I'm trying to write simple code for that problem. If I get an array and number I need to find the 3 numbers that their sum are close to the number that's given.
I've thought about first to pop out the last digit (the first number)
then I'll have a new array without this digit. So now I look for the second number who needs to be less the sum target. so I take only the small numbers that it's smaller them the second=sum-first number (but I don't know how to choose it.
The last number will be third=sum-first-second
I tried to write code but it's not working and it's very basic
def f(s,target):
s=sorted(s)
print(s)
print(s[0])
closest=s[0]+s[1]+s[2]
m=s[:-1]
print(m)
for i in range(len(s)):
for j in range(len(m)):
if (closest<=target-m[0]) and s[-1] + m[j] == target:
print (m[j])
n = m[:j] + nums[j+1:]
for z in range (len(z)):
if (closest<target-n[z]) and s[-1]+ m[j]+n[z] == target:
print (n[z])
s=[4,2,12,3,4,8,14]
target=20
f(s,target)
if you have idea what to change here. Please let me know
Thank you
Here is my solution I tried to maximize the performance of the code to not repeat any combinations. Let me know if you have any questions.
Good luck.
def find_3(s,target):
to_not_rep=[] #This list will store all combinations without repetation
close_to_0=abs(target - s[0]+s[1]+s[2]) #initile
There_is_one=False #False: don't have a combination equal to the target yet
for s1,first_n in enumerate(s):
for s2,second_n in enumerate(s):
if (s1==s2) : continue #to not take the same index
for s3,third_n in enumerate(s):
if (s1==s3) or (s2==s3) : continue #to not take the same index
val=sorted([first_n,second_n,third_n]) #sorting
if val in to_not_rep :continue #to not repeat the same combination with diffrent positions
to_not_rep.append(val)#adding all the combinations without repetation
sum_=sum(val) #the sum of the three numbers
# Good one
if sum_==target:
print(f"Found a possibility: {val[0]} + {val[1]} + {val[2]} = {target}")
There_is_one = True
if There_is_one is False: #No need if we found combination equal to the target
# close to the target
# We know that (target - sum) should equal to 0 otherwise :
# We are looking for the sum of closet combinations(in abs value) to 0
pos_n=abs(target-sum_)
if pos_n < close_to_0:
closet_one=f"The closet combination to the target is: {val[0]} + {val[1]} + {val[2]} = {sum_} almost {target} "
close_to_0=pos_n
# Print the closet combination to the target in case we did not find a combination equal to the target
if There_is_one is False: print(closet_one)
so we can test it :
s =[4,2,3,8,6,4,12,16,30,20,5]
target=20
find_3(s,target)
#Found a possibility: 4 + 4 + 12 = 20
#Found a possibility: 2 + 6 + 12 = 20
#Found a possibility: 3 + 5 + 12 = 20
another test :
s =[4,2,3,8,6,4,323,23,44]
find_3(s,target)
#The closet combination to the target is: 4 + 6 + 8 = 18 almost 20
This is a simple solution that returns all possibilites.
For your case it completed in 0.002019 secs
from itertools import combinations
import numpy as np
def f(s, target):
dic = {}
for tup in combinations(s, 3):
try:
dic[np.absolute(np.sum(tup) - target)].append(str(tup))
except KeyError:
dic[np.absolute(np.sum(tup) - target)] = [tup]
print(dic[min(dic.keys())])
Use itertools.combinations to get all combinations of your numbers without replacement of a certain length (three in your case). Then take the three-tuple for which the absolute value of the difference of the sum and target is minimal. min can take a key argument to specify the ordering of the iterable passed to the function.
from typing import Sequence, Tuple
def closest_to(seq: Sequence[float], target: float, length: int = 3) -> Tuple[float]:
from itertools import combinations
combs = combinations(seq, length)
diff = lambda x: abs(sum(x) - target)
return min(combs, key=diff)
closest_to([4,2,12,3,4,8,14], 20) # (4, 2, 14)
This is not the fastest or most efficient way to do it, but it's conceptionally simple and short.
Something like this?
import math
num_find = 1448
lst_Results = []
i_Number = num_find
while i_Number > 0:
num_Exp = math.floor(math.log(i_Number) / math.log(2))
lst_Results.append(dict({num_Exp: int(math.pow(2, num_Exp))}))
i_Number = i_Number - math.pow(2, num_Exp)
print(lst_Results)
In a sequence of numbers: for example 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, etc ...
