Now consider the Perceptron algorithm with Offset. Whenever there is a "mistake" $($or equivalently, whenever y$^($i$)(\backslash$theta $$ x$^($i$)+\backslash$theta $\_0)<=0$ i$.$e$.$ when the label y$^$i and h$($x$)$ do not match$),$ perceptron updates $\backslash$theta with $\backslash$theta $+$y$^($i$)$ x$^($i$)$ and $\backslash$theta $\_0$ with $\backslash$theta $\_0+$y$^($i$)$ More formally, the Perceptron Algorithm with Offset is defined as follows: Perceptron $(\{($x$^($i$),$ y$^($i$)),$ i$=1,...,$ n$\},$ T$)$ : initialize $\backslash$theta $=0($vector$)$; $\backslash$theta $\_0=0($scalar$)$ for t$=1,...,$ T do for i$=1,...,$ n do if y$^($i$)(\backslash$theta $$ x$^($i$)+\backslash$theta $\_0)<=0$ then update $\backslash$theta $=\backslash$theta $+$y$^($i$)$ x$^($i$)$ update $\backslash$theta $\_0=\backslash$theta $\_0+$y$^($i$)$ In the next set of problems, we will try to understand why such an update is a reasonable one. When a mistake is spotted, do the updated values of $\backslash$theta and $\backslash$theta $\_0$ provide a better prediction? In other words, is y$^($i$)((\backslash$theta $+$y$^($i$)$ x$^($i$))$ x$^($i$)+\backslash$theta $\_0+$y$^($i$))$ always greater than or equal to y$^($i$)(\backslash$theta $$ x$^($i$)+\backslash$theta $\_0)$ Yes, because $\backslash$theta $+$y$^($i$)$ x$^($i$)$ is always larger than $\backslash$theta Yes, because $($y$^($i$))^2$x$^($i$)^2+($y$^($i$))^2>=0$ No$,$ because $($y$^($i$))^2$x$^($i$)^2-($y$^$i$)^2<=0$ No$,$ because $\backslash$theta $+$y$^($i$)$ x$^($i$)$ is always larger than $\backslash$theta $/$r$/$n For a given example i$,$ we defined the training error as $1$ if y$^($i$)(\backslash$theta $$ x$^($i$)+\backslash$theta $\_0)<=0,$ and $0$ otherwise: $\backslash$epsi $\_$i$(\backslash$theta $,\backslash$theta $\_0)=[[$y$^($i$)(\backslash$theta $$ x$^($i$)+\backslash$theta $\_0)<=0]]$ Say we have a linear classifier given by $\backslash$theta $,\backslash$theta $\_0.$ After the perceptron update using example i$,$ the training error $\backslash$epsi $\_$i$(\backslash$theta $,\backslash$theta $\_0)$ for that example can $($select all those apply$)$: Increase Stay the same Decrease