ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \\ \mathrm{F}\left(\mathrm{x},\mathrm{y}\right) = \left\{\begin{array}{l}\mathrm{F}\left(1,\mathrm{y}-1\right) , \mathrm{x}=0, \mathrm{y}\ne 0\to \boxed{1} \\ \mathrm{x}+1 , \mathrm{y}=0\to \boxed{2}\\ \mathrm{F}\left(\mathrm{F}\left(\mathrm{x}-1,\mathrm{y}\right),\mathrm{y}-1\right) , \mathrm{x}\ne 0 , \mathrm{y}\ne 0\to \boxed{3}\end{array}\right. \end{gathered}$$

$$\begin{gathered} \mathrm{ลองคำนวณ} \mathrm{F}\left(\mathrm{x},0\right) \mathrm{จาก} \boxed{2} \\ \mathrm{เมื่อ} \mathrm{x}=1 \mathrm{จะได้} \mathrm{F}\left(1,0\right)=1+1=2 \\ \mathrm{x}=2 ; \mathrm{F}\left(2,0\right)=2+1=3 \\ \mathrm{x}=3 ; \mathrm{F}\left(3,0\right)=3+1=4 \\ \mathrm{x}=4 ; \mathrm{F}\left(4,0\right)=4+1=5 \end{gathered}$$

$$\begin{gathered} \mathrm{ลองคำนวณ} F(x,1) \\ \mathrm{เมื่อ} \mathrm{x}=0 \mathrm{จะได้} \mathrm{F}\left(0,1\right)=\mathrm{F}\left(1,0\right)=2 \\ \mathrm{x}=1 ; \mathrm{F}\left(1,1\right)=\mathrm{F}\left(\mathrm{F}\left(0,1\right),0\right)=\mathrm{F}\left(2,0\right)=3 \\ \mathrm{x}=2 ; \mathrm{F}\left(2,1\right)=\mathrm{F}\left(\mathrm{F}\left(1,1\right),0\right)=\mathrm{F}\left(3,0\right)=4 \\ \mathrm{x}=3 ; \mathrm{F}\left(3,1\right)=\mathrm{F}\left(\mathrm{F}\left(2,1\right),0\right)=\mathrm{F}\left(4,0\right)=5 \end{gathered}$$

$$\begin{gathered} \mathrm{พิจารณา} \mathrm{F}\left(1,2\right)=? \\ \mathrm{จาก} \boxed{3} \mathrm{จะได้} \mathrm{F}\left(1,2\right)=\mathrm{F}\left(\mathrm{F}\left(0,2\right),1\right)\to \boxed{4} \\ \mathrm{จาก} \boxed{1} \mathrm{จะได้} \mathrm{F}\left(0,2\right)=\mathrm{F}\left(1,1\right)=3 \\ \mathrm{จาก} \boxed{4} \mathrm{จะได้} \mathrm{F}\left(1,2\right)=\mathrm{F}\left(3,1\right)=5 \\ \mathrm{ดังนั้น} \mathrm{F}\left(1,2\right)+\mathrm{F}\left(3,1\right)=5+5 \\ =10 \end{gathered}$$