 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Logidoku-8 (Posted on 2007-01-24) The numbers 1 to 8 are to appear once in every row, column, long diagonal, shaded rectangle and irregular shape.

 1 1 8 7 5 4 7 3 3 6 5
From the given starters, can you finish the puzzle?

 See The Solution Submitted by Josie Faulkner Rating: 4.4615 (13 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Step by step walkthrough | Comment 8 of 10 | Oh man, I always forget the long diagonals clue! Almost every time I do a puzzle that includes the long diagonal clue, I chug along until I get stuck. I fret about if I made a mistake so far, and I wonder if there might be an error in the puzzle (but when they are Josie�s I am confident that�s not the case). I go into a little more in depth "tracking" of where some numbers go to see if any non-obvious things pop up.  FINALLY I say to myself, hmmm, maybe there is a clue I forgot about and then I smack myself on the head. You�d think I would have learned by now, but no. I keep getting thrown off by those long diagonals. (If you�re wondering why I shared this pointless story, it�s so I could possibly commiserate with my fellow long diagonal forgetters, assuming I�m not the only one.)

I will be calling the columns A-H (left to right) and the rows 1-8 (top to bottom).
I will be calling the rectangles J-Q. J-M are in the top row (left to right) and N-Q (left to right).

I will be calling the irregular shapes S-Z in the following manner:
Shape S is mostly in rectangle J (with 1 and 4 given in it)
Shape T is mostly in rectangle K (with 1 and 8 given in it)
Shape U is mostly in rectangle L (with 7 given in it)
Shape V is mostly in rectangle M (with no starters given in it)
Shape W is mostly in rectangle N (with 5 and 3 given in it)
Shape X is mostly in rectangle O (with 7 given in it)
Shape Y is (technically) mostly in rectangle P (with 6 and 5 given in it)
Shape Z is mostly in rectangle Q (with 3 given in it)

Column A: Shape S already has a 1, so the 1 for column A cannot go in any of the cells that are also part of shape S. So A8 is 1.
Shape Y: Row 8 already has a 1, so E7 is 1.
Shape X: Columns D and rectangle P already have a 1. So C5 is 1.
Rectangle Q: Rows 5, 7 and 8 already have a 1. So G6 is 1.
Rectangle L: Column E, row 2, and the / diagonal already have a 1. So F4 is 1.
Rectangle M: Rows 1, 2 and 4 already have a 1. So H3 is 1.

Rectangle K: Column C and shape U already have a 7. So D4 is 7.
Rectangle L: Shape U and row 4 already have a 7. So F1 is 7.
Rectangle J: Rows 1, 3 and 4 already have a 7. So A2 is 7.
Rectangle P: Shape X and column F already have a 7. So E8 is 7.
Rectangle N: Row 6 and the / diagonal already have a 7. So B5 is 7.
Rectangle Q: Column G and row 5 already have a 7. So H7 is 7.

Shape W: Column B already has a 6. So C7 is 6.
Rectangle K: Column C already has a 6. So D2 is 6.
Rectangle L: Shape U already has a 6. So E4 is 6.
Shape X: Columns D and E already have a 6. So F6 is 6.
Rectangle J: Column B, row 4, and the \ diagonal already have a 6. So A3 is 6.
Rectangle M: Rows 2, 3 and 4 and the / diagonal already have a 6. So G1 is 6.
Rectangle Q: Column G already has a 6. So H5 is 6.

Rectangle N: Column B already has a 5. So A6 is 5.
Rectangle M: Column H and row 4 already have a 5. So G2 is 5.
Rectangle O: Rows 6 and 8, and the / diagonal already have a 5. So D7 is 5.
Shape Z: Column G and row 7 already have a 5. So F5 is 5.
Rectangle K: Rows 2 and 4, and the \ diagonal already have a 5. So C1 is 5.
Rectangle L: Rows 1 and 2, and column F already have a 5. So E3 is 5.

Rectangle J: Shape S already has a 4. So B3 is 4.
\ Diagonal: A1 can�t be 4 since there is a 4 in column A. C3 can�t be 4 since there is a 4 in row 3. E5 can�t be 4 since there is a 4 in row 5. So that leaves G7 is 4.
/ Diagonal: B7 can�t be 4 since there is a 4 in column B. D5 can�t be 4 since there is a 4 in row 5. F3 can�t be 4 since there is a 4 in row 3. So H1 is 4.
Row 4: Columns A, G and H already have a 4. So C4 is 4.

Rectangle O: Column D already has an 8. So C8 is 8.
/ Diagonal: F3 can�t be 8 since there is an 8 in row 3. D5 can�t be 8 since there is an 8 in column D. So B7 is 8.
Shape Z: Row 7 already has an 8. So G5 is 8.
Shape X: Column D and row 5 already have an 8. So E6 is 8.
Rectangle L: Column E and row 3 already have an 8. So F2 is 8.
Rectangle M: Column G and row 2 already have an 8. So H4 is 8.
Rectangle J: Column B and row 4 already have an 8. So A1 is 8.

Rectangle P: Rows 5 and 7 already have a 4. So F8 is 4.
Rectangle O: Rows 5 and 8 already have a 4. So D6 is 4.
Rectangle L: Rows 1 and 3 already have a 4. So E2 is 4.

Rectangle J: Column A already has a 3. So B1 is 3.
Rectangle L: Row 1 already has a 3. So F3 is 3.
Rectangle K: Row 3 already has a 3. So C2 is 3.
Rectangle M: Row 2 already has a 3. So G4 is 3.
Rectangle P: Row 7 already has a 3. So E5 is 3.
Rectangle O: Row 5 already has a 3. So D8 is 3.

So that leaves all the remaining cells to be 2. That is, a 2 is in A4, B6, C3, D5, E1, F7, G8, and H2.

Cool puzzle!

 Posted by nikki on 2008-01-17 16:24:18 Please log in:

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