algorithms applied to n-tuples of numbers to end up with $(0,0,...,0)$ I came to the following problem in Problem Solving Strategies by Arthur Engel on page 18 :

Start with a sequence $S=(a,b,c,d) $ of positive integers and find the derived sequence $ S_1=T(S)=(|a-b|,|b-c|,|c-d|,|d-a|)$. Does the sequence $ S, S_1, S_2=T(S_1), S_3=T(S_2) ,... $ always end up with $(0,0,0,0)$?

Then he makes some claims :
After at most four steps, all four terms of the sequence become even and in general after $4k$ steps at most become divisible by $2^k$.
let max $S$ be the maximal element of $S$. Then max $S_{i+1}\leq$ max $S_i$ , and max $S_{i+4}<$ max $S_i$ as long as max $S_i>0$.
How one should attempt to prove these, what are ideas involved and are there other functions to apply to the first sequence and still having the ending with $(0,0,0,0)$ property? the book refers to problem by shrinking squares, what is the relation?

"We have a square, each of its vertices are labeled with a value", I think it's called shrinking squares because you get a sequence of squares and the values on the vertices shrink to $0$.
To prove the claim about parity note there are only $16$ possible combinations for parity so in the worse case scenario you just have to draw the digraph and check it has a sink node, of course you can skip some steps:
If there are $3$ nodes of one parity and one node of another parity then after one iteration you end up in  configuration  $(1,0,1,0)$ or $(0,1,0,1)$ and after this you get to $(1,1,1,1)$ and after that $(0,0,0,0)$.
So now you just need to check the configurations with two consecutive values of one parity and two consecutive of the other parity, but these always go to $(1,0,1,0)$ or $(0,1,0,1)$. So indeed after $4$ steps you'll reach $(0,0,0,0)$ ( which stabilizes).
The claim about $S_i \geq S_{i+1}$ is straightforward since the values are always non-negative.
The claim that $S_{i+4} < S_{i}$ seems a bit harder to me.
We can assume that not all of $(a,b,c,d)$ are even.
If $S_i$ is odd then by the previous result at one of the next $4$ iterations all values are even, so the maximum decreases.
If $S_i$ is even then by the previous result at one of the next $4$ iterations all values are odd, so the maximum decreases.

The previous solution (it's wrong because $x-x$ can be $0$ ):
If there is a value in the range $(0,S_i)$ then in the next iteration both this value and the previous one (ciclically) will also be in the range $(0,S_{i})$, so after $4$ steps all values will be in range $(0,S_i)$.
So you can assume all values are $0$ or $S_i$, and then you get the same diagram as for the previous claim ( the parity one). And with the same analysis we observe all configurations of only $0$ and $S_i$ reach  $(0,0,0,0)$ after $4$ iterations.
