I encountered the following algorithmic question which has strict constraints on runtime (<10s and no large memory footprint) and I am stumped. My approach fails half the test cases.
Question
A box contains a number of items that can only br removed 1 at a time or 3 at a time.
How many ways can the box be emptied? the answer can be very large so return it as modulo of 10^9+7.
for example,there are n=7 items initially.They can be removed nine ways,as follows:
1.(1,1,1,1,1,1,1)
2.(1.1.1.1.3)
3.(1,1,1,3,1)
4.(1,1,3,1,1)
5.(1,3,1,1,1)
6.(3,1,1,1,1)
7.(1,3,3)
8.(3,1,3)
9.(3,3,1)
So the function should return 9.
Function Description:
Your function must take in a parameter, n for the number of items, and return an integer which denotes the number of ways to empty the box.
Constraints: 1<=n<=10^8
Sample cases :
Input: 1
Sample OutPut: 1
Explanation: There is only 1 way to remove 1 item. Answer=(1%1000000007)=1
Input: 7
Sample OutPut: 9
There is only 9 ways to remove 7 items
My Approach
This leads to a standard recurrence relation where f(n) = f(n-3) + f(n-1) for n > 2, so i did it as follows
def memoized_number_of_ways(dic, n):
if n not in dic:
dic[n] = memoized_number_of_ways(dic, n-3) + memoized_number_of_ways(dic, n-1)
return dic[n]
def numberOfWays(n):
# Write your code here
memoize = {1:1,2:1,3:2}
import math
ans = memoized_number_of_ways(memoize,n)
return ans % (math.pow(10,9) + 7)
However this fails on any case where n > 10**2. How can you do this problem while accomodating n up to 10^8 and in less than 10s with not much memory?
Just write your recurrence using matrices (pardon my way of writing matrices, StackOverflow doesn't allow LaTeX).
[f(n) ] = [1 0 1] [f(n-1) ]
[f(n-1)] = [1 0 0] [f(n-2) ]
[f(n-2)] = [0 1 0] [f(n-3) ]
Now all you have to do is raise a 3x3 matrix (modulo fixed constant) to power n (or n-3 or something like that, depending on your "base case column vector", fill in the details), and then multiply it by a "base case column vector". This can be done in time O(logn).
PS: You may want to lookup matrix exponentiation.
Three solutions, fastest takes about 31 μs for n=108 on tio.run (which has medium-fast computers).
A matrix power solution like described by advocateofnone that takes about 1 millisecond (Try it online!):
import numpy as np
from time import time
class ModInt:
def __init__(self, x):
self.x = x % (10**9 + 7)
def __add__(a, b):
return ModInt(a.x + b.x)
def __mul__(a, b):
return ModInt(a.x * b.x)
def __str__(self):
return str(self.x)
def solve(n):
O = ModInt(0)
I = ModInt(1)
A = np.matrix([[O,I,O], [O,O,I], [I,O,I]])
return (A**n)[2,2]
for _ in range(3):
t0 = time()
print(solve(10**8), time() - t0)
Output (result and time in seconds for n=108, three attempts):
109786077 0.0010712146759033203
109786077 0.0010180473327636719
109786077 0.0009677410125732422
Another, taking about 0.5 milliseconds (Try it online!):
import numpy as np
from time import time
def solve(n):
A = np.matrix([[0,1,0], [0,0,1], [1,0,1]])
power = 1
mod = 10**9 + 7
while n:
if n % 2:
power = power * A % mod
A = A**2 % mod
n //= 2
return power[2,2]
for _ in range(3):
t0 = time()
print(solve(10**8), time() - t0)
One based on #rici's solution in the comments, takes about 31 μs (Try it online!):
from timeit import repeat
def solve(n):
m = 10**9 + 7
def abc(n):
if n == 0:
return 0, 1, 0
a, b, c = abc(n // 2)
d = a + c
e = b + d
A = 2*a*b + c*c
C = 2*b*c + d*d
E = 2*c*d + e*e
D = A + C
B = E - D
if n % 2:
A, B, C = B, C, D
return A%m, B%m, C%m
return sum(abc(n)) % m
n = 10**8
print(solve(n))
for _ in range(3):
t = min(repeat(lambda: solve(n), 'gc.enable()', number=1000)) / 1000
print('%.1f μs' % (t * 1e6))
Explanation: Looking at the matrix powers from my previous solutions, I noticed they only actually contain five different values, and they're consecutive result numbers from our desired sequence. For example, A**19 is:
[[277 189 406]
[406 277 595]
[595 406 872]]
I gave them names in increasing order:
| b a c |
| c b d |
| d c e |
Squaring that matrix results in a matrix for larger n, with entries A/B/C/D/E. And if you square the above matrix, you'll find the relationships A = 2*a*b + c*c etc.
