R"09"^0 * S"13579" does not work because R"09"^0 will consume all digits and S"13579" will have nothing to match.
Normally, odd <- [0-9] odd / [1357] is correct enough to match small numbers. Alternatively, you could first check it's not last digit of an odd number, before consuming it:
odd <- (!([1357] ![0-9]) [0-9])* [1357]
Related
So I'm using the String.sub function and I'm wondering if there's any way to get all the characters in a string until the end of the string. Since String.sub takes in a string, int (the index of where to start getting chars), and then a number of how many chars, I'm not sure what the easiest way of doing all chars since we want a possibly positive infinite amount
For String.sub you just have to subtract from the length of the string. There's no simpler way, i.e., there's no value for length that has a special meaning. A value larger than the remainder of the string is an error in OCaml (which tends to be strict when checking parameters).
Assume i is >= 0 and < length of the string:
String.sub s i (String.length s - i)
You can use Str.last_chars, but you still need to know how many characters you want. I.e., you still have to subtract from the length of the string.
Str.last_chars s (String.length s - i)
Suppose we are given two strings s1 and s2(both lowercase). We have two find the minimal lexographic string that can be formed by merging two strings.
At the beginning , it looks prettty simple as merge of the mergesort algorithm. But let us see what can go wrong.
s1: zyy
s2: zy
Now if we perform merge on these two we must decide which z to pick as they are equal, clearly if we pick z of s2 first then the string formed will be:
zyzyy
If we pick z of s1 first, the string formed will be:
zyyzy which is correct.
As we can see the merge of mergesort can lead to wrong answer.
Here's another example:
s1:zyy
s2:zyb
Now the correct answer will be zybzyy which will be got only if pick z of s2 first.
There are plenty of other cases in which the simple merge will fail. My question is Is there any standard algorithm out there used to perform merge for such output.
You could use dynamic programming. In f[x][y] store the minimal lexicographical string such that you've taken x charecters from the first string s1 and y characters from the second s2. You can calculate f in bottom-top manner using the update:
f[x][y] = min(f[x-1][y] + s1[x], f[x][y-1] + s2[y]) \\ the '+' here represents
\\ the concatenation of a
\\ string and a character
You start with f[0][0] = "" (empty string).
For efficiency you can store the strings in f as references. That is, you can store in f the objects
class StringRef {
StringRef prev;
char c;
}
To extract what string you have at certain f[x][y] you just follow the references. To udapate you point back to either f[x-1][y] or f[x][y-1] depending on what your update step says.
It seems that the solution can be almost the same as you described (the "mergesort"-like approach), except that with special handling of equality. So long as the first characters of both strings are equal, you look ahead at the second character, 3rd, etc. If the end is reached for some string, consider the first character of the other string as the next character in the string for which the end is reached, etc. for the 2nd character, etc. If the ends for both strings are reached, then it doesn't matter from which string to take the first character. Note that this algorithm is O(N) because after a look-ahead on equal prefixes you know the whole look-ahead sequence (i.e. string prefix) to include, not just one first character.
EDIT: you look ahead so long as the current i-th characters from both strings are equal and alphabetically not larger than the first character in the current prefix.
I have a string of numbers separated by spaces and I need to store them in a table but for some reason negative symbol is not getting recognize.
cord = "-53 2 -21"
map = {}
for num in cord:gmatch("%w+") do
table.insert(map, num)
end
map[1], map[2], map[3] = tonumber(map[1]), tonumber(map[2]), tonumber(map[3])
print(map[1])
print(map[2])
print(map[3])
This is the output I'm getting:
53
2
21
I think the problem is with the pattern I'm using, what should I change?
The pattern "%w" is for alphanumeric characters, which doesn't include -, use this pattern instead:
"%-?%w+"
or better:
"%-?%d+"
since numbers are all you need.
%w+ does not attempt to mach only numbers, so try %S+ to get all "words", that is, all sequences of non-zero characters.
If you want to match only numbers, try %-?%d+. Note the optional minus sign in the pattern. Note also that you must escape the minus sign.
I'm just playing around with Lua trying to make a calculator that uses string manipulation. Basically I take two numbers out of a string, then do something to them (+ - * /). I can successfully take a number out of x, but taking a number out of y always returns nil. Can anyone help?
local x = "5 * 75"
function calculate(s)
local x, y =
tonumber(s:sub(1, string.find(s," ")-1)),
tonumber(s:sub(string.find(s," ")+3), string.len(s))
return x * y
end
print(calculate(x))
You have a simple misplaced parenthesis, sending string.len to tonumber instead of sub.
local x, y =
tonumber(s:sub(1, string.find(s," ")-1)),
tonumber(s:sub(string.find(s," ")+3, string.len(s)))
You actually don't need the string.len, as end of string is the default value for sub if nothing is given.
EDIT:
You can actually do what you want to do way shorter by using string.match instead.
local x,y = string.match(s,"(%d+).-(%d+)")
Match looks for tries to match the string with the pattern given and returns the captured values, in this case the numbers. This pattern translates to "One or more digits, then as few as possible of any character, then one or more digits". %d is 1 digit, + means one or more. . means any character and - means as few as possible. The values within the parentheses are captured, which means that they are returned.
I'm fairly new to Haskell programming and I'm having trouble understanding why I'm receiving this error in my code.
My problem is as follows: Any positive integer i can be expressed as i = 2^n*k, where k is odd, that is, as a power of 2 times an odd number. We call n the exponent of 2 in i. For example, the exponent of 2 in 40 is 3 (because 40 = 2^3*5) whereas the exponent of 2 in 42 is 1. If i itself is odd, then n is zero. If, on the other hand, i is even, that means it can be divided by 2. Write a function exponentOfTwo for finding the exponent of 2 in its argument.
I understand the psuedocode and it seems fairly simple: recursively divide i by 2 until result is odd, the number of times the division happens is n
here is my code (line 31-32):
exponentOfTwo :: Int -> Int
exponentOfTwo i = if odd i then 0 else 1 + exponentOfTwo (i 'div' 2)
I'm receiving the error "lexical error in string/character literal at character 'i'" on line 32 column 62.
I've tried searching for a solution to this error everywhere and so far I've had no luck.
To use a function in infix for, surround it with backticks (`), not with single quotes ('). The latter are for character literals, which, well are only one character long.
Are the characters around div backquotes rather than normal quotes? They need to be to allow a function name to be used as an infix operator. I changed that in your definition and the code worked for me.