The sum of the previous numbers is never greater than the next. This gives us the possibility of combinations, for example:
The number: 1448, there is no other combination than the sum of the previous numbers: 8 + 32 + 128 + 256 + 1024
Then you find the numbers whose sum is close to the number provided
I'm trying to implement a filter with Python to sort out the points on a point cloud generated by Agisoft PhotoScan. PhotoScan is a photogrammetry software developed to be user friendly but also allows to use Python commands through an API.
Bellow is my code so far and I'm pretty sure there is better way to write it as I'm missing something. The code runs inside PhotoScan.
Objective:
Selecting and removing 10% of points at a time with error within defined range of 50 to 10. Also removing any points within error range less than 10% of the total, when the initial steps of selecting and removing 10% at a time are done. Immediately after every point removal an optimization procedure should be done. It should stop when no points are selectable or when selectable points counts as less than 1% of the present total points and it is not worth removing them.
Draw it for better understanding:
Actual Code Under Construction (3 updates - see bellow for details):
import PhotoScan as PS
import math
doc = PS.app.document
chunk = doc.chunk
# using float with range and that by setting i = 1 it steps 0.1 at a time
def precrange(a, b, i):
if a < b:
p = 10**i
sr = a*p
er = (b*p) + 1
p = float(p)
for n in range(sr, er):
x = n/p
yield x
else:
p = 10**i
sr = b*p
er = (a*p) + 1
p = float(p)
for n in range(sr, er):
x = n/p
yield x
"""
Determine if x is close to y:
x relates to nselected variable
y to p10 variable
math.isclose() Return True if the values a and b are close to each other and
False otherwise
var is the tolerance here setted as a relative tolerance:
rel_tol is the relative tolerance – it is the maximum allowed difference
between a and b, relative to the larger absolute value of a or b. For example,
to set a tolerance of 5%, pass rel_tol=0.05. The default tolerance is 1e-09,
which assures that the two values are the same within about 9 decimal digits.
rel_tol must be greater than zero.
"""
def test_isclose(x, y, var):
if math.isclose(x, y, rel_tol=var): # if variables are close return True
return True
else:
False
# 1. define filter limits
f_ReconstUncert = precrange(50, 10, 1)
# 2. count initial point number
tiePoints_0 = len(chunk.point_cloud.points) # storing info for later
# 3. call Filter() and init it
f = PS.PointCloud.Filter()
f.init(chunk, criterion=PS.PointCloud.Filter.ReconstructionUncertainty)
a = 0
"""
Way to restart for loop!
should_restart = True
while should_restart:
should_restart = False
for i in xrange(10):
print i
if i == 5:
should_restart = True
break
"""
restartLoop = True
while restartLoop:
restartLoop = False
for count, i in enumerate(f_ReconstUncert): # for each threshold value
# count points for every i
tiePoints = len(chunk.point_cloud.points)
p10 = int(round((10 / 100) * tiePoints, 0)) # 10% of the total
f.selectPoints(i) # selects points
nselected = len([p for p in chunk.point_cloud.points if p.selected])
percent = round(nselected * 100 / tiePoints, 2)
if nselected == 0:
print("For threshold {} there´s no selectable points".format(i))
break
elif test_isclose(nselected, p10, 0.1):
a += 1
print("Threshold found in iteration: ", count)
print("----------------------------------------------")
print("# {} Removing points from cloud ".format(a))
print("----------------------------------------------")
print("# {}. Reconstruction Uncerntainty:"
" {:.2f}".format(a, i))
print("{} - {}"
" ({:.1f} %)\n".format(tiePoints,
nselected, percent))
f.removePoints(i) # removes points
# optimization procedure needed to refine cameras positions
print("--------------Optimizing cameras-------------\n")
chunk.optimizeCameras(fit_f=True, fit_cx=True,
fit_cy=True, fit_b1=False,
fit_b2=False, fit_k1=True,
fit_k2=True, fit_k3=True,
fit_k4=False, fit_p1=True,
fit_p2=True, fit_p3=False,
fit_p4=False, adaptive_fitting=False)
# count again number of points in point cloud
tiePoints = len(chunk.point_cloud.points)
print("= {} remaining points after"
" {} removal".format(tiePoints, a))
# reassigning variable to get new 10% of remaining points
p10 = int(round((10 / 100) * tiePoints, 0))
percent = round(nselected * 100 / tiePoints, 2)
print("----------------------------------------------\n\n")
# restart loop to investigate from range start
restartLoop = True
break
else:
f.resetSelection()
continue # continue to next i
else:
f.resetSelection()
print("for loop didnt work out")
print("{} iterations done!".format(count))
tiePoints = len(chunk.point_cloud.points)
print("Tiepoints 0: ", tiePoints_0)
print("Tiepoints 1: ", tiePoints)
Problems:
A. Currently I'm stuck on an endless processing because of a loop. I know it's about my bad coding. But how do I implement my objective and get away with the infinite loops? ANSWER: Got the code less confusing and updated above.
