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The lucky numbers are the positive integers whose decimal representations contain only the digits 4 or 7 .enter code here`
For example, numbers 47 , 474 , 4 are lucky and 3 , 13 , 567 are not
if there is no such no then output should -1.
input is sum of digits.
i have written this code:
int main(){
long long int s,no=0,minimum=999999999999999999999;
cin>>s;
for(int i=0; i<=s; i++){
for(int j=0; j<=s; j++){
if(i*4+j*7==s){no=0;
for(int k=0; k<i; k++){
no=no*10+4;
}
for(int l=0; l<j; l++){
no=no*10+7;
}if(no<minimum){
minimum=no;}
}
}
}if(minimum==999999999999999999999){cout<<-1;}
else {cout<<minimum;}
}
it is working fine smaller sum values but input is large then no formed is large due to which i am not able to compare them, the constraints for sum is 1<=n<=10^6
This answer shows a process, one of refinement to develop an efficient solution. The most efficient answer can be found in the paragraphs at the bottom, starting with the text "Of course, you can be even more clever ...".
I've left the entire process in so you can understand the sort of thinking that goes into algorithm development. So, let's begin.
First, I wouldn't, in this case, try to compare large numbers, it's totally unnecessary and limits the sort of ranges you want to handle. Instead, you simply have some number of fours and some number of sevens, which you can easily turn into a sum with:
sum = numFours * 4 + numSevens * 7
In addition, realising that the smallest number is, first and foremost, the one with the least number of digits, you want the absolute minimum number of fours and maximum number of sevens. So start with no fours and as many sevens as needed until you're at or just beyond the required sum.
Then, as long as you're not at the sum, perform the following mutually exclusive steps:
If you're over the desired sum, take away a seven if possible. If there are no sevens to take away, you're done, and there's no solution. Log that fact and exit.
Otherwise (i.e., if you're under the sum), add a four.
At this point, you will have a solution (no solution possible means that you would have already performed an exit in the first bullet point above).
Hence you now have a count of the fours and sevens that sum to the desired number, so the lowest number will be the one with all the fours at the left (for example 447 is less than any of {474, 744}). Output that, and you're done.
By doing it this way, the limitation (say, for example, an unsigned 32-bit int) is no longer the number you use (about four billion, so nine digits), instead it is whatever number of fours you can hold in four billion (about a billion digits).
That's an increase of about 11 billion percent, hopefully enough of an improvement for you, well beyond the 106 maximum sum specified.
In reality, you won't get that many fours since any group of seven fours can always be replaced with four sevens, giving a smaller number (a7777b will always be less than a4444444b, where a is zero or more fours and b is zero or more sevens, same counts in both numbers), so the maximum count of fours will always be six.
Here's some pseudo-code (Python code, actually) to show it in action. I've chosen Python, even though you stated C++, for the following reasons:
This is almost certainly an educational question (there's very little call for this sort of program in the real world). That means you're better off doing the heavy lifting of writing the code yourself, to ensure you understand and also to ensure you don't fail for just copying code off the net.
Python is the most awesome pseudo-code language ever. It can easily read like normal English pseudo-code but has the added benefit that a computer can actually run it for testing and validation purposes :-)
The Python code is:
import sys
# Get desired sum from command line, with default.
try:
desiredSum = int(sys.argv[1])
except:
desiredSum = 22
# Init sevens to get at or beyond sum, fours to zero, and the sum.
(numSevens, numFours) = ((desiredSum + 6) // 7, 0)
thisSum = numSevens * 7 + numFours * 4
# Continue until a solution is found.
while thisSum != desiredSum:
if thisSum > desiredSum:
# Too high, remove a seven. If that's not possible, exit.
if numSevens == 0:
print(f"{desiredSum}: no solution")
sys.exit(0)
numSevens -= 1
thisSum -= 7
else:
# Too low, add a four.
numFours += 1
thisSum += 4
# Only get here if solution found, so print lowest
# possible number that matches four/seven count.
print(f"{desiredSum}: answer is {'4' * numFours}{'7' * numSevens}")
And here's a transcript of it in action for a small sample range:
pax:~> for i in {11..20} ; do ./test47.py ${i} ; done
11: answer is 47
12: answer is 444
13: no solution
14: answer is 77
15: answer is 447
16: answer is 4444
17: no solution
18: answer is 477
19: answer is 4447
20: answer is 44444
And here's the (rough) digit count for a desired sum of four billion, well over half a billion digits:
pax:~> export LC_NUMERIC=en_US.UTF8
pax:~> printf "%'.f\n" $(./test47.py 4000000000 | wc -c)
571,428,597
If you really need a C++ solution, see below. I wouldn't advise using this if this is course-work, instead suggesting you convert the algorithm shown above into your own code (for reasons previously mentioned). This is provided just to show the similar approach in C++:
#include <iostream>
int main(int argc, char *argv[]) {
// Get desired sum from command line, defaulting to 22.
