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clm
Posted on: Fri Jan 13, 2012 11:47 pm
Posts: 690 Joined: Fri May 13, 2011 6:51 pm

SudoCalcudoku
Here is another experiment: I haved named this 9x9 puzzle SudoCalcudoku (or SudoNewdoku) because it looks like a hybrid between the Calcudoku and the Sudoku. It may contain individual cells and cages (with different operators) that would behave like in the Calcudoku, but it must additionally respond to the rules of the Sudoku, that is, the 9 coloured 3x3 boxes must contain once the numbers 1 thru 9. This particular example is not very difficult and I believe the solution is unique.




starling
Posted on: Sat Jan 14, 2012 2:10 am
Posts: 175 Joined: Fri May 13, 2011 2:11 am

Re: SudoCalcudoku
Turned into a sudoku way too fast. That is, everything that was a true cage solved almost instantly. Other than that, this is really cool.
I'm not positive it has a unique solution, though. I got to a point where I couldn't conclusively prove anything, but this seemed to stem from the fact that I haven't done any sudoku in months. I could guess and test, and the trees I tried and didn't use didn't work, which makes me think it's likely the situation is unique, I just can't prove that because I had to guess and test.




clm
Posted on: Sat Jan 14, 2012 3:09 pm
Posts: 690 Joined: Fri May 13, 2011 6:51 pm

Re: SudoCalcudoku
starling wrote: Turned into a sudoku way too fast. That is, everything that was a true cage solved almost instantly. Other than that, this is really cool.
I'm not positive it has a unique solution, though. I got to a point where I couldn't conclusively prove anything, but this seemed to stem from the fact that I haven't done any sudoku in months. I could guess and test, and the trees I tried and didn't use didn't work, which makes me think it's likely the situation is unique, I just can't prove that because I had to guess and test. Thank you for your comments. I think that if these type of puzzles were created with a software, more complex puzzles than this one of course, certainly, earlier or later in the solution process, the puzzle will turn into a pure sudoku. In this example we have "information" about 28 cells; if I am right (and I am not a very expert "sudoker") a really good and difficult sudoku has no more than 20 or 21 individual cells "given" (I would like to hear the opinion of the advanced "sudokers"  and they may also give an opinion about the unicity of this example). In those conditions and since the "given" numbers are usually separated and distributed over the 9x9 grid, it would be very difficult to group them in order to form cages with operations and results. With that purpose in mind the number of individual "given" numbers should decrease much in favor of the numbers grouped in cages, some compromise should be adopted. In the sudoku, the numbers 1 to 9 are independent and not linked to any operation, no mathematical knowledge (arithmetic or other operators) is required, and this is probably why the sudoku is attractive to a great variety of persons, but perhaps something intermediate could be realized (not in this site, of course) and that "SudoCalcudoku" would be in our borderland.




starling
Posted on: Sun Jan 15, 2012 1:25 am
Posts: 175 Joined: Fri May 13, 2011 2:11 am

Re: SudoCalcudoku
I had no problem solving 60 of the 81 cells. I was missing A46, I46, D1, E1, F1, D9, E9, F9, and the middle section, and had no places where I could eliminate candidates.




clm
Posted on: Sun Jan 15, 2012 3:15 pm
Posts: 690 Joined: Fri May 13, 2011 6:51 pm

Re: SudoCalcudoku
starling wrote: I had no problem solving 60 of the 81 cells. I was missing A46, I46, D1, E1, F1, D9, E9, F9, and the middle section, and had no places where I could eliminate candidates. Hi, starling, I feel there is a 99% of probability of arriving to the situation you describe, so I am sure you have that part correct. The "difficulty" comes from this point. Try this: 1) A4 = 1 and complete the segment A4A6, then necessarily D5 = 9 and now there is no valid number for F4, so A4 <> 1. 2) Place then the 1 in its correct position and complete the segment A4A6. Place the 7 of row 6 in its proper position (it's inmediate). Now we have two possible places for the 6 in row 6. Consider D6 = 6 > I4 = 6 and, at the same time, D9 = 9 and now there is no place for a 9 in row 4, so D6 <> 6, the 6 of row 6 must go the other possible place in row 6. Completing the puzzle is now automatic. This process "looks" similar to the "trial and error" but, in some way, any logic process contains some "trial and error" inherently I think. (I will send the full solution anyway in a few days, perhaps any other puzzler wants to try it or find a different way)




starling
Posted on: Sun Jan 15, 2012 10:07 pm
Posts: 175 Joined: Fri May 13, 2011 2:11 am

