutions for the 10 year anniversary?

Patrick has never had a time limit on other book puzzles that I am aware of

That's correct, there's no time limit.

So you can submit them all when the site turns 20

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ddarthez wrote:

Just a question for the time pressed users among us (like me at the moment): is there any time limit to submit solutions for the 10 year anniversary?

Patrick has never had a time limit on other book puzzles that I am aware of

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Quote:

very surprised and still wondering why (this will require me some time in the future, ... ). The michaele's considerations on cell h7 (which is the only cell with four possibilities, [5,6,7,8], are very interesting) (we could also start by f7 where the 5 is possible only once while 7 and 8 have both three possibilities, ... ), but my confusion is mainly related to the global concept of permutations in these situations.

I think the quickest way to find the number of solutions is to start with the cell with the most pencil marks, if you start by looking at f7 (5 or 6) it does not help very much trying either 5 or 6 leaves several cells unsolved.

My previous post was a bit vague, so below is a step by step method for finding all solutions.

The steps to determine the number of possible solutions are:

1. Find the cell with the most values, cell h7 has 4 values (5,6,7,8) this indicates that there are at least 4 possible solutions.

2. Try each value:

h7=5 gives a solution

h7=6 gives a solution

h7=7 does not provide a single solution

h7=8 does not provide a single solution

3. There is no solution for 7 & 8, so do the same as steps 1&2 to find the number of possible solutions:

h7=7 cells h4 and d7 have 3 values (this indicates a minimum of 3 solutions if h7=7), choose one of these cells (it does not matter which cell you use), and try each value, this will give 3 solutions

h7=8 several cells have 2 values (this indicates a minmum of 2 solutions if h7=8), choose one of them and try each value, this will give 2 solutions.

This give a total of 7 solutions.

This is a little different to what I posted before, previously I had made testing h7=7 more complicated than it needed to be.

This is quite quick to do, it seems like a lot, but there are only 14 cells with pencil marks so each possible value is quick and easy to test.

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paulv66 wrote:

michaele wrote:

I also thought that 7 possible solutions seemed an unlikely number, I thought it would have to be an even number.

So I solved it again and I found 7 solutions, even though I was sure it would not be 7.

I did not want to give too much away for other users who might not have solved the puzzle yet.

So I solved it again and I found 7 solutions, even though I was sure it would not be 7.

I did not want to give too much away for other users who might not have solved the puzzle yet.

Hi Michaele

I see that you arrived at the same conclusion as me.

I take your point about not giving away too much so as not to spoil it for other users. I guess my earlier posts were a bit more specific than they needed to be, but I figured that anyone who had got that far had essentially solved the puzzle to all intents and purposes.

Apologies if I went too far.

Paul

Thank you, michaele, for your cooperation. I have seen in detail the Paul's spreadsheet and it's correct, it contains an additional solution that I had initially missed (due to swapping of numbers 5 and 6 in e7-e8 and h7-h8 ... ).

So the number of solutions must be 7 (I have played with the options but I do no find any other). You both paulv66 and michaele are right (great analysts ). However, as you both, I am very surprised and still wondering why (this will require me some time in the future, ... ). The michaele's considerations on cell h7 (which is the only cell with four possibilities, [5,6,7,8], are very interesting) (we could also start by f7 where the 5 is possible only once while 7 and 8 have both three possibilities, ... ), but my confusion is mainly related to the global concept of permutations in these situations.

Best, clm.

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michaele wrote:

I also thought that 7 possible solutions seemed an unlikely number, I thought it would have to be an even number.

So I solved it again and I found 7 solutions, even though I was sure it would not be 7.

I did not want to give too much away for other users who might not have solved the puzzle yet.

Hi Michaele

I see that you arrived at the same conclusion as me.

I take your point about not giving away too much so as not to spoil it for other users. I guess my earlier posts were a bit more specific than they needed to be, but I figured that anyone who had got that far had essentially solved the puzzle to all intents and purposes.

Apologies if I went too far.

Paul

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