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Diagonal surprise

 
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Ruud
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Joined: 30 Dec 2005
Posts: 601

PostPosted: Sun Nov 11, 2007 6:30 pm    Post subject: Diagonal surprise Reply with quote

Not a V2, but a brand new challenge with unorthodox (unorthogonal) cage shapes.



3x3::k:2816:4609:1538:5379:5124:5379:5124:1799:3336:5385:2816:4609:1538:5379:5124:1799:3336:3601:4370:5385:2816:4609:1538:1799:3336:3601:2330:3611:4370:5385:4894:2335:2592:3601:2330:5155:4370:3611:4894:4391:2592:2335:2592:5155:3628:3611:1838:3631:4894:4391:2592:4659:3628:5155:1838:3631:3384:3129:5690:4923:2620:4659:3628:3631:3384:3129:4674:3139:5690:4923:2620:4659:3384:3129:4674:3139:4674:3139:5690:4923:2620:

For those who are familiar with old Dutch traditions, you know where those candy colors refer to...

Enjoy,
Ruud
_________________
“If the human brain were so simple that we could understand it, we would be so simple that we couldn't.” - Emerson M Pugh
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Caida
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Joined: 03 Nov 2007
Posts: 38
Location: Canada

PostPosted: Sun Nov 11, 2007 8:15 pm    Post subject: Reply with quote

This looks really cool.

I have a 5.5 hour flight tonight - hope the battery on my computer lasts long enough to get a good start.

I know that I could use pencil & paper - but that would likely mean taking off my shoes and socks to help me with counting.

My fellow passengers might not appreciate that Shocked

Caida

Ruud wrote:
For those who are familiar with old Dutch traditions, you know where those candy colors refer to...

a quick search on google failed to illuminate me on the Dutch candy color tradition - could it be related to Sinterklaas? Question
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mhparker
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Joined: 20 Jan 2007
Posts: 345
Location: Germany

PostPosted: Mon Nov 12, 2007 9:30 pm    Post subject: Reply with quote

Hi folks,

Good job Caida's got no internet access at the moment. Allows me to get my WT in first! Smile

It was a very enjoyable puzzle, although not too difficult. Rating probably around 1.25.

Thanks to Ruud for another interesting new idea. BTW, Para's toroidal killer looks fascinating (albeit mind-boggling!), too. Nothing like being kept on one's toes... Very Happy

Edit: Couple of typos fixed. Thanks, Andrew!

Diagonal Surprise Walkthrough

Prelims:

a) 11(3)n1 = {128/137/146/236/245} (no 9)
b) 6(3)n12 = {123} -> no 1,2,3 in r1c456 (CPE)
c) 21(3)n2 = {489/579/678} (no 1..3)
d) 20(3)n23 = {389/479/569/578} (no 1,2)
e) 7(3)n23 = {124} -> no 1,2,4 in r3c789 (CPE)
f) 21(3)n14 = {489/579/678} (no 1..3)
g) 9(2)n36 = [36/54/63/72/81]
h) 17(2)n5 = {89}, locked for n5
i) 19(3)n45 = {379/469/478/568} (no 1,2) (Note: {289} unplaceable due to no 8,9 in r6c4)
j) 10(4)n56 = {1234} -> no 1..4 in r5c6 (CPE)
k) 9(2)n5 = [27/36/45] (no 1,5..9 in r4c5)
l) 20(3)n6 = {389/479/569/578} (no 1,2)
m) 7(2)n47 = {16/25/34} (no 7..9)
n) 22(3)n89 = {589/679} -> no 9 in r9c456 (CPE)
o) 19(3)n89 = {289/379/469/478/568} (no 1)
p) 10(3)n9 = {127/136/145/235} (no 8,9)

1. Hidden triple (HT) at r46c4+r5c6 = {567} (no 3,4)
1a. -> 19(3)n45 = {568} (last combo)
1b. -> r5c3 = 8
1c. r46c4 = {56}, locked for c4 and n5

2. 17(2)n5 = [98]

