EventHorizon
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Posts posted by EventHorizon
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The title is a hint and "arrangement" is referring to wolfgang's method.
I didn't initially have the second question, but the title is also a hint for it.
So why did I include "again and again" in the title if I originally only had 1 question?
Look how similar the orderings of 7 items are to wolfgang's ordering for 7.
The results for the smaller ones for the first method as well as observations:
1 (Surprise!)
2,1 (same as for wolfgang's)
1,3,2 (same as for wolfgang's)
2,3,1,4 (oddly enough, the reverse of the ordering for the second method)
3,2,4,1,5 (position of 3 and 4 are the same in the second method)
5,4,2,6,3,1 (position of 2 is the same as in the second method)
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Inspired by I thought I would give my own.
2,4,3,1,6,5,7
3,7,5,6,1,8,2,4
2,7,6,3,9,4,8,5,1
1) What would it be for 10 numbers? What method am I using?
Here's another one (but it is (hopefully) harder, so I'll give all of them up to 9):
1
1,2
1,2,3
4,1,3,2
3,1,4,5,2
6,3,2,5,4,1
1,4,3,2,6,5,7
5,7,8,6,4,1,3,2
4,5,8,6,2,7,3,1,9
2) What is the order for 10 numbers? What is the method?
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4,9,10,1,3,6,8,2,5,7
1
2,1
1,3,2
2,4,3,1
5,3,2,1,4
4,2,5,1,3,6
For the next number, start at the position of the last number, and moving circularly to the right skip over a number of empty spots equal to the number of letters in the name of the number to be placed.
1 is always in the 4th spot (if it exists) because it gets first choice and has 3 letters in the name 'one.'
2 is always in the 8th spot (if it exists) because it is 4 spots from spot 4, which means 3 need to be skipped which is the number of letters in 'two.'
etc.
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First, an excerpt from the wikipedia page for "decimal."
Consider those rational numbers which have only the factors 2 and 5 in the denominator, i.e., which can be written as p/(2a5b). In this case there is a terminating decimal representation. For instance, 1/1 = 1, 1/2 = 0.5, 3/5 = 0.6, 3/25 = 0.12 and 1306/1250 = 1.0448. Such numbers are the only real numbers which do not have a unique decimal representation, as they can also be written as a representation that has a recurring 9, for instance 1 = 0.99999…, 1/2 = 0.499999…, etc. The number 0 = 0/1 is special in that it has no representation with recurring 9.
This leaves the irrational numbers. They also have unique infinite decimal representations, and can be characterised as the numbers whose decimal representations neither terminate nor recur.
So in general the decimal representation is unique, if one excludes representations that end in a recurring 9.
The same trichotomy holds for other base-n positional numeral systems:
- Terminating representation: rational where the denominator divides some nk
- Recurring representation: other rational
- Non-terminating, non-recurring representation: irrational
A version of this even holds for irrational-base numeration systems, such as golden mean base representation.
And another excerpt from the page for "0.999..."
Every nonzero, terminating decimal has an equal twin representation with trailing 9s, such as 8.32 and 8.31999... The terminating decimal representation is almost always preferred, contributing to the misconception that it is the only representation. The same phenomenon occurs in all other bases or in any similar representation of the real numbers.The equality of 0.999... and 1 is closely related to the absence of nonzero infinitesimals in the real number system, the most commonly used system in mathematical analysis. Some alternative number systems, such as the hyperreals, do contain nonzero infinitesimals. In most such number systems, the standard interpretation of the expression 0.999... makes it equal to 1, but in some of these number systems, the symbol "0.999..." admits other interpretations that contain infinitely many 9s while falling infinitesimally short of 1.
So since decimal places are the digits after the decimal point, it seems there is one pair of equal decimal expansion representations for each positive integer and each negative integer. And this is exactly the set of all the numbers that fit the OP. Any other terminating decimal expansion will have the pair of representations equal at one or more decimal places.
It is interesting that 0 fails in this regard. The method of decimal representation we use has no way of approaching 0 with repeating 9's due to the fact that adding value in a decimal place always increases the distance from 0.
If you change the OP to require a different digit in each place (not just decimal place, and can't simply not have a value explicitely put in a place (ie, both 0's in hundreds place of 34 and 67)), you find that at least one number needs to be infinite. You obviously cannot have an infinite number equal to a finite one. But what about two infinite numbers?
Sure, if you subtract the decimal representations of two infinite numbers created by an infinite decimal expansion, you could end up with a finite decimal expansion. But it seems it isn't possible in the case where the digits are different at every position. If you try the "trailing 9's" approach you end up with ....99999.9999... = ...00000.0000..., but the number on the right is 0, while the other is infinite. But are all infinite numbers created by this decimal expansion method equal anyway? For instance, is ...22222.2222... = ...11111.1111... ? You can keep subtracting 1 from both and never reach a negative number (or even a finite number). I'm thinking every infinite number created using this method of decimal expansion is equal (and equal to aleph-naught), and so the subtraction of the decimal expansions doesn't really apply since all are the same number/concept... and the difference is simply undefined. Sure, there are higher degrees of infinity, but they are essentially 2^(one of these infinite number) or more in "value."
Hyperreals sound interesting. I can't, at the moment, understand how there can be a non-zero infinitesimal. I'll quickly look that up. It seems that they handle it essentially like the different degrees of infinity, and just call all the infinitesimals nonzero. It seems to me like they are kept as two separate concepts, and when you order values you order them first by its "standard part" and then by the infinitesimals included.
Anyone have thoughts on any of this or disagree vehemently on a point?
- Terminating representation: rational where the denominator divides some nk
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Let the mouse start at (0,0) running up.
The cat starts at (1,0).
Let g be the golden ratio.
The equation for the path is as follows:
y = (1/2g) * xg - (1/(4-2g)) * x2-g + 1
Derivation:
y' = (t-y)/(0-x)
-xy' = t-y
y-xy' = t
Since t is equal to the arc length divided by the speed of the cat...
y-xy' = (1/g) * definite integral from x to 1 of (sqrt(1+y'2))
gy-gxy' = definite integral from x to 1 of (sqrt(1+y'2))
Take the derivative of both sides (Notice the definite integral above is some constant minus the value of the integral at x, so we'll need to negate it).
gy'-gy'-gxy'' = -sqrt(1+y'2)
0 = sqrt(1+y'2) - gxy''
0 = (1/gx) - (y''/sqrt(1+y'2))
Now integrate...
c = (ln x)/g - sinh-1(y')
sinh-1(y') = -c + (ln x)/g
y' = sinh(-c + (ln x)/g)
y' = (e-cx1/g - ecx-1/g)/2
y' = (ax1/g - (1/a)x-1/g)/2 (substituted in a for e-c)
use y'(1) = 0 to find out a.
0 = (a - (1/a)) / 2, so a = 1.
y' = .5 * (x1/g - x-1/g)
Integrate again...
y = .5 * ( (1/(1+1/g)) * x1+1/g - (1/(1-1/g)) * x1-1/g ) + c
Find the value for the introduced constant using y(1) = 0.
0 = .5 * ( (1/(1+1/g)) - (1/(1-1/g)) ) + c
0 = .5 * ( (1/(1+g-1)) - (1/(1-(g-1))) ) + c
0 = .5 * ( (1/g) - (1/(2-g)) ) + c
0 = .5 * ( g-1 - g2 ) + c
0 = .5 * ( g-1 - (1+g) ) + c
0 = .5 * (-2) + c
c = 1
I substituted in prettier values (1/g = g-1) and distributed the 1/2 through
y = (1/2g) * xg - (1/(4-2g)) * x2-g + 1
done.
