New Killer from Miyuki Misawa
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New Killer from Miyuki Misawa
by jf27 » Tue Oct 11, 2005 10:09 pm
Has anyone tried the interesting new Killer from Miyuki Misawa's website, the last one on this page: http://www7a.biglobe.ne.jp/~sumnumberplace/22326860/ (NEW 08/10/2005)? If so, could someone please give me a clue as to how to get the first number?
- jf27
- Posts: 34
- Joined: 18 September 2005
by Sue De Coq » Wed Oct 12, 2005 12:22 pm
Some observations:
i. Box 5 contains six cells that sum to 35 - therefore the other three must sum to 10. There are four ways to make three numbers sum to 10:
7+2+1
6+3+1
5+4+1
5+3+2
ii. It's possible to fill the dotted area to the bottom-left that sums to 28 in two ways:
9+8+6+5
9+8+7+4
iii. It's possible to fill the dotted area to the top-right that sums to 27 in three ways:
9+7+6+5
9+8+7+3
9+8+6+4
iv. The dotted area to the bottom-right has five elements that sum to 32. Since the maximum sum it's possible to generate from four numbers is 9+8+7+6=30, it's clear that the dotted area cannot contain a 1, as then the four other elements would have to sum to an impossible 31.
Put it all together:
The three dotted areas that sum to 27, 28 and 32 cannot possibly contain the figure 1 - therefore, the three digits in Box 5 that sum to 10 must be 5, 3 and 2. Furthermore, we see that the 5 must go into the 28 area (r6c4), the 3 must go into the 27 area (r4c6) and the 2 into the 32 area (r6c6).
i. Box 5 contains six cells that sum to 35 - therefore the other three must sum to 10. There are four ways to make three numbers sum to 10:
7+2+1
6+3+1
5+4+1
5+3+2
ii. It's possible to fill the dotted area to the bottom-left that sums to 28 in two ways:
9+8+6+5
9+8+7+4
iii. It's possible to fill the dotted area to the top-right that sums to 27 in three ways:
9+7+6+5
9+8+7+3
9+8+6+4
iv. The dotted area to the bottom-right has five elements that sum to 32. Since the maximum sum it's possible to generate from four numbers is 9+8+7+6=30, it's clear that the dotted area cannot contain a 1, as then the four other elements would have to sum to an impossible 31.
Put it all together:
The three dotted areas that sum to 27, 28 and 32 cannot possibly contain the figure 1 - therefore, the three digits in Box 5 that sum to 10 must be 5, 3 and 2. Furthermore, we see that the 5 must go into the 28 area (r6c4), the 3 must go into the 27 area (r4c6) and the 2 into the 32 area (r6c6).
- Sue De Coq
- Posts: 93
- Joined: 01 April 2005
by CathyW » Wed Oct 12, 2005 7:23 pm
Hmm - a tricky one indeed
I've made a little progress but now stuck - have completed the 27 and 28 cages of 4, worked out the pairs in the 32 W shape and have a 1,2,3 triple remaining to complete column 9. Also have r2,c8; r8,c8; r9,c8 and r8,c6 but spent more than half an hour so far.
Didn't want to post actual numbers just yet to avoid spoiling the fun for others.
Do tell if you've got further than this.
Cathy
Everything will be OK in the end. If it's not OK, it's not the end!
I've made a little progress but now stuck - have completed the 27 and 28 cages of 4, worked out the pairs in the 32 W shape and have a 1,2,3 triple remaining to complete column 9. Also have r2,c8; r8,c8; r9,c8 and r8,c6 but spent more than half an hour so far.Didn't want to post actual numbers just yet to avoid spoiling the fun for others.
Do tell if you've got further than this.
Cathy
Everything will be OK in the end. If it's not OK, it's not the end!
- CathyW
- Posts: 316
- Joined: 20 June 2005
by Sue De Coq » Thu Oct 13, 2005 2:22 am
I've just completed the puzzle and I'd like to congratulate Misawa-san on his excellent work. (The chap who solved it in 29 minutes must be pretty hot too).
