voider

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"interesting". This means the puzzle is so good that it got each one of us to think differently. Thank you for the brain teaser, ujjagrawal.
Although after going through the posts, I am now convinced that bonanova has nailed the answer.
Personally I find it bizarre. I've noted that the "signal to noise ratio" (if signal means correct/true/good reasoning/answer/process, noise means bad/incorrect/false) on brainden is worse than most ad hoc forums or messages, even for the trivially easy puzzles. One of these forms is where people give answers that don't mean a thing to anyone else; the words and logic is practically gibberish in English. How does this happen? Even for most people to acknowledge a correct logical answer seems to be impossible here. If this place lacks common sense, you have to wonder what "communal" value there is given the objectives of contributing here.
In this case, the fact that we have completely different solutions means most of us are completely wrong, and yet believe we are all on the right track.
I have no problem accepting I could be wrong, but I'm the only one who has given a proof, and I have implicitly disproved all other answers. But in my experience, proofs and disproofs mean nothing to most people.

Got to say, it's __________ that five of us have completely different solutions, and no working in common.
Someone find the right word for me

The probability of four same color (thus first chair winning):
8/8 * 3/7 * 2/6 * 1/5 = 1/35. Same answer results from using combinations.
There are 70 (ordered) permutations of the 8 stamps, each with equal chance.
There are 16 (ordered) permutations of 4 of the 8 stamps, not with equal chance.
If this doesn't occur, then first chair cannot know the answer.
Second chair now knows that first chair does not see: YYXXXX
Specifically it leaves 68 permutations with equal chance.
Then, if he sees XXYYXX he wins. As in the first case, there are two permutations that satisfy this.
So he wins this way with 1/34 chance.
I believe he learns nothing else.
Third chair now knows that second chair does not see: XXYYXX
This eliminates 2 possibilities, leaving 66 with equal chance.
If he sees XXXXYY he wins, 2/66 chance.
Also if he sees XXYY?? then he wins knowing he has XY (since the others haven't won yet).
This could happen as BBWW**** or WWBB**** where **** is permutations of BBWW. 2 * 6 = 12 ways
However XXYY?? intersects with XX??XX at XXYYXX so there are 8 ways left.
So if he didn't win by XXXXYY he wins this with 8/64.
Equivalent overall: 2/66 + 64/66 (8/64) === 10/66 = 5/33
I believe he learns nothing else.
First chair:
He's survived this far, so there are 56 possibilities left. YYXX?? is not possible.
He applies same thing as third chair, he wins if he sees ??XXYY where he must have XY. Still 12 possibilities, minus 4 intersections.
Wins with chance 8/54 = 4/27.
I can't see him learning anything else.
Second chair:
48 possibilities left, YYXX?? and ??XXYY are excluded.
If he sees XX??YY he must have XY and he wins. 8 ways => 8/48 = 1/6 chance of winning like this.
If he doesn't win like that, he knows the XX, YY, XY pairs aren't there (no one sees them). XXYYXY and XXYYXX and XXYYYY aren't there, it must mean XX YY isn't there. What about XX XY? (XX XY YY, XX XY XX can't be). One mixed pair doesn't exist, this means there are two or three mixed pairs.
Thus if he sees XX??MM or MM??XX where MM = mixed pair, then he knows he has XY.
XXMMMM has 8 forms: 2x for swapping X with Y, 4x for two MM pairs as XY or YX. No more multipliers because the rest must be YY.
MMMMXX has 8 ways also.
He wins with 16/40 chance now = 2/5
Overall: 8/48 + 40/48 (16/40) = 24/48 = 1/2 chance.
Third chair: 24 left.
So I think there are 2 or 3 mixed pairs. Second chair would have won if one of 1st or 3rd pair was not mixed, therefore they must both be mixed.
I believe possibilities left are MMMMMM or MMXXMM. This would mewan third must have XY no matter what.
Can check this by seeing how many possibilities MMMMMM and MMXXMM form:
MMXXMM: 2x for XY or YX on third pair, 2x for XY/YX on 1st pair, 2x for identify of X, 1x for the rest must be YY. 8 ways.
MMMMMM: 8x for three pairs XY/YX. Remaining is also a pair, 2x for that. 16 ways.
Totals 24 so correct.
Patterns:
1 XXXX??, 2 ways
2 XX??XX, 2 ways
3 ??XXXX, 2 ways
3 XXYY??, 8 new ways
1 ??XXYY, 8 new ways
2 XX??YY, 8 new ways
2 XXMMMM or MMMMXX, 16 ways
3 MMXXMM or MMMMMM, 24 ways
By intuition or otherwise, the order of events is independent of the ways of winning (must be better way to explain). Anyhow, you can just add up the number of ways.
First chair: 10 ways
Second chair: 26 ways
Third chair: 34 ways
So the logical choice is third chair, with 34/70 chance of winning, assuming the other two students are not (color)blind and are equally logical.