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if you have the mouse at (1,0) moving up at a speed of 1 and the cat at the origin moving at speed s (now found to be the golden ratio) this is what I got but couldn't as of yet verify... so it could be way off. If I can verify it, I'll post how I got it.
y = (e^(x/s)+e^(-x/s)-2)/(e^(1/s)+e^(-1/s)-2)
equivalently,
y = (cosh(x/s)-1)/(cosh(1/s)-1)
Which is pretty close to what bonanova guessed in post 3.
I found a mistake in the derivation, so my equation is wrong. I accidently omitted something which made it much easier to solve, but incorrect.
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(Hint: don't look for the curve, just examine the distance between them and how it changes with time)
Let the mouse start at (0,w) moving right with speed v.
Let (x,y) be the position of the cat that starts at the origin and moves at speed s.
Let z be the angle from the cat to the mouse (0 being straight right as usual).
Let D be the distance between the cat and the mouse.
dx/dt = s*cos z (eq1)
dD/dt = v*cos z - s (eq2)
rearranging (eq1) gives cos z = (1/s) dx/dt
plugging into (eq2) gives dD/dt = (v/s) dx/dt - s (eq3)
Integrating (eq3) gives D = (v/s) * x - st + c.
At time t=0, D=w and x=0. So w = 0-0+c, so c=w.
This means the equation for the distance between the cat and the mouse is...
D = (v/s) * x - st + w.
We need D=0 when x=vt=w.
0 = (v/s) * w - s*(w/v) + w
0 = (v/s) - (s/v) + 1 (divide by w)
0 = (s/v)^2 - (s/v) -1 (multiply by -s/v, and rearrange)
(s/v) = (-(-1) +/- sqrt(1-4(1)(-1)))/2(1) (quadratic equation)
(s/v) = (1 +/- sqrt(5))/2
Since (1-sqrt(5))/2 is negative.... the ratio is (1+sqrt(5)) / 2 = 1.618033989...
Looks like the cat is golden.
I'll try (again) to find the equation for the path in a bit.
Edit: I thought I'd include the last equation I got, but kept failing to verify. So I don't know it if is correct.
if you have the mouse at (1,0) moving up at a speed of 1 and the cat at the origin moving at speed s (now found to be the golden ratio) this is what I got but couldn't as of yet verify... so it could be way off. If I can verify it, I'll post how I got it.
y = (e^(x/s)+e^(-x/s)-2)/(e^(1/s)+e^(-1/s)-2)
equivalently,
y = (cosh(x/s)-1)/(cosh(1/s)-1)
Which is pretty close to what bonanova guessed in post 3.
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Let x be the distance to the center of the tunnel.
Let z = the angle between the direction to the center of the earth and the center of the tunnel.
This means cos z = x / r2, where r2 is the distance to the center of the earth.
x / cos z = r2, by just moving things around.
The gravity at r2 is r2*9.8/r in m/s2, where r is the radius of the earth.
So when you are x away from the center of the tunnel, the gravity downwards (towards the center of the earth) is (9.8/r)* x / cos z.
To get the gravity in the direction of the tunnel, multiply by cos z, resulting in (9.8/r)*x.
So, same characteristic equation, same roots, and the resulting function's period is the same as before.
So it doesn't matter that the tunnel doesn't go through the center of the earth. It'll still take the same time.
What this means is that in a quazi-futuristic world where there are (essentially frictionless and perfectly straight) tunnels through the earth everywhere, it'll take about 42.2 minutes to get from anywhere to anywhere else via a tunnel (assuming the presence of the many tunnels doesn't effect gravity). So even if you want to move 3 feet to the left... if just by gravity, it'll take 42.2 minutes.
Also, if you sit on the edge of a completely flat and frictionless ice rink (that is exactly perpendicular to the direction to the center of the earth at the center), you will move from one side to the other side of the rink in 42.2 minutes by just sitting there. Or if you get the right initial momentum... you could orbit the center of the rink in the same time (Edit:42.2 minutes to get to the other side, 84.4 for one whole orbit) (as is obvious from the math so far, the gravity in the x direction simply depends on the distance away from 'center' of the x axis, which also applies to a frictionless orbit of the whole earth).
Kinda cool.
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Many many years ago, a friend asked me the very same question. When I 'passed', he gave an answer pretty-well the same as EventHorizon's.
His explanation was along the lines:
Travelling through the earth governed by gravity is much the same as orbitting the earth: the same principles apply. As Yuri Gagarin's first orbit was only about 200km (ave) above the surface, it should take about the same time to go through the earth as he took to do a half orbit.
I've checked the times out, and they do seem to tally.
Comments anybody?
It takes the space shuttle 90 minutes to orbit the earth. Half of this orbit is 45 minutes, which would be the time to get to the other side of the earth. Removing the slight atmosphere encountered in LEO may make up for the 3 minute disparity (Edit: not to mention the other assumptions and imprecise numbers used).
As an aside, I read that because the density of the earth increases as you go down, you'd actually experience more gravity for a little while before it decreases, but we assumed homogeneity.
"when would you next see the light of day?"
It just came to me that if it was day on one side of the earth, it would be night on the other, so you'd need to make the return trip too. Unless it was 12 noon and the sun is directly over the tunnel... then it would be shining all the way through
. It would be interesting to see a tunnel like that at night with the sun shining from beneath you.Or it could just be the 42.2 minutes if the tunnel is perpendicular to the direction to the sun.
Or you could just say you always see the light of day, but in the form of the "light at the end of the tunnel."
So now we just need to determine if it matters if the frictionless tunnel goes through the center of the earth or not.
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Yeah, both of those questions work. Nice!
Are there others? If Jaimie's response is taken as a clue, for example.
If I ask a question for which a yes makes me guess 2 and a no makes me guess 1, which response will let me guess your number?
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Davey.
How about the question "If I pick my own number that is a positive integer less than 3, will it be greater than or equal to your number?"
Answer of no: His number is 3. My number can be 1 or 2, neither of which are greater than or equal to 3.
Answer of yes: His number is 1. My number can be 1 or 2, both of which are greater than or equal to 1.
Answer of maybe/could be/i don't know/i cannot answer: His number is 2. If my number is 1, the answer is no, but if my number is 2, the answer is yes... so it could potentially be either since he doesn't know my number.
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2
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One useful result is that inside a hollow shell, you feel no gravity. Here's a description of a proof of it (taken from
http://www.merlyn.de...vity1.htm#FiSSh):This means that you can completely ignore the parts of the earth further than you from the center of the earth, so effectively you feel the gravity of standing on a smaller earth.
The effective mass of this smaller earth would be d*(4/3)pi*r2^3, where d is the density of the earth (assumed in the OP as homogeneous) and r2 is your distance from the center of the earth. Notice that this effective mass is equal to the mass of the earth multiplied by r2^3/r^3.
Plugging the new mass and radius into the equation to calculate gravitation gives
f2 = GmE(r2^3/r^3)/r2^2 = (r2/r)*GmE/r^2 = (r2/r)*9.8m/s^2
So now we've got the gravity we can plug into some calculus.
Here's the setup for a tunnel through the center of the earth.
d2x/dt2= -9.8*x/r
dx/dt (0) = 0
x(0) = r
where r = (40*10^6)/2pi.
(time to pull out my old diff eq textbook... since I remember essentially nothing from that class.)
x''+(9.8/r)x=0
roots of the characteristic equation are +/- sqrt(9.8/r) * i.
x = c1* cos(sqrt(9.8/r)*t) + c2* sin(sqrt(9.8/r)*t)
The equation goes through half an oscillation (the answer we're looking for) in pi/(sqrt(9.8/r))
=pi*sqrt®/sqrt(9.8).