Cathy - here's how I progressed from the last post:
We have the digits 6, 7, 8 and 9 in the cells r7c3-r7c6 and r3c7-r6c7. Consider Box 9. The digits 6 to 9 have to go into the four cells r8c8-r9c9. Consider the area in Box 9 that has three cells that sum to 22. We have two possibilities:
9+8+5
9+7+6
The first possibility has to be scrapped because we don't have a 5 - therefore, 9, 7 and 6 fill this area and 8 goes into r8c8. Consider the area that sums to 24 that lies mostly in Box 9. Of the three possible ways:
1+2+3+5+6+7
1+2+3+4+6+8
1+2+3+4+5+9
we have to take the second, because we have an 8 to accommodate. The only cell in the area that could possible house the 6 is r8c6, which forces a 6 into r7c3. Now consider the 6 in Box 9, which could go into r9c8 or r9c9. Suppose it were to go into r9c8. The digits 9 and 7 would be forced into column 9, which, combined with the presence of the 8 at r6c3 or at one of r5c7 or r6c7, would make it impossible to find three digits to sum to 18 higher up Column 9. Therefore, we put the 6 into r9c9. It quickly becomes clear that, in order to fill the 18 area, we have to put the 9 into r9c8, the 7 into r8c9, a 9 into r5c9, a 4 into r6c9 and a 5 into r7c9. The 9 in Column 7 is forced to r3c7, which gives us an 8 in r3c6, a 7 in r4c7, a 9 in r6c3 and an 8 in r7c4.
Since the 24 area must contain a 4 (in cells r7c7, r7c8 or r8c7) cell r9c7 must contain a 1, 2 or 3. Consider Box 8. The digits 9, 8, 7 and 6 are already accounted for, so the only way to fill the 12 area is to have 5 and 4 in r9c5 and r9c6 in some order and a 3 in r9c7. Furthermore, the 24 area restriction means that a 3 has to go into r6c8. We now have that cells r4c8, r4c9 and r5c8 contain 1, 2 and 5, which means that cells r2c9, r3c8 and r3c9 must sum to 18 (=26-(1+2+5)). Given the cells already in place, there's just the one way to do this - 8 into r2c9, 7 into r3c8 and 3 into r3c9. The fact that Box 3 sums to 45 gives us that r1c7 is 2, after which the 24 area tells us that r7c8 is 2, which forces r4c9 9 to 2 and row1c9 to 1. Since a 5 appears in r4c8 or r5c8, we have to have a 5 in r2c7.
Similar tricks work in Boxes 7 and 8. We have 1, 2 and 3 in r8c4, r8c5 and r9c4, so the remaining 3 cells in the 25 area must some to 19. Given that the 9 isn't available, we're forced to choose 19=8+7+4. The fact that we already have 7 and 8 in Row 8 forces the 4 into r8c3, which forces a 4 into r7c7 and a 1 into r8c7. Moreover, we had already restricted the 1 to 3 cells in Box 8 - now it has to go into r9c4. The fact that Box sums to 45 gives us r7c1 is 1. Given that 2 and 3 are in r8c4 and r8c5, their only cells in Box 7 are r9c1 and r7c2.
Consider the 29 area mostly in Box 4. We have to make five cells sum to 28, without using 9. There are two ways:
8+7+6+5+2
8+7+6+4+3
Note that a 1 definitely won't appear in the area - so the 1 in Box 4 belongs at r4c2. The other two cells in Box 4 outside of the 29 area will be 3 and 4 (if the first sum is used) or 2 and 5 (if the second). Of course, the 2 at r9c1 means the first sum will have to be used. A 4 goes into r4c1 and a 3 into r5c1.
From here on in, traditional Sudoku starts to take over. E.g., Row 4 lacks a 5, 6, 8 and 9. Consider r4c8. Column 8 already has an 8 and 9 and there must be a 6 in r2c8 or r3c8, so a 5 belongs in r4c8 and an 8 in r4c3. We have 1s in r5c8 and r6c5. Traditional analysis of Row 6 gives us an 8 in r6c7, which gives us a 6 in r5c7. We have 8s in r9c2, r5c5 and r1c1 and a 7 in r9c3. The 1 in Column 6 belongs in r2c6. To complete the 28 area, we must have four figures that sum to 25. the only acceptable solution is 9+7+5+4. The 4 goes into r3c4 and the 5 into r3c5. We have a 4 in r9c5 and a 5 in r9c6. The 4 in Column 6 belongs in r5c6, which forces the 7 in Box 5 into r5c4.
We've now garnered all of the available information from the summed areas - so it's straightforward Sudoku fro here on in. Congratulations once more to Misawa-san for such a magnificent puzzle.
