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I have two ordered lists of consecutive integers m=0, 1, ... M and n=0, 1, 2, ... N. Each value of m has a probability pm, and each value of n has a probability pn. I am trying to find the ordered list of unique values r=n/m and their probabilities pr. I am aware that r is infinite if n=0 and can even be undefined if m=n=0.
In practice, I would like to run for M and N each be of the order of 2E4, meaning up to 4E8 values of r - which would mean 3 GB of floats (assuming 8 Bytes/float).
For this calculation, I have written the python code below.
The idea is to iterate over m and n, and for each new m/n, insert it in the right place with its probability if it isn't there yet, otherwise add its probability to the existing number. My assumption is that it is easier to sort things on the way instead of waiting until the end.
The cases related to 0 are added at the end of the loop.
I am using the Fraction class since we are dealing with fractions.
The code also tracks the multiplicity of each unique value of m/n.
I have tested up to M=N=100, and things are quite slow. Are there better approaches to the question, or more efficient ways to tackle the code?
Timing:
M=N=30: 1 s
M=N=50: 6 s
M=N=80: 30 s
M=N=100: 82 s
import numpy as np
from fractions import Fraction
import time # For timiing
start_time = time.time() # Timing
M, N = 6, 4
mList, nList = np.arange(1, M+1), np.arange(1, N+1) # From 1 to M inclusive, deal with 0 later
mProbList, nProbList = [1/(M+1)]*(M), [1/(N+1)]*(N) # Probabilities, here assumed equal (not general case)
# Deal with mn=0 later
pmZero, pnZero = 1/(M+1), 1/(N+1) # P(m=0) and P(n=0)
pNaN = pmZero * pnZero # P(0/0) = P(m=0)P(n=0)
pZero = pmZero * (1 - pnZero) # P(0) = P(m=0)P(n!=0)
pInf = pnZero * (1 - pmZero) # P(inf) = P(m!=0)P(n=0)
# Main list of r=m/n, P(r) and mult(r)
# Start with first line, m=1
rList = [Fraction(mList[0], n) for n in nList[::-1]] # Smallest first
rProbList = [mProbList[0] * nP for nP in nProbList[::-1]] # Start with first line
rMultList = [1] * len(rList) # Multiplicity of each element
# Main loop
for m, mP in zip(mList[1:], mProbList[1:]):
for n, nP in zip(nList[::-1], nProbList[::-1]): # Pick an n value
r, rP, rMult = Fraction(m, n), mP*nP, 1
for i in range(len(rList)-1): # See where it fits in existing list
if r < rList[i]:
rList.insert(i, r)
rProbList.insert(i, rP)
rMultList.insert(i, 1)
break
elif r == rList[i]:
rProbList[i] += rP
rMultList[i] += 1
break
elif r < rList[i+1]:
rList.insert(i+1, r)
rProbList.insert(i+1, rP)
rMultList.insert(i+1, 1)
break
elif r == rList[i+1]:
rProbList[i+1] += rP
rMultList[i+1] += 1
break
if r > rList[-1]:
rList.append(r)
rProbList.append(rP)
rMultList.append(1)
break
# Deal with 0
rList.insert(0, Fraction(0, 1))
rProbList.insert(0, pZero)
rMultList.insert(0, N)
# Deal with infty
rList.append(np.Inf)
rProbList.append(pInf)
rMultList.append(M)
# Deal with undefined case
rList.append(np.NAN)
rProbList.append(pNaN)
rMultList.append(1)
print(".... done in %s seconds." % round(time.time() - start_time, 2))
print("************** Final list\nr", 'Prob', 'Mult')
for r, rP, rM in zip(rList, rProbList, rMultList): print(r, rP, rM)
print("************** Checks")
print("mList", mList, 'nList', nList)
print("Sum of proba = ", np.sum(rProbList))
print("Sum of multi = ", np.sum(rMultList), "\t(M+1)*(N+1) = ", (M+1)*(N+1))
Based on the suggestion of #Prune, and on this thread about merging lists of tuples, I have modified the code as below. It's a lot easier to read, and runs about an order of magnitude faster for N=M=80 (I have omitted dealing with 0 - would be done same way as in original post). I assume there may be ways to tweak the merge and conversion back to lists further yet.
# Do calculations
data = [(Fraction(m, n), mProb(m) * nProb(n)) for n in range(1, N+1) for m in range(1, M+1)]
data.sort()
# Merge duplicates using a dictionary
d = {}
for r, p in data:
if not (r in d): d[r] = [0, 0]
d[r][0] += p
d[r][1] += 1
# Convert back to lists
rList, rProbList, rMultList = [], [], []
for k in d:
rList.append(k)
rProbList.append(d[k][0])
rMultList.append(d[k][1])
I expect that "things are quite slow" because you've chosen a known inefficient sort. A single list insertion is O(K) (later list elements have to be bumped over, and there is added storage allocation on a regular basis). Thus a full-list insertion sort is O(K^2). For your notation, that is O((M*N)^2).
If you want any sort of reasonable performance, research and use the best-know methods. The most straightforward way to do this is to make your non-exception results as a simple list comprehension, and use the built-in sort for your penultimate list. Simply append your n=0 cases, and you're done in O(K log K) time.
I the expression below, I've assumed functions for m and n probabilities.
This is a notational convenience; you know how to directly compute them, and can substitute those expressions if you wish.
data = [ (mProb(m) * nProb(n), Fraction(m, n))
for n in range(1, N+1)
for m in range(0, M+1) ]
data.sort()
data.extend([ # generate your "zero" cases here ])
I want to get the length of a string including a part of the string that represents its own length without padding or using structs or anything like that that forces fixed lengths.
So for example I want to be able to take this string as input:
"A string|"
And return this:
"A string|11"
On the basis of the OP tolerating such an approach (and to provide an implementation technique for the eventual python answer), here's a solution in Java.
final String s = "A String|";
int n = s.length(); // `length()` returns the length of the string.
