I have long integers numbers like this: 5291658276538691055
How could I programmatically convert this number to a 4-6 capital letters only that is a unique combination that can also be reversed to get back to the number?
For example using OBJ-C.
There are 26 capital letters;
6 of them could represent 26 ^ 6 numbers (308915776);
So, no. You are trying to map a much larger range of numbers into a much smaller range, it cannot be reversible.
Also, log 5291658276538691055 / log 26 is less than 14, so if 14 letters is good for you, just transform the number into 26-based and map the digits to letters.
And one more thing - if the range of numbers is small enough, you could do some manipulation on the numbers (e.g., just subtract the min) and encode it, which will cost you less digits.
You will need to convert the numbers to Base 26 (Hexavigesimal - snappy name!)
The Wikipedia article on Hexavigesimal gives example code in Java - you should be able to adapt this pretty easily.
NB: You cannot get the long number you mentioned down to 4-6 capital letters only using a conversion algorithm (your example in Base 26 is BCKSATKEBRYBXJ). If you need conversion that short, you only have two options:
Lookup tables (store mappings, e.g. 5291658276538691055 = ABCDEF). Obviously only useful if you have a discrete set of numbers.
Including additional characters (e.g. lower case + numbers).
Related
The server sends alphanumerical ids for a list of items. At the same time, recycler view getItemId (required for has stable ids) must return Long. How to encode string to unique long?
Short answer: you probably can't. Not unless the IDs are guaranteed to be short.
A Long uses 8 bytes, so it can hold 2⁶⁴ (about 1.8×10¹⁹) different values. So it could only represent that number of strings. (A result of the pigeonhole principle.)
However, if the IDs contain only basic ASCII letters (let's assume upper case) and digits — 36 possibilities — and are 13 characters long, then there are 36¹³ (about 1.7×10²⁰) different strings. That's an order of magnitude more than 2⁶⁴, so some of them will have to map to the same Long value.
(In fact, each Long would map to about 10 IDs on average — and even more if you include strings with fewer characters, and/or a greater range of characters.)
So unless the range of IDs is limited, you'll have to find another approach.
I am making an anagram solver in Visual Basic that gives you every possible combination when you enter a string. I need to work out how many combinations there are depending on the amount of characters in the string and how many different characters there are.
E.G.
Sample string:
abc
Total characters: 3, Different Characters: 3
Possible combinations: 6
abc, acb, bac, bca, cab, cba
I need an equation (using the number of characters and different characters) to link this to a string that contains a different amount of characters.
I've been using trial and error to try and figure is out, but I can't quite get my head around it. So far I have:
((letters - 1) ^ (different letters - 1)) + (letters - 1)
which works for a few different letter counts but now for all.
Help please???
I'll lead you to the answer, but I'll try to explain along the way. Let's say you had 10 different letters. You'd have 10 choices for the first, 9 for the second, 8 for the third, etc. Ultimately, there would be 10*9*8*7*6...*2*1 = 10! possibilities. However, sometimes you'll have multiple instances of the same letter. For example, using that for the string "aaabcd" would overcount possibilities, because it counts each of the a's as distinct letters, even though they're not. To correct for that, you would have to divide by the factorial of the number of repeated letters. A good way to calculate the total number of possibilities would be (total number of letters factorial)/ (product of the factorials of the number of repeated instances of each letter).
For example:
There are 6!/(3!) ways to arrange the letters in "aaabcd"
There are 6! ways to arrange the letters is "abcdef"
There are 6!/(3!*2!) ways to arrange the letters in "aaabbc"
There are 10!/(5!*3!*2!) ways to arrange the letters in "aaaaabbbcc"
I hope this helps.
For the possible counting number, it's exactly the same as computing Multinomial Coefficient
A simple explanation is that, for no repeating characters,
It's simply permutation = n!
(It is easy to understand if you draw a tree diagram, with first character has n choices, second character has n-1choices...etc.)
However as you may have repeating characters, you will double count many of them.
Let's see an simple example: for aaa, how many possible arrangements IF WE COUNT EVEN THE OUTCOME IS THE SAME?
Answer is 3!(aaa,aaa,aaa,aaa,aaa,aaa)
This gives us an idea that, when we have a character appearing for m times, we will count m! instead of 1
So the counting is just n!(all possible arrangements, including same outcome) / m! (a character appear for m times)
Same for more characters repeating: n!/a!b!c!.. (first character appear a times, another appear for b times...)
