$P = NP$, what am I missing? First post here so hope I'm not missing too many guidelines.
I've had this idea for a few weeks now and I can't myself see where I'm going wrong with it, hope it makes some sense to you and thanks in advance for the help.
Here we go:
Assuming we could solve a NP-complete problem in logn SPACE, e.g. TSP. Then by reachability we could show that $P = NP$.
The machine will keep track of this information:

$\log n$ nodes for a path walked, call this memory pathMem
$\log n$ nodes for a reconstructed path walked, call this memory pathMemTemp
$\log n$ numbers indicating the chosen path, call this chosenPathNums
$\log n$ nodes for ending nodes of each path, call this endMem
some representation of the cost of the walk.

Step 1: Walk $\log n$ steps, note down the $\log n$ walked nodes in pathMem
Step 2: Check that there are no duplicates in pathMem (that means that you've walked $\log n$ unique nodes).
Step 3: Use a modified version of Savitch that counts the number of possible paths between start node and last node in pathMem. This should be at most $(\log(n)!)$ paths, represented in that number using binary and we get $O(log(log(n)!))$
Step 4: Find which of these paths that is in pathMem. Save that number in chosenPathNums at pos 1.
This gives us a deterministic way of refinding a specific path.
Step 5: Move last node in pathMem to endMem pos 1, clear pathMem.
Step 6: Walk $\log n$ steps from node at endMem pos 1, note down walked nodes in pathMem.
Step 7: Check that there are no duplicates in pathMem (that means that you've walked $\log n$ unique nodes).
Step 8: Reconstruct path that was taken from start node to node at endMem pos 1 by using Savitch's and the chosenPathNum, save it in pathMemTemp.
Step 9: Check that pathMem and pathMemTemp has no overlap. If there is overlap, halt with no, otherwise continue.
Step 10: Find the path number that is in pathMem. Save that number in chosenPathNums at pos 2.
Step 11: Move last node in pathMem to endMem pos 2, clear pathMem.
Step 12: Rinse and repeat, now checking pathMem vs start-&gt;(endMem pos 1) and then (endMem pos 1)-&gt;(endMem pos 2) etc., until you've walked $n$ nodes, thus having $\log n$ nodes in endMem and $\log n$ numbers in chosenPathNums.
Step 13: Calculate distance walked, check if less or equal to threshold.
This algorithm could be turned deterministic by looping every possible $\log n$ walk that is found from a node to another node.
It is not clear what is your question, but I will guess that it is &quot;what is my mistake in my proof of $P = NP$?&quot;
In addition to the fact that the algorithm does not guarantee to find the minimal path (because the algorithm cant deterministicaly choose the endMem nodes), the number of possible $\log n$-walks given $\log n$ nodes is not polynomial: $(\log n)! \sim \left(\frac{\log n}{e}\right)^{\log n}\sqrt{2\pi\log n}\sim \frac{\sqrt{2\pi\log n}}{n}e^{\log n \log \log n}\sim \sqrt{2\pi\log n}\times n^{\log\log n - 1}$.
As for the space complexity, to check there are no overlap between two $\log n$-walks, you need to keep track of the whole path (and not only two consecutive $\log n$-walks). That needs $\Omega(n\log n)$ space ($n$ nodes with a binary encoding of size $\log n$). So the corresponding Turing Machine use polynomial space, not logarithmic space.