beaker wrote:

80 users and 79 solves ...... and I am the non-solver.....a bit depressing

No, 79 people solved it, I don't know how many people got the puzzle (== 80 or more).

I could find out how many did, but that's not straight-forward (requires some digging through log files..).

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beaker wrote:

If you look up at previous posts, 80 users have been given this bonus puzzle and all (but one:me) plus clm have completed it successfully......time to put this experience in past but have learned more about these puzzles but, frankly, hope to never see one again.

Maybe I'm interpreting Patrick's post differently to how you're interpreting it. He started off by saying 79 people had been given the puzzle and then clarified by saying that 79 people had solved it. Anyway, I didn't know how to access the first few bonus puzzles I qualified for and then became quite paranoid about doing them as soon as they became available for fear I would leave it too late.

I'll take your advice and will treat future -1 to 4 puzzles with a great deal of caution!

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beaker wrote:

80 users and 79 solves ...... and I am the non-solver.....a bit depressing

There could be a lot of other non solvers. I had a look at the puzzle and it's quite tricky!

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pnm wrote:

beaker wrote:

Curious to know from other users if they have had one of these as a bonus configuration (-1 to +4)

So far, 79 people have had this specific puzzle as a bonus puzzle

(well 79 people solved it, rather)(including

When solving this type of Calcudoku (or any other type), by writing in the "old" books or in an independent piece of paper, I usually verify carefully the final solution. This time I made a terrible mistake (one always learns something): I was busy doing other things so, in parallel, I took a ballpoint pen and quickly solve the puzzle in a paper handkerchief. Usually, in the daily task, I am accustomed to think that after my analytical process the final verification is not really necessary, apart of the very useful "Continuous error checking" feature. Now I am more humble, and assumed something essential: that one must

"Robotically", I considered the total sum of two columns as being 20 , instead of 18, in this -1 to 4 case [I do not make this mistake when solving similar book puzzles (let's say, -5 to 4, -1 to 7, -2 to 3, etc.) or one that we have every thursday, the symmetrical -3 to 3, where the sum is 0].

Anyway, and thanks to beaker, this topic has given me the opportunity of doing a step by step solution (without complementary graphics) for a -1 to 4 6x6 Calcudoku.

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beaker wrote:

Curious to know from other users if they have had one of these as a bonus configuration (-1 to +4)

So far, 79 people have had this specific puzzle as a bonus puzzle

(well 79 people solved it, rather)(including

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First, we observe that cages a1-a2 and b2-b3 can only have -1 and 1 as candidates (the only way of having a sum of 0).

With all positive numbers (including the 0) and no 1 in, cage 8+ only admits (4,4,0), (3,3,2) or (2,2,4). The case (4,4,0) is inmediately eliminated since you would have a 0 in b6 and then the 2 in a5 is not possible. The case (2,2,4) is not possible since the 3 goes to a6 (>>> b6 = 0 and then a 0 is not possible in a5). Consequently 8+ must be (3,3,2).

The cage 1- in f3-f4 should be (-1,0), (0,1), (1,2), (2,3) or (3,4) (distance of 1 between the two candidates). Obviously, since the 0 is not possible (cage 0: up), only (1,2) is valid, then f3=2 and f4 = 1.

The cage 3+ (bottom right) can not contain a 0 or a 2. Additionally the 4 is not valid inside (because it would require -1 and 0 to add to +3), then only -1, 1 and 3 are inside (all different, it's not possible to repeat any of them), so we place a 1 in e6 and the pair (-1,3) as candidates in f5-f6 >>> cage 0: is (0,4).

Adding numbers in columns d and e (total sum of 18) we find that d4-d5 must sum -1, so this pair is (-1,0).

Cage 6+, not having 0's inside, could be (1,1,4) (which is not valid due to e6 = 1) or (1,2,3) or (4,3,-1).

Now we observe that d3 is not a 1 because it would require d2-e3 to be either (2,4) or (3,3) both impossible. Similarly d2 can not be a 1 since it would require d3-e3 to be (2,4). Than d1 = 1, so cage 6+ is (1,2,3) and e1-e2 = (2,3).

We can fill now a1 = -1, a2 = 1, b2 = -1, b3 = 1, d2 = 3, d3 = 4, e3 = 0 and c3 = -1. Also cage 3+ is (-1,4), and certainly e1 = 3 and e2 = 2.

In row 6, f6 = -1 and f5 = 3.