=pi*sqrt(40*10^6/(2pi)))/sqrt(9.8)
=2532.0769 seconds.
=42.2 minutes
I think I'll wait a bit to look at a tunnel not going through the center of the earth.... because I'm not sure I have much confidence in what I just did.
But if it is right, falling through a vacuum to the center of the earth would take 20 minutes. That's kinda interesting.
Does that sound plausible to anyone?
Lets see what constant gravity would be required to get to the center of the earth in the 21.1 minutes I figured it would take to get there.
40*106/2pi=.5g(2532/2)2
g=7.944 m/s2
You'd be moving slow through areas of higher gravity so you'd be accelerated quicker and be effected less by the lower gravity near the center of the earth since you'll be going quickly at that point. I'd think the equivalent constant gravity would be somewhat significantly more than half of surface gravity. ~80% doesn't seem to me like it would be too far off.
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One useful result is that inside a hollow shell, you feel no gravity. Here's a description of a proof of it (taken from
http://www.merlyn.de...vity1.htm#FiSSh):Consider an arbitrary point X inside a uniformly-charged spherical surface, and consider both an arbitrary small element of "charged" surface area A and that second small element of area B which is marked out by straight lines from the edge of the first area through the point to meet the sphere again on the opposite side of the point.The solid angles subtended by A and B at the point X are equal, and the directions XA and XB are opposite. Simple geometry makes it clear that the areas A and B (which have the same "tilt") and hence the charges are proportional to the squares of their distances from the point X. Their fields therefore cancel at that point; so there will be no net field from the whole set of such areas which covers the whole sphere, for any point X.
The argument is correct; I think I have expressed it correctly. Ramsey (s.3.2) says that Newton used it, in Principia Proposition LXX; indeed, it is in Hawking's cited book, p.880.
This means that you can completely ignore the parts of the earth further than you from the center of the earth, so effectively you feel the gravity of standing on a smaller earth.
The effective mass of this smaller earth would be d*(4/3)pi*r2^3, where d is the density of the earth (assumed in the OP as homogeneous) and r2 is your distance from the center of the earth. Notice that this effective mass is equal to the mass of the earth multiplied by r2^3/r^3.
Plugging the new mass and radius into the equation to calculate gravitation gives
f2 = GmE(r2^3/r^3)/r2^2 = (r2/r)*GmE/r^2 = (r2/r)*9.8m/s^2
So now we've got the gravity we can plug into some calculus.
Here's the setup for a tunnel through the center of the earth.
d2x/dt2= -9.8*x/r
dx/dt (0) = 0
x(0) = r
where r = (40*10^6)/2pi.
(time to pull out my old diff eq textbook... since I remember essentially nothing from that class.)
x''+(9.8/r)x=0
roots of the characteristic equation are +/- sqrt(9.8/r) * i.
x = c1* cos(sqrt(9.8/r)*t) + c2* sin(sqrt(9.8/r)*t)
The equation goes through half an oscillation (the answer we're looking for) in pi/(sqrt(9.8/r))
=pi*sqrt®/sqrt(9.8).
=pi*sqrt(40*10^6/(2pi)))/sqrt(9.8)
=2532.0769 seconds.
=42.2 minutes
I think I'll wait a bit to look at a tunnel not going through the center of the earth.... because I'm not sure I have much confidence in what I just did.
But if it is right, falling through a vacuum to the center of the earth would take 20 minutes. That's kinda interesting.
Does that sound plausible to anyone?
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Add them all and divide by 4 for the sum of the 5 weights (since each person is weighed with four others).
This gives a total weight of 303.
Subtracting away the lightest pairing (two lightest people) and heaviest pairing (two heaviest people) gives the middle person's weight.
303-114-129 = 60, so the middle person weighs 60.
The difference between the two lightest pairings gives the difference between the second and third people. 116-114=2, so the second lightest person weighs 58.
114-58 = 56, so the lightest person weighs 56.
The difference between the two heaviest pairings gives the difference between the third and fourth people. 129-125=4, so the second heaviest person weighs 64.
129-64 = 65, so the heaviest person weighs 65.
So the weights of all the students are {56, 58, 60, 64, 65}.
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If 2 is true, 1 is false. (statement 2)
If 2 is false, 1 cannot be true (statement 2)
So 1 is false. (previous two lines)
Since 1 is false, so are 9 and 10 (statement 1)
6 is true (assuming 6 is false says 6 is the last true statement = paradox)
If 2 is false, then 3 must be true (if 1,2,and 3 are false, there are 3 consecutive false, so 3 is true =paradox), and if 3 is false, 2 must be true. So how about we assume 3 is false and see what happens.
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Assume 3 is false
2 is true since if 1,2,and 3 are false then 3 is true (statement 3)
8 is true since if 8,9,and 10 are false then 3 is true (statement 3)
5 is false since 8 is true and sum of numbers from 1 to 8 is 36, if 5 was true 8 would be false (percentage would need to be simultaneously greater than 40 and less than 36).
4 is true (statement 3)
x has a factor of 6. (statement 4)
7 is true (10 is false)
The true statements are 2,4,6,7,8. So x=50. (50% are true and statement 8)
But 50 does not have a factor of 6
PARADOX
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3 is true (previous block)
8 is false (the three consecutive false statements (from 3) must include 8 now)
7 is true (statement 6)
5 is false (since 7 is true and 6 and 7 are factors of x, but adding up 1 through 7 is less than 42)
2 and 4 are true. (There needs to be 3 consecutive true statements or 10 would need to be true)
5 is a factor of x (statement 4)
2,3,4,6 and 7 are factors of x (statement 7)
Since 2,3,4,5,6, and 7 are factors of x, x is a multiple of 420. (least common multiple)
The factors of 420 (besides 1 and 420), are:
2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, and 210.
There are 22 of them.
Adding the numbers of the true statements also gives 22.
If x was any multiple of 420 other than 420 itself, it would make statement 9 true by having too many factors.
x=420
Looks like I agree with Molly Mae
So it looks like I had paradoxes on the mind when drafting that post. What I meant (in each of the three mentions of the word paradox) was contradiction. I hate when I read through one of my posts and find misused words. Too bad I'm no longer a VIP or I'd be able to fix them.
Edit: Hey, I'm a VIP again, thanks rookie1ja.
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Got it in 6 with a1,b3,c5,d8,g4,h2
Didn't notice psykomakia's post. That's what i get for making dinner while puzzling.
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7.
Obviously 8 along the diagonal will cover them all. But you can remove one from one of the corners and still have them all covered.
I'll consider the problem more carefully now.
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If 2 is true, 1 is false. (statement 2)
If 2 is false, 1 cannot be true (statement 2)
So 1 is false. (previous two lines)
Since 1 is false, so are 9 and 10 (statement 1)
6 is true (assuming 6 is false says 6 is the last true statement = contradiction)
If 2 is false, then 3 must be true (if 1,2,and 3 are false, there are 3 consecutive false, so 3 is true =contradiction), and if 3 is false, 2 must be true. So how about we assume 3 is false and see what happens.
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Assume 3 is false
2 is true since if 1,2,and 3 are false then 3 is true (statement 3)
8 is true since if 8,9,and 10 are false then 3 is true (statement 3)
5 is false since 8 is true and sum of numbers from 1 to 8 is 36, if 5 was true 8 would be false (percentage would need to be simultaneously greater than 40 and less than 36).