3. Naked single (NS) at r5c6 = 7
3a. -> r4c5 = 2
3b. cleanup: no 7 in r3c9

4. Hidden single (HS) in 10(4)n56 at r5c7 = 2
(Note: also derivable as outie, n5)

5. 13(3)n3 = {139/157/238/256/346} = {(1/2/4)..}
(Note: {148/247} combos blocked by r1c8+r2c7)
5a. -> r1c8, r2c7 and 13(3)n3 form killer triple on {124} within n3
5b. -> no 1,2,4 elsewhere in n3

6. 14(3)n36 = {158/167/347/356} (no 9)
6a. min. r2c9+r3c8 = 8
6b. -> no 7,8 in r4c7

7. HS in r4/n6 at r4c9 = 8
7a. -> r5c8+r6c9 = [39/57]
7b. cleanup. no 1 in r4c8

8. {79} in n6 locked in r6 -> not elsewhere in r6

9. Hidden killer triple on {789} in n3 at r1c7, 13(3)n3 and 14(3)n36
9a. -> r1c7 = {789},
9b. {256/346} combos blocked for 13(3)n3 = {139/157/238} (no 4,6), and
9c. {356} combo blocked for 14(3)n36 = {158/167/347} = {(1/4)..}
9d. {14} of 14(3)n36 only available in r4c7
9e. -> r4c7 = {14}
9f. 8 only available in r3c8
9g. -> no 5 in r3c8

10. 13(3)n3 requires 1 of {37} (step 9b)
10a. -> {347} combo blocked for 14(3)n36 (step 9c) = {1(58/67)} (no 3,4)
10b. -> r24c7 = [41]

11. Innie/outie (I/O) diff. n3: r1c7 = r3c6 + r4c8 + 3
11a. -> no 6 in r4c8
11b. cleanup: no 3 in r3c9

12. 3 in n3 locked in 13(3) = {139/238} (no 5,7)

13. 5 in n3 locked in c9 -> not elsewhere in c9

14. NP on {34} at r4c68 -> no 3,4 elsewhere in r4

15. Naked triple (NT) on {123} at r2c4+r3c56 -> no 1,2,3 elsewhere in n2
15a. 3 in n2 locked in 6(3)n12
15b. -> no 3 in r1c3
15c. 2 in 6(3)n12 locked in r1c3+r2c4
15d. -> no 2 in r2c23 (CPE)

16. NP on {12} at r1c38 -> no 1,2 elsewhere in r1

17. Hidden pair (HP) in n3 at r12c8 = {12}, locked for c8

18. 10(3)n9 = {136/145/235} (no 7)
18a. must have 1 of {12}, only available in r9c9
18b. -> r9c9 = {12}

19. 20(3)n23 = [497/569/659/758]
(Note: [587] blocked by 21(3)n2 (Prelim c))
19a. -> no 9 in r1c5, no 8 in r2c6

20. r1c3 blocks {128} combo for 11(3)n1 = {137/146/236/245} (no 8) = {(1/2)..}
20a. r1c3 and 11(3)n1 form killer pair on {12} -> no 1,2 elsewhere in n1

21. HS in r2 at r2c1 = 8
21a. split 13(2) at r3c2+r4c3 = [49/67/76] (no 5)
21b. no 9 in r3c2

22. 2 in n1 locked in c3 -> not elsewhere in c3

23. 21(3)n2 = {489/579/678} (Prelim c)
23a. -> either 21(3)n2 contains a 7, or...
23b. ...both of {48} -> r3c4 = 7
23c. -> 7 in n2 locked in r1c4+r2c5+r3c4
23d. -> no 7 in r1c5
23e. Furthermore, either 21(3)n2 contains a 9, or...
23f. ...both of {78} -> r3c4 = 4
23e. -> 21(3)n2 and r3c4 require at least 1 of {49}
23f. -> 20(3)n23 cannot contain both of {49} within n2
23g. -> (from steps 23d and 23f) 20(3)n23 = [569/659]
23h. -> r1c7 = 9
23i. r1c5+r2c6 = {56}, locked for n2

Rest is just singles and cage sums.

