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Congratulations on the site's 10 year anniversary!
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Posted on: Sun Feb 17, 2019 2:09 pm

Posts: 749
Joined: Fri May 13, 2011 6:51 pm
Re: Congratulations on the site's 10 year anniversary!
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.

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.

Posted on: Sun Feb 17, 2019 7:43 pm

Posts: 41
Joined: Fri Jan 13, 2017 2:51 pm
Re: Congratulations on the site's 10 year anniversary!
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?

Posted on: Mon Feb 18, 2019 1:53 am

Posts: 64
Joined: Sun Jan 31, 2016 7:52 am
Re: Congratulations on the site's 10 year anniversary!
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.

Posted on: Mon Feb 18, 2019 6:49 pm

Posts: 400
Joined: Wed Apr 16, 2014 9:20 pm
Re: Congratulations on the site's 10 year anniversary!
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

Posted on: Mon Feb 18, 2019 10:57 pm

Posts: 2565
Joined: Thu May 12, 2011 11:58 pm
Re: Congratulations on the site's 10 year anniversary!
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

Posted on: Tue Feb 19, 2019 11:11 am

Posts: 749
Joined: Fri May 13, 2011 6:51 pm
Re: Congratulations on the site's 10 year anniversary!
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 " ... if you start by looking at f7 (5 or 6) it does not help ... " you are meaning e7 (since f7 has 3 possibilities: 5,7,8).
Best, clm.

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