我很难找到解决这个问题的方法。
输入输出序列如下：
**input1 :** aaagctgctagag
**output1 :** a3gct2ag2
**input2 :** aaaaaaagctaagctaag
**output2 :** a6agcta2ag
输入 nsequence 可以是 10^6 个字符，并且将考虑最大的连续模式。
例如，对于 input2，“agctaagcta”输出将不是“agcta2gcta”，而是“agcta2”。
任何帮助表示赞赏。
问问题
5224 次
3 回答
10
算法说明：
- 有一个带有符号 s(1), s(2),…, s(N) 的序列 S。
- 令 B(i) 为具有元素 s(1)、s(2)、…、s(i) 的最佳压缩序列。
- 因此，例如，B(3) 将是 s(1)、s(2)、s(3) 的最佳压缩序列。
- 我们想知道的是B(N)。
为了找到它，我们将通过归纳进行。我们要计算 B(i+1)，知道 B(i), B(i-1), B(i-2), ..., B(1), B(0)，其中 B(0) 为空序列，并且 B(1) = s(1)。同时，这构成了解决方案是最优的证明。;)
为了计算 B(i+1)，我们将在候选中选择最佳序列：
最后一个块有一个元素的候选序列：
B(i )s(i+1)1 B(i-1)s(i+1)2 ; 仅当 s(i) = s(i+1) B(i-2)s(i+1)3 时；仅当 s(i-1) = s(i) 且 s(i) = s(i+1) ... B(1)s(i+1)[i-1] 时；仅当 s(2)=s(3) and s(3)=s(4) and … and s(i) = s(i+1) B(0)s(i+1)i = s(i +1)我; 仅当 s(1)=s(2) and s(2)=s(3) and … and s(i) = s(i+1)
最后一个块有 2 个元素的候选序列：
B(i-1)s(i)s(i+1)1 B(i-3)s(i)s(i+1)2 ; 仅当 s(i-2)s(i-1)=s(i)s(i+1) B(i-5)s(i)s(i+1)3 时；仅当 s(i-4)s(i-3)=s(i-2)s(i-1) 和 s(i-2)s(i-1)=s(i)s(i+1 ) …</p>
最后一个块有 3 个元素的候选序列：
…</p>
最后一个块有 4 个元素的候选序列：
…</p>
…</p>
最后一个块有 n+1 个元素的候选序列：
s(1)s(2)s(3)………s(i+1)
对于每一种可能性，当序列块不再重复时，算法就会停止。就是这样。
该算法在伪 c 代码中将是这样的：
B(0) = “”
for (i=1; i<=N; i++) {
// Calculate all the candidates for B(i)
BestCandidate=null
for (j=1; j<=i; j++) {
Calculate all the candidates of length (i)
r=1;
do {
Candidadte = B([i-j]*r-1) s(i-j+1)…s(i-1)s(i) r
If ( (BestCandidate==null)
|| (Candidate is shorter that BestCandidate))
{
BestCandidate=Candidate.
}
r++;
} while ( ([i-j]*r <= i)
&&(s(i-j*r+1) s(i-j*r+2)…s(i-j*r+j) == s(i-j+1) s(i-j+2)…s(i-j+j))
}
B(i)=BestCandidate
}
希望这可以提供更多帮助。
下面给出了执行所需任务的完整 C 程序。它在 O(n^2) 中运行。中心部分只有 30 行代码。
编辑我已经对代码进行了一些重组，更改了变量的名称并添加了一些注释，以便更具可读性。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
// This struct represents a compressed segment like atg4, g3, agc1
struct Segment {
char *elements;
int nElements;
int count;
};
// As an example, for the segment agagt3 elements would be:
// {
// elements: "agagt",
// nElements: 5,
// count: 3
// }
struct Sequence {
struct Segment lastSegment;
struct Sequence *prev; // Points to a sequence without the last segment or NULL if it is the first segment
int totalLen; // Total length of the compressed sequence.
};
// as an example, for the sequence agt32ta5, the representation will be:
// {
// lastSegment:{"ta" , 2 , 5},
// prev: @A,
// totalLen: 8
// }
// and A will be
// {
// lastSegment{ "agt", 3, 32},
// prev: NULL,
// totalLen: 5
// }
// This function converts a sequence to a string.
// You have to free the string after using it.
// The strategy is to construct the string from right to left.
char *sequence2string(struct Sequence *S) {
char *Res=malloc(S->totalLen + 1);
char *digits="0123456789";
int p= S->totalLen;
Res[p]=0;
while (S!=NULL) {
// first we insert the count of the last element.
// We do digit by digit starting with the units.
int C = S->lastSegment.count;
while (C) {
p--;
Res[p] = digits[ C % 10 ];
C /= 10;
}
p -= S->lastSegment.nElements;
strncpy(Res + p , S->lastSegment.elements, S->lastSegment.nElements);
S = S ->prev;
}
return Res;
}
// Compresses a dna sequence.