_________________
Cheers,
Mike


Last edited by mhparker on Thu Nov 15, 2007 7:57 am; edited 1 time in total
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goooders
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Joined: 18 Feb 2007
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Location: london

PostPosted: Tue Nov 13, 2007 9:43 am    Post subject: Reply with quote

i enjoyed it too just like a normal killer once it started to crack it fell over very quickly i thought it wouldnt because of the weird cage structure
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Andrew
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Joined: 11 Aug 2006
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Location: Lethbridge, Alberta

PostPosted: Thu Nov 15, 2007 7:05 am    Post subject: Reply with quote

An enjoyable puzzle. As goooders said, it looked as if it might be difficult. There appeared to be very few useful 45s. Fortunately there was one on N3 that mattered, used in slightly different ways by Mike and myself. I felt very early that the interaction between the 21(3) and 20(3) cages in R12 would probably be needed but I delayed using it as long as I could.

I think this is the first time that I've used a Naked Quad as the first step after the Prelims. Mike got the same result with a Hidden Triple as his step 1.

I agree with Mike's rating of 1.25. Diagonal cages are always harder for human solvers so, for that reason, I won't rate it any lower.

Here is my walkthrough for Diagonal Surprise.

Prelims

a) 11(3) cage at R1C1 = {128/137/146/236/245}, no 9
b) 6(3) cage at R1C3 = {123}, CPE no 1,2,3 in R1C456
c) 21(3) cage at R1C4 = {489/579/678}, no 1,2,3
d) 20(3) cage at R1C5 = {389/479/569/578}, no 1,2
e) 7(3) cage at R1C8 = {124}, CPE no 1,2,4 in R3C789
f) 21(3) cage at R2C1 = {489/579/678}, no 1,2,3
g) 9(2) cage at R3C9 = {36}/[54/72/81], no 9, no 5,7,8 in R4C8
h) 19(3) cage at R4C4 = {289/379/469/478/568}, no 1
i) 9(2) cage at R4C5 = {18/27/36/45}, no 9
j) 10(4) cage at R4C6 = {1234}, CPE no 1,2,3,4 in R5C46, clean-up: no 5,6,7,8 in R4C5
k) 20(3) cage at R4C9 = {389/479/569/578}, no 1,2
l) 17(2) cage at R5C4 = {89}, locked for N5, clean-up: no 1 in R4C5
m) 7(2) cage at R6C2 = {16/25/34}, no 7,8,9
n) 22(3) cage at R7C5 = 9{58/67}, CPE no 9 in R9C456
o) 19(3) cage at R7C6 = {289/379/469/478/568}, no 1
p) 10(3) cage at R7C7 = {127/136/145/235}, no 8,9

1. Naked quad {1234} in R4C56 + R5C5 + R6C6, locked for N5
1a) 1 in N5 locked in 10(4) cage -> no 1 in R5C7

2. 19(3) cage at R4C4 = {568} (only remaining combination), cannot be {289/379/469/478} because 2,3,4,8,9 only in R5C3) -> R5C3 = 8, R5C4 = 9, R6C5 = 8
2a. Naked pair {56} in R46C4, locked for C4 and N5 -> R5C6 = 7, R4C5 = 2, clean-up: no 7 in R3C9

3. R5C7 = 2 (only remaining cell for 2 in 10(4) cage)

4. 6(3) cage at R1C3, 2 in R1C3 + R2C4 -> CPE no 2 in R2C23

5. 45 rule on N3 2 innies R1C7 + R3C9 – 11 = 2 outies R3C6 + R4C7
5a. Max R1C7 + R3C9 = 17 -> max R3C6 + R4C7 = 6 -> max R4C7 = 5
5b. Min R3C6 + R4C7 = 2 -> min R1C7 + R3C9 = 13, no 3, clean-up: no 6 in R4C8

6. R4C9 = 8 (hidden single in R4), clean-up: no 1 in R4C8
6a. R5C8 + R6C9 = 12 = [39/57]
6b. 7,9 in R6 locked in R6C789, locked for R6

7. R1C7 + R3C9 – 11 (step 5) = R3C6 + R4C7
7a. Max R1C7 + R3C9 = 15 -> max R3C6 + R4C7 = 4, no 4,5
7b. Min R1C7 + R3C9 = 13 (step 5b) -> min R1C7 = 7
7c. 4 in 7(3) cage at R1C8 locked in R1C8 + R2C7, locked for N3

