For a ti-84 program I used the quadratic equation and stored the root with the plus sign as N and the root with the negative sign as R. I then want to only use the positive root as the value for the rest of the program (if both are positive it doesn't matter I just want to check one is positive), so I did as follows:
If N>0 --> U
ElseR-->U
But it didn't work. Is it not possible to use store as a command in an if-else statement? Or is there another way to only choose the positive variable and store that as U?
Okay, listen. Blazo's answer is wrong. Every If statement that has an Else also needs a Then and an End. This is the correct way to write it without piecewise expressions (piecewise expressions are usually the best way to go, though, so I suggest you check out dohaqatar's answer as well):
If N>0
Then
N->U
Else
R->U
End
With two separate lone If's, you can reduce the size of the code by one byte to 17 bytes:
If N>0
N->U
If N<=0
R->U
You can compress it even more, saving 5 more bytes to make 12 bytes:
R->U
If N>0
N->U
You can also use a piecewise expression such as dohaqatar's below, also 12 bytes:
N<0:RAns+Nnot(Ans->U
The best part is, an algorithm change reduces code size to 6 bytes:
max(R,N->U
This can be quite easily accomplished through your standard If, Else statements; however, A quicker way of doing this is through piece-wise functions. In TI-Basic, every Boolean expression evaluates to either 1 or 0, representing true and false respectively. This fact can be exploited to make certain conditional expressions much shorter.
Your code using standard If Else syntax:
If N>0
Then
N→U
Else
R→U
End
This option is 18 bytes long.
Using piece-wise expressions, your code can be compressed to this:
N<0:RAns+Nnot(Ans→U
Resulting in a line of code only 12 bytes long.
Related
I want to write a program to convert hexadecimal numbers into their decimal forms without using a variable of fixed length to store the result because that would restrict the range of inputs that my program can work with.
Let's say I were to use a variable of type long long int to calculate, store and print the result. Doing so would limit the range of hexadecimal numbers that my program can handle to between 8000000000000001 and 7FFFFFFFFFFFFFFF. Anything outside this range would cause the variable to overflow.
I did write a program that calculates and stores the decimal result in a dynamically allocated string by performing carry and borrow operations but it runs much slower, even for numbers that are as big as 7FFFFFFFF!
Then I stumbled onto this site which could take numbers that are way outside the range of a 64 bit variable. I tried their converter with numbers much larger than 16^65 - 1 and still couldn't get it to overflow. It just kept on going and printing the result.
I figured that they must be using a much better algorithm for hex to decimal conversion, one that isn't limited to 64 bit values.
So far, Google's search results have only led me to algorithms that use some fixed-length variable for storing the result.
That's why I am here. I wanna know if such an algorithm exists and if it does, what is it?
Well, it sounds like you already did it when you wrote "a program that calculates and stores the decimal result in a dynamically allocated string by performing carry and borrow operations".
Converting from base 16 (hexadecimal) to base 10 means implementing multiplication and addition of numbers in a base 10x representation. Then for each hex digit d, you calculate result = result*16 + d. When you're done you have the same number in a 10-based representation that is easy to write out as a decimal string.
There could be any number of reasons why your string-based method was slow. If you provide it, I'm sure someone could comment.
The most important trick for making it reasonably fast, though, is to pick the right base to convert to and from. I would probably do the multiplication and addition in base 109, so that each digit will be as large as possible while still fitting into a 32-bit integer, and process 7 hex digits at a time, which is as many as I can while only multiplying by single digits.
For every 7 hex digts, I'd convert them to a number d, and then do result = result * (16^7) + d.
Then I can get the 9 decimal digits for each resulting digit in base 109.
This process is pretty easy, since you only have to multiply by single digits. I'm sure there are faster, more complicated ways that recursively break the number into equal-sized pieces.
What is the typical approach in LUA (before the introduction of integers in 5.3) for dealing with calculated range values in for loops? Mathematical calculations on the start and end values in a numerical for loop put the code at risk of bugs, possibly nasty latent ones as this will only occur on certain values and/or with changes to calculation ordering. Here's a concocted example of a loop not producing the desire output:
a={"a","b","c","d","e"}
maybethree = 3
maybethree = maybethree / 94
maybethree = maybethree * 94
for i = 1,maybethree do print(a[i]) end
This produces the unforuntate output of two items rather than the desired three (tested on 5.1.4 on 64bit x86):
a
b
Programmers unfamiliar with this territory might be further confused by print() output as that prints 3!
The application of a rounding function to the nearest whole number could work here. I understand the approximatation with FP and why this fails, I'm interested in what the typical style/solution is for this in LUA.
