The Square-Sum Problem - Numberphile
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- čas přidán 10. 01. 2018
- Matt Parker discusses a problem involving Square Sums. Go deeper with extra footage: • The Square-Sum Problem...
More links & stuff in full description below ↓↓↓
More Matt Parker on Numberphile: bit.ly/Matt_Videos
Matt's projects and other stuff: standupmaths.com
This problem is discussed in Matt's book: amzn.to/2mksdD5
Thanks to Charlie Turner - more from her in Part 2: • The Square-Sum Problem...
Parker Square T-Shirts: bit.ly/ParkerSquareTshirt
Discuss on Brady's subreddit: redd.it/7pnbqm
Numberphile is supported by the Mathematical Sciences Research Institute (MSRI): bit.ly/MSRINumberphile
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I would like to thank Robert Gerbicz for his solution to the conjecture in the video, and HexagonVideo for explaining it well in video form. Cheers everyone!
Came here to say something similar so will instead just add my voice to support this comment.
It's an especially elegant proof, the idea of transforming one sequence into another and preserving its structure, though the bit from 4900 and onwards is a beyond me
Ninja pairs ftw
square in title and parker in thumbnail do not go very well together
xD
Still want me a Parker square t-shirt! (And a Parker circle one - remember that?!)
ಠ_ಠ
Treleleleleoellelele
The 'Give it a go' screen has a Parker Square in the background.
0:59 suspicious Parker square...
Parker square is all about giving things a go and not getting upset that you failed. :P
I don't appreciate Matt starting us off with a Parker sequence of numbers. It was almost right when he told us to give it a go.
Ahem , *The Parker Square-Sum Problem*
I hadn't even seen the video, but was gonna write this. An hour too late I guess.
To be fair it can be related to it.
We all were thinking that.
I actually thought that the video was going to state that they found a magic square with perfect squares.
I believe the aborted beginning (8, 1, 3, 6, 10) he gives you is known as the Parker Square-Sum.
Matt's really playing with fire here, he needs to stay away from the square topics
What would that help? The comments for videos he's in are asinine no matter what he's discussing.
@@KyleJMitchell not sure if you caught the whole Parker square thing
Why?
@@joshyoung1440 i think he did
Silence, he's in his element
One of my favorite logic puzzles in video games is apparently basically finding hamiltonians. In Oracle of Ages, there's a few rooms where you're expected to walk over every tile (turning it a different color), and in the Minish Cap as well. Something similar in Link's Awakening, where you push some strange tile machine around in turtle rock, filling up all the holes to get keys. Not really numbery in those games unlike this, but for some reason I just really like those puzzles.
There's one of those in pokemon
I solved it basically in the same way, but by tabulating the different ways each square number could be made. I then counted the number of times each number appeared. 8 and 9 appeared only once each, so they must go on the ends of the line. 1 and 3 appeared 3 times, but they can only touch 2 others if on a line, so we must ignore the pairing {1,3} to make 4. This leaves one unique chain.
You can also just go through them and look at the "square neighbours" they have - it's really easy to check since the only reachable squares are 4,9,16,25. Then you will see that 8 and 9 only have a single neighbour. For any set of numbers where exactly 2 numbers only have one neighbour, this thing is possible. So you don't even need to draw graphs or guess random ways through them. :)
You can't rule out there being a closed cycle at some point in the future though. That wouldn't be solvable and also could still have 2 numbers that are alone.
True, I didnt think about that! Fortunately it looks like from 14 upwards no closed cycles show up - the only way to introduce that would be if bigger numbers didn't connect in any way to the previous ones but only to themselves, and that seems unlikely... although I can't prove that for now.
Solved it the same way
I randomly guessed a few times, and got it right so I didn't have to do any of that lol... Though thats what I would have resorted to...
that sneaky parker start
The thumbnail spoiled it!! I wouldn't have immediately thought of finding a Hamiltonian path but the graph in the thumbnail gave it away :P
I see Matt Parker, I click the video
Well now I know how to Rick Roll you!!!
How to subtly give ideas for April Fools
a clickbait for math geeks
Now isn't that the main reason why we're all here? :P
Would that be a slightly not perfect lure ... a parker lure including parker unable to not include and therefor not wrong. This sentence is false. o.O
1:01 I see what you did there.
PARKER SQUARE!!!!!
Does anyone know the name of the music - it's so soothing!
