I have an array: A B AB BAB ABBAB BABABBAB ... The number of each term of the array is base on the Fibonacci number.Put the n-th string and the n+1-th string together, then producing the n+2-th string,eg.BABABBAB = BAB + ABBAB.
Then is the 10^16-th letter is A or B?
Let f[n] be the nth fibonacci number.
Assume we want to find the value at position x in the string obtained by concatenating f[1], f[2], ..., f[n].
Find the lowest k such that f[1] + f[2] + ... + f[k] >= x. So position x belongs to word that has f[k] characters (at least in the concatenation it does. But since all words are made up from a and b, we'll try to reduce the problem to just those two).
To find the position corresponding to x in the term f[k], subtract f[1] + ... + f[k - 1] from x.
if k = 1 print a, if k = 2 print b, else go to step 4.
if f[k - 2] < x, then the position we're looking for is in the word corresponding to (with length) f[k - 1]. Subtract 1 from k and f[k - 2] from x and go to step 3.
Else, the position we're looking for is in the word corresponding to f[k - 2]. Subtract 2 from k, leave x unchanged and go to step 3.
This doesn't require generating the actual words, just their lengths, which are the basic fibonacci numbers.
Worked example
- note that I'm only using the actual words for illustration purposes, they are not needed.
n f[n] corresponding word
1 1 a
2 1 b
3 2 ab
4 3 bab
5 5 abbab
6 8 bababbab
Concatenating all these we get: ababbababbabbababbab. Let's ask ourselves what's at position 10 (it's b).
1. f[1] + f[2] + f[3] + f[4] + f[5] >= 10, so k = 5
2. x = 10, f[1] + f[2] + f[3] + f[4] = 7, so subtract 7 from x, giving x = 3
3'. k isn't 1 or 2, skip this.
4'. f[k - 2 = 3] = 2 < x. k = 4, x = 1
3''. still ignoring this.
4'' f[k - 2 = 2] = 1 >= x. k = 2, x = 1
3'''. k = 2, print 'b'.
please don't take this answer too serious:
i never was good and math and this sounds like this task should be too heavy to calculate without a freaky algorithm, so my solution would be to simply have a guess. to choose between A and B, i would write a very simple programm like this in php to print out the first 15 or 20 "lines":
<?php
$var1 = "B";
$var2 = "A";
for($i=3;$i<=15;$i++){
$tmp = $var2;
$var2 = $var1;
$var1 = $tmp.$var1;
echo $i." ".$var1."<br>";
}
echo strlen(implode('',explode('B',$var1)));
echo "<br>";
echo strlen(implode('',explode('A',$var1)));
?>
the result shows there are always less As (38%) than Bs (62%) - so the chance to be right when guessing B aren't that bad.
Related
can comeone help to understand in MAtlab this:
k=2
n = (0:-1:-4)+k
the result; 2 1 0 -1 -2
how it works?
You are dealing with a colon operator and a vectorized sum at the same time. Let's split the problem into smaller, stand-alone problems:
In Matlab, if you add or subtract between a scalar value to a matrix, the arithmetic operation is performed on all the elements of the matrix, in a vectorized way. Example:
A = [1 2; 3 4]; % 2-by-2 matrix
S1 = A + 2 % output: S1 = [3 4; 5 6]
B = [1 2 3 4] % 1-by-5 matrix, also called column vector
S2 = B - 5 % output: S2 = [3 4 5 6]
The column operator in Matlab can be used in many situation: indexing, for iterations and vector creation. In your case, its purpose is the third one and it's syntax is START(:STEP):END. The default STEP, if not specified, is 1. The START and END parameters are never exceeded. Example:
A = 1:5 % output: A = [1 2 3 4 5]
B = -2.5:2.5:6 % output: B = [-2.5 0 2.5 5]
C = 1:-1:-5 % output: C = [1 0 -1 -2 -3 -4 -5]
D = -4:-2:0 % output: D = []
In all the programming languages, an operator precedence criterion is defined so that a one-liner calculation that uses multiple operators is atomized into smaller calculations that respect the given priority, unless parentheses are used to redefine the default criterion... just like in common maths. Example
A = 2 * 5 + 3 % output: A = 13
B = 2 * (5 + 3) % output: B = 16
Let's put all this together to provide you an explaination:
n = (0:-1:-4) + k
% vector creation has parentheses, so it's executed first
% then, the addition is executed on the result of the first operation
Let's subdivide the calculation into intermediate steps:
n_1 = 0:-1:-4 % output: n_1 = [0 -1 -2 -3 -4]
n_2 = n_1 + k % output: n_2 = [2 1 0 -1 -2]
n = n_2
Want to see what happens without parentheses?