4 is true (statement 3)
x has a factor of 6. (statement 4)
7 is true (10 is false)
The true statements are 2,4,6,7,8. So x=50. (50% are true and statement 8)
But 50 does not have a factor of 6
Contradiction!
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3 is true (previous block)
8 is false (the three consecutive false statements (from 3) must include 8 now)
7 is true (statement 6)
5 is false (since 7 is true and 6 and 7 are factors of x, but adding up 1 through 7 is less than 42)
2 and 4 are true. (There needs to be 3 consecutive true statements or 10 would need to be true)
5 is a factor of x (statement 4)
2,3,4,6 and 7 are factors of x (statement 7)
Since 2,3,4,5,6, and 7 are factors of x, x is a multiple of 420. (least common multiple)
The factors of 420 (besides 1 and 420), are:
2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, and 210.
There are 22 of them.
Adding the numbers of the true statements also gives 22.
If x was any multiple of 420 other than 420 itself, it would make statement 9 true by having too many factors.
x=420
Looks like I agree with Molly Mae
Edit: Fixed terminology.
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If the s1 lever is initially up:
1 + 21*2 + 1 resets are needed = 44, but once all 22 counters set the lever twice the reset count will be 45.
If the s1 lever is initially down:
0 + 21*2 + 1 resets are needed = 43, but once all 22 counters set the lever twice the reset count will be 44.
If you declare everyone has visited at 43 resets in the case that the s1 lever was initially up, it is possible one person hasn't visited. If you wait until 44, it's guaranteed everyone's been there regardless of the initial state of s1.
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First, make the stamps square. Make however many fan-like folds necessary to have them end up square, being careful not to have the fan folds large enough to overlap into the area of another stamp. (Notice that the stamps need not all be the same size initially, just that the perforation needs to always be aligned rectilinearly and along the whole sheet. You would just end up needing to make all the stamps a square the size of the smallest individual length of perforation (whether it is a height or width of a stamp). This means you could have made the puzzle even harder
)Second, fan along the diagonal. Fold the top left stamp in half forwards diagonally (top left corner to bottom right). Then you will make a backwards fold that goes along the diagonal that goes through the two stamps touching that stamp (the stamp one to the right and the stamp one down). Continue through the whole sheet. This will leave you with something that looks like the following (but with 45 and 90 degree angles):
|\ | \ | \ | /| | / | |/ | |\ | | \ | | \| | /| | / | |/ | |\ | | \ | | \| | /| | / | |/ | |\ | | \ | | \| | / | / |/
Third, fold again using a fan-like method as shown in the diagram below, alternating forwards and backwards folds.
|\ | \ | \ | /| | / | |/ | |\ | | \ | | \| | /| | / | |/ | |\ | | \ | | \| | /| | / | |/ | |\ | | \ | | \| | / | / |/ to ___ <--folded back | /| | / | |/ | |\ | | \ | | \| | /| | / | |/ | |\ | | \ | | \| | /| | / | |/ | |\ | | \ | | \| | / | / |/ to ___ <--folded forward |\ | | \ | |__\| | /| | / | |/ | |\ | | \ | | \| | /| | / | |/ | |\ | | \ | | \| | / | / |/ to ___ <--folded back | /| | / | |/ | |\ | | \ | | \| | /| | / | |/ | |\ | | \ | | \| | / | / |/ to ___ <-- folded forward |\ | | \ | |__\| | /| | / | |/ | |\ | | \ | | \| | / | / |/ to ___ <-- folded back | /| | / | |/ | |\ | | \ | | \| | / | / |/ to ___ <-- folded forward |\ | | \ | |__\| | / | / |/ to ___ <--folded back | /| | / | |/__|
Now all the perforation and only the perforation is along that diagonal... so letta rip.
-
1
-
-
One tear. Explanation to come (wanted to post answer first
). -
Good analysis all, but it's actually faster than the answers given so far.
An object presenting a frontal area of F and a top area of T
traveling a distance x at a speed s
in a rain of density d and speed sr
will intercept a number of raindrops D given by
D = d x [F + T sr/s] which is minimal when s tends to infinity:
Dmin = d x F = 1,000,000 F [for the values in the OP]
D = 2 Dmin when F = T sr/s, or when s = sr T/F
The OP tells us the man does not lean, so from the given dimensions we know that T = 1.
The OP does not dictate whether the man faces forward, sideways or somewhere in between.
His forward orientation is not in the list of things that can be assumed.
F when facing forward is 12.
F has extreme values of 6 x sqrt[4.25] = 12.37 and 3.
The OP asks: what speed should you run to get only twice as wet as optimal?
One stays driest by minimizing F and running infinitely fast: Dmin = 3,000,000 drops.
The speed that doubles the optimal drop count is 10 x 1/3 = 3.33... feet/second.
If you cannot assume...
If you cannot assume an orientation in determining the optimal case, why assume one in the answer? After all, it wasn't stated that we were to look for the minimum speed needed. It seems the answer is a function of the desired run orientation... which would make the speeds possible to become twice as wet anywhere from 3.33 ft/s to still infinite (to impossible for many orientations).
(Just for fun... it's 10/( 6 - 12*|cos x| - 3*|sin x| ) ft/s, where x is the angle away from straight forward. Impossible for negative values, infinite speed for denomiator of 0.)
"Let's assume there is a speed that keeps you the driest,
but you don't have the stamina for it. My question is what
speed should you run to get only twice as wet as optimal?"
This paragraph seems to me to indicate speed is the only variable intentionally being scrutinized / optimized over. Perhaps one interpretation for the answer is dependant on the orientation by finding the optimal speed (always infinite) for that orientation, and then finding the speed to get twice as wet as that keeping that same orientation.
(Just for fun... it's 10/( 12*|cos x| + 3*|sin x| ) ft/s where x is the angle away from straight forward. x=0 degrees gives the answer Bushindo and I got, and x=90 degrees gives the answer given by bonanova)
If you cannot assume a static orientation (spin infinitely fast a possibility), constant run speed, constant wind speed / angle of rain (is that 10 ft/s its absolute speed or just its downward velocity or both?), you soak up every drop of rain touched (so no limit on "wetness") and don't drip any, or even the path taken (still a fixed distance as with Bushindo's puzzle?) to the shelter... things can get tricky quick.
Did I miss any applicable assumptions? Besides crazy ones like assuming there is not a person who will throw a bucket of water if you take too long, etc.
Could examining any of these (or other such assumptions) provide an interesting puzzle? Perhaps "what is the most wet you can get given a minimum speed, always moving directly towards the shelter, staying a box shape, and the other given assumptions?" Perhaps the runner is like the Wicked Witch of the West or Gumby and will melt according to how wet they get... "oh, what a world!" Maybe Gumby and the Wicked Witch of the West are having a race in the rain... that may make a funny picture. Two green streaks ending in puddles, one with eyes (and perhaps some clay clumps) and the other with black clothes, black hat, and broom...
-
Since you are (essentially) a box that's always in a standing position and moving through rain (assumed to be) falling straight down, you will get wet on top and on the side pointing the direction you are moving (assumed to be the largest side... no running sideways).
Like Bushindo started with, if you run infinitely fast, only the side of the box will get wet... your head will remain dry. If you imagine this face of the box as carving a space out through time, you'll see the volume will be 6 feet * 2 feet * 1000 feet = 12000 ft3 regardless of how long it takes to get there. This is because if you skew a rectangular prism parallel to two sides, the volume doesn't change.
