How does this code work?
;''
6666,-2%{2+.2/#*\/10.3??2*+}*
`1000<~\;
It seem to use an array #* and a cycle {/**/}, but what is 6666? what is \/?
The first three characters; ;'', are unneeded for the program to function. They simply discard all input and replace it with an empty string, in case your compiler needs an input necessarily.
6666, prints out an array 6666 elements long, each of which are the numbers 0-6665.
-2% is a mapping function. It reverses the function and deletes every two elements. You now you have an array that is 3333 elements long, and it goes [6665 6663 6661 … 5 3 1]
{foo}* is a folding block call. For every element, do the following to the combination of elements. For example, 5,{+}* would add together the numbers 0-4.
So, let's see what we're doing in this folding block call.
2+ add two to the element.
. duplicate the element.
2/ halve it. Your sub-stack looks like this; (n+2),((n+2)/2)
# pulls the third element to the top.
This is the first function we cannot do, since our original stack is only two tall. We'll get back to this later.
*\/ will be skipped for now, we'll get back to it once we discuss folding more.
10.3?? Duplicate 10, then push a 3. [10 10 3]. ? is exponentiation, so we have [10 1000], then again gives us a 1 with 1000 zeroes afterwards.
2* Multiply it by two. So now we have a 2 with 1000 zeroes after.
+ Adds the rest of our math to 2e(1e3)
So, let's go back to that pesky #.
#*\/ will grab the third element and bring it to the top, then multiply it by the next top element ((n+2)/2), then we divide n by this number.
This is an expansion of the Leibniz Series.
\`1000< turns the int into a string, then throws a decimal after the 3.
~ dumps the string into a number again.
\; deleted the rest of the stack.
To answer your specific questions;
6666 was chosen, since half is 3333 (length of array), and we want more than pi times the number of digits of accuracy we want. We could make it smaller if we wanted, but 6666 is a cute number to use.
\/ Is the "inverse division" pair. Take a, take b, then calculate b/a. This is because the \ changes the order of the top two elements in the array, and / divides them.
Related
In my code, legends are running within a loop, and I am trying to show a graph with
0-10%
10-20%
and so on. The problem is when I write this code
legend->AddEntry(gr[i], Form("%d0-%d0 %%",i+0,i+1), "lep");
It shows
00-10%
10-20% etc
So how to not show 00, but 0 in the first line?
A small adaptation of the shown statement should be enough; use:
legend->AddEntry(gr[i], Form("%d-%d %%", i*10 , (i+1)*10), "lep");
Explanation:
Form("%d0-%d0 %%",i+0,i+1) seems to be some kind of string formatting, and i your loop variable which runs from 0 to 9, right? The shown Form statement just appends "0" hard-coded to the single digit in i; instead, you can multiply i by 10, resulting in the actual numbers you want printed; and since 10*0 is still 0, this will be a single digit still; so, replace the previous Form(...) call with Form("%d-%d %%", i*10, (i+1)*10) and you should have the result you want!
In case you're worrying that printing i*10 is "less efficient" than printing i with "0" suffix - don't. The formatting and output of the string is most probably orders of magnitude slower than the multiplication anyway, so any overhead of doing multiple multiplications is negligible.
(Using Python 3)
Given this list named numList: [1,1,2,2,3,3,3,4].
I want to remove exactly one instance of “1” and “3” from numList.
In other words, I want a function that will turn numList into: [1,2,2,3,3,4].
What function will let me remove an X number of elements from a Python list once per element I want to remove?
(The elements I want to remove are guaranteed to exist in the list)
For the sake of clarity, I will give more examples:
[1,2,3,3,4]
Remove 2 and 3
[1,3,4]
[3,3,3]
Remove 3
[3,3]
[1,1,2,2,3,4,4,4,4]
Remove 2, 3 and 4
[1,1,2,4,4,4]
I’ve tried doing this:
numList=[1,2,2,3,3,4,4,4]
remList = [2,3,4]
for x in remList:
numList.remove(x)
This turns numList to [1,2,3,4,4] which is what I want. However, this has a complexity of:
O((len(numList))^(len(remList)))
This is a problem because remList and numList can have a length of 10^5. The program will take a long time to run. Is there a built-in function that does what I want faster?
