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# phil1882

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1. ## Lanterns of Eden.

this game seems to be impossible for the angel to win. if 50% or more lanterns are dark, the devil flips a dark lantern. otherwise, leave it alone.

3. ## golf payout problem

edit: so my dad was playing with three other people in golf for 9 holes. The bet is each player puts in \$5 dollars. There are 2 games being played. \$3 dollars of the \$5 goes to who has the most points and \$2 of the \$5 goes to the trash pot. At he conclusion of the 9 holes my dad was the only winner of the points pot and Larry was the only winner of the trash pot. The points pot should have \$12 dollars and the trash pot should have \$8. However my dad didn't have correct change nor did Larry so the pot was left with \$10 dollars. Dad took \$6 of the \$10 and Larry took \$4. The next time they played Larry said we did not distribute the money correctly. Larry said Dad should have gotten \$7 and Larry \$3. Larry's logic is if everyone contributed to the pot, the trash pot should have \$8 dollars in it. After subtracting out his contribution of \$5, he should have a net winnings of \$3 dollars not \$4 dollars, and dad \$7 dollars instead of 6. So which way is right???
4. ## golf payout problem

so my dad was playing with three other people in golf. you get 3 dollars for every hole you win, and 2 dollars for odd or unusual shots each player put in 5 dollars, but my dad didn't have the money nor did one of his fellow golfers. my dad won all the holes for 12 dollars all together, and the buddy that couldn't pay won all the money from the odd shots. so because neither he nor his partner contributed to the hole pot, my dad gets 6 dollars from the remaining two players, and his buddy gets 4 dollars for his. however, anther way to look at is, my dad won 12-5 = 7, and the buddy won 8-5 = 3 for his. what do you think?
5. ## shuffling algorithm for letters in a word

there is a way however to permute through all possibilities by swapping two letters. ABC -> ACB -> BCA -> BAC -> CAB -> CBA ABCD -> ABDC -> ACDB -.> ACBD -> ADBC -> ADCB -> BDCA -> BDAC -> BADC -> BACD -> BCAD -> BCDA -> CBDA -> CBAD -> CABD -> CADB -> CDAB -> CDBA-> DCBA -> DCAB -> DBAC -> DBCA -> DACB -> DABC
6. ## 1 + 2 = 4?

i'm not fully sure what you are asking, yes, no object is perfectly identical to another, but that doesn't mean they are necessarily worth less/more. if i have two apples, one shaped like a pear, the other like a cucumber, i still have two apples.

hi phil

9. ## Cutting pizza

your description still doesn't make sense. how can all cuts be both vertical and in any direction and no horizontal. which is it, all vertical (same direction) or any direction?
10. ## Cutting pizza

i dont really understand the question. your picture seem contrary to your description. your picture seems to suggest you can cut in any direction, but your description suggests all cuts have to be in the same direction. i'll solve both. for lines in any direction: http://mathworld.wolfram.com/CircleDivisionbyLines.html 1/2*(n^2 +n +2), n=500; 125251 5000000 = 1/2*(n^2 +n +2) n=3162 for cuts in the same direction: each cut adds 1 new region. for 500 cuts, that 501 rejoins to get 5000000 peices 4999999 cuts are needed

i think so.
12. ## Who can go the lowest?

i'd probably pick

14. ## Westworld Mafia Signups

count me in, though i may be rather slow in posting.
15. ## A bug problem

lets do several steps and see if we can develop such a function. 1 H 1 S 1 N 3 H 1 S 2 H 3 S 3 N 9 H 5 S 2 N 12 H 11 S 9 N 31 H 21 S 12 N 54 H 43 S 31 N so here are our rules. Hn = 2*Sn-1 +Nn-1 Sn = Hn-1 +Nn-1 Nn = Hn-1 so, combining we have... Sn = Nn +Nn-1 then Sn-1 = Nn-1 +Nn-2 thus Hn = 2*(Hn-2 +Hn-3) +Hn-2 Hn = 3*Hn-2 +2*Hn-3 so... F(x) = 1 +3x^2 +2x^3 +9x^4 +12x^5 ... 3x^2*F(x) = +3x^2 +9x^4 +6x^5 2x^3*F(x) = +2x^3 +6x^5 F(x)*(1 -3x^2 -2x^3) = 1 F(x) =1/(1 -3x^2 -2x^3) thus it will be the nth term of this series, whatever that is.
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