CORRECTION: I MADE AN ERROR IN THIS VIDEO. the error is at 43:37. I said that, after scenario 2, you can track your moves starting from 3/2, then 5/3, and etcetera... But thats wrong. For example, at t=3/2, which would be the "first" move after scenario 2 is reached, what would you see? In theory, you should see a peg at the end of the line jumping over another peg. But thats impossible because the line is infinite. Its still correct to say that you reach scenario 2 at time 1 and scenario 3 at time 2, but the error is in the method used to get from scenario 2 to scenario 3. The correct way to do it is to start from scenario 3 (at t = 2) and then go to time t = 3/2, which is going to be the move right before scenario 3 is reached. Then you keep going backwards in time: the move before that is at t = 4/3, the move before that is at t = 5/4, and so on, until in an infinite amount of time you reach time t = 1. I apologize for the mistake! Luckily its the only one in the video
@@TaranovskiAlex If you allow reaching scenario 2 in infinite time "forward" from scenario 1 (e.g. using times (2-1/2), (2-1/4), (2-1/8), ...) and "backward" from scenario 3 (e.g. using times (2+1/2),(2+1/4),(2+1/8),...) then you get that after "double" inifinite number of moves you get from scenario 1 to scenario 3 and it is correct. The questionable trick is that in any time interval of length greater than zero that includes 2, i.e., time of scenario 2, you are doing infinite number of moves. There is really no difference if it is "forward" infinity or "backward" infinity, either you allow it or not. The questionable part is that you cannot do _single_ move from scenario 2 in either direction. But there is nothing in the formalized rules that you need to get to or from a scenario in a single move. Let's define a finite valid set of scenarios as a set of scenarios where you get from first scenario to last scenario. Then you can define valid infinite set of scenarios that include scenario 1, scenario 2 and all scenarios in between as a limit of a sequence of finite valid set of scenarios. Same as you define value of infinite sum as a limit of sequence of partial (finite) sums. Thus there exist valid set of scenarios from starting position to winning position only as infinite valid set of scenarios that is constructed by taking limit of some other infinite valid sets of scenarios which are constructed as taking limits of finite valid sets of scenarios. (Maybe there is one or two more levels of limits, but that does not matter.) Anyways these construction allow only allow sets of scenarios that have same cardinality as ℕ, i.e. as set of natural numbers, because ℕ⨯ℕ has same cardinality as ℕ. So the number of steps is countably infinite.
As TaranovskiAlex is saying, the conclusion about scenario 2 is entirely flawed. That extra spacing between the previously moved peg and the peg that will be jumped next will never go away, regardless of the line being infinite. You can't just move backwards from an assumed possible position to fix this, because you are literally pulling that scenario 3 peg out of thin air to do it. The only way to get a peg close enough to another peg in a line of evenly spaced pegs to be able to jump them all is to pull it in from one of the sides, which will obviously interfere with the so-called "mega whoosh".
@@tomasebenlendr6440 , you cannot just "allow" that scenario 2 is achievable from both scenarios and say it works, because coming from the two directions only provide a similar result, not the same result. Moving forward from scenario 1 always has a spot that is two consecutive spaces, while moving backward from scenario 3 always has a spot that is two consecutive pegs. Because there is no way over infinite time for two consecutive spaces to transform into two consecutive pegs, there is no transition that exists between scenario 1 and scenario 3.
