Zig-Zag

Logic Level 5

Let us say that Z n Z_{n} is the maximum number of regions which can be determined by n n lines containing exactly one sharp turn (see image of Z 1 Z_{1} and Z 2 Z_{2} for examples).

Find Z 100 Z_{100} .


The answer is 19901.

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2 solutions

Arjen Vreugdenhil
Dec 11, 2015

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 Z_1 = 2 and Z 2 = 7 Z_2 = 7 , and a quick sketch showed me that Z 3 = 16 Z_3 = 16 . Thus the increment is Δ Z 1 , 2 = 5 \Delta Z_{1,2} = 5 and Δ Z 2 , 3 = 9 \Delta 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 99 , 100 = 5 , 9 , , 397 ( = 5 + 98 4 ) . \Delta Z_{1,2}, \Delta Z_{2,3}, \dots, \Delta Z_{99,100} = 5, 9, \dots, 397\ \ (= 5+98\cdot 4). (The progression is confirmed by the fact that Δ Z 0 = 1 \Delta Z_0 = 1 so that Δ Z 0 , 1 = 1 \Delta Z_{0,1} = 1 .)

To find the final number of regions, we must add all these increments to the initial value Z 1 = 2 Z_1 = 2 . Note that the average value of the 99 increments is ( 5 + 397 ) / 2 = 201 (5 + 397)/2 = 201 , so that the answer is 2 + 99 201 = 19901 . 2 + 99\cdot 201 = \boxed{19901}. Note: since the increments Δ Z n , n + 1 \Delta Z_{n,n+1} are a linear function of n n , the number of regions is a quadratic function of n n . It is not difficult to see that Z n = 2 n 2 n + 1. Z_n = 2n^2 - n + 1.

Mithil Dani
Dec 10, 2015

The general expression is nC2+1 where n is number of 'straight' lines 200 C 2 + 1 = 19901 200C2+1=19901

first lets look at Z 1 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 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 100 Z_{100} , 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 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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