Solved today
This was a really fun problem to get my brain going at 6am! I made a list from 1 - 15 and wrote next to them all of the possible combinations that would equal 4, 9, 16 and/or 25 and saw that 8 and 9 only had one possible combination so I knew they had to go at the end. It was pretty quick to fill in the rest although I got stuck going from the 9 end at the number 3 and had to go from the 8 end (remembering that 1 had to go with 8). I ended up with the correct order but backwards from what was later shown in the video haha. I also like my list of numbers a little more than the graph since that looks like it'll get pretty messy once you start crossing lines and making curved ones and such. It's a really cool visualisation, though!
Thanks for the video and the little puzzle ^^
I started with the fact that 15 has to be between 1 and 10 because it can only sum up to 16 or 25. I continued on a chain from 10 using the only possible choices. Once I had 3 used, I picked 8 next to 1 as the only possible choice and continued on from 3 again (since 8 is a dead end). And it worked out:
9,7,2,14,11,5,4,12,13,3,6,10,15,1,8
I love how Parker Square is now a thing :)
Normie
A thing 2+ years running
4+ years now.
5+ years now
YES!! Matt's back! I've been making my way through his numberphile playlist for the past week or so
Fear not for I am always with you.
I managed to figure it out with 10 mins of concerted effort, I figured that he would give a false start, so I just wrote out 1-15 chose 8 as a starting place (it was in the middle) and went on from there. I did attempt it at first with 1 at the start, which was not a great idea, and led to many minutes of just staring angrily at the paper...
Matt gave the same problem when he met our school! Thnx for the amazing day Matt.
I got Matt's book for Christmas. It was my favourite gift :)
I went with the way I intuitively thought it would work, if it worked at all. So I started with 1 and then took the highest possible number to pair up with it, then the lowest possible number to pair up with that one, then the highest again, and so forth. This gave me 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9. Then I slapped the remaining 8 in front and Bob's your uncle.
Didn't watch the rest of the video yet and I don't know if this is significant in any way, but I notice that the squares form a pattern: 9, 16, 25, 16, 9, 16, 25, 16, 9, 16, 25, 16, 9, 16.
Yeah I got the same sequence
The pattern of squares should change in interesting ways as the available integer set becomes larger.
Not only the squares. And not only a pattern. When you make an ordered list of possibles [1,3][2,2][1,3] in rows for squares 4 9 16 25 etc there is oscillations both when just listed 1 to 25 or listed in solution order. Also as you approach a point of it failing the pattern breaks.
When I paused the video I wrote 1 through 15 in a circle clockwise and then added the edges. The result is approximately a square grid (if you ignore the [1, 3] link, which isn't used anyway). The horizontal edges are the ones that sum to 16, while the verticals are 25 on the left, 9 on the right. As you trace the path, you have to alternate horizontal with vertical, and left-side with right-side. So there's at least some geometric significance to the pattern. Try it!
DaTux91 I did it by getting rid of all the square numbers first. So I crammed them in as soon as I could. My sequence was 9,7,2,14,11,5,4,12,13,3,6,10,15,1,8. I got it right the first time itself using this method. I left out 1 for later, though, because it's pretty easy to link various numbers using 1. My pattern was 16 9 16 25 16 9 16 25...
I almost got it but a couple of numbers were in the wrong place. I called it The Parker Sequence.
Fun puzzle! I always love these videos. Keep up the great work!
Parker always gives great exposition, enthusiastic and enlightening
The first time I tried this, I started with 8, 1,15 - and so because there was only one path-I got it on the first go, this is really cool!
I can tell what will happen in the comments just by looking at the thumbnail and the title of this video.
love matt in these numberphile vids, such a cheerful maths guy
I would bave never come up with such a solution. Brilliant.
That sneaky Parker square
Something really cool happens if you use Fibonacci numbers instead of Square numbers. You can string together all of the numbers from 1 to Fn -1 and the pair sums will just F(n-1), Fn and F(n+1). For example, up to 20 is 17,4,9,12,1,20,14,7,6,15,19,2,11,10,3,18,16,5,8,13 and the pair sums will be 21,13,21,13,21,34, etc.
Lets list the first Fibonacci numbers 1,1,2,3,5,8,13,21,34 so that we can check it.
Can you prove that?
Yes, the last two numbers are Fibonacci numbers. I can add them to find the next Fibonacci number. So I can extend this list by the following Fibonacci numbers.