@@SgtSupaman Scenario 2 does not have two consecutive pegs nor two consecutive spaces, thus by your logic there must be no sequence of transitions from any state before secario 2 and scenario 2. Or from scenario 2 to any state after scenario 2. Moreover for every scenario before scenario 2 there exist a peg after which there are no spaces and for every scenario after scenario 2 there exist a space after which there is no peg. Scenario 2 does not have this property. This is what dealing with infinity is about, some property that you observe along all elements in sequence may or may not hold for limit of that sequence. Let's take set of times where we do a move, i.e., {t: |{x,y: f(x,y,t)=0.5}| = 3}. If we say that in time 1 we are in scenario 1, and in time (2-3/4) we do first move, so that in time (2-1/2) we are in one move behind scenario 1, in time (2-3/8) we do second move so that in time (2-1/4) we are in scenario that is two moves behind scenario 1, then our set of times is sequence: (2-3/4),(2-3/8),(2-3/16),....,...., (2+3/16),(2+3/8),(2+3/4). This sequence does _not_ contain number 2 (moreover there is no number in this sequenece "last before 2" and no "first after 2". This sequence is specially crafted that it has property that is observed in set of real numbers (which have strictly greater cardinality than set of natural numbers) yet the sequence has cardinality of natural numbers. From this point of view is scenario 2 in some sense unachievable, as there is no move that goes into that scenario and there is no move that goes from that from that scenario. Let's formalize the set of steps: Scenario 1 has pegs at {(0,y): y = 0, integer }. At time 2-3/(2^(n+2)) we have: f(0,y)=0.5 for y in {1-n, -n, -1-n} f(0,y)=1 if y is not in {1-n, -n, -1-n} and y
Wouldn’t it be more accurate to say that you will always approach scenario 2 as you approach t=1, but you will never actually reach scenario 2? It’s essentially a limit on a graph. This proof essentially shows that as you make moves you can get infinitely close to level 5, but will never reach level 5. Still a very interesting video either way.
Please use "arbitrarily" instead of "randomly", because "randomly" usually means according to some predefined random distribution, i.e., crucial property of random selection is the probability of selecting the item from some subset of all possible selections. Whereas "arbitrarily" usually means that the selection was not driven by any (relevant) condition.
This was an excellent video! I have watched many other videos about Conway's Soldiers, but none of them showed this infinite solution. Some of them didn't even consider the infinite case (and just declared or proved the finite case). Also, this process can be completed in finite time, but only if you also accept the mathematical idea of Zeno-ian supertasks.
Note: For the resolution of the infinite series in 21:00, this only proves that if the series is convergent than S would be equal to Phi^2 but it doesn't prove the convergence itself I am sure the series has been properly shown to be convergent but it's always risky to present this type of resolution of series to a general public without also quickly mentioning that it only works for convergent series Otherwise you can end up with very wrong result like the sum of the natural number being equal to -1/12 ;-p
What felt wrong about the transition from scenario 2 to 3 is the fact that there is a difference between transitioning from 0 to ω and transitioning from ω to 0. It's like I have a rule in my head that says the positions of any game must be well-ordered. I honestly didn't find any other way to describe how I feel 😅 The other thing I want to say is that the time steps to transition from scenario 2 to 3 are dense near 1 and not near 2, something like 1, ...1+1/2^3, 1+1/2^2, 1+1/2^1, 1+1/2^0 = 2 .
@@RUclips_username_not_found , scenario 2 is entirely flawed and will never lead to scenario 3, regardless of infinite time/pegs. The space between the previously moved peg and the peg that will be jumped next will always be two spaces, for the entire infinite line, meaning that it will never be possible to move a peg up the line and achieve the second level that is being shown by the "whoosh". All the "proof" does is just ignore those spaces after some time, which isn't how infinity works.
@@SgtSupaman I see what you mean. Perhaps you are right. I don't know. To be honest, I skipped the rigorous rules because they were so cumbersome I couldn't keep up with them. I am not in a position that allows me to make a judgement.
40:19 seems to be not very well-defined I guess we say something like "if for positions A and B, for all N we have positions A1 and B1 such that A1 is reachable from A, B is reachable from B1 and A1 and B1 don't differ at any x, y < N, then we call A and B connected"? Am I getting that right? Or is it just that we can plug ordinal values for t into the function. I am fairly certain that there is no set of all ordinals, which would mean in order to properly define it, we would need to first set some boundary I guess in the video we use fractional times, which would correspond to 1, 2, 3, ..., omega, omega + 1, omega + 2, ..., 2omega, 2omega + 1, ... I get that ordinals are outside of the scope of the video, but it's kinda important that the allowed times form an order, such that there is no infinite decreasing sequence of times, otherwise the invariant logic does not hold problem is, in the woosh there is an infinite decreasing sequence (as well as an increasing sequence), so would the correct way of formulating it be "we allow ordinal times for the forwards variant and for the backwards variant, also if from A we can reach B forwards and from C we can reach B backwards, then we also say that we can reach C from A"? This would solve the invariant problem, so long as we only make finitely many such connections, at least. Except in the video we make countably many... so we must allow ordinals for that too, it seems. A mega whoosh seems to be something on the order of omega cubed? ...and I think even my iterpretation/explanation may have holes oh well
on second thought maybe the invariant preservation thing follows from the fact that no cell gets changed more than N times, but that seems non-trivial to prove
One "quirk" is that to get from a row to a peg, you'll need a single piece which starts from the "end" of the row to start compressing the alternating column to a single peg. You can't get that with the previously mentioned method. Since we're dealing with infinity, the rules are weird and although your method did follow the rules you set, is there any method which doesn't use such a hacky approach?
