in the cell immediately to its left. Taking the leftmost such entry yields the diagram on the right in Figure \[fig-ex2-simple-no-132-after-max\]. In this diagram we see that the entries of $\pi$ lying in the box bounded by $\{b,c,d\}$ (including those three entries) form a proper interval, contradicting the simplicity of $\pi$.

[ccc]{}

(0,0)(50,40) (0,20)(10,20)(10,30)(0,30) (10,0)(20,0)(20,10)(10,10) (10,20)(20,20)(20,30)(10,30) (10,30)(20,30)(20,40)(10,40) (20,0)(30,0)(30,10)(20,10) (20,30)(30,30)(30,40)(20,40) (30,0)(40,0)(40,10)(30,10) (30,10)(40,10)(40,20)(30,20) (30,30)(40,30)(40,40)(30,40) (40,20)(50,20)(50,30)(40,30) (10,40)[2.0]{} (20,10)[2.0]{} (20,10)[$a$]{} (30,30)[2.0]{} (30,30)[$c$]{} (40,20)[2.0]{} (40,20)[$b$]{} (0,0)(0,40) (0,0)(50,0) (10,0)(10,40) (0,10)(50,10) (20,0)(20,40) (0,20)(50,20) (30,0)(30,40) (0,30)(50,30) (40,0)(40,40) (0,40)(50,40) (50,0)(50,40)

&

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&

(0,0)(60,50) (0,20)(10,20)(10,30)(0,30) (0,30)(10,30)(10,40)(0,40) (10,0)(20,0)(20,10)(10,10) (10,20)(20,20)(20,30)(10,30) (10,30)(20,30)(20,40)(10,40) (10,40)(20,40)(20,50)(10,50) (20,0)(30,0)(30,10)(20,10) (20,20)(30,20)(30,30)(20,30) (20,30)(30,30)(30,40)(20,40) (20,40)(30,40)(30,50)(20,50) (30,0)(40,0)(40,10)(30,10) (30,10)(40,10)(40,20)(30,20) (30,40)(40,40)(40,50)(30,50) (40,0)(50,0)(50,10)(40,10) (40,10)(50,10)(50,20)(40,20) (40,40)(50,40)(50,50)(40,50) (50,20)(60,20)(60,30)(50,30) (50,30)(60,30)(60,40)(50,40) (10,50)[2.0]{} (20,10)[2.0]{} (20,10)[$a$]{} (30,30)[2.0]{} (30,30)[$d$]{} (40,40)[2.0]{} (40,40)[$c$]{} (50,20)[2.0]{} (50,20)[$b$]{} (0,0)(0,50) (0,0)(60,0) (10,0)(10,50) (0,10)(60,10) (20,0)(20,50) (0,20)(60,20) (30,0)(30,50) (0,30)(60,30) (40,0)(40,50) (0,40)(60,40) (50,0)(50,50) (0,50)(60,50) (60,0)(60,50)

We can now argue in a similar fashion that the entries preceding the maximum entry of $\pi$ form an increasing sequence, i.e., that there cannot be a $21$ pattern before the maximum entry of $\pi$. Suppose to the contrary that there were one, and choose such a pattern $ba$ where $b$ is as high as possible, and $a$ is as low as possible (for the chosen $b$). Now the cell defined by $\{a,b\}$ must be 