String t; // the result
do {
t = s + n; // append the stringified n to the original string
if (n == t.length()){
return t; // string length no longer changing; we're good.
}
n = t.length(); // n must hold the total length
} while (true); // round again
The problem of, course, is that in appending n, the string length changes. But luckily, the length only ever increases or stays the same. So it will converge very quickly: due to the logarithmic nature of the length of n. In this particular case, the attempted values of n are 9, 10, and 11. And that's a pernicious case.
A simple solution is :
def addlength(string):
n1=len(string)
n2=len(str(n1))+n1
n2 += len(str(n2))-len(str(n1)) # a carry can arise
return string+str(n2)
Since a possible carry will increase the length by at most one unit.
Examples :
In [2]: addlength('a'*8)
Out[2]: 'aaaaaaaa9'
In [3]: addlength('a'*9)
Out[3]: 'aaaaaaaaa11'
In [4]: addlength('a'*99)
Out[4]: 'aaaaa...aaa102'
In [5]: addlength('a'*999)
Out[5]: 'aaaa...aaa1003'
Here is a simple python port of Bathsheba's answer :
def str_len(s):
n = len(s)
t = ''
while True:
t = s + str(n)
if n == len(t):
return t
n = len(t)
This is a much more clever and simple way than anything I was thinking of trying!
Suppose you had s = 'abcdefgh|, On the first pass through, t = 'abcdefgh|9
Since n != len(t) ( which is now 10 ) it goes through again : t = 'abcdefgh|' + str(n) and str(n)='10' so you have abcdefgh|10 which is still not quite right! Now n=len(t) which is finally n=11 you get it right then. Pretty clever solution!
It is a tricky one, but I think I've figured it out.
Done in a hurry in Python 2.7, please fully test - this should handle strings up to 998 characters:
import sys
orig = sys.argv[1]
origLen = len(orig)
if (origLen >= 98):
extra = str(origLen + 3)
elif (origLen >= 8):
extra = str(origLen + 2)
else:
extra = str(origLen + 1)
final = orig + extra
print final
Results of very brief testing
C:\Users\PH\Desktop>python test.py "tiny|"
tiny|6
C:\Users\PH\Desktop>python test.py "myString|"
myString|11
C:\Users\PH\Desktop>python test.py "myStringWith98Characters.........................................................................|"
myStringWith98Characters.........................................................................|101
Just find the length of the string. Then iterate through each value of the number of digits the length of the resulting string can possibly have. While iterating, check if the sum of the number of digits to be appended and the initial string length is equal to the length of the resulting string.
def get_length(s):
s = s + "|"
result = ""
len_s = len(s)
i = 1
while True:
candidate = len_s + i
if len(str(candidate)) == i:
result = s + str(len_s + i)
break
i += 1
This code gives the result.
I used a few var, but at the end it shows the output you want:
def len_s(s):
s = s + '|'
b = len(s)
z = s + str(b)
length = len(z)
new_s = s + str(length)
new_len = len(new_s)
return s + str(new_len)
s = "A string"
print len_s(s)
Here's a direct equation for this (so it's not necessary to construct the string). If s is the string, then the length of the string including the length of the appended length will be:
L1 = len(s) + 1 + int(log10(len(s) + 1 + int(log10(len(s)))))
The idea here is that a direct calculation is only problematic when the appended length will push the length past a power of ten; that is, at 9, 98, 99, 997, 998, 999, 9996, etc. To work this through, 1 + int(log10(len(s))) is the number of digits in the length of s. If we add that to len(s), then 9->10, 98->100, 99->101, etc, but still 8->9, 97->99, etc, so we can push past the power of ten exactly as needed. That is, adding this produces a number with the correct number of digits after the addition. Then do the log again to find the length of that number and that's the answer.
To test this:
from math import log10
def find_length(s):
L1 = len(s) + 1 + int(log10(len(s) + 1 + int(log10(len(s)))))
return L1
# test, just looking at lengths around 10**n
for i in range(9):
for j in range(30):
L = abs(10**i - j + 10) + 1
s = "a"*L
x0 = find_length(s)
new0 = s+`x0`
if len(new0)!=x0:
print "error", len(s), x0, log10(len(s)), log10(x0)
I am writing some quiz game and need computer to solve 1 game in the quiz if players fail to solve it.
Given data :
List of 6 numbers to use, for example 4, 8, 6, 2, 15, 50.
Targeted value, where 0 < value < 1000, for example 590.
Available operations are division, addition, multiplication and division.
Parentheses can be used.
Generate mathematical expression which evaluation is equal, or as close as possible, to the target value. For example for numbers given above, expression could be : (6 + 4) * 50 + 15 * (8 - 2) = 590
My algorithm is as follows :
Generate all permutations of all the subsets of the given numbers from (1) above
For each permutation generate all parenthesis and operator combinations
Track the closest value as algorithm runs
I can not think of any smart optimization to the brute-force algorithm above, which will speed it up by the order of magnitude. Also I must optimize for the worst case, because many quiz games will be run simultaneously on the server.
Code written today to solve this problem is (relevant stuff extracted from the project) :
from operator import add, sub, mul, div
import itertools
ops = ['+', '-', '/', '*']
op_map = {'+': add, '-': sub, '/': div, '*': mul}
# iterate over 1 permutation and generates parentheses and operator combinations
def iter_combinations(seq):
if len(seq) == 1:
yield seq[0], str(seq[0])
else:
for i in range(len(seq)):
left, right = seq[:i], seq[i:] # split input list at i`th place
# generate cartesian product
for l, l_str in iter_combinations(left):
for r, r_str in iter_combinations(right):
for op in ops:
if op_map[op] is div and r == 0: # cant divide by zero
continue
else:
yield op_map[op](float(l), r), \
('(' + l_str + op + r_str + ')')
numbers = [4, 8, 6, 2, 15, 50]
target = best_value = 590
best_item = None
for i in range(len(numbers)):
for current in itertools.permutations(numbers, i+1): # generate perms
for value, item in iter_combinations(list(current)):
if value < 0:
continue
if abs(target - value) < best_value:
best_value = abs(target - value)
best_item = item
print best_item
It prints : ((((4*6)+50)*8)-2). Tested it a little with different values and it seems to work correctly. Also I have a function to remove unnecessary parenthesis but it is not relevant to the question so it is not posted.
