How can I decode this strange text?
+[----->+++<]>++.+++.+++.+++++++.+[->+++<]>.[--->+<]>----.--------.[--->+<]>.---------.-----------.++.+++++++++++.+++[->+++<]>++.+.
This text should be a program.
The program outputs 'iloveskypegrab' when ran. It's a extremely simplified programming language called brainfuck.
As Torkoal's answer says, it does indeed print iloveskypegrab when run. Let me explain what this program does, exactly:
Brainf**k (which will be referred to as BF) operates on an array of memory cells, also referred to as the tape, each initially set to zero. There is a pointer, initially pointing to the first memory cell. The commands are:
> Move the pointer to the right
< Move the pointer to the left
+ Increment the memory cell under the pointer
- Decrement the memory cell under the pointer
. Output the character signified by the cell at the pointer
, Input a character and store it in the cell at the pointer
[ Jump past the matching ] if the cell under the pointer is 0
] Jump back to the matching [ if the cell under the pointer is nonzero
All characters other than ><+-.,[] are comments and are ignored by the BF interpreter. I won't explain too much here. Let's split up the program, piece by piece:
+[----->+++<]
This is an example of a simple loop. It adds one to the current cell, then (because it's non-zero) goes into the brackets. It subtracts 5 from that cell, moves the pointer right one space, adds 3, and then moves back.
>++.
This moves the pointer right one, adds 2, and outputs the current cell. It makes i.
+++.+++.
Add 3, output, than add 3 more, and output again. This outputs the lo.
+++++++.
Add 7, and output the character. This returns v.
+[->+++<]
Add 1, and then inside the loop, subtract 1, then move right, add 3, and move back. Because the cell isn't zero until everything has been subtracted, it runs multiple times, and is a useful way of adding large numbers to cells quickly.
>.
Output the cell we added to in the last loop. This returns e.
[--->+<]
Keep subtracting 3 from the cell with e in it. While it's not zero, add one to the cell to the right.
>----.
Move right, subtract 4, and output. This is the s.
--------.
Subtract 8 from that cell, and output it again. This prints a k.
[--->+<]
Same as before: Keep subtracting 3, and while that isn't zero, add one to the cell.
>.
Still the same: move right and print the cell. This prints y.
---------.
Subtract 9. Output. This makes the p.
----------.
Subtract 10, output. Prints e.
++.+++++++++++.
By now this should become clear. Prints gr.
+++[->+++<]
More loops - add 3, then continually subtract 1, and add 3 to the cell to the right.
>++.+.
Move right. Add 2, output, add one, output. Outputs ab.
I hope this helped you! Let me know if something is wrong.
Related
I'm parsing strings in excel, and I need to return everything through the last number. For example:
Input: A00XX
Output: A00
In my case, I know the last number will be between index 3 and 5, so I'm brute-forcing it with:
=LEFT([#Point],
IF(SUM((MID([#Point],5,1)={"0","1","2","3","4","5","6","7","8","9"})+0),5,
IF(SUM((MID([#Point],4,1)={"0","1","2","3","4","5","6","7","8","9"})+0),4,
IF(SUM((MID([#Point],3,1)={"0","1","2","3","4","5","6","7","8","9"})+0),3,
))))
Unfortunately, I've run into some edge cases where the numbers extended beyond index 5. Is there a generic way to find the last number in a string using excel formulas?
Note:
I've tried =MAX(SEARCH(... but it returns the index of the first number, not the last.
As a starting point: if we know the position of the last number, we can use LEFT to get the string to that point. Suppose that the position is 5:
=LEFT(A1, 5)
But, we don't know the position of the last number. Now, what if the only valid number was 0, and it only appeared once: then we could use FIND to locate the position of the number:
=LEFT(A1, FIND(0, A1))
But, we have more than one valid number. Suppose that we had all the numbers from 0 through 9, but each number could only appear once — then we could use MAX on a FIND array, to tell us which of the numbers is the last one:
=LEFT(A1, MAX(FIND({0,1,2,3,4,5,6,7,8,9}, A1)))
Unfortunately, FIND will throw a #VALUE! error any number doesn't appear, which will then make MAX return the same error. So, we need to fix that with IFERROR:
=LEFT(A1, MAX(IFERROR(FIND({0,1,2,3,4,5,6,7,8,9}, A1), 0)))
However, numbers can appear more than once. As such, we need a method to find the last occurrence of a value in a string (since FIND and SEARCH will, by default, return the first occurrence).
The SUBSTITUTE function has 3 mandatory arguments — Initial String, Value to be Replaced, Value to Replace with — and one Optional argument — the occurrence to replace. Normally, this is omitted, so that all occurrences are replaced. But, if we know how many times a character appears in a string, then we can replace just the last instance with a special/uncommon sub-string to search for.