int desiredSum = 22;
if (argc >= 2) desiredSum = atoi(argv[1]);
// Init sevens to get at or beyond desired sum, fours to zero,
// and the sum based on that.
int numSevens = (desiredSum + 6) / 7, numFours = 0;
int thisSum = numSevens * 7 + numFours * 4;
// Continue until a solution is found.
while (thisSum != desiredSum) {
if (thisSum > desiredSum) {
// Too high, remove a seven if possible, exit if not.
if (numSevens == 0) {
std::cout << desiredSum << ": no solution\n";
return 0;
}
--numSevens; thisSum -= 7;
} else {
// Too low, add a four.
++numFours; thisSum += 4;
}
}
// Only get here if solution found, so print lowest
// possible number that matches four / seven count.
std::cout << desiredSum << ": answer is ";
while (numFours-- > 0) std::cout << 4;
while (numSevens-- > 0) std::cout << 7;
std::cout << '\n';
}
Of course, you can be even more clever when you realise that the maximum number of fours will be six, and that you can add one to the sum-of-digits by removing one seven and adding two fours.
So simply:
work out the number of sevens required to get at or just below the desired sum;
add a single four if that will still keep you at or below the desired sum;
then adjust by enough actions of "remove one seven and add two fours" until you get to that desired sum (keeping in mind you may already be there). This will be done exactly once for each unit the shortfall in your current sum (how far it is below the desired sum) so, if the shortfall was two, you would remove two sevens and add four fours (- 14 + 16 = 2). That means you can use a simple mathematical formula rather than a loop.
if that formula results in a negative count of sevens, there was no solution, otherwise use the counts as previously mentioned to form the lowest number (fours followed by sevens).
Just Python for this solution, given how easy it is:
import sys
# Get desired number.
desiredNum = int(sys.argv[1])
# Work out seven and four counts as per description in text.
numSevens = int(desiredNum / 7) # Now within six of desired sum.
shortFall = desiredNum - (numSevens * 7)
numFours = int(shortFall / 4) # Now within three of desired sum.
shortFall = shortFall - numFours * 4
# Do enough '+7-4-4's to reach desired sum (none if already there).
numSevens = numSevens - shortFall
numFours = numFours + shortFall * 2
# Done, output solution, if any.
if numSevens < 0:
print(f"{desiredNum}: No solution")
else:
print(f"{desiredNum}: {'4' * numFours}{'7' * numSevens}")
That way, no loop is required at all. It's all mathematical reasoning.
If I understand the question correctly, you are searching for the smallest number x which contains only the numbers 4 and 7 and the sum of its digits N. The smallest number is for sure written as:
4...47...7
and consists of m times 4 and n times 7. So we know that N = n · 4 + m · 7.
Here are a couple of rules that apply:
(n + m) · 7 ≥ N :: This is evident, just replace all 4's by 7's.
(n + m) · 4 ≤ N :: This is evident, just replace all 7's by 4's.
(n + m) · 7 − N = m · (7 − 4) :: in other words (m+n) · 7 − N needs to be divisible by 7 − 4
So with these two conditions, we can now write the pseudo-code very quickly:
# always assume integer division
j = N/7 # j resembles n+m (total digits)
if (N*7 < N) j++ # ensure rule 1
while ( (j*4 <= N) AND ((j*7 - N)%(7-4) != 0) ) j++ # ensure rule 2 and rule 3
m = (j*7 - N)/(7-4) # integer division
n = j-m
if (m>=0 AND n>=0 AND N==m*4 + n*7) result found
Here is a quick bash-awk implementation:
$ for N in {1..30}; do
awk -v N=$N '
BEGIN{ j=int(N/7) + (N%7>0);
while( j*4<=N && (j*7-N)%3) j++;
m=int((j*7-N)/3); n=j-m;
s="no solution";
if (m>=0 && n>=0 && m*4+n*7==N) {
s=""; for(i=1;i<=j;++i) s=s sprintf("%d",(i<=m?4:7))
}
print N,s
}'
done
1 no solution
2 no solution
3 no solution
4 4
5 no solution
6 no solution
7 7
8 44
9 no solution
10 no solution
11 47
12 444
13 no solution
14 77
15 447
16 4444
17 no solution
18 477
19 4447
20 44444
21 777
22 4477
23 44447
24 444444
25 4777
26 44477
27 444447
28 7777
29 44777
30 444477
The constraints for sum are 1 ≤ n ≤ 106
It means that you might have to find and print numbers with more than 105 digits (106 / 7 ≅ 142,857). You can't store those in a fixed-sized integral type like long long, it's better to directly generate them as std::strings composed by only 4 and 7 characters.