Problem is that this runs very slowly because of all this permutations, combinations and evaluations. On my mac book air it runs for a few minutes for 1 example. I would like to make it run in a few seconds tops on the same machine, because many quiz game instances will be run at the same time on the server. So the questions are :
Can I speed up current algorithm somehow (by orders of magnitude)?
Am I missing on some other algorithm for this problem which would run much faster?
You can build all the possible expression trees with the given numbers and evalate them. You don't need to keep them all in memory, just print them when the target number is found:
First we need a class to hold the expression. It is better to design it to be immutable, so its value can be precomputed. Something like this:
class Expr:
'''An Expr can be built with two different calls:
-Expr(number) to build a literal expression
-Expr(a, op, b) to build a complex expression.
There a and b will be of type Expr,
and op will be one of ('+','-', '*', '/').
'''
def __init__(self, *args):
if len(args) == 1:
self.left = self.right = self.op = None
self.value = args[0]
else:
self.left = args[0]
self.right = args[2]
self.op = args[1]
if self.op == '+':
self.value = self.left.value + self.right.value
elif self.op == '-':
self.value = self.left.value - self.right.value
elif self.op == '*':
self.value = self.left.value * self.right.value
elif self.op == '/':
self.value = self.left.value // self.right.value
def __str__(self):
'''It can be done smarter not to print redundant parentheses,
but that is out of the scope of this problem.
'''
if self.op:
return "({0}{1}{2})".format(self.left, self.op, self.right)
else:
return "{0}".format(self.value)
Now we can write a recursive function that builds all the possible expression trees with a given set of expressions, and prints the ones that equals our target value. We will use the itertools module, that's always fun.
We can use itertools.combinations() or itertools.permutations(), the difference is in the order. Some of our operations are commutative and some are not, so we can use permutations() and assume we will get many very simmilar solutions. Or we can use combinations() and manually reorder the values when the operation is not commutative.
import itertools
OPS = ('+', '-', '*', '/')
def SearchTrees(current, target):
''' current is the current set of expressions.
target is the target number.
'''
for a,b in itertools.combinations(current, 2):
current.remove(a)
current.remove(b)
for o in OPS:
# This checks whether this operation is commutative
if o == '-' or o == '/':
conmut = ((a,b), (b,a))
else:
conmut = ((a,b),)
for aa, bb in conmut:
# You do not specify what to do with the division.
# I'm assuming that only integer divisions are allowed.
if o == '/' and (bb.value == 0 or aa.value % bb.value != 0):
continue
e = Expr(aa, o, bb)
# If a solution is found, print it
if e.value == target:
print(e.value, '=', e)
current.add(e)
# Recursive call!
SearchTrees(current, target)
# Do not forget to leave the set as it were before
current.remove(e)
# Ditto
current.add(b)
current.add(a)
And then the main call:
NUMBERS = [4, 8, 6, 2, 15, 50]
TARGET = 590
initial = set(map(Expr, NUMBERS))
SearchTrees(initial, TARGET)
And done! With these data I'm getting 719 different solutions in just over 21 seconds! Of course many of them are trivial variations of the same expression.
24 game is 4 numbers to target 24, your game is 6 numbers to target x (0 < x < 1000).
That's much similar.
Here is the quick solution, get all results and print just one in my rMBP in about 1-3s, I think one solution print is ok in this game :), I will explain it later:
def mrange(mask):
#twice faster from Evgeny Kluev
x = 0
while x != mask:
x = (x - mask) & mask
yield x
def f( i ) :
global s
if s[i] :
#get cached group
return s[i]
for x in mrange(i & (i - 1)) :
#when x & i == x
#x is a child group in group i
#i-x is also a child group in group i
fk = fork( f(x), f(i-x) )
s[i] = merge( s[i], fk )
return s[i]
def merge( s1, s2 ) :
if not s1 :
return s2
if not s2 :
return s1
for i in s2 :
#print just one way quickly
s1[i] = s2[i]
#combine all ways, slowly
# if i in s1 :
# s1[i].update(s2[i])
# else :
# s1[i] = s2[i]
return s1
def fork( s1, s2 ) :
d = {}
#fork s1 s2
for i in s1 :
for j in s2 :
if not i + j in d :
d[i + j] = getExp( s1[i], s2[j], "+" )
if not i - j in d :
d[i - j] = getExp( s1[i], s2[j], "-" )
if not j - i in d :
d[j - i] = getExp( s2[j], s1[i], "-" )
if not i * j in d :
d[i * j] = getExp( s1[i], s2[j], "*" )
if j != 0 and not i / j in d :
d[i / j] = getExp( s1[i], s2[j], "/" )
if i != 0 and not j / i in d :
d[j / i] = getExp( s2[j], s1[i], "/" )
return d
def getExp( s1, s2, op ) :
exp = {}
for i in s1 :
for j in s2 :
exp['('+i+op+j+')'] = 1
#just print one way
break
#just print one way
break
return exp
def check( s ) :
num = 0
for i in xrange(target,0,-1):
if i in s :
if i == target :
print numbers, target, "\nFind ", len(s[i]), 'ways'
for exp in s[i]:
print exp, ' = ', i
else :
print numbers, target, "\nFind nearest ", i, 'in', len(s[i]), 'ways'
for exp in s[i]:
print exp, ' = ', i
break
print '\n'
def game( numbers, target ) :
global s
s = [None]*(2**len(numbers))
for i in xrange(0,len(numbers)) :
numbers[i] = float(numbers[i])
n = len(numbers)
for i in xrange(0,n) :
s[2**i] = { numbers[i]: {str(numbers[i]):1} }
for i in xrange(1,2**n) :
#we will get the f(numbers) in s[2**n-1]
s[i] = f(i)
check(s[2**n-1])
numbers = [4, 8, 6, 2, 2, 5]
s = [None]*(2**len(numbers))
target = 590
game( numbers, target )
numbers = [1,2,3,4,5,6]
target = 590
game( numbers, target )
Assume A is your 6 numbers list.