i get what you're saying. When passing from scenario 2 to scenario 3 of the whoosh, its like you're creating a peg at the end of the infinite line that wasn't there before. It definitely feels hacky. But the point is that, in mathematics, it doesn't matter if it feels wrong. If it doesn't go against the rules, then its right. Also, if you think about it logically, it all seems to confirm that the move its right. First of all, think about the values. Lets say you have an infinite line which starts at the point with value x^5 (so the line goes down through x^6, x^7 and so on). That means that using a whoosh you can convert that line into a single point in x^3 (two spaces over x^5). if you use a calculator, you will see that x^3 = (1/golden ratio)^3 = 0.23 then, you can find out that the geometric series x^5 + x^6 + x^7 + ... is also equal to 0.23 so the line and the peg have the same identical value. The invariant is not changing. I hope it feels a bit less wrong now And also theres the fact that, at the end of the proof, you reach level 5 with only one peg, like the decreasing monovariant proved. So these two things confirm that the move isn't cheating
@@4real-e9m I understand that it is right and why it is right, but is it possible (even theoretically) to do it without doing this sort of hacky thing? Is there an alternative solution for conway's soldiers?
I made a 3D and 4D version of Conway's Soldiers a couple years ago but I can't seem to find my project file! It used to be on my itch io but I guess I must have deleted it at some point for some odd reason. I always thought it might be interesting to know what happens as you add dimensions so if I can find the project I'll upload it again haha
as mentioned in the video, the sum of a line (1 + x + x + x^2 + x^2 + x^3 + x^3 + ... = phi^3 for x = 1/phi) so, if you were to sum up the lines that form a line by first summing up each individual line and then them all, you would get phi^3 times the same sum, or phi^6 this generalizes to f(n)=phi^(3n) where n is the dimention of the thing you want to sum up (1 for a line, 2 for a plane, etc) the semi invariant at the starting position would then be I=Sf(n)phi^(-dst)=phi^2phi^(3n)phi^(-dst)=phi^(3n+2-dst) note that n = N - 1 where N is the dimension of the field and you get I=phi^(3N-dst-1), which gives you dst
yes, before my next big video i was planning to make a couple of shorter videos about variants of conway soldiers; like adding diagonal jumps, or conway soldiers in 3d. But it will take some time
I want make you suggestions for more good example of invariant. We have grid with white squares only in origin we have black square. In one move you can delete black square and make black square top and right of the cell question it is posibble to get to some configuration of black and white square s
I was trying to figure out the order of infinity. With whooshes and mega whooshes it is not easy to determine. A single whoosh is a basic order infinity also called aleph null. Have you given this some thought?
Great video but the 43:39 feels weird, the ,,Woosh" thing and going from scenario 2, to scenario 3 it seems like you're supposed to get a peg which is out of the ,,order" in order to reach scenario 3, which doesn't seems like it should work. I'm not a professional but it's just weird.
Btw if you're interested i have a few mathemathical problems which i think are kinda similar to Conway's Soldier, if you're willing to know about them or maybe even create a video i would be happy to share those with you
thank you for your comment. I noticed that i made an error in the video, which was what you said, and corrected it in the pinned comment. I apologize for the mistake. And yes, i would be happy if you could share those problems with me. This channel is not going to be only about maths, so maybe i won't talk about them immediately; but sooner or later...
The "whoosh" really bothers me, specifically the part starting at 40:40 which implies that a semi-infinite sequence of odd numbers can be converted to the next higher odd number following the rule that an odd number and an adjacent even number can always be converted to the next higher odd number, which begs the question "but where did you get the even number from"? Saying "we got it from infinity" feels like you're really sweeping something under the rug.
This is overall a great video, but you spend way too long talking about things that are common knowledge to anyone watching this kind of video, like the quadratic formula and the golden ratio. Also, why do you not just call it phi like a normal person!?
manspains algebra. made me click off. if you want to see a video on a proof similar to this, there are plenty of other videos available. wouldn't watch if I were you.