My helper function abc(n) computes the entries a/b/c of the n-th matrix power. For n=0, that's the identity matrix, so my a/b/c are 0/1/0 there. And in the end I, return the e-value (computed as e=b+d=a+b+c).
Here's a simple iterative O(n) time / O(1) space solution whose optimized version takes 6 seconds on a medium-fast machine (unoptimized takes 15 seconds there).
Unoptimized (Try it online!):
def solve(n):
mod = 10**9 + 7
a = b = c = 1
for _ in range(n):
a, b, c = b, c, (a+c) % mod
return a
print(solve(7))
print(solve(10**8))
Optimized (Try it online!):
def solve(n):
mod = 10**9 + 7
a = b = c = 1
for _ in range(n // 300):
for _ in range(100):
a += c
b += a
c += b
a %= mod
b %= mod
c %= mod
for _ in range(n % 300):
a, b, c = b, c, (a+c) % mod
return a
Your solution is on the right track and the bug is not related to your algorithm (Yay).
The problem is when you are performing operations on some big numbers you lose precision. Notice that you can apply the mod 10 ** 9 + 7 along your code since addition is not affected by it. By doing so you keep all your numbers below a certain size and you will not have any floating point precision errors:
import math
def memoized_number_of_ways(dic, n):
if n not in dic:
dic[n] = (memoized_number_of_ways(dic, n-3) + memoized_number_of_ways(dic, n-1)) % (math.pow(10,9) + 7)
return dic[n]
def numberOfWays(n):
memoize = {1:1,2:1,3:2}
ans = memoized_number_of_ways(memoize,n)
return ans
Note that for you to be able to answer the question for n > 1000 you need to solve this recursion error problem.
Unfortunately even a very efficient solution (hint: you don't really need more than 3 items in your dict at any moment) will not solve the question for n ~ 10 ** 9 under a second. And you will need to find another way - a great option is the second answer here :)
I once got the following as an interview question:
I'm thinking of a positive integer n. Come up with an algorithm that can guess it in O(lg n) queries. Each query is a number of your choosing, and I will answer either "lower," "higher," or "correct."
This problem can be solved by a modified binary search, in which you listing powers of two until you find one that exceeds n, then run a standard binary search over that range. What I think is so cool about this is that you can search an infinite space for a particular number faster than just brute-force.
The question I have, though, is a slight modification of this problem. Instead of picking a positive integer, suppose that I pick an arbitrary rational number between zero and one. My question is: what algorithm can you use to most efficiently determine which rational number I've picked?
Right now, the best solution I have can find p/q in at most O(q) time by implicitly walking the Stern-Brocot tree, a binary search tree over all the rationals. However, I was hoping to get a runtime closer to the runtime that we got for the integer case, maybe something like O(lg (p + q)) or O(lg pq). Does anyone know of a way to get this sort of runtime?
I initially considered using a standard binary search of the interval [0, 1], but this will only find rational numbers with a non-repeating binary representation, which misses almost all of the rationals. I also thought about using some other way of enumerating the rationals, but I can't seem to find a way to search this space given just greater/equal/less comparisons.
Okay, here's my answer using continued fractions alone.
First let's get some terminology here.
Let X = p/q be the unknown fraction.
Let Q(X,p/q) = sign(X - p/q) be the query function: if it is 0, we've guessed the number, and if it's +/- 1 that tells us the sign of our error.
The conventional notation for continued fractions is A = [a0; a1, a2, a3, ... ak]
= a0 + 1/(a1 + 1/(a2 + 1/(a3 + 1/( ... + 1/ak) ... )))
We'll follow the following algorithm for 0 < p/q < 1.
Initialize Y = 0 = [ 0 ], Z = 1 = [ 1 ], k = 0.
Outer loop: The preconditions are that:
Y and Z are continued fractions of k+1 terms which are identical except in the last element, where they differ by 1, so that Y = [y0; y1, y2, y3, ... yk] and Z = [y0; y1, y2, y3, ... yk + 1]
(-1)k(Y-X) < 0 < (-1)k(Z-X), or in simpler terms, for k even, Y < X < Z and for k odd, Z < X < Y.
Extend the degree of the continued fraction by 1 step without changing the values of the numbers. In general, if the last terms are yk and yk + 1, we change that to [... yk, yk+1=∞] and [... yk, zk+1=1]. Now increase k by 1.
Inner loops: This is essentially the same as #templatetypedef's interview question about the integers. We do a two-phase binary search to get closer:
Inner loop 1: yk = ∞, zk = a, and X is between Y and Z.