We define f(A) is all result that can calculate by all A numbers, if we search f(A), we will find if target is in it and get answer or the closest answer.
We can split A to two real child groups: A1 and A-A1 (A1 is not empty and not equal A) , which cut the problem from f(A) to f(A1) and f(A-A1). Because we know f(A) = Union( a+b, a-b, b-a, a*b, a/b(b!=0), b/a(a!=0) ), which a in A, b in A-A1.
We use fork f(A) = Union( fork(A1,A-A1) ) stands for such process. We can remove all duplicate value in fork(), so we can cut the range and make program faster.
So, if A = [1,2,3,4,5,6], then f(A) = fork( f([1]),f([2,3,4,5,6]) ) U ... U fork( f([1,2,3]), f([4,5,6]) ) U ... U stands for Union.
We will see f([2,3,4,5,6]) = fork( f([2,3]), f([4,5,6]) ) U ... , f([3,4,5,6]) = fork( f([3]), f([4,5,6]) ) U ..., the f([4,5,6]) used in both.
So if we can cache every f([...]) the program can be faster.
We can get 2^len(A) - 2 (A1,A-A1) in A. We can use binary to stands for that.
For example: A = [1,2,3,4,5,6], A1 = [1,2,3], then binary 000111(7) stands for A1. A2 = [1,3,5], binary 010101(21) stands for A2. A3 = [1], then binary 000001(1) stands for A3...
So we get a way stands for all groups in A, we can cache them and make all process faster!
All combinations for six number, four operations and parenthesis are up to 5 * 9! at least. So I think you should use some AI algorithm. Using genetic programming or optimization seems to be the path to follow.
In the book Programming Collective Intelligence in the chapter 11 Evolving Intelligence you will find exactly what you want and much more. That chapter explains how to find a mathematical function combining operations and numbers (as you want) to match a result. You will be surprised how easy is such task.
PD: The examples are written using Python.
I would try using an AST at least it will
make your expression generation part easier
(no need to mess with brackets).
http://en.wikipedia.org/wiki/Abstract_syntax_tree
1) Generate some tree with N nodes
(N = the count of numbers you have).
I've read before how many of those you
have, their size is serious as N grows.
By serious I mean more than polynomial to say the least.
2) Now just start changing the operations
in the non-leaf nodes and keep evaluating
the result.
But this is again backtracking and too much degree of freedom.
This is a computationally complex task you're posing. I believe if you
ask the question as you did: "let's generate a number K on the output
such that |K-V| is minimal" (here V is the pre-defined desired result,
i.e. 590 in your example) , then I guess this problem is even NP-complete.
Somebody please correct me if my intuition is lying to me.
So I think even the generation of all possible ASTs (assuming only 1 operation
is allowed) is NP complete as their count is not polynomial. Not to talk that more
than 1 operation is allowed here and not to talk of the minimal difference requirement (between result and desired result).
1. Fast entirely online algorithm
The idea is to search not for a single expression for target value,
but for an equation where target value is included in one part of the equation and
both parts have almost equal number of operations (2 and 3).
Since each part of the equation is relatively small, it does not take much time to
generate all possible expressions for given input values.
After both parts of equation are generated it is possible to scan a pair of sorted arrays
containing values of these expressions and find a pair of equal (or at least best matching)
values in them. After two matching values are found we could get corresponding expressions and
join them into a single expression (in other words, solve the equation).
To join two expression trees together we could descend from the root of one tree
to "target" leaf, for each node on this path invert corresponding operation
('*' to '/', '/' to '*' or '/', '+' to '-', '-' to '+' or '-'), and move "inverted"
root node to other tree (also as root node).
This algorithm is faster and easier to implement when all operations are invertible.
So it is best to use with floating point division (as in my implementation) or with
rational division. Truncating integer division is most difficult case because it produces same result for different inputs (42/25=1 and 25/25 is also 1). With zero-remainder integer division this algorithm gives result almost instantly when exact result is available, but needs some modifications to work correctly when approximate result is needed.
See implementation on Ideone.
2. Even faster approach with off-line pre-processing
As noticed by #WolframH, there are not so many possible input number combinations.
Only 3*3*(49+4-1) = 4455 if repetitions are possible.
Or 3*3*(49) = 1134 without duplicates. Which allows us to pre-process
all possible inputs off-line, store results in compact form, and when some particular result
is needed quickly unpack one of pre-processed values.
Pre-processing program should take array of 6 numbers and generate values for all possible
expressions. Then it should drop out-of-range values and find nearest result for all cases
where there is no exact match. All this could be performed by algorithm proposed by #Tim.
His code needs minimal modifications to do it. Also it is the fastest alternative (yet).
Since pre-processing is offline, we could use something better than interpreted Python.
One alternative is PyPy, other one is to use some fast interpreted language. Pre-processing
all possible inputs should not take more than several minutes.
Speaking about memory needed to store all pre-processed values, the only problem are the
resulting expressions. If stored in string form they will take up to 4455*999*30 bytes or 120Mb.
But each expression could be compressed. It may be represented in postfix notation like this:
arg1 arg2 + arg3 arg4 + *. To store this we need 10 bits to store all arguments' permutations,
10 bits to store 5 operations, and 8 bits to specify how arguments and operations are
interleaved (6 arguments + 5 operations - 3 pre-defined positions: first two are always
arguments, last one is always operation). 28 bits per tree or 4 bytes, which means it is only
20Mb for entire data set with duplicates or 5Mb without them.
3. Slow entirely online algorithm
There are some ways to speed up algorithm in OP:
Greatest speed improvement may be achieved if we avoid trying each commutative operation twice and make recursion tree less branchy.
Some optimization is possible by removing all branches where the result of division operation is zero.
Memorization (dynamic programming) cannot give significant speed boost here, still it may be useful.