8. Naked triple {134} in R4C678, locked for R4

9. Naked triple {123} in R2C4 + R3C56, locked for N2
9a. 3 in N2 locked in R2C4 + R3C5 for 6(3) cage -> no 3 in R1C3

10. 14(3) cage at R2C9 = {158/167/239/356} (cannot be {257} because R4C7 only contains 1,3)
10a. 1 of {158/167} must be in R4C7 -> no 1 in R2C9
10b. 2 of {239} must be in R2C9 -> no 9 in R2C9
10c. 3 of {239/356} must be in R4C7 -> no 3 in R2C9 + R3C8

11. 3 in N3 locked in 13(3) cage = 3{19/28}, no 5,6,7

12. Hidden killer triple 7,8,9 for N3 in R1C7, 13(3) cage and 14(3) cage -> 14(3) cage at R2C9 (step 10) must contain 7/8/9 = {158/167/239} (cannot be {356})
12a. 8 of {158} must be in R3C8 -> no 5 in R3C8
12b. 5 in N3 locked in R23C9, locked for C9

13. Hidden killer pair 5,6 for N3 in R3C9 and 14(3) cage -> 14(3) cage at R2C9 (step 12) must contain 5/6 = {158/167} (cannot be {239}), no 2,3,9 -> R4C7 = 1, R2C7 = 4
13a. Naked pair {12} in R1C38, locked for R1

14. 45 rule on R123 2 remaining outies R4C38 – 4 = 1 innie R3C1, min R4C38 = 8 -> min R3C1 = 4

15. 20(3) cage at R1C5 = {479/569/578}
15a. 4 of {479} must be in R1C5
15b. 9 of {569} must be in R1C7
15c. -> no 9 in R1C5

16. 21(3) cage at R1C4 = {489/579/678}
16a. 7 of {579} must be in R1C4 with {59} in R1C6 + R2C5 -> cannot place any of the remaining combinations for 20(3) at R1C5 (step 15)
16b. -> 21(3) cage at R1C4 = {489/678} = 8{49/67}, no 5, 8 locked in R1C46 for R1 and N2
16c. 9 of {489} must be in R2C5 -> no 9 in R1C6

17. 5 in N2 locked in 20(3) cage at R1C5 (step 15) = {569} (only remaining combination) -> R1C7 = 9, clean-up: no 1 in 13(3) cage in N3 (step 11)
17a. Naked pair {56} in R1C5 + R2C6, locked for N2, clean-up: no 7 in 21(3) cage (step 16b) -> R2C5 = 9
17b. Naked pair {48} in R1C46, locked for R1 and N2 -> R3C4 = 7

18. R1C9 = 3 (naked single), R3C7 = 8, R2C8 = 2, R3C8 = 6, R23C9 = [75] , R4C8 = 4, R4C6 = 3, R5C9 = 6, R6C9 = 9, R5C8 = 3 (step 6a), R1C8 = 1, R3C6 = 2, R1C3 = 2

19. Naked pair {57} in R6C78, locked for R6 -> R6C4 = 6, R4C4 = 5, clean-up: no 1,2 in R7C1

20. Naked pair {57} in R68C8, locked for C8

21. R8C6 = 9 (only remaining place in 22(3) cage), R7C5 + R9C7 = {67} (only remaining combination), no 5
21a. CPE no 6,7 in R7C7 + R9C5

22. 10(3) cage at R7C7 = {235} (only remaining combination, cannot be {127/145} because 1,2,4 only in R9C9), locked for N9 -> R9C9 = 2, R8C8 = 5, R7C7 = 3, R6C78 = [57], R7C9 = 1, R8C9 = 4, R7C8 = 9, R9C8 = 8, clean-up: no 4 in R6C2
22a. R9C8 = 8 -> R7C6 + R8C7 = 11 = [47/56]