Double Z's last term: Compute M = Z but with mk = 2*a = 2*zk.
Query the unknown number: q = Q(X,M).
If q = 0, we have our answer and go to step 17 .
If q and Q(X,Y) have opposite signs, it means X is between Y and M, so set Z = M and go to step 5.
Otherwise set Y = M and go to the next step:
Inner loop 2. yk = b, zk = a, and X is between Y and Z.
If a and b differ by 1, swap Y and Z, go to step 2.
Perform a binary search: compute M where mk = floor((a+b)/2, and query q = Q(X,M).
If q = 0, we're done and go to step 17.
If q and Q(X,Y) have opposite signs, it means X is between Y and M, so set Z = M and go to step 11.
Otherwise, q and Q(X,Z) have opposite signs, it means X is between Z and M, so set Y = M and go to step 11.
Done: X = M.
A concrete example for X = 16/113 = 0.14159292
Y = 0 = [0], Z = 1 = [1], k = 0
k = 1:
Y = 0 = [0; ∞] < X, Z = 1 = [0; 1] > X, M = [0; 2] = 1/2 > X.
Y = 0 = [0; ∞], Z = 1/2 = [0; 2], M = [0; 4] = 1/4 > X.
Y = 0 = [0; ∞], Z = 1/4 = [0; 4], M = [0; 8] = 1/8 < X.
Y = 1/8 = [0; 8], Z = 1/4 = [0; 4], M = [0; 6] = 1/6 > X.
Y = 1/8 = [0; 8], Z = 1/6 = [0; 6], M = [0; 7] = 1/7 > X.
Y = 1/8 = [0; 8], Z = 1/7 = [0; 7]
--> the two last terms differ by one, so swap and repeat outer loop.
k = 2:
Y = 1/7 = [0; 7, ∞] > X, Z = 1/8 = [0; 7, 1] < X,
M = [0; 7, 2] = 2/15 < X
Y = 1/7 = [0; 7, ∞], Z = 2/15 = [0; 7, 2],
M = [0; 7, 4] = 4/29 < X
Y = 1/7 = [0; 7, ∞], Z = 4/29 = [0; 7, 4],
M = [0; 7, 8] = 8/57 < X
Y = 1/7 = [0; 7, ∞], Z = 8/57 = [0; 7, 8],
M = [0; 7, 16] = 16/113 = X
--> done!
At each step of computing M, the range of the interval reduces. It is probably fairly easy to prove (though I won't do this) that the interval reduces by a factor of at least 1/sqrt(5) at each step, which would show that this algorithm is O(log q) steps.
Note that this can be combined with templatetypedef's original interview question and apply towards any rational number p/q, not just between 0 and 1, by first computing Q(X,0), then for either positive/negative integers, bounding between two consecutive integers, and then using the above algorithm for the fractional part.
When I have a chance next, I will post a python program that implements this algorithm.
edit: also, note that you don't have to compute the continued fraction each step (which would be O(k), there are partial approximants to continued fractions that can compute the next step from the previous step in O(1).)
edit 2: Recursive definition of partial approximants:
If Ak = [a0; a1, a2, a3, ... ak] = pk/qk, then pk = akpk-1 + pk-2, and qk = akqk-1 + qk-2. (Source: Niven & Zuckerman, 4th ed, Theorems 7.3-7.5. See also Wikipedia)
Example: [0] = 0/1 = p0/q0, [0; 7] = 1/7 = p1/q1; so [0; 7, 16] = (16*1+0)/(16*7+1) = 16/113 = p2/q2.
This means that if two continued fractions Y and Z have the same terms except the last one, and the continued fraction excluding the last term is pk-1/qk-1, then we can write Y = (ykpk-1 + pk-2) / (ykqk-1 + qk-2) and Z = (zkpk-1 + pk-2) / (zkqk-1 + qk-2). It should be possible to show from this that |Y-Z| decreases by at least a factor of 1/sqrt(5) at each smaller interval produced by this algorithm, but the algebra seems to be beyond me at the moment. :-(
Here's my Python program:
