I have a string numbers which contains 13 digits, e.g.:
string numbers = "01234567890123456781";
and I have to multiply them:
0 * 1 * 2 * 3 * ... * 7 * 8 * 1
I've a got a solution for adding the digits together:
0 + 1 + 2 + 3 + ... + 7 + 8 + 1
I convert number string to List<int> numb and then sum items of the list with a help of Linq Sum() method:
string numbers = "01234567890123456781";
List<int> numb = new List<int>();
for (int i = 0; i < numbers.Length; i++)
{
int num = int.Parse(numbers[i].ToString());
numb.Add(num);
}
for (int b = 0; b <numb.Count()-12; b++)
{
var x = numb.Skip(b).Take(13);
int a = x.Sum(); // <- it's easy to sum, but what shall I do to multiply?
Console.WriteLine(a);
}
However, Linq doesn't have any Mult() or alike method, so what can I do to multiply the items together?
Providing that the numbers contains digits [0..9] only, a simple Linq will do:
// since 0 * anything == 0, I've removed all 0's to obtain non-zero result:
// 1 * 2 * 3 * ... * 7 * 8 * 1
string numbers = "123456789123456781";
// long: int is not large enough
long result = numbers
.Select(c => (long) (c - '0'))
.Aggregate((s, a) => s * a);
Outcome:
// 14631321600
Console.Write(result);
Aggregate (custom aggregation) will do if you want to change IEnumerable<T>.Sum() from summation (as in your current code) to multiplication:
...
var x = numb.Skip(b).Take(13); //TODO: get rid of magic number 13
// instead of summation
// int a = x.Sum();
// let's use multiplication as a custom aggregation:
// please, mind overflow: int is not enough when multiplying items
long a = x
.Select(item => (long) item)
.Aggregate((s, item) => s * item);
I have a number. For instance, my number is 19 . Then I want to populate my drop down with range in multiplication of 5. So my dropdownlist will consist of items of:
1-5
6-10
11-15
16-19
I tried modulus and division, however, I can't seems to get the range. Is there a fixed method?
Sample code
List<string> range = new List<string>();
int number = 19;
int numOfOccur = (19/5);
for (int i = 1; i < numOfOccur ; i++)
{
range.Add(i + " - " + (i * 5))
}
Sometime I think that old school code, without fancy linq is a bit more clear
int maximum = 19;
int multiple = 5;
int init = 1;
while (init + multiple <= maximum )
{
string addToDDL = init.ToString() + "-" + (init + multiple - 1).ToString();
Console.WriteLine(addToDDL);
init += multiple;
}
if(init <= maximum)
{
string last = init.ToString() + "-" + maximum.ToString();
Console.WriteLine(last);
}
Linq solution (modern techs allow us to put it consize):
int number = 19;
int div = 5;
List<string> range = Enumerable
.Range(0, number / div + (number % div == 0 ? 0 : 1))
.Select(i => $"{i * div + 1} - {Math.Min((i + 1) * div, number)}")
.ToList();
Test
Console.Write(string.Join(Environment.NewLine, range));
Returns
1 - 5
6 - 10
11 - 15
16 - 19
When using modulo arithmetics, do not forget about remainders: you have an error in int numOfOccur = (19/5); line. It should be
int numOfOccur = 19 / 5 + (19 % 5 == 0 ? 0 : 1);
for the last incomplete 16 - 19 range to be proceeded.
Add this package to your project : https://www.nuget.org/packages/System.Interactive/
Then you can do this:
IEnumerable<IList<int>> buffers2 = Enumerable.Range(1, 19).Buffer(5);
IList<int>[] result2 = buffers2.ToArray();
// { { 1, 2, 3, 4, 5 }, { 6, 7, 8, 9, 10 }, ...
Don't forget to add System.Interactive namespace to your using block.
Summary:
I'm beginning with some details about alignment algorithms, and at the end, I ask my question. If you know about alignment algorithm pass the beginning.
