Author |
Message |
billa 
Posted on: Sun Jan 05, 2025 3:20 pm
Posts: 3 Joined: Sat Dec 23, 2023 10:48 pm
|
 Difficult 6x6
I present to the forum a 6x6 Calcudoku from Simon Tatham's puzzle page which I am unable to solve. You may play it from this link.  I have spent a long time on this puzzle and not gotten very far. I've been writing my own logical solver but it is not able to solve this yet - Andrew Stuart's solver can only make the most trivial cage eliminations. I got a bit further than the program did and will briefly explain my eliminations. In columns 5 & 6: Rule of 720: the cells in a column must multiply to 720 (1*2*3*4*5*6), so the cells in 2 columns must multiply to 720^2=518,400. The big cages multiply to 12*60*120 = 86,400, which must be multiplied by 6 to get 518,400. Therefore the remaining 2 cells must have a product of 6, which eliminates 45 from both of them, then eliminates 12 from r1c5. The possible sums for this pair are 2+3=5 or 1+6=7. They are each contained in addition cages, which together sum to 18: the remainder must be 11 or 13. The remaining cells in these cages are all in column 4, so the remaining cells in this column must add to 21-11=10 or 21-13=8. This eliminates 1 from r56c4. I can't get any further than this - parity arguments in c1&c2 are fruitless and I cannot find any more useful combinations of cages. It just seems every angle of attack comes up empty. If anyone here has an easy solution I will be embarrassed but grateful - if the solution is difficult, well done and thank you ![Tongue [tongue]](./images/smilies/msp_tongue.gif)
_________________ ▄▀ ▀▀▀
|
|
|
|
 |
pnm 
Posted on: Sun Jan 05, 2025 3:31 pm
Posts: 3471 Joined: Thu May 12, 2011 11:58 pm
|
 Re: Difficult 6x6
for whatever it's worth, my solver rates this puzzle as "difficult" ![Scared [scared]](./images/smilies/msp_scared.gif)
|
|
|
 |
clm 
Posted on: Sun Jan 05, 2025 11:30 pm
Posts: 879 Joined: Fri May 13, 2011 6:51 pm
|
 Re: Difficult 6x6
billa wrote: I present to the forum a 6x6 Calcudoku from Simon Tatham's puzzle page which I am unable to solve. You may play it from I have spent a long time on this puzzle and not gotten very far. I've been writing my own logical solver but it is not able to solve this yet - Andrew Stuart's solver can only make the most trivial cage eliminations. I got a bit further than the program did and will briefly explain my eliminations. In columns 5 & 6: Rule of 720: the cells in a column must multiply to 720 (1*2*3*4*5*6), so the cells in 2 columns must multiply to 720^2=518,400. The big cages multiply to 12*60*120 = 86,400, which must be multiplied by 6 to get 518,400. Therefore the remaining 2 cells must have a product of 6, which eliminates 45 from both of them, then eliminates 12 from r1c5. The possible sums for this pair are 2+3=5 or 1+6=7. They are each contained in addition cages, which together sum to 18: the remainder must be 11 or 13. The remaining cells in these cages are all in column 4, so the remaining cells in this column must add to 21-11=10 or 21-13=8. This eliminates 1 from r56c4. I can't get any further than this - parity arguments in c1&c2 are fruitless and I cannot find any more useful combinations of cages. It just seems every angle of attack comes up empty. If anyone here has an easy solution I will be embarrassed but grateful - if the solution is difficult, well done and thank you ![Tongue [tongue]](./images/smilies/msp_tongue.gif) It is difficult, I agree with pnm, but it can be solved analytically. In my case I have found the parity rule useful in some parts of the solution process. I have arrived to the solution, but for the moment I will not provide it, since there are probably puzzlers that want to try it. Anyway, I give you some tips. Cage "60x" in columns a and b have only two possible combinations: [1,3,4,5] (which sum is 13) and [1,2,5,6] (which sum is 14), since [2,2,3,5} is not valid. The reason for this is that if you place a 2 in a3 >>> a6 =1 and then since b2-b3-b4 = [2,3,5], you cannot complete cage "10+" with [3,6] or [4,5]. Next. Cage "1-" in column can only be [4,5] or [5,6]: [1,2] >>> "2:" = [3,6] so the sum of "60x" and b1 = 20 (total sum of 42 in columns a and b) which is impossible, 13 + 7 (out of range) or 14 + 6 (a third 6 in columns a and b). [2,3] makes impossible cage "2:" in column a. And [3,4] >>> "2:" = [1,2] so the sum of "60x" and b1 = 22, that is 13 + 9 or 14 + 8, obviously impossible. Since e1 x e4 = 6 (using the product of all numbers in the last two columns, as you well say), either e1 = 3, e4 = 2 or e1 = 6, e4 = 1. First, we eliminate e1 = 3, e4 = 2 (>>> d4 = 4). We continue from this point finding that "3-" in b1-c1 cannot be [5,2] (b1 = 5, c1 = 2, no more 5's in columns a or b) or b1 = 2, c1 = 5 (it's easy to see using the parity rule). But "3-" = [1,4] is also impossible: If c1 =1 >>> "2:" is even = [2,4] >>> "1-" in c6-d6 has a sum of 13 >>> [6,7] (out or range). And if c1 =4, the sum of "2:" in column c and "1-" in c6-d6 = 16, which is impossible, 1 + 2 + 6 + 7 (out of range) or 3 + 6 + 3+ 4 (three 4's in columns c and d). Then, e1 = 6, e4 = 1 and d4 = 5. You can finish the puzzle from this point with similar analysis.
