An even number of odd handshakes
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4 members have voted
December 3rd, 2017 at 9:53:43 PM
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At the 2018 WoV Spring Fling there were many handshakes. The total number of people who shook hands an odd number of times was even. Prove this to be true.
Free beer to the first acceptable proof.
Free beer to the first acceptable proof.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 3rd, 2017 at 10:22:25 PM
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There were 4 people. They all shook hands with every other person, once each. So 4 people shook 3 hands each.
#Easy
#Actually,SeemsTooEasy
#Meh
#ShipTheSherbert
#Covfefe
#Easy
#Actually,SeemsTooEasy
#Meh
#ShipTheSherbert
#Covfefe
December 3rd, 2017 at 10:35:13 PM
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Quote: WizardAt the 2018 WoV Spring Fling there were many handshakes.
If you can beat games using time travel, is that AP or cheating?
100% risk of ruin
December 3rd, 2017 at 11:12:26 PM
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Quote: RogerKintQuote: WizardAt the 2018 WoV Spring Fling there were many handshakes.
If you can beat games using time travel, is that AP or cheating?
Well we know what he did with the proceeds form the sale of the site,
he went out and bought himself one of those Deloreans with the flux capacitor.
December 4th, 2017 at 6:21:25 AM
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Quote: RogerKintQuote: WizardAt the 2018 WoV Spring Fling there were many handshakes.
If you can beat games using time travel, is that AP or cheating?
It has happened already with disastrous results
https://wizardofvegas.com/forum/off-topic/general/646-a-fool-proof-system-to-beat-roulette/
For Whom the bus tolls; The bus tolls for thee
December 4th, 2017 at 6:33:04 AM
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Quote: RSThere were 4 people. They all shook hands with every other person, once each. So 4 people shook 3 hands each.
Okay, there's one situation where it works but you have to show it is true of every situation.
#NoBeerForYou
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 4th, 2017 at 7:32:59 AM
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I am a physicist not a mathematician, which must be why I have just written the must garbled and poorly-written mathematical proof in the history of humanity.
Even-ness can be defined as being congruent with 0mod(2), odd-ness can be defined as congruence with 1mod(2)
Every handshake involves two "half-shakes" -one "half-shake" per person. Therefore, no matter how many handshakes there are, the total number of "half-shakes" must be an even number because
if n = total number of handshakes, then the total number of 'half-shakes' =2*n, and 2*n must be an even number, mathematically 2n | 0mod(2) (2n is congruent with 0mod(2) )
Note that when each nerd at the event shakes hands, they are only counting their own "half-shake"
Now, let k =number of people who shook hands an even number of times, 0mod(2)
m=number of people who shook hands an odd number of times, 1mod(2)
the condition that must be satisfied, because 2n | 0mod(2), is:
0mod(2) = k * 0mod(2)+m * 1mod(2) (Eq. 1)
No matter whether k is even or odd, the expression (k * 0mod(2) ) must be even, i.e., (k * 0mod(2) ) | 0mod(2), . So this means that the second term ( m * 1mod(2) ) must also be even, i.e. it must congruent with 0mod(2) to satisfy Equation 1. The only way for ( m * 1mod(2) ) to be even is if m is even, or if m | 0mod(2).
Thus m, the number of people having an odd number of handshakes, must be even.
In English
if you ask every person how many handshakes they experienced, the sum of their answers must be even (because there are two people for each handshake.). Now separate all the people into two groups; Group 1: those who had an odd number of handshakes and Group 2: those who had an even number of handshakes.
In Group 2, the sum of the handshakes will always be even, because any number of people, whether odd or even, times an even number of handshakes will yield a total number that is an even number.
Thus, the total number of handshakes in group 1 must also be even. Since everyone in group 1 had an odd number of handshakes, the only way the sum of their handshakes can be even is if there is an even number of people in Group 1.
