当b7=1时, 计算(00000010)* (b7b6b5b4b3b2b1b0 )= ( A ) ⊕( B ),其中⊕表示异或加,A与B为8位二进制序列,求解A与B
时间: 2024-03-02 20:49:28 浏览: 39
根据题意,当b7=1时,(b7b6b5b4b3b2b1b0)的值为一个8位二进制数,因此,可以将它转换成十进制数,记为n,则n的值为2的几次幂加上其中某些位的值,其中幂的值为7,6,5,4,3,2,1,0,对应的系数为1,b6,b5,b4,b3,b2,b1,b0。例如,当(b7b6b5b4b3b2b1b0)的值为00001101时,n=2^3+2^2+2^0=13。
因此,(00000010) * (b7b6b5b4b3b2b1b0)可以表示为:
```
(00000010) * (b7b6b5b4b3b2b1b0) = 2 * n
```
将2 * n表示成二进制数,可以得到:
```
2 * n = (b8)1(b6b5b4b3b2b1b0)0
```
对于A和B,根据异或加的定义,可以得到:
```
A = (00000010) ⊕ (b8)1(b6b5b4b3b2b1b0)0 = (b8)1(b6b5b4b3b2b1b0)0 + 00000010
B = (00000010) & (b8)1(b6b5b4b3b2b1b0)0 = 00000000
```
因此,A为一个8位二进制数,其最高位为b8,次高位为1,其余位为(b6b5b4b3b2b1b0)0加上00000010的结果。B为一个8位二进制数,其所有位均为0。
相关问题
为什么将 b7b6b5b4b3b2b1b0 替换为 11011010呢,可以讲的具体些吗
好的,我理解了您的问题。首先,我们来看一下题目中的式子:
(00000010)* (b7b6b5b4b3b2b1b0 )= ( A ) ⊕( B )
其中,b7b6b5b4b3b2b1b0 表示一个 8 位二进制数,每位上可能是 0 或 1。而题目中又给出了一个条件:b7=1。因此,我们需要将 b7b6b5b4b3b2b1b0 中的 b7 替换成 1,也就是得到了 11011010。
接下来,我们按照乘法的规则进行计算,例如:
```
00000010
* 11011010
----------
```
我们从右往左对 00000010 中的每一位与 11011010 中的每一位进行乘法运算。由于 00000010 中只有第 2 位为 1,因此我们只需要将 11011010 左移一位,得到 110110100,然后将它与 00000010 相加,得到 110110110。这个结果的右边 8 位与之前的结果无关,因此直接忽略,只保留左边的 8 位,得到 00000000。
接下来,我们对 00000010 中的第 1 位与 11011010 进行乘法运算。由于 00000010 中只有第 1 位为 0,因此这次的乘积直接为 0,无需进行任何计算,因此得到的结果依然是 00000000。
最后,我们将两次计算的结果相加,得到:
```
00000000
+ 00000000
----------
00000000
```
因此,最终的结果 A 为 00000000,B 也为 00000000。
但是,题目要求的是在 b7=1 的条件下进行计算,因此我们需要将 b7b6b5b4b3b2b1b0 中的 b7 替换为 1,即得到 11011010,然后再按照上述步骤进行计算,才能得到符合题目要求的结果。
Use the ALC-Worlds algorithm to decide the satisfiability of the concept name B0 w.r.t. the simple Tbox,Draw the recursion tree of a successful run and of an unsuccessful run. Does the algorithm return a positive or negative result on this input? B0 ≡ B1\bigsqcapB2 B1 ≡ ∃r.B3 B2 ≡ B4 \bigsqcap B5 B3 ≡ P B4 ≡ ∃r.B6 B5 ≡ B7 \sqcap B8 B6 ≡ Q B7 ≡ ∀r.B4 B8 ≡ ∀r.B9 B9 ≡ ∀r.B10 B10 ≡ ¬P
To use the ALC-Worlds algorithm, we first need to convert the Tbox and the concept name B0 into their ALC syntax:
B0 ≡ B1 ⊓ B2
B1 ≡ ∃r.B3
B2 ≡ B4 ⊓ B5
B3 ≡ P
B4 ≡ ∃r.B6
B5 ≡ B7 ⊓ B8
B6 ≡ Q
B7 ≡ ∀r.B4
B8 ≡ ∀r.B9
B9 ≡ ∀r.B10
B10 ≡ ¬P
Now, we can start the algorithm:
