Can you buy some percent of bitcoin at different rates and merge them to form one bitcoin? Is it like you buy at different rates so you will get different exchange rates for particular percent
I’m installing LineageOS 17.1 on my new phone. It works fine, but I can’t get to data in TWRP.
As usual, to install, I wipe everything, drop into TWRP, format (not just wipe) data, wipe cache/davlik/system, and flash the LineageOS installer file. When the system starts first time, it gives a message “encrypting data”, which doesn’t have a “skip” option.
When I drop back into TWRP it asks for a decryption password to access data, but I don’t have one because I haven’t entered one in the first place.
How do I get LOS to not encrypt data, or alternatively, what decryption password should I use in TWRP?
Note: this isn’t a duplicate question. Other questions about decrypting data involve corrupt data or system issues that prevent data being accessed because of encryption, or trying to recover files from data. In my case, it’s a clean install with no prior data content, but I can’t figure how to access data in TWRP because it seem to encrypt it automatically but there’s no hint what key to enter in TWRP.
A while ago I made a linux partition on my MBP (2017), but it didn’t work so well so I needed to delete the partition. After a lot of trouble trying to delete it, I found this answer; I used the second part to free the space on the old linux partition, and then I resized it to 0. Only, it didn’t actually resize. The partition is still there (though it is no longer showing up in Disk Utility):
$ diskutil list /dev/disk0 (internal): #: TYPE NAME SIZE IDENTIFIER 0: GUID_partition_scheme 500.3 GB disk0 1: EFI EFI 314.6 MB disk0s1 2: Apple_APFS Container disk1 276.5 GB disk0s2 3: EFI NO NAME 629.1 MB disk0s3 /dev/disk1 (synthesized): #: TYPE NAME SIZE IDENTIFIER 0: APFS Container Scheme - +276.5 GB disk1 Physical Store disk0s2 1: APFS Volume Macintosh HD 246.9 GB disk1s1 2: APFS Volume Preboot 45.9 MB disk1s2 3: APFS Volume Recovery 510.8 MB disk1s3 4: APFS Volume VM 1.1 GB disk1s4
I think it is this line that is the culprit, though I am not sure:
2: Apple_APFS Container disk1 276.5 GB disk0s2
As I say, I have run
diskutil apfs resizeContainer disk0s2 0
As per the linked answer, but nothing changed. Even in recovery mode, Disk Utility will only show the ~277 GB I have available on the main partition.
How can I reclaim this space, which is no longer actually showing up?
P.S., this is what I had before I ran those commands:
$ diskutil list /dev/disk0 (internal): #: TYPE NAME SIZE IDENTIFIER 0: GUID_partition_scheme 500.3 GB disk0 1: EFI EFI 314.6 MB disk0s1 2: Apple_APFS Container disk1 276.5 GB disk0s2 3: EFI NO NAME 629.1 MB disk0s3 4: Apple_HFS Untitled 222.7 GB disk0s4 5: Apple_Boot Boot OS X 134.2 MB disk0s5 /dev/disk1 (synthesized): #: TYPE NAME SIZE IDENTIFIER 0: APFS Container Scheme - +276.5 GB disk1 Physical Store disk0s2 1: APFS Volume Macintosh HD 249.7 GB disk1s1 2: APFS Volume Preboot 45.9 MB disk1s2 3: APFS Volume Recovery 510.8 MB disk1s3 4: APFS Volume VM 6.4 GB disk1s4 /dev/disk3 (disk image): #: TYPE NAME SIZE IDENTIFIER 0: NO NAME +10.7 GB disk3
NO NAME was just a USB drive, by the way.
Is it possible to find, locate, or install the windows 10 boot manager through the grub 2 command line?
If so, what are those commands?
I cannot boot into windows of course and i can not enter ubuntu either. If the questions above are not possible, what are my options? I only want/need windows.
I have a dual boot (windows 10 & Ubuntu) installation on my laptop HDD, and no longer wanted to keep ubuntu on my hdd. I wanted to erase all trace of ubuntu so that i could safely transfer data to an ssd.
So Naturally, as someone new to linux, i thought it made sense to simply delete the linux partition from Windows’ Disk Manager since windows was (pre)installed first. Then migrate the data over to the clean ssd.