Every handshake involves two "half-shakes" -one "half-shake" per person. Therefore, no matter how many handshakes there are, the total number of "half-shakes" must be an even number because
if n = total number of handshakes, then the total number of 'half-shakes' =2*n, and 2*n must be an even number, mathematically 2n | 0mod(2) (2n is congruent with 0mod(2) )
Note that when each nerd at the event shakes hands, they are only counting their own "half-shake"
Now, let k =number of people who shook hands an even number of times, 0mod(2)
m=number of people who shook hands an odd number of times, 1mod(2)
the condition that must be satisfied, because 2n | 0mod(2), is:
0mod(2) = k * 0mod(2)+m * 1mod(2) (Eq. 1)
No matter whether k is even or odd, the expression (k * 0mod(2) ) must be even, i.e., (k * 0mod(2) ) | 0mod(2), . So this means that the second term ( m * 1mod(2) ) must also be even, i.e. it must congruent with 0mod(2) to satisfy Equation 1. The only way for ( m * 1mod(2) ) to be even is if m is even, or if m | 0mod(2).
Thus m, the number of people having an odd number of handshakes, must be even.
In English
if you ask every person how many handshakes they experienced, the sum of their answers must be even (because there are two people for each handshake.). Now separate all the people into two groups; Group 1: those who had an odd number of handshakes and Group 2: those who had an even number of handshakes.
In Group 2, the sum of the handshakes will always be even, because any number of people, whether odd or even, times an even number of handshakes will yield a total number that is an even number.
Thus, the total number of handshakes in group 1 must also be even. Since everyone in group 1 had an odd number of handshakes, the only way the sum of their handshakes can be even is if there is an even number of people in Group 1.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
December 4th, 2017 at 8:49:55 AM
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I think this has been explained already, but I'll try to do a simple one.
If there are N people, each person will shake hands N-1 times to shake hands with everybody.
In order to shake hands an odd number of times, N must be an even number, so that N -1 (total handshakes per person) is odd.
Since N is an even number, and for every person in N there are an odd number of handshakes, the number of people with an odd number of handshakes is also N, which is an even number.
If there are N people, each person will shake hands N-1 times to shake hands with everybody.
In order to shake hands an odd number of times, N must be an even number, so that N -1 (total handshakes per person) is odd.
Since N is an even number, and for every person in N there are an odd number of handshakes, the number of people with an odd number of handshakes is also N, which is an even number.
December 4th, 2017 at 8:56:26 AM
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Quote: gordonm888I am a physicist not a mathematician, which must be why I have just written the must garbled and poorly-written mathematical proof in the history of humanity.
I believe your proof only works in a perfect vacuum.
My goal of being well informed conflicts with my goal of remaining sane.
December 4th, 2017 at 9:07:31 AM
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Let N be # of people where N = a non-negative integer. N people shake hands with N-1 people.
If N is odd then N-1 is even. Thus 0 people shake hands an odd # of times. Zero is an EVEN number
If N is even then N-1 is odd. Thus in either case the assumption is true.
My goal of being well informed conflicts with my goal of remaining sane.
December 4th, 2017 at 9:41:38 AM
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Quote: Dalex64I think this has been explained already, but I'll try to do a simple one.
If there are N people, each person will shake hands N-1 times to shake hands with everybody.
In order to shake hands an odd number of times, N must be an even number, so that N -1 (total handshakes per person) is odd.
Since N is an even number, and for every person in N there are an odd number of handshakes, the number of people with an odd number of handshakes is also N, which is an even number.
That would work if everyone shook every other hand. However, not necessarily every two people will shake.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 4th, 2017 at 9:45:08 AM
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Quote: gordonm888I am a physicist not a mathematician, which must be why I have just written the must garbled and poorly-written mathematical proof in the history of humanity.