Re: SudoCalcudoku
clm wrote: starling wrote: I had no problem solving 60 of the 81 cells. I was missing A46, I46, D1, E1, F1, D9, E9, F9, and the middle section, and had no places where I could eliminate candidates. Hi, starling, I feel there is a 99% of probability of arriving to the situation you describe, so I am sure you have that part correct. The "difficulty" comes from this point. Try this: 1) A4 = 1 and complete the segment A4A6, then necessarily D5 = 9 and now there is no valid number for F4, so A4 <> 1. 2) Place then the 1 in its correct position and complete the segment A4A6. Place the 7 of row 6 in its proper position (it's inmediate). Now we have two possible places for the 6 in row 6. Consider D6 = 6 > I4 = 6 and, at the same time, D9 = 9 and now there is no place for a 9 in row 4, so D6 <> 6, the 6 of row 6 must go the other possible place in row 6. Completing the puzzle is now automatic. This process "looks" similar to the "trial and error" but, in some way, any logic process contains some "trial and error" inherently I think. (I will send the full solution anyway in a few days, perhaps any other puzzler wants to try it or find a different way) This was exactly what I actually ended up doing, and was what I was referring to as Trial and error. Reason why I think this is more trial and error was because I only did this after I tried a 7 in A4, and realized it was in error. I realize we've had the whole "Eventually everything boils into guessing that isn't a single cage," discussion, but past a certain point it's not really a question to me that it's trial and error.




clm
Posted on: Wed Apr 04, 2012 1:55 pm
Posts: 690 Joined: Fri May 13, 2011 6:51 pm

Re: SudoCalcudoku
clm wrote: Here is another experiment: I haved named this 9x9 puzzle SudoCalcudoku (or SudoNewdoku) because it looks like a hybrid between the Calcudoku and the Sudoku. It may contain individual cells and cages (with different operators) that would behave like in the Calcudoku, but it must additionally respond to the rules of the Sudoku, that is, the 9 coloured 3x3 boxes must contain once the numbers 1 thru 9. This particular example is not very difficult and I believe the solution is unique.
... The solution (unique ?) for the "experiment":




jaek
Posted on: Wed Apr 04, 2012 11:39 pm
Posts: 218 Joined: Fri Jun 17, 2011 8:15 pm

Re: SudoCalcudoku
My experience was exactly the same as Starling's. I got the same 60 before stalling out. Maybe if you included the center value as a onecell cage? It definitely fell into place when I 'gave' myself the 5 there.
Or maybe if you included 0,,e4e5e6? (result,operation,cells in case I am forgetting the standard notation.) [2,4,6] and [2,5,7] are the possible solutions once the first 60 cells are filled, but e1 prevents [2,4,6].
I didn't go back and try solving with either of those additional pieces of information so it may make the whole solution too easy.
The one overarching drawback to me was that I mostly solved the few calcudoku cells before finishing the rest as sudoku. Almost like I had to solve the calcudoku parts in order to fill in the initial sudoku values. I think it may have been more enjoyable if there was more switching back and forth between the two styles.




clm
Posted on: Thu Apr 05, 2012 2:05 pm
Posts: 690 Joined: Fri May 13, 2011 6:51 pm

Re: SudoCalcudoku
jaek wrote: My experience was exactly the same as Starling's. I got the same 60 before stalling out. Maybe if you included the center value as a onecell cage? It definitely fell into place when I 'gave' myself the 5 there.
Or maybe if you included 0,,e4e5e6? (result,operation,cells in case I am forgetting the standard notation.) [2,4,6] and [2,5,7] are the possible solutions once the first 60 cells are filled, but e1 prevents [2,4,6].
I didn't go back and try solving with either of those additional pieces of information so it may make the whole solution too easy.
The one overarching drawback to me was that I mostly solved the few calcudoku cells before finishing the rest as sudoku. Almost like I had to solve the calcudoku parts in order to fill in the initial sudoku values. I think it may have been more enjoyable if there was more switching back and forth between the two styles. Thank you, your comments are very interesting. But if we gave more information (like values in e4 or e5 or e6, or for a cage "e4e5e6", etc.) the full puzzle would become very easy, since with the actual info it can be solved (with a unique solution I believe; in the next days I will try to provide the full solution process, if I find some time ). You are right that with more switching back and forth between the two styles it would be more enjoyable; in this "prototype" I tried to make it attractive to the "calcudokers" by solving initially the calcudoku part (very easy, unfortunately the "sudokers" do not read this page). To have more interaction we would need a software (I prepared this "manually" and it was just an "experiment"). Really, the difficulty is assigned to the "sudoku" part itself, and, as I commented to starling, the probability of arriving to those 60 cells is maximum.




clm
Posted on: Fri Apr 06, 2012 10:24 pm
Posts: 690 Joined: Fri May 13, 2011 6:51 pm