To get twice as wet, we need the face of the box representing your head to carve out an equal area in the falling rain. I'll simplify this by ignoring the last 6 inches of the journey (your thickness), and the result is simply another skewed rectanglular prism (but one for which the distance separating the two parallel lines you are skewing along is dependant on time). So the volume/time will be 2 feet * 6 inches * 10 ft/sec = 10 ft3/sec. Since we want this volume to equal 12000 ft3, we divide them. This is 12000/10 = 1200 seconds. The needed speed is 1000 ft / 1200 seconds = 5/6 feet per second.
So it looks like Bushindo was right.
If we want to include the time it takes to get into the shelter (your thickness further will be needed, and the rain falling on your head will carve out a rectanglular prism cut diagonally in half). This will result in a total volume of 1000 ft / speed * 10 ft3/sec + .5 ft / speed * .5 * 10ft3/sec.
12000 ft3 = 1000 ft / speed * 10 ft3/sec + .5 ft / speed * .5 * 10ft3/sec.
12000 ft3 * speed = 1000 ft * 10 ft3/sec + .5 ft * .5 * 10ft3/sec.
12000 ft3 * speed = 10000 ft4/sec + 2.5 ft4/sec.
12000 ft3 * speed = 10002.5 ft4/sec.
speed = 10002.5 / 12000 ft/sec = 0.8335416666666 ft/sec
So you need to go unnoticably faster than 5/6 feet/sec.
What speed would I need to run sideways to get just as wet as running infinitely fast running normally?
We already found the area carved out running infinitely fast is 12000 ft3/sec.
The area carved out by the side of the box would simply be a quarter of the 12000 ft3/sec, so 3000ft3/sec. This leaves 9000 ft3/sec for the rain falling on your head.
So using the volume/time found previously, 10 ft3/sec, we just need to divide the volume by that.
9000/10 = 900 seconds. This leaves the speed at 1000/900 ft/sec = 1.111111 ft/sec. So going a little bit faster (.277777777 ft/sec, which means 33% faster) and moving sideways towards your destination will have you get half as wet.
Conclusion: Always run sideways in the rain.
Hmm... or maybe crawl on your side! -
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <memory.h>
#include <vector>
using namespace std;
#define GRIDSIZE 8
enum dir
{
dir_front = 0,
dir_back = 1,
dir_up = 2,
dir_down = 3,
dir_right = 4,
dir_left = 5,
num_dirs = 6,
dir_error = 7,
dir_undo = 8
};
dir facing;
dir move;
vector<int> hist;
int displacement = {0,0,-GRIDSIZE,GRIDSIZE,1,-1,0,0};
int startpos;
dir startdir;
dir tiltarray[] = {dir_error,dir_error,dir_up,dir_down,dir_right,dir_left,
dir_error,dir_error,dir_down,dir_up,dir_left,dir_right,
dir_error,dir_error,dir_back,dir_front,dir_up,dir_up,
dir_error,dir_error,dir_front,dir_back,dir_down,dir_down,
dir_error,dir_error,dir_right,dir_right,dir_back,dir_front,
dir_error,dir_error,dir_left,dir_left,dir_front,dir_back,
dir_error,dir_error,dir_error,dir_error,dir_error,dir_error,
dir_error,dir_error,dir_error,dir_error,dir_error,dir_error};
dir tilt(dir cur, dir tiltdir)
{
int val = cur*num_dirs+tiltdir;
return tiltarray;
}
void printdir(dir cur)
{
switch (cur)
{
case dir_front:
cout << "1";
break;
case dir_back:
cout << "6";
break;
case dir_up:
cout << "^";
break;
case dir_down:
cout << "v";
break;
case dir_right:
cout << ">";
break;
case dir_left:
cout << "<";
break;
default:
cout << " ";
break;
}
}
void drawgrid()
{
for(int i = 0; i < GRIDSIZE*2+1; i++) cout << "_";
cout << endl;
for(int j = 0; j < GRIDSIZE; j++)
{
cout << "|";
for(int i = 0; i < GRIDSIZE; i++)
{
printdir(facing);
//printdir(move);
if (i <= GRIDSIZE-2)
{
if ((move==dir_right) || (move==dir_left))
{
cout << "=";
}
else
{
cout << " ";
}
}
}
cout << "|" << endl;
if (j <= GRIDSIZE-2)
{
cout << "|";
for(int i = 0; i < GRIDSIZE; i++)
{
if ((move==dir_down) || (move==dir_up))
{
cout << "|";
}
else
{
cout << " ";
}
if (i < GRIDSIZE-1) cout << "X";
}
cout << "|" << endl;
}
}
for(int i = 0; i < GRIDSIZE*2+1; i++) cout << "-";
cout << endl;
}
int validmove(dir tiltdir, int verbose)
{
if (((tiltdir==dir_up)&&(hist.back()<GRIDSIZE)) ||
((tiltdir==dir_down)&&(hist.back()>=(GRIDSIZE-1)*GRIDSIZE)) ||
((tiltdir==dir_right)&&(hist.back()%GRIDSIZE==GRIDSIZE-1)) ||
((tiltdir==dir_left)&&(hist.back()%GRIDSIZE==0)))
{
if (verbose) cout << "Invalid move: Boundary hit." << endl;
return 0;
}
if (facing] != dir_error)
{
if (verbose) cout << "Invalid move: Repeating a square." << endl;
return 0;
}
if ((tilt(facing,tiltdir)==dir_front) && (hist.size() != GRIDSIZE*GRIDSIZE-1))
{
if (verbose) cout << "Invalid move: 1 would be facing up." << endl;
return 0;
}
return 1;
}
int solved()
{
if ((hist.size()==GRIDSIZE*GRIDSIZE) &&
(facing == dir_front))
return 1;
return 0;
}
int validmovecircuit(dir tiltdir)
{
if (((tiltdir==dir_up)&&(hist.back()<GRIDSIZE)) ||
((tiltdir==dir_down)&&(hist.back()>=(GRIDSIZE-1)*GRIDSIZE)) ||
((tiltdir==dir_right)&&(hist.back()%GRIDSIZE==GRIDSIZE-1)) ||
((tiltdir==dir_left)&&(hist.back()%GRIDSIZE==0)))
return 0;
if (facing] != dir_error)
return 0;
if (tilt(facing,tiltdir)==dir_front)
return 0;
return 1;
}
int solvedcircuit(dir tiltdir)
{
dir nextdir = tilt(facing,tiltdir);
int nextpos = hist.back()+displacement;
if ((hist.size()==GRIDSIZE*GRIDSIZE) &&
(nextdir == startdir) &&
(nextpos == startpos))
return 1;
return 0;
}
int main(int argc, char *argv[])
{
hist.push_back(0);
for(int i = 0; i < GRIDSIZE*GRIDSIZE; i++)
{
facing = dir_error;
move = dir_error;
}
facing = dir_front;
drawgrid();
int ch = 0;
while (1)
{
if (ch != 10) cout << "Enter command (wasd,k-solve,o-circuit,' '-undo): ";
ch = getchar();
dir tiltdir = dir_error;
switch (ch)
{
case 'w'://up
tiltdir = dir_up;