After enhancing OP's approach with these ideas, approximately 30x speedup is achieved:
from itertools import combinations
numbers = [4, 8, 6, 2, 15, 50]
target = best_value = 590
best_item = None
subsets = {}
def get_best(value, item):
global best_value, target, best_item
if value >= 0 and abs(target - value) < best_value:
best_value = abs(target - value)
best_item = item
return value, item
def compare_one(value, op, left, right):
item = ('(' + left + op + right + ')')
return get_best(value, item)
def apply_one(left, right):
yield compare_one(left[0] + right[0], '+', left[1], right[1])
yield compare_one(left[0] * right[0], '*', left[1], right[1])
yield compare_one(left[0] - right[0], '-', left[1], right[1])
yield compare_one(right[0] - left[0], '-', right[1], left[1])
if right[0] != 0 and left[0] >= right[0]:
yield compare_one(left[0] / right[0], '/', left[1], right[1])
if left[0] != 0 and right[0] >= left[0]:
yield compare_one(right[0] / left[0], '/', right[1], left[1])
def memorize(seq):
fs = frozenset(seq)
if fs in subsets:
for x in subsets[fs].items():
yield x
else:
subsets[fs] = {}
for value, item in try_all(seq):
subsets[fs][value] = item
yield value, item
def apply_all(left, right):
for l in memorize(left):
for r in memorize(right):
for x in apply_one(l, r):
yield x;
def try_all(seq):
if len(seq) == 1:
yield get_best(numbers[seq[0]], str(numbers[seq[0]]))
for length in range(1, len(seq)):
for x in combinations(seq[1:], length):
for value, item in apply_all(list(x), list(set(seq) - set(x))):
yield value, item
for x, y in try_all([0, 1, 2, 3, 4, 5]): pass
print best_item
More speed improvements are possible if you add some constraints to the problem:
If integer division is only possible when the remainder is zero.
If all intermediate results are to be non-negative and/or below 1000.
Well I don't will give up. Following the line of all the answers to your question I come up with another algorithm. This algorithm gives the solution with a time average of 3 milliseconds.
#! -*- coding: utf-8 -*-
import copy
numbers = [4, 8, 6, 2, 15, 50]
target = 590
operations = {
'+': lambda x, y: x + y,
'-': lambda x, y: x - y,
'*': lambda x, y: x * y,
'/': lambda x, y: y == 0 and 1e30 or x / y # Handle zero division
}
def chain_op(target, numbers, result=None, expression=""):
if len(numbers) == 0:
return (expression, result)
else:
for choosen_number in numbers:
remaining_numbers = copy.copy(numbers)
remaining_numbers.remove(choosen_number)
if result is None:
return chain_op(target, remaining_numbers, choosen_number, str(choosen_number))
else:
incomming_results = []
for key, op in operations.items():
new_result = op(result, choosen_number)
new_expression = "%s%s%d" % (expression, key, choosen_number)
incomming_results.append(chain_op(target, remaining_numbers, new_result, new_expression))
diff = 1e30
selected = None
for exp_result in incomming_results:
exp, res = exp_result
if abs(res - target) < diff:
diff = abs(res - target)
selected = exp_result
if diff == 0:
break
return selected
if __name__ == '__main__':
print chain_op(target, numbers)
Erratum: This algorithm do not include the solutions containing parenthesis. It always hits the target or the closest result, my bad. Still is pretty fast. It can be adapted to support parenthesis without much work.
Actually there are two things that you can do to speed up the time to milliseconds.
You are trying to find a solution for given quiz, by generating the numbers and the target number. Instead you can generate the solution and just remove the operations. You can build some thing smart that will generate several quizzes and choose the most interesting one, how ever in this case you loose the as close as possible option.
Another way to go, is pre-calculation. Solve 100 quizes, use them as build-in in your application, and generate new one on the fly, try to keep your quiz stack at 100, also try to give the user only the new quizes. I had the same problem in my bible games, and I used this method to speed thing up. Instead of 10 sec for question it takes me milliseconds as I am generating new question in background and always keeping my stack to 100.
What about Dynamic programming, because you need same results to calculate other options?
From Section 15.2 of Programming Pearls
The C codes can be viewed here: http://www.cs.bell-labs.com/cm/cs/pearls/longdup.c
When I implement it in Python using suffix-array:
example = open("iliad10.txt").read()
def comlen(p, q):
i = 0
for x in zip(p, q):
if x[0] == x[1]:
i += 1
else:
break
return i
suffix_list = []
example_len = len(example)
idx = list(range(example_len))
idx.sort(cmp = lambda a, b: cmp(example[a:], example[b:])) #VERY VERY SLOW
max_len = -1
for i in range(example_len - 1):
this_len = comlen(example[idx[i]:], example[idx[i+1]:])
print this_len
if this_len > max_len:
max_len = this_len
maxi = i
I found it very slow for the idx.sort step. I think it's slow because Python need to pass the substring by value instead of by pointer (as the C codes above).
The tested file can be downloaded from here
The C codes need only 0.3 seconds to finish.
time cat iliad10.txt |./longdup
On this the rest of the Achaeans with one voice were for
respecting the priest and taking the ransom that he offered; but
not so Agamemnon, who spoke fiercely to him and sent him roughly
away.
real 0m0.328s
user 0m0.291s
sys 0m0.006s
But for Python codes, it never ends on my computer (I waited for 10 minutes and killed it)
Does anyone have ideas how to make the codes efficient? (For example, less than 10 seconds)
My solution is based on Suffix arrays. It is constructed by Prefix doubling the Longest common prefix. The worst-case complexity is O(n (log n)^2). The file "iliad.mb.txt" takes 4 seconds on my laptop. The longest_common_substring function is short and can be easily modified, e.g. for searching the 10 longest non-overlapping substrings. This Python code is faster than the original C code from the question, if duplicate strings are longer than 10000 characters.
from itertools import groupby
from operator import itemgetter
def longest_common_substring(text):
"""Get the longest common substrings and their positions.
