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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Quote:

In my opinion 7 is an strange number of solutions (perhaps 8, in this case we are missing another solution), ... .

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 wasn't actually that difficult to find the number of solutions, I solved the puzzle as much as possible, this left several cells unsolved, one cell had 4 possible values, so I tried each value. 2 of the values provided a solution, another value gave 2 possible solutions, and the other value gave 3 possible solutions, a total of 7 possible solutions. The reason that an odd number is possible is because one of the values had 3 possible solutions becuase when I tried that value in the cell it still did not give a solution, so I tried one possibile way to solve and that gave

a soluition, and when i tried the other possible way it still did not give a solution and another 2 possible solutions appeared.

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

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clm wrote:

paulv66 wrote:

Hi clm

I know that 7 seems an unlikely number, but I've set it up on a spreadsheet and gone through all possible combinations changing the 14 cells that are in question (I think we're both agreed that there is an unique solution for the other 86 cells).

While I understand your reasoning about factorials and it would appear intuitively obvious, I think the apparent anomaly arises because the effect of swapping two cells can be asymmetric. For example, in the case of this puzzle, having an 8 in d4 causes f4 to equal 6 and h4 to equal 1. This then gives unique results for 7 of the remaining 11 cells and the only remaining issue is the positioning of 7-8 in f5-f7 and h5-h7.

However, having a 6 in d4 allows 8 to be in either f4 or h4, with 1 in the other cell. When 8 is in f4, then we are left with two possibilites for row 5 - namely 5-7-8 and 8-5-7 - each of which leads to an unique solution. Similarly, when 1 is in f4, there are two possibilities for row 5 - namely 8-7-5 and 5-8-7. However, only 8-7-5 leads to a unique solution. With 5-8-7, you have 5-6 in h7-h8, which can be reversed with the 5-6 in e7-e8.

I'm happy to send you a spreadsheet containing all 7 solutions if you want to pm me your email address. I'm 99.99% sure I've exhausted all possibilities, but I'm happy to be proved wrong if you can find one that I've overlooked.

Best wishes

Paul

I know that 7 seems an unlikely number, but I've set it up on a spreadsheet and gone through all possible combinations changing the 14 cells that are in question (I think we're both agreed that there is an unique solution for the other 86 cells).

While I understand your reasoning about factorials and it would appear intuitively obvious, I think the apparent anomaly arises because the effect of swapping two cells can be asymmetric. For example, in the case of this puzzle, having an 8 in d4 causes f4 to equal 6 and h4 to equal 1. This then gives unique results for 7 of the remaining 11 cells and the only remaining issue is the positioning of 7-8 in f5-f7 and h5-h7.

However, having a 6 in d4 allows 8 to be in either f4 or h4, with 1 in the other cell. When 8 is in f4, then we are left with two possibilites for row 5 - namely 5-7-8 and 8-5-7 - each of which leads to an unique solution. Similarly, when 1 is in f4, there are two possibilities for row 5 - namely 8-7-5 and 5-8-7. However, only 8-7-5 leads to a unique solution. With 5-8-7, you have 5-6 in h7-h8, which can be reversed with the 5-6 in e7-e8.

I'm happy to send you a spreadsheet containing all 7 solutions if you want to pm me your email address. I'm 99.99% sure I've exhausted all possibilities, but I'm happy to be proved wrong if you can find one that I've overlooked.

Best wishes

Paul

Hi, paulv66, I have sent you, via pm, my email address.

I am curious about your solutions (it will take me some time), I can comment now that it has been a great surprise for me when you commence saying "

Hi clm

Sorry, I got the 1 and 6 backwards in my post above. It should say 1 in f4 and 6 in h4. I'll send on the spreadsheet.

Cheers

Paul

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paulv66 wrote:

Hi clm

I know that 7 seems an unlikely number, but I've set it up on a spreadsheet and gone through all possible combinations changing the 14 cells that are in question (I think we're both agreed that there is an unique solution for the other 86 cells).

While I understand your reasoning about factorials and it would appear intuitively obvious, I think the apparent anomaly arises because the effect of swapping two cells can be asymmetric. For example, in the case of this puzzle, having an 8 in d4 causes f4 to equal 6 and h4 to equal 1. This then gives unique results for 7 of the remaining 11 cells and the only remaining issue is the positioning of 7-8 in f5-f7 and h5-h7.

However, having a 6 in d4 allows 8 to be in either f4 or h4, with 1 in the other cell. When 8 is in f4, then we are left with two possibilites for row 5 - namely 5-7-8 and 8-5-7 - each of which leads to an unique solution. Similarly, when 1 is in f4, there are two possibilities for row 5 - namely 8-7-5 and 5-8-7. However, only 8-7-5 leads to a unique solution. With 5-8-7, you have 5-6 in h7-h8, which can be reversed with the 5-6 in e7-e8.

I'm happy to send you a spreadsheet containing all 7 solutions if you want to pm me your email address. I'm 99.99% sure I've exhausted all possibilities, but I'm happy to be proved wrong if you can find one that I've overlooked.

Best wishes

Paul

Hi, paulv66, I have sent you, via pm, my email address.

I am curious about your solutions (it will take me some time), I can comment now that it has been a great surprise for me when you commence saying "

]]>

I know that 7 seems an unlikely number, but I've set it up on a spreadsheet and gone through all possible combinations changing the 14 cells that are in question (I think we're both agreed that there is an unique solution for the other 86 cells).

While I understand your reasoning about factorials and it would appear intuitively obvious, I think the apparent anomaly arises because the effect of swapping two cells can be asymmetric. For example, in the case of this puzzle, having an 8 in d4 causes f4 to equal 6 and h4 to equal 1. This then gives unique results for 7 of the remaining 11 cells and the only remaining issue is the positioning of 7-8 in f5-f7 and h5-h7.