The video explains too much and sometimes its unnecessary. Manhattan geometry is like a name you made for children and you over explaind a simple concept. This video could easily be 15 minutes at max
CORRECTION: I MADE AN ERROR IN THIS VIDEO.
the error is at 43:37. I said that, after scenario 2, you can track your moves starting from 3/2, then 5/3, and etcetera... But thats wrong. For example, at t=3/2, which would be the "first" move after scenario 2 is reached, what would you see? In theory, you should see a peg at the end of the line jumping over another peg. But thats impossible because the line is infinite. Its still correct to say that you reach scenario 2 at time 1 and scenario 3 at time 2, but the error is in the method used to get from scenario 2 to scenario 3. The correct way to do it is to start from scenario 3 (at t = 2) and then go to time t = 3/2, which is going to be the move right before scenario 3 is reached. Then you keep going backwards in time: the move before that is at t = 4/3, the move before that is at t = 5/4, and so on, until in an infinite amount of time you reach time t = 1.
I apologize for the mistake! Luckily its the only one in the video
@@TaranovskiAlex If you allow reaching scenario 2 in infinite time "forward" from scenario 1 (e.g. using times (2-1/2), (2-1/4), (2-1/8), ...) and "backward" from scenario 3 (e.g. using times (2+1/2),(2+1/4),(2+1/8),...) then you get that after "double" inifinite number of moves you get from scenario 1 to scenario 3 and it is correct.
The questionable trick is that in any time interval of length greater than zero that includes 2, i.e., time of scenario 2, you are doing infinite number of moves. There is really no difference if it is "forward" infinity or "backward" infinity, either you allow it or not. The questionable part is that you cannot do _single_ move from scenario 2 in either direction.
But there is nothing in the formalized rules that you need to get to or from a scenario in a single move.
Let's define a finite valid set of scenarios as a set of scenarios where you get from first scenario to last scenario.
Then you can define valid infinite set of scenarios that include scenario 1, scenario 2 and all scenarios in between as a limit of a sequence of finite valid set of scenarios. Same as you define value of infinite sum as a limit of sequence of partial (finite) sums.
Thus there exist valid set of scenarios from starting position to winning position only as infinite valid set of scenarios that is constructed by taking limit of some other infinite valid sets of scenarios which are constructed as taking limits of finite valid sets of scenarios. (Maybe there is one or two more levels of limits, but that does not matter.)
Anyways these construction allow only allow sets of scenarios that have same cardinality as ℕ, i.e. as set of natural numbers, because ℕ⨯ℕ has same cardinality as ℕ. So the number of steps is countably infinite.
As TaranovskiAlex is saying, the conclusion about scenario 2 is entirely flawed. That extra spacing between the previously moved peg and the peg that will be jumped next will never go away, regardless of the line being infinite. You can't just move backwards from an assumed possible position to fix this, because you are literally pulling that scenario 3 peg out of thin air to do it. The only way to get a peg close enough to another peg in a line of evenly spaced pegs to be able to jump them all is to pull it in from one of the sides, which will obviously interfere with the so-called "mega whoosh".
@@tomasebenlendr6440 , you cannot just "allow" that scenario 2 is achievable from both scenarios and say it works, because coming from the two directions only provide a similar result, not the same result. Moving forward from scenario 1 always has a spot that is two consecutive spaces, while moving backward from scenario 3 always has a spot that is two consecutive pegs. Because there is no way over infinite time for two consecutive spaces to transform into two consecutive pegs, there is no transition that exists between scenario 1 and scenario 3.
@@SgtSupaman Scenario 2 does not have two consecutive pegs nor two consecutive spaces, thus by your logic there must be no sequence of transitions from any state before secario 2 and scenario 2. Or from scenario 2 to any state after scenario 2. Moreover for every scenario before scenario 2 there exist a peg after which there are no spaces and for every scenario after scenario 2 there exist a space after which there is no peg. Scenario 2 does not have this property. This is what dealing with infinity is about, some property that you observe along all elements in sequence may or may not hold for limit of that sequence.
