Version of Perceptron
Old question, but in case anyone is still interested in an answer...
In the perceptron algorithm a point $x$ has a label $y$ equal to 1 or -1. The predicted label for a point is $w\cdot x$. The goal is to separate the points with an hyperplane orthogonal to w in such a way that the points with label 1 are on one side and the points with label -1 are on the other side. Mathematically, the &quot;sides&quot; of the hyperplane are characterised by the equations $w\cdot x&gt;0$ and $w\cdot x&lt;0$, respectively. ($w\cdot x=0$ is the equation of the hyperplane itself).
Therefore $x$ point is correctly labelled (it is on the right side of the hyperplane orthogonal to $w$) if

$y=1$ and $w\cdot x&gt;0$, or
$y=-1$ and $w\cdot x&lt;0$

In both cases, a correctly labelled point corresponds to $yw\cdot x &gt; 0$, while an incorrectly labelled points corresponds to $yw\cdot x \leq 0$.
The whole algorithm and the proof of its convergence are based on this simple observation.
If we change the condition for update to $yw\cdot x &lt; 1$, the algorithm would not be able to find a separating hyperplane. For example, consider the simple dataset $x_1 = (2,0), y_1= 1$ , $x_2 = (-2,1), y_2= -1$.

Start: $w=(0, 0)$.
Step 1: $y_1w\cdot x_1 = 0 &lt; 1$ therefore $w \leftarrow w + y_1x_1 = (2, 0)$
Step 2: $y_2w\cdot x_2 = 4 &gt; 1$, do not update $w$.
Step 3: $y_1w\cdot x_1 = 4 &gt; 1$, do not update $w$. We went over all points without updating $w$, therefore exit the loop.
Result: the hyperplane orthogonal to $w=(2, 0)$ is the y-axis, which clearly does not separate $x_1$ from $x_2$.

TLDR;  Update conditions different from $yw\cdot x \leq 0$ do not offer any guarantees for finding a separating hyperplane.