Mathematica 生成带锁定位的二进制数
Mathematica Generate Binary Numbers with Locked Bits
我有一个非常具体的 Mathematica 问题。我正在尝试围绕某些 'locked' 位生成所有二进制数。我正在使用字符串值列表来表示哪些位被锁定,例如{"U","U,"L","U"},其中 U 是一个 "unlocked" 可变位,L 是一个 "locked" 不可变位。我从一个临时的开始已格式化为先前列表的随机二进制数列表,例如 {0, 1, 1, 0},其中 1是锁定位,我需要找到1[的所有剩余二进制数=22=] 位是不变的。我已经递归地、迭代地解决了这个问题,并且两者结合都没有结果。这是我在大学里做的研究。
我正在构建一个以 10 为基数的二进制数列表。我意识到这段代码是完全错误的。这只是一次尝试。
Do[
If[bits[[pos]] == "U",
AppendTo[returnList, myFunction[bits, temp, pos, returnList]]; ],
{pos, 8, 1}]
myFunction[bits_, bin_, pos_, rList_] :=
Module[{binary = bin, current = Length[bin], returnList = rList},
If[pos == current,
Return[returnList],
If[bits[[current]] == "U",
(*If true*)
If[! MemberQ[returnList, FromDigits[binary, 2]],
(*If true*)
AppendTo[returnList, FromDigits[binary, 2]];
binary[[current]] = Abs[binary[[current]] - 1],
(*If false*)
binary[[current]] = 0;
current = current - 1]; ,
(*If false*)
current = current - 1];
returnList = myFunction[bits, binary, pos, returnList];
Return[returnList]]]
myFunction[bits_] := Module[{length, num, range, all, pattern},
length = Length[bits];
num = 2^length;
range = Range[0, num - 1];
all = PadLeft[IntegerDigits[#, 2], length] & /@ range;
pattern = bits /. {"U" -> _, "L" -> 1};
Cases[all, pattern]]
bits = {"U", "U", "L", "U"};
myFunction[bits]
{{0, 0, 1, 0}, {0, 0, 1, 1}, {0, 1, 1, 0}, {0, 1, 1, 1},
{1, 0, 1, 0}, {1, 0, 1, 1}, {1, 1, 1, 0}, {1, 1, 1, 1}}
您可以使用 Tuples
和 Fold
只生成您感兴趣的位集。
bits = {"U", "U", "L", "U"};
Fold[
Function[{running, next},
Insert[running, 1, next]], #, Position[bits, "L"]] & /@ Tuples[{0, 1}, Count["U"]@bits]
(*
{{0, 0, 1, 0}, {0, 0, 1, 1}, {0, 1, 1, 0}, {0, 1, 1, 1},
{1, 0, 1, 0}, {1, 0, 1, 1}, {1, 1, 1, 0}, {1, 1, 1, 1}}
*)
希望对您有所帮助。
in = IntegerDigits[Round[ Pi 10^9 ], 2];
mask = RandomSample[ConstantArray["L", 28]~Join~ConstantArray["U", 4],32];
subs[in_, mask_] := Module[ {p = Position[mask, "U"]} ,
ReplacePart[in, Rule @@@ Transpose[{p, #}]] & /@
Tuples[{0, 1}, Length@p]]
subs[in, mask]
{{1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1,
0, 0, 1, 0, 0, 1, 0, 1, 0}, {1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0,
0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0}, {1, 0, 1,
1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0,
1, 0, 0, 1, 0, 1, 0}, {1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0,
0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0}, ...
FromDigits[#, 2] & /@ %
{3108030026, 3108030030, 3108038218, 3108038222, 3108095562,
3108095566, 3108103754, 3108103758, 3141584458, 3141584462,
3141592650, 3141592654, 3141649994, 3141649998, 3141658186,
3141658190}
我有一个非常具体的 Mathematica 问题。我正在尝试围绕某些 'locked' 位生成所有二进制数。我正在使用字符串值列表来表示哪些位被锁定,例如{"U","U,"L","U"},其中 U 是一个 "unlocked" 可变位,L 是一个 "locked" 不可变位。我从一个临时的开始已格式化为先前列表的随机二进制数列表,例如 {0, 1, 1, 0},其中 1是锁定位,我需要找到1[的所有剩余二进制数=22=] 位是不变的。我已经递归地、迭代地解决了这个问题,并且两者结合都没有结果。这是我在大学里做的研究。
我正在构建一个以 10 为基数的二进制数列表。我意识到这段代码是完全错误的。这只是一次尝试。
Do[
If[bits[[pos]] == "U",
AppendTo[returnList, myFunction[bits, temp, pos, returnList]]; ],
{pos, 8, 1}]
myFunction[bits_, bin_, pos_, rList_] :=
Module[{binary = bin, current = Length[bin], returnList = rList},
If[pos == current,
Return[returnList],
If[bits[[current]] == "U",
(*If true*)
If[! MemberQ[returnList, FromDigits[binary, 2]],
(*If true*)
AppendTo[returnList, FromDigits[binary, 2]];
binary[[current]] = Abs[binary[[current]] - 1],
(*If false*)
binary[[current]] = 0;
current = current - 1]; ,
(*If false*)
current = current - 1];
returnList = myFunction[bits, binary, pos, returnList];
Return[returnList]]]
myFunction[bits_] := Module[{length, num, range, all, pattern},
length = Length[bits];
num = 2^length;
range = Range[0, num - 1];
all = PadLeft[IntegerDigits[#, 2], length] & /@ range;
pattern = bits /. {"U" -> _, "L" -> 1};
Cases[all, pattern]]
bits = {"U", "U", "L", "U"};
myFunction[bits]
{{0, 0, 1, 0}, {0, 0, 1, 1}, {0, 1, 1, 0}, {0, 1, 1, 1}, {1, 0, 1, 0}, {1, 0, 1, 1}, {1, 1, 1, 0}, {1, 1, 1, 1}}
您可以使用 Tuples
和 Fold
只生成您感兴趣的位集。
bits = {"U", "U", "L", "U"};
Fold[
Function[{running, next},
Insert[running, 1, next]], #, Position[bits, "L"]] & /@ Tuples[{0, 1}, Count["U"]@bits]
(*
{{0, 0, 1, 0}, {0, 0, 1, 1}, {0, 1, 1, 0}, {0, 1, 1, 1},
{1, 0, 1, 0}, {1, 0, 1, 1}, {1, 1, 1, 0}, {1, 1, 1, 1}}
*)
希望对您有所帮助。
in = IntegerDigits[Round[ Pi 10^9 ], 2];
mask = RandomSample[ConstantArray["L", 28]~Join~ConstantArray["U", 4],32];
subs[in_, mask_] := Module[ {p = Position[mask, "U"]} ,
ReplacePart[in, Rule @@@ Transpose[{p, #}]] & /@
Tuples[{0, 1}, Length@p]]
subs[in, mask]
{{1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0}, {1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0}, {1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0}, {1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0}, ...
FromDigits[#, 2] & /@ %
{3108030026, 3108030030, 3108038218, 3108038222, 3108095562, 3108095566, 3108103754, 3108103758, 3141584458, 3141584462, 3141592650, 3141592654, 3141649994, 3141649998, 3141658186, 3141658190}