Consider we have two strings like:
ACCGAATCGA
ACCGGTATTAAC
There is some algorithms like: Smith-Waterman Or Needleman–Wunsch, that align this two sequence and create a matrix. take a look at the result in the following section:
Smith-Waterman Matrix
§ § A C C G A A T C G A
§ 0 0 0 0 0 0 0 0 0 0 0
A 0 4 0 0 0 4 4 0 0 0 4
C 0 0 13 9 4 0 4 3 9 4 0
C 0 0 9 22 17 12 7 3 12 7 4
G 0 0 4 17 28 23 18 13 8 18 13
G 0 0 0 12 23 28 23 18 13 14 18
T 0 0 0 7 18 23 28 28 23 18 14
A 0 4 0 2 13 22 27 28 28 23 22
T 0 0 3 0 8 17 22 32 27 26 23
T 0 0 0 2 3 12 17 27 31 26 26
A 0 4 0 0 2 7 16 22 27 31 30
A 0 4 4 0 0 6 11 17 22 27 35
C 0 0 13 13 8 3 6 12 26 22 30
Optimal Alignments
A C C G A - A T C G A
A C C G G A A T T A A
Question:
My question is simple, but maybe the answer is not easy as it looks. I want to use a group of character as a single one like: [A0][C0][A1][B1]. But in these algorithms, we have to use individual characters. How can we achieve that?
P.S. Consider we have this sequence: #read #write #add #write. Then I convert this to something like that: #read to A .... #write to B.... #add to C. Then my sequence become to: ABCB. But I have a lot of different words that start with #. And the ASCII table is not enough to convert all of them. Then I need more characters. the only way is to use something like [A0] ... [Z9] for each word. OR to use numbers.
P.S: some sample code for Smith-Waterman is exist in this link
P.S: there is another post that want something like that, but what I want is different. In this question, we have a group of character that begins with a [ and ends with ]. And no need to use semantic like ee is equal to i.
I adapted this Python implementation (GPL version 3 licensed) of both the Smith-Waterman and the Needleman-Wunsch algorithms to support sequences with multiple character groups:
#This software is a free software. Thus, it is licensed under GNU General Public License.
#Python implementation to Smith-Waterman Algorithm for Homework 1 of Bioinformatics class.
#Forrest Bao, Sept. 26 <http://fsbao.net> <forrest.bao aT gmail.com>
# zeros() was origianlly from NumPy.
# This version is implemented by alevchuk 2011-04-10
def zeros(shape):
retval = []
for x in range(shape[0]):
retval.append([])
for y in range(shape[1]):
retval[-1].append(0)
return retval
match_award = 10
mismatch_penalty = -5
gap_penalty = -5 # both for opening and extanding
gap = '----' # should be as long as your group of characters
space = ' ' # should be as long as your group of characters
def match_score(alpha, beta):
if alpha == beta:
return match_award
elif alpha == gap or beta == gap:
return gap_penalty
else:
return mismatch_penalty
def finalize(align1, align2):
align1 = align1[::-1] #reverse sequence 1
align2 = align2[::-1] #reverse sequence 2
i,j = 0,0
#calcuate identity, score and aligned sequeces
symbol = []
found = 0
score = 0
identity = 0
for i in range(0,len(align1)):
# if two AAs are the same, then output the letter
if align1[i] == align2[i]:
symbol.append(align1[i])
identity = identity + 1
score += match_score(align1[i], align2[i])
# if they are not identical and none of them is gap
elif align1[i] != align2[i] and align1[i] != gap and align2[i] != gap:
score += match_score(align1[i], align2[i])
symbol.append(space)
found = 0
#if one of them is a gap, output a space
elif align1[i] == gap or align2[i] == gap:
symbol.append(space)
score += gap_penalty
identity = float(identity) / len(align1) * 100
print 'Identity =', "%3.3f" % identity, 'percent'
print 'Score =', score
print ''.join(align1)
# print ''.join(symbol)
print ''.join(align2)