However, having a 6 in d4 allows 8 to be in either f4 or h4, with 1 in the other cell. When 8 is in f4, then we are left with two possibilites for row 5 - namely 5-7-8 and 8-5-7 - each of which leads to an unique solution. Similarly, when 1 is in f4, there are two possibilities for row 5 - namely 8-7-5 and 5-8-7. However, only 8-7-5 leads to a unique solution. With 5-8-7, you have 5-6 in h7-h8, which can be reversed with the 5-6 in e7-e8.

I'm happy to send you a spreadsheet containing all 7 solutions if you want to pm me your email address. I'm 99.99% sure I've exhausted all possibilities, but I'm happy to be proved wrong if you can find one that I've overlooked.

Best wishes

Paul

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paulv66 wrote:

clm wrote:

I have just downlodad the corrected booklet of the "Ten 10 Years of Calcudoku 10x10 Puzzles", which includes the necessary modifications (for the unique solution) of the y2018 puzzle (I have sent the solution this morning). Anyway I have some comments. The original y2009 had 7 cells with possible permutations, located in columns d, f and h. The original y2018 had 14 cells with possible permutations (d4-d5-d7, e7-e8, f4-f5-f6-f7 and h4-h5-h6-h7-h8) located in columns d, e, f and h. It would have been very interesting to know how many solutions were possible (i.e., not spending too much time, I have manually found 6).

I originally thought there were only 5 possible solutions, but I had another look at the puzzle and I realised that I had excluded the possibility of h7-h8 being 5-6, which introduces two further possible solutions, bringing the total to 7. I'm pretty sure that's it - 2 possible solutions with an 8 in d4, 2 with 6 in d4 and 8 in f4 and 3 with 6 in d4 and 1 in f4.

Hi, paulv66, without the help of the Patrick's software it will be very difficult (and time consuming) to manually determine the exact number of solutions. In my opinion 7 is an strange number of solutions (perhaps 8, in this case we are missing another solution), ... .

If we only have two numbers swapped in a non-unique solution puzzle, we will have 2 different solutions (obviously those two numbers will swap in a parallel line), that is 2! = 2 solutions.

Permutaions imply the factorial, that is, if we permute 3 numbers we have a total of 3! = 6 permutations, ... . But the factorial is always even (except in the "trivial" cases 0! = 1 or 1! = 1 ), so I do not think that an odd number (7 in this case) of solutions is possible. Combining the groups of numbers involved, that is, the groups of permutations, we will have n! x p! x q! ... , which is always an even number.

In these conditions, or either 8 is the total number of solutions (2! x 2! x 2!) or we must go to the next even, 12 possible solutions (2! x 3!), since 10 solutions, for instance, being even, would not be acceptable (10 cann't be broken into a product of factorials).

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clm wrote:

I have just downlodad the corrected booklet of the "Ten 10 Years of Calcudoku 10x10 Puzzles", which includes the necessary modifications (for the unique solution) of the y2018 puzzle (I have sent the solution this morning). Anyway I have some comments. The original y2009 had 7 cells with possible permutations, located in columns d, f and h. The original y2018 had 14 cells with possible permutations (d4-d5-d7, e7-e8, f4-f5-f6-f7 and h4-h5-h6-h7-h8) located in columns d, e, f and h. It would have been very interesting to know how many solutions were possible (i.e., not spending too much time, I have manually found 6).

I originally thought there were only 5 possible solutions, but I had another look at the puzzle and I realised that I had excluded the possibility of h7-h8 being 5-6, which introduces two further possible solutions, bringing the total to 7. I'm pretty sure that's it - 2 possible solutions with an 8 in d4, 2 with 6 in d4 and 8 in f4 and 3 with 6 in d4 and 1 in f4.

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paulv66 wrote:

pnm wrote:

paulv66 wrote:

Partick, I will be happy to send on all five solutions to you if you would find this helpful. I've saved them in a spreadsheet.

Yes, can you send them?

Then I can also "fix" the y2018 ...

I've done that just now.

Cheers

Paul

I have just downlodad the corrected booklet of the "Ten 10 Years of Calcudoku 10x10 Puzzles", which includes the necessary modifications (for the unique solution) of the y2018 puzzle (I have sent the solution this morning). Anyway I have some comments. The original y2009 had 7 cells with possible permutations, located in columns d, f and h. The original y2018 had 14 cells with possible permutations (d4-d5-d7, e7-e8, f4-f5-f6-f7 and h4-h5-h6-h7-h8) located in columns d, e, f and h. It would have been very interesting to know how many solutions were possible (i.e., not spending too much time, I have manually found 6).

In the other hand, ¿How many wrong attempts are possible when sending a book solution?. ¿Is there no limit?.

It looks like the risk of permutations is greater when the very big cages have a "shape" with large columns (this is the case with the "35+" and the "75+" cages in the y2009 puzzle, and with the "41+" and the "78+" cages in the y2018 puzzle).

Particularly the cage "78+" has 9 (out of 14) cells affected by permutations, so it seems that a more exhaustive analysis (and longer time) is required, in this type of puzzles, before concluding that the solution is unique and that's my interest in knowing the total numbers of possible solutions in this particular case. Obviously these problems can be avoided by creating "stepped cages" instead, but then the difficulty decreases, in fact the only ****** puzzle of this booklet is just the y2018 (probably due to all those permutations).

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pnm wrote:

paulv66 wrote:

Partick, I will be happy to send on all five solutions to you if you would find this helpful. I've saved them in a spreadsheet.

Yes, can you send them?

Then I can also "fix" the y2018 ...

I've done that just now.

Cheers

Paul

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