// Returns a string with the in sequence compressed.
// The returned string must be freed after using it.
char *dnaCompress(char *in) {
int i,j;
int N = strlen(in);; // Number of elements of a in sequence.
// B is an array of N+1 sequences where B(i) is the best compressed sequence sequence of the first i characters.
// What we want to return is B[N];
struct Sequence *B;
B = malloc((N+1) * sizeof (struct Sequence));
// We first do an initialization for i=0
B[0].lastSegment.elements="";
B[0].lastSegment.nElements=0;
B[0].lastSegment.count=0;
B[0].prev = NULL;
B[0].totalLen=0;
// and set totalLen of all the sequences to a very HIGH VALUE in this case N*2 will be enougth, We will try different sequences and keep the minimum one.
for (i=1; i<=N; i++) B[i].totalLen = INT_MAX; // A very high value
for (i=1; i<=N; i++) {
// at this point we want to calculate B[i] and we know B[i-1], B[i-2], .... ,B[0]
for (j=1; j<=i; j++) {
// Here we will check all the candidates where the last segment has j elements
int r=1; // number of times the last segment is repeated
int rNDigits=1; // Number of digits of r
int rNDigitsBound=10; // We will increment r, so this value is when r will have an extra digit.
// when r = 0,1,...,9 => rNDigitsBound = 10
// when r = 10,11,...,99 => rNDigitsBound = 100
// when r = 100,101,.,999 => rNDigitsBound = 1000 and so on.
do {
// Here we analitze a candidate B(i).
// where the las segment has j elements repeated r times.
int CandidateLen = B[i-j*r].totalLen + j + rNDigits;
if (CandidateLen < B[i].totalLen) {
B[i].lastSegment.elements = in + i - j*r;
B[i].lastSegment.nElements = j;
B[i].lastSegment.count = r;
B[i].prev = &(B[i-j*r]);
B[i].totalLen = CandidateLen;
}
r++;
if (r == rNDigitsBound ) {
rNDigits++;
rNDigitsBound *= 10;
}
} while ( (i - j*r >= 0)
&& (strncmp(in + i -j, in + i - j*r, j)==0));
}
}
char *Res=sequence2string(&(B[N]));
free(B);
return Res;
}
int main(int argc, char** argv) {
char *compressedDNA=dnaCompress(argv[1]);
puts(compressedDNA);
free(compressedDNA);
return 0;
}
于 2013-06-03T18:11:53.400 回答
2
忘记乌康宁吧。动态规划就是这样。带3维表：
- 序列位置
- 子序列大小
- 段数
术语：例如a = "aaagctgctagag"，序列位置坐标将从 1 运行到 13。在序列位置 3（字母“g”），子序列大小为 4，子序列将是“gctg”。明白了吗？至于段数，则将 a 表示为“aaagctgctagag1”由 1 个段（序列本身）组成。将其表示为“a3gct2ag2”由 3 个部分组成。“aaagctgct1ag2”由 2 个段组成。“a2a1ctg2ag2”将由 4 个段组成。明白了吗？现在，有了这个，你开始填充一个 13 x 13 x 13 的 3 维数组，所以你的时间和内存复杂度似乎在n ** 3为了这。你确定你可以处理百万碱基序列吗？我认为贪婪的方法会更好，因为大的 DNA 序列不太可能完全重复。而且，我建议您将作业范围扩大到近似匹配，并且可以直接在期刊上发表。
无论如何，您将开始填充从某个位置（维度 1）开始的压缩子序列表，其长度等于维度 2 坐标，最多具有维度 3 段。因此，您首先填充第一行，表示长度为 1 的子序列的压缩，最多包含 1 个段：
a a a g c t g c t a g a g
1(a1) 1(a1) 1(a1) 1(g1) 1(c1) 1(t1) 1(g1) 1(c1) 1(t1) 1(a1) 1(g1) 1(a1) 1(g1)
数字是字符成本（对于这些简单的 1 字符序列，始终为 1；数字 1 不计入字符成本），并且在括号中，您有压缩（对于这个简单的情况也很简单）。第二行仍然很简单：
2(a2) 2(a2) 2(ag1) 2(gc1) 2(ct1) 2(tg1) 2(gc1) 2(ct1) 2(ta1) 2(ag1) 2(ga1) 2(ag1)
只有一种方法可以将 2 个字符的序列分解为 2 个子序列——1 个字符 + 1 个字符。如果它们相同，则结果为a + a = a2。如果它们不同，例如a + g，那么，因为只有 1 段序列是可接受的，所以结果不能是a1g1，而必须是ag1。第三行最终会更有趣：
2(a3) 2(aag1) 3(agc1) 3(gct1) 3(ctg1) 3(tgc1) 3(gct1) 3(cta1) 3(tag1) 3(aga1) 3(gag1)
在这里，您始终可以在 2 种组合压缩字符串的方式之间进行选择。例如，aag可以组成为aa + g或a + ag。但同样，我们不能有 2 个段，如aa1g1or a1ag1，所以我们必须满足aag1，除非两个组件包含相同的字符，如aa + a=> a3，字符成本为 2。我们可以继续第 4 行：
4(aaag1) 4(aagc1) 4(agct1) 4(gctg1) 4(ctgc1) 4(tgct1) 4(gcta1) 4(ctag1) 4(taga1) 3(ag2)
在这里，在第一个位置，我们不能使用a3g1，因为在这一层只允许 1 个段。但是在最后一个位置，压缩到字符成本 3 是由ag1 + ag1 = ag2. 这样，一个人可以将整个一级表一直填充到 13 个字符的单个子序列，并且每个子序列将在与其关联的最多 1 个段的一级约束下具有其最佳字符成本和压缩.