B. How do I start over (or restart) my search for valid threshold values in the range(50, 20) after finding one of them? ANSWER: Stack Exchange: how to restart a for loop
C. How do I turn the code more pythonic?
IMPORTANT UPDATE 1: altered above
Using a better range with float solution adapted from stackoverflow: how-to-use-a-decimal-range-step-value
# using float with range and that by setting i = 1 it steps 0.1 at a time
def precrange(a, b, i):
if a < b:
p = 10**i
sr = a*p
er = (b*p) + 1
p = float(p)
return map(lambda x: x/p, range(sr, er))
else:
p = 10**i
sr = b*p
er = (a*p) + 1
p = float(p)
return map(lambda x: x/p, range(sr, er))
# some code
f_ReconstUncert = precrange(50, 20, 1)
And also using math.isclose() to determine if selected points are close to the 10% selected points instead of using a manual solution through assigning new variables. This was implemented as follows:
"""
Determine if x is close to y:
x relates to nselected variable
y to p10 variable
math.isclose() Return True if the values a and b are close to each other and
False otherwise
var is the tolerance here setted as a relative tolerance:
rel_tol is the relative tolerance – it is the maximum allowed difference
between a and b, relative to the larger absolute value of a or b. For example,
to set a tolerance of 5%, pass rel_tol=0.05. The default tolerance is 1e-09,
which assures that the two values are the same within about 9 decimal digits.
rel_tol must be greater than zero.
"""
def test_threshold(x, y, var):
if math.isclose(x, y, rel_tol=var): # if variables are close return True
return True
else:
False
# some code
if test_threshold(nselected, p10, 0.1):
# if true then a valid threshold is found
# some code
UPDATE 2: altered on code under construction
Minor fixes and got to restart de for loop from beginning by following guidance from another Stack Exchange post on the subject. Have to improve the range now or alter the isclose() to get more values.
restartLoop = True
while restartLoop:
restartLoop = False
for i in range(0, 10):
if condition:
restartLoop = True
break
UPDATE 3: Code structure to achieve listed objectives:
threshold = range(0, 11, 1)
listx = []
for i in threshold:
listx.append(i)
restart = 0
restartLoop = True
while restartLoop:
restartLoop = False
for idx, i in enumerate(listx):
print("do something as printing i:", i)
if i > 5: # if this condition restart loop
print("found value for condition: ", i)
del listx[idx]
restartLoop = True
print("RESTARTING LOOP\n")
restart += 1
break # break inner while and restart for loop
else:
# continue if the inner loop wasn't broken
continue
else:
continue
print("restart - outer while", restart)
The following Python program flips a coin several times, then reports the longest series of heads and tails. I am trying to convert this program into a program that uses functions so it uses basically less code. I am very new to programming and my teacher requested this of us, but I have no idea how to do it. I know I'm supposed to have the function accept 2 parameters: a string or list, and a character to search for. The function should return, as the value of the function, an integer which is the longest sequence of that character in that string. The function shouldn't accept input or output from the user.
import random
print("This program flips a coin several times, \nthen reports the longest
series of heads and tails")
cointoss = int(input("Number of times to flip the coin: "))
varlist = []
i = 0
varstring = ' '
while i < cointoss:
r = random.choice('HT')
varlist.append(r)
varstring = varstring + r
i += 1
print(varstring)
print(varlist)
print("There's this many heads: ",varstring.count("H"))
print("There's this many tails: ",varstring.count("T"))
print("Processing input...")
i = 0
longest_h = 0
longest_t = 0
inarow = 0
prevIn = 0
while i < cointoss:
print(varlist[i])
if varlist[i] == 'H':
prevIn += 1
if prevIn > longest_h:
longest_h = prevIn
print("",longest_h,"")
inarow = 0
if varlist[i] == 'T':
inarow += 1
if inarow > longest_t:
longest_t = inarow
print("",longest_t,"")
prevIn = 0
i += 1
print ("The longest series of heads is: ",longest_h)
print ("The longest series of tails is: ",longest_t)
If this is asking too much, any explanatory help would be really nice instead. All I've got so far is:
def flip (a, b):
flipValue = random.randint
but it's barely anything.
