I saw ~.2$<##if this code being called comparing code.
It does work, but I know # means
Rotates the top 3 elements of the stack so that the 3rd down is now on top.
and I can't understand how it works.
Can anyone please explain what # is in this code?
Thank You.
Now I know, # is literally rotating.
.2$< returns 1 if the left arg is bigger, and 0 if the right one is bigger.
## makes the input a b to 1 a b or 0 a b.
The code would say a when the condition is 1, or b when 0.
Related
I come to this site in need of help, after struggling with this problem for a few days now. I am trying to program a poem that accepts some data from standard input and then outputs a poem based on that data.
The code seems to be working, but it is not correct! It is giving me the wrong index of the array I am using. I would love extra eyes to help me with my code and let me know what I am doing wrong.
ALSO! For some reason, I am not able to access the third array of the char array... I tried to place "SIZE - 1" in there but it prints nothing... Would love to understand why this is. Does this look right?
// Program that accepts some data from standard input,
#include <iostream>
#include <cstring>
//here... extracted.
for (int sign = 0; sign < poem[line]; sign++)
{
if (line > word_count)
{
std::cout << " ";
print_poem(seed);
}
else
{
print_poem(seed);
}
You haven't mentioned what exactly the task is but I can at least explain to you parts of the problem.
Are the correct syllables being printed?
Let's assess if the correct syllables are being printed. I ran your code on my machine (with the input you provided that is "100 3 1 5 7 5") and got:
nahoewachi
tetsunnunoyasa
munahohuke
The syllable count of each line is fine (5,7,5) so that's not a problem.
The first syllable you have a problem with is chi in nahoewachi. I'm only illustrating why this syllable is being printed. You can apply the same logic to the rest.
Initially, the seed is 100. Before processing the first row, you apply generate_prnd, which gives 223. Before calculating chi, you print 4 other syllables (na, ho, e and wa). This means that you have applied generate_prnd 8 times before calculating the fifth syllable.
Applying generate_prnd 8 times on 223 gives 711. Applying one more time (to get row) gives 822.
822%9 = 3rd row (0 indexed).
Applying one more time (to get column) gives 361. 361%5 = 1st column.
Therefore the index for the fifth syllable is (3,1). The string at the (3,1)th index is "chi". Therefore, the correct syllable is being printed. The indexing is correct. There's a problem with your logic if you want a different syllable to be printed.
Now, let's assess why there aren't any spaces in your output.
In the example you provided, num_lines=3. The word_counts (actually syllable counts) are 5, 7 and 5. You are applying a space when line (which is always less than num_lines) is greater than word_count.
However, line is always less than word_count since the maximum value of line is 2 (num_lines - 1). Therefore, a space will never be printed.
P.S. If you are allocating memory using new, don't forget to deallocate using delete later.
I initially learned that if I want to see if a cell has any contents to use if(A1<>"",.... But as I receive more and more assistance on SO, it seems most people use if(LEN(A1),... Is there a difference? Did I learn the wrong information? Should I ever opt for one over the other or just always use LEN from now on?
pretty much the same result. difference is:
LEN(A1) - checks if A1 has a length
A1<>"" - checks if A1 is not equal to "empty"
then there is a length of the formula itself (some prefer to save 1 extra character space):
A1<>"" has 6 characters compared to LEN(A1) 7 characters
the superiority of LEN comes when you need to check for character count like:
=IF(LEN(A1)=4, TRUE, FALSE)
eg. output TRUE only if A1 value has exactly 4 characters
I can't find an example close enough to this one on StackOverflow so here goes:
I want to return a message "Type?" if cell X is blank and cell Y has any text. But I'm trying to nestle it into an existing set of IFs.
Existing :
=IF($G241="Evo";M241*L241;IF($G241="Free";M241*L241;IF($G241="GN";M241*L241))))
Nestling this into the above:
=IF(AND(NOT(ISBLANK($J234));ISBLANK(G234));"Type?";"OK")
I tried this but it returns FALSE, maybe due to the AND I'm using, which I need since I'm creating a return based on two cells two cells.
