For the given fen string -
k7/1R1RN3/p3p3/P3P2p/1PP4P/3K1PP1/8/8 b KQkq h3 0 1
Chess.js does not detect and show chess.moves() gives you ['0-0-0'], which is queen side castling, Not possible without a queen.
Any idea?
Turns out to be a problem with the fen string, The castling flags were not set right!
The fen should be :
k7/1R1RN3/p3p3/P3P2p/1PP4P/3K1PP1/8/8 b - h3 0 1
Related
I have column in a df with a size range with different sizeformats.
artikelkleurnummer size
6725 0161810ZWA B080
6726 0161810ZWA B085
6727 0161810ZWA B090
6728 0161810ZWA B095
6729 0161810ZWA B100
in the sizerange are also these other size formats like XS - XXL, 36-50 , 36/38 - 52/54, ONE, XS/S - XL/XXL, 363-545
I have tried to get the prefix '0' out of all sizes with start with a letter in range (A:K). For exemple: Want to change B080 into B80. B100 stays B100.
steps:
1 look for items in column ['size'] with first letter of string in range (A:K),
2 if True change second position in string into ''
for range I use:
from string import ascii_letters
def range_alpha(start_letter, end_letter):
return ascii_letters[ascii_letters.index(start_letter):ascii_letters.index(end_letter) + 1]
then I've tried a for loop
for items in df['size']:
if df.loc[df['size'].str[0] in range_alpha('A','K'):
df.loc[df['size'].str[1] == ''
message
SyntaxError: unexpected EOF while parsing
what's wrong?
You can do it with regex and the pd.Series.str.replace -
df = pd.DataFrame([['0161810ZWA']*5, ['B080', 'B085', 'B090', 'B095', 'B100']]).T
df.columns = "artikelkleurnummer size".split()
replacement = lambda mpat: ''.join(g for g in mpat.groups() if mpat.groups().index(g) != 1)
df['size_cleaned'] = df['size'].str.replace(r'([a-kA-K])(0*)(\d+)', replacement)
Output
artikelkleurnummer size size_cleaned
0 0161810ZWA B080 B80
1 0161810ZWA B085 B85
2 0161810ZWA B090 B90
3 0161810ZWA B095 B95
4 0161810ZWA B100 B100
TL;DR
Find a pattern "LetterZeroDigits" and change it to "LetterDigits" using a regular expression.
Slightly longer explanation
Regexes are very handy but also hard. In the solution above, we are trying to find the pattern of interest and then replace it. In our case, the pattern of interest is made of 3 parts -
A letter in from A-K
Zero or more 0's
Some more digits
In regex terms - this can be written as r'([a-kA-K])(0*)(\d+)'. Note that the 3 brackets make up the 3 parts - they are called groups. It might make a little or no sense depending on how exposed you have been to regexes in the past - but you can get it from any introduction to regexes online.
Once we have the parts, what we want to do is retain everything else except part-2, which is the 0s.
The pd.Series.str.replace documentation has the details on the replacement portion. In essence replacement is a function that takes all the matching groups as the input and produces an output.
In the first part - where we identified three groups or parts. These groups are accessed with the mpat.groups() function - which returns a tuple containing the match for each group. We want to reconstruct a string with the middle part excluded, which is what the replacement function does
sizes = [{"size": "B080"},{"size": "B085"},{"size": "B090"},{"size": "B095"},{"size": "B100"}]
def range_char(start, stop):
return (chr(n) for n in range(ord(start), ord(stop) + 1))
for s in sizes:
if s['size'][0].upper() in range_char("A", "K"):
s['size'] = s['size'][0]+s['size'][1:].lstrip('0')
print(sizes)
Using a List/Dict here for example.
I have to draw a DFA that accepts set of all strings containing 1011 as a substring in it. I tried but could not come up with one. Can anyone help me please?
Thanks
The idea for a DFA that does this is simple: keep track of how much of that substring we have seen on the end of the input we've seen so far. If you eventually get to a point where the input you've seen so far ends with that substring, then you accept the whole input. If you get to the end of input before ever seeing a prefix that ends with your substring, you don't accept.
We can create the DFA by adding states as necessary to represent differing levels of match against the target substring. All DFAs need at least one state: let's call it q0.
---->q0
The implied alphabet of your language is {0, 1}, so we need transitions for both of these symbols on the state q0. Let's think about how much of the substring we will have seen in state q0. We can get to q0 with the empty string; that is, before consuming any input at all. After seeing the empty string we have seen zero of the four symbols that make up our substring. So, q0 should correspond to the case "the input I've seen up until now ends with a string that matches 0/4 of the target substring".