This whole video is so interesting to me. When the puzzle was first explained at the beginning, I really didn't think it would be that difficult. But once I thought about it and tried doing it out in my head I realized how difficult it really is. I think it's so cool how at first glance it really doesn't seem all that challenging when in reality it actually takes a lot of dedication and it must be perfect. What confused me the most about this is how it works up to any number, not just 15. This video inspired me to try solving this puzzle which I quickly gave up on out of frustration.
I solved this in about 5 minutes. Before I watch the rest of the video, I'd just like to say how I did it. So I first added 14 and 15, the two largest numbers and got 29. Therefore the largest square number that can possibly appear is 25. I then took a random number from the list, e.g. 11, then went, well, 11 can form 25 or 16 (both greater than itself, of course) with two other numbers, which will be 5 and 14. This means 5 and 14 will be on either side of 11. Then I did the same thing for 5 and 14, finding the two numbers that they will be in between, one of which will be 11. Repeating this process is quite easy and the chain quickly formed. There are two particular numbers that I came across when doing this, which are 3 and 9 (1 also, but by the time I got to 1 there was only 8 left); 9 only worked with 7 for 16, and since there's no 0 or 16, it must be on one end of the chain, or string. Eventually things pieced together and gave me the answer (I think that's right, I'm gonna hope he doesn't say that it's actually impossible and I did or understood something wrong). So now I'll go finish the video and see if they did it the same way :)
Edit: I meant 8, 3 was already used so it wouldn't be the other end. Idiot.
i just came for the Parker Square jokes..
"I deliberately and meanly gave you a -- umm -- a starting point that does not work."
"Why would you do that?!"
"Because I am angry at the world about my hairline."
Hahaha
Amazing channel! Thanks guys!
I've literally done this yesterday after reading in Matts book.
@Numberphile Where is the video on the new biggest known prime number?
There is no biggest prime number.
TaiFerret 'known'
@@janeerland6449 there is no biggest known prime. There are mathematical formulas that give a prime number for any positive integer input (though none yet that list every prime).
He means the biggest prime that's currently been found. We know that they keep going, but mathematicians (and this very channel) frequently like to discuss when the new largest "known" prime is determined.
@Keks 257 any prime above 3 can be described by either 6x+1 or 6x-1
This problem has now been solved!🥳
yes, see video by HexagonVideos
I solved it by writing a program to iterate permutations of (1,2,...,14,15) with early pruning. Associating numbers with lists of numbers that can be added to them to make squares occurred to me only as an optimized alternative to checking each remaining element; I didn't realize you could just make a graph out of it and find solutions as paths
It is much faster than to go through all permutations.
I wrote a script to generate square sums graphs in OmniGraffle when I read about this in your book, and I started work on a script to try to find a Hamiltonian path through any given OmniGraffle graph, but I got distracted by another project. I did solve some big square sums graphs but it didn’t work on all the ones where it’s possible because I didn’t get around to adding proper backtracking.
Cycles are clearly more fun than path. Give us a value with an hamilonian cycle!
I'm going to ignore the fact that it is believed to work ad infintum past 25 and focus exclusively on the fact that it works for 42.
I started by listing the possible squares: 4, 9, 16, 25. These are the only options that could result from adding two numbers between 1-15. Then I looked at possible pairings and I realized that 8 can ONLY pair with 1. The only way to get to another square using 1-15 would be to add 8 to itself, which isn't allowed. Based on that, I knew that the number line had to start 8, 1... After that, there was only one question: Does 1 pair with 3 to make 4 or 15 to make 16. I tried 3 first and ran out when I hit 9 (8 1 3 13 12 4 5 11 14 2 7 9). Since that didn't work, the only other option was to pair 1 with 15.
This is actually a very similar topic to what I did my dissertation on.
I was looking for patterns amongst numbers such that
a + b^2 = c^2
a - b^2 = d^2 where a,b,c,d are all natural numbers and b^2 is the next square number below a such that no number (we'll call e) exists where b^2 < e^2 < a
The thumbnail gave it a bit away how you can solve it (even though it had different numbers):
9,7,2,14,11,5,4,12,13,3,6,10,15,1,8
Der Mathze omg , is there only one way to solve ? I find it same
Technically 2 but you can reverse it.
Where's the next calculator unboxing video at
Two Matt Parker videos in one day? Awesome.
The new Parker Square update is looking great!
Well guess what Matt merry Christmas the problem's been solved
Something something Parker Square something something.