If you understand the concept behind, then you will find that, actually for those "non-repeating" characters, it's just dividing an 1!. For eg, character (multi)set = {a,a,a,b,b,c}, #a = 3, #b = 2, #c = 1, so the answer (without repeating count) is (3+2+1)!/3!2!1! and fraction of this format is named multinomial coefficient as stated above.
In programming point of view, you can just pre-compute all factorials (with a pretty small n though as n~30 is already too large for a variable to store) with simple for loop
declare frac = array(n);
frac[0] = 1;
FOR i=1; i<=n;i++
frac[i] = i*frac[i-1]
For a larger n, you may just calculate double/float division on the fly in the loop to avoid overflow..you may face precision problem though.
If you further need to output the different strings, you may use DFS to backtrack all the possible outcomes. Or if you could use another language like C++, you can use built-in function like next_permutation() after sort the character set.
I'm trying to create a method that creates a random string consisting of 32 characters. This method will generate a random number using arc4random_uniform(62) to choose a number between 0 and 61 and then chose a character from a string that holds numbers from 0 to 9 and alphabet letters both small and capital letters, respectively. For an instance, if arc4random_uniform(62) returns 10, the chosen character will be a, if it returns 61, the chosen character will be Z). The method will do this for 32 times to create the final generated string.
I was wondering when this approach will fail to generate a unique String and result in a repeated one. I searched about this topic and didn't find a satisfying answer. I hope that you will help with me this since I am trying to use this method to generate unique IDs for use in my app.
This method will generate a random number using arc4random_uniform(62) to choose a number between 0 and 61 and then chose a character from a string that holds numbers from 0 to 9 and alphabet letters both small and capital letters, respectively.
You could create an array with a string for all the characters you want to include, and randomly pick values. Or, alternatively you could take advantage of the ASCII encoding has mostly sequential character positions and you can fairly easily convert an ascii number to an NSString.
An integer between 48 and 57 is the numbers 0-9 in ASCII, 65 to 90 is A-Z and 97 to 122 is a-z: https://en.wikipedia.org/wiki/Ascii_table#ASCII_printable_code_chart
I was wondering when this approach will fail to generate a unique String and result in a repeated one. I searched about this topic and didn't find a satisfying answer.
It's often referred to as the "birthday problem". As long as your value is reasonably long (say, 20 characters), it is effectively impossible to have a collision. The world is more likely to be destroyed in the next 2 seconds than your app ever creating a collision.
I hope that you will help with me this since I am trying to use this method to generate unique IDs for use in my app.
Apple provides an API for generating unique IDs. You should use that instead of inventing your own system:
NSString *id = [NSUUID UUID].UUIDString;
That will give you a value like D19B40AA-322C-4ADF-BEF6-2EC4D4CE7BA8. It conforms to "Version 4" of the UUID standard — according to Wikipedia if you generate 1 billion UUIDs every second for the next 100 years, there is a 50% chance of getting two IDs that are the same.
If the UUID is longer than you want, you could grab a smaller part part of the string. Beware that the 4 at the start of the third block means this is a "version 4" UUID and is not a random value. Also the first character at the start of the 4th block is only has four possible values — so avoid or strip off those two characters if you want to grab a smaller part of the string for use as your random ID. See the wikipedia page on UUIDs for more detail.
I have tried to find information and how.
But it did not contain any information that would help.
With T-SQL, I want to convert negative decimal to binary
and convert it back.
Sample value: -9223372036854775543
I try in convert with Calculater this value to Bin result is ...
1000000000000000000000000000000000000000000000000000000100001001
and Convert back to Dec. It's OK.
How i can Convert like this with T-SQL(SQL2008) Script/Function ?
Long time to find information for how to.
Anyone who knows about this, please help.
There is no build in functionality.
for INT and BIGINT you can use CONVERT(VARCHAR(100),CAST(3 AS VARBINARY(100)),2) to get the hex representation as a string. then you can do a simple search replace as every hex digit represents exactly 4 binary digits. However, with values outside of the BIGINT range there is no standard as to how they are represented internally. You might get the right result or not and that behavior might even change between versions.
There is also no standard as to how negative numbers are represented. Most implementations of integers use the two's-complement representation. In that representation the top most bit indicates the sign of the number. How many bits you have is a metter of convention and fully dependent on your environment.
In mathematics -3 woud be -11 in binary and not 11111101.
To solve your problem you can either use a CLR function or you go through your number the old fashioned way:
Is it odd? -> output a 1
Is it even? -> output a 0
integer divide by 2
repeat until the value is 0
This will give you the digits in opposite order, so you have to flip the result. To get the two's-complement representation of a negative number n calculate 1-n, convert the result to binary using the above algorithm but with reversed digits (0 instead of 1 and vice versa). After flipping the digits into the right order prepend with enough 1s to fill your "box".