To count how many times a character appears in a String, just start with the length of the String, then subtract the length when you SUBSTITUTE all copies of that character for Nothing:
=LEN(A1) - LEN(SUBSTITUTE(A1, 0, ""))
This means we can now replace the last occurrence of the character with, for example, ">¦<", and then FIND that:
=FIND(">¦<", SUBSTITUTE(A1, 0, ">¦<", LEN(A1) - LEN(SUBSTITUTE(A1, 0, ""))))
Of course, we want to do this for all the numbers from 0 to 9, and take the MAX value (remembering our IFERROR), so we need to put the Array of values back in:
=MAX(IFERROR(FIND(">¦<", SUBSTITUTE(A1, {0,1,2,3,4,5,6,7,8,9}, ">¦<", LEN(A1) - LEN(SUBSTITUTE(A1, {0,1,2,3,4,5,6,7,8,9}, "")))), 0))
Then, we plug that all back into our initial LEFT function:
=LEFT(A1, MAX(IFERROR(FIND(">¦<", SUBSTITUTE(A1, {0,1,2,3,4,5,6,7,8,9}, ">¦<", LEN(A1) - LEN(SUBSTITUTE(A1, {0,1,2,3,4,5,6,7,8,9}, "")))), 0)))
An alternative, assuming that the length of the string in question will never be more than 9 characters (which seems a safe assumption based on your description):
=LEFT(A1,MATCH(0,0+ISERR(0+MID(A1,{1;2;3;4;5;6;7;8;9},1))))
This, depending on your version of Excel, may or may not require committing with CTRL+SHIFT+ENTER.
Note also that the separator within the array constant {1;2;3;4;5;6;7;8;9} is the semicolon, which, for English-language versions of Excel, represents the row-separator. This may require amending if you are using a non-English-language version.
Of course, we can replace this static constant with a dynamic construction. However, since we are already making the assumption that 9 is an upper limit on the number of characters for the string in question, this would not seem to be necessary.
If you have the newest version of Excel, you can try something like:
=LEFT(D1,
LET(x, SEQUENCE(LEN(D1)),
MAX(IF(ISNUMBER(NUMBERVALUE(MID(D1, SEQUENCE(LEN(D1)), 1))), x))))
For example:
Wanted to check if CONCATENATE is the one to use (not sure if my excel has TEXTJOIN), and how to show just the text that has empty value in the cells.
For example in my attachment below, I want the intended result shown like in B2 and B3, where the texts shown with delimiter, when the values are false (empty).
If I were to use CONCATENATE like in Row 10 and Row 11, it's rather manual and it only capture "positive values" as in non-blank cells.
Purpose: To show pending tasks (empty/blank status cells)
Use MID with CONCATENATED IFS:
=MID(IF(C2="","/"&$C$1,"")&IF(D2="","/"&$D$1,"")&IF(E2="","/"&$E$1,"")&IF(F2="","/"&$F$1,"")&IF(GC2="","/"&$G$1,"")&IF(H2="","/"&$H$1,""),2,999)
I would use TEXJOIN and FILTER if you have the newest version of Excel.
For example: =TEXTJOIN("/",1,FILTER($E$2:$I$2, ISBLANK(E3:I3)))
EDIT: For older versions, a temporary workaround is as follows:
make a temporary array the same size as your original dataframe where each value is determined by a formula such as =IF(ISBLANK(E3), E$2&"/","")
Use something like =LEFT(CONCAT(E15:J15), LEN(CONCAT(E15:J15))-1) to get the desired result (where E15:J15 is where I elected to store the first row of the temporary array created in step 1).
I am not sure of your Excel version, but I think this would work in older versions (formatted for readability - will work if you paste it directly into cell B2 and copy down):
=LEFT(CONCAT( INDEX( CHOOSE({1;2;3},$C$1:$H$1,{"/","/","/","/","/","/"},{"","","","","",""}),
INDEX( IF(ISBLANK(C2:H2),{1;2},{3;3}),
MOD(COLUMN(A1:INDEX(1:1,,12))-1,2)+1,
(COLUMN(A1:INDEX(1:1,,12))-1)/2+1 ),
(COLUMN(A1:INDEX(1:1,,12))-1)/2+1 ) ),
SUM(7*ISBLANK(C2:H2))-1 )
Notes
As this is an array formula, you may have to enter it with CTRL + SHIFT + ENTER with an older version of Excel.
The stat labels must all have a length of 6 characters as shown in your post. If not, then they must at least have the same length and the last line SUM(7*ISBLANK(C2:H2))-1 must be changed to replace the 7 with the string length + 1, e.g. a length of 9 would be SUM(10*ISBLANK(C2:H2))-1.
If they don't have the same length, the LEFT( can be removed along with the SUM(10*ISBLANK(C2:H2))-1) at the end. You will end up having a trailing / delimiter at the end. You could fix that for the case of stat F being the last part by changing {"/","/","/","/","/","/"} to {"/","/","/","/","/",""}, but the other cases would still have a trailing /. Another approach is much more complex, but the component SUM(10*ISBLANK(C2:H2))-1) could be shaped to identify what to cut off or maybe a helper column could be built - in any case, let's hope your situation is that the stat labels all have the same length.
The delimiter "/" can be changed, but must always be a single character. If not, then then last line must be changed to SUM( [label length + delimiter length] *ISBLANK(C2:H2))-1.
