updates
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@@ -59,6 +59,14 @@
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### Post-Quantum (PQ) crypto
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- Fundamentally different computation paradigm than "classical" von Neumann or dataflow models
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- Relies on properties of quantum physics to solve problems efficiently
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- Superposition: state of quantum bit ("qubit") expressed by probability model over continuous range of values (vs. classic bit: 0 or 1 only)
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- Like being able to operate on all possible bit combos of a register simultaneously, instead of operating on only one among all possibilities
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- Entanglement: operating on one qubit affects others
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### Zero-Knowledge (ZK) proofs
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### Homomorphic encryption
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@@ -38,10 +38,14 @@ Context: computing stack
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- 2. Complete mediation (reference monitor)
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- 3. Correct
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Isolating User Processes from Each Other
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Isolating OS from Untrusted User Code
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- How do we meet the first requirement of a TCB (e.g., isolation or tamper-proofness)?
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- Hardware support for memory protection
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- Processor execution modes (system AND user modes, execution rings)
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- Privileged instructions which can only be executed in system mode
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- System calls used to transfer control between user and system code
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- How do we meet the user/user isolation and separation?
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- OS uses hardware support for memory protection to ensure this.
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System Calls: Going from User to OS Code
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@@ -50,16 +54,107 @@ System Calls: Going from User to OS Code
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- The processor execution mode or privilege ring changes when call and return happen.
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- x86 `sysenter` / `sysexit` instructions
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## Isolating OS from Untrusted User Code
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Isolating User Processes from Each Other
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- How do we meet the first requirement of a TCB (e.g., isolation or tamper-proofness)?
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- Hardware support for memory protection
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- Processor execution modes (system AND user modes, execution rings)
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- Privileged instructions which can only be executed in system mode
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- System calls used to transfer control between user and system code
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- How do we meet the user/user isolation and separation?
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- OS uses hardware support for memory protection to ensure this.
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Virtualization
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- OS is large and complex, even different operating systems may be desired by different customers
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- Compromise of an OS impacts all applications
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Complete Mediation: The TCB
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- Make sure that no protected resource (e.g., memory page or file) could be accessed without going through the TCB
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- TCB acts as a reference monitor that cannot be bypassed
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- Privileged instructions
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Limiting the Damage oa a Hacked OS
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Use: Hypervisor, virtual machines, guest OS and applications
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Compromise of OS in VM1 only impacts applications running on VM1
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### Secure boot and Root of Trust (RoT)
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Goal: create chain of trust back to hardware-stored cryptographic keys
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#### Secure enclave: overview (Intel SGX)
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Goal: keep sensitive data within hardware-isolated encrypted environment
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### Access control
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Controlling Accesses to Resources
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- TCB (reference monitor) sees a request for a resource, how does it decide whether it should be granted?
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- Example: Should John's process making a request to read a certain file be allowed to do so?
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- Authentication establishes the source of a request (e.g., John's UID)
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- Authorization (or access control) answers the question if a certain source of a request (User ID) is allowed to read the file
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- Subject who owns a resource (creates it) should be able to control access to it (sometimes this is not true)
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- Access control
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- Basically, it is about who is allowed to access what.
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- Two parts
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- Part I - Policy: decide who should have access to certain resources (access control policy)
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- Part II - Enforcement: only accesses defined by the access control policy are granted.
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- Complete mediation is essential for successful enforcement
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Discretionary Access Control
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- In discretionary access control (DAC), owner of a resource decides how it can be shared
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- Owner can choose to give read or write access to other users
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- Two problems with DAC:
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- You cannot control if someone you share a file with will not further share the data contained in it
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- Cannot control "information flow"
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- In many organizations, a user does not get to decide how certain type of data can be shared
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- Typically the employer may mandate how to share various types of sensitive data
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- Mandatory Access Control (MAC) helps address these problems
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Mandatory Access Control (MAC) Models
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- User works in a company and the company decides how data should be shared
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- Hospital owns patient records and limits their sharing
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- Regulatory requirements may limit sharing
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- HIPAA for health information
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#### Example: Linux system controls
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Unix file access control list
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- Each file has owner and group
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- Permissions set by owner
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- Read, write, execute
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- Owner, group, other
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- Represented by vector of four octal values
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- Only owner, root can change permissions
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- This privilege cannot be delegated or shared
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- Setid bits -- Discuss in a few slides
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Process effective user id (EUID)
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- Each process has three IDs (+ more under Linux)
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- Real user ID (RUID)
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- Same as the user ID of parent (unless changed)
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- Used to determine which user started the process
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- Effective user ID (EUID)
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- From set user ID bit on the file being executed, or sys call
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- Determines the permissions for process
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- File access and port binding
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- Saved user ID (SUID)
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- So previous EUID can be restored
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- Real group ID, effective group ID used similarly
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#### Weaknesses in Unix isolation, privileges
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- Shared resources
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- Since any process can create files in `/tmp` directory, an untrusted process may create files that are used by arbitrary system processes
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- Time-of-Check-to-Time-of-Use (TOCTTOU), i.e. race conditions
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- Typically, a root process uses system call to determine if initiating user has permission to a particular file, e.g. `/tmp/X`.
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- After access is authorized and before the file open, user may change the file `/tmp/X` to a symbolic link to a target file `/etc/shadow`.
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### Hazard: race conditions
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@@ -29,6 +29,8 @@ $f|_{B^2}$ is a continuous map from $B^2\to \mathbb{R}^2-\{0\}$.
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$f|_{S^1=\partial B^2}:S^1\to \mathbb{R}-\{0\}$ **is nulhomotopic**.
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> Recall that: Any map $g:S^1\to Y$ is nulhomotopic whenever it extends to a continuous map $G:B^2\to Y$.
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Construct a homotopy between $f|_{S^1}$ and $g$
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$$
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@@ -57,7 +59,7 @@ Therefore $f$ must have a root in $B^2$.
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<details>
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<summary>Proof: part 2</summary>
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If \|a_{n-1}\|+\|a_{n-2}\|+\cdots+\|a_0\|< R$ has a root in the disk $B^2_R$. (and $R\geq 1$, otherwise follows part 1)
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If $\|a_{n-1}\|+\|a_{n-2}\|+\cdots+\|a_0\|< R$ has a root in the disk $B^2_R$. (and $R\geq 1$, otherwise follows part 1)
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Consider $\tilde{f}(x)=f(Rx)$.
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public/CSE4303/Intel_SGX.png
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public/CSE4303/Intel_SGX.png
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