Some mathematical properties may help in finding a suitable algorithm.
We know that n = i * 4 + j * 7.
Of all the possible numbers generated by each combination of i digits four and j digits seven, the minimum is the one with all the fours at left of all the sevens. E.g. 44777 < 47477 < 47747 < ... < 77744.
The minimal lucky number has at max six 4 digits, because, even if the sum of their digits is equal, 4444444 > 7777.
Now, let's introduce s = n / 7 (integer division) and r = n % 7 (the remainder).
If n is divisible by 7 (or when r == 0), the lucky number is composed only by exactly s digits (all 7).
If the remainder is not zero, we need to introduce some 4. Note that
If r == 4, we can just put a single 4 at the left of s sevens
Every time we substitute (if we can) a single 7 with two 4s, the sum of the digits increases by 1.
We can calculate exactly how many 4 digits we need (6 at max) without a loop.
This is enough to write an algorithm.
#include <string>
struct lucky_t
{
long fours, sevens;
};
// Find the minimum lucky number (composed by only 4 and 7 digits)
// that has the sum of digits equal to n.
// Returns it as a string, if exists, otherwise return "-1".
std::string minimum_lucky(long n)
{
auto const digits = [multiples = n / 7L, remainder = n % 7L] {
return remainder > 3
? lucky_t{remainder * 2 - 7, multiples - remainder + 4}
: lucky_t{remainder * 2, multiples - remainder};
} ();
if ( digits.fours < 0 || digits.sevens < 0 )
{
return "-1";
}
else
{
std::string result(digits.fours, '4');
result.append(digits.sevens, '7');
return result;
}
}
Tested here.
In spite of the fact that there are online plenty of algorithms and functions for generating unique combinations of any size from a list of unique items, there is none available in case of a list of non-unique items (i.e. list containing repetitions of same value.)
The question is how to generate ON-THE-FLY in a generator function all
the unique combinations from a non-unique list without the
computational expensive need of filtering out duplicates?
I consider combination comboA to be unique if there is no other combination comboB for which sorted lists for both combinations are the same. Let's give an example of code checking for such uniqueness:
comboA = [1,2,2]
comboB = [2,1,2]
print("B is a duplicate of A" if sorted(comboA)==sorted(comboB) else "A is unique compared to B")
In the above given example B is a duplicate of A and the print() prints B is a duplicate of A.
The problem of getting a generator function capable of providing unique combinations on-the-fly in case of a non-unique list is solved here: Getting unique combinations from a non-unique list of items, FASTER?, but the provided generator function needs lookups and requires memory what causes problems in case of a huge amount of combinations.
The in the current version of the answer provided function does the job without any lookups and appears to be the right answer here, BUT ...
The goal behind getting rid of lookups is to speed up the generation of unique combinations in case of a list with duplicates.
I have initially (writing the first version of this question) wrongly assumed that code which doesn't require creation of a set used for lookups needed to assure uniqueness is expected to give an advantage over code needing lookups. It is not the case. At least not always. The code in up to now provided answer does not using lookups, but is taking much more time to generate all the combinations in case of no redundant list or if only a few redundant items are in the list.
Here some timings to illustrate the current situation:
-----------------
k: 6 len(ls): 48
Combos Used Code Time
---------------------------------------------------------
12271512 len(list(combinations(ls,k))) : 2.036 seconds
12271512 len(list(subbags(ls,k))) : 50.540 seconds
12271512 len(list(uniqueCombinations(ls,k))) : 8.174 seconds
12271512 len(set(combinations(sorted(ls),k))): 7.233 seconds
---------------------------------------------------------
12271512 len(list(combinations(ls,k))) : 2.030 seconds
1 len(list(subbags(ls,k))) : 0.001 seconds
1 len(list(uniqueCombinations(ls,k))) : 3.619 seconds
1 len(set(combinations(sorted(ls),k))): 2.592 seconds
Above timings illustrate the two extremes: no duplicates and only duplicates. All other timings are between this two.
