Congratulations on the site's 10 year anniversary!
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pnm
Posted on: Wed Feb 13, 2019 12:42 pm
Posts: 2636 Joined: Thu May 12, 2011 11:58 pm

Re: Congratulations on the site's 10 year anniversary!
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 ...




paulv66
Posted on: Wed Feb 13, 2019 3:08 pm
Posts: 609 Location: Dublin, Ireland Joined: Tue Mar 01, 2016 10:03 pm

Re: Congratulations on the site's 10 year anniversary!
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




clm
Posted on: Thu Feb 14, 2019 12:42 pm
Posts: 762 Joined: Fri May 13, 2011 6:51 pm

Re: Congratulations on the site's 10 year anniversary!
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 (d4d5d7, e7e8, f4f5f6f7 and h4h5h6h7h8) 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).




paulv66
Posted on: Thu Feb 14, 2019 6:10 pm
Posts: 609 Location: Dublin, Ireland Joined: Tue Mar 01, 2016 10:03 pm

Re: Congratulations on the site's 10 year anniversary!
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 (d4d5d7, e7e8, f4f5f6f7 and h4h5h6h7h8) 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 h7h8 being 56, 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.




clm
Posted on: Fri Feb 15, 2019 1:54 pm
Posts: 762 Joined: Fri May 13, 2011 6:51 pm

Re: Congratulations on the site's 10 year anniversary!
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 (d4d5d7, e7e8, f4f5f6f7 and h4h5h6h7h8) 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 h7h8 being 56, 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 nonunique 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).




paulv66
Posted on: Fri Feb 15, 2019 7:08 pm
Posts: 609 Location: Dublin, Ireland Joined: Tue Mar 01, 2016 10:03 pm

Re: Congratulations on the site's 10 year anniversary!
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 78 in f5f7 and h5h7.
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 578 and 857  each of which leads to an unique solution. Similarly, when 1 is in f4, there are two possibilities for row 5  namely 875 and 587. However, only 875 leads to a unique solution. With 587, you have 56 in h7h8, which can be reversed with the 56 in e7e8.
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




clm
Posted on: Fri Feb 15, 2019 8:26 pm
Posts: 762 Joined: Fri May 13, 2011 6:51 pm

Re: Congratulations on the site's 10 year anniversary!
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 78 in f5f7 and h5h7.
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 578 and 857  each of which leads to an unique solution. Similarly, when 1 is in f4, there are two possibilities for row 5  namely 875 and 587. However, only 875 leads to a unique solution. With 587, you have 56 in h7h8, which can be reversed with the 56 in e7e8.
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 " having an 8 in d4 causes f4 to equal 6 and h4 to equal 1", because I have f10 = 6 as one of my 86 cells with a unique possibility, so a 6 is already in column f.




paulv66
Posted on: Fri Feb 15, 2019 8:59 pm
Posts: 609 Location: Dublin, Ireland Joined: Tue Mar 01, 2016 10:03 pm

Re: Congratulations on the site's 10 year anniversary!
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 78 in f5f7 and h5h7.
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 578 and 857  each of which leads to an unique solution. Similarly, when 1 is in f4, there are two possibilities for row 5  namely 875 and 587. However, only 875 leads to a unique solution. With 587, you have 56 in h7h8, which can be reversed with the 56 in e7e8.
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 " having an 8 in d4 causes f4 to equal 6 and h4 to equal 1", because I have f10 = 6 as one of my 86 cells with a unique possibility, so a 6 is already in column f. 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




michaele
Posted on: Sun Feb 17, 2019 10:19 am
Posts: 64 Joined: Sun Jan 31, 2016 7:52 am

Re: Congratulations on the site's 10 year anniversary!
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.




paulv66
Posted on: Sun Feb 17, 2019 1:18 pm
Posts: 609 Location: Dublin, Ireland Joined: Tue Mar 01, 2016 10:03 pm

Re: Congratulations on the site's 10 year anniversary!
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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