Also, I would prefer the optimum function which can do this job in terms of space and time because the program needs to run in less than a second and the size of the list is large.
Your approach:
for x in rem_list:
num_list.remove(x)
is intuitative and unless the lists are going to be very large I might do that because it is easy to read.
One alternative would be:
result = []
for x in num_list:
if x in rem_list:
rem_list.remove(x)
else:
result.append(x)
This would be O(len(rem_list) ^ len(num_list)) and faster than the first solution if len(rem_list) < len(num_list).
If rem_list was guaranteed to not contain any duplicates (as per your examples) you could use a set instead and the complexity would be O(len(num_list)).
For an AI project of mine, I need to apply to a factored state all rules that apply to its partial components. This needs to be done very frequently so I'm looking for a way to make this as fast as possible.
I'm going to describe my problem with strings, however the true problem works in the same way with vectors of unsigned integers.
I have a bunch of entries (of length N) like this which I need to store in some way:
__a_b
c_e__
___de
abcd_
fffff
__a__
My input is a single entry ciede to which I must find, as fast as possible, all stored entries which match to it. For example in this case the matches would be c_e__ and ___de. Removal and adding of entries should be supported, however I don't care how slow it is. What I would like to be as fast as possible is:
for ( const auto & entry : matchedEntries(input) )
My problem, as I said, is one where each letter is actually an unsigned integer, and the vector is of an unspecified (but known) length. I have no requirements for how entries should be stored, or what type of metadata is going to be associated with them. The naive algorithm of matching all is O(N), is it possible to do better? The number of reasonable entries I need stored is <=100k.
I'm thinking some kind of sorting might help, or some weird looking tree structure, but I can't seem to figure out a good way to approach this problem. It also looks like something word processers already need to do, so someone might be able to help.
The easiest solution is to build a trie containing your entries. When searching the trie, you start in the root and recursively follow an edge, that matches character from your input. There will be at most two of those edges in each node, one for the wildcard _ and one for the actual letter.
In the worst case you have to follow two edges from each node, which would add up to O(2^n) complexity, where n is the length of the input, while the space complexity is linear.
A different approach would be to preprocess the entries, to allow for linear search. This is basically what compiling regular expressions does. For your example, consider following regular expression, which matches your desired input:
(..a.b|c.e..|...de|abcd.|fffff|..a..)
This expression can be implemented as a nondeterministic finite state automaton, with initial state having ε-moves to a deterministic automaton for each of the single entries. This NFSA can then be turned to a deterministic FSA, using the standard powerset construction.
Although this construction can increase the number of states substantially, searching the input word can then be done in linear time, simply simulating the deterministic automaton.
Below is an example for entries ab, a_, ba, _a and __. First start with a nondeterministic automaton, which upon removing ε-moves and joining equivalent states is actually a trie for the set.
Then turn it into a deterministic machine, with states corresponding to subsets of states of the NFSA. Start in the state 0 and for each edge, other than _, create the next state as the union of the states in the original machine, that are reachable from any state in the current set.
For example, when DFSA is in state 16, that means the NFSA could be either in state 1 or 6. Upon transition on a, the NFSA could get to states 3 (from 1), 7 or 8 (from 6) - that will be your next state in the DFSA.
The standard construction would preserve the _-edges, but we can omit them, as long as the input does not contain _.
Now if you have a word ab on the input, you simulate this automaton (i.e. traverse its transition graph) and end up in state 238, from which you can easily recover the original entries.
Store the data in a tree, 1st layer represents 1st element (character or integer), and so on. This means the tree will have a constant depth of 5 (excluding the root) in your example. Don't care about wildcards ("_") at this point. Just store them like the other elements.
When searching for the matches, traverse the tree by doing a breadth first search and dynamically build up your result set. Whenever you encounter a wildcard, add another element to your result set for all other nodes of this layer that do not match. If no subnode matches, remove the entry from your result set.