Related questions:
Lua for loop does not do all iterations
Lua: converting from float to int
The solution is to avoid this reliance on floating-point math where floating-point precision may become an issue. Or, more realistically, just be aware of when you are using FP and be mindul of the precision issue. This isn’t a Lua problem that requires a Lua-specific solution.
maybethree is a misnomer: it is never three. Your code above is deterministic. It will always print just a and b. Since the maybethree variable is less than three, of course the for loop would not execute 3 times.
The print function is also behaving as defined/expected. Use string.format to show thr FP number in all its glory:
print(string.format("%1.16f", maybethree)) -- 2.9999999999999996
Still need to use calculated values to control your for loop? Then you already mentioned the answer: implement a rounding function.
The code examples are gonna be in Lua, but the question is rather general - it's just an example.
for k=0,100 do
::again::
local X = math.random(100)
if X <= 30
then
-- do something
else
goto again
end
end
This code generates 100 pseudorandom numbers between 0-30. It should do it between 0-100, but doesn't let the loop go on if any of them is larger than 30.
I try to do this task without goto statement.
for k=0,100 do
local X = 100 -- may be put behind "for", in some cases, the matter is that we need an 'X' variable
while X >= 30 do --IMPORTANT! it's the opposite operation of the "if" condition above!
X = math.random(100)
end
-- do the same "something" as in the condition above
end
Instead, this program runs the random number generation until I get a desired value. In general, I put all the codes here that was between the main loop and the condition in the first example.
Theoretically, it does the same as the first example, only without gotos. However, I'm not sure in it.
Main question: are these program codes equal? They do the same? If yes, which is the faster (=more optimized)? If no, what's the difference?
It is bad practice to use Goto. Please see http://xkcd.com/292/
Anyway, I'm not much into Lua, but this looks simple enough;
For your first code: What you are doing is starting a loop to repeat 100 times. In the loop you make a random number between 0 and 100. If this number is less than or equal to 30, you do something with it. If this number is greater than 30, you actually throw it away and get another random number. This continues until you have 100 random numbers which will ALL be less than or equal to thirty.
The second code says: Start a loop from 0 to 100. Then you set X to be 100. Then you start another loop with this condition: As long as X is greater than 30, keep randomizing X. Only when X is less than 30 will your code exit and perform some action. When it has performed that action 100 times, the program ends.
Sure, both codes do the same thing, but the first one uses a goto - which is bad practice regardless of efficiency.
The second code uses loops, but is still not efficient - there are 2 levels of loops - and one is based on psuedo-random generation which can be extremely inefficient (maybe the CPU generates only numbers between 30-100 for a trillion iterations?) Then things get very slow. But this is also true for you're first piece of code - it has a 'loop' that is based on psuedo-random number generation.
TLDR; strictly speaking about efficiency, I do not see one of those being more efficient than the other. I could be wrong but it seems the same things is going on.
you can directly use math.random(lower, upper)
for k=0,100 do
local X = math.random(0, 30)
end
even faster.
As I see this pieces of code do the same, but using goto always isn't the best choice (in any programming language). For lua see details here
This may not be a programming question but it's a problem that arised recently at work. Some background: big C development with special interest in performance.
I've a set of integers and want to test the membership of another given integer. I would love to implement an algorithm that can check it with a minimal set of algebraic functions, using only a integer to represent the whole space of integers contained in the first set.
I've tried a composite Cantor pairing function for instance, but with a 30 element set it seems too complicated, and focusing in performance it makes no sense. I played with some operations, like XORing and negating, but it gives me low estimations on membership. Then I tried with successions of additions and finally got lost.
Any ideas?
For sets of unsigned long of size 30, the following is one fairly obvious way to do it:
store each set as a sorted array, 30 * sizeof(unsigned long) bytes per set.
to look up an integer, do a few steps of a binary search, followed by a linear search (profile in order to figure out how many steps of binary search is best - my wild guess is 2 steps, but you might find out different, and of course if you test bsearch and it's fast enough, you can just use it).
So the next question is why you want a big-maths solution, which will tell me what's wrong with this solution other than "it is insufficiently pleasing".
I suspect that any big-math solution will be slower than this. A single arithmetic operation on an N-digit number takes at least linear time in N. A single number to represent a set can't be very much smaller than the elements of the set laid end to end with a separator in between. So even a linear search in the set is about as fast as a single arithmetic operation on a big number. With the possible exception of a Goedel representation, which could do it in one division once you've found the nth prime number, any clever mathematical representation of sets is going to take multiple arithmetic operations to establish membership.