Well, it is a low-level question
Suppose I store a number (of course computer store number in binary format)
How can I print it in decimal format. It is obvious in high-level program, just print it and the library does it for you.
But how about in a very low-level situation where I don't have this library.
I can just tell what 'character' to output. How to convert the number into decimal characters?
I hope you understand my question. Thank You.
There are two ways of printing decimals - on CPUs with division/remainder instructions (modern CPUs are like that) and on CPUs where division is relatively slow (8-bit CPUs of 20+ years ago).
The first method is simple: int-divide the number by ten, and store the sequence of remainders in an array. Once you divided the number all the way to zero, start printing remainders starting from the back, adding the ASCII code of zero ('0') to each remainder.
The second method relies on the lookup table of powers of ten. You define an array of numbers like this:
int pow10 = {10000,1000,100,10,1}
Then you start with the largest power, and see if you can subtract it from the number at hand. If you can, keep subtracting it, and keep the count. Once you cannot subtract it without going negative, print the count plus the ASCII code of zero, and move on to the next smaller power of ten.
If integer, divide by ten, get both the result and the remainder. Repeat the process on the result until zero. The remainders will give you decimal digits from right to left. Add 48 for ASCII representation.
Basically, you want to tranform a number (stored in some arbitrary internal representation) into its decimal representation. You can do this with a few simple mathematical operations. Let's assume that we have a positive number, say 1234.
number mod 10 gives you a value between 0 and 9 (4 in our example), which you can map to a character¹. This is the rightmost digit.
Divide by 10, discarding the remainder (an operation commonly called "integer division"): 1234 → 123.
number mod 10 now yields 3, the second-to-rightmost digit.
continue until number is zero.
Footnotes:
¹ This can be done with a simple switch statement with 10 cases. Of course, if your character set has the characters 0..9 in consecutive order (like ASCII), '0' + number suffices.
It doesnt matter what the number system is, decimal, binary, octal. Say I have the decimal value 123 on a decimal computer, I would still need to convert that value to three characters to display them. Lets assume ASCII format. By looking at an ASCII table we know the answer we are looking for, 0x31,0x32,0x33.
If you divide 123 by 10 using integer math you get 12. Multiply 12*10 you get 120, the difference is 3, your least significant digit. we go back to the 12 and divide that by 10, giving a 1. 1 times 10 is 10, 12-10 is 2 our next digit. we take the 1 that is left over divide by 10 and get zero we know we are now done. the digits we found in order are 3, 2, 1. reverse the order 1, 2, 3. Add or OR 0x30 to each to convert them from integers to ascii.
change that to use a variable instead of 123 and use any numbering system you like so long as it has enough digits to do this kind of work
You can go the other way too, divide by 100...000, whatever the largest decimal you can store or intend to find, and work your way down. In this case the first non zero comes with a divide by 100 giving a 1. save the 1. 1 times 100 = 100, 123-100 = 23. now divide by 10, this gives a 2, save the 2, 2 times 10 is 20. 23 - 20 = 3. when you get to divide by 1 you are done save that value as your ones digit.
here is another given a number of seconds to convert to say hours and minutes and seconds, you can divide by 60, save the result a, subtract the original number - (a*60) giving your remainder which is seconds, save that. now take a and divide by 60, save that as b, this is your number of hours. subtract a - (b*60) this is the remainder which is minutes save that. done hours, minutes seconds. you can then divide the hours by 24 to get days if you want and days and then that by 7 if you want weeks.
A comment about divide instructions was brought up. Divides are very expensive and most processors do not have one. Expensive in that the divide, in a single clock, costs you gates and power. If you do the divide in many clocks you might as well just do a software divide and save the gates. Same reason most processors dont have an fpu, gates and power. (gates mean larger chips, more expensive chips, lower yield, etc). It is not a case of modern or old or 64 bit vs 8 bit or anything like that it is an engineering and business trade off. the 8088/86 has a divide with a remainder for example (it also has a bcd add). The gates/size if used might be better served than for a single instruction. Multiply falls into that category, not as bad but can be. If operand sizes are not done right you can make either instruction (family) not as useful to a programmer. Which brings up another point, I cant find the link right now but a way to avoid divides but convert from a number to a string of decimal digits is that you can multiply by .1 using fixed point. I also cant find the quote about real programmers not needing floating point related to keeping track of the decimal point yourself. its the slide rule vs calculator thing. I believe the link to the article on dividing by 10 using a multiply is somewhere on stack overflow.