This formula is fixed to 6 stat columns. If you need for it to accommodate more, it is possible by extending the {"/","/","/","/","/","/"} and {"","","","","",""} (one element for each new column) and replacing every 12 with 2 times the number of columns. Also, obviously, the references $C$1:$H$1 and C1:H2 must be changed to read in your new columns.
I would like to use the command text to type numbers within 57 hexagons. I want to use a loop:
for mm=1:57
text(x(m),y(m),'m')
end
where x(m) and y(m) are the coordinates of the text .
The script above types the string "m" and not the value of m. What am I doing wrong?
Jubobs pretty much told you how to do it. Use the num2str function. BTW, small typo in your for loop. You mean to use mm:
for mm=1:57
text(x(mm),y(mm),num2str(mm));
end
The reason why I've even decided to post an answer is because you can do this vectorized without a loop, which I'd also like to write an answer for. What you can do place each number into a character array where each row denotes a unique number, and you can use text to print out all numbers simultaneously.
m = sprintfc('%2d', 1:57);
d = reshape([m{:}], 2, 57).';
text(x, y, d);
The (undocumented!) function sprintfc takes a formatting specifier and an array and creates a cell array of strings where each cell is the string version of each element in the array you supply. In order to ensure that the character array has the same number of columns per row, I ensure that each string takes up 2 characters, and so any number less than 10 will have a blank space at the beginning. I then convert the cell array of strings into a character array by converting the cell array into a comma-separated list of strings and I reshape the matrix into an acceptable form, and then I call text with all of the pairs of x and y, with the corresponding labels in m together on the screen.
I've been working on this programming challenge: http://www.codeabbey.com/index/task_view/summing-up
Which basically states:
Input data has two values A and B in the single line.
Output should have the sum A+B printed into it.
Additionally after the stop the program should have values A, B, A+B in the cells 0, 1 and 2 respectively.
So for example input would look like this:
9 26
Now, I think I be misunderstanding either the problem or the solution because I believe the solution is supposed to be 9 26 35 where 9, 26, and 35 are all in their own cells.
My solution returns 9 26 35 and I believe in the correct cells (0,1, and 2 respectfully) but I am getting the answer wrong. Can anyone please look at the problem and my code and tell me where I am going wrong?
Code:
;:>;:><[-<+>]<:
I tried plugging this into a couple of online brainfuck interpreters. There is one here:
http://copy.sh/brainfuck/
and another here:
http://esoteric.sange.fi/brainfuck/impl/interp/i.html
In both cases, I needed to change your character set slightly --> : becomes . and ; becomes ,
The output from both was
9 Y
Notice that 35 - 9 = 24, and Y is the 24th letter of the alphabet. I think you are outputting the number "35" and having it interpretted as a letter.
I would try changing the program so that your output is literally single digits of the answer -- ie, output a 3, then output a 5, instead of outputting a numerical "35" (but leave the numerical value in cell 2 at the end). In other words, your text output should be a formatted version of the values in memory, rather than just outputting the numerical values directly.
It sounds like the output should only have A+B printed, not A, B, and A+B, as you're doing with :.
And your result seems like it'll have A+B in cell 0, and 0 in cell 1 (essentially the same as the example code).
>< is just cancelling itself out.
Given an IP in string form (XX.XXX.YYY.XX) is it possible to check the values of YYY and copy them into an adjacent cell?
The value will either be 2 or 3 characters long and will always come after the 2nd period.
Answered below by Jerry. Thank you again Jerry!
You can extract YYY into a cell with the formula:
=MID(A1,FIND("#",SUBSTITUTE(A1,".","#",2))+1,FIND(".",A1,FIND("#",SUBSTITUTE(A1,".","#",2))+1)-FIND("#",SUBSTITUTE(A1,".","#",2))-1)
Assuming the IP is in A1.
It works for whatever length of IP you have.
EDIT: Some details:
There's some bits of recycled formulas here. SUBSTITUTE(A1,".","#",2) returns XX.XXX#YYY.XX (substitute the 2nd occurrence of the dot in A1 by #) We'll use this in the big formula and let's call it R for the time being.
This turns the formula into:
=MID(A1,FIND("#",R)+1,FIND(".",A1,FIND("#",R)+1)-FIND("#",R)-1)
^-----------^ ^--------------------------------------^
Start | 1 | | 2 |
Length
Much better!
The starting position part:
FIND("#",R)+1 returns the position of the character just after #, so that MID starts with the first Y. Here, the position becomes 7.
The length position part:
FIND(".",A1,FIND("#",R)+1) There's a formula we already used here, FIND("#",R)+1 is 7 so that we have: FIND(".",A1,7). This finds the position of the dot in A1 which is after or at the 7th character. This one gets the value 10.
This one should be familiar and gets the position of # in R, which is 6.
10-6 gives 4, which is one character longer than what we're looking for. (because we're dealing with ranked positions; e.g. Length of the string between 1st and 3nd character of a string is 1 but 3-1 gives 2)
Hence why there's the final -1 part.