After deleting the linux partition and extending the c drive to take over unallocated space, I can no longer boot windows. As soon as i turn on my laptop, the GNU GRUB ver 2.04 shows up. In the command line it displays: “grub> ”
(It is Not grub rescue)
Typing “exit” simply brings me back to the same screen. The UEFI menu that i have doesnt have any option change the boot back to windows bootmgr. I dont have a windows installation disk, or any external windows backup device that was recommended. And I dont currently have but could make another ubuntu live usb, if necessary. But it would be cool if i could do this in grub
Neither the hdd nor the ssd can boot windows nor ubuntu. But both devices bring up the GRUB> command line. I now know that data migration moves everything except the boot manager itself. Ive also learned that installing ubuntu gave the booting reigns to ubuntu and took that control away from windows. Which is why uninstalling ubuntu from windows caused this issue..
So my question is, since ubuntu is gone, is windows boot manager also erased? Or is it possible to locate it through GRUB?
I have SSD and HDD. I’ve Windows OS installed in HDD. After when I bought SSD, I installed Windows OS to my SSD too. But didn’t remove old Windows OS from HDD. Later I installed Ubuntu 20.04 to my SSD as second OS, but /boot/efi mounted to HDD efi partition even I’ve created additional EFI partition in SSD too. Now I’m going to remove all things from my HDD. But I don’t know how to change /boot/efi partition to EFI partition from SSD.
BTW I’ve created in my SSD additional one more efi partition for Ubuntu in my SSD, does I need it or I can override to my Windows EFI partiton?
/dev/sda my HDD and
/dev/sdb my SSD
/dev/sda1 2048 923647 921600 450M Windows recovery environment /dev/sda2 923648 1128447 204800 100M EFI System /dev/sda3 1128448 1161215 32768 16M Microsoft reserved /dev/sda4 1161216 408525630 407364415 194.3G Microsoft basic data /dev/sda5 408526848 409599999 1073152 524M Windows recovery environment /dev/sda6 409602048 976771071 567169024 270.5G Microsoft basic data
/dev/sdb1 2048 1085439 1083392 529M Windows recovery environment /dev/sdb2 1085440 1290239 204800 100M EFI System /dev/sdb3 1290240 1323007 32768 16M Microsoft reserved /dev/sdb4 1323008 266242047 264919040 126.3G Microsoft basic data /dev/sdb5 266242048 266436607 194560 95M EFI System /dev/sdb6 573442048 976771071 403329024 192.3G Microsoft basic data /dev/sdb7 266436608 297687039 31250432 14.9G Linux swap /dev/sdb8 297687040 573442047 275755008 131.5G Linux filesystem
Let’s assume one has created an encrypted partition, e. g. with the LUKS standard. Then one creates a (virtual) disk image, e. g. for use by a virtual machine, containing an encrypted partition created by the same method and using the same encryption password. The disk image is stored inside the outer encrypted partition. I assume that a symmetric encryption is used.
Is it possible that the parts of the real disk which are occupied by the encrypted partion of the inside disk image are visible in plain text (or something close to that) as if no encryption was used (due to applying the same symmetric encryption method twice)?
If yes, in which particular configuration?
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From what I’ve been reading the guideline regarding manually changing the size of a Bitlocker encrypted partition is: (a) decrypt the partition. (b) resize the partition. (c) re-encrypt the partition.
Now, since there’re certain malwares that can hide themselves in a hidden partition patch (for example on a USB drive) I was wondering whether it’s theoretically possible for a malware to resize a partition encrypted using Bitlocker (I specifically refer to Bitlocker utilizing software encryption, not hardware encryption, since this is the default and more secure option) without decrypting it first or would it render the partition unusable?
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In the Partition problem, there is a set of integers, and the goal is to decide whether it can be partitioned into two sets of equal sum. This problem is known to be NP-complete.
Suppose we are given an instance and we know that it admits an equal-sum partition. Can this partition be found in polynomial time (assuming $Pneq NP$)?
Let’s call this problem “GuaranteedPartition”. On one hand, apparently one can prove that GuaranteedPartition is NP-hard, by reduction from Partition: if we could solve GuaranteedPartition with $n$ numbers in $T(n)$ steps, where $T(n)$ is a polynomial function, then we could give it as input any instance of Partition and stop it after $T(n)$ steps: if it returns an equal-sum partition the answer is “yes”, otherwise the answer is “no”.
On the other hand, the GuaranteedPartition is apparently in the class TFNP, whose relation to NP is currently not known.
Which of the above arguments (if any) is correct, and what is known about the GuaranteedPartition problem?