def needle(seq1, seq2):
m, n = len(seq1), len(seq2) # length of two sequences
# Generate DP table and traceback path pointer matrix
score = zeros((m+1, n+1)) # the DP table
# Calculate DP table
for i in range(0, m + 1):
score[i][0] = gap_penalty * i
for j in range(0, n + 1):
score[0][j] = gap_penalty * j
for i in range(1, m + 1):
for j in range(1, n + 1):
match = score[i - 1][j - 1] + match_score(seq1[i-1], seq2[j-1])
delete = score[i - 1][j] + gap_penalty
insert = score[i][j - 1] + gap_penalty
score[i][j] = max(match, delete, insert)
# Traceback and compute the alignment
align1, align2 = [], []
i,j = m,n # start from the bottom right cell
while i > 0 and j > 0: # end toching the top or the left edge
score_current = score[i][j]
score_diagonal = score[i-1][j-1]
score_up = score[i][j-1]
score_left = score[i-1][j]
if score_current == score_diagonal + match_score(seq1[i-1], seq2[j-1]):
align1.append(seq1[i-1])
align2.append(seq2[j-1])
i -= 1
j -= 1
elif score_current == score_left + gap_penalty:
align1.append(seq1[i-1])
align2.append(gap)
i -= 1
elif score_current == score_up + gap_penalty:
align1.append(gap)
align2.append(seq2[j-1])
j -= 1
# Finish tracing up to the top left cell
while i > 0:
align1.append(seq1[i-1])
align2.append(gap)
i -= 1
while j > 0:
align1.append(gap)
align2.append(seq2[j-1])
j -= 1
finalize(align1, align2)
def water(seq1, seq2):
m, n = len(seq1), len(seq2) # length of two sequences
# Generate DP table and traceback path pointer matrix
score = zeros((m+1, n+1)) # the DP table
pointer = zeros((m+1, n+1)) # to store the traceback path
max_score = 0 # initial maximum score in DP table
# Calculate DP table and mark pointers
for i in range(1, m + 1):
for j in range(1, n + 1):
score_diagonal = score[i-1][j-1] + match_score(seq1[i-1], seq2[j-1])
score_up = score[i][j-1] + gap_penalty
score_left = score[i-1][j] + gap_penalty
score[i][j] = max(0,score_left, score_up, score_diagonal)
if score[i][j] == 0:
pointer[i][j] = 0 # 0 means end of the path
if score[i][j] == score_left:
pointer[i][j] = 1 # 1 means trace up
if score[i][j] == score_up:
pointer[i][j] = 2 # 2 means trace left
if score[i][j] == score_diagonal:
pointer[i][j] = 3 # 3 means trace diagonal
if score[i][j] >= max_score:
max_i = i
max_j = j
max_score = score[i][j];
align1, align2 = [], [] # initial sequences
i,j = max_i,max_j # indices of path starting point
#traceback, follow pointers
while pointer[i][j] != 0:
if pointer[i][j] == 3:
align1.append(seq1[i-1])
align2.append(seq2[j-1])
i -= 1
j -= 1
elif pointer[i][j] == 2:
align1.append(gap)
align2.append(seq2[j-1])
j -= 1
elif pointer[i][j] == 1:
align1.append(seq1[i-1])
align2.append(gap)
i -= 1
finalize(align1, align2)
If we run this with the following input:
seq1 = ['[A0]', '[C0]', '[A1]', '[B1]']
seq2 = ['[A0]', '[A1]', '[B1]', '[C1]']
print "Needleman-Wunsch"
needle(seq1, seq2)
print
print "Smith-Waterman"
water(seq1, seq2)
We get this output:
Needleman-Wunsch
Identity = 60.000 percent
Score = 20
[A0][C0][A1][B1]----
[A0]----[A1][B1][C1]
Smith-Waterman
Identity = 75.000 percent
Score = 25
[A0][C0][A1][B1]
[A0]----[A1][B1]
For the specific changes I made, see: this GitHub repository.
Imagine we have a log file with alphabetic sequences. Like something you said, I converted sequences to A0A1... . For example, if there was a sequence like #read #write #add #write, it converted to A0A1A2A1. Every time, I read two character and compare them but keep score matrix like before. Here is my code in C# for smith-waterman string alignment.
Notice that Cell is a user defined class.
