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It might help to start at this configuration and then manipulate this to get other forms too:|dw:1439845619600:dw| But yeah idk other than just rearranging within these diagonals we still won't cover all of them, cause like you've shown @dan815 we have some others.
the arragenment u have there is also since its increasing wrt to the sum of the indices too
how about a coming with bounds like
difference of 2 cannot exist for top and bottom
as 2 cannot be placed so we know that one side must always have a +1 to it
now its about the separation that is possible we can have 1 side with +1 and the other side can go up to a max of the row length i think
Hmmm one bound I see is that for a 3x3 matrix we have: \(b_1 \ge 1\), \(b_2 \ge 3\), \(b_3 \ge 1\)
lets come back to this -.- lets do this http://prntscr.com/85qvgt
Hmmm I think this is related to triangular numbers isn't it?
its all related to something alright
how about building it up from smaller, like whats the max intersections for 3 lines, are there some number of intersections not possible
3 lines either intersect 0, 1, 2, or 3 times.
okay right and for this question 0 is not possible becase of the no 3 concurrent rule
how does that scale with more lines
Is that what concurrent means? Parallel?
ya i googled it one of the synonyms said parallel so i went with it lol
yes concurrent = parallel
it was written by some dude that had a notion of a line
Waste of a word imo
Then 3 lines must intersect: 1 or 3 points only.
wait 2 is possible since we are allowed 2 concurrent just not 3
how about we just try to figure that out for a 100 lines, how many lines can be have with same slopes or not same slopes
the least number of slopes present can be 50 the max is 100 then we can consider the different intercepts vs same intercepts
now the problem looks more algebraic
the case of the 3 line intersections can be broken down into number of different slopes number of different intercepts to determine the number of intersections
f(m,b)=intersections(m,b) the number of intersections is some function of the the different number of slopes m and the different number of intercepts b lets try to come up with an equation like this one
we can be solving like a general problem then!
f(n,m,b)=intersections(n,m,b) let n be the number of lines m be the number of unique slopes b be the number of unique intercepts
the last one doesnt make sense, i dunno but something like this
how about this different slopes and different intersections points like how 3 slopes must not share the same slope, and no 3 lines can share the same intersection point
y-y1 = m1*(x-x1) let there be 100 lines of this form max number of mi =2 and we can see the max number of (xi,yi)s so that 2002 intercsecs exist or if possible
base case all same (xi,yi) means everything has different slope and 1 intersetion for 1 same slope, we must have atleast 1 different (xi,yi), other wise a line would coincide and we have infinte intersections