I assume she will randomly choose the stamps because e.g. if you chose the first chair + the four stamps you see are the same color. So you want to maximise your chances.

The best strategy for N is to yell STOP only and as soon as he or she recieves something with property M...
You can't say STOP after they've already won.

Consider the decision tree.
You can calculate the probability that you will win, when you predetermine on which round you will say STOP (if you survive until that round).
The first round probability is 4/12=0.33
Second is 0.397979...
This increases, until some round to say STOP where the probability of winning will begin to drop.
Obviously the "general solution" is to say STOP on the round that produces the overall highest probability of winning relative to the beginning of the decision tree.
My solution would be a computation, rather than a calculation. There are multiple ways of presenting it, but none of them would look nice.

It looks strange but it's too easy. Might be harder if you have to construct it in your mind with your eyes closed.

133 is correct, = 7C2 * 6 + 7
37766040
There are several subproblems that have to be combined optimally:
1. Iterating through order 4^6 operator combinations
2. Iterating through order 2^10 postfix expressions
3. Iterating through order 7! variable combinations
While this method is faster than brute force it is still very slow. If anyone finds an optimization based on a revelational observation that changes the need to "try almost everything" please let me know.


abortoperation
d?????????????
day in monday
izj
sday in tuesday
oizq
day in days
iaq
Does the alphabet/number map change after using a letter?

1. 1 is on
2. primes are off as two divisors
3. squares of primes are on as three divisors
4. squares of squares of primes are on as odd divisors
5. nonsquare composites (the rest) are off as even divisors
6. 40000 = 200^2 so on
7. check: 40000 has 35 divisors so on
Oops correction:
4. Squares of anything are on as odd divisors.

1. 1 is on
2. primes are off as two divisors
3. squares of primes are on as three divisors
4. squares of squares of primes are on as odd divisors
5. nonsquare composites (the rest) are off as even divisors
6. 40000 = 200^2 so on
7. check: 40000 has 35 divisors so on

Boolean algebra: 1 + 1 = 1
Galois field 2: 1 + 1 = 0
Subspaces: 1 subspace + 1 subspace = 1 subspace
Sand: 1 pile + 1 pile = 1 pile
Speed of light: 1 c + 1 c = 1 c
Networks: power of 1 component + power of 1 component < power of (1 component + 1 component)
Problems: 1 small problem + 1 small problem = RAGE
Memorisation: time taken to learn something twice as long

I've looked at a few patterns, none of which seem to fit perfectly, but my intuition definitely favours D and rejects A and B overall.
The more obvious observations are the nearsymmetry along one or both diagonals, that (3, 2) is symmetrical, (3, 1) is a rotation of (3, 2), that each block has 3 of each colour, that the columns of the first row blocks have all colours.
Look at the pattern of blocks having rows or columns having two colours.
B = Black, G = Grey, W = White
1. First row blocks, horizontals  BW WG GB
2. Second row blocks, verticals  GB BW WG
3. Third row blocks
horizontals GB GB umm maybe GB again?
verticals WG GB probably BW
This would leave B and D.
This doesn't quite fit enough to my own liking... Could be a misleading coincidence?