Note also that there are two different reasons you might care about the performance of "look up an integer in a set":
You are looking up lots of different integers in a single set, in which case you might be able to go faster by constructing a custom lookup function for that data. Of course in C that means you need either (a) a simple virtual machine to execute that "function", or (b) runtime code generation, or (c) to know the set at compile time. None of which is necessarily easy.
You are looking up the same integer in lots of different sets (to get a sequence of all the sets it belongs to), in which case you might benefit from a combined representation of all the sets you care about, rather than considering each set separately.
I suppose that very occasionally, you might be looking up lots of different integers, each in a different set, and so neither of the reasons applies. If this is one of them, you can ignore that stuff.
One good start is to try Bloom Filters.
Basically, it's a probabilistic data structure that gives you no false negative, but some false positive. So when an integer matches a bloom filter, you then have to check if it really matches the set, but it's a big speedup by reducing a lot the number of sets to check.
if i'd understood your correctly, python example:
>>> a=[1,2,3,4,5,6,7,8,9,0]
>>>
>>>
>>> len_a = len(a)
>>> b = [1]
>>> if len(set(a) - set(b)) < len_a:
... print 'this integer exists in set'
...
this integer exists in set
>>>
math base: http://en.wikipedia.org/wiki/Euler_diagram
[Background Story]
I am working with a 5 year old user identification system, and I am trying to add IDs to the database. The problem I have is that the system that reads the ID numbers requires some sort of checksum, and no-one working here now has ever worked with it, so no-one knows how it works.
I have access to the list of existing IDs, which already have correct checksums. Also, as the checksum only has 16 possible values, I can create any ID I want and run it through the authentication system up to 16 times until I get the correct checksum (but this is quite time consuming)
[Question]
What methods can I use to help guess the checksum algorithm of used for some data?
I have tried a few simple methods such as XORing and summing, but these have not worked.
So my question is: if I have data (in hexadecimal) like this:
data checksum
00029921 1
00013481 B
00026001 3
00004541 8
What methods can I use work out what sort of checksum is used?
i.e. should I try sequential numbers such as 00029921,00029922,00029923,... or 00029911,00029921,00029931,... If I do this what patterns should I look for in the changing checksum?
Similarly, would comparing swapped digits tell me anything useful about the checksum?
i.e. 00013481 and 00031481
Is there anything else that could tell me something useful? What about inverting one bit, or maybe one hex digit?
I am assuming that this will be a common checksum algorithm, but I don't know where to start in testing it.
I have read the following links, but I am not sure if I can apply any of this to my case, as I don't think mine is a CRC.
stackoverflow.com/questions/149617/how-could-i-guess-a-checksum-algorithm
stackoverflow.com/questions/2896753/find-the-algorithm-that-generates-the-checksum
cosc.canterbury.ac.nz/greg.ewing/essays/CRC-Reverse-Engineering.html
[ANSWER]
I have now downloaded a much larger list of data, and it turned out to be simpler than I was expecting, but for completeness, here is what I did.
data:
00024901 A
00024911 B
00024921 C
00024931 D
00042811 A
00042871 0
00042881 1
00042891 2
00042901 A
00042921 C
00042961 0
00042971 1
00042981 2
00043021 4
00043031 5
00043041 6
00043051 7
00043061 8
00043071 9
00043081 A
00043101 3
00043111 4
00043121 5
00043141 7
00043151 8
00043161 9
00043171 A
00044291 E
From these, I could see that when just one value was increased by a value, the checksum was also increased by the same value as in:
00024901 A
00024911 B
Also, two digits swapped did not change the checksum:
00024901 A
00042901 A
This means that the polynomial value (for these two positions at least) must be the same
Finally, the checksum for 00000000 was A, so I calculated the sum of digits plus A mod 16:
( (Σxi) +0xA )mod16
And this matched for all the values I had. Just to check that there was nothing sneaky going on with the first 3 digits that never changed in my data, I made up and tested some numbers as Eric suggested, and those all worked with this too!
Many checksums I've seen use simple weighted values based on the position of the digits. For example, if the weights are 3,5,7 the checksum might be 3*c[0] + 5*c[1] + 7*c[2], then mod 10 for the result. (In your case, mod 16, since you have 4 bit checksum)
To check if this might be the case, I suggest that you feed some simple values into your system to get an answer:
1000000 = ?
0100000 = ?
0010000 = ?
... etc. If there are simple weights based on position, this may reveal it. Even if the algorithm is something different, feeding in nice, simple values and looking for patterns may be enlightening. As Matti suggested, you/we will likely need to see more samples before decoding the pattern.