Even-ness can be defined as being congruent with 0mod(2), odd-ness can be defined as congruence with 1mod(2)
Every handshake involves two "half-shakes" -one "half-shake" per person. Therefore, no matter how many handshakes there are, the total number of "half-shakes" must be an even number because
if n = total number of handshakes, then the total number of 'half-shakes' =2*n, and 2*n must be an even number, mathematically 2n | 0mod(2) (2n is congruent with 0mod(2) )
Note that when each nerd at the event shakes hands, they are only counting their own "half-shake"
Now, let k =number of people who shook hands an even number of times, 0mod(2)
m=number of people who shook hands an odd number of times, 1mod(2)
the condition that must be satisfied, because 2n | 0mod(2), is:
0mod(2) = k * 0mod(2)+m * 1mod(2) (Eq. 1)
No matter whether k is even or odd, the expression (k * 0mod(2) ) must be even, i.e., (k * 0mod(2) ) | 0mod(2), . So this means that the second term ( m * 1mod(2) ) must also be even, i.e. it must congruent with 0mod(2) to satisfy Equation 1. The only way for ( m * 1mod(2) ) to be even is if m is even, or if m | 0mod(2).
Thus m, the number of people having an odd number of handshakes, must be even.
In English
if you ask every person how many handshakes they experienced, the sum of their answers must be even (because there are two people for each handshake.). Now separate all the people into two groups; Group 1: those who had an odd number of handshakes and Group 2: those who had an even number of handshakes.
In Group 2, the sum of the handshakes will always be even, because any number of people, whether odd or even, times an even number of handshakes will yield a total number that is an even number.
Thus, the total number of handshakes in group 1 must also be even. Since everyone in group 1 had an odd number of handshakes, the only way the sum of their handshakes can be even is if there is an even number of people in Group 1.
I'm satisfied with this. I owe you a beer. Since you are a physics man, here is one for you.
Two glasses of beer are put in a freezer. The two glasses of beer are identical in every way except one starts at a warmer temperature. Is it possible that the warm beer can freeze first? If yes, and I claim it is, how is that possible?
Note that I don't claim it is possible in every situation but carefully contrived it could be.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 4th, 2017 at 9:46:09 AM
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Start with zero handshakes. Everybody has shaken hands an even number (zero) of times, so the number of people with an odd number of handshakes is zero, which is even.
For each subsequent handshake, one of three things is true:
1. Prior to the shake, both had an even number, so now both have an odd number; the number of people with an odd number of handshakes increases by two.
2. Prior to the shake, both had an odd number, so now both have an even number; the number of people with an odd number of handshakes decreases by two.
3. Prior to the shake, one had an odd number, which becomes even, and the other had an even number, which becomes odd, so the number of people with an odd number of handshakes stays the same.
The number of people with an odd number of handshakes never changes by an odd number, and since it starts with an even number (zero), it will always be even.
December 4th, 2017 at 9:47:16 AM
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Quote: ThatDonGuy
Start with zero handshakes. Everybody has shaken hands an even number (zero) of times, so the number of people with an odd number of handshakes is zero, which is even.
For each subsequent handshake, one of three things is true:
1. Prior to the shake, both had an even number, so now both have an odd number; the number of people with an odd number of handshakes increases by two.
2. Prior to the shake, both had an odd number, so now both have an even number; the number of people with an odd number of handshakes decreases by two.
3. Prior to the shake, one had an odd number, which becomes even, and the other had an even number, which becomes odd, so the number of people with an odd number of handshakes stays the same.
The number of people with an odd number of handshakes never changes by an odd number, and since it starts with an even number (zero), it will always be even.
Nice one! I like it. Too late for the beer though.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 4th, 2017 at 9:49:48 AM
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Quote: WizardAt the 2018 WoV Spring Fling there were many handshakes. The total number of people who shook hands an odd number of times was even. Prove this to be true.
First, I contest the claim that "At the 2018 WoV Spring Fling there were many handshakes." At least not as of today, since the 2018 WoV Spring Fling has not yet occurred.
Second, if we were to assume that this is just a typo and that it is a reference to the 2017 WoV Spring Fling, I have these thoughts:
Of all the attendees, there was just that one weirdo who looked as if he would have a really odd "handshake". I don't think anyone wanted to risk touching him, so he went around just shaking his hand out in front of himself.
Does that really constitute an even number of odd handshakes?
And no, I can't testify to that situation as fact, because I missed the 2017 WoV Spring Fling.