Last edited by clm on Mon Jan 06, 2025 7:44 pm, edited 1 time in total.
|
|
|
|
 |
pnm 
Posted on: Sun Jan 05, 2025 11:44 pm
Posts: 3471 Joined: Thu May 12, 2011 11:58 pm
|
 Re: Difficult 6x6
For reference, the puzzle with "calcudoku.org" style coordinates: 
|
|
|
 |
clm 
Posted on: Mon Jan 06, 2025 12:12 am
Posts: 879 Joined: Fri May 13, 2011 6:51 pm
|
 Re: Difficult 6x6
clm wrote: ... Cage "60x" in columns b and c have only two possible combinations: [1,3,4,5] (which sum is 13) and [1,2,5,6] (which sum is 14), since [2,2,3,5} is not valid...+ 2 + 6 + 7 (out of range) or 3 + 6 + 3+ 4 (three 4's in columns c and d).
It must say: ... Cage "60x" in columns a and b... This has been corrected in the original post.
Last edited by clm on Mon Jan 06, 2025 7:44 pm, edited 1 time in total.
|
|
|
|
 |
pnm 
Posted on: Mon Jan 06, 2025 1:07 am
Posts: 3471 Joined: Thu May 12, 2011 11:58 pm
|
 Re: Difficult 6x6
clm wrote: It must say: ... Cage "60x" in columns a and b... Near the top right of your post there will be a button "Edit", that allows you to edit your own posts.
|
|
|
 |
clm 
Posted on: Mon Jan 06, 2025 1:31 am
Posts: 879 Joined: Fri May 13, 2011 6:51 pm
|
 Re: Difficult 6x6
pnm wrote: clm wrote: It must say: ... Cage "60x" in columns a and b... Near the top right of your post there will be a button "Edit", that allows you to edit your own posts. Thank you, I was probably thinking in "The Magic Kings"  .
|
|
|
|
 |
pnm 
Posted on: Mon Jan 06, 2025 1:14 pm
Posts: 3471 Joined: Thu May 12, 2011 11:58 pm
|
 Re: Difficult 6x6
clm wrote: pnm wrote: clm wrote: It must say: ... Cage "60x" in columns a and b... Near the top right of your post there will be a button "Edit", that allows you to edit your own posts. Thank you, I was probably thinking in "The Magic Kings"  . ? in any case, the Edit function allows you to correct the typo in your original post if you want.
|
|
|
 |
billa 
Posted on: Mon Jan 06, 2025 6:44 pm
Posts: 3 Joined: Sat Dec 23, 2023 10:48 pm
|
 Re: Difficult 6x6
clm wrote: It is difficult, I agree with pnm, but it can be solved analytically. In my case I have found the parity rule useful in some parts of the solution process. I have arrived to the solution, but for the moment I will not provide it, since there are probably puzzlers that want to try it. Anyway, I give you some tips. Thank you for your expert analysis clm. I had neglected to consider the 10+ cage as I wrote it off as having too many combinations but this was an error as it is central to your reasoning. I realised after posting that the 3- cage cannot contain {36} because it would clear out e1, an obvious step I had missed. Your deduction that the 1- cage contains {45} or {56} narrows down the possible combinations of the 10+ cage to only 4; checking these shows that a3 is always 1, unless a6 is 1, in which case it is 3. Therefore the 2: cage cannot be {12}. Also, in all possibilities, r1c2 is either 1 or 2. Not as elegant as your solution but it worked for me! I don't quite follow your reasoning that r1c5 must be 2, I will have to do my own analysis using columns 3 and 4. Cheers and have a happy new year.
_________________ ▄▀ ▀▀▀
|
|
|
|
 |
clm 
Posted on: Mon Jan 06, 2025 8:11 pm
Posts: 879 Joined: Fri May 13, 2011 6:51 pm
|
 Re: Difficult 6x6
billa wrote: ...
I don't quite follow your reasoning that r1c5 must be 2, I will have to do my own analysis using columns 3 and 4.
Cheers and have a happy new year. Hi, billa. Thank you for your comments. I think there is a little misunderstanding perhaps due to our different notation. I guess that your notation r1c5 stands for row 1 column 5, while in the coordinates notation we normally use in the site this corresponds to e1, that is, column e row 1 (as in the drawing provided by Patrick, certainly the curious thing is that with this notation we place first the column and secondly the row inversely to the algebra typical notation ![Smile [smile]](./images/smilies/msp_smile.gif) ). But what I say in the reasoning is that this cell cannot be a 3 (and consequently e4 = 2) and I later demonstrate why. Of course, if e1 = 2 >>> e4 = 3, but this would have been impossible since we have a "6+" cage in cells c4-d4. So, necessarily, e1 = 6 (and e4 = 1) in the final solution since the product of both cells is 6. Happy new year too for you and your relatives. ![Smile [smile]](./images/smilies/msp_smile.gif)
|
|
|
 |
|
|
You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum
|
|