Rob_Gandy
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Posts posted by Rob_Gandy
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This is my first try at this so I hope I do good.
The following is a ciphered quote and the name of the speaker.
MQJLGTBEVNWBGSXSBYAAJGLVKZAKDXWMSHWJENHMXIYBGIQGSBSBNLLMIJLUJVPHFNVLYTSSRPG
The key/for this, my/mystery,/lies only/in an important key word.
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I can do it with 16 knights. I'm sure that's not the minimum, but it's a start. 4 squares of 4 with the knights in front of the 2nd and 3rd pawns.
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Davie got the bullseye.
Alex: 25,20,20,3,2,1
Davie: 50,10,5,3,2,1
Jamie: 25,20,10,10,5,1
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2. Four Aces. (11*11=121 'four aces' very nice)
. . . 9 8 7
. . . 1 2 1
. . . -----
. . . 9 8 7
. 1 9 7 4
. 9 8 7
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1 1 9 4 2 7
At first I thought the title meant there were 4 1's in the factors.
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1. Easy as A-B-C.
. . . 2 8 6
. . . 8 2 6
. . -------
. . 1 7 1 6
. . 5 7 2
2 2 8 8
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2 3 6 2 3 6
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. .. .7 7 5
. . .. . 3 3
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. .2 3 2 5
2 3 2 5
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2 5 5 7 5
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5 triangles.bmp Triangles are ACF, BCD, DEG, FHI, and GJH.
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1
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1) lemon
2) lime
3) apple
4) pear
5) grape
6) orange
7) plum
8) date
9) fig
10) peach
11) mango
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What is the missing number in the sequence?
0, 1, 6, 105, 27 030,_______________, 7 608 434 000 728 254 870
The sequence
here may seem
unusual, but
examination
may reveal
obvious steps
required to
solve for each
element in it.
The method to find each term is actually pretty 'basic.'
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I was thinking along the same lines as jim. Using your definitions you cannot create an infinite monotone subsequence from a sequence of one repeated number.
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I tried this:
(12-8)-(10-8)=2
(4-4)-(6-4) =-2
(9-6)-(7-6) =2
(11-6)-(13-6)=-2
but this role does not work with 5%15=2
otherwise, the missing number should be =0
but an interresting relationship between the numbers. Unfortunately 0 isn't correct, but you can get the correct answer using your method.
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well... i guess my interpretation is not inconsistent with original problem statement, if 1-on-1 is interpreted as a one-to-one mapping i.e. a pair.
"Assume all encounters are 1-on-1 and random."
i can definitely see how one would see this and assume that in any "round" only 2 people are selected from the population, and said to have "met".
oh well, twas a good exercise of thought.
Looking back at your answer posts
(n-k)/n for for chance of survival in a given round. k being number of killers and n being number of outcomes affecting survival.
I always enjoy seeing the way people work out problems. That's one of the great things about being a teacher, seeing how people understand.
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chances of survival are probably not very high either way.
but, i showed mathematically in a prior post (unless someone finds a flaw in my reasoning, which is of course very possible), that the chance of surviving a single encounter (assuming everyone is paired together in such an encounter, with the exception of maybe one odd man out), increases as the total number of pacifists increase while the number of killers remain the same.
it is thus in your best interest to make any possible decision you can to result in as many pacifists being alive in each round as possible.
it's like the evolutionary strategy of a school of fish - band together so that although the chances of a shark getting a meal are large, within the school itself, the likelihood of any particular fish getting eaten in a single shark encounter decreases.
being a killer erodes your "protection" faster.
you may likely die either way, but you have a slightly better chance of surviving if you don't kill others, letting them live in order to take a bullet for you later in the future.
Pacifist
You: assuming everyone is paired together in such an encounter, with the exception of maybe one odd man out
Me: I worked under the assumption...And that meetings occur one at a time.
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Let's say there were 2 killers left, and 100,000 pacifists (I know, can't be more than 10 -- have patience).
Do you really think your chances of dying here are
(2-1)/(2+1) = 1/3?
Of all the other people out there they could kill, why would the killers be so likely to kill you?
Answer:
Because your chance of dying is not (n-1)/(n+1), but depends on number of non-killers out there too.
And yes I do still think that my chances of survival are still 1/3 in that situation.
My chance of dying in any given encounter does depend on the total number of people, but, if we are thinking end game, it doesn't matter to my final survival what happens to those 100,000 pacifists. My survival depends on the two killers meeting each other before one of them meets me. So the only meetings that affect my survival are (K1,K2), (K1,Me) and (K2,Me). So based on those meetings I have a 1/3 chance or survival.
I worked under the assumption that there comes a point where either there is a single killer left or no killers and some number of pacifists and thus no more meetings are important. And that meetings occur one at a time.
Also, I only considered meetings that affected my survival as important and thats how I came up with (n-1)/(n+1).
Am I on the right track or way off?
Edit:In red.
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completely irrelelvant.
(Assuming that when I leave the town the only possible meetings left involve no deaths and all meetings occur one at a time.)
No matter which I choose to be I have about a 9% chance of surviving. That is without counting any meeting that didn't involve me or a killer dying. Because none of the other meetings affect my chance of surviving at the end.
Basically it comes down to me vs. killers. Whether I'm a killer or a pacifist I cannot meet a killer or else I die.
So I need there to be 5 meetings of killers for me to survive. For each set of possible meetings, that count, my chance of survival is (n-1)/(n+1) where n is the number of killers left that aren't me. Which gives us...
9/11 * 7/9 * 5/7 * 3/5 * 1/3 = 1/11 or .0909... or 9.0909...%
And that is my take on it.
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I remember seeing a couple posts similar to this one some time back and thought I'd try my hand at one.