Next is to see that, in column c, c5 = 1 and c6 = 3. And this forces to select d5 = 0 (and not -1) to comply with the 3-cell cage 2-. Consequently d4 = -1, e4 = 4 and e5 = -1.

An the rest is straight forward: a5 = 4, b5 = 2, a6 = 0, b6 = 4, b1 = 0, f1 = 4, f2 = 0, c1 = 2 and c2 = 4.

The solution, then, is:

-1 0 2 1 3 4

1 -1 4 3 2 0

3 1 -1 4 0 2

2 3 0 -1 4 1

4 2 1 0 -1 3

0 4 3 2 1 -1

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beaker wrote:

Has any one followed the logic of clm.......only one question is that if d1 = -1 and 2e = 3 and e2 = 2, then how is it possible to add 3,2,-1 and get 6.....or did I read the sequence wrong

Yes, you are right, thank you, beaker, the development is wrong from "Adding numbers in columns d and e ... ", the reason being that two columns do not sum 20, in this case, due to the presence of a -1, so the sum is 9 per column, that is, 18 in total.

I did it to quickly, I apologize for that, , tomorrow I will correct from that point on.

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beaker wrote:

where is the Patrick instructions......I am not that computer literate......too late as the 4 timed [color=#BF0000]points put me past point of no return.[/color].....someday I will have to find out the "imgur" procedure.....I have spent over 6 hours in the last 24 with this puzzle......am glad it is gone and I can quit obsessing about it (my 75 year old brain just would not let go of it).......sigh of relief

Hi, beaker.

Here is the full solution.

First, we observe that cages a1-a2 and b2-b3 can only have -1 and 1 as candidates (the only way of having a sum of 0).

With all positive numbers (including the 0) and no 1 in, cage 8+ only admits (4,4,0), (3,3,2) or (2,2,4). The case (4,4,0) is inmediately eliminated since you would have a 0 in b6 and then the 2 in a5 is not possible. The case (2,2,4) is not possible since the 3 goes to a6 (>>> b6 = 0 and then a 0 is not possible in a5). Consequently 8+ must be (3,3,2).

The cage 1- in f3-f4 should be (-1,0), (0,1), (1,2), (2,3) or (3,4) (distance of 1 between the two candidates). Obviously, since the 0 is not possible (cage 0: up), only (1,2) is valid, then f3=2 and f4 = 1.

The cage 3+ (bottom right) can not contain a 0 or a 2. Additionally the 4 is not valid inside (because it would require -1 and 0 to add to +3), then only -1, 1 and 3 are inside (all different, it's not possible to repeat any of them), so we place a 1 in e6 and the pair (-1,3) as candidates in f5-f6 >>> cage 0: is (0,4).

Adding numbers in columns d and e (total sum of 20) we find that d4-d5 must sum 1, so this pair is (0,1) and then d5 = 1 and d4 = 0.

Now cage 3+ in e4-e5 has the candidates (-1,4) [only this option, because (0,3) or (1,2) are not valid].

A 3 in d1 is not possible since to have a sum of 3 in e1-e2 you would require (-1,4) or (0,3) or (1,2) and no one is possible. Then d2 = 3 and now, to complete 7+, d3 = 4 and e3 = 0. Consequently d1 = -1, e1 = 3 and e2 = 2. An we complete the two cages 0+ with a1 = 1 (a2 = -1) and b2 = 1 (b3 = -1). Then c3 = 1.

Next is to see that a 3 must be inside the 3-cell cage 2-, so a 0 must be the third candidate (it's quickly seen that other numbers are not possible: 4, 2 or -1).

With (0,3) in c5-c6 >>> b1 = 0 >>> a6 = 0 >>> a5 = 4 >>> b5 = 2. Adittionally b6 = 4. And f2 = 0, f1 = 4, c1 =2, c2 = 4.

An the rest is straight forward: c4 = -1, e4 = 4, e5 = -1, f5 = 3, f6 = -1, c5 = 0 and c6 = 3.

Now, let's see if either Patrick or yourself can upload the puzzle

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beaker wrote:

It is still available........I can print it out but then I do not know how to post the screen shot

I'm a little confused by your wording. If you took a screen shot then you can go to imgur.com, click on "new post" and then upload the picture. Patrick linked to more complete instructions for this earlier today.

If you need help taking a screen shot then I'd suggest doing a search for how to take a screen shot with your operating system.

I know you want to sort this out before you lose out on your next possible bonus becomes available so good luck.

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