MattTheCatThatShatInTheHat I had a good laugh at that
Please stop
I made a chart of the “factors” of each number, then I used those to make a “factor tree” and I got my answer!
Did it! Paused at 0:58
9,7,2,14,11,5,4,12,13,3,6,10,15,1,8.
So the pattern of sums goes: 16, 9, 16, 25, 16, 9, 16, 25... and so on
What if instead of squared number it's a cubed number
Callum Williams I've gone up to 100 and it's not worked thus far, apart from a list 1 number long. Cuz, you know- 1. It looks as though it's either going to be a pretty massive number or impossible. I have no proofs or anything though. :/
It's pretty boring; there aren't a lot of connections for any of the lower cubes, and eventually, they may get added in, but like I said... boring.
I'm actually adding them by pairs: 27 is 1+26 or 2+25 or 3+24 or ... , so it may make the whole thing harder.
8 27 64 125 I would try with 124 numbers does that work?
Ooh, I want to do the prime one now...
cara cara orange it doesn't unfortunately. They just hang together in little clumps of 4s or so.
Hamiltonian?
I'm not throwing away my, plot!
I'm not throwing away my, plot!
kujmous 😂😂😂 omg HAHAHAHA
such a badass problem and awesome solution
For once I solved one before watching through! my order is 9-7-2-14-11-5-4-12-13-3-6-10-15-1-8. I found it by making a grid, with 1-15 on one side, and 1, 4, 9, 16, and 25 on the other (as 36 is greater than 29, or 14+15) and writing in the number required to sum to the top square with the left number. crossing out all cases where the number was outside of 1-15, or the number was the same as the side number (2+2=4), i was left with one case where the number could only sum with one number : 9, with 7. I then made a tree diagram, using the numbers as a choose-your-own-adventure book guide. where there were two possible choices, i followed them both until one terminated (by not having an option that was not already used.)
0:08 Best editing I've ever seen
And the most useful one!
When will you make a video on the Parker Square-sum problem?! :D
Andrew S Please stop. Youre not clever or funny.
Broken Wave look in a mirror
Hi Brady,
I think it would be awesome for guys like Matt or James etc. to mention
(the link in the description to)
their CZcams channels in the actual video to support them.
Excellent problem, clear explanation. What more could you ask for on a foggy Friday morning, right?! Cheers!
It's actually pretty easy. I started with 15 because it only makes a square with 1 and 10, and I just went in both directions and branched out from each next number to all possible "partners". Took me about ten minutes.
Edit: Just realised that I sould've started with 9 cause it makes a square with only 7.
Ivan Miletic or you could start with 8 and 1
I found all the possible sums to make a square for each number and that left me knowing that 8 and 9 only added with one other number to make a square. This allowed me to put those at the ends, then work my way inwards with the other numbers. I finished in the same amount of time. It's cool to see how many diff ways people went about solving this.
Feels similar to the 7 bridges of Konnsburg
Check the bonus video, he mentions that.
It's easier though. Took me under 5 minutes, most of that was just creating a table, to list for each of the numbers, which of the other numbers add to a square. Figuring out the solution took about a minute after that list was done, as there are no choices, no trial and error...
Love the SET game on the shelf in the background! :-)
Solved it in 10-12 minutes! Here's how: I wrote down every number from 15 down to 1, and alongside it I wrote any other number(s) which would make it a square sum. (e.g. 15: 1, 10; 14: 2, 11; 13: 3, 12; etc.). Two numbers (8 and 9) had only one pair (1 and 7, respectively), so I decided to use one of those as the starting point of the sequence. Starting with 8, I put its only pair, 1, next to it. I then looked back at my chart to see in which other lines did 1 appear. The only other place it appeared was in the first line (15: 1, 10). And out of those three numbers, the only one that would make a square sum was 15, so I used that to continue the sequence. So then, 15 worked with both 1 and 10, but since 1 had already been used, I chose 10. In turn, 10 worked with both 6 and 15, but since 15 had already been taken, I chose 6. I followed this pattern until I used up all the numbers exactly once. This resulted in the finished sequence.
Goes through all the vertices
ALEXANDER HAMILTONian path
A jacksfilms + numberphile viewer? Is this for real?! 😂
me me big boy
oldcowbb me me math boy
Me me number boy
XD
I spy a utilities mug in the background. #shamelessproductplacement #gocheckoutmathsgear
That's been around in the videos for ages. It took me a while to request what it was.