import math
# Return a function that returns Q(p0/q0,p/q)
# = sign(p0/q0-p/q) = sign(p0q-q0p)*sign(q0*q)
# If p/q < p0/q0, then Q() = 1; if p/q < p0/q0, then Q() = -1; otherwise Q()=0.
def makeQ(p0,q0):
def Q(p,q):
return cmp(q0*p,p0*q)*cmp(q0*q,0)
return Q
def strsign(s):
return '<' if s<0 else '>' if s>0 else '=='
def cfnext(p1,q1,p2,q2,a):
return [a*p1+p2,a*q1+q2]
def ratguess(Q, doprint, kmax):
# p2/q2 = p[k-2]/q[k-2]
p2 = 1
q2 = 0
# p1/q1 = p[k-1]/q[k-1]
p1 = 0
q1 = 1
k = 0
cf = [0]
done = False
while not done and (not kmax or k < kmax):
if doprint:
print 'p/q='+str(cf)+'='+str(p1)+'/'+str(q1)
# extend continued fraction
k = k + 1
[py,qy] = [p1,q1]
[pz,qz] = cfnext(p1,q1,p2,q2,1)
ay = None
az = 1
sy = Q(py,qy)
sz = Q(pz,qz)
while not done:
if doprint:
out = str(py)+'/'+str(qy)+' '+strsign(sy)+' X '
out += strsign(-sz)+' '+str(pz)+'/'+str(qz)
out += ', interval='+str(abs(1.0*py/qy-1.0*pz/qz))
if ay:
if (ay - az == 1):
[p0,q0,a0] = [pz,qz,az]
break
am = (ay+az)/2
else:
am = az * 2
[pm,qm] = cfnext(p1,q1,p2,q2,am)
sm = Q(pm,qm)
if doprint:
out = str(ay)+':'+str(am)+':'+str(az) + ' ' + out + '; M='+str(pm)+'/'+str(qm)+' '+strsign(sm)+' X '
print out
if (sm == 0):
[p0,q0,a0] = [pm,qm,am]
done = True
break
elif (sm == sy):
[py,qy,ay,sy] = [pm,qm,am,sm]
else:
[pz,qz,az,sz] = [pm,qm,am,sm]
[p2,q2] = [p1,q1]
[p1,q1] = [p0,q0]
cf += [a0]
print 'p/q='+str(cf)+'='+str(p1)+'/'+str(q1)
return [p1,q1]
and a sample output for ratguess(makeQ(33102,113017), True, 20):
p/q=[0]=0/1
None:2:1 0/1 < X < 1/1, interval=1.0; M=1/2 > X
None:4:2 0/1 < X < 1/2, interval=0.5; M=1/4 < X
4:3:2 1/4 < X < 1/2, interval=0.25; M=1/3 > X
p/q=[0, 3]=1/3
None:2:1 1/3 > X > 1/4, interval=0.0833333333333; M=2/7 < X
None:4:2 1/3 > X > 2/7, interval=0.047619047619; M=4/13 > X
4:3:2 4/13 > X > 2/7, interval=0.021978021978; M=3/10 > X
p/q=[0, 3, 2]=2/7
None:2:1 2/7 < X < 3/10, interval=0.0142857142857; M=5/17 > X
None:4:2 2/7 < X < 5/17, interval=0.00840336134454; M=9/31 < X
4:3:2 9/31 < X < 5/17, interval=0.00379506641366; M=7/24 < X
p/q=[0, 3, 2, 2]=5/17
None:2:1 5/17 > X > 7/24, interval=0.00245098039216; M=12/41 < X
None:4:2 5/17 > X > 12/41, interval=0.00143472022956; M=22/75 > X
4:3:2 22/75 > X > 12/41, interval=0.000650406504065; M=17/58 > X
p/q=[0, 3, 2, 2, 2]=12/41
None:2:1 12/41 < X < 17/58, interval=0.000420521446594; M=29/99 > X
None:4:2 12/41 < X < 29/99, interval=0.000246366100025; M=53/181 < X
4:3:2 53/181 < X < 29/99, interval=0.000111613371282; M=41/140 < X
p/q=[0, 3, 2, 2, 2, 2]=29/99
None:2:1 29/99 > X > 41/140, interval=7.21500721501e-05; M=70/239 < X
None:4:2 29/99 > X > 70/239, interval=4.226364059e-05; M=128/437 > X
4:3:2 128/437 > X > 70/239, interval=1.91492009996e-05; M=99/338 > X
p/q=[0, 3, 2, 2, 2, 2, 2]=70/239
None:2:1 70/239 < X < 99/338, interval=1.23789953207e-05; M=169/577 > X
None:4:2 70/239 < X < 169/577, interval=7.2514738621e-06; M=309/1055 < X
4:3:2 309/1055 < X < 169/577, interval=3.28550190148e-06; M=239/816 < X
p/q=[0, 3, 2, 2, 2, 2, 2, 2]=169/577
None:2:1 169/577 > X > 239/816, interval=2.12389981991e-06; M=408/1393 < X
None:4:2 169/577 > X > 408/1393, interval=1.24415093544e-06; M=746/2547 < X
None:8:4 169/577 > X > 746/2547, interval=6.80448470014e-07; M=1422/4855 < X
None:16:8 169/577 > X > 1422/4855, interval=3.56972657711e-07; M=2774/9471 > X
16:12:8 2774/9471 > X > 1422/4855, interval=1.73982239227e-07; M=2098/7163 > X
12:10:8 2098/7163 > X > 1422/4855, interval=1.15020646951e-07; M=1760/6009 > X
10:9:8 1760/6009 > X > 1422/4855, interval=6.85549088053e-08; M=1591/5432 < X