Let's take set of times where we do a move, i.e., {t: |{x,y: f(x,y,t)=0.5}| = 3}. If we say that in time 1 we are in scenario 1, and in time (2-3/4) we do first move, so that in time (2-1/2) we are in one move behind scenario 1, in time (2-3/8) we do second move so that in time (2-1/4) we are in scenario that is two moves behind scenario 1, then our set of times is sequence: (2-3/4),(2-3/8),(2-3/16),....,...., (2+3/16),(2+3/8),(2+3/4). This sequence does _not_ contain number 2 (moreover there is no number in this sequenece "last before 2" and no "first after 2". This sequence is specially crafted that it has property that is observed in set of real numbers (which have strictly greater cardinality than set of natural numbers) yet the sequence has cardinality of natural numbers. From this point of view is scenario 2 in some sense unachievable, as there is no move that goes into that scenario and there is no move that goes from that from that scenario.
Let's formalize the set of steps:
Scenario 1 has pegs at {(0,y): y = 0, integer }.
At time 2-3/(2^(n+2)) we have:
f(0,y)=0.5 for y in {1-n, -n, -1-n}
f(0,y)=1 if y is not in {1-n, -n, -1-n} and y
Wouldn’t it be more accurate to say that you will always approach scenario 2 as you approach t=1, but you will never actually reach scenario 2? It’s essentially a limit on a graph. This proof essentially shows that as you make moves you can get infinitely close to level 5, but will never reach level 5. Still a very interesting video either way.
good video but like why did u shorten golden ratio to gr like just say phi,,,
,,,
Its Pi!!!
Pi and Phi are different constants.
Pi is the ratio of a circles radius and its circumference.
Phi is the golden ratio (sqrt(5)+1)/2
X: why is it possible?
4real: YOU HAVE TO MEGA WOOSH-
Please use "arbitrarily" instead of "randomly", because "randomly" usually means according to some predefined random distribution, i.e., crucial property of random selection is the probability of selecting the item from some subset of all possible selections. Whereas "arbitrarily" usually means that the selection was not driven by any (relevant) condition.
🤓👆
@@catiniya
it’s a fucking math video.
13:06 "And you're left with 1 - 5 which is equal to 4" 💀
This was an excellent video! I have watched many other videos about Conway's Soldiers, but none of them showed this infinite solution. Some of them didn't even consider the infinite case (and just declared or proved the finite case).
Also, this process can be completed in finite time, but only if you also accept the mathematical idea of Zeno-ian supertasks.
really nice video! looking forward to seeing more
Note: For the resolution of the infinite series in 21:00, this only proves that if the series is convergent than S would be equal to Phi^2 but it doesn't prove the convergence itself
I am sure the series has been properly shown to be convergent but it's always risky to present this type of resolution of series to a general public without also quickly mentioning that it only works for convergent series
Otherwise you can end up with very wrong result like the sum of the natural number being equal to -1/12 ;-p
Conway creates the most fascinating proofs & problems I've never seen any person do something more abstract
Really high quality video! My man deserves more subs!
4real
@@Ескендір-б5рLOL
I am the 100th sub
Here’s a game I’d need infinite time to win: tear down my house half at a time. Tear down half, then half of that and half of that and so on.
Infinity finally given justice
Awesome video, keep up the great work!
I feel like at 43:32 the scenario is not well defined at t= 3/2, maybe I am missing something.
Yeah, they probably meant
Scenario 1:
1/2, 2/3, 3/4,… [formula 1-1/n]
Scenario 2:
1, …, 5/4, 4/3, 3/2 [formula 1+1/n]
Scenario 3:
2
yes, you're correct. I made an error; i corrected it in the pinned comment
Underrated
I love the VVVVVV music playing.
Predestined Fate
@@stormfury1143 thank you for writing the song name
The algorithm brought me in and it went very well.
What felt wrong about the transition from scenario 2 to 3 is the fact that there is a difference between transitioning from 0 to ω and transitioning from ω to 0. It's like I have a rule in my head that says the positions of any game must be well-ordered.
I honestly didn't find any other way to describe how I feel 😅
The other thing I want to say is that the time steps to transition from scenario 2 to 3 are dense near 1 and not near 2, something like 1, ...1+1/2^3, 1+1/2^2, 1+1/2^1, 1+1/2^0 = 2 .
yeah you're right, i made a mistake. Sorry for that; i corrected it in the pinned comment
@@TaranovskiAlex What part are you referring to specifically? What is wrong with the proof?
