MATLAB: Extracting subsequences of binary string

Statistics and Machine Learning Toolboxsubsequences

as would be the code for the following string have the next subsequences ?
 STRING
1(1), 0(2), 1(3), 1(4), 0(5), 0(6), 1(7), 0(8), 0(9), 1(10), 1(11), 1(12), 1(13), 0(14), 0(15), 0(16), 1(17), 1(18), 1(19), 0(20)
SUBSEQUENCES
01: 1(01), 0(02), 1(03), 1(04) -> [1,0,1,1],
02: 1(01), 1(03), 0(05), 1(07) -> [1,1,0,1],
03: 1(01), 1(04), 1(07), 1(10) -> [1,1,1,1],
04: 1(01), 0(05), 0(09), 1(13) -> [1,0,0,1],
05: 1(01), 0(06), 1(11), 0(16) -> [1,0,1,0],
06: 1(01), 1(07), 1(13), 1(19) -> [1,1,1,1],
07: 0(02), 1(03), 1(04), 0(05) -> [0,1,1,0],
08: 0(02), 1(04), 0(06), 0(08) -> [0,1,0,0],
09: 0(02), 0(05), 0(08), 1(11) -> [0,0,0,1],
10: 0(02), 0(06), 1(10), 0(14) -> [0,0,1,0],
11: 0(02), 1(07), 1(12), 1(17) -> [0,1,1,1],
12: 0(02), 0(08), 0(14), 0(20) -> [0,0,0,0],
13: 1(03), 1(04), 0(05), 0(06) -> [1,1,0,0],
14: 1(03), 0(05), 1(07), 0(09) -> [1,0,1,0],
15: 1(03), 0(06), 0(09), 1(12) -> [1,0,0,1],
16: 1(03), 1(07), 1(11), 0(15) -> [1,1,1,0],
17: 1(03), 0(08), 1(13), 1(18) -> [1,0,1,1],
18: 1(04), 0(05), 0(06), 1(07) -> [1,0,0,1],
19: 1(04), 0(06), 0(08), 1(10) -> [1,0,0,1],
20: 1(04), 1(07), 1(10), 1(13) -> [1,1,1,1],
21: 1(04), 0(08), 1(12), 0(16) -> [1,0,1,0],
22: 1(04), 0(09), 0(14), 1(19) -> [1,0,0,1],
23: 0(05), 0(06), 1(07), 0(08) -> [0,0,1,0],
24: 0(05), 1(07), 0(09), 1(11) -> [0,1,0,1],
25: 0(05), 0(08), 1(11), 0(14) -> [0,0,1,0],
26: 0(05), 0(09), 1(13), 1(17) -> [0,0,1,1],
27: 0(05), 1(10), 0(15), 0(20) -> [0,1,0,0],
28: 0(06), 1(07), 0(08), 0(09) -> [0,1,0,0],
29: 0(06), 0(08), 1(10), 1(12) -> [0,0,1,1],
30: 0(06), 0(09), 1(12), 0(15) -> [0,0,1,0],
31: 0(06), 1(10), 0(14), 1(18) -> [0,1,0,1],
32: 1(07), 0(08), 0(09), 1(10) -> [1,0,0,1],
33: 1(07), 0(09), 1(11), 1(13) -> [1,0,1,1],
34: 1(07), 1(10), 1(13), 0(16) -> [1,1,1,0],
35: 1(07), 1(11), 0(15), 1(19) -> [1,1,0,1],
36: 0(08), 0(09), 1(10), 1(11) -> [0,0,1,1],
37: 0(08), 1(10), 1(12), 0(14) -> [0,1,1,0],
38: 0(08), 1(11), 0(14), 1(17) -> [0,1,0,1],
39: 0(08), 1(12), 0(16), 0(20) -> [0,1,0,0],
40: 0(09), 1(10), 1(11), 1(12) -> [0,1,1,1],
41: 0(09), 1(11), 1(13), 0(15) -> [0,1,1,0],
42: 0(09), 1(12), 0(15), 1(18) -> [0,1,0,1],
43: 1(10), 1(11), 1(12), 1(13) -> [1,1,1,1],
44: 1(10), 1(12), 0(14), 0(16) -> [1,1,0,0],
45: 1(10), 1(13), 0(16), 1(19) -> [1,1,0,1],
46: 1(11), 1(12), 1(13), 0(14) -> [1,1,1,0],
47: 1(11), 1(13), 0(15), 1(17) -> [1,1,0,1],
48: 1(11), 0(14), 1(17), 0(20) -> [1,0,1,0],
49: 1(12), 1(13), 0(14), 0(15) -> [1,1,0,0],
50: 1(12), 0(14), 0(16), 1(18) -> [1,0,0,1],
51: 1(13), 0(14), 0(15), 0(16) -> [1,0,0,0],
52: 1(13), 0(15), 1(17), 1(19) -> [1,0,1,1],
53: 0(14), 0(15), 0(16), 1(17) -> [0,0,0,1],
54: 0(14), 0(16), 1(18), 0(20) -> [0,0,1,0],
55: 0(15), 0(16), 1(17), 1(18) -> [0,0,1,1],
56: 0(16), 1(17), 1(18), 1(19) -> [0,1,1,1],
57: 1(17), 1(18), 1(19), 0(20) -> [1,1,1,0],

Best Answer

N = 20;
n = 4;
A = hankel(1:N-n+1,N-n+1:N);
k = 0:n-1;
idx = [];
for ii = 1:size(A,1)
    p = A(ii,:);
    while p(end,end) + k(end) <= N
        p = [p;p(end,:)+k];
    end
    idx=[idx;p];
end
or
N = 20;
n = 4;
A = hankel(1:N-n+1,N-n+1:N);
k = 0:n-1;
c = ceil((N - A(:,end) + 1)/k(end));
i2 = cumsum(c);
i1 = i2 - c + 1;
idx = zeros(i2(end),n);
for jj = 1:N-n+1
    idx(i1(jj):i2(jj),:) = bsxfun(@plus,A(jj,:),(0:c(jj)-1)'*k);
end
ADD
s = [1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0];
[j1,j2,j2] = unique(s(idx),'rows')
out = [j1, histc(j2,1:max(j2))/i2(end)]; % This row corrected
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