mod(x,2) is zero if and only if x is an integer multiple of 2; otherwise, mod(x,2) is a number larger than 0. z = mod(x,2)-1 thus reaches its minimum, -1 if x is an integer multiple of 2 and reaches its maximum (2-delta)-1 = 1-delta at limit(delta->0, delta>0, x+delta is an integer multiple of 2). -1 <= z < 1 .
abs(z) is thus 1 if and only if x is an integer multiple of 2, and is less than 1 otherwise. Hence abs(z)<1 is false (0) if and only if x is an integer multiple of 2, and is true (1) otherwise.
abs(z) decreases from 1 as x increases away from an integer multiple of 2, reaching minimum of 0 at x-1 being an integer multiple of 2, and increasing again to 1-delta as x approaches the next multiple of 2.
Hence 1-abs(z) increases from 0 as x increases away from an integer multiple of 2, reaching 1 when x-1 is an integer multiple of 2, and decreases again to delta as x approaches the next multiple of 2. None of these values are negative so abs(1-abs(z)) has the same behavior.
The effect of multiplying the above expression by a number that is 0 when x is an integer multiple of 2 is nothing, as the value is already 0 when that is the case. The overall expression can thus be reduced to
1-abs(mod(x,2)-1)
As analyzed above, this expresion is 0 when x is an integer multiple of 2, is mod(x,2) when mod(x,2) <= 1, and is 1-mod(x,2) for mod(x,2) > 1
This function is not 1-periodic as is required by the problem definition paper: it is 2-periodic instead. You would have to work with z = mod(2*x,2)-1 instead to make it 1-periodic, and if you are going to do that you might as well rebuild the function from scratch.