n = 0:-1:-4+k % output: n = [0 -1 -2]
Why? Because the addition has priority over the colon operator. It's like writing n = 0:-1:(-4+k) and adding k to the END parameter of the colon operator. Let's subdivide the calculation into intermediate steps:
n_1 = -4 + k % output: n_1 = -2
n_2 = 0:-1:n_1 % output: n_2 = [0 -1 -2]
n = n_2
Basic Matlab syntax, you're dealing with range operators. There are two patterns:
[start]:[end]
and
[start]:[step]:[end]
Patterns like this result in arrays / vectors / "1D matrices".
In your example, you will get a vector first, stepping through the numbers 0 to -4 (step == -1). Then, you are adding k == 2 to all numbers in this vector.
octave:1> k = 2
k = 2
octave:2> n = (0:-1:-4)+k
n =
2 1 0 -1 -2
octave:3> 0:-1:-4
ans =
0 -1 -2 -3 -4
The parenthesizes expression determines an array. The the first number there is the first element, the second is the step and the last one is the last element. So the parenthesizes returns 0 -1 -2 -3 -4. Next we add k=2 to each element that results in 2 1 0 -1 -2
I have two arrays:
array1: [0 1 2 3]
array2: [0 1 2 3 4 5 6 7]
array3: [0 1 2 3 4 5]
I want to find combination from each array like 1 element from each array such that the sum >= m eg.3 and sum <=r eg. 6
Example:
array1 array2 array3 elements
1 1 1 (sum is 3)
1 5 0 (sum is 6)
1 6 1 (wrong result sum is greater than 6)
Thanks in advance.
I don't understand how to solve this problem . A pseudocode will be very helpful.
As I have mentioned in my comment; compared to the brute force, in JS by a dynamical programming approach you may get the results much faster, like in total around 10ms for your test case, as follows;
function getCombos(arr,n,m){
function getBinaryMap(d){
return Array(a.length).fill()
.reduce((p,c,i) => p.concat(d >> i & 1),[]);
}
var a = arr.reduce((p,c) => p.concat(c.filter(e => e !== 0 && e <= m)),[]);
t = [0],
r = [];
a.forEach(e => t = t.concat(t.map((f,i) => (f + e >= n && f + e <= m && r.push(getBinaryMap(t.length + i)), f + e))));
return r.map(e => e.reduce((p,c,i) => c ? p.concat(a[i]) : p,[]));;
}
var arrays = [[0, 1, 2, 3],[0, 1, 2, 3, 4, 5, 6, 7],[0, 1, 2, 3, 4, 5]];
result = getCombos(arrays,3,6);
console.log(JSON.stringify(result));
Let there be N=3 sorted arrays A(size = p), B(size = q) and C(size=r).
Choose a number say X and try finding a sum of N=3 numbers x1 + x2 + x3
such that X = x1 + x2 + x3. Finding x1, x2 and x3 can be done using dynamic programming.
Now,
Consider the sequence ABC and use binary search to find x1 in array A, x2 in array B and x3 in array C.
Consider the sequence ACB and use binary search to find x1 in array A, x2 in array C and x3 in array B.
Consider the sequence BAC and use binary search to find x1 in array B, x2 in array A and x3 in array C.
You will find that there are N! number of such sequences.
You will be required to repeat the above process for all X such that [m]<=[X]<=[r].
Total Complexity would be (r - m + 1) * (N!) * ((log p) + (log q) + (log r)).
Total Complexity does not include the dynamic programming part.
I've just found out how to solve this in O(n^2 log n) time (assuming each array has the same length):
for each A[i]:
for each B[j]:
if A[i] + B[j] + C.binarySearch(S - A[i] - B[j]) == S:
return (i, j, k)
Is there any way to solve this in O(n^2) time or to improve the above algorithm?
The algorithm you have ain't bad. Relative to n^2, log(n) grows so slowly that it can practically be considered a constant. For example, for n = 1000000, n^2 = 1000000000000 and log(n) = 20. Once n becomes large enough for log(n) to have any significant influence, n^2 will already be so big that the result cannot be computed anyway.