If you put the numbers in a circle, it’s easier to visualize the path as it bounces around and around... very neat problem! Thx for sharing! :D (I love it when I can actually solve these... so often I get stumped or run out of patience, but this was a fun little puzzler!)
I went about a different way actually! Looking at which numbers fit with just one other number (with the original 15). I realized that 8 only pairs with 1 and 9 works solely with 7. Knowing that, I went off starting with 8 and came up with the same order as in the video. Neat puzzle! I’m going to have to challenge my pals with this one to see if they can solve it.
Best I did was 15,10,6,3,13,12,4,5,11 :(
You're on the right track, keep going! :)
Parker Square Number!
the misleading sequenced kinda helped me, cause once i was stuck starting from your sequence, i immediately figured out that 8 and 9 must be on the sides, and it's a downslope from there
Love the Parker Square background
It's very easy because 9 must be at one endpoint, then all the other numbers are uniquely determined. So you can even conclude there are only two such sequences.
More than that, 8 must go on the other end.
That's included in what I said - all the other numbers are uniquely determined.
nice
Yeah, that's how I solved it in a minute or two
Aleph Null It's not really included in what you said. You could have a series in which a number with 2 possible neighbours needs to go to the other end out of necessity.
I've apparently forgotten some important bits of my graph theory course during my computer science degree, because I'm now wondering if it's possible to efficiently calculate (a) whether a Hamiltonian path exists for any given graph and (b) what one example of such a path is for that graph. I know the TSP is NP-complete, but that's specifically looking for the *shortest* Hamiltonian; I don't remember if there was a verdict on calculating *any* Hamiltonian...
all hamiltonian paths are the same length
"In general, the problem of finding a Hamiltonian path is NP-complete (Garey and Johnson 1983, pp. 199-200), so the only known way to determine whether a given general graph has a Hamiltonian path is to undertake an exhaustive search." - Wolfram MathWorld, "Hamiltonian Path"
TSP is looking for a Hamiltonian cycle, not a path.
Hamiltonian paths aren't the same length on a weighted graph.
@Numberphile I have a question on the execution of the code of finding hamming path. It is known that the problem is NP-Complete and as you shown your program can compute the Ham-path quite quickly. I used standard SAT solvers, for the same problem, but could not reach any closer to the speed you are showing. Can you provide the tools you used?
Took me about a minute. I noticed that starting from the ends (1 and 15) and working inward, every number except for eight could be used to add to 16. Similarly, all numbers eight and below could be used to make nine and all of the ones greater than nine could be used to make 25. I then started with eight and matched up the 16-pairs like dominoes.
8,(1,15),(10,6),(3,13),(12,4),(5,11),(14,2),(7,9)
i got that too!
#ParkerSolution
Because you don't have a way to go to every path, you can't go through 3 and 1.
8,1,15,10,6,3,13,12,4,5,11,14,2,7,9
If you reverse the order it's still a valid solution because addition is commutative
Same
After 2.43 minutes
After 3 it started do find the correct number automatically
When I first saw this in Matt's book, I wanted to see if I could do it with the numbers 1-25. I did, and I was surprised to see that with the way I had written the puzzle, it ended with my birthday 7-18 !
Before watching Parker's method:
8,1,15,10,6,3,13,12,4,5,11,14,2,7,9
I made a 15x15 addition table in excel, and highlighted all the cells which added to a square. Then I eliminated the cells which were along the main diagonal, to get rid of a+a squares. I observed that 8 and 9 had to be on the edge of the list, because they only add up to a square with one other number each. So I picked one of them arbitrarily and did a manual depth first search until I found a successful list.
After watching Parker's method: In my method I created the adjacency matrix for the graph he created. I prefer my method because it makes it harder to miss squares, and you can use the properties of addition to draw diagonal lines perpendicular to the main diagonal to make connections easier to create.
so, is this a parker square-sum problem?
This joke is so annoying
I just had to give it a go!
Kolly.G Yeah...its so used up and far from funny. People see the word "square" now and they think its so clever to make the same joke as everyone else.
More like
Parker square sum ( someone had to do it )
And since literally millions of people already have, you didn't need to.
I see Matt Parker, I tune in for a good mornings working out
You guys should do a video on the Thompson Problem!
...so you could actually do 0-17 !! Just add the zero behind the 16 :-P
Parker sum
My method:
I created a spreadsheet with the square numbers along one axis and the numbers 1 to 15 along the other, with the square number minus the other number in each cell of the array. Anywhere the result was more than 15 or less than 1, I marked red. I also marked the 2 in the 2 row and the 8 in the 8 row red (because you can't put a number next to itself.