My interpretation of the results above is that a pure Python function (not using any C-compiled modules) can be extremely faster, but it can be also much slower depending on how many duplicates are in a list. So there is probably no way around writing C/C++ code for a Python .so extension module providing the required functionality.
Instead of post-processing/filtering your output, you can pre-process your input list. This way, you can avoid generating duplicates in the first place. Pre-processing involves either sorting (or using a collections.Counter on) the input. One possible recursive realization is:
def subbags(bag, k):
a = sorted(bag)
n = len(a)
sub = []
def index_of_next_unique_item(i):
j = i + 1
while j < n and a[j] == a[i]:
j += 1
return j
def combinate(i):
if len(sub) == k:
yield tuple(sub)
elif n - i >= k - len(sub):
sub.append(a[i])
yield from combinate(i + 1)
sub.pop()
yield from combinate(index_of_next_unique_item(i))
yield from combinate(0)
bag = [1, 2, 3, 1, 2, 1]
k = 3
i = -1
print(sorted(bag), k)
print('---')
for i, subbag in enumerate(subbags(bag, k)):
print(subbag)
print('---')
print(i + 1)
Output:
[1, 1, 1, 2, 2, 3] 3
---
(1, 1, 1)
(1, 1, 2)
(1, 1, 3)
(1, 2, 2)
(1, 2, 3)
(2, 2, 3)
---
6
Requires some stack space for the recursion, but this + sorting the input should use substantially less time + memory than generating and discarding repeats.
The current state-of-the-art inspired initially by a 50 than by a 100 reps bounties is at the moment (instead of a Python extension module written entirely in C):
An efficient algorithm and implementation that is better than the obvious (set + combinations) approach in the best (and average) case, and is competitive with it in the worst case.
It seems to be possible to fulfill this requirement using a kind of "fake it before you make it" approach. The current state-of-the-art is that there are two generator function algorithms available for solving the problem of getting unique combinations in case of a non-unique list. The below provided algorithm combines both of them what becomes possible because it seems to exist a threshold value for percentage of unique items in the list which can be used for appropriate switching between the two algorithms. The calculation of the percentage of uniqueness is done with so tiny amount of computation time that it even doesn't clearly show up in the final results due to common variation of the taken timing.
def iterFastUniqueCombos(lstList, comboSize, percUniqueThresh=60):
lstListSorted = sorted(lstList)
lenListSorted = len(lstListSorted)
percUnique = 100.0 - 100.0*(lenListSorted-len(set(lstListSorted)))/lenListSorted
lstComboCandidate = []
setUniqueCombos = set()
def idxNextUnique(idxItemOfList):
idxNextUniqueCandidate = idxItemOfList + 1
while (
idxNextUniqueCandidate < lenListSorted
and
lstListSorted[idxNextUniqueCandidate] == lstListSorted[idxItemOfList]
): # while
idxNextUniqueCandidate += 1
idxNextUnique = idxNextUniqueCandidate
return idxNextUnique
def combinate(idxItemOfList):
if len(lstComboCandidate) == sizeOfCombo:
yield tuple(lstComboCandidate)
elif lenListSorted - idxItemOfList >= sizeOfCombo - len(lstComboCandidate):
lstComboCandidate.append(lstListSorted[idxItemOfList])
yield from combinate(idxItemOfList + 1)
lstComboCandidate.pop()
yield from combinate(idxNextUnique(idxItemOfList))
if percUnique > percUniqueThresh:
from itertools import combinations
allCombos = combinations(lstListSorted, comboSize)
for comboCandidate in allCombos:
if comboCandidate in setUniqueCombos:
continue
yield comboCandidate
setUniqueCombos.add(comboCandidate)
else:
yield from combinate(0)
#:if/else
#:def iterFastUniqueCombos()
The below provided timings show that the above iterFastUniqueCombos() generator function provides a clear advantage
over uniqueCombinations() variant in case the list has less than 60 percent of unique elements and is not worse as
the on (set + combinations) based uniqueCombinations() generator function in the opposite case where it gets much faster than the iterUniqueCombos() one (due to switching between
the (set + combinations) and the (no lookups) variant at 60% threshold for amount of unique elements in the list):
=========== sizeOfCombo: 6 sizeOfList: 48 noOfUniqueInList 1 percUnique 2
Combos: 12271512 print(len(list(combinations(lst,k)))) : 2.04968 seconds.
Combos: 1 print(len(list( iterUniqueCombos(lst,k)))) : 0.00011 seconds.
Combos: 1 print(len(list( iterFastUniqueCombos(lst,k)))) : 0.00008 seconds.