Cathy - here's how I progressed from the last post:
We have the digits 6, 7, 8 and 9 in the cells r7c3-r7c6 and r3c7-r6c7. Consider Box 9. The digits 6 to 9 have to go into the four cells r8c8-r9c9. Consider the area in Box 9 that has three cells that sum to 22. We have two possibilities:
9+8+5
9+7+6
The first possibility has to be scrapped because we don't have a 5 - therefore, 9, 7 and 6 fill this area and 8 goes into r8c8. Consider the area that sums to 24 that lies mostly in Box 9. Of the three possible ways:
1+2+3+5+6+7
1+2+3+4+6+8
1+2+3+4+5+9
we have to take the second, because we have an 8 to accommodate. The only cell in the area that could possible house the 6 is r8c6, which forces a 6 into r7c3. Now consider the 6 in Box 9, which could go into r9c8 or r9c9. Suppose it were to go into r9c8. The digits 9 and 7 would be forced into column 9, which, combined with the presence of the 8 at r6c3 or at one of r5c7 or r6c7, would make it impossible to find three digits to sum to 18 higher up Column 9. Therefore, we put the 6 into r9c9. It quickly becomes clear that, in order to fill the 18 area, we have to put the 9 into r9c8, the 7 into r8c9, a 9 into r5c9, a 4 into r6c9 and a 5 into r7c9. The 9 in Column 7 is forced to r3c7, which gives us an 8 in r3c6, a 7 in r4c7, a 9 in r6c3 and an 8 in r7c4.
Since the 24 area must contain a 4 (in cells r7c7, r7c8 or r8c7) cell r9c7 must contain a 1, 2 or 3. Consider Box 8. The digits 9, 8, 7 and 6 are already accounted for, so the only way to fill the 12 area is to have 5 and 4 in r9c5 and r9c6 in some order and a 3 in r9c7. Furthermore, the 24 area restriction means that a 3 has to go into r6c8. We now have that cells r4c8, r4c9 and r5c8 contain 1, 2 and 5, which means that cells r2c9, r3c8 and r3c9 must sum to 18 (=26-(1+2+5)). Given the cells already in place, there's just the one way to do this - 8 into r2c9, 7 into r3c8 and 3 into r3c9. The fact that Box 3 sums to 45 gives us that r1c7 is 2, after which the 24 area tells us that r7c8 is 2, which forces r4c9 9 to 2 and row1c9 to 1. Since a 5 appears in r4c8 or r5c8, we have to have a 5 in r2c7.
Similar tricks work in Boxes 7 and 8. We have 1, 2 and 3 in r8c4, r8c5 and r9c4, so the remaining 3 cells in the 25 area must some to 19. Given that the 9 isn't available, we're forced to choose 19=8+7+4. The fact that we already have 7 and 8 in Row 8 forces the 4 into r8c3, which forces a 4 into r7c7 and a 1 into r8c7. Moreover, we had already restricted the 1 to 3 cells in Box 8 - now it has to go into r9c4. The fact that Box sums to 45 gives us r7c1 is 1. Given that 2 and 3 are in r8c4 and r8c5, their only cells in Box 7 are r9c1 and r7c2.
Consider the 29 area mostly in Box 4. We have to make five cells sum to 28, without using 9. There are two ways:
8+7+6+5+2
8+7+6+4+3
Note that a 1 definitely won't appear in the area - so the 1 in Box 4 belongs at r4c2. The other two cells in Box 4 outside of the 29 area will be 3 and 4 (if the first sum is used) or 2 and 5 (if the second). Of course, the 2 at r9c1 means the first sum will have to be used. A 4 goes into r4c1 and a 3 into r5c1.
From here on in, traditional Sudoku starts to take over. E.g., Row 4 lacks a 5, 6, 8 and 9. Consider r4c8. Column 8 already has an 8 and 9 and there must be a 6 in r2c8 or r3c8, so a 5 belongs in r4c8 and an 8 in r4c3. We have 1s in r5c8 and r6c5. Traditional analysis of Row 6 gives us an 8 in r6c7, which gives us a 6 in r5c7. We have 8s in r9c2, r5c5 and r1c1 and a 7 in r9c3. The 1 in Column 6 belongs in r2c6. To complete the 28 area, we must have four figures that sum to 25. the only acceptable solution is 9+7+5+4. The 4 goes into r3c4 and the 5 into r3c5. We have a 4 in r9c5 and a 5 in r9c6. The 4 in Column 6 belongs in r5c6, which forces the 7 in Box 5 into r5c4.
We've now garnered all of the available information from the summed areas - so it's straightforward Sudoku fro here on in. Congratulations once more to Misawa-san for such a magnificent puzzle.
- Sue De Coq
- Posts: 93
- Joined: 01 April 2005
by CathyW » Thu Oct 13, 2005 7:24 pm
Thanks for the tips Sue - I had tried to get further on my own and thought I was making good progress but evidently I went wrong somewhere. Will try again
Cathy
Everything will be OK in the end. If it's not OK, it's not the end!
Cathy
Everything will be OK in the end. If it's not OK, it's not the end!
- CathyW
- Posts: 316
- Joined: 20 June 2005
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