December 4th, 2017 at 11:54:29 AM
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Quote: WizardNice one! I like it. Too late for the beer though.
I think I am already at the point where I can't alky my handlehol...
December 4th, 2017 at 2:04:21 PM
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My interpretation of the problem is that there is no guarantee that each person will shake the hand of every other person; e.g., some people may have 0 or 1 handshakes because they are shy or whatever..
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
December 4th, 2017 at 2:33:10 PM
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Quote: gordonm888My interpretation of the problem is that there is no guarantee that each person will shake the hand of every other person; e.g., some people may have 0 or 1 handshakes because they are shy or whatever..
That is a correct assumption. Perhaps I should have used a larger group to make the problem more clear. Shall we say the first day of the Global Gaming Expo.
The thing about making it the 2018 WoV spring fling was meant to make the point that the even number of odd handshakes will work with any size with with any handshaking propensity, including an unknown future group.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 4th, 2017 at 2:41:42 PM
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Quote: Wizard
I'm satisfied with this. I owe you a beer. Since you are a physics man, here is one for you.
Two glasses of beer are put in a freezer. The two glasses of beer are identical in every way except one starts at a warmer temperature. Is it possible that the warm beer can freeze first? If yes, and I claim it is, how is that possible?
Note that I don't claim it is possible in every situation but carefully contrived it could be.
I have a friend who wrote a masters thesis in which he designed a nuclear reactor that is cooled with beer rather than with water. Unfortunately, I do not know the hydrodynamic properties of Beer, but will assume they are very similar to water.
1. Water has a maximum density at about 4oC and is less dense when it is colder or hotter. (this is why solid ice floats on top of cold water.) Therefore, a glass of beer/water at 4oC, will actually contain more fluid than a glass of beer that is warmer, and might possibly (I imagine) take longer to freeze given certain initial conditions.
or
2. Heat is kinetic energy, but a water molecule also has potential energy. I seem to remember there is a threshold temperature above which the covalent oxygen bonds in the water molecule will stretch/relax and contain less potential energy. As you try to cool such a relaxed (warm) H2O molecule, the covalent oxygen bond compresses, thus converting kinetic energy (heat) into potential energy (internal chemical energy) which would rapidly lower the temperature of the glass of beer - and an identical but colder water molecule that started out with the higher potential energy level would not have the benefit of this kind of cooling.
or
3. Sleeping Beauty drills a cylindrical hole in a sphere, cuts it in half and puts the two parts in two envelopes . . .
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
December 4th, 2017 at 4:38:19 PM
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Quote: gordonm888I have a friend who wrote a masters thesis in which he designed a nuclear reactor that is cooled with beer rather than with water. Unfortunately, I do not know the hydrodynamic properties of Beer, but will assume they are very similar to water.
1. Water has a maximum density at about 4oC and is less dense when it is colder or hotter. (this is why solid ice floats on top of cold water.) Therefore, a glass of beer/water at 4oC, will actually contain more fluid than a glass of beer that is warmer, and might possibly (I imagine) take longer to freeze given certain initial conditions.
or
2. Heat is kinetic energy, but a water molecule also has potential energy. I seem to remember there is a threshold temperature above which the covalent oxygen bonds in the water molecule will stretch/relax and contain less potential energy. As you try to cool such a relaxed (warm) H2O molecule, the covalent oxygen bond compresses, thus converting kinetic energy (heat) into potential energy (internal chemical energy) which would rapidly lower the temperature of the glass of beer - and an identical but colder water molecule that started out with the higher potential energy level would not have the benefit of this kind of cooling.
or
3. Sleeping Beauty drills a cylindrical hole in a sphere, cuts it in half and puts the two parts in two envelopes . . .
Very good answer, thank you. I hope our resident science man, Doc, will give some thoughtful comments about it but the physics goes mostly over my head. That said, this question is usually phrased as filling an ice tray with water and sticking it in the freezer. I read somewhere that if warmer water is put in, it may actually freeze faster, because more water evaporates and thus there is less water needed to freeze.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
December 4th, 2017 at 5:48:05 PM
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The by-far-more-common explanation (from decades ago) as to why a warm/hot ice tray would lead more quickly to frozen cubes:Quote: WizardI read somewhere that if warmer water is put in, it may actually freeze faster, because more water evaporates and thus there is less water needed to freeze.