p/q=[0, 3, 2, 2, 2, 2, 2, 2, 9]=1591/5432
None:2:1 1591/5432 < X < 1760/6009, interval=3.06364213998e-08; M=3351/11441 < X
p/q=[0, 3, 2, 2, 2, 2, 2, 2, 9, 1]=1760/6009
None:2:1 1760/6009 > X > 3351/11441, interval=1.45456726663e-08; M=5111/17450 < X
None:4:2 1760/6009 > X > 5111/17450, interval=9.53679318849e-09; M=8631/29468 < X
None:8:4 1760/6009 > X > 8631/29468, interval=5.6473816179e-09; M=15671/53504 < X
None:16:8 1760/6009 > X > 15671/53504, interval=3.11036635336e-09; M=29751/101576 > X
16:12:8 29751/101576 > X > 15671/53504, interval=1.47201634215e-09; M=22711/77540 > X
12:10:8 22711/77540 > X > 15671/53504, interval=9.64157420569e-10; M=19191/65522 > X
10:9:8 19191/65522 > X > 15671/53504, interval=5.70501257346e-10; M=17431/59513 > X
p/q=[0, 3, 2, 2, 2, 2, 2, 2, 9, 1, 8]=15671/53504
None:2:1 15671/53504 < X < 17431/59513, interval=3.14052228667e-10; M=33102/113017 == X
Since Python handles biginteger math from the start, and this program uses only integer math (except for the interval calculations), it should work for arbitrary rationals.
edit 3: Outline of proof that this is O(log q), not O(log^2 q):
First note that until the rational number is found, the # of steps nk for each new continued fraction term is exactly 2b(a_k)-1 where b(a_k) is the # of bits needed to represent a_k = ceil(log2(a_k)): it's b(a_k) steps to widen the "net" of the binary search, and b(a_k)-1 steps to narrow it). See the example above, you'll note that the # of steps is always 1, 3, 7, 15, etc.
Now we can use the recurrence relation qk = akqk-1 + qk-2 and induction to prove the desired result.
Let's state it in this way: that the value of q after the Nk = sum(nk) steps required for reaching the kth term has a minimum: q >= A*2cN for some fixed constants A,c. (so to invert, we'd get that the # of steps N is <= (1/c) * log2 (q/A) = O(log q).)
Base cases:
k=0: q = 1, N = 0, so q >= 2N
k=1: for N = 2b-1 steps, q = a1 >= 2b-1 = 2(N-1)/2 = 2N/2/sqrt(2).
This implies A = 1, c = 1/2 could provide desired bounds. In reality, q may not double each term (counterexample: [0; 1, 1, 1, 1, 1] has a growth factor of phi = (1+sqrt(5))/2) so let's use c = 1/4.
Induction:
for term k, qk = akqk-1 + qk-2. Again, for the nk = 2b-1 steps needed for this term, ak >= 2b-1 = 2(nk-1)/2.
So akqk-1 >= 2(Nk-1)/2 * qk-1 >= 2(nk-1)/2 * A*2Nk-1/4 = A*2Nk/4/sqrt(2)*2nk/4.
Argh -- the tough part here is that if ak = 1, q may not increase much for that one term, and we need to use qk-2 but that may be much smaller than qk-1.
Let's take the rational numbers, in reduced form, and write them out in order first of denominator, then numerator.
1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 5/6, ...
Our first guess is going to be 1/2. Then we'll go along the list until we have 3 in our range. Then we will take 2 guesses to search that list. Then we'll go along the list until we have 7 in our remaining range. Then we will take 3 guesses to search that list. And so on.
In n steps we'll cover the first 2O(n) possibilities, which is in the order of magnitude of efficiency that you were looking for.
Update: People didn't get the reasoning behind this. The reasoning is simple. We know how to walk a binary tree efficiently. There are O(n2) fractions with maximum denominator n. We could therefore search up to any particular denominator size in O(2*log(n)) = O(log(n)) steps. The problem is that we have an infinite number of possible rationals to search. So we can't just line them all up, order them, and start searching.