Combos: 1 print(len(list( uniqueCombinations(lst,k)))) : 3.61812 seconds.
========== sizeOfCombo: 6 sizeOfList: 48 noOfUniqueInList 48 percUnique 100
Combos: 12271512 print(len(list(combinations(lst,k)))) : 1.99383 seconds.
Combos: 12271512 print(len(list( iterUniqueCombos(lst,k)))) : 49.72461 seconds.
Combos: 12271512 print(len(list( iterFastUniqueCombos(lst,k)))) : 8.07997 seconds.
Combos: 12271512 print(len(list( uniqueCombinations(lst,k)))) : 8.11974 seconds.
========== sizeOfCombo: 6 sizeOfList: 48 noOfUniqueInList 27 percUnique 56
Combos: 12271512 print(len(list(combinations(lst,k)))) : 2.02774 seconds.
Combos: 534704 print(len(list( iterUniqueCombos(lst,k)))) : 1.60052 seconds.
Combos: 534704 print(len(list( iterFastUniqueCombos(lst,k)))) : 1.62002 seconds.
Combos: 534704 print(len(list( uniqueCombinations(lst,k)))) : 3.41156 seconds.
========== sizeOfCombo: 6 sizeOfList: 48 noOfUniqueInList 31 percUnique 64
Combos: 12271512 print(len(list(combinations(lst,k)))) : 2.03539 seconds.
Combos: 1114062 print(len(list( iterUniqueCombos(lst,k)))) : 3.49330 seconds.
Combos: 1114062 print(len(list( iterFastUniqueCombos(lst,k)))) : 3.64474 seconds.
Combos: 1114062 print(len(list( uniqueCombinations(lst,k)))) : 3.61857 seconds.
Let say I've a system that distribute 8820 values into 96 values, rounding using Banker's Round (call them pulse). The formula is:
pulse = BankerRound(8820 * i/96), with i[0,96[
Thus, this is the list of pulses:
0
92
184
276
368
459
551
643
735
827
919
1011
1102
1194
1286
1378
1470
1562
1654
1746
1838
1929
2021
2113
2205
2297
2389
2481
2572
2664
2756
2848
2940
3032
3124
3216
3308
3399
3491
3583
3675
3767
3859
3951
4042
4134
4226
4318
4410
4502
4594
4686
4778
4869
4961
5053
5145
5237
5329
5421
5512
5604
5696
5788
5880
5972
6064
6156
6248
6339
6431
6523
6615
6707
6799
6891
6982
7074
7166
7258
7350
7442
7534
7626
7718
7809
7901
7993
8085
8177
8269
8361
8452
8544
8636
8728
Now, suppose the system doesn't send to me these pulses directly. Instead, it send these pulse in 8820th (call them tick):
tick = value * 1/8820
The list of the ticks I get become:
0
0.010430839
0.020861678
0.031292517
0.041723356
0.052040816
0.062471655
0.072902494
0.083333333
0.093764172
0.104195011
0.11462585
0.124943311
0.13537415
0.145804989
0.156235828
0.166666667
0.177097506
0.187528345
0.197959184
0.208390023
0.218707483
0.229138322
0.239569161
0.25
0.260430839
0.270861678
0.281292517
0.291609977
0.302040816
0.312471655
0.322902494
0.333333333
0.343764172
0.354195011
0.36462585
0.375056689
0.38537415
0.395804989
0.406235828
0.416666667
0.427097506
0.437528345
0.447959184
0.458276644
0.468707483
0.479138322
0.489569161
0.5
0.510430839
0.520861678
0.531292517
0.541723356
0.552040816
0.562471655
0.572902494
0.583333333
0.593764172
0.604195011
0.61462585
0.624943311
0.63537415
0.645804989
0.656235828
0.666666667
0.677097506
0.687528345
0.697959184
0.708390023
0.718707483
0.729138322
0.739569161
0.75
0.760430839
0.770861678
0.781292517
0.791609977
0.802040816
0.812471655
0.822902494
0.833333333
0.843764172
0.854195011
0.86462585
0.875056689
0.88537415
0.895804989
0.906235828
0.916666667
0.927097506
0.937528345
0.947959184
0.958276644
0.968707483
0.979138322
0.989569161
Unfortunately, between these ticks it sends to me also fake ticks, that aren't multiply of original pulses. Such as 0,029024943, which is multiply of 256, which isn't in the pulse lists.
How can I find from this list which ticks are valid and which are fake?