Clarification:
The operators are binary.
If there was one variable...
With two variables...
I was not clear on this.
What I meant was, still assuming there are 7 variables:
The number of expressions with exactly one variable is 7: a, b, c, d, e, f, g
The number of expressions with two or less variables (i.e. including the 7) is 133: a, b, c, d, e, f, g, a + b, a  b, a * b, a / b, ..., f / g, ... g / a.
I have an answer now.

If you have 7 integer variables, e.g. a, b, c, d, e, f, g, and you can use each at most once in an expression with the operators +, , *, / (integer division), how many unique expressions can be formed?
E.g. (a  b) * c + f is mathematically the same as:
c * (a  b) + f
f + (a  b) * c
f + c * (a  b)
so they all correspond to a single unique expression.
If there was one variable, there are 7 unique expressions.
With two variables, I think there are 133 unique expressions.
I don't know the answer (yet), and I very much doubt anyone can find a closed form formula for the general problem. About the context where this problem originated, I suspect it was intended to be virtually unsolvable... (and deceptively mediocrelooking)

I was gonna say
Mr B is a student teacher so you can ask him.
but = W = has the right answer.
Let F(x) = Fib(x) correspond to the usual Fibonnaci sequence.
Let G(x) correspond to the sequence seeded by Al and Bob.
G(0) = Al's number, G(1) = Bob's number.
If you draw up a table decomposing G(x) to a combination of G(0) and G(1) you will see
G(6) = F(5) * G(0) + F(6) * G(1)
and
Sum from n = 0 to n = 9 of G(n) = F(11) * G(0) + (F(12)  1) * G(1)
Now it just so 'happens' that...
<you know>
So you can have the answer even though you cannot determine the sequence.

Į̴̵̻̪̖̮̺̱͎̎͋ͧ̍͆̅̑̆̔̚͠I̹̙̗͔͈̭̙̺̯̼̞̥̹̪̻̞͚͖ͥ͗ͤ̓̀͊̄̎͡͠͞ ̧̧̛̞̮̯͇̫͎͓̝͈͖̥̮̰̣̗̻͋̂ͯ̉̈́ͯ̎ͤ̔ͭ̑̎͝b̡͇͇̻̹̲͕̀̎͊͛̂́̀e̯͓̺͚̪̱͕͍̫̻̭̞͙̲̟͈̭̭͈ͧ̀̆̐̿ͯ̃ͯ̕͠t̸̴̵̞̝̩͖̱͉̝͔̙̳̮̻̭̝̖̜͒́͊̒̊̕͟ ̷̡̡́ͤ̌͂̎͑̿̍ͦ̇̂ͤͬ̈́͗ͫ̚͏̥̠̝͓̻̣͎̝̦͖͚͚̙͓̹ţ̸̵̼̱̭̻͈̬͙̺͎ͣ̿ͫ̄̉̃̃͜͟ḣ̵̸̨̪̳̹̮͖̜͓̖̻̼͎̭͋͋̍̐a̸̫̖̟̝̞̲̯̮̦̞̱̮̱͔̲͓͍̦̔̍̎̆̓͂̽͒ͬͣ̆̃ͨ͠t̵̢̪̰̘̞̗̰̦͙̤͔̯͙̘̫̙͕̪̙̮ͪ͋͌̊̏ͫ̈́ͫ́͗̄͌͑͆̊̿̍̀́̕ ̡̛̖̻̦͓̱ͯͣ̐͗ͫ͑̊͆̅̇ͥ̌̚̚y̷̵̴̩̥̜̤ͭ̓̃͒ͯͭ͊͂̋̍ͮ̉̽ͥͫ̏͌̅͠o̵̸̧ͭ̏ͣͬ҉̼͓͙͇͇̱̻͇̟̩̹͔̩̗u̴͚͈̰̺̖̖̰͇ͪͨ͌̌̀̕͢ ̛͔̪̳͕̞̈̃̔̒ͬ̓c̴̢͇͎͎͍̘̬͍̝̼͕̹̫͔ͤ̄̍͗ͫ̀͑̓ͧ͌̑̈́͆ͭͭͣͣ̍̓a̵̠͙̗̩͙͍̹̒̇ͫ̒̐͞ͅņ̴͍͔̼̪͓̘ͪͤ͒̅ͤ͆ͭ̂͆̽ͯ͂̐̽̎̎̅͟͞'̶̲̖̭̦̪͙̫̭͈̱͍̤̣̌͌̇͒̓̇͊́̚̕ţ́̋̉ͩ̓ͦ̃ͩ̋̌ͯͨ͒̊̆̅ͬ͏͙̙̺̦̘͈͔̣͕̞̼̖̮̗̤̺͓̖́ ̷͓̳͚͎̣̻̞̩̥̹̪͙̠̬͖̣ͣ̂ͮͬ̀͑̓̌̈́ͥ̉͐̆̇̍̓͜͜r̜̦̯͇̞̗̼̟̠͉̠̣ͧͫ͒ͦ͊̄͒̾ͩͯ̽̓̌̎ͥ̈̀̕͢͟͡e̴͙͍̳̣̱̖̙͕̬͉̰̗̟ͪ̔͌̓ͭ̓̉ͣ̀͋̊ͫ̊ͤ̐ͨͯ́̕͘͠ą͎̤̥͔̤̣̘͙̝͓̎͗̌̉́͘d͑ͤ͑̍̅ͨ̀̌͑ͦͩ͗ͩ͛̌ͯ̄̿́͠͏̣͙̖͎ͅ ̷̴̢̗̳̜͉̙̣̭̞̙͙͆̆̐̔̃͐̋͋͋̕̕ṱ̡̢͉͔̟̐̒ͫͮ͂͋ͬͬ͑̊̀ͮ̋̔ͮ̚͘͟͝h̶̛͕̗̦̯̰̥͈̫͓͈̖̙͒͛̈̇̋ͭ͗͒̍͌ͩ͋ͪ̊͝͞͡į̛̮̫̻̜̻̦̤̮͔̩͔̟͍̬͕ͭͥͨ͌͗̽̚͞͝s̡ͯͩ̈̓ͥ͆̑̀̓̇͂̑̓̌̈́̐ͨ̀͏̭͉͔̦͡͞
How does your signature work?
I googled the first character and there was exactly one result: this page.

