I don’t hold anything against homeless people but I do hold things against unkept creepy people
I’m fine with unkept people actually it’s mostly creepy people
God that freaks me out mentally
I do not like that man my survival instincts were kicking in he kept chuckling afyer anything he ever said
see y’all friday ;)
3 Likes
First episode is solid.
I need Morihito to shut up a bit.
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hey atlas when do you sleep
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uhh, good morning straight people, i guess 
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well that’s a bit of a personal question
if you must know i let a hamster type all my posts so it can make up for my deficit when i do
5 Likes
Good limbo to you too.
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I’d do the glitch text, but Orange yelled at me last time.
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the what
2 Likes
Zalgo
T̴̨̬̰̹̮̰̼̯̼͇̹͈͖̰̥̪̻͂̈́̈̈́̓͒̐́̍̍̈́́̒͜͝ͅḩ̶̛͍͉͕͙̩͈̥̦͍͈̐̿͐ͅȩ̶̧͇̜̘̘͈͇̥̩̻̜̣̼̳̘̲͇͔̪͇̄̂̀ ̸̡̧̜̠̹̝̣̭͎̫̆̅͒̑̃̒̒̓̈́̅̀̇́̕̚̕͜͠g̷̡̞̙͉͇̏̄̋̈́̉͋͑͐͂̈́̆̒̃̇̋̔̅̓͘͝͝͠͝l̸̨̢̙̼͔̗͍͒͋̑͐̔͒̄̀͝i̵̡̤̣̣͍̻̦̪̜̩̜̙̳̟̩̾͊̈́́̉̑͒̏ͅt̴̛̛̩͔͌̂̃̔̈̊̈̆̃̿̒̾͊͊̎̈́̎͜͠͠c̴̡̨̢̢̧̙̜̮͙̫̟̬̝̗̰̻̙̆̌̊̌̄͛͛̐̕͝ḩ̴̡̺̞͚̞̫̯͕̝̗̙̞̰̹̗͈̼̞̪̙̎̍͗̇͜͠ͅ ̶̡̨̤͙̘̬͙̦̯͈̤͔̙̻̣̳̝͋͌̌͂̊͋̇͋̈̚͘͝t̵̯̞̗͇̠̟̟̮̟̲͎̊͂̅́̊̌̽̀̇̃͘͘e̵̡̡̨͇̰̘̺̰̤͕̩̠͔̞̘̼̩̪̫̹̤̙͉͍͙͌̋̋̍̓͒̍̃͒̍͘͝ͅx̶̧̢̢̡̨̢̥̱̫̰̮̹̫͕̱̟͓̠͇̻̗͙̥̹̯̫̪̉̐́̂͑͘͘t̵̛̝̺̪̰̼̟̥̻̎̐̎̎̊͗̂̿͒͝.̷̧̛͉͓̪͇͖͇͎̲͍͙͈̠͑̃͆͆̈͐͆̅̂̊́̚͝͠͝ͅͅ
good night will wood & atlas
4 Likes
~23 minutes, no hard brute force needed.
this part needs a 'soft' brute force
(Box 8 had 2 possible cage sums (5 or 15) in the late solution, but 15 makes box 2 and 8 impossible. It forces a 29 in c6r1-2, but that makes the row 9 cage impossible, there is already a 86 cage in box 5 col 6 at that point)
Anything before that is just simple (sudoku) logic.
Would people use a second-hand thread? Where you can like post what you’re looking for and maybe become a preferential buyer if someone was to sell?
The only concern is anonymity and privacy really
1 Like
@Zone_Q11
Step by step solution.
long spoiler
Box 2, 4 and six have a caged sum between 35 and 45 and needs to be able to be divided by 4, there are only 3 of thos numbers (36, 40, 44) so those are 9, 10 and 11 sum cages. (And the last digits are 159 in thos boxes)
Box 1, 3, 7, 9 have a min sum of 21 (1 to 6) and a max sum od 39. (4 to 9)
21/3=7 (min cage sum)
39/3=13 (max cage sum)
9-11 is already taken so the sums are 7,8,12,13
7,8 are bot discarding 8 and 9, and 12,13 are both discarding 1,2.
So the remaining numbers in those boxes are 123, 126, 987 and 984. The five is forced in box 4 and 6 into col 1 and 9, which makes the box 4 into a 10 sum cage. The 5 into r6c1 and the col 1 cage is either 37 or 46. This eliminates 5 from r4c9 (otherwhise the remaining 2 digits makes a 10 sum cage, which isn’t allowed by the rules.)
Box 5 has to have a cage sum which can be divided be 3, and it has to be higher than than 39, (it can’t be 45, so only 42 is acceptable, this makes the 3 cage into a 347, and the outside of cage numbers into a 12, this forces the last five in box 4-5-6 into r4c4, inviting a 9 next to itself, wich makes the 9 in box 4 into row 6 and since 9 cant be in r4c8 (none of the c9 number would make it into a 9 or 11 sum, it needs to be in r4c7 which makes the 6 box into a 9 cage sum) forcing the 1 in row 4 into col6 (it doesn’t have space anywhere else) forcing a 2 in r6c4.
Col 4 eliminates the 6 sum in box 8. (With the 2, 5 pair in box 5) which leaves with a 5 or 15 sum. Which I already showed the solution for in the previous post.
The rest is mostly sudoku.
1 Like