I don't have the pulse list to compare with during the process, since 8820 will change during the time, so I don't have a list to compare step by step. I need to deduce it from ticks at each iteration.
What's the best math approch to this? Maybe reasoning only in tick and not pulse.
I've thought to find the closer error between nearest integer pulse and prev/next tick. Here in C++:
double pulse = tick * 96.;
double prevpulse = (tick - 1/8820.) * 96.;
double nextpulse = (tick + 1/8820.) * 96.;
int pulseRounded=round(pulse);
int buffer=lrint(tick * 8820.);
double pulseABS = abs(pulse - pulseRounded);
double prevpulseABS = abs(prevpulse - pulseRounded);
double nextpulseABS = abs(nextpulse - pulseRounded);
if (nextpulseABS > pulseABS && prevpulseABS > pulseABS) {
// is pulse
}
but for example tick 0.0417234 (pulse 368) fails since the prev tick error seems to be closer than it: prevpulseABS error (0.00543795) is smaller than pulseABS error (0.0054464).
That's because this comparison doesn't care about rounding I guess.
NEW POST:
Alright. Based on what I now understand, here's my revised answer.
You have the information you need to build a list of good values. Each time you switch to a new track:
vector<double> good_list;
good_list.reserve(96);
for(int i = 0; i < 96; i++)
good_list.push_back(BankerRound(8820.0 * i / 96.0) / 8820.0);
Then, each time you want to validate the input:
auto iter = find(good_list.begin(), good_list.end(), input);
if(iter != good_list.end()) //It's a match!
cout << "Happy days! It's a match!" << endl;
else
cout << "Oh bother. It's not a match." << endl;
The problem with mathematically determining the correct pulses is the BankerRound() function which will introduce an ever-growing error the higher values you input. You would then need a formula for a formula, and that's getting out of my wheelhouse. Or, you could keep track of the differences between successive values. Most of them would be the same. You'd only have to check between two possible errors. But that falls apart if you can jump tracks or jump around in one track.
OLD POST:
If I understand the question right, the only information you're getting should be coming in the form of (p/v = y) where you know 'y' (that's each element in your list of ticks you get from the device) and you know that 'p' is the Pulse and 'v' is the Values per Beat, but you don't know what either of them are. So, pulling one point of data from your post, you might have an equation like this:
p/v = 0.010430839
'v', in all the examples you've used thus far, is 8820, but from what I understand, that value is not a guaranteed constant. The next question then is: Do you have a way of determining what 'v' is before you start getting all these decimal values? If you do, you can work out mathematically what the smallest error can be (1/v) then take your decimal information, multiply it by 'v', round it to the nearest whole number and check to see if the difference between its rounded form and its non-rounded form falls in the bounds of your calculated error like so:
double input; //let input be elements in your list of doubles, such as 0.010430839
double allowed_error = 1.0 / values_per_beat;
double proposed = input * values_per_beat;
double rounded = std::round(proposed);
if(abs(rounded - proposed) < allowed_error){cout << "It's good!" << endl;}
If, however, you are not able to ascertain the values_per_beat ahead of time, then this becomes a statistical question. You must accumulate enough data samples, remove the outliers (the few that vary from the norm) and use that data. But that approach will not be realtime, which, given the terms you've been using (values per beat, bpm, the value 44100) it sounds like realtime might be what you're after.
Playing around with Excel, I think you want to multiply up to (what should be) whole numbers rather than looking for closest pulses.
Tick Pulse i Error OK
Tick*8820 Pulse*96/8820 ABS( i - INT( i+0.05 ) ) Error < 0.01
------------ ------------ ------------- ------------------------ ------------
0.029024943 255.9999973 2.786394528 0.786394528 FALSE
0.0417234 368.000388 4.0054464 0.0054464 TRUE
0 0 0 0 TRUE
0.010430839 91.99999998 1.001360544 0.001360544 TRUE
0.020861678 184 2.002721088 0.002721088 TRUE
0.031292517 275.9999999 3.004081632 0.004081632 TRUE
0.041723356 367.9999999 4.005442176 0.005442176 TRUE
0.052040816 458.9999971 4.995918336 0.004081664 TRUE
0.062471655 550.9999971 5.99727888 0.00272112 TRUE
0.072902494 642.9999971 6.998639424 0.001360576 TRUE
0.083333333 734.9999971 7.999999968 3.2E-08 TRUE
The table shows your two "problem" cases (the real wrong value, 256, and the one your code gets wrong, 368) followed by the first few "good" values.
If both 8820s vary at the same time, then obviously they will cancel out, and i will just be Tick*96.
