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  • 5 state

5 state · Changes

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update state, and read trace doc authored Oct 19, 2023 by geruiwang's avatar geruiwang
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zkevm-docs/5-state.markdown
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TODO
# 布局
如下几列。注意,和LookupEntry的State基本一致。
```rust
pub struct StateCircuitConfig<F> {
q_enable: Selector,
/// Type of value, one of stack, memory, storage, call context, call data or return data
/// A `BinaryNumberConfig` can return the indicator by method `value_equals`
tag: BinaryNumberConfig<Tag, LOG_NUM_STATE_TAG>,
/// Stamp that increments for each state operation, unique for each row
stamp: Column<Advice>,
/// High 128-bit value of the row
value_hi: Column<Advice>,
/// Low 128-bit value of the row
value_lo: Column<Advice>,
/// Call id (other types) or contract address (storage type only)
call_id_contract_addr: Column<Advice>,
/// High 128-bit of the key (storage type only)
pointer_hi: Column<Advice>,
/// Low 128-bit of the key (storage type only) or call context tag
/// Or stack pointer or memory address or data index (call data and return data)
pointer_lo: Column<Advice>,
/// Whether it is write or read, binary value
is_write: Column<Advice>,
_marker: PhantomData<F>,
}
```
其中,tag我们没有简单的使用一个advice列,而是用了一个电路小工具名叫“BinaryNumberConfig/BinaryNumberChip”。它既可以表示一个tag,也可以当做动态选择器,用于启用约束。例如,我们想tag的值为stack(1)时启动约束1,为memory(0)时启动约束2。那么,示例代码如下
```rust
let condition_1 = tag.value_equals(1);
let condition_2 = tag.value_equals(0);
vec![condition_1 * constraint_1, condition_2 * constraint_2]
```
排序:在当做电路的输入之前,需要将这个table排序。按元组排序:先按tag,再按callid, pointer hi, pointer, stamp的顺序排序。
# 约束
为了保证读写一致性,我们需要排序,按照如下顺序排序的:先按tag,再按callid, pointer hi, pointer lo, stamp的顺序排序。
验证排序的方式:将元组编码为16bit的limbs,假设有L个。(可以bit更大但是那样比较难做range proof)。上下两行的limbs作差,有L个diff。第一个非零的diff,index记为first_diff_limb。要求diff[first_diff_limb]==0即可。
在开发的代码中,例如Rust语言,排序十分简单。难的是在电路中增加约束保证整个Witness的行是符合排序规则的。参考scroll的电路,我们使用如下方法。
## 约束排序的方式
具体做法:用binary表示first_diff_limb. 需要用log_2 L个变量。再定义变量limb_diff表示diff[first_diff_limb]。再定义inv表示前者的逆。
约束1:建立一个expr数组v,是diff的累计RLC:[diff[0], diff[0]+r diff[1], diff[0]+r diff[1]+r' diff[2]...]。约束first_diff_limb.equal(i) * v[i] == 0。即,前first_diff_limb是否都为0。注:equal是binary表示的一种工具。
将tag, callid, pointer hi, pointer lo, stamp元组编码为16-bit的limbs,假设有L个limbs。(可以bit更大但是那样比较难做range proof)。上下两行的limbs作差,有L个diff。第一个非零的diff,index记为first_diff_limb。要求diff[first_diff_limb]==0即可。
具体做法:
- 用BinaryNumberChip表示变量first_diff_limb。需要用log_2(L)个变量。
- 再定义变量limb_diff表示diff[first_diff_limb]。
- 再定义inv表示前者的逆。
- 约束1:建立一个expr数组v,是diff的累计RLC:[diff[0], diff[0]+r diff[1], diff[0]+r diff[1]+r' diff[2]...]。约束first_diff_limb.value_equals(i) * v[i] == 0。即,前first_diff_limb是否都为0。注:value_equals(x)方法是BinaryNumberChip中用来判断值是否等于x的,若等于输出1,否则输出0。
注:也许可以把RLC改成^2
约束2:first_diff_limb.equal(i) * diff[i] - limb_diff。即,limb_diff值是否正确
约束3:limb_diff存在inv. 1-limb_diff * inv == 0。即它非零
约束4:limb_diff in range u16
\ No newline at end of file
- 约束2:first_diff_limb.value_equals(i) * diff[i] - limb_diff。即,limb_diff值是否按照定义等于diff[first_diff_limb]。
- 约束3:limb_diff存在inv。 1-limb_diff * inv == 0。即它非零。
- 约束4:limb_diff in range u16,即[0,2^16-1]。因为两个limb如果都属于u16,大的limb(下一行的)减小的limb(上一行的)的diff必然也是u16。反之,小的limb减大的limb的diff必然不是u16。
## 不同state类型的约束
stack的约束:TODO
memory的约束:TODO。。。
\ No newline at end of file
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