Therefore my idea was to line up a few, search, line up more, search, and so on. Each time we line up more we line up about double what we did last time. So we need one more guess than we did last time. Therefore our first pass uses 1 guess to traverse 1 possible rational. Our second uses 2 guesses to traverse 3 possible rationals. Our third uses 3 guesses to traverse 7 possible rationals. And our k'th uses k guesses to traverse 2k-1 possible rationals. For any particular rational m/n, eventually it will wind up putting that rational on a fairly big list that it knows how to do a binary search on efficiently.
If we did binary searches, then ignored everything we'd learned when we grab more rationals, then we'd put all of the rationals up to and including m/n in O(log(n)) passes. (That's because by that point we'll get to a pass with enough rationals to include every rational up to and including m/n.) But each pass takes more guesses, so that would be O(log(n)2) guesses.
However we actually do a lot better than that. With our first guess, we eliminate half the rationals on our list as being too big or small. Our next two guesses don't quite cut the space into quarters, but they don't come too far from it. Our next 3 guesses again don't quite cut the space into eighths, but they don't come too far from it. And so on. When you put it together, I'm convinced that the result is that you find m/n in O(log(n)) steps. Though I don't actually have a proof.
Try it out: Here is code to generate the guesses so that you can play and see how efficient it is.
#! /usr/bin/python
from fractions import Fraction
import heapq
import readline
import sys
def generate_next_guesses (low, high, limit):
upcoming = [(low.denominator + high.denominator,
low.numerator + high.numerator,
low.denominator, low.numerator,
high.denominator, high.numerator)]
guesses = []
while len(guesses) < limit:
(mid_d, mid_n, low_d, low_n, high_d, high_n) = upcoming[0]
guesses.append(Fraction(mid_n, mid_d))
heapq.heappushpop(upcoming, (low_d + mid_d, low_n + mid_n,
low_d, low_n, mid_d, mid_n))
heapq.heappush(upcoming, (mid_d + high_d, mid_n + high_n,
mid_d, mid_n, high_d, high_n))
guesses.sort()
return guesses
def ask (num):
while True:
print "Next guess: {0} ({1})".format(num, float(num))
if 1 < len(sys.argv):
wanted = Fraction(sys.argv[1])
if wanted < num:
print "too high"
return 1
elif num < wanted:
print "too low"
return -1
else:
print "correct"
return 0
answer = raw_input("Is this (h)igh, (l)ow, or (c)orrect? ")
if answer == "h":
return 1
elif answer == "l":
return -1
elif answer == "c":
return 0
else:
print "Not understood. Please say one of (l, c, h)"
guess_size_bound = 2
low = Fraction(0)
high = Fraction(1)
guesses = [Fraction(1,2)]
required_guesses = 0
answer = -1
while 0 != answer:
if 0 == len(guesses):
guess_size_bound *= 2
guesses = generate_next_guesses(low, high, guess_size_bound - 1)
#print (low, high, guesses)
guess = guesses[len(guesses)/2]
answer = ask(guess)
required_guesses += 1
if 0 == answer:
print "Thanks for playing!"
print "I needed %d guesses" % required_guesses
elif 1 == answer:
high = guess
guesses[len(guesses)/2:] = []
else:
low = guess
guesses[0:len(guesses)/2 + 1] = []
As an example to try it out I tried 101/1024 (0.0986328125) and found that it took 20 guesses to find the answer. I tried 0.98765 and it took 45 guesses. I tried 0.0123456789 and it needed 66 guesses and about a second to generate them. (Note, if you call the program with a rational number as an argument, it will fill in all of the guesses for you. This is a very helpful convenience.)
I've got it! What you need to do is to use a parallel search with bisection and continued fractions.
Bisection will give you a limit toward a specific real number, as represented as a power of two, and continued fractions will take the real number and find the nearest rational number.
How you run them in parallel is as follows.
At each step, you have l and u being the lower and upper bounds of bisection. The idea is, you have a choice between halving the range of bisection, and adding an additional term as a continued fraction representation. When both l and u have the same next term as a continued fraction, then you take the next step in the continued fraction search, and make a query using the continued fraction. Otherwise, you halve the range using bisection.
Since both methods increase the denominator by at least a constant factor (bisection goes by factors of 2, continued fractions go by at least a factor of phi = (1+sqrt(5))/2), this means your search should be O(log(q)). (There may be repeated continued fraction calculations, so it may end up as O(log(q)^2).)
Our continued fraction search needs to round to the nearest integer, not use floor (this is clearer below).
