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Today's noop (8 Nov 2015)
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beaker
Posted on: Sun Nov 08, 2015 8:36 pm
Posts: 530 Location: Ladysmith, BC, Canada Joined: Fri May 13, 2011 1:37 am

Today's noop (8 Nov 2015)
Today's noop proved to be the toughest one yet (at least for me)........must have tried it at least 20 times but kept getting same result........no green screen and kept trying different combinations of numbers but just not seeing the solution and finally I had the puzzle about 80% done and then I realized what the solution had to be! ........any others have problems with this one.........and "clm" was there an easy solution??? Edit by pnm: for reference, here is the puzzle:




bram
Posted on: Sun Nov 08, 2015 10:49 pm
Posts: 193 Joined: Tue May 24, 2011 4:55 pm

Re: Today's noop
This topic belongs in the section "Specific puzzles / your own puzzles" I think I have a straightforward solution to today's noop (sorry, no pictures). Let's start with the observation that any "1" cage consisting of only two cells in a regular calcudoku puzzle with only positive integers can't be "+", "x" or ":" because "+" and "x" would yield a number larger than 1 while the two numbers in the cage would have to be the same for ":" to yield 1. That leaves us with "" as the only possible operator for both of the "1" cages to the left side of the puzzle. Turning our attention to the second row from the top, the "11" cage can't be "x" because 11 is a prime, so it must be "+" and contain [5,6]. The "120" cage must be "x" and contain [4,5,6], meaning that its lower cell must contain 4. As 1 can't be in the rightmost cell of the row, the "1" cage must contain [1,2] and the rightmost cell 3. The "7" cage is "+" (for the same reason as the "11" cage) and its lower cell must therefore contain 4, which is a bit of information that is going to come in handy soon. In the rightmost column, the "12" cage must be "x" and contain [2,6]. The uppermost cell in the column thus contains 5, meaning that 6 and 4 are both in the part of the "120x" cage that is in the second column from right. That means that the "10" cage in the bottom row can't be "+" (because it would then have to contain [4,6]) but must instead be "x" and contain [2,5]. It's also obvious that the "72" and "36" cages must be "x" while the "14" cage must be "+". The "72x" cage can't contain [6,6,2] because of [2,5] in the "10x" cage, leaving [6,3,4] as the only possible combination. (*)If the "14+" cage didn't contain 6, the combination would have to be [5,5,4], which is impossible because of the 4 in the lower cell of the "7+" cage. So the "14+" cage contains a 6 but only one because there is already a 6 in the "72x" cage. The combination must therefore be either [6,5,3] or [6,4,4], the latter of which is excluded by the 4 in the lower cell of the "7+" cage. As the "14+" and "72x" cages both contain the numbers 3 and 6, both of those numbers must be present in the cages' rightmost cells (in the third column from left). The 6 in the "11+" cage must then be in fourth column from left, meaning that the "36+" cage can't contain 6, which leaves [3,3,4] as the only possible combination. There follows a series of easy steps where you don't need to calculate much (or, for that matter, determine any further operators) to find out which numbers go where. After those steps are completed, all that remains is to distribute the remaining 1s, 2s and 5s correctly while deciding on suitable operators for the "3" and "2" cages. (*) Even without taking the 2 in the "10x" cage into account, it is possible to determine that the combination in the "72x" cage must be [3,4,6]. As those three numbers are the only ones not yet placed in the lower row, two of them must necessarily go in the lower cells of the "72x" cage. And as their product is 72, the number in the upper cell of the cage must necessarily be the one of the three factors that didn't go in the lower cells. This is a practical tool for "x" and "+" cages, and clm or somebody else may have covered it in the section "Solving strategies and tips", but I don't have the time to look it up




clm
Posted on: Mon Nov 09, 2015 12:31 am
Posts: 700 Joined: Fri May 13, 2011 6:51 pm

Re: Today's noop
bram wrote: This topic belongs in the section "Specific puzzles / your own puzzles" I think I have a straightforward solution to today's noop (sorry, no pictures). ... Let's start with the observation that any "1" cage consisting of only two cells in a regular calcudoku puzzle with only positive integers can't be "+", "x" or ":" because "+" and "x" would yield a number larger than 1 while the two numbers in the cage would have to be the same for ":" to yield 1. That leaves us with "" as the only possible operator for both of the "1" cages to the left side of the puzzle. Turning our attention to the second row from the top, the "11" cage can't be "x" because 11 is a prime, so it must be "+" and contain [5,6]. The "120" cage must be "x" and contain [4,5,6], meaning that its lower cell must contain 4. As 1 can't be in the rightmost cell of the row, the "1" cage must contain [1,2] and the rightmost cell 3. The "7" cage is "+" (for the same reason as the "11" cage) and its lower cell must therefore contain 4, which is a bit of information that is going to come in handy soon. In the rightmost column, the "12" cage must be "x" and contain [2,6]. The uppermost cell in the column thus contains 5, meaning that 6 and 4 are both in the part of the "120x" cage that is in the second column from right. That means that the "10" cage in the bottom row can't be "+" (because it would then have to contain [4,6]) but must instead be "x" and contain [2,5]. It's also obvious that the "72" and "36" cages must be "x" while the "14" cage must be "+". The "72x" cage can't contain [6,6,2] because of [2,5] in the "10x" cage, leaving [6,3,4] as the only possible combination. (*) ... " Thanks, bram. Hi, beaker, I followed exactly the same way as bram until this point, having completed first the candidates in row 1: The 4 of row 1 is not in cage 3, that is, [1,4] with a  as operator since [2,3] have no possible operation to produce a 2 in cage c1d1. So c1d1 = [2,4] and a1b1 = [1,3]. Now I continued with the 5 of column 1: It cannot be in cage 2 (a3a4) because then the other number should be a 3 and the operator a "", but then a 6 would not be possible in cage a5a6, so a5 = 5. And I continued simply with the 1's. Row 5: c5 =1 (obviously it is not possible in d5 because it would require two 6's in d4 and e5), or in e5 (same reason). For the same reason, in row 4, a 1 is not possible in d4 so e4 = 1 and, in row 3, d3 = 1. Now, for instance: If a6 = 4, a 6 would be in cage 2 (a3a4) so now [3,6] but in that case b1 + b2 = 4 and the "outies" of column 2: c3 + c6 = 4 + 14 + 13  21 = 10 which is impossible having 4's in f3 and a6, so a6 = 6 (>>> b5 = 6 and b6c6 = [3,4]). Then a4 = 4 and a3 =2. The 5 of row 4: b4 = 5 (unique) then b3 + c3 = 9 >>> b3c3 = [3,6], consequently b3 = 3, c3 = 6 and e3 = 5 (>>> e6 = 2, e5 =3), producing a "" operator in the 3cell Lshape cage "3". The rest comes automatically.




beaker
Posted on: Mon Nov 09, 2015 12:48 am
Posts: 530 Location: Ladysmith, BC, Canada Joined: Fri May 13, 2011 1:37 am

Re: Today's noop
Am wondering how many puzzlers are going to see these 2 posts and say "aha, now I can get it"!!!






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