Posts: 13 Country : United States | RJH0723 posted @ 2010-11-07 9:08 PM Could somebody please post a starting point for how to solve the last puzzle in the Hot Pot. I could only get a couple numbers when I attempted it. A full step by step solution will be great, but only a starting spot is necessary. Thanks in advance! |
@ 2010-11-08 8:32 AM (#2414 - in reply to #2409) (#2414) Top | |
Posts: 14 Country : China | cpickerel posted @ 2010-11-08 8:32 AM Step 1 The three digits in Row 6 have sum 8, so they can only be 125 or 134. Using the skyscraper information on the left and the sum information on the right we can deduct the above four digits for Row 6. The 9 in Row 4 can also be deducted directly from the skyscraper information. Step 2 According to the requirement for Column 6, the closest odd digit to the bottom is 7, and the distance between 9 and 1 in this column is 3, therefore R3C6=1. Since the sum of all digits between 1 and 9 in Row 3 is 0, and 9 in Column 7 is in Box 9, so R3C5=9. Then Step 3 Now all digits in Box 7 can be deducted. With the information outside Boxes 8 and 9, we have R7C5=4, R7C9=2, R9C9=1. Step 4 Back to the skyscraper in Row 6. With the 6 we know R6C1 R1C1 can't be 2, R2C1=1. Since the sum of all digits between 1 and 9 in Row 2 is 27, and in Column 7 the digit 9 is in Box 9, R2C8=9, R2C9=8. Step 5 With the information for Row 1, the distance between 1 and 9 should be 6 or 7; since there are already 1’s in Columns 1 and 9, R1C2=9, R1C8=1. R1C1+R1C9=7. And since there are 1’s and 2’s in Columns 1 and 9, and 3 in Column 1, R1C1=4, R1C9=3. Step 6 According to the information outside Column 2, we have R4C2=3. All digits in Row 6 can then be deducted. Step 7 Using standard Sudoku algorithm and the information outside Box 9, we have the above digits. Step 8 With the information for Box 8 Column 4, R9C4=2. With the sum requirement for Box 6, R5C7+R5C8+R5C9=22. Using the known digits in Box 9, we have R5C7=5, R5C8=8. The rest can be solved using standard Sudoku algorithm. sneaky trick On the other hand, a sneaky trick can be applied at the beginning: Following Step 1, using the information for skyscrapers, R6C1=2, R6C2=5, or R6C1=3, R6C2=4. With the information for Column 2, either R3C2=4, or R4C2=3. No matter which one holds, R6C1=3 and R6C2=4 will lead to contradiction. Thus R6C1=2, R6C2=5. The algorithm can then be continued by applying information for Column 1. |
@ 2010-11-08 8:59 AM (#2416 - in reply to #2414) (#2416) Top | |
Posts: 14 Country : China | cpickerel posted @ 2010-11-08 8:59 AM This is the author's algorithm; other players are more than welcome to contribute to additional solutions. |
@ 2010-11-08 10:34 AM (#2418 - in reply to #2416) (#2418) Top | |
Posts: 460 Country : India | purifire posted @ 2010-11-08 10:34 AM cpickerel - 2010-11-08 8:59 AM This is the author's algorithm; other players are more than welcome to contribute to additional solutions. Lovely explanation Chen Cen.... However, when I solved, I deviated from step 7 onwards... Steps 1 to 6 are exactly as you explained... I will continue from step 7 onwards an alternate route to solve this.... -------------------------------------------- R9C5 has to be an odd number since the 4 is in R7C5 and no other even number can precede it .... 1,5,9 are eliminated due to normal sudoku rules... 7 has to be a part of R8C6 or R9C6 since R6C6 is a 9 and no odd number can precede 7 in column six via the constraints of box 8. Hence R9C5 is a 3. Similarly 2 has to be in R9C4. (OStep7.png) Attachments ---------------- OStep7.png (32KB - 1 downloads) |
@ 2010-11-08 10:44 AM (#2419 - in reply to #2418) (#2419) Top | |
Posts: 460 Country : India | purifire posted @ 2010-11-08 10:44 AM Now in Box 9, the numbers 8 and 9 are in R7C7 or R8C7 or R9C7.... this allows for a naked single 7 in R9C8. Based on this we fill up the 8 and 9 also in Row 9. R9C6=8 and R9C7=9. Entire Box 9 can be filled now. R8C7=8, R7C7=3 and R8C8=4 by normal sudoku rules. This allows for R7C8=6 as per constraints of Box 9 and finally R8C9=5. (OStep8.png) Attachments ---------------- OStep8.png (36KB - 0 downloads) |
@ 2010-11-08 11:02 AM (#2420 - in reply to #2419) (#2420) Top | |
Posts: 460 Country : India | purifire posted @ 2010-11-08 11:02 AM Again based on normal sudoku rules, we also fill up the entire Box 8. R7C4=9, R7C6=5, R8C6=7, R8C4=6. (OStep9.png) Attachments ---------------- OStep9.png (37KB - 0 downloads) |
@ 2010-11-08 11:07 AM (#2421 - in reply to #2420) (#2421) Top | |
Posts: 460 Country : India | purifire posted @ 2010-11-08 11:07 AM Now in Box 6, the middle row has a sum of 22 which can only be either 9,6,7 or 9,8,5. since 6 and 7 are already present in Column 8, the middle row has to be 9,8,5. By normal sudoku rules, R5C8=8,R5C7=5,R3C8=5,R4C3=8, R4C8=2. By constraints of Box 4, R5C1 has to be a 7. (OStep10.png) Attachments ---------------- OStep10.png (40KB - 1 downloads) |
@ 2010-11-08 11:11 AM (#2422 - in reply to #2421) (#2422) Top | |
Posts: 460 Country : India | purifire posted @ 2010-11-08 11:11 AM After step 10, since all external constraints are satisfied, the present grid can be solved logically as any other classic sudoku. Hope this was helpful. Rishi (purifire) |
@ 2010-11-08 4:52 PM (#2425 - in reply to #2409) (#2425) Top | |
Country : India | debmohanty posted @ 2010-11-08 4:52 PM Thanks cpickerel and Rishi for the detailed writeup with images. RJH0723, hope that was useful to you. |
@ 2010-11-09 2:30 AM (#2433 - in reply to #2409) (#2433) Top | |
Posts: 13 Country : United States | RJH0723 posted @ 2010-11-09 2:30 AM Thanks Guys. That helped a lot. The key thing I missed was the R6 skyscraper constraint. That would have enabled me to finish it. You guys are the best! |
@ 2010-11-09 11:10 PM (#2449 - in reply to #2433) (#2449) Top | |
Posts: 460 Country : India | purifire posted @ 2010-11-09 11:10 PM RJH0723 - 2010-11-09 2:30 AM Thanks Guys. That helped a lot. The key thing I missed was the R6 skyscraper constraint. That would have enabled me to finish it. You guys are the best! Glad this was helpful.... Rishi |