My interpretation of "interior diagonals": It has n sides where n is even.
Under this interpretation, solve n / 2 = (1 + 1/30) * (n / 2  1)
yields n = 62
I know I know my paint skills are pretty good

Turns out I was wrong.
The answer is
1622125426984879983056127 = 1.622125426984879983056127 * 10^24
How? I used Anza's code and changed all longs to BigIntegers.
Moral of the story?
 (Generally) Don't switch to C++ to get 2^64  1 instead of 2^63  1
 Only Java, of contest languages, offers unlimited precision through BigInteger/BigDecimal
 In case you still don't have a high precision calculator handy, http://www.wolframal.../?i=2%5E1000000 . Python also works as a calculator, but of course Wolfram is the ultimate genius for anything.

s(100000000000000000) = 2^{80} + 2^{70} + 986*2^{60} + 2^{50} + 190*2^{40} + 2^{30} + 224*2^{20} + 2^{10} + 255
Like I said, without reading your program I can guarantee it is wrong.
s[10^17] has to be greater than 2^80 + 2^78.
superprismatic has a good estimate, but the exact breakdown is proving prohibitively expensive yet.

Everyone seems to be making several assumptions which may or may not have been intended by Cool Javelin but are unrealistic in the real world.
Assumptions are intended; he said it's a calculus problem.
I did a complete calculation of the volume (V=((pi) r square)H where (pi)=3.14159..., r = average radius of log segment and H= height = 2ft. However, since H and (pi) are constant and we are only comparing relative amounts, it is only necessary to use r square , just as we can ignore exact density, another factor in the mass.
An example of why I'm accusing everyone of calculating volume incorrectly. It's not a standard cylinder.