A solution, inspired by #YvesDaoust, though I'm not sure if it's exactly the same:
For every A[i], calculate the remainder R = S - A[i] which should be a combination of some B[j] and C[k];
Let j = 0 and k = |C|-1 (the last index in C);
If B[j] + C[k] < R, increase j;
If B[j] + C[k] > R, decrease k;
Repeat the two previous steps until B[j] + C[k] = R or j >= |B| or k < 0
I suggest not to complicate the algorithm too much with micro-optimizations. For any reasonably small set of numbers and it will be fast enough. If the arrays become too large for this approach, your problem would make a good candidate for Machine Learning approaches such as Hill Climbing.
if the arrays are of non-negatives
* you can trim all 3 arrays to at S => A[n] > S
* similarly, dont bother checking array C if A[aIdx] + B[bIdx] > S
prepare:
sort each array ascending +O(N.log(n))
implement binary search on each array ?O(log(N))
compute:
i=bin_search(smallest i that A[i]+B[0]+C[0]>=S); for (;i<Na;i++) { if (A[i]+B[0]+C[0]>S) break;
j=bin_search(smallest j that A[i]+B[j]+C[0]>=S); for (;j<Nb;j++) { if (A[i]+B[j]+C[0]>S) break;
ss=S-A[i]-B[j];
if (ss<0) break;
k=bin_search(ss);
if (k found) return; // found solution is: i,j,k
}
}
if I see it right and: N=max(Na,Nb,Nc), M=max(valid intervals A,B,C) ... M<=N
it is (3*N.log(N)+log(N)+M*log(N)*M*log(N)) -> O((M^2)*log(N))
the j binary search can be called just once and then iterate +1 if needed
the complexity is the same but the N has changed
for average conditions is this much much faster because M<<N
The O(N²) solution is very simple.
First consider the case of two arrays, finding A[i] + B[j] = S'.
This can be rewritten as A[i] = S' - B[j] = B'[j]: you need to find equal values in two sorted arrays. This is readily done in linear time with a merging process. (You can explicitly compute the array B' but this is unnecessary, just do it on the fly: instead of fetching B'[j], get S' - B[NB-1-j]).
Having established this procedure, it suffices to use it for all elements of C, in search of S - C[k].
Here is Python code that does that and reports all solutions. (It has been rewritten to be compact and symmetric.)
for k in range(NC):
# Find S - C[k] in top-to-tail merged A and B
i, j= 0, NB - 1
while i < NA and 0 <= j:
if A[i] + B[j] + C[k] < S:
# Move forward A
i+= 1
elif A[i] + B[j] + C[k] > S:
# Move back B
j-= 1
else:
# Found
print A[i] + B[j] + C[k], "=", A[i], "+", B[j], "+", C[k]
i+= 1; j-= 1
Execution with
A= [1, 2, 3, 4, 5, 6, 7]; NA= len(A)
B= [2, 3, 5, 7, 11]; NB= len(B)
C= [1, 1, 2, 3, 5, 7]; NC= len(C)
S= 15
gives
15 = 3 + 11 + 1
15 = 7 + 7 + 1
15 = 3 + 11 + 1
15 = 7 + 7 + 1
15 = 2 + 11 + 2
15 = 6 + 7 + 2
15 = 1 + 11 + 3
15 = 5 + 7 + 3
15 = 7 + 5 + 3
15 = 3 + 7 + 5
15 = 5 + 5 + 5
15 = 7 + 3 + 5
15 = 1 + 7 + 7
15 = 3 + 5 + 7
15 = 5 + 3 + 7
15 = 6 + 2 + 7
I am trying to create all possible 1xM vectors (word) from a 1xN vector (alphabet) in MATLAB. N is > M. For example, I want to create all possible 2x1 "words" from a 4x1 "alphabet" alphabet = [1 2 3 4];
I expect a result like:
[1 1]
[1 2]
[1 3]
[1 4]
[2 1]
[2 2]
...
I want to make M an input to my routine and I do not know it beforehand. Otherwise, I could easily do this using nested for-loops. Anyway to do this?
Try
[d1 d2] = ndgrid(alphabet);
[d2(:) d1(:)]
To parameterize on M:
d = cell(M, 1);
[d{:}] = ndgrid(alphabet);
for i = 1:M
d{i} = d{i}(:);
end
[d{end:-1:1}]
In general, and in languages that don't have ndgrid in their library, the way to parameterize for-loop nesting is using recursion.