然后进入第二层，允许有 2 个段...再次，从下往上，通过比较所有可能的方法来确定在给定级别的段数约束下每个表坐标的最佳成本和压缩使用已经计算的位置组成子序列，直到您完全填满表格并因此计算全局最优值。有一些细节需要解决，但对不起，我不会为你编写代码。
于 2013-06-02T05:26:09.337 回答
2
在尝试了我自己的方式一段时间后，我对 jbaylina 的漂亮算法和 C 实现表示敬意。这是我在 Haskell 中尝试的 jbaylina 算法的版本，下面是我对线性时间算法的尝试的进一步开发，该算法试图以一个接一个的方式压缩包含重复模式的段：
import Data.Map (fromList, insert, size, (!))
compress s = (foldl f (fromList [(0,([],0)),(1,([s!!0],1))]) [1..n - 1]) ! n
where
n = length s
f b i = insert (size b) bestCandidate b where
add (sequence, sLength) (sequence', sLength') =
(sequence ++ sequence', sLength + sLength')
j' = [1..min 100 i]
bestCandidate = foldr combCandidates (b!i `add` ([s!!i,'1'],2)) j'
combCandidates j candidate' =
let nextCandidate' = comb 2 (b!(i - j + 1)
`add` ((take j . drop (i - j + 1) $ s) ++ "1", j + 1))
in if snd nextCandidate' <= snd candidate'
then nextCandidate'
else candidate' where
comb r candidate
| r > uBound = candidate
| not (strcmp r True) = candidate
| snd nextCandidate <= snd candidate = comb (r + 1) nextCandidate
| otherwise = comb (r + 1) candidate
where
uBound = div (i + 1) j
prev = b!(i - r * j + 1)
nextCandidate = prev `add`
((take j . drop (i - j + 1) $ s) ++ show r, j + length (show r))
strcmp 1 _ = True
strcmp num bool
| (take j . drop (i - num * j + 1) $ s)
== (take j . drop (i - (num - 1) * j + 1) $ s) =
strcmp (num - 1) True
| otherwise = False
输出：
*Main> compress "aaagctgctagag"
("a3gct2ag2",9)
*Main> compress "aaabbbaaabbbaaabbbaaabbb"
("aaabbb4",7)
线性时间尝试：
import Data.List (sortBy)
group' xxs sAccum (chr, count)
| null xxs = if null chr
then singles
else if count <= 2
then reverse sAccum ++ multiples ++ "1"
else singles ++ if null chr then [] else chr ++ show count
| [x] == chr = group' xs sAccum (chr,count + 1)
| otherwise = if null chr
then group' xs (sAccum) ([x],1)
else if count <= 2
then group' xs (multiples ++ sAccum) ([x],1)
else singles
++ chr ++ show count ++ group' xs [] ([x],1)
where x:xs = xxs
singles = reverse sAccum ++ (if null sAccum then [] else "1")
multiples = concat (replicate count chr)
sequences ws strIndex maxSeqLen = repeated' where
half = if null . drop (2 * maxSeqLen - 1) $ ws
then div (length ws) 2 else maxSeqLen
repeated' = let (sequence,(sequenceStart, sequenceEnd'),notSinglesFlag) = repeated
in (sequence,(sequenceStart, sequenceEnd'))
repeated = foldr divide ([],(strIndex,strIndex),False) [1..half]
equalChunksOf t a = takeWhile(==t) . map (take a) . iterate (drop a)
divide chunkSize b@(sequence,(sequenceStart, sequenceEnd'),notSinglesFlag) =
let t = take (2*chunkSize) ws
t' = take chunkSize t
in if t' == drop chunkSize t
then let ts = equalChunksOf t' chunkSize ws
lenTs = length ts
sequenceEnd = strIndex + lenTs * chunkSize
newEnd = if sequenceEnd > sequenceEnd'
then sequenceEnd else sequenceEnd'
in if chunkSize > 1
then if length (group' (concat (replicate lenTs t')) [] ([],0)) > length (t' ++ show lenTs)
then (((strIndex,sequenceEnd,chunkSize,lenTs),t'):sequence, (sequenceStart,newEnd),True)
else b
else if notSinglesFlag
then b
else (((strIndex,sequenceEnd,chunkSize,lenTs),t'):sequence, (sequenceStart,newEnd),False)
else b
addOne a b
| null (fst b) = a
| null (fst a) = b
| otherwise =