Given this, what transitions should we add for 0 and 1? If we see a 0 in state q0, that doesn't help at all, since the substring we're looking for begins with 1; so, seeing a 0 in q0 doesn't change the fact that the input we've seen so far matches 0/4 symbols. This means we can have the transition from q0 on 0 return to q0.
/-\
0 | |
V /
---->q0
What about if we see a 1 in q0? Well, if we see a 1, then the input we've seen so far ends with a string that matches 1011 in exactly 1/4 places (the first 1); so, we need another state to represent the fact we're a little closer to the goal. Let's call this state q1.
/-\
0 | |
V /
---->q0---->q1
1
We repeat the process now for state q1. If we see a 0 in q1, we get a little closer to our target of 1011, so we can go to a new state, q2. If we see a 1 in q1, we don't get any closer to our goal, but we also don't fall back.
0 1
/-\ /-\
| | | |
V / V /
---->q0---->q1---->q2
1 0
If we see a 0 in q2, that means we've seen the substring 00; that doesn't appear in 1011 at all, which means we are totally back to square one and must return to q0. If we see a 1 though, we get a little closer to our goal and must move to a new state; let's call this q3:
0 1
/-\ /-\
| | | |
V / V /
---->q0---->q1---->q2---->q3
^ 1 0 | 1
| |
\-------------/
0
If we see a 0 in q3 then our input has ended with the substring 10, which puts us back at q2; if we see a 1, then we have seen the whole target 1011 and need to go to a new state to remember this fact.
0 1 0
/-\ /-\ /------\
| | | | | |
V / V / V |
---->q0---->q1---->q2---->q3---->q4
^ 1 0 | 1 1
| |
\-------------/
0
Finally, in state q4, no matter what we see, we know we must accept the input since we've already seen the substring 1011 somewhere in the input. This means we should make q4 accepting and have both transitions go back to q4:
0 1 0
/-\ /-\ /------\ /---\
| | | | | | | |
V / V / V | V | 0,1
---->q0---->q1---->q2---->q3---->[q4]--/
^ 1 0 | 1 1
| |
\-------------/
0
You can check some samples to convince yourself that this DFA accepts the language you want. We built it one state at a time by asking ourselves where the transitions had to go. We stopped adding new states when new transitions didn't demand them anymore.
We want to construct a DFA for a string which contains 1011 as a substring which means it language contain
L={0,1}
which means the strings may be
{0111011,001011,11001011,........}
A string must contains 1011 has a substring.
As we observed in the transition diagram at initial state if q0 accepts 1 then move to next state otherwise remains in the same state.
If q1 accepts 0 then move to next state q2 otherwise remains in the same state.
If q2 accepts 1 then move to q3 else move to q0 because we want to substring which starts with 1 not with 0.
If q3 accepts 1 then move to q4 else which is a final state if system reaches to a final state it means a string is accepted because it contains a 1011 as a substring , if q3 accepts then back to q2.
After reaching the final state a string may not end with 1011 but it have some more words or string to be taken like in 001011110 110 is left which have to accept that's why at q4 if it accepts 0 or 1 it remains in the same state.
DFA for accepting strings with a substring 1011.
They are four transitions A,B,C,D in every construction of DFA we have to check each transaction must have both transactions otherwise it is not a DFA so that it is given to construct a DFA that accept string of odd 0's and 1's that was as shown below
A is the initial state on transition of 0 it will goes to C and
On transition 1 it will give to B
B is another state gives transition of D on 0 and A on 1 and C is a state that will give transition of D and A on transition of 1 and 0
D is final state will give transition of B and C on 0 and 1
This is the process is been done on the below figure let us check the DFA with example 1011 it has odd no of 1'sand odd no of 0 so A on 1 it will give B and B on 0 it will give D and D on 1 it will give C and C on 1 it will give D hence it is the required DFA.
You’re given a chess board with dimension n x n. There’s a king at the bottom right square of the board marked with s. The king needs to reach the top left square marked with e. The rest of the squares are labeled either with an integer p (marking a point) or with x marking an obstacle. Note that the king can move up, left and up-left (diagonal) only. Find the maximum points the king can collect and the number of such paths the king can take in order to do so.
Input Format
The first line of input consists of an integer t. This is the number of test cases. Each test case contains a number n which denotes the size of board. This is followed by n lines each containing n space separated tokens.
Output Format
For each case, print in a separate line the maximum points that can be collected and the number of paths available in order to ensure maximum, both values separated by a space. If e is unreachable from s, print 0 0.
Sample Input
3
3
e 2 3
2 x 2
1 2 s
3
e 1 2
1 x 1
2 1 s
3
e 1 1
x x x
1 1 s
Sample Output
7 1
4 2
0 0
Constraints
1 <= t <= 100
2 <= n <= 200
1 <= p <= 9
I think this problem could be solved using dynamic-programing. We could use dp[i,j] to calculate the best number of points you can obtain by going from the right bottom corner to the i,j position. We can calculate dp[i,j], for a valid i,j, based on dp[i+1,j], dp[i,j+1] and dp[i+1,j+1] if this are valid positions(not out of the matrix or marked as x) and adding them the points obtained in the i,j cell. You should start computing from the bottom right corner to the left top, row by row and beginning from the last column.