Re: SudoCalcudoku
SudoCalcudoku solution process. Step 1. We solve the calcudoku part. Since c6 = 9 (b6 = 5) >>> b3 = 9 and b4 = 4, that is, the 9 of cage “36x” must go to a different 3x3 box (sudoku rules). Obviously h6 = 1, h7 = 9; and c8 = 4, d8 = 7. And, consequently, “24x” = [3,8] >>> f8 = 8, g8 = 3. Since “210x” = [5,6,7] and there is a 5 in “5x” (d2d3) >>> g2 = 5 >>> h3 = 6, h4 = 5; and then f2 = 6, f3 = 7. We finish the “calcudoku” part with d3 = 5, d2 = 1. Step 2. Now we have simply a sudoku with 28 given values. Let’s name, for simplicity, 1, 2, 3, …, 9, the 3x3 boxes from left to right and from top to bottom, i.e., the box 4 would be the a4b4c4a5b5c5a6b6c6. (Above graphic, numbers in blue colour): Inmediately we can see that g3 = 4 (observing the other numbers in row 3, column g and box 3). Similarly g7 = 6 (observing the other numbers in row 7, column g and box 9). And h8 = 2 (the only left number in the cross point of row 8 and column h). In row 1, c1 = 5 (due to 5’s in d3 and g2 and b6 = 5). In row 7, e7 = 3 (due to 3’s in a9 and g8). In row 8, g9 = 1 (due to 1’s in i1 and h6 and b7 = 1). In column a, a8 = 9 (due to 9’s in b3, c6 and h7). (In green colour): The 1 in g9 forces e8 = 1 (the only place for a 1 in box 8). In row 7 we have two possible places for a 7, a7 and c7, but a7 is not valid since it would force the 7 of box 4 to be in b5 or c5 entering in conflict with the 7 of box 6 that must be in g5 or h5 (as a consequence of i9 = 7). Then c7 = 7. Now since b6 = 5 >>> i8 = 5 and b8 = 6. (In orange colour): We complete row 7 with a7 = 5, i7 = 8 and box 9 with h9 = 4. We also complete box 7: b9 = 8 and c8 = 2. We can also complete column c: c3 = 1 (due to h3 = 6) and c5 = 6. We have arrived to this stuation: Step 3 (Next graphic, in violet colour): a3 = 2 (all the other numbers are present in row 3, column 1 and box 1; anyway we will also see inmediately that due to c2 = 8 and i7 = 8 >>> e3 = 8 >>> i3 = 3 and again a3 = 2). This forces b5 = 2. (In green colour): Now the 4 of box1 to a3 (due to b4 = 4). We need a 3 and a 7 in b1b2 (box 1). But b1 <> 3 otherwise a 3 would not be possible inside box 2 due to e7 = 3 so b2 = 3 and b1 = 7. (In brown colour): The 7’s in b1 and i7 force h2 = 7 (box 3). And this implies g5 = 7 and, consequently, g1= 9 (being the 9 the only left number for column g). Then e2 = 9 and i2 = 2. And, as commented, e3 = 8 and i3 = 3 what forces (in blue colour) h5 = 3 and h1 = 8. We have arrived to the drawing referred to in the previous posts, with 60 cells defined and 21 pending: d1e1f1, a4a5a6, i4i5i6, d9e9f9 besides the central box 5 (9 cells). To continue from this pont (next graphic) we will write some candidates, for instance, in a4a5a6, d5 and f4 (being relevant a4 = [1,7], a5 = [1,8], d5 = [8,9], f4 = [1,9]). Step 4. We see that a 1 is not possible in a4 due to a5 = 8, d5 = 9 and then there is no number to place in f4, consequently (next graphic, in blue colour) it must be a5 = 1, a6 = 8, a4 = 7. And, obviously, from this point, f4 = 1, d5 = 8, e6 = 7. Again (next graphic) we write the candidates, for instance, for d1, d6 and e4 (d1 = [2,3], d6 = [3,6], e4 = [2,6]). Step 5. We have two initial places for a 2 in column d, d1 or d4. But d4 = 2 is not possible since d1 = 3 >>> d6 = 6 and there is no number to be placed in e4. Consequently, d1 = 2; now the only position for a 3 in column d is d6 = 3; now f2 = 6 >>> i6 = 6, f6 = 4; and then i5 = 4, i4 = 9. Completing the row 4: e4 = 2 and d4 = 6 >>> d9 = 9. Completing the puzzle (numbers in green colour) is now straight forward. Row 1: f1 = 3, e1 = 4. Row 5: f5 = 9, e5 = 5. Row 9: f9 = 5, e9 = 6.






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