The Error term is the difference between the calculated i and the nearest integer; if this less than 0.01, then it is a "good" value.
NOTE: the 0.05 and 0.01 values were chosen somewhat arbitrarily (aka inspired first time guess based on the numbers): adjust if needed. Although I've only shown the first few rows, all the 96 "good" values you gave show as TRUE.
The code (completely untested) would be something like:
double pulse = tick * 8820.0 ;
double i = pulse * 96.0 / 8820.0 ;
double error = abs( i - floor( i + 0.05 ) ) ;
if( error < 0.05 ) {
// is pulse
}
I assume your initializing your pulses in a for-loop, using int i as loop variable; then the problem is this line:
BankerRound(8820 * i/96);
8820 * i / 96 is an all integer operation and the result is integer again, cutting off the remainder (so in effect, always rounding towards zero already), and BankerRound actually has nothing to round any more. Try this instead:
BankerRound(8820 * i / 96.0);
Same problem applies if you are trying to calculate prev and next pulse, as you actually subtract and add 0 (again, 1/8820 is all integer and results in 0).
Edit:
From what I read from the commments, the 'system' is not – as I assumed previously – modifiable. Actually, it calculates ticks in the form of n / 96.0, n ∊ [0, 96) in ℕ
however including some kind of internal rounding appearently independent from the sample frequency, so there is some difference to the true value of n/96.0 and the ticks multiplied by 96 do not deliver exactly the integral values in [0, 96) (thanks KarstenKoop). And some of the delivered samples are simply invalid...
So the task is to detect, if tick * 96 is close enough to an integral value to be accepted as valid.
So we need to check:
double value = tick * 96.0;
bool isValid
= value - floor(value) < threshold
|| ceil(value) - value < threshold;
with some appropriately defined threshold. Assuming the values really are calculated as
double tick = round(8820*i/96.0)/8820.0;
then the maximal deviation would be slightly greater than 0.00544 (see below for a more exact value), thresholds somewhere in the sizes of 0.006, 0.0055, 0.00545, ... might be a choice.
Rounding might be a matter of internally used number of bits for the sensor value (if we have 13 bits available, ticks might actually be calculated as floor(8192 * i / 96.0) / 8192.0 with 8192 being 1 << 13 &ndash and floor accounting to integer division; just a guess...).
The exact value of the maximal deviation, using 8820 as factor, as exact as representable by double, was:
0.00544217687075132516838493756949901580810546875
The multiplication by 96 is actually not necessary, you can compare directly with the threshold divided by 96, which would be:
0.0000566893424036596371706764330156147480010986328125
I have this program that is supposed to search for perfect numbers.
(X is a perfect number if the sum of all numbers that divide X, divided by 2 is equal to X)
sum/2 = x
Now It has found the first four, which were known in Ancient Greece, so it's not really a anything awesome.
The next one should be 33550336.
I know it is a big number, but the program has been going for about 50 minutes, and still hasn't found 33550336.
Is it because I opened the .txt file where I store all the perfect numbers while the program was running, or is it because I don't have a PC fast enough to run it*, or because I'm using Python?
*NOTE: This same PC factorized 500 000 in 10 minutes (while also running the perfect number program and Google Chrome with 3 YouTube tabs), also using Python.
Here is the code to the program:
i = 2
a = open("perfect.txt", 'w')
a.close()
while True:
sum = 0
for x in range(1, i+1):
if i%x == 0:
sum += x
if sum / 2 == i:
a = open("perfect.txt", 'a')
a.write(str(i) + "\n")
a.close()
i += 1
The next one should be 33550336.
Your code (I fixed the indentation so that it does in principle what you want):
i = 2
a = open("perfect.txt", 'w')
a.close()
while True:
sum = 0
for x in range(1, i+1):
if i%x == 0:
sum += x
if sum / 2 == i:
a = open("perfect.txt", 'a')
a.write(str(i) + "\n")
a.close()
i += 1
does i divisions to find the divisors of i.
So to find the perfect numbers up to n, it does
2 + 3 + 4 + ... + (n-1) + n = n*(n+1)/2 - 1
divisions in the for loop.
Now, for n = 33550336, that would be
Prelude> 33550336 * (33550336 + 1) `quot` 2 - 1
562812539631615
roughly 5.6 * 1014 divisions.
Assuming your CPU could do 109 divisions per second (it most likely can't, 108 is a better estimate in my experience, but even that is for machine ints in C), that would take about 560,000 seconds. One day has 86400 seconds, so that would be roughly six and a half days (more than two months with the 108 estimate).