[result] = function cartesian(alphabet, M)
if M <= 1
result = alphabet;
else
recursed = cartesian(alphabet, M-1)
N = size(recursed,1);
result = zeros(M, N * numel(alphabet));
for i=1:numel(alphabet)
result(1,1+(i-1)*N:i*N) = alphabet(i);
result(2:M,1+(i-1)*N:i*N) = recursed; % in MATLAB, this line can be vectorized with repmat... but in MATLAB you'd use ndgrid anyway
end
end
end
To get all k-letter combinations from an arbitrary alphabet, use
n = length(alphabet);
aux = dec2base(0:n^k-1,n)
aux2 = aux-'A';
ind = aux2<0;
aux2(ind) = aux(ind)-'0'
aux2(~ind) = aux2(~ind)+10;
words = alphabet(aux2+1)
The alphabet may consist of up to 36 elements (as per dec2base). Those elements may be numbers or characters.
How this works:
The numbers 0, 1, ... , n^k-1 when expressed in base n give all groups of k numbers taken from 0,...,n-1. dec2base does the conversion to base n, but gives the result in form of strings, so need to convert to the corresponding number (that's part with aux and aux2). We then add 1 to make the numbers 1,..., n. Finally, we index alphabet with that to use the real letters of numbers of the alphabet.
Example with letters:
>> alphabet = 'abc';
>> k = 2;
>> words
words =
aa
ab
ac
ba
bb
bc
ca
cb
cc
Example with numbers:
>> alphabet = [1 3 5 7];
>> k = 2;
>> words
words =
1 1
1 3
1 5
1 7
3 1
3 3
3 5
3 7
5 1
5 3
5 5
5 7
7 1
7 3
7 5
7 7
use ndgrid function in Matlab
[a,b] = ndgrid(alphabet)
I have the following problem. Suppose I have to arrays X and Y whose entries consist of integers in the range 0 through 9 like
X = [1 2 3]
Y = [7 0 9]
I want to think of these arrays as being the digits of base-10 numbers, so X represents the number 123 and Y represents the number 709. I want to write a program which outputs the digits of the sum of these integers. So in the example given, it should output the array
Z = [8 3 2]
since 123 + 709 = 832. For the sake of this question it suffices to assume that X and Y have the same length, i.e., that the numbers they represent have the same number of digits. While I am doing this, I also want to keep track of carries which were performed during the addition. So in the example, I'd also want to output
C = [0 0 1]
which represents that I had to carry a 1 when I added the digits 9 + 3 = 12, but that the additions of the digits in other positions went through without carries. So my main question is
Does anyone know of a simple way to achieve this goal using MATLAB?
So far, what I've come up with is the following code which is given the two numbers as X and Y
clear all; clc;
C = zeros(1, length(X));
for j=length(X):-1:1
if X(j) + Y(j) > 9
C(j) = 1;
end
Z(j) = mod(X(j) + Y(j), 10);
if j < length(X)
if Z(j) + C(j+1) < 9
Z(j) = Z(j) + C(j+1);
else
Z(j) = mod(Z(j) + C(j+1), 10);
C(j) = 1;
end
end
end
if C(1) == 1
Z = [1 Z];
end
Z
C
The problem is that the code only works sometimes. For example, it has no problem working through the example I gave above, that 123 + 709 = 832 with a "carry array" of [0 0 1]. But the input X = 52514 and Y = 41525 does not yield the correct results. So my follow up question is
Does anyone see the bug in the code above/how can I fix it?
You need to change the line
if Z(j) + C(j+1) < 9
to
if Z(j) + C(j+1) <= 9
Then it should work.
You can take advantage of dec2base, str2num and num2str functions:
function StackOverflow
x = [1 2 3];
y = [4 5 6];
numX = DigitsToNum(x);
numY = DigitsToNum(y);
digits = NumToDigits(numX+numY);
disp(digits);
end
function digits = NumToDigits(x)
digits = double( dec2base(x,10) - '0');
end
function num = DigitsToNum(x)
st = strrep(num2str(x),' ','');
num = str2num(st); %#ok<ST2NM>
end
Another way to implement the two functions above:
function digits = NumToDigits(x)
digits = [];
while x > 0
digits(end+1) = mod(x,10); %#ok<AGROW>
x = floor(x/10);
end
end
function num = DigitsToNum(x)
baseMult = power(10,0: (numel(x)-1));
num = sum( baseMult.*x );
end