Remember that in those days, we did not have ice makers in our refrigerators and had to make cubes from trays of water (with dividers) placed in the freezer compartment. Most folks did not have frost-free freezers. The typical freezer compartment involved a steel bottom plus a steel shelf, both with embedded cooling coils, but with both likely having a significant layer of frost that inhibited the transfer of heat from the water in the tray to the cooling coils. By using hot water in the tray, you could melt the frost layer, getting much better surface contact and lower thermal resistance, which gave much higher rates of heat transfer, leading to frozen cubes in a shorter period of time.
I never conducted that experiment myself, but I heard from several people (including a heat-transfer professor) who claimed to have saved the day at parties where there was a shortage of ice.
Edit: BTW, if you really did get the ice cubes frozen faster because some of the hot water had evaporated (original explanation), that would mean that you wound up with less ice and might need to freeze another tray of cubes. Not a very attractive way to save time. You could easily accomplish that route by just putting less water in the tray.
December 4th, 2017 at 8:08:07 PM
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Hmmm. I figured the question had to do with the fact that cold air sinks and hot air rises. The relative coolness of the top shelf and bottom shelf of the freezer could be dofferent enough to freeze the bottom ice tray faster, even if it started out warmer.
If the House lost every hand, they wouldn't deal the game.
December 4th, 2017 at 8:40:55 PM
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I don't think temperature variations in different regions of the freezer are likely to be a significant factor, so long as the ice trays have good contact with the metal surfaces containing the refrigerant.
In a modern freezer using ice cube trays, I think there is relatively little heat transfer directly between the air (convection heat transfer) and the water and only a modest portion by convection between the air and the sides of the metal tray. Natural convection of heat to or from air usually involves a much higher thermal resistance than conduction between metal surfaces in intimate contact.
Most of the heat transfer will be from the water (by convection) to the tray surface, followed by conduction through the tray wall and the shelf surface, and finally convection to the refrigerant in the cooling coils inside the metal of the freezer bottom or shelf. In the absence of frost buildup, those modes of heat transfer are rather effective.
Convection transfer between a surface and water is also much more effective (lower thermal resistance) than between the surface and air. (Imagine the options of exposing your bare hand to -10° air vs. -10° water and you should get the idea.)
In a modern freezer using ice cube trays, I think there is relatively little heat transfer directly between the air (convection heat transfer) and the water and only a modest portion by convection between the air and the sides of the metal tray. Natural convection of heat to or from air usually involves a much higher thermal resistance than conduction between metal surfaces in intimate contact.
Most of the heat transfer will be from the water (by convection) to the tray surface, followed by conduction through the tray wall and the shelf surface, and finally convection to the refrigerant in the cooling coils inside the metal of the freezer bottom or shelf. In the absence of frost buildup, those modes of heat transfer are rather effective.
Convection transfer between a surface and water is also much more effective (lower thermal resistance) than between the surface and air. (Imagine the options of exposing your bare hand to -10° air vs. -10° water and you should get the idea.)
December 4th, 2017 at 9:09:42 PM
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https://www.sciencealert.com/does-hot-water-really-freeze-faster-than-cold-water
DUHHIIIIIIIII HEARD THAT!
December 5th, 2017 at 7:35:00 AM
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Yeah, just Google it.
h + ttps://medium.com/the-physics-arxiv-blog/why-hot-water-freezes-faster-than-cold-physicists-solve-the-mpemba-effect-d8a2f611e853
h + ttps://medium.com/the-physics-arxiv-blog/why-hot-water-freezes-faster-than-cold-physicists-solve-the-mpemba-effect-d8a2f611e853
Never make a bet that you wouldn't take, yourself.
