Z n is the maximum number of regions which can be determined by n lines containing exactly one sharp turn (see image of Z 1 and Z 2 for examples).
Let us say thatFind Z 1 0 0 .
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The general expression is nC2+1 where n is number of 'straight' lines 2 0 0 C 2 + 1 = 1 9 9 0 1
first lets look at Z 1 , 2 lines divide the space into 2 regions: inner and outer. (2C2+1=2) but this doesn't explain it all.
So lets look at Z 2 , now if we select any 2 lines from 4 [4C2] we will get a unique inner region corresponding to each. (regions 2,3,4,5,6,7) also one outer region will be mutual to all (region 1) So the solution becomes 4C2+1=7
for Z 1 0 0 , we will be selecting 2 lines from 200 lines for corresponding inner region (200C2) and one mutual outer region. 200C2+1=19901
NOTE: If there is a confusion that when we select the 2 lines (those equivalent to 2 lines in Z 1 ) 3 regions are formed (6,7,4) then consider it this way, those 2 lines form only region 6. region 4 will be formed when we select the "vertical right" line and "horizontal lower" line. Similarly, region 3 will be formed when we select the "vertical right" line and "horizontal upper" line, region 7 by "vertical left" line and "horizontal lower" line, region 2 by "vertical left" line and "horizontal upper" line, region 5 by 2 vertical lines.
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And here is my purely intuitive (but correct) guess: In problems like this, the number of extra regions (or intersections, etc.) created in each step tends to be a linear function of the number of lines present.
We already have Z 1 = 2 and Z 2 = 7 , and a quick sketch showed me that Z 3 = 1 6 . Thus the increment is Δ Z 1 , 2 = 5 and Δ Z 2 , 3 = 9 . Thus I assumed that the increments form an arithmetic progression with step size 4: Δ Z 1 , 2 , Δ Z 2 , 3 , … , Δ Z 9 9 , 1 0 0 = 5 , 9 , … , 3 9 7 ( = 5 + 9 8 ⋅ 4 ) . (The progression is confirmed by the fact that Δ Z 0 = 1 so that Δ Z 0 , 1 = 1 .)
To find the final number of regions, we must add all these increments to the initial value Z 1 = 2 . Note that the average value of the 99 increments is ( 5 + 3 9 7 ) / 2 = 2 0 1 , so that the answer is 2 + 9 9 ⋅ 2 0 1 = 1 9 9 0 1 . Note: since the increments Δ Z n , n + 1 are a linear function of n , the number of regions is a quadratic function of n . It is not difficult to see that Z n = 2 n 2 − n + 1 .