=IF($G240="Evo";M240*L240;IF(AND(NOT(ISBLANK($J240));ISBLANK(G240);"Type?";"OK");IF($G240="Free";M240*L240;IF($G240="GN";M240*L240))))
getting Error:
AND expects boolean values. But 'Type?' is a text and cannot be coerced to a boolean.
IF(and(isblank(cell x),iferror(isstring(cell y),false)),"Type?","OK")
That should do it for you I think. you will need to replace cell x and cell y with the appropriate references. The iferror statement is there to catch what happens when evaluating a blank cell y.
The problem with this formula
=IF($G240="Evo";M240*L240;IF(AND(NOT(ISBLANK($J240));ISBLANK(G240);"Type?";"OK");IF($G240="Free";M240*L240;IF($G240="GN";M240*L240))))
is you are trying to check G240 for different values when it cant. Lets simplify your formula. We will replace your empty cell check with FORMULA 1
=If($G240="EVO", Do True Condition, Do Formula 1, IF(G$240=Free, Do Free True Condition, Do Free False Condition)
The problem is since you already did something (Formula 1) when G240 = "EVO", you cant start another check on what G240 after the fact with the way you have embedded your formula. a batter way of thinking of it is how to do a second check when G240="EVO" is false. Remember the general format of an if statement is:
IF(CONDITION,True Result, False Result)
There are only 3 things that go into an if statement. you tried putting in 3.
Try rearranging to this:
=If($G240="EVO", Do True Condition, IF(SOME CHECK to determine DO FOMULA 1 or CHECK for G240 = FREE, Do Formula 1, IF(G$240=Free, Do Free True Condition, Do Free False Condition)))
Basically break down what you want to check for in G240 and do it in sequence with your IF statement. Right now with what you have written, I cant tell how you want to determine if you want to run your formula 1 or if you want to check if G240="free" since you have two different outcomes if G240="Free"/
OK I think i found the issue. The IF(AND(NOT(ISBLANK works on it's own since there are no other IFs in the formula. I do want to test two different cells for text(letters) in order to show a warning if one cell was blank while the other not. But as soon as you insert the (AND into a string of multiple IFs it doesn't work.
Simply removing the (AND was all I needed to do. Another way to achieve a test for more than one blank cell was to simply add multiple IF(ISBLANKs.
EG: =IF(ISBLANK(A1)+IF(ISBLANK(A2)>2;condition true;condition false)
ForwardEd thanks very much for your help!
Regards
I would like to know how can I construct a regex to know if a number in base 2 (binary) is multiple of 3. I had read in this thread Check if a number is divisible by 3 but they dont do it with a regex, and the graph someone drew is wrong(because it doesn't accept even numbers). I have tried with: ((1+)(0*)(1+))(0) but it doesn't works for some values. Hope you can help me.
UPDATE:
Ok, thanks all for your help, now I know how to draw the NFA, here I left the graph and the regular expresion:
In the graph, the states are the number in base 10 mod 3.
For example: to go to state 1 you have to have 1, then you can add 1 or 0, if you add 1, you would have 11(3 in base 10), and this number mod 3 is 0 then you draw the arc to the state 0.
((0*)((11)*)((1((00) *)1) *)(101 *(0|((00) *1 *) *0)1) *(1(000)+1*01)*) *
And the other regex works, but this is shorter.
Thanks a lot :)
I know this is an old question, but an efficient answer is yet to be given and this question pops up first for "binary divisible by 3 regex" on Google.
Based on the DFA proposed by the author, a ridiculously short regex can be generated by simplifying the routes a binary string can take through the DFA.
The simplest one, using only state A, is:
0*
Including state B:
0*(11)*0*
Including state C:
0*(1(01*0)*1)*0*
And include the fact that after going back to state A, the whole process can be started again.