This gave two rows where the number could only be next to one other number: 1 next to 8 and 7 next to 9. These had to be the ends of the sequence. With the exception of 1 and 3, every other number had only two possible neighbours, so it was simply a matter of looking up 7's other neighbour (2) and adding it to the list, looking up 2's other neighbour (14), and so on.
The result: 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9
i just read about this in his book about a week ago, now here we are
Matt noes da wae
I didn't see the answer yet, but: 9,7,2,14,11,5,4,12,13,3,6,10,15,1,8
I generated a list of pair of numbers that summed give a square number between 4 (the lowest square number you make up) and 25 (the higher square number you make up). The code to do this on Python is:
i=1
while i
Neat I got a different order...
8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9
If you turn it around, you will see it's the same order, just inverted.
I put the numbers in Excel (more convenient for me than coding something), and made a list: For every number from 1 to 15, which of the other number(-s) can you use, to add up to any square number. This quickly shows that 8 and 9 only have one "partner", so they can't be anywhere in the middle.
Then I worked my way in - for the first step there was only one possibility from either end, but from the end starting 8-1, you have 2 choices. So I left that open, went from the other end, and voila: There was only one choice at any given point, until it was finished, because for the numbers that connect to 3 others, one of these 3 had been used on a previous step, one came immediately before it, so only one was left to follow.
Took about 5 minutes, I think, to prove by example that there is one and ONLY one way.
Knowing that it works from 25 on is mindblowing.
Absolutely nailed it on my first go; figured out Matt was trying to pull me a Parker Square :P
I just tried by myself, the game can actually be extended to 1-17
Fml I didn't finish the video I'm sorry
That's ok: you gave it a go and you got excited!
standupmaths Hey!
Encouraging experimentation and discovery regardless of what others may have previously found is the best thing you could be doing. You're a hero of mine, Matt Parker.
8 1 15 10 6 3 13 12 4 5 11 14 2 7 9
This is what I got I hope I'm right
I DID IT! I paused at "Give it a go" and I found a working order within a few minutes. Took some careful thinking to work my way around it though! I'll put it below a bit of a spoiler buffer. Awesome puzzle, onward to the rest of the video now!
.
.
.
.
.
.
.
8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9
I noticed that 8 can only sum up to 9 with 1 (because 8 + 8 is 16, but there's only one 8, and 8 + 17 is 25, but 17 is not a valid number) so it is an endpoint.
9 also can only sum up to 16 with 7, because 0 is not one of the numbers, making it the other endpoint.
Ultimately though, it helped to write down the combinations that could sum up to 25, since there are only 3, and use those as many times as possible to make the most of the space and spread the high numbers out a little.
What about finding one that cycles for higher numbers? So the first and last also look around to add a square number?
Please tell someone tell Matt to shave his head
ligamentless123 he actually did that
He already did, this was probably just filmed some time ago
@ligamentless123
Why? Because he is balding? Why do all men have to shave their head because they are starting to get bald? Men should embrace the baldness instead.
HiS oThEr HeAd
Uhm... you're embracing it by shaving your head though, aren't you?
I was told in school that the path which Matt calls Hamiltonian is called Eulerian...
A Hamiltonian path visits each vertex exactly once.
A Eulerian path follows each edge exactly once.
They sound very similar, but the existence of one does not imply the existence of the other.
Is then the Eulerian path of a graph the Hamiltonian path of its dual, and vise versa?
Yes fede it is exactly
On second thought , not quite... The inverse of a graph replaces the "faces" with vertices and then connects the new vertices with new edges... I see why you would think the eulerian/Hamiltonian/inverse graph connection as I fell for it too
Ah, right! Dual is the facesvertices change.
Is there a name for a graph that changes verticesedges?
I know I've seen that thing somewhere... Category Theory maybe? :/
Thanks!
(Commenting at the break before watching the rest)
After a bit of pondering, I realized that 8 had to be on one end of the line. The only allowable sums are 4, 9, 16, and 25, and 8 can only add to another number on the allowed list to get to 9.
Once I had that, the rest was easy: 8 1 15 10 6 3 13 12 4 5 11 14 2 7 9
Even with the thumbnail, I didn't consider trying to find a Hamiltonian path on a graph as a solution. Nice video and explanation!