break;
case 's'://down
tiltdir = dir_down;
break;
case 'd'://right
tiltdir = dir_right;
break;
case 'a'://left
tiltdir = dir_left;
break;
case ' '://undo
tiltdir = dir_undo;
break;
case 'o'://circuit
{
int xpos,ypos,dirtemp;
cout << "X (0-" << (GRIDSIZE-1) << "): ";
cin >> xpos;
cout << "Y (0-" << (GRIDSIZE-1) << "): ";
cin >> ypos;
cout << "dir (0-f,1-b,2-u,3-d,4-r,5-l):";
cin >> dirtemp;
startdir = (dir) dirtemp;
startpos = ypos*GRIDSIZE+xpos;
hist.clear();
hist.push_back(startpos);
for(int i = 0; i < GRIDSIZE*GRIDSIZE; i++)
{
facing = dir_error;
move = dir_error;
}
facing = startdir;
hist.reserve(GRIDSIZE*GRIDSIZE+1);
while(hist.size() > 0)
{
dir nextone = dir_error;
switch(move)
{
case dir_error: nextone = dir_up; break;
case dir_up: nextone = dir_down; break;
case dir_down: nextone = dir_right; break;
case dir_right: nextone = dir_left; break;
case dir_left: nextone = dir_undo; break;
default:
cout << "Should never get here!!" << endl;
}
if (nextone == dir_undo)
{
if (hist.size() <= 1) break;
facing = dir_error;
move = dir_error;
hist.pop_back();
continue;
}
move = nextone;
if (validmovecircuit(nextone))
{
dir temp = tilt(facing,nextone);
hist.push_back(hist.back()+displacement);
facing = temp;
move = dir_error;
if (hist.size()==GRIDSIZE*GRIDSIZE)
{
dir dirs = {dir_up,dir_down,dir_right,dir_left,dir_undo};
for(int i = 0; i < 4; i++)
{
if (solvedcircuit(dirs))
{
move = dirs;
cout << "Found one!" << endl;
drawgrid();
move = dir_error;
}
}
}
}
}
cout << "Done!" << endl;
break;
}
case 'k'://solve it
{
int xpos,ypos;
cout << "X (0-" << (GRIDSIZE-1) << "): ";
cin >> xpos;
cout << "Y (0-" << (GRIDSIZE-1) << "): ";
cin >> ypos;
startpos = ypos*GRIDSIZE+xpos;
hist.clear();
hist.push_back(startpos);
for(int i = 0; i < GRIDSIZE*GRIDSIZE; i++)
{
facing = dir_error;
move = dir_error;
}
facing = dir_front;
hist.reserve(GRIDSIZE*GRIDSIZE+1);
while(hist.size() > 0)
{
dir nextone = dir_error;
switch(move)
{
case dir_error: nextone = dir_up; break;
case dir_up: nextone = dir_down; break;
case dir_down: nextone = dir_right; break;
case dir_right: nextone = dir_left; break;
case dir_left: nextone = dir_undo; break;
default:
cout << "Should never get here!!" << endl;
}
if (nextone == dir_undo)
{
if (hist.size() <= 1) break;
facing = dir_error;
move = dir_error;
hist.pop_back();
continue;
}
move = nextone;
if (validmove(nextone,0))
{
dir temp = tilt(facing,nextone);
hist.push_back(hist.back()+displacement);
facing = temp;
move = dir_error;
if (solved())
{
cout << "Found one!" << endl;
drawgrid();
}
}
}
cout << "Done!" << endl;
break;
}
default:
break;
}
if (tiltdir == dir_error) continue;
if (tiltdir == dir_undo)
{
if (hist.size() <= 1) continue;
facing = dir_error;
hist.pop_back();
move = dir_error;
drawgrid();
continue;
}
if (validmove(tiltdir,1))
{
move = tiltdir;
dir temp = tilt(facing,tiltdir);
hist.push_back(hist.back()+displacement);
facing = temp;
}
drawgrid();
}
}
I will put the commands used to find them in the title of the spoiler (or just listed if nothing was found with those commands).
I'll also remove solutions that are reflections and/or rotations of other solutions when found during the same command, but I'll leave the duplicates found with different commands (but they'll be in different spoiler boxes anyway).
Found one!
_________________
|1=>=6=< >=6 6=<|
| X X X|X|X|X|X||
|>=6=< < > ^=^ <|
||X X|X|X|X X X||
|> 6=< < > v=v <|
||X|X X|X|X|X|X||
|> ^ 6=< >=6 6 <|
||X|X|X X X X|X||
|> 1 ^=^=^=^=^ <|
||X X X X X X X||
|> v=v=v v=v=v <|
||X|X X|X|X X|X||
|> 6=< 6 6 >=6 <|
||X X|X|X|X|X X||
|>=6=< ^=^ >=6=<|
-----------------
Found one!
_________________
|1=>=6=< >=6=<=1|
| X X X|X|X X X |
|>=6 6=< >=6 6=<|
||X|X|X X X|X|X||
|> ^=^ v=v ^=^ <|
||X X X|X|X X X||
|> >=6 6 6 6=< <|
||X|X|X|X|X|X|X||
|> > ^=^ ^=^ < <|
||X|X X X X X|X||
|> >=6=< >=6=< <|
||X X X|X|X X X||
|> v=v < > v=v <|
||X|X|X|X|X|X|X||
|>=6 6=< >=6 6=<|
-----------------
Done!
k,1,0: None for a 1 at (1,0).
Found one!
_________________
|v=v 1=> v=v=v=v|
||X|X X|X|X X X||
|6 6=< >=6 >=6 6|
||X X|X X X|X|X||
|^ 6=< 6=< > ^=^|
||X|X X|X|X|X X |
|1 ^=^=^ < >=6=<|
| X X X X|X X X||
|v=v=v=v < >=6=<|
||X X X|X|X|X X |
|6=< >=6 < > v=v|
| X|X|X X|X|X|X||
|6=< >=6=< >=6 6|
||X X X X X X X||
|^=^=^=^=^=^=^=^|
-----------------
Found one!
_________________
|>=6 1=>=6 >=6=<|
||X|X X X|X|X X||
|> ^=^=^=^ >=6 <|
||X X X X X X|X||
|> >=6=< ^=^=^ <|
||X|X X|X|X X X||
|> > 6=< 1 6=< <|
||X|X|X X X|X|X||
|> > ^=^=^=^ < <|
||X|X X X X X|X||
|> >=6=< >=6=< <|
||X X X|X|X X X||
|> v=v < > v=v <|
||X|X|X|X|X|X|X||
|>=6 6=< >=6 6=<|
-----------------
Done!
Found one!
_________________
|>=6=< 1=>=6 6=<|
||X X|X X X|X|X||
|> 6=< 6=< ^=^ <|
||X|X X|X|X X X||
|> ^=^=^ < >=6=<|
||X X X X|X|X X |
|>=6=< 6=< >=6 1|
| X X|X|X X X|X||
|>=6=< ^=^=^=^ v|
||X X X X X X X||
|> v=v=v v=v >=6|
||X|X X|X|X|X|X |
|> 6=< 6 6 6 >=6|
||X X|X|X|X|X X||
|>=6=< ^=^ ^=^=^|
-----------------
Found one!
_________________
|>=6=< 1=>=6 6=<|
||X X|X X X|X|X||
|> 6=< 6=< ^=^ <|
||X|X X|X|X X X||
|> ^=^=^ < >=6=<|
||X X X X|X|X X |
|>=6 6=< < >=6=<|
| X|X|X|X|X X X||
|1 ^=^ < < >=6=<|
||X X X|X|X|X X |
|v >=6=< < > v=v|
||X|X X X|X|X|X||
|6 >=6 6=< >=6 6|
||X X|X|X X X X||
|^=^=^ ^=^=^=^=^|
-----------------
Found one!
_________________
|>=6=< 1=>=6 6=<|
||X X|X X X|X|X||
|> 6=< 6=< ^=^ <|
||X|X X|X|X X X||
|> ^=^=^ < >=6=<|
||X X X X|X|X X |
|>=6=<=1 < >=6=<|
| X X X X|X X X||
|v=v=v=v < >=6=<|
||X X X|X|X|X X |
|6=< >=6 < > v=v|
| X|X|X X|X|X|X||
|6=< >=6=< >=6 6|
||X X X X X X X||
|^=^=^=^=^=^=^=^|
-----------------
Found one!