For the number of ways you can add a new matrix ways and use it to store the number of ways.
This is an example code to show the idea:
dp[i,j] = dp[i+1,j+1] + board[i,j]
ways[i,j] = ways[i+1,j+1]
if dp[i,j] < dp[i+1,j] + board[i,j]:
dp[i,j] = dp[i+1,j] + board[i,j]
ways[i,j] = ways[i+1,j]
elif dp[i,j] == dp[i+1,j] + board[i,j]:
ways[i,j] += ways[i+1,j]
# check for i,j+1
This assuming all positions are valid.
The final result is stored in dp[0,0] and ways[0,0].
Brief Overview:
This problem can be solved through recursive method call, starting from nn till it reaches 00 which is the king's destination.
For the detailed explanation and the solution for this problem,check it out here -> https://www.callstacker.com/detail/algorithm-1
Using VB.Net
When i add a the number with zeros means, it is showing exact result without zero's
For Example
Dim a, b, c as int32
a = 001
b = 5
c = a + b
a = 009
b = 13
c = a + b
Showing output as 6 instead of 006, 22 instead of 022
Expected output
006
022
How to do this.
Need vb.net code help
You need to store a number as a string if you want to store the exact number of zeros. Then addition won't work though.
If you just want to display the number with 3 digits, you can store it as an integer and format the result when you print it.
c.ToString("D3")
zero is nothing.. If you do a regular mathematical calculation of 001 + 5 the result is still 6. I would suggest you check out string padding.
Not sure how else to ask this but, I want to search for a term within several string elements. Here's what my code looks like (but wrong):
inplay = vector(length=nrow(des))
for (ii in 1:nrow(des)) {
if (des[ii] = 'In play%')
inplay[ii] = 1
else inplay[ii] = 0
}
des is a vector that stores strings such as "Swinging Strike", "In play (run(s))", "In play (out(s) recorded)" and etc. What I want inplay to store is a 1s and 0s vector corresponding with the des vector, with the 1s in inplay indicating that the des value had "In play%" in it and 0s otherwise.
I believe the 3rd line is incorrect, because all this does is return a vector of 0s with a 1 in the last element.
Thanks in advance!
The data.table package has syntax that is often similar to SQL. The package includes %like%, which is a "convenience function for calling regexpr". Here is an example taken from its help file:
## Create the data.table:
DT = data.table(Name=c("Mary","George","Martha"), Salary=c(2,3,4))
## Subset the DT table where the Name column is like "Mar%":
DT[Name %like% "^Mar"]
## Name Salary
## 1: Mary 2
## 2: Martha 4
The R analog to SQL's LIKE is just R's ordinary indexing syntax.
The 'LIKE' operator selects data rows from a table by matching string values in a specified column against a user-supplied pattern
> # create a data frame having a character column
> clrs = c("blue", "black", "brown", "beige", "berry", "bronze", "blue-green", "blueberry")
> dfx = data.frame(Velocity=sample(100, 8), Colors=clrs)
> dfx
Velocity Colors
1 90 blue
2 94 black
3 71 brown
4 36 beige
5 75 berry
6 2 bronze
7 89 blue-green
8 93 blueberry
> # create a pattern to use (the same as you would do when using the LIKE operator)
> ptn = '^be.*?' # gets beige and berry but not blueberry
> # execute a pattern-matching function on your data to create an index vector
> ndx = grep(ptn, dfx$Colors, perl=T)
> # use this index vector to extract the rows you want from the data frome:
> selected_rows = dfx[ndx,]
> selected_rows
Velocity Colors
4 36 beige
5 75 berry
In SQL, that would be:
SELECT * FROM dfx WHERE Colors LIKE ptn3
Something like regexpr?
> d <- c("Swinging Strike", "In play (run(s))", "In play (out(s) recorded)")
> regexpr('In play', d)
[1] -1 1 1
attr(,"match.length")
[1] -1 7 7
>
or grep
> grep('In play', d)
[1] 2 3
>
Since stringr 1.5.0, you can use str_like, which follows the structure of SQL's LIKE:
library(stringr)
fruit <- c("apple", "banana", "pear", "pineapple")
str_like(fruit, "app%")
#[1] TRUE FALSE FALSE FALSE
Not only does it include %, but also several other operators (see ?str_like).
Must match the entire string
_ matches a single character (like .)
% matches any number of characters (like .*)
% and _ match literal % and _
The match is case insensitive by default