Here is the situation:
While scanning space for signs of life a group of scientists receive a series of messages the last of which requires a response. The messages are translated as follows:
12 % 10 = 8
4 % 6 = 4
9 % 7 = 6
11 % 13 = 6
5 % 15 = 2
14 % 16 = ___
What number do we need to respond with?
you have to figure out what operation(s) are done with the two numbers to get the last number.
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9999. No matter how I try to pair them up I have to have 3645 with a good right eye and at least 6354 with a bad right eye (or 4563 with a good left eye and at least 5436 with a bad left eye). Each pair of numbers adds to 9999.
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if neither spun the cylinder the bullets would both fire on the sixth shot.
Now, to me, spinning the cylinder a turn means a full turn to me putting it back in the same spot. So we have the same situation as the initial condition meaning that both die on the sixth shot.
Now lets assume that spinning the cylinder a turn means the next chamber is now on top. If they are both given the exact same revolver the cylinder would come out on the same side for both guys meaning that they could only spin it one way, pushing from the top down.
If the cylinder comes out on the right side of the gun then they spin it clockwise making the chamber one clockwise from the bullet on top meaning that both guns will fire on the first shot since the next cylinder is the 'active' one.
If the cylinder comes out on the left side of the gun then they spin it counterclockwise making the champer on counterclockwise from the bullet on top meaning that both guns will fire on the fifth shot.
So in all cases it is a lose/lose situation.
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A funky looking piece of electrical material has a length of 2 cm on the x-axis, which is the axis of current flow.
Assume a coordinate system where the object is sitting between x = -1 cm and x = 1 cm.
The cross-sectional shape of the object is circular, with radius varying as a function of x like so:
r(x) = 0.04 + 0.4*x^2+x^4
r(x) is in cm
A differential volume of this material has a resistance of R ohms in the the direction of current flow, where R varies with distance from the x-axis. If the distance of the differential element from the x-axis is p, then R(p) = 2*sqrt(p)
R(p) is in ohms
What is the total resistance of the whole object for current flowing along the x-axis?
Resistance R of a regular homogeneous solid is given by its resistivity rho multiplied by its length L (in the direction of current flow) divided by its cross-sectional area A:
R = rho x L / A. The unit of resistivity is ohm-cm.
R of a unit cube is thus numerically equal to rho.
If the material were homogeneous, we would slice the object into circular "steaks" of thickness (length) dL and area A(x) = pi r(x)2, compute, and add the resistances. The material is not homogeneous. Its resistivity depends on distance r from the x-axis. So there is an added step. Instead of "steaks" we have a collection of "onion rings" where each ring has a resistance commensurate with the same length dL, but area A = 2 pi r dr and resistivity rho[r]. The resistances of the rings are added in parallel. That is their conductances are added and the reciprocal of the sum is taken to get the resistance of the onion slice. The resistances of the onion slices are then summed to get the resistance of the object.
You could do this numerical procedure with a program to arbitrary accuracy using aribitrarily small values of dL and dr. Or, you could perform a double integral to get the answer exactly.
That's the procedure. The answer is left to another solver.
Does R(p) give me rho?
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That taking away one measly percentage point in the water proportion corresponds to losing 50 pounds of water?
I know right? It seems quite odd and slightly unbelievable, but, as I'm sure you know, since the water makes up such a large portion of the total mass 1% less is a significant change.
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Well it all depends on how you look at the data. If we think about totals 60/160 or 37.5% of the people taking A recovered and 65/230 or about 28.3% of the people taking B recovered, so A looks better. If we look at all four groups separately 20/100 or 20% of men recovered with A but 50/210 or 23.8% men recovered with B and 40/60 or 66.7% of women recovered with A but 15/20 or 75% or women recovered with B, so B looks better. I'd say B for now, but I'm still looking at the numbers.
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Thanks for the clarification. If that's the case,
Assuming no net gain or net loss of residents, the probability of call from floor 0 is 1/2, while probability of calls from each remaining floor is 1/24. The 2 elevators should idle at floor 0 and floor 8.
Floor 0 and Floor 9. It would result in the exact same average wait time.
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Yay for over analyzing!
Start with where you want to end and work backwards. n, n-1, n-2,...,1.
Step 1: Set aside n-1 and add one to all the other numbers n+1,n-1,n-2,...,2. You now have n-1 piles.
Step 2: Set aside the first number you have and add one to all the other numbers. If you don't have enough piles to equal the number you set aside add ones until you do.
Step 3: Repeat step 2 until you can no longer create a new set of piles.
Step 4: Count your steps.
Example of the first few iterations with 45 cards:
9 8 7 6 5 4 3 2 1 (Final Config)
9 7 6 5 4 3 2 1 8->10 8 7 6 5 4 3 2 (Step 1) I have a set of eight piles.
10 8 7 6 5 4 3 2 (This is where I end after Step 1)
8 7 6 5 4 3 2 10->9 8 7 6 5 4 3 1 1 1 (Step 2 I added 3 ones so that I would have 10 total piles.
9 8 7 6 5 4 3 1 1 1 (This is where I end after Step 2)
8 7 6 5 4 3 1 1 1 9->9 8 7 6 5 4 2 2 2 (I don't need to add any ones because I have 9 piles already)
9 8 7 6 5 4 2 2 2 (End of last step.)
8 7 6 5 4 2 2 2 9->9 8 7 6 5 3 3 3 1 (Add 1 one to make 9 piles.)
After working backward 72 steps I get to 8 8 7 6 5 4 3 2 1 1 which is what Bonanova stated was the most dificult starting position. I actually worked the problem forwards and noticed that this was happening afterwards. Oh and I hope it makes sense. >.<
Obscure Jokes
in Jokes
Posted
NaCl
------
NaOH
The base is under assault!