_________________
|>=6=<=1 1=>=6=<|
||X X X X X X X||
|> v=v=v=v=v=v <|
||X|X X X X X|X||
|> 6 6=< >=6 6 <|
||X|X|X|X|X|X|X||
|> ^=^ < > ^=^ <|
||X X X|X|X X X||
|> >=6=< > >=6=<|
||X|X X X|X|X X |
|> >=6=< > >=6=<|
||X X X|X|X X X||
|> v=v < > v=v <|
||X|X|X|X|X|X|X||
|>=6 6=< >=6 6=<|
-----------------
Found one!
_________________
|>=6=<=1 1=>=6=<|
||X X X X X X X||
|> v=v=v=v=v=v <|
||X|X X X X X|X||
|> 6 6=< >=6 6 <|
||X|X|X|X|X|X|X||
|> ^=^ < > ^=^ <|
||X X X|X|X X X||
|> >=6=< >=6=< <|
||X|X X X X X|X||
|> >=6=< >=6=< <|
||X X X|X|X X X||
|> v=v < > v=v <|
||X|X|X|X|X|X|X||
|>=6 6=< >=6 6=<|
-----------------
Found one!
_________________
|>=6=<=1 1=>=6=<|
||X X X X X X X||
|> v=v=v=v=v=v <|
||X|X X X X X|X||
|> 6 6=< >=6 6 <|
||X|X|X|X|X|X|X||
|> ^=^ < > ^=^ <|
||X X X|X|X X X||
|>=6=< < > >=6=<|
| X X|X|X|X|X X |
|>=6=< < > >=6=<|
||X X X|X|X X X||
|> v=v < > v=v <|
||X|X|X|X|X|X|X||
|>=6 6=< >=6 6=<|
-----------------
Found one!
_________________
|>=6=<=1 1=>=6=<|
||X X X X X X X||
|> v=v=v v=v=v <|
||X|X X|X|X X|X||
|>=6 >=6 6=< 6=<|
| X X|X X X|X X |
|>=6 >=6 6=< 6=<|
||X|X X|X|X X|X||
|> ^=^=^ ^=^=^ <|
||X X X X X X X||
|> v=v=v v=v=v <|
||X|X X|X|X X|X||
|> 6=< 6 6 >=6 <|
||X X|X|X|X|X X||
|>=6=< ^=^ >=6=<|
-----------------
Found one!
_________________
|>=6=<=1 >=6 6=<|
||X X X X|X|X|X||
|>=6 6=< > ^=^ <|
| X|X|X|X|X X X||
|1 ^=^ < > v=v <|
||X X X|X|X|X|X||
|v=v 6=< >=6 6 <|
| X|X|X X X X|X||
|>=6 ^=^=^=^=^ <|
||X X X X X X X||
|> v=v=v v=v=v <|
||X|X X|X|X X|X||
|> 6=< 6 6 >=6 <|
||X X|X|X|X|X X||
|>=6=< ^=^ >=6=<|
-----------------
Found one!
_________________
|>=6=<=1 >=6 6=<|
||X X X X|X|X|X||
|>=6=< 1 > ^=^ <|
| X X|X|X|X X X||
|>=6=< v > v=v <|
||X X X|X|X|X|X||
|>=6=< 6 >=6 6 <|
| X X|X|X X X|X||
|>=6=< ^=^=^=^ <|
||X X X X X X X||
|> v=v=v v=v=v <|
||X|X X|X|X X|X||
|> 6=< 6 6 >=6 <|
||X X|X|X|X|X X||
|>=6=< ^=^ >=6=<|
-----------------
Done!
Found one!
_________________
|>=6=< v=v=v=v=v|
||X X|X|X X X X||
|> 1 < 6=< >=6 6|
||X|X|X X|X|X|X||
|> v < 1 < > ^=^|
||X|X|X|X|X|X X |
|> 6=< v < >=6=<|
||X X X|X|X X X||
|>=6=< 6=< >=6=<|
| X X|X X X|X X |
|v=v < v=v > v=v|
||X|X|X|X|X|X|X||
|6 6=< 6 6 >=6 6|
||X X X|X|X X X||
|^=^=^=^ ^=^=^=^|
-----------------
Done!
Found one!
_________________
|v=v=v=v=v=v=v=v|
||X X X X X X X||
|6=< 1=>=6=< >=6|
| X|X X X X|X|X |
|6=< 6=< 6=< >=6|
||X X|X|X|X X X||
|^=^=^ < ^=^=^=^|
| X X X|X X X X |
|v=v=v < v=v=v=v|
||X X|X|X|X X X||
|6=< 6=< 6=< >=6|
| X|X X X X|X|X |
|6=< 1=>=6=< >=6|
||X X X X X X X||
|^=^=^=^=^=^=^=^|
-----------------
Found one!
_________________
|v=v=v=v v=v=v=v|
||X X X|X|X X X||
|6=<=1 6 6 >=6 6|
| X X X|X|X|X|X||
|>=6=< ^=^ > ^=^|
||X X|X X X|X X |
|> 6=< 6=< >=6=<|
||X|X X|X|X X X||
|> ^=^=^ < >=6=<|
||X X X X|X|X X |
|> 1 v=v < > v=v|
||X|X|X|X|X|X|X||
|> v 6 6=< >=6 6|
||X|X|X X X X X||
|>=6 ^=^=^=^=^=^|
-----------------
Found one!
_________________
|v=v=v=v=v=v=v=v|
||X X X X X X X||
|6=<=1 v=v=v >=6|
| X X X|X X|X|X |
|>=6=< 6=< 6 >=6|
||X X|X X|X|X X||
|> 6=< 6=< ^=^=^|
||X|X X|X X X X |
|> ^=^=^ 1=>=6=<|
||X X X X X X X||
|> v=v=v v=v=v <|
||X|X X|X|X X|X||
|> 6=< 6 6 >=6 <|
||X X|X|X|X|X X||
|>=6=< ^=^ >=6=<|
-----------------
Done!
Found one!
_________________
|>=6=<=1 >=6 6=<|
||X X X X|X|X|X||
|>=6=< 1 > ^=^ <|
| X X|X|X|X X X||
|>=6=< v > v=v <|
||X X X|X|X|X|X||
|>=6=< 6 >=6 6 <|
| X X|X|X X X|X||
|>=6=< ^=^=^=^ <|
||X X X X X X X||
|> v=v=v v=v=v <|
||X|X X|X|X X|X||
|> 6=< 6 6 >=6 <|
||X X|X|X|X|X X||
|>=6=< ^=^ >=6=<|
-----------------
Found one!
_________________
|v=v=v=v=v=v=v 1|
||X X X X X X|X||
|6 6=< 1=>=6 6 v|
||X|X|X X X|X|X||
|^=^ < >=6 ^=^ 6|
| X X|X|X|X X X||
|>=6=< > ^=^=^=^|
||X X X|X X X X |
|>=6=< > v=v=v=v|
| X X|X|X|X X X||
|v=v < > 6=< >=6|
||X|X|X|X X|X|X |
|6 6=< >=6=< >=6|
||X X X X X X X||
|^=^=^=^=^=^=^=^|
-----------------
Found one!
