Calcudoku 7×7 with undetermined numbers
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sica
Posted on: Sun Dec 28, 2014 10:02 am
Posts: 10 Joined: Wed Oct 15, 2014 12:42 pm

Re: Calcudoku 7×7 with undetermined numbers
kozibrada wrote: firefly wrote: The only solution I can see is if you've allowed the use of duplicate numbers. So the solution to the 2x2 you posted would be [2, 2]; [2, 2].
That's the only thing that makes this puzzle make sense for me. Exactly so. Now it makes sense in relation to the “7×7” (already possible to see in my today’s sketch)… This solution is violating a basic rule of calcudoku (no duplicate numbers on rows or columns). If we accept that rule violation we have another problems:  how many times we accept this violations in a puzzle?  in this case is the solution unique (another basic rule)? Do we speak about a calcudoku in this case?




firefly
Posted on: Sun Dec 28, 2014 10:22 am
Posts: 37 Joined: Mon May 05, 2014 4:59 am

Re: Calcudoku 7×7 with undetermined numbers
sica wrote: This solution is violating a basic rule of calcudoku (no duplicate numbers on rows or columns). If we accept that rule violation we have another problems:  how many times we accept this violations in a puzzle?  in this case is the solution unique (another basic rule)?
Do we speak about a calcudoku in this case? Well, the rows/columns would have to have the same set of numbers each, so if one row had two 2's, then all rows/columns would have two 2's, etc. So it adds the additional step of trying to figure out the set before you can finish solving the puzzle. For instance, in this one we know certain things right off the bat. The set must include at least one each of 0, 1, 2, 5. Then it has to have one from [3, 6] and [4, 8]. That's 6 of the 7 numbers, just for starters. Since we've determined that there's at least one duplicate, and that it must be somewhere in the 03 range to make row 3 make sense, things become easier. On top of that, the 7 cage means that at least one number must be 7+, so the 8 is in and the 4 is out. And row 7 shows us that there can't be two 0's. That leaves us with 0, 1, 2, 5, 8, [3, 6], [1, 2, 3]. And kozibrada's hint in the sketch shows us that, based on the 3: cage, those last two numbers are either 1 and 3 or 2 and 6, leaving us with 0, 1, 2, 5, 8, [3, 6], [1, 2]. With the 5 and 8 known in the 240x cage, the only options left for the two cells are 1 and 6 or 2 and 3. The 1 and 6 would cause a double 1 in either column D or E, which can't happen, since a 6 should mean a double 2 instead of a double 1. So the set would appear to be 0, 1, 1, 2, 3, 5, 8. The rest is mostly calcudoku as normal. Edit: Definitely a unique solution here. Interesting solve. Edit: Solution:




sica
Posted on: Sun Dec 28, 2014 12:08 pm
Posts: 10 Joined: Wed Oct 15, 2014 12:42 pm

Re: Calcudoku 7×7 with undetermined numbers
Very nice solution. We couldn't find it because of missing information. "this 7×7 calcudoku contains some of numbers 0–9 (integers). And a hint: F…" is not enogh to find this unique solution. I'm sorry that we haven't the chance to find it.




firefly
Posted on: Sun Dec 28, 2014 1:03 pm
Posts: 37 Joined: Mon May 05, 2014 4:59 am

Re: Calcudoku 7×7 with undetermined numbers
I gotta say, I wouldn't have guessed Fibonacci from simply "F..." or even the portrait you posted earlier.
I do, of course, recognize the Fibonacci sequence in hindsight.
Maybe if I'd bothered to check out the portrait's filename, I guess...




kozibrada
Posted on: Mon Dec 29, 2014 7:03 am
Posts: 37 Location: Karviná Joined: Sun Feb 03, 2013 3:25 am

Re: Calcudoku 7×7 with undetermined numbers
To firefly: Well done. In the sketch, I skipped explanation of impossibility of numbers 3–6 in the “7−” cage and numbers 7, 9 and 8’s in the “C” column (for theoretical, fast solving by the “3 mod” cage: 74; 85; 96 don’t work → 3?). But I find your deduction by the “240×” cage ([1, 6] vs. [2, 3]) better if anybody doesn’t know the modulo operator. To firefly and sica: How do I write? I’ve tried to expand my last topic (arith. sequence from 2nd December 2014) quietly, and test your patience (not only both of you, of course). The “F…” clue should be rather confusing. I’ve relied on difficulty of change “conventional thinking” about (usual) calcudoku rules. But still it should have the basic rule (in other case, solving wouldn’t make sense ): the same numbers in each rows/columns. Finally, I’m not going to use similar “innovation” – due to another misunderstandings and conservating of calcudoku originality…




firefly
Posted on: Mon Dec 29, 2014 9:17 am
Posts: 37 Joined: Mon May 05, 2014 4:59 am

Re: Calcudoku 7×7 with undetermined numbers
kozibrada wrote: Finally, I’m not going to use similar “innovation” – due to another misunderstandings and conservating of calcudoku originality… You know, it's funny. Once I realized that you weren't just using random numbers for your set, but that you could also duplicate numbers, I figured "meh, this is too annoying to bother to solve." Even when I started writing that last post, I hadn't planned on really figuring out the full set or solving it, but as I was describing how you could do it, I just sort of did it. All in all it was actually surprisingly interesting to solve, once I decided to actually do so. (And once I knew about the duplicate numbers.) Then again, I pretty much only ever do FAS (full analytical solution) solving. I really, really can't stand having to resort to TAE (trial and error). I tend to look at a lot of the changes to calcudoku as making them less structured, more vague, and therefore more likely to lead to a TAE situation. But this puzzle, at least, was eminently solvable with logic alone. I, for one, would be interested in trying a few more just to see if that's consistent or if this was just a "lucky" one. :P Edit: The 240x path is just what I happened to notice as I was writing the post. A few minutes later, as I was actually solving the rest of the puzzle, I realized that you could have done the same thing with the 3mod cage. But, yeah, I generally have to look up the mod tables each time, so it's not what I tend to look for first.




sica
Posted on: Mon Dec 29, 2014 9:58 am
Posts: 10 Joined: Wed Oct 15, 2014 12:42 pm

Re: Calcudoku 7×7 with undetermined numbers
kozibrada wrote: To firefly: Well done. In the sketch, I skipped explanation of impossibility of numbers 3–6 in the “7−” cage and numbers 7, 9 and 8’s in the “C” column (for theoretical, fast solving by the “3 mod” cage: 74; 85; 96 don’t work → 3?). But I find your deduction by the “240×” cage ([1, 6] vs. [2, 3]) better if anybody doesn’t know the modulo operator. To firefly and sica: How do I write? I’ve tried to expand my last topic (arith. sequence from 2nd December 2014) quietly, and test your patience (not only both of you, of course). The “F…” clue should be rather confusing. I’ve relied on difficulty of change “conventional thinking” about (usual) calcudoku rules. But still it should have the basic rule (in other case, solving wouldn’t make sense ): the same numbers in each rows/columns. Finally, I’m not going to use similar “innovation” – due to another misunderstandings and conservating of calcudoku originality… Please don't give up. This kind of calcudoku is very interesting and chalanging. I'm sorry that I didn't have enough elements to solve it as you think it. I've found another solution without duplicates numbers except the 1^ cage. I don't know what could you write for us to understand this exceptions.






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