You should also skip reduntant entries when building up the tree: In your example, __a_b is reduntant, because whenever it matches, __a__ also matches.
I've got an algorithm in mind which I plan to implement and benchmark, but I'll describe the approach already. It needs n_templates * template_length * n_symbols bits of storage (so for 100k templates of length 100 and 256 distinct symbols needs 2.56 Gb = 320 MB of RAM. This does not scale nicely to large number of symbols unless succinct data structure is used.
Query takes O(n_templates * template_length * n_symbols) time but should perform quite well thanks to bit-wise operations.
Let's say we have the given set of templates:
__a_b
c_e__
___de
abcd_
_ied_
bi__e
The set of symbols is abcdei, for each symbol we pre-calculate a bit mask indicating whether the template differs from the symbol at that location or not:
aaaaa bbbbb ccccc ddddd eeeee iiiii
....b ..a.. ..a.b ..a.b ..a.b ..a.b
c.e.. c.e.. ..e.. c.e.. c.... c.e..
...de ...de ...de ....e ...d. ...de
.bcd. a.cd. ab.d. abc.. abcd. abcd.
.ied. .ied. .ied. .ie.. .i.d. ..ed.
bi..e .i..e bi..e bi..e bi... b...e
Same tables expressed in binary:
aaaaa bbbbb ccccc ddddd eeeee iiiii
00001 00100 00101 00101 00101 00101
10100 10100 00100 10100 10000 10100
00011 00011 00011 00001 00010 00011
01110 10110 11010 11100 11110 11110
01110 01110 01110 01100 01010 00110
11001 01001 11001 11001 11000 10001
These are stored in columnar order, 64 templates / unsigned integer. To determine which templates match ciede we check the 1st column of c table, 2st column from i, 3rd from e and so forth:
ciede ciede
__a_b ..a.b 00101
c_e__ ..... 00000
___de ..... 00000
abcd_ abc.. 11100
_ied_ ..... 00000
bi__e b.... 10000
We find matching templates as rows of zeros, which indicates that no differences were found. We can check 64 templates at once, and the algorithm itself is very simple (python-like code):
for i_block in range(n_templates / 64):
mask = 0
for i in range(template_length):
# Accumulate difference-indicating bits
mask |= tables[i_block][word[i]][i]
if mask == 0xFFFFFFFF:
# All templates differ, we can stop early
break
for i in range(64):
if mask & (1 << i) == 0:
print('Match at template ' + (i_block * 64 + i))
As I said I haven't yet actually tried implementing this, so I have no clue how fast it is in practice.
I am aware that next_permutations() in the STL library gives you ALL the possible permutations. However, how can I get specific permutations with a fixed position.
An Example:
Suppose I have the bit format XXXXX(where X is 1 or 0). Can I do something where I keep the second index 1, say (now being X1XXX), and want to find all other permutations for that specific case? Basically, I want the final result as the permutations of ALL other positions when the second index is 1.
Something like (2nd element is always fixed as 1):
01000
01001
...
11000
..
An emirp (prime spelled backwards) is a pime number whose reversal is also prime. Ex. 17 & 71. I have to write a program that displays the first 100 emirps. It has to display 10 numbers per line and align the numbers properly:
2 3 5 7 11 13 17 31 37 71
73 79 97 101 107 113 131 149 151 157.
I have no cue what I am doing and would love if anyone could dump this down for me.
It sounds like there are two general problems:
Finding the emirps.
Formatting the output as required.
Break down your tasks into smaller parts, and then you'll be able to see more clearly how to get the whole task done.
To find the emirps, first write some helper functions:
is_prime() to determine whether a number is prime or not
reverse_digits() to reverse the digits of any number
Combining these two functions, you can imagine a loop that finds all the numbers that are primes both forward and reversed. Your first task is complete when you can simply generate a list of those numbers, printing them to the console one per line.
Next, work out what format you want to use (it looks like a fixed format of some number of character spaces per number is what you need). You know that you have 100 numbers, 10 per line, so working out how to format the numbers should be straightforward.