import random
def Main():
numOfFlips=getFlips()
outcome=flipping(numOfFlips)
print(outcome)
def getFlips():
Flips=int(input("Enter number if flips:\n"))
return Flips
def flipping(numOfFlips):
longHeads=[]
longTails=[]
Tails=0
Heads=0
for flips in range(0,numOfFlips):
flipValue=random.randint(1,2)
print(flipValue)
if flipValue==1:
Tails+=1
longHeads.append(Heads) #recording value of Heads before resetting it
Heads=0
else:
Heads+=1
longTails.append(Tails)
Tails=0
longestHeads=max(longHeads) #chooses the greatest length from both lists
longestTails=max(longTails)
return "Longest heads:\t"+str(longestHeads)+"\nLongest tails:\t"+str(longestTails)
Main()
I did not quite understand how your code worked, so I made the code in functions that works just as well, there will probably be ways of improving my code alone but I have moved the code over to functions
First, you need a function that flips a coin x times. This would be one possible implementation, favoring random.choice over random.randint:
def flip(x):
result = []
for _ in range(x):
result.append(random.choice(("h", "t")))
return result
Of course, you could also pass from what exactly we are supposed to take a choice as a parameter.
Next, you need a function that finds the longest sequence of some value in some list:
def longest_series(some_value, some_list):
current, longest = 0, 0
for r in some_list:
if r == some_value:
current += 1
longest = max(current, longest)
else:
current = 0
return longest
And now you can call these in the right order:
# initialize the random number generator, so we get the same result
random.seed(5)
# toss a coin a hundred times
series = flip(100)
# count heads/tails
headflips = longest_series('h', series)
tailflips = longest_series('t', series)
# print the results
print("The longest series of heads is: " + str(headflips))
print("The longest series of tails is: " + str(tailflips))
Output:
>> The longest series of heads is: 8
>> The longest series of heads is: 5
edit: removed the flip implementation with yield, it made the code weird.
Counting the longest run
Let see what you have asked for
I'm supposed to have the function accept 2 parameters: a string or list,
or, generalizing just a bit, a sequence
and a character
again, we'd speak, generically, of an item
to search for. The function should return, as the value of the
function, an integer which is the longest sequence of that character
in that string.
My implementation of the function you are asking for, complete of doc
string, is
def longest_run(i, s):
'Counts the longest run of item "i" in sequence "s".'
c, m = 0, 0
for el in s:
if el==i:
c += 1
elif c:
m = m if m >= c else c
c = 0
return m
We initialize c (current run) and m (maximum run so far) to zero,
then we loop, looking at every element el of the argument sequence s.
The logic is straightforward but for elif c: whose block is executed at the end of a run (because c is greater than zero and logically True) but not when the previous item (not the current one) was not equal to i. The savings are small but are savings...
Flipping coins (and more...)
How can we simulate flipping n coins? We abstract the problem and recognize that flipping n coins corresponds to choosing from a collection of possible outcomes (for a coin, either head or tail) for n times.
As it happens, the random module of the standard library has the exact answer to this problem
In [52]: random.choices?
Signature: choices(population, weights=None, *, cum_weights=None, k=1)
Docstring:
Return a k sized list of population elements chosen with replacement.
If the relative weights or cumulative weights are not specified,
the selections are made with equal probability.
File: ~/lib/miniconda3/lib/python3.6/random.py
Type: method
Our implementation, aimed at hiding details, could be
def roll(n, l):
'''Rolls "n" times a dice/coin whose face values are listed in "l".
E.g., roll(2, range(1,21)) -> [12, 4] simulates rolling 2 icosahedron dices.
'''
from random import choices
return choices(l, k=n)
Putting this together
def longest_run(i, s):
'Counts the longest run of item "i" in sequence "s".'
c, m = 0, 0
for el in s:
if el==i:
c += 1
elif c:
m = m if m >= c else c
c = 0
return m
def roll(n, l):
'''Rolls "n" times a dice/coin whose face values are listed in "l".
E.g., roll(2, range(1,21)) -> [12, 4] simulates rolling 2 icosahedron dices.
'''
from random import choices
return choices(l, k=n)
N = 100 # n. of flipped coins
h_or_t = ['h', 't']
random_seq_of_h_or_t = flip(N, h_or_t)
max_h = longest_run('h', random_seq_of_h_or_t)
max_t = longest_run('t', random_seq_of_h_or_t)
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?