The above is kind of handwavy. Let's use a concrete example of r = 1/31:
l = 0, u = 1, query = 1/2. 0 is not expressible as a continued fraction, so we use binary search until l != 0.
l = 0, u = 1/2, query = 1/4.
l = 0, u = 1/4, query = 1/8.
l = 0, u = 1/8, query = 1/16.
l = 0, u = 1/16, query = 1/32.
l = 1/32, u = 1/16. Now 1/l = 32, 1/u = 16, these have different cfrac reps, so keep bisecting., query = 3/64.
l = 1/32, u = 3/64, query = 5/128 = 1/25.6
l = 1/32, u = 5/128, query = 9/256 = 1/28.4444....
l = 1/32, u = 9/256, query = 17/512 = 1/30.1176... (round to 1/30)
l = 1/32, u = 17/512, query = 33/1024 = 1/31.0303... (round to 1/31)
l = 33/1024, u = 17/512, query = 67/2048 = 1/30.5672... (round to 1/31)
l = 33/1024, u = 67/2048. At this point both l and u have the same continued fraction term 31, so now we use a continued fraction guess.
query = 1/31.
SUCCESS!
For another example let's use 16/113 (= 355/113 - 3 where 355/113 is pretty close to pi).
[to be continued, I have to go somewhere]
On further reflection, continued fractions are the way to go, never mind bisection except to determine the next term. More when I get back.
I think I found an O(log^2(p + q)) algorithm.
To avoid confusion in the next paragraph, a "query" refers to when the guesser gives the challenger a guess, and the challenger responds "bigger" or "smaller". This allows me to reserve the word "guess" for something else, a guess for p + q that is not asked directly to the challenger.
The idea is to first find p + q, using the algorithm you describe in your question: guess a value k, if k is too small, double it and try again. Then once you have an upper and lower bound, do a standard binary search. This takes O(log(p+q)T) queries, where T is an upper bound for the number of queries it takes to check a guess. Let's find T.
We want to check all fractions r/s with r + s <= k, and double k until k is sufficiently large. Note that there are O(k^2) fractions you need to check for a given value of k. Build a balanced binary search tree containing all these values, then search it to determine if p/q is in the tree. It takes O(log k^2) = O(log k) queries to confirm that p/q is not in the tree.
We will never guess a value of k greater than 2(p + q). Hence we can take T = O(log(p+q)).
When we guess the correct value for k (i.e., k = p + q), we will submit the query p/q to the challenger in the course of checking our guess for k, and win the game.
Total number of queries is then O(log^2(p + q)).
Okay, I think I figured out an O(lg2 q) algorithm for this problem that is based on Jason S's most excellent insight about using continued fractions. I thought I'd flesh the algorithm out all the way right here so that we have a complete solution, along with a runtime analysis.
The intuition behind the algorithm is that any rational number p/q within the range can be written as
a0 + 1 / (a1 + 1 / (a2 + 1 / (a3 + 1 / ...))
For appropriate choices of ai. This is called a continued fraction. More importantly, though these ai can be derived by running the Euclidean algorithm on the numerator and denominator. For example, suppose we want to represent 11/14 this way. We begin by noting that 14 goes into eleven zero times, so a crude approximation of 11/14 would be
0 = 0
Now, suppose that we take the reciprocal of this fraction to get 14/11 = 1 3/11. So if we write
0 + (1 / 1) = 1
We get a slightly better approximation to 11/14. Now that we're left with 3 / 11, we can take the reciprocal again to get 11/3 = 3 2/3, so we can consider
0 + (1 / (1 + 1/3)) = 3/4
Which is another good approximation to 11/14. Now, we have 2/3, so consider the reciprocal, which is 3/2 = 1 1/2. If we then write
0 + (1 / (1 + 1/(3 + 1/1))) = 5/6
We get another good approximation to 11/14. Finally, we're left with 1/2, whose reciprocal is 2/1. If we finally write out
0 + (1 / (1 + 1/(3 + 1/(1 + 1/2)))) = (1 / (1 + 1/(3 + 1/(3/2)))) = (1 / (1 + 1/(3 + 2/3)))) = (1 / (1 + 1/(11/3)))) = (1 / (1 + 3/11)) = 1 / (14/11) = 11/14
which is exactly the fraction we wanted. Moreover, look at the sequence of coefficients we ended up using. If you run the extended Euclidean algorithm on 11 and 14, you get that
11 = 0 x 14 + 11 --> a0 = 0
14 = 1 x 11 + 3 --> a1 = 1
11 = 3 x 3 + 2 --> a2 = 3
3 = 2 x 1 + 1 --> a3 = 2
It turns out that (using more math than I currently know how to do!) that this isn't a coincidence and that the coefficients in the continued fraction of p/q are always formed by using the extended Euclidean algorithm. This is great, because it tells us two things:
There can be at most O(lg (p + q)) coefficients, because the Euclidean algorithm always terminates in this many steps, and
Each coefficient is at most max{p, q}.