private void alignment()
{
string strSeq1;
string strSeq2;
string strTemp1;
string strTemp2;
scoreMatrix = new int[Log.Length, Log.Length];
// Lists That Holds Alignments
List<char> SeqAlign1 = new List<char>();
List<char> SeqAlign2 = new List<char>();
for (int i = 0; i<Log.Length; i++ )
{
for (int j=i+1 ; j<Log.Length; j++)
{
strSeq1 = "--" + logFile.Sequence(i);
strSeq2 = "--" + logFile.Sequence(j);
//prepare Matrix for Computing optimal alignment
Cell[,] Matrix = DynamicProgramming.Intialization_Step(strSeq1, strSeq2, intSim, intNonsim, intGap);
// Trace back matrix from end cell that contains max score
DynamicProgramming.Traceback_Step(Matrix, strSeq1, strSeq2, SeqAlign1, SeqAlign2);
this.scoreMatrix[i, j] = DynamicProgramming.intMaxScore;
strTemp1 = Reverse(string.Join("", SeqAlign1));
strTemp2 = Reverse(string.Join("", SeqAlign2));
}
}
}
class DynamicProgramming
{
public static Cell[,] Intialization_Step(string Seq1, string Seq2,int Sim,int NonSimilar,int Gap)
{
int M = Seq1.Length / 2 ;//Length+1//-AAA //Changed: /2
int N = Seq2.Length / 2 ;//Length+1//-AAA
Cell[,] Matrix = new Cell[N, M];
//Intialize the first Row With Gap Penality Equal To Zero
for (int i = 0; i < Matrix.GetLength(1); i++)
{
Matrix[0, i] = new Cell(0, i, 0);
}
//Intialize the first Column With Gap Penality Equal To Zero
for (int i = 0; i < Matrix.GetLength(0); i++)
{
Matrix[i, 0] = new Cell(i, 0, 0);
}
// Fill Matrix with each cell has a value result from method Get_Max
for (int j = 1; j < Matrix.GetLength(0); j++)
{
for (int i = 1; i < Matrix.GetLength(1); i++)
{
Matrix[j, i] = Get_Max(i, j, Seq1, Seq2, Matrix,Sim,NonSimilar,Gap);
}
}
return Matrix;
}
public static Cell Get_Max(int i, int j, string Seq1, string Seq2, Cell[,] Matrix,int Similar,int NonSimilar,int GapPenality)
{
Cell Temp = new Cell();
int intDiagonal_score;
int intUp_Score;
int intLeft_Score;
int Gap = GapPenality;
//string temp1, temp2;
//temp1 = Seq1[i*2].ToString() + Seq1[i*2 + 1]; temp2 = Seq2[j*2] + Seq2[j*2 + 1].ToString();
if ((Seq1[i * 2] + Seq1[i * 2 + 1]) == (Seq2[j * 2] + Seq2[j * 2 + 1])) //Changed: +
{
intDiagonal_score = Matrix[j - 1, i - 1].CellScore + Similar;
}
else
{
intDiagonal_score = Matrix[j - 1, i - 1].CellScore + NonSimilar;
}
//Calculate gap score
intUp_Score = Matrix[j - 1, i].CellScore + GapPenality;
intLeft_Score = Matrix[j, i - 1].CellScore + GapPenality;
if (intDiagonal_score<=0 && intUp_Score<=0 && intLeft_Score <= 0)
{
return Temp = new Cell(j, i, 0);
}
if (intDiagonal_score >= intUp_Score)
{
if (intDiagonal_score>= intLeft_Score)
{
Temp = new Cell(j, i, intDiagonal_score, Matrix[j - 1, i - 1], Cell.PrevcellType.Diagonal);
}
else
{
Temp = new Cell(j, i, intDiagonal_score, Matrix[j , i - 1], Cell.PrevcellType.Left);
}
}
else
{
if (intUp_Score >= intLeft_Score)
{
Temp = new Cell(j, i, intDiagonal_score, Matrix[j - 1, i], Cell.PrevcellType.Above);
}
else
{
Temp = new Cell(j, i, intDiagonal_score, Matrix[j , i - 1], Cell.PrevcellType.Left);
}
}
if (MaxScore.CellScore <= Temp.CellScore)
{
MaxScore = Temp;
}
return Temp;
}
public static void Traceback_Step(Cell[,] Matrix, string Sq1, string Sq2, List<char> Seq1, List<char> Seq2)
{
intMaxScore = MaxScore.CellScore;
while (MaxScore.CellPointer != null)
{
if (MaxScore.Type == Cell.PrevcellType.Diagonal)
{
Seq1.Add(Sq1[MaxScore.CellColumn * 2 + 1]); //Changed: All of the following lines with *2 and +1
Seq1.Add(Sq1[MaxScore.CellColumn * 2]);
Seq2.Add(Sq2[MaxScore.CellRow * 2 + 1]);
Seq2.Add(Sq2[MaxScore.CellRow * 2]);
}
if (MaxScore.Type == Cell.PrevcellType.Left)
{
Seq1.Add(Sq1[MaxScore.CellColumn * 2 + 1]);
Seq1.Add(Sq1[MaxScore.CellColumn * 2]);
Seq2.Add('-');
}
if (MaxScore.Type == Cell.PrevcellType.Above)
{
Seq1.Add('-');
Seq2.Add(Sq2[MaxScore.CellRow * 2 + 1]);
Seq2.Add(Sq2[MaxScore.CellRow * 2]);
}
MaxScore = MaxScore.CellPointer;
}
}
}
I'm just struggled with question about that:
I need to write a program that get 3 positive digits from the user and print all the 3 digits numbers that can be created from them. I'm not allowed to use recursion.. any ideas?