Now all that's left is to calculate s(10^{17}f[80]) and s(10^{17}f[81]) which according to my Java program are 4905601217378542463 and 6081262013183160703 so final answer is:
2^{80}1 +
4905601217378542463 +
6081262013183160703
I did not read your program and I don't know what s(10^17  f[80]) is considering I doubt you calculated 10^17 entries of s.
Anyway you claim 4905601217378542463 = 4.9 * 10^18 to be the number of sums involving f[80] with the sum less than 10^17.
This is not correct, since the sum from f[0] to f[77] = f[79]  2, which is less than 10^17  f[80].
Significance? Consider the sums with f[80] plus combinations of the set of f[0] to f[77]. All of them are less than 10^17 so there are 2^78  1 = 302231454903657293676543 = 3.0 * 10^23.
This shows that your program is wrong, but more importantly how are you going to compute the real number? Obviously a full dynamic programming table is not feasible. What course is this homework for anyway?

zeno considered there to be four possiblities:
space is discrete; time is discrete (discrete meaning there is a smallest unit which cannot be divided further.)
space is continuous, time is continuous (no smallest unit, infinitely divisible this is his famous tortoise achilles paradox.)
space is discrete, time is continuous.
space is continuous, time is discrete.
he came up with a seeming paradox for each scenario.
http://www.suitcaseo...ox_Achilles.htm gives a brief glimpse into the depth of the paradox.
personally, i don't really see the paradox. if you cut the amount of time traveled in half, of course the distance traveled will be halved.but if you let the amount of time traveled be constant, the way we normally view motion, achilles will most certianly pass the tortoise.
Sometimes Greek philosophers get too much credit? But in this case if you think there's nothing in it then you are just reading the paradox without knowing what its subject is (if you thought about your own first paragraph?).
As the link points out clearly, the discussion is not about whether Achilles reaches the tortoise (of course he does so in a finite amount of time, any statement saying "therefore he never catches up" is just for dramatic effect). I think the relativity argument is a bit over the top (plus I don't buy it).
To be honest I don't know what Zeno thought about space/time discreteness/continuity and I would like to know more.
Which Chair you will choose ?
in New Logic/Math Puzzles
Posted · Edited by voider
Referring to the website, not this particular thread: (Maybe in the wrong section of the forum, but it is a reply.)
Firstly I agree on the value of collaboration and openness to creativity. But I believe in the context of logic puzzles there is usually a distinction between "something I haven't thought about" and "something that makes no sense, and is based on assumptions that are difficult to intelligently comprehend". Obviously I must have some pride in what I believe to be logical or at least reasonable thinking (even if it is inaccurate) because it affects me when others can't see this distinction, or get it the wrong way round. In fact, this is the first thing I discovered on brainden, other than that it had puzzles and problems on it. Of 200+ comments on the Epimenides Paradox, I dare say 20% or less showed accurate comprehension of its meaning and solution. I discussed this with rookie and proposed a way to improve people's thinking, not just define an answer, a rare opportunity. It didn't come together. I know that generally you can't change people nor the way they think, and that's why my prior post is just an expression of my observations. If it is assumed that nothing can be done about it, and no one has tried, then what does it say about an appropriate environment on brainden? Is it appropriate that most people don't "get" one of the most famous paradoxes, doesn't it virtually encourage the notion that "it doesn't matter what's true, as long as others agree with you", or even "it doesn't matter what's true, as long as you agree with yourself [, and have fun in the process]" (deliberate hint of Hedonism, it's hard to initiate a discussion on that level) ? I would imagine a problemsolving context to do a bit better than that.
Can you express whatever you think about a logic puzzle on brainden? Yes
Should you be able to express it without second thought, and expect it to be accepted for what it is? I don't know.
Do all expressions have value to the "system"? I don't believe so. Spam, redundancy, etc.
Although not directly relevant, I will point out that the ideal logic problem solver would probably read problems, solve alone (never give up) and never come to a forum. But if I'm communicating any sense to you, you'll see that I think the worth of the community with respect to its social and objective values (if any) is a conundrum. You can't have everything, and I'm observing what brainden doesn't have.