let (((start,end,patLen,lenS),sequence):rest,(sStart,sEnd)) = a
(((start',end',patLen',lenS'),sequence'):rest',(sStart',sEnd')) = b
in if sStart' < sEnd && sEnd < sEnd'
then let c = ((start,end,patLen,lenS),sequence):rest
d = ((start',end',patLen',lenS'),sequence'):rest'
in (c ++ d, (sStart, sEnd'))
else a
segment xs baseIndex maxSeqLen = segment' xs baseIndex baseIndex where
segment' zzs@(z:zs) strIndex farthest
| null zs = initial
| strIndex >= farthest && strIndex > 0 = ([],(0,0))
| otherwise = addOne initial next
where
next@(s',(start',end')) = segment' zs (strIndex + 1) farthest'
farthest' | null s = farthest
| otherwise = if start /= end && end > farthest then end else farthest
initial@(s,(start,end)) = sequences zzs strIndex maxSeqLen
areExclusive ((a,b,_,_),_) ((a',b',_,_),_) = (a' >= b) || (b' <= a)
combs [] r = [r]
combs (x:xs) r
| null r = combs xs (x:r) ++ if null xs then [] else combs xs r
| otherwise = if areExclusive (head r) x
then combs xs (x:r) ++ combs xs r
else if l' > lowerBound
then combs xs (x: reduced : drop 1 r) ++ combs xs r
else combs xs r
where lowerBound = l + 2 * patLen
((l,u,patLen,lenS),s) = head r
((l',u',patLen',lenS'),s') = x
reduce = takeWhile (>=l') . iterate (\x -> x - patLen) $ u
lenReduced = length reduce
reduced = ((l,u - lenReduced * patLen,patLen,lenS - lenReduced),s)
buildString origStr sequences = buildString' origStr sequences 0 (0,"",0)
where
buildString' origStr sequences index accum@(lenC,cStr,lenOrig)
| null sequences = accum
| l /= index =
buildString' (drop l' origStr) sequences l (lenC + l' + 1, cStr ++ take l' origStr ++ "1", lenOrig + l')
| otherwise =
buildString' (drop u' origStr) rest u (lenC + length s', cStr ++ s', lenOrig + u')
where
l' = l - index
u' = u - l
s' = s ++ show lenS
(((l,u,patLen,lenS),s):rest) = sequences
compress [] _ accum = reverse accum ++ (if null accum then [] else "1")
compress zzs@(z:zs) maxSeqLen accum
| null (fst segment') = compress zs maxSeqLen (z:accum)
| (start,end) == (0,2) && not (null accum) = compress zs maxSeqLen (z:accum)
| otherwise =
reverse accum ++ (if null accum || takeWhile' compressedStr 0 /= 0 then [] else "1")
++ compressedStr
++ compress (drop lengthOriginal zzs) maxSeqLen []
where segment'@(s,(start,end)) = segment zzs 0 maxSeqLen
combinations = combs (fst $ segment') []
takeWhile' xxs count
| null xxs = 0
| x == '1' && null (reads (take 1 xs)::[(Int,String)]) = count
| not (null (reads [x]::[(Int,String)])) = 0
| otherwise = takeWhile' xs (count + 1)
where x:xs = xxs
f (lenC,cStr,lenOrig) (lenC',cStr',lenOrig') =
let g = compare ((fromIntegral lenC + if not (null accum) && takeWhile' cStr 0 == 0 then 1 else 0) / fromIntegral lenOrig)
((fromIntegral lenC' + if not (null accum) && takeWhile' cStr' 0 == 0 then 1 else 0) / fromIntegral lenOrig')
in if g == EQ
then compare (takeWhile' cStr' 0) (takeWhile' cStr 0)
else g
(lenCompressed,compressedStr,lengthOriginal) =
head $ sortBy f (map (buildString (take end zzs)) (map reverse combinations))
输出：
*Main> compress "aaaaaaaaabbbbbbbbbaaaaaaaaabbbbbbbbb" 100 []
"a9b9a9b9"
*Main> compress "aaabbbaaabbbaaabbbaaabbb" 100 []
"aaabbb4"
于 2013-06-02T05:59:48.350 回答