23. R1C6 = 8 (hidden single in C6), R1C4 = 4

24. R1C1 = 7 (hidden single in R1) -> R2C4 + R3C5 = 4 = {13}
24a. Naked pair {13} in R2C2 + R3C3, locked for N1
24b. Naked pair {56} in R1C2 + R2C3, locked for N1 -> R2C1 = 8

25. R2C1 = 8 -> R3C2 + R4C3 = 13 = [49], R3C1 = 9, R4C12 = [67], R5C1 = 1, R5C5 = 4, R6C6 = 1, R5C2 = 5, R6C1 = 3, R1C2 = 6, R2C3 = 5, R1C5 = 5, R2C6 = 6, R6C2 = 2, R6C3 = 4, R7C2 = 8, R8C1 = 2, R7C4 = 2, clean-up: no 4 in R7C1

26. Naked pair {13} in R9C45, locked for R9 and N8 -> R8C4 = 8, R9C2 = 9, R8C3 = 1

and the rest is naked singles and cage sums
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Caida
Hooked
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Joined: 03 Nov 2007
Posts: 38
Location: Canada

PostPosted: Wed Nov 28, 2007 5:33 am    Post subject: walkthrough for Diagonal Surprise Reply with quote

I finally had time to finish this walkthrough. Doing it on the plane didn't work at all - my laptop battery didn't last and my mental math abilities are not as strong as they should be - and I haven't had time to look at it again until today.

I quite liked this puzzle - but kept slowing down looking to use 45 rules and only found application.


Here's my walkthrough (I've gone through it a couple times and made some adjustments to my nomenclature based on my learnings so far - but as always welcome any suggestions!)

Cheers,

Caida



Diagonal Surprise Walkthrough

Prelims:

a. 11(3)n1 = {128/137/146/236/245} (no 9)
b. 6(3)n12 = {123} (no 4..9) (no 1,2,3 in r1c456 – common peers)
c. 21(3)n14 and n2 = {489/579/678} (no 1..3)
d. 20(3)n23 and n6 = {389/479/569/578} (no 1,2)
e. 7(3)n32 = {124} (no 3,5..9) (no 1,2,4 in r3c789 – common peers)
f. 9(2)n36 and n5 = {18/27/36/45} (no 9)
g. 7(2)n47 = {16/25/34} (no 7..9)
h. 19(3)n54 and n89 = {289/379/469/478/568} (no 1)
i. 17(2)n5 = {89} (no 1..7) (8,9 locked for n5)
j. 10(4)n56 = {1234} (no 5..9) (no 1,2,3,4 in r5c6 – common peers)
k. 22(3)n89 = {589/679} (no 1..4) (no 9 in r9c456 – common peers)
l. 10(3)n9 = {127/136/145/235} (no 8,9)

m. cleanup: 9(2)n5; r4c5 no 1,5,6,7
n. cleanup: 9(2)n36: r4c8 no 5,7,8

1. 19(3)n45 no {89} combo available
1a. -> no 2 in 19(3)n45
1b. -> 19(3)45 requires either 8 or 9; only available in r5c3
1c. -> no 3..7 in r5c3
1d. -> {89} locked for r5 in c34
1e. -> no [45] in 9(2)n5 - > this would make n5 of 10(4)n56 equal {123} and would make n5 of 19(3)n45 equal {67}; no possible combination of 19(3) with {67}
1f. -> 5 locked in 19(3)n45 in n5 in r46c4
1g. -> 19(3)n45 = {568} (only possible combination with 5)
1h. -> 6 locked in 19(3)n45 in n5 in r46c4
1i. -> r5c3 = 8
1j. -> r5c4 = 9; r4c5 = 8
1k. -> r5c6 = 7; r6c6 = 2
1l. -> HS r6c6 = 2 (only 2 in 10(4)n56)
1m. cleanup 9(2)n46: r3c9 no 7

2. 2 in 6(3)n12 locked in r1c3 and r2c4
2a. -> no 2 in r2c3 (common peer)

3. 20(3)n6: r46c9 no 3 or 4 as no {789} in r5c8 (combos {389/479} only possibilities for 3 or 4)