@@RUclips_username_not_found , scenario 2 is entirely flawed and will never lead to scenario 3, regardless of infinite time/pegs. The space between the previously moved peg and the peg that will be jumped next will always be two spaces, for the entire infinite line, meaning that it will never be possible to move a peg up the line and achieve the second level that is being shown by the "whoosh". All the "proof" does is just ignore those spaces after some time, which isn't how infinity works.
@@SgtSupaman
I see what you mean.
Perhaps you are right. I don't know.
To be honest, I skipped the rigorous rules because they were so cumbersome I couldn't keep up with them.
I am not in a position that allows me to make a judgement.
Tim conway was a genius
Everything connects together! The golden ratio stuff, the invariants, and the game states.
Great proof!👍
Conway’s a math gamer
30:20 yes I want to know why. math the video! lol
I had to grapple with the woosh for a few minutes to accept that it made sense given an infinite number of moves.
40:19 seems to be not very well-defined
I guess we say something like "if for positions A and B, for all N we have positions A1 and B1 such that A1 is reachable from A, B is reachable from B1 and A1 and B1 don't differ at any x, y < N, then we call A and B connected"?
Am I getting that right?
Or is it just that we can plug ordinal values for t into the function. I am fairly certain that there is no set of all ordinals, which would mean in order to properly define it, we would need to first set some boundary
I guess in the video we use fractional times, which would correspond to 1, 2, 3, ..., omega, omega + 1, omega + 2, ..., 2omega, 2omega + 1, ...
I get that ordinals are outside of the scope of the video, but it's kinda important that the allowed times form an order, such that there is no infinite decreasing sequence of times, otherwise the invariant logic does not hold
problem is, in the woosh there is an infinite decreasing sequence (as well as an increasing sequence), so would the correct way of formulating it be "we allow ordinal times for the forwards variant and for the backwards variant, also if from A we can reach B forwards and from C we can reach B backwards, then we also say that we can reach C from A"? This would solve the invariant problem, so long as we only make finitely many such connections, at least. Except in the video we make countably many... so we must allow ordinals for that too, it seems. A mega whoosh seems to be something on the order of omega cubed?
...and I think even my iterpretation/explanation may have holes
oh well
on second thought maybe the invariant preservation thing follows from the fact that no cell gets changed more than N times, but that seems non-trivial to prove
One "quirk" is that to get from a row to a peg, you'll need a single piece which starts from the "end" of the row to start compressing the alternating column to a single peg. You can't get that with the previously mentioned method. Since we're dealing with infinity, the rules are weird and although your method did follow the rules you set, is there any method which doesn't use such a hacky approach?
i get what you're saying. When passing from scenario 2 to scenario 3 of the whoosh, its like you're creating a peg at the end of the infinite line that wasn't there before. It definitely feels hacky. But the point is that, in mathematics, it doesn't matter if it feels wrong. If it doesn't go against the rules, then its right. Also, if you think about it logically, it all seems to confirm that the move its right.
First of all, think about the values. Lets say you have an infinite line which starts at the point with value x^5 (so the line goes down through x^6, x^7 and so on). That means that using a whoosh you can convert that line into a single point in x^3 (two spaces over x^5).
if you use a calculator, you will see that x^3 = (1/golden ratio)^3 = 0.23
then, you can find out that the geometric series x^5 + x^6 + x^7 + ... is also equal to 0.23
so the line and the peg have the same identical value. The invariant is not changing. I hope it feels a bit less wrong now
And also theres the fact that, at the end of the proof, you reach level 5 with only one peg, like the decreasing monovariant proved. So these two things confirm that the move isn't cheating
@@4real-e9m I understand that it is right and why it is right, but is it possible (even theoretically) to do it without doing this sort of hacky thing? Is there an alternative solution for conway's soldiers?
@@OmPatil-zj9uo As for now, there isn't any known alternative to the whoosh
@@4real-e9m Interesting. Crazy proof either way though
unrelated the soundtrack slaps :)
The video looks awesome and is interesting. What did you use for making it? You earned a sub!
i used adobe premiere pro and a bit of manim in some parts. Thank you
How do you not have at least 10k subscribers??
Because this is the channel's first video
Couldn't believe when I saw the sub count. Learned something new.
Numberphile has a good video on this topic too, but this is probably more polished/easier to grasp
This is the video ive been waiting for. ive never understood this proof!!