thanks
Providing that a, b, c are given digits, e.g.
int a = 1;
int b = 2;
int c = 3;
the implementation (C#) could be
String report = String.Join(Environment.NewLine,
new HashSet<int>() {
100 * a + 10 * b + c,
100 * a + 10 * c + b,
100 * b + 10 * a + c,
100 * b + 10 * c + a,
100 * c + 10 * a + b,
100 * c + 10 * b + a,
});
Console.Write(report);
The output is
123
132
213
231
312
321
Note, that for (a = 1, b = 2 and c = 1) you'll get only
121
112
211
I doubt if this solution will be accepted by your professor (even if it doesn't have any recursion), but you can use it as a test when elaborating your own routine.
I am looking for an algorithm that can iterate through a grid and transform it into another grid with the indexes in a new order.
Basically, given a grid of size n*m:
1_1 1_2 1_3 ... 1_n
2_1 2_2 2_3 ... 2_n
.
.
.
m_1 m_2 m_3 ... m_m
How could I transform it to:
1_1 1_2 1_4 ...
1_3 1_5 ...
1_6 ...
...
.
.
.
Assume, you iterate through the first grid, going left to right in the top row, then
left to right in the second row, all the way to, left to right in the bottom row.
Basically I pushing the elements into an upper triangle.
Another problem is how do I figure out the length and width of the grid used to store the triangle just by knowing what n and m is?
Is there a formula for that?
For example, a grid of 5*6, gets changed to 8*7...
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
26 27 28 29 30
becomes:
1 2 4 7 11 16 22 29
3 5 8 12 17 23 30
6 9 13 18 24
10 14 19 25
15 20 26
21 27
28
The following seems to work for me:
public static T[,] ConvertToUpperTriangle<T>(T[,] arr)
{
// calculate the dimensions
int elements = arr.GetLength(0) * arr.GetLength(1);
double rows = 0.5 * (Math.Sqrt(8 * elements + 1) - 1);
int newHeight = (int)rows;
int newWidth = (int)Math.Ceiling(rows);
// create the new array
var arr2 = new T[newHeight, newWidth];
int j = 0;
int i = 0;
foreach (T element in arr)
{
arr2[j, i] = element;
i--;
j++;
if (i < 0)
{
i = j;
j = 0;
}
}
return arr2;
}
The 0.5 * (Math.Sqrt(8 * elements + 1) - 1) comes from running sum from 1 to n of n and then solve a = 0.5 * n * (n + 1) for n through Wolfram Alpha.
Edit:
You can get the indices i and j for a given index as follows:
int rows = (int)(0.5 * (Math.Sqrt(8 * index + 1) - 1));
int bottomLeft = (int)(0.5 * rows * (rows + 1));
int difference = index - bottomLeft;
int i;
int j;
if (bottomLeft == index)
{
i = 0;
j = rows - 1;
}
else
{
i = rows + 1 - difference;
j = difference - 1;
}
Let's define the "ordinal position" O(i,j) of each grid element (i,j) in a starting grid NxM, which is the function (i,j) -> i*N + j.
Now for the largest triangular number less than O(i,j), call it T == (k(k+1)/2 for some k, the new grid position for our (i,j) will be:
(i,j) -> ( O(i,j) - T, k + T - O(i,j) )
Now substitute for O(i,j) and T to get:
(i,j) -> ( i*N + j - k(k+1)/2, k + (k+1)(k+2)/2 - i*N + j)
= ( i*N + j - k(k+1)/2, (k+1)(k+2)/2 - i*N + j)
That's as far as I can get it just now.
Update:
Note again that k is the side-length for the triangualr number T == k(k+1)/2.