>>> longest_common_substring('banana')
{'ana': [1, 3]}
>>> text = "not so Agamemnon, who spoke fiercely to "
>>> sorted(longest_common_substring(text).items())
[(' s', [3, 21]), ('no', [0, 13]), ('o ', [5, 20, 38])]
This function can be easy modified for any criteria, e.g. for searching ten
longest non overlapping repeated substrings.
"""
sa, rsa, lcp = suffix_array(text)
maxlen = max(lcp)
result = {}
for i in range(1, len(text)):
if lcp[i] == maxlen:
j1, j2, h = sa[i - 1], sa[i], lcp[i]
assert text[j1:j1 + h] == text[j2:j2 + h]
substring = text[j1:j1 + h]
if not substring in result:
result[substring] = [j1]
result[substring].append(j2)
return dict((k, sorted(v)) for k, v in result.items())
def suffix_array(text, _step=16):
"""Analyze all common strings in the text.
Short substrings of the length _step a are first pre-sorted. The are the
results repeatedly merged so that the garanteed number of compared
characters bytes is doubled in every iteration until all substrings are
sorted exactly.
Arguments:
text: The text to be analyzed.
_step: Is only for optimization and testing. It is the optimal length
of substrings used for initial pre-sorting. The bigger value is
faster if there is enough memory. Memory requirements are
approximately (estimate for 32 bit Python 3.3):
len(text) * (29 + (_size + 20 if _size > 2 else 0)) + 1MB
Return value: (tuple)
(sa, rsa, lcp)
sa: Suffix array for i in range(1, size):
assert text[sa[i-1]:] < text[sa[i]:]
rsa: Reverse suffix array for i in range(size):
assert rsa[sa[i]] == i
lcp: Longest common prefix for i in range(1, size):
assert text[sa[i-1]:sa[i-1]+lcp[i]] == text[sa[i]:sa[i]+lcp[i]]
if sa[i-1] + lcp[i] < len(text):
assert text[sa[i-1] + lcp[i]] < text[sa[i] + lcp[i]]
>>> suffix_array(text='banana')
([5, 3, 1, 0, 4, 2], [3, 2, 5, 1, 4, 0], [0, 1, 3, 0, 0, 2])
Explanation: 'a' < 'ana' < 'anana' < 'banana' < 'na' < 'nana'
The Longest Common String is 'ana': lcp[2] == 3 == len('ana')
It is between tx[sa[1]:] == 'ana' < 'anana' == tx[sa[2]:]
"""
tx = text
size = len(tx)
step = min(max(_step, 1), len(tx))
sa = list(range(len(tx)))
sa.sort(key=lambda i: tx[i:i + step])
grpstart = size * [False] + [True] # a boolean map for iteration speedup.
# It helps to skip yet resolved values. The last value True is a sentinel.
rsa = size * [None]
stgrp, igrp = '', 0
for i, pos in enumerate(sa):
st = tx[pos:pos + step]
if st != stgrp:
grpstart[igrp] = (igrp < i - 1)
stgrp = st
igrp = i
rsa[pos] = igrp
sa[i] = pos
grpstart[igrp] = (igrp < size - 1 or size == 0)
while grpstart.index(True) < size:
# assert step <= size
nextgr = grpstart.index(True)
while nextgr < size:
igrp = nextgr
nextgr = grpstart.index(True, igrp + 1)
glist = []
for ig in range(igrp, nextgr):
pos = sa[ig]
if rsa[pos] != igrp:
break
newgr = rsa[pos + step] if pos + step < size else -1
glist.append((newgr, pos))
glist.sort()
for ig, g in groupby(glist, key=itemgetter(0)):
g = [x[1] for x in g]
sa[igrp:igrp + len(g)] = g
grpstart[igrp] = (len(g) > 1)
for pos in g:
rsa[pos] = igrp
igrp += len(g)
step *= 2
del grpstart
# create LCP array
lcp = size * [None]
h = 0
for i in range(size):
if rsa[i] > 0:
j = sa[rsa[i] - 1]
while i != size - h and j != size - h and tx[i + h] == tx[j + h]:
h += 1
lcp[rsa[i]] = h
if h > 0:
h -= 1
if size > 0:
lcp[0] = 0
return sa, rsa, lcp
I prefer this solution over more complicated O(n log n) because Python has a very fast list sorting algorithm (Timsort). Python's sort is probably faster than necessary linear time operations in the method from that article, that should be O(n) under very special presumptions of random strings together with a small alphabet (typical for DNA genome analysis). I read in Gog 2011 that worst-case O(n log n) of my algorithm can be in practice faster than many O(n) algorithms that cannot use the CPU memory cache.
The code in another answer based on grow_chains is 19 times slower than the original example from the question, if the text contains a repeated string 8 kB long. Long repeated texts are not typical for classical literature, but they are frequent e.g. in "independent" school homework collections. The program should not freeze on it.
I wrote an example and tests with the same code for Python 2.7, 3.3 - 3.6.
The translation of the algorithm into Python:
from itertools import imap, izip, starmap, tee
from os.path import commonprefix
def pairwise(iterable): # itertools recipe
a, b = tee(iterable)
next(b, None)
return izip(a, b)
def longest_duplicate_small(data):
suffixes = sorted(data[i:] for i in xrange(len(data))) # O(n*n) in memory
return max(imap(commonprefix, pairwise(suffixes)), key=len)
buffer() allows to get a substring without copying:
def longest_duplicate_buffer(data):
n = len(data)
sa = sorted(xrange(n), key=lambda i: buffer(data, i)) # suffix array
def lcp_item(i, j): # find longest common prefix array item
start = i
while i < n and data[i] == data[i + j - start]:
i += 1
return i - start, start
size, start = max(starmap(lcp_item, pairwise(sa)), key=lambda x: x[0])
return data[start:start + size]
It takes 5 seconds on my machine for the iliad.mb.txt.
In principle it is possible to find the duplicate in O(n) time and O(n) memory using a suffix array augmented with a lcp array.