I made a 3D and 4D version of Conway's Soldiers a couple years ago but I can't seem to find my project file!
It used to be on my itch io but I guess I must have deleted it at some point for some odd reason.
I always thought it might be interesting to know what happens as you add dimensions so if I can find the project I'll upload it again haha
Please do, It would be very interesting
as mentioned in the video, the sum of a line (1 + x + x + x^2 + x^2 + x^3 + x^3 + ... = phi^3 for x = 1/phi)
so, if you were to sum up the lines that form a line by first summing up each individual line and then them all, you would get phi^3 times the same sum, or phi^6
this generalizes to f(n)=phi^(3n) where n is the dimention of the thing you want to sum up (1 for a line, 2 for a plane, etc)
the semi invariant at the starting position would then be I=Sf(n)phi^(-dst)=phi^2phi^(3n)phi^(-dst)=phi^(3n+2-dst)
note that n = N - 1 where N is the dimension of the field and you get I=phi^(3N-dst-1), which gives you dst
Nice bro
WDYM 150 SUBS THIS IS SO GOOD
the setup dragged on awhile but the ending was satisfying
Wow, this video is growing fast! From around 2.4 thousand views to 7.2 thousand views in less than a day.
why did you shorten golden ratio to gr instead of phi
Never heard of general relativity would be my guess.
@deltalima6703 no
I would not pick an acronym that is already in use at least. I would just say phi or use the symbol.
What if you allow diagonal jumps? Can you make a video about that?
yes, before my next big video i was planning to make a couple of shorter videos about variants of conway soldiers; like adding diagonal jumps, or conway soldiers in 3d. But it will take some time
@@4real-e9m also, I forgot to ask but what if you ONLY allow diagonal jumps?
This means you can do diagonal jumps but not normal jumps.
@3141minecraft Isn't that isomorphic to two separate normal grids that are interlocked with eachother?
@@taektiek526 I am not sure
@@taektiek526no because starting position? not sure
I want make you suggestions for more good example of invariant. We have grid with white squares only in origin we have black square. In one move you can delete black square and make black square top and right of the cell question it is posibble to get to some configuration of black and white square s
I was trying to figure out the order of infinity.
With whooshes and mega whooshes it is not easy to determine.
A single whoosh is a basic order infinity also called aleph null.
Have you given this some thought?
Great video but the 43:39 feels weird, the ,,Woosh" thing and going from scenario 2, to scenario 3 it seems like you're supposed to get a peg which is out of the ,,order" in order to reach scenario 3, which doesn't seems like it should work. I'm not a professional but it's just weird.
Btw if you're interested i have a few mathemathical problems which i think are kinda similar to Conway's Soldier, if you're willing to know about them or maybe even create a video i would be happy to share those with you
thank you for your comment. I noticed that i made an error in the video, which was what you said, and corrected it in the pinned comment. I apologize for the mistake. And yes, i would be happy if you could share those problems with me. This channel is not going to be only about maths, so maybe i won't talk about them immediately; but sooner or later...
HUH? 84 SUBSCRIBERS?? bro you are seriously underrated
In just 22 hours, the subscriber count exploded from 84 to 221. So, it takes some time to explode for real, given that this video is relatively new.
The "whoosh" really bothers me, specifically the part starting at 40:40 which implies that a semi-infinite sequence of odd numbers can be converted to the next higher odd number following the rule that an odd number and an adjacent even number can always be converted to the next higher odd number, which begs the question "but where did you get the even number from"? Saying "we got it from infinity" feels like you're really sweeping something under the rug.
How does this video only have 332 views and the person 21 subscribers
you italian?
Ye
This is overall a great video, but you spend way too long talking about things that are common knowledge to anyone watching this kind of video, like the quadratic formula and the golden ratio. Also, why do you not just call it phi like a normal person!?
manspains algebra. made me click off. if you want to see a video on a proof similar to this, there are plenty of other videos available. wouldn't watch if I were you.
The video explains too much and sometimes its unnecessary. Manhattan geometry is like a name you made for children and you over explaind a simple concept. This video could easily be 15 minutes at max
Manhattan geometry is a real thing mathematicians study, he didn't invent it for this video
Read a textbook then.
Bro what do you mean that’s what makes this video good. None of the parts are rushed so I was able to understand all of it.
88 subs? for a full, working channel?!