I run an openSUSE 15.1 server with samba shares. I've always been able to connect to my shared folders using Windows 10. However, after the Win10 2004 (May 2020) update I now get this:  Error code: 0x80070043 The network name cannot be found.  I can ping the server and even connect to it using another PC w/o the update, so it appears something is now broken. Has anyone else experienced this? EDIT: My smb.conf file contains win protocol = SMB2. Changing to SMB3 had no effect. Purpose of this post is to see if anyone else who upgraded to 2004 is also running into this problem. EDIT #2: I verified that other Windows 10 computers on my network (same version) can still access the server shares so not sure what's up with this one PC. It's doing some other strange things as well, like not connecting to port 80 on this same server when using its public DNS.
I have a virtual machine running ubuntu and I am trying to host some files on ubuntu and map the drive in windows PC.  Following the steps , I am able to see (from windows) the "work" folder after typing in \ubuntu_ip. But when trying to access or map it with "svc_acct" from windows, I get "0x80070043 error the network name cannot be found". I guess I am not accessing correctly from windows.   SAMBA config below (svc_acct is an account set up under linux). When accessing the folder from windows, I get prompted for username/password. Not sure what credential to type in exactly.  samba config: [work] path = /home/svc_acct/work valid users = svc_acct available = yes read only = no browsable = yes public = yes writable = yes force user = svc_acct create mask = 0777 directory mask = 0777   UPDATE:  All I did to resolve this is by changing samba config's path from path = /home/svc_acct/work to path = /home/svc_acct following  post  (last sentence). Then no more prompt for password in windows when I try to click into the folder. Thanks to @sancho.s for suggesting the link.
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From the perspective of information theory, I have such an intuitive understanding:  Say there are two ensembles $A$ and $B$ which are composed of the same set of elements labeled by $x$. $p(x)$ and $q(x)$ are different probability distributions over ensemble $A$ and $B$ respectively.   From the perspective of information theory, $\log_{2}(P(x))$ is the least amount of bits that required for recording an element $x$ for ensemble $A$. So that the expectation $$\sum_{x \in  ensemble}-p(x)\ln(p(x))$$ can be interpreted as at least how many bits that we need for recording an element in $A$ on average.   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