4. 17(3)n14: r3c1 and r4c2 no 1 as no 7 or 9 in r5c1

5. 18(3)n78: r9c3 and r8c4 no 1 as no {89} in r9c5
5a. options for 18(3)n78: (order r8c4,r9c35) = [8]{19}/[297]/[396]/[8]{37}/[495]/[7]{56}
5b. -> r9c35 no 2,4

6. 13(3)n3 = {139/157/238/256/346}; ({148/247} combos blocked by 7(3)n23)
6a. -> {124} locked in n3 in 13(3)n3 and 7(3)n23
6b. -> no 1,2,4 anywhere else in n3

7. 14(3)n36 = {158/167/347/356} (no 9)
7a. min. r2c9+r3c8 = 8
7b. -> no 7,8 in r4c7
7c. Hidden Single r4c9 = 8

8. 20(3)n6 = [8][39/57]
8a. -> r5c8 no 4,6; r6c9 no 5,6
8b. 7 locked in n4 in r4c123
8c. 9 locked in n4 in r4c123
8d. cleanup: r4c8 no 1

9. Hidden killer triple on {789} in n3 at r1c7, 13(3)n3 and 14(3)n36
9a. -> r1c7 = {789},
9b. 13(3)n3 = {139/157/238} (no 4,6)
9c. 14(3)n36 = {158/167/347};
9d. -> r4c7 = {14} (no 3,5,6)
9e. -> r3c8 no 5 (8 only available in r3c8)
9f. -> 14(3)n36 combo {347} blocked by 13(3)n3 (no 3,4)
9g. -> r4c7 = 1
9h. single r2c7 = 4

10. Innies and Outies n3: r1c7 + r3c9 – r3c6 = 12
10a. r1c7 + r3c9 = either 13 or 14
10b. r3c9 no 3
10c. cleanup: r4c8 no 6

11. naked pair {34} in r4c68 -> no 3,4 elsewhere in r4

12. 3 in n3 locked in 13(3)n3 = {139/238} (no 5,7)

14. 5 in n3 locked in r23c9 -> no 5 elsewhere in c9

15. naked triple {123} in r2c4 and r3c56 -> no 1,2,3 elsewhere in n2
15a. 3 in n2 locked in 6(3)n12; -> no 3 in r1c3
15b. naked pair {12} locked in r1c38 -> no 1,2 elsewhere in r1
15c. hidden pair {12} locked in r23c8 -> no 1,2 elsewhere c8
15d. 2 in 6(3)n12 locked in r1c3 and r2c4; no 2 in r2c2 (CPE)

16. 10(3)n9 = {136/145/235} (no 7)
16a. -> r9c9 = {12}

17. 14(3)n69; min r5c9+r6c8 = 7
17a. -> r7c9 no 9
17b. -> r7c9 no 6 (r5c9 and r6c8 cannot be {35} blocked by r5c8)
17c. -> r6c8 no 6 (no combo of r57c9 = 8)

18. 11(3)n1 = {137/146/236/245}
18a. -> combo {128} blocked by r1c3 (no 8)
18b. -> r3c3 no 5 (requires 2 in either r1c1 or r2c2)
18c. killer pair {12} in n1 locked for r1c3 and 11(3)n1
18d. 2 locked in n1 in r13c3 -> no 2 elsewhere in c3

19. 20(3)n23 = [497/569/659/758] (combo [587] blocked by 21(3)n2)
19a. no 9 in r1c5, no 8 in r2c6
19b. hidden single in r2; r2c1 = 8

20. 21(3)n14 = 8 + 13(2)n14 = [49/67/76] (no 5)
20a. r3c2 no 9

21. 21(3)n2 = {489/579/678}
21a. 7 locked for n2 in either 21(3)n2 or r3c4
21b. -> r1c5 no 7
21c. cleanup r1c7 no 8 (step 19)
21d. 8 locked in n3 in r3c78 -> no 8 elsewhere in r3
21e. r2c5 no 5 (blocked by r1c7)
21f. 20(3)n23 = [569/659] other combinations blocked by 21(3)n2 and r3c4
21g. -> {56} locked in r1c5 and r2c6 for n2
21h. -> r1c7 = 9
21i. hidden single r2c5 = 9; r3c1 = 9

Now it is all over – just singles and sums
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