Note: *_memoryview() is deprecated by *_buffer() version
More memory efficient version (compared to longest_duplicate_small()):
def cmp_memoryview(a, b):
for x, y in izip(a, b):
if x < y:
return -1
elif x > y:
return 1
return cmp(len(a), len(b))
def common_prefix_memoryview((a, b)):
for i, (x, y) in enumerate(izip(a, b)):
if x != y:
return a[:i]
return a if len(a) < len(b) else b
def longest_duplicate(data):
mv = memoryview(data)
suffixes = sorted((mv[i:] for i in xrange(len(mv))), cmp=cmp_memoryview)
result = max(imap(common_prefix_memoryview, pairwise(suffixes)), key=len)
return result.tobytes()
It takes 17 seconds on my machine for the iliad.mb.txt. The result is:
On this the rest of the Achaeans with one voice were for respecting
the priest and taking the ransom that he offered; but not so Agamemnon,
who spoke fiercely to him and sent him roughly away.
I had to define custom functions to compare memoryview objects because memoryview comparison either raises an exception in Python 3 or produces wrong result in Python 2:
>>> s = b"abc"
>>> memoryview(s[0:]) > memoryview(s[1:])
True
>>> memoryview(s[0:]) < memoryview(s[1:])
True
Related questions:
Find the longest repeating string and the number of times it repeats in a given string
finding long repeated substrings in a massive string
The main problem seems to be that python does slicing by copy: https://stackoverflow.com/a/5722068/538551
You'll have to use a memoryview instead to get a reference instead of a copy. When I did this, the program hung after the idx.sort function (which was very fast).
I'm sure with a little work, you can get the rest working.
Edit:
The above change will not work as a drop-in replacement because cmp does not work the same way as strcmp. For example, try the following C code:
#include <stdio.h>
#include <string.h>
int main() {
char* test1 = "ovided by The Internet Classics Archive";
char* test2 = "rovided by The Internet Classics Archive.";
printf("%d\n", strcmp(test1, test2));
}
And compare the result to this python:
test1 = "ovided by The Internet Classics Archive";
test2 = "rovided by The Internet Classics Archive."
print(cmp(test1, test2))
The C code prints -3 on my machine while the python version prints -1. It looks like the example C code is abusing the return value of strcmp (it IS used in qsort after all). I couldn't find any documentation on when strcmp will return something other than [-1, 0, 1], but adding a printf to pstrcmp in the original code showed a lot of values outside of that range (3, -31, 5 were the first 3 values).
To make sure that -3 wasn't some error code, if we reverse test1 and test2, we'll get 3.
Edit:
The above is interesting trivia, but not actually correct in terms of affecting either chunks of code. I realized this just as I shut my laptop and left a wifi zone... Really should double check everything before I hit Save.
FWIW, cmp most certainly works on memoryview objects (prints -1 as expected):
print(cmp(memoryview(test1), memoryview(test2)))
I'm not sure why the code isn't working as expected. Printing out the list on my machine does not look as expected. I'll look into this and try to find a better solution instead of grasping at straws.
This version takes about 17 secs on my circa-2007 desktop using totally different algorithm:
#!/usr/bin/env python
ex = open("iliad.mb.txt").read()
chains = dict()
# populate initial chains dictionary
for (a,b) in enumerate(zip(ex,ex[1:])) :
s = ''.join(b)
if s not in chains :
chains[s] = list()
chains[s].append(a)
def grow_chains(chains) :
new_chains = dict()
for (string,pos) in chains :
offset = len(string)
for p in pos :
if p + offset >= len(ex) : break
# add one more character
s = string + ex[p + offset]
if s not in new_chains :
new_chains[s] = list()
new_chains[s].append(p)
return new_chains
# grow and filter, grow and filter
while len(chains) > 1 :
print 'length of chains', len(chains)
# remove chains that appear only once
chains = [(i,chains[i]) for i in chains if len(chains[i]) > 1]
print 'non-unique chains', len(chains)
print [i[0] for i in chains[:3]]
chains = grow_chains(chains)
The basic idea is to create a list of substrings and positions where they occure, thus eliminating the need to compare same strings again and again. The resulting list look like [('ind him, but', [466548, 739011]), (' bulwark bot', [428251, 428924]), (' his armour,', [121559, 124919, 193285, 393566, 413634, 718953, 760088])]. Unique strings are removed. Then every list member grows by 1 character and new list is created. Unique strings are removed again. And so on and so forth...
I'm programming a spellcheck program in Python. I have a list of valid words (the dictionary) and I need to output a list of words from this dictionary that have an edit distance of 2 from a given invalid word.
I know I need to start by generating a list with an edit distance of one from the invalid word(and then run that again on all the generated words). I have three methods, inserts(...), deletions(...) and changes(...) that should output a list of words with an edit distance of 1, where inserts outputs all valid words with one more letter than the given word, deletions outputs all valid words with one less letter, and changes outputs all valid words with one different letter.
I've checked a bunch of places but I can't seem to find an algorithm that describes this process. All the ideas I've come up with involve looping through the dictionary list multiple times, which would be extremely time consuming. If anyone could offer some insight, I'd be extremely grateful.
The thing you are looking at is called an edit distance and here is a nice explanation on wiki. There are a lot of ways how to define a distance between the two words and the one that you want is called Levenshtein distance and here is a DP (dynamic programming) implementation in python.
def levenshteinDistance(s1, s2):
if len(s1) > len(s2):
s1, s2 = s2, s1
distances = range(len(s1) + 1)
for i2, c2 in enumerate(s2):
distances_ = [i2+1]
for i1, c1 in enumerate(s1):
if c1 == c2:
distances_.append(distances[i1])
else:
distances_.append(1 + min((distances[i1], distances[i1 + 1], distances_[-1])))
distances = distances_
return distances[-1]
And a couple of more implementations are here.
difflib in the standard library has various utilities for sequence matching, including the get_close_matches method that you could use. It uses an algorithm adapted from Ratcliff and Obershelp.