Your algorithm is just too slow to reach that in reasonable time.
If you don't want to use number-theory (even perfect numbers have a very simple structure, and if there are any odd perfect numbers, those are necessarily huge), you can still do better by dividing only up to the square root to find the divisors,
i = 2
a = open("perfect.txt", 'w')
a.close()
while True:
sum = 1
root = int(i**0.5)
for x in range(2, root+1):
if i%x == 0:
sum += x + i/x
if i == root*root:
sum -= x # if i is a square, we have counted the square root twice
if sum == i:
a = open("perfect.txt", 'a')
a.write(str(i) + "\n")
a.close()
i += 1
that only needs about 1.3 * 1011 divisions and should find the fifth perfect number in a couple of hours.
Without resorting to the explicit formula for even perfect numbers (2^(p-1) * (2^p - 1) for primes p such that 2^p - 1 is prime), you can speed it up somewhat by finding the prime factorisation of i and computing the divisor sum from that. That will make the test faster for all composite numbers, and much faster for most,
def factorisation(n):
facts = []
multiplicity = 0
while n%2 == 0:
multiplicity += 1
n = n // 2
if multiplicity > 0:
facts.append((2,multiplicity))
d = 3
while d*d <= n:
if n % d == 0:
multiplicity = 0
while n % d == 0:
multiplicity += 1
n = n // d
facts.append((d,multiplicity))
d += 2
if n > 1:
facts.append((n,1))
return facts
def divisorSum(n):
f = factorisation(n)
sum = 1
for (p,e) in f:
sum *= (p**(e+1) - 1)/(p-1)
return sum
def isPerfect(n):
return divisorSum(n) == 2*n
i = 2
count = 0
out = 10000
while count < 5:
if isPerfect(i):
print i
count += 1
if i == out:
print "At",i
out *= 5
i += 1
would take an estimated 40 minutes on my machine.
Not a bad estimate:
$ time python fastperf.py
6
28
496
8128
33550336
real 36m4.595s
user 36m2.001s
sys 0m0.453s
It is very hard to try and deduce why this has happened. I would suggest that you run your program either under a debugger and test several iteration manually to check if the code is really correct (I know you have already calculated 4 numbers but still). Alternatively it would be good to run your program under a python profiler just to see if it hasn't accidentally blocked on a lock or something.
It is possible, but not likely that this is an issue related to you opening the file while it is running. If it was an issue, there would have probably been some error message and/or program close/crash.
I would edit the program to write a log-type output to a file every so often. For example, everytime you have processed a target number that is an even multiple of 1-Million, write (open-append-close) the date-time and current-number and last-success-number to a log file.
You could then Type the file once in a while to measure progress.
I'm pretty new to Python (just started teaching myself a week ago), so my debugging skills are weak right now. I tried to make a program that would ask a user-submitted number of randomly-generated multiplication questions, with factors between 0 and 12, like a multiplication table test.
import math
import random
#establish a number of questions
questions = int(input("\n How many questions do you want? "))
#introduce score
score = 1
for question in range(questions):
x = random.randrange(0,13)
y = random.randrange(0,13)
#make the numbers strings, so they can be printed with strings
abc = str(x)
cba = str(y)
print("What is " + abc + "*" + cba +"?")
z = int(input("Answer here: "))
print z
a = x*y
#make the answer a string, so it can be printed if you get one wrong
answer = str(a)
if z > a or z < a:
print ("wrong, the answer is " + answer)
print("\n")
#this is the line that's being skipped
score = score - 1/questions
else:
print "Correct!"
print ("\n")
finalscore = score*100
finalestscore = str(finalscore)
print (finalestscore + "%")
The idea was that every time the user gets a question wrong, score (set to 1) goes down by 1/question,so when multiplied by 100 it gives a percentage of questions wrong. However, no matter the number of questions or the number gotten wrong, score remains 1, so finalestscore remains 100. Line 26 used to be:
if math.abs(z)-math.abs(a) != 0:
but 2.7.3 apparently doesn't acknowledge that math has an abs function.
Such a simple accumulator pattern doesn't seem like it would be an issue, even for an older version of Python. Help?
Try score = score - 1.0/questions
The problem is that you're doing integer division, which truncates to the nearest integer, so 1/questions will always give 0.
The problem is that you are using integers for all of your calculations. In particular, when you calculate 1/questions, it truncates (rounds down) to an integer because both values in the calculation are integers.
To avoid this, you could instead use 1.0/questions to make the calculations use floating point numbers instead (and not truncate)