1. Initialize the set of worlds W to contain the empty interpretation {}.
2. For each concept name and role name, initialize its interpretation in each world to the empty set.
3. For each concept name A that appears in the Tbox, add the pair (A, {}) to the interpretation of A in each world.
4. For each role name r that appears in the Tbox, add the pair (r, {}) to the interpretation of r in each world.
5. For each concept name A that appears in the definition of a concept or role in the Tbox, propagate its interpretation to all worlds that satisfy that concept or role.
6. For each concept name A that appears in the definition of a concept or role in the Tbox, repeat step 5 until no more interpretations can be propagated.
7. For each world w in W, check if w satisfies B0. If so, return "satisfiable" and the model consisting of all interpretations in w. If not, mark w as unsatisfiable.
8. If all worlds are marked as unsatisfiable, return "unsatisfiable".
Recursion tree for a successful run:
```
W={}
|
W1={r={}, B3={}}
|
W2={r={}, B3={}, B6={}}
|
W3={r={}, B3={}, B6={Q}}
|
W4={r={}, B3={P}, B6={Q}}
|
W5={r={r1:W4}, B3={P}, B6={Q}}
|
W6={r={r1:W4, r2:W5}, B3={P}, B6={Q}}
|
W7={r={r1:W4, r2:W5, r3:W6}, B3={P}, B6={Q}}
|
W8={r={r1:W4, r2:W5, r3:W6, r4:W7}, B3={P}, B6={Q}}
|
W9={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8}, B3={P}, B6={Q}}
|
W10={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8, r6:W9}, B3={P}, B6={Q}}
|
W11={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8, r6:W9}, B3={}, B6={Q}}
|
W12={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8, r6:W9}, B3={P}, B6={}}
|
W13={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8, r6:W9, r7:W12}, B3={P}, B6={}}
|
W14={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8, r6:W9, r7:W12, r8:W13}, B3={P}, B6={}}
|
W15={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8, r6:W9, r7:W12, r8:W13, r9:W14}, B3={P}, B6={}}
|
W16={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8, r6:W9, r7:W12, r8:W13, r9:W14, r10:W15}, B3={P}, B6={¬P}}
The algorithm returns a positive result (i.e., "satisfiable") with the model consisting of all interpretations in W16.
Recursion tree for an unsuccessful run:
```
W={}
|
W1={r={}, B3={}}
|
W2={r={}, B3={}, B6={}}
|
W3={r={}, B3={}, B6={Q}}
|
W4={r={}, B3={P}, B6={Q}}
|
W5={r={r1:W4}, B3={P}, B6={Q}}
|
W6={r={r1:W4, r2:W5}, B3={}, B6={Q}}
|
W7={r={r1:W4, r2:W5, r3:W6}, B3={P}, B6={Q}}
|
W8={r={r1:W4, r2:W5, r3:W6, r4:W7}, B3={P}, B6={Q}}
|
W9={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8}, B3={P}, B6={Q}}
|
W10={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8, r6:W9}, B3={}, B6={Q}}
|
W11={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8, r6:W9, r7:W10}, B3={P}, B6={Q}}
|
W12={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8, r6:W9, r7:W10, r8:W11}, B3={P}, B6={Q}}
|
W13={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8, r6:W9, r7:W10, r8:W11, r9:W12}, B3={P}, B6={Q}}
|
W14={r={r1:W4, r2:W5, r3:W6, r4:W7, r5:W8, r6:W9, r7:W10, r8:W11, r9:W12, r10:W13}, B3={P}, B6={¬P}}
The algorithm returns a negative result (i.e., "unsatisfiable") as all worlds are marked as unsatisfiable.