From the docs
>>> from difflib import get_close_matches
>>> get_close_matches('appel', ['ape', 'apple', 'peach', 'puppy'])
['apple', 'ape']
Here is my version for Levenshtein distance
def edit_distance(s1, s2):
m=len(s1)+1
n=len(s2)+1
tbl = {}
for i in range(m): tbl[i,0]=i
for j in range(n): tbl[0,j]=j
for i in range(1, m):
for j in range(1, n):
cost = 0 if s1[i-1] == s2[j-1] else 1
tbl[i,j] = min(tbl[i, j-1]+1, tbl[i-1, j]+1, tbl[i-1, j-1]+cost)
return tbl[i,j]
print(edit_distance("Helloworld", "HalloWorld"))
#this calculates edit distance not levenstein edit distance
word1="rice"
word2="ice"
len_1=len(word1)
len_2=len(word2)
x =[[0]*(len_2+1) for _ in range(len_1+1)]#the matrix whose last element ->edit distance
for i in range(0,len_1+1): #initialization of base case values
x[i][0]=i
for j in range(0,len_2+1):
x[0][j]=j
for i in range (1,len_1+1):
for j in range(1,len_2+1):
if word1[i-1]==word2[j-1]:
x[i][j] = x[i-1][j-1]
else :
x[i][j]= min(x[i][j-1],x[i-1][j],x[i-1][j-1])+1
print x[i][j]
Using the SequenceMatcher from Python built-in difflib is another way of doing it, but (as correctly pointed out in the comments), the result does not match the definition of an edit distance exactly. Bonus: it supports ignoring "junk" parts (e.g. spaces or punctuation).
from difflib import SequenceMatcher
a = 'kitten'
b = 'sitting'
required_edits = [
code
for code in (
SequenceMatcher(a=a, b=b, autojunk=False)
.get_opcodes()
)
if code[0] != 'equal'
]
required_edits
# [
# # (tag, i1, i2, j1, j2)
# ('replace', 0, 1, 0, 1), # replace a[0:1]="k" with b[0:1]="s"
# ('replace', 4, 5, 4, 5), # replace a[4:5]="e" with b[4:5]="i"
# ('insert', 6, 6, 6, 7), # insert b[6:7]="g" after a[6:6]="n"
# ]
# the edit distance:
len(required_edits) # == 3
I would recommend not creating this kind of code on your own. There are libraries for that.
For instance the Levenshtein library.
In [2]: Levenshtein.distance("foo", "foobar")
Out[2]: 3
In [3]: Levenshtein.distance("barfoo", "foobar")
Out[3]: 6
In [4]: Levenshtein.distance("Buroucrazy", "Bureaucracy")
Out[4]: 3
In [5]: Levenshtein.distance("Misisipi", "Mississippi")
Out[5]: 3
In [6]: Levenshtein.distance("Misisipi", "Misty Mountains")
Out[6]: 11
In [7]: Levenshtein.distance("Buroucrazy", "Born Crazy")
Out[7]: 4
Similar to Santoshi's solution above but I made three changes:
One line initialization instead of five
No need to define cost alone (just use int(boolean) 0 or 1)
Instead of double for loop use product, (this last one is only cosmetic, double loop seems unavoidable)
from itertools import product
def edit_distance(s1,s2):
d={ **{(i,0):i for i in range(len(s1)+1)},**{(0,j):j for j in range(len(s2)+1)}}
for i, j in product(range(1,len(s1)+1), range(1,len(s2)+1)):
d[i,j]=min((s1[i-1]!=s2[j-1]) + d[i-1,j-1], d[i-1,j]+1, d[i,j-1]+1)
return d[i,j]
Instead of going with Levenshtein distance algo use BK tree or TRIE, as these algorithms have less complexity then edit distance. A good browse over these topic will give a detailed description.
This link will help you more about spell checking.
You need Minimum Edit Distance for this task.
Following is my version of MED a.k.a Levenshtein Distance.
def MED_character(str1,str2):
cost=0
len1=len(str1)
len2=len(str2)
#output the length of other string in case the length of any of the string is zero
if len1==0:
return len2
if len2==0:
return len1
accumulator = [[0 for x in range(len2)] for y in range(len1)] #initializing a zero matrix
# initializing the base cases
for i in range(0,len1):
accumulator[i][0] = i;
for i in range(0,len2):
accumulator[0][i] = i;
# we take the accumulator and iterate through it row by row.
for i in range(1,len1):
char1=str1[i]
for j in range(1,len2):
char2=str2[j]
cost1=0
if char1!=char2:
cost1=2 #cost for substitution
accumulator[i][j]=min(accumulator[i-1][j]+1, accumulator[i][j-1]+1, accumulator[i-1][j-1] + cost1 )
cost=accumulator[len1-1][len2-1]
return cost
Fine tuned codes based on the version from #Santosh and should address the issue brought up by #Artur Krajewski; The biggest difference is replacing an effective 2d matrix
def edit_distance(s1, s2):
# add a blank character for both strings
m=len(s1)+1
n=len(s2)+1
# launch a matrix
tbl = [[0] * n for i in range(m)]
for i in range(m): tbl[i][0]=i
for j in range(n): tbl[0][j]=j
for i in range(1, m):
for j in range(1, n):
#if strings have same letters, set operation cost as 0 otherwise 1
cost = 0 if s1[i-1] == s2[j-1] else 1
#find min practice
tbl[i][j] = min(tbl[i][j-1]+1, tbl[i-1][j]+1, tbl[i-1][j-1]+cost)
return tbl
edit_distance("birthday", "Birthdayyy")
following up on #krassowski's answer
from difflib import SequenceMatcher
def sequence_matcher_edits(word_a, word_b):
required_edits = [code for code in (
SequenceMatcher(a=word_a, b=word_b, autojunk=False).get_opcodes()
)
if code[0] != 'equal'
]
return len(required_edits)
print(f"sequence_matcher_edits {sequence_matcher_edits('kitten', 'sitting')}")
# -> sequence_matcher_edits 3