_________________
|v=v=v=v=v >=6=<|
||X X X X|X|X X||
|6 6=<=1 6 >=6 <|
||X|X X X|X X|X||
|^=^ v=v ^=^=^ <|
| X X|X|X X X X||
|>=6 6 6 1=>=6 <|
||X|X|X|X X X|X||
|> ^=^ ^=^=^=^ <|
||X X X X X X X||
|> v=v=v v=v=v <|
||X|X X|X|X X|X||
|> 6=< 6 6 >=6 <|
||X X|X|X|X|X X||
|>=6=< ^=^ >=6=<|
-----------------
Done!
k,2,2: None for a 1 at (2,2).
Found one!
_________________
|>=6=< v=v=v=v=v|
||X X|X|X X X X||
|> 1 < 6=< >=6 6|
||X|X|X X|X|X|X||
|> v < 1 < > ^=^|
||X|X|X|X|X|X X |
|> 6=< v < >=6=<|
||X X X|X|X X X||
|>=6=< 6=< >=6=<|
| X X|X X X|X X |
|v=v < v=v > v=v|
||X|X|X|X|X|X|X||
|6 6=< 6 6 >=6 6|
||X X X|X|X X X||
|^=^=^=^ ^=^=^=^|
-----------------
Done!
Found one!
_________________
|v=v=v=v >=6 6=<|
||X X X|X|X|X|X||
|6 6=< 6 > ^=^ <|
||X|X|X|X|X X X||
|^=^ < ^ > v=v <|
| X X|X|X|X|X|X||
|1 6=< 1 >=6 6 <|
||X|X X X X X|X||
|v ^=^=^=^=^=^ <|
||X X X X X X X||
|6=< v=v v=v=v <|
| X|X|X|X|X X|X||
|6=< 6 6 6 >=6 <|
||X X|X|X|X|X X||
|^=^=^ ^=^ >=6=<|
-----------------
Found one!
_________________
|v=v=v=v >=6 6=<|
||X X X|X|X|X|X||
|6 6=< 6 > ^=^ <|
||X|X|X|X|X X X||
|^=^ < ^ >=6=< <|
| X X|X|X X X|X||
|>=6=< 1 >=6=< <|
||X X X X|X X X||
|>=6=< 1=> >=6=<|
| X X|X X X|X X |
|v=v < v=v > v=v|
||X|X|X|X|X|X|X||
|6 6=< 6 6 >=6 6|
||X X X|X|X X X||
|^=^=^=^ ^=^=^=^|
-----------------
Found one!
_________________
|v=v=v=v >=6 6=<|
||X X X|X|X|X|X||
|6 6=< 6 > ^=^ <|
||X|X|X|X|X X X||
|^=^ < ^ > >=6=<|
| X X|X|X|X|X X |
|>=6=< 1 > >=6=<|
||X X X X|X X X||
|>=6=< 1=> >=6=<|
| X X|X X X|X X |
|v=v < v=v > v=v|
||X|X|X|X|X|X|X||
|6 6=< 6 6 >=6 6|
||X X X|X|X X X||
|^=^=^=^ ^=^=^=^|
-----------------
Found one!
_________________
|v=v=v=v=v=v=v=v|
||X X X X X X X||
|6 6=< 6=< >=6 6|
||X|X|X|X|X|X|X||
|^=^ < ^ < > ^=^|
| X X|X|X|X|X X |
|>=6=< 1 < >=6=<|
||X X X X|X X X||
|> 1 >=6=< >=6=<|
||X|X|X X X|X X |
|> v >=6=< > v=v|
||X|X X X|X|X|X||
|> 6=< 6=< >=6 6|
||X X|X|X X X X||
|>=6=< ^=^=^=^=^|
-----------------
Found one!
_________________
|v=v=v=v=v=v=v=v|
||X X X X X X X||
|6 6=< v=v=v >=6|
||X|X|X|X X|X|X |
|^=^ < 6=< 6 >=6|
| X X|X X|X|X X||
|6=< <=1 < ^=^=^|
||X|X X X|X X X |
|^ < v=v < v=v=v|
||X|X|X|X|X|X X||
|1 < 6 6=< 6 >=6|
| X|X|X X X|X|X |
|6=< ^=^=^=^ >=6|
||X X X X X X X||
|^=^=^=^=^=^=^=^|
-----------------
Found one!
_________________
|>=6=< v=v >=6=<|
||X X|X|X|X|X X||
|> 6=< 6 6 >=6 <|
||X|X X|X|X X|X||
|> ^=^=^ ^=^=^ <|
||X X X X X X X||
|>=6=<=1 v=v=v <|
| X X X X|X X|X||
|v=v=v >=6 >=6 <|
||X X|X|X X|X X||
|6=< 6 >=6 >=6=<|
| X|X|X X|X X X |
|6=< ^=^=^ 1=>=6|
||X X X X X X X||
|^=^=^=^=^=^=^=^|
-----------------
Done!
Well, that was interesting. There were quite a few solutions in general, and a few spots with none.
Note: All reflections are shown twice by the code due to two directions to start moving.
o,0,0,0: None for a 1 at (0,0).
o,1,0,0: None for a 1 at (1,0).
o,2,0,0: None for a 1 at (2,0).
o,3,0,0: None for a 1 at (3,0).
o,1,1,0: None for a 1 at (1,1).
Two of them.Found one!
_________________
|v=v=v=v=v >=6=<|
||X X X X|X|X X||
|6=<=1=> 6 >=6 <|
| X X X|X|X X|X||
|>=6=< > ^=^=^ <|
||X X|X|X X X X||
|> 6=< >=6 >=6=<|
||X|X X X|X|X X |
|> ^=^=^=^ >=6=<|
||X X X X X X X||
|> v=v=v v=v=v <|
||X|X X|X|X X|X||
|> 6=< 6 6 >=6 <|
||X X|X|X|X|X X||
|>=6=< ^=^ >=6=<|
-----------------
Found one!
_________________
|v=v=v=v=v=v=v=v|
||X X X X X X X||
|6=<=1 >=6=< >=6|
| X X|X|X X|X|X |
|v=v=v > 6=< >=6|
||X X X|X|X X X||
|6 6=< > ^=^=^=^|
||X|X|X|X X X X |
|^=^ < > v=v=v=v|
| X X|X|X|X X X||
|v=v < > 6=< >=6|
||X|X|X|X X|X|X |
|6 6=< >=6=< >=6|
||X X X X X X X||
|^=^=^=^=^=^=^=^|
-----------------
Done!
o,3,1,0: None for a 1 at (3,1).
o,2,2,0: None for a 1 at (2,2).
o,3,2,0: None for a 1 at (3,2).
o,3,3,0: None for a 1 at (3,3).
It turns out that there is only one circuit (unique ignoring reflections and rotations) without any 1's. (found setting die in top left corner to point right (easy to show it cannot be anything but right or down and still complete a circuit)).
_________________
|>=6=< v=v >=6=<|
||X X|X|X|X|X X||
|> 6=< 6 6 >=6 <|
||X|X X|X|X X|X||
|> ^=^=^ ^=^=^ <|
||X X X X X X X||
|> v=v=v=v=v=v <|
||X|X X X X X|X||
|> 6 6=< >=6 6 <|
||X|X|X|X|X|X|X||
|> ^=^ < > ^=^ <|
||X X X|X|X X X||
|> v=v < > v=v <|
||X|X|X|X|X|X|X||
|>=6 6=< >=6 6=<|
-----------------
So there are 2 solutions (unique ignoring reflections and rotations) for a circuit containing one 1, and they both happen with the 1 a knights move from a corner.
And there exists a single circuit (unique ignoring reflections and rotations) without a 1.
I was thinking there would be more.
An Odd Fact
in New Logic/Math Puzzles
Posted