Break the problem down into simpler sub-problems:
Firstly, you need to check whether a number is prime. This is such a common task that you can easily Google it - or try a naive implementation yourself, which may be better given that this is homework.
Secondly, you need to reverse the digits of a number. I'd suggest you figure out an algorithm for this on a piece of paper first, then implement it in code.
Put the two together - it shouldn't be that hard.
Format the results properly. Printing 10 numbers per line is something you should be able to figure out easily once the rest is done.
Once you have a simple version working you might be able to optimise it in some way.
A straight forward way of checking if a number is prime is by trying all known primes less than it and seeing if it divides evenly into that number.
Example: To find the first couple of primes
Start off with the number 2, it is prime because the only divisors are itself and 1, meaning the only way to multiple two numbers to get 2 is 2 x 1. Likewise for 3.
So that starts us off with two known primes 2 and 3. To check the next number we can check if 4 modulo 2 equals 0. This means when divide 2 into 4 there is no remainder, which means 2 is a factor of 4. Specifically we know 2 x 2 = 4. Thus 4 is not prime.
Moving on to the next number: 5. To check five we try 5 modulo 2 and 5 modulo 3, both of which equals one. So 5 is prime, add it to our list of known primes and then continue on checking the next number. This rather tedious process is great for a computer.
So on and so forth - check the next number by looping through all previous found primes and check if they divide evenly, if all previously found primes don't divide evenly, you have a new prime. Repeat.
You can speed this up by counting by 2's, since all even numbers are divisible by two. Also, another nice trick is you don't have to check any primes greater than the square root of the number, since anything larger would need a smaller prime factor. Cuts your loops in half.
So that is your algorithm to generate a large list of primes.
Collect a good chunk of them in an array, say the first 10,000 or so. And then loop through them, reverse the numbers and see if the result is in your array. If so you have a emirp, continue until you get the first 100 emirps
If the first 10,000 primes don't return 100 emirps. Move on to the next 10,000. Repeat.
For homework, I would use a fairly simplistic isPrime function, pseudo-code along the lines of:
def isPrime (num):
set testDiv1 to 2
while testDiv1 multiplied by testDiv1 is less than or equal to num:
testDiv2 = integer part of (num divided by testDiv1)
if testDiv1 multiplied by testDiv2 is equal to num:
return true
Add 1 to testDiv1
return false
This basically checks whether the number is evenly divisible by any number between 2 and the square root of the number, a primitive primality check. The reson you stop at the square root is because you would have already found a match below it if there was one above it.
For example 100 is 2 times 50, 4 times 25, 5 time 20 and 10 times 10. The next one after that would be 20 times 5 but you don't need to check 20 since it would have been found when you checked 5. Any positive number can be expressed as a product of two other positive numbers, one below the square root and one above (other than the exact square root case of course).
The next tricky bit is the reversal of digits. C has some nice features which will make this easier for you, the pseudo-code is basically:
def reverseDigits (num):
set newNum to zero
while num is not equal to zero:
multiply newnum by ten
add (num modulo ten) to newnum
set num to the integer part of (num divided by ten)
return newNum
In C, you can use int() for integer parts and % for the modulo operator (what's left over when you divide something by something else - like 47 % 10 is 7, 9 % 4 is 1, 1000 % 10 is 0 and so on).
The isEmirp will be a fairly simplistic:
def isEmirp (num):
if not isPrime (num):
return false
num2 = reverseDigits (num)
if not isPrime (num2):
return false
return true
Then at the top level, your code will look something like:
def mainProg:
create array of twenty emirps
set currEmirp to zero
set tryNum to two
while currEmirp is less than twenty
if isEmirp (tryNum):
put tryNum into emirps array at position currEmirp
add 1 to currEmirp
for currEmirp ranging from 0 to 9:
print formatted emirps array at position currEmirp
print new line
for currEmirp ranging from 10 to 19:
print formatted emirps array at position currEmirp
print new line
Right, you should be able to get some usable code out of that, I hope. If you have any questions of the translation, leave a comment and I'll provide pointers for you, rather than solving it or doing the actual work.
You'll learn a great deal more if you try yourself, even if you have a lot of trouble initially.