0*((1(01*0)*1)*0*)*
Using some basic regex rules, this simplifies to
(1(01*0)*1|0)*
Have a nice day.
If I may plug my solution for this code golf question! It's a piece of JavaScript that generates regexes (probably inefficiently, but does the job) for divisibility for each base.
This is what it generates for divisibility by 3 in base 2:
/^((((0+)?1)(10*1)*0)(0(10*1)*0|1)*(0(10*1)*(1(0+)?))|(((0+)?1)(10*1)*(1(0+)?)|(0(0+)?)))$/
Edit: comparing to Asmor's, probably very inefficient :)
Edit 2: Also, this is a duplicate of this question.
For some who is learning and searching how to do this:
see this video:
https://www.youtube.com/watch?v=SmT1DXLl3f4&t=138s
write state quations and solve them with Axden's Theorem
The way I did is visible in the image-result is the same as pointed out by user #Kert Ojasoo. I hope i did it corretly because i spent 2 days to solve it...
n+2n = 3n. Thus, 2 adjacent bits set to 1 denote a multiple of 3. If there are an odd number of adjacent 1s, that would not be 3.
So I'd propose this regex:
(0*(11)?)+
Buttons
Each cell of an N x N grid is either a 0 or a 1. You are given two such N x N grids, the initial grid and the final grid. There is a button against each row and each column of the initial N x N grid. Pressing a row-button toggles the values of all the cells in that row, and pressing a column-button toggles the values of all the cells in that column.
You are required to find the minimum number of button presses required to transform the grid from the initial configuration to the final configuration, and the buttons that must be pressed in order to make this transformation.
When the initial and the final configurations are the same, print "0".
Input
The first line contains t, the number of test cases (about 10). Then t test cases follow.
Each test case has the following form:
The first line contains n, the size of the board (1 ≤ n ≤ 1000).
n lines follow. The ith line contains n space separated integers representing the ith row of the initial grid. Each integer is either a 0 or a 1.
n lines follow, representing the final grid, in the same format as above.
Output
For each test case, output the number of row-button presses, followed by the row buttons that must be pressed. Print the number of column-button presses next, followed by 0-indexed indices of the column buttons that must be pressed. The total number of button presses must be minimized.
Output "-1" if it is impossible to achieve the final configuration from the initial configuration. If there is more than one solution, print any one of them.
Input:
1
3
0 0 0
1 1 0
1 1 0
1 1 0
1 1 1
1 1 1
Output:
1
0
1
2
Though it works absolutely fine on my machine,it doesnt accept a solution at codechef and gives me a wrong answer.Can anyone guide me what to do pls pls pls??
Code has been written in C++ and compiled using g++ compiler.
In the code posted, I would revise the code after you calculate "matrixc". I find it very difficult to follow beyond that point, so I'm going to stop looking at the code and talk about the problem. For those without the code, matrix C = initial matrix - final matrix. The matrices are over the binary field.
In problems like these, look at the symmetries in the solution. There are three symmetries. One is the order of the buttons does not matter. If you take a valid solution and rearrange it, you get another valid solution. Another symmetry is that pressing a button twice is the same as not pressing it at all. The last symmetry is that if you take the complement of a valid solution, you get another valid solution. For example, in a 3x3 grid, if S = { row1, row3, col1 } is a solution, then S' = { row2, col2, col3 } is also a solution.
So all you need to do is find one solution, then exploit the symmetry. Since you only need to find one, just do the easiest thing you can think of. I would just look at column 1 and row 1 to construct the solution, then check the solution against the whole matrix. If this solution gives you more than N buttons to press for an NxN grid, then take the solution's complement and you'll end up with a smaller one.
Symmetry is a very important concept in computer science and it comes up almost everywhere. Understanding the symmetries of this problem is what allows you to solve it without checking every possible solution.
P.S. You say this code is C++, but it is also perfectly valid C if you remove #include <iostream> from the top. It might take a lot less time to compile if you compile it as C.