Given these two facts, we can come up with an algorithm to recover any rational number p/q, not just those between 0 and 1, by applying the general algorithm for guessing arbitrary integers n one at a time to recover all of the coefficients in the continued fraction for p/q. For now, though, we'll just worry about numbers in the range (0, 1], since the logic for handling arbitrary rational numbers can be done easily given this as a subroutine.
As a first step, let's suppose that we want to find the best value of a1 so that 1 / a1 is as close as possible to p/q and a1 is an integer. To do this, we can just run our algorithm for guessing arbitrary integers, taking the reciprocal each time. After doing this, one of two things will have happened. First, we might by sheer coincidence discover that p/q = 1/k for some integer k, in which case we're done. If not, we'll find that p/q is sandwiched between 1/(a1 - 1) and 1/a0 for some a1. When we do this, then we start working on the continued fraction one level deeper by finding the a2 such that p/q is between 1/(a1 + 1/a2) and 1/(a1 + 1/(a2 + 1)). If we magically find p/q, that's great! Otherwise, we then go one level down further in the continued fraction. Eventually, we'll find the number this way, and it can't take too long. Each binary search to find a coefficient takes at most O(lg(p + q)) time, and there are at most O(lg(p + q)) levels to the search, so we need only O(lg2(p + q)) arithmetic operations and probes to recover p/q.
One detail I want to point out is that we need to keep track of whether we're on an odd level or an even level when doing the search because when we sandwich p/q between two continued fractions, we need to know whether the coefficient we were looking for was the upper or the lower fraction. I'll state without proof that for ai with i odd you want to use the upper of the two numbers, and with ai even you use the lower of the two numbers.
I am almost 100% confident that this algorithm works. I'm going to try to write up a more formal proof of this in which I fill in all of the gaps in this reasoning, and when I do I'll post a link here.
Thanks to everyone for contributing the insights necessary to get this solution working, especially Jason S for suggesting a binary search over continued fractions.
Remember that any rational number in (0, 1) can be represented as a finite sum of distinct (positive or negative) unit fractions. For example, 2/3 = 1/2 + 1/6 and 2/5 = 1/2 - 1/10. You can use this to perform a straight-forward binary search.
Here is yet another way to do it. If there is sufficient interest, I will try to fill out the details tonight, but I can't right now because I have family responsibilities. Here is a stub of an implementation that should explain the algorithm:
low = 0
high = 1
bound = 2
answer = -1
while 0 != answer:
mid = best_continued_fraction((low + high)/2, bound)
while mid == low or mid == high:
bound += bound
mid = best_continued_fraction((low + high)/2, bound)
answer = ask(mid)
if -1 == answer:
low = mid
elif 1 == answer:
high = mid
else:
print_success_message(mid)
And here is the explanation. What best_continued_fraction(x, bound) should do is find the last continued fraction approximation to x with the denominator at most bound. This algorithm will take polylog steps to complete and finds very good (though not always the best) approximations. So for each bound we'll get something close to a binary search through all possible fractions of that size. Occasionally we won't find a particular fraction until we increase the bound farther than we should, but we won't be far off.
So there you have it. A logarithmic number of questions found with polylog work.
Update: And full working code.
#! /usr/bin/python
from fractions import Fraction
import readline
import sys
operations = [0]
def calculate_continued_fraction(terms):
i = len(terms) - 1
result = Fraction(terms[i])
while 0 < i:
i -= 1
operations[0] += 1
result = terms[i] + 1/result
return result
def best_continued_fraction (x, bound):
error = x - int(x)
terms = [int(x)]
last_estimate = estimate = Fraction(0)
while 0 != error and estimate.numerator < bound:
operations[0] += 1
error = 1/error
term = int(error)
terms.append(term)
error -= term
last_estimate = estimate
estimate = calculate_continued_fraction(terms)
if estimate.numerator < bound:
return estimate
else:
return last_estimate
def ask (num):
while True:
print "Next guess: {0} ({1})".format(num, float(num))
if 1 < len(sys.argv):
wanted = Fraction(sys.argv[1])
if wanted < num:
print "too high"
return 1
elif num < wanted:
print "too low"
return -1
else:
print "correct"
return 0
answer = raw_input("Is this (h)igh, (l)ow, or (c)orrect? ")
if answer == "h":
return 1
elif answer == "l":
return -1
elif answer == "c":
return 0
else:
print "Not understood. Please say one of (l, c, h)"
ow = Fraction(0)
high = Fraction(1)
bound = 2
answer = -1
guesses = 0
while 0 != answer:
mid = best_continued_fraction((low + high)/2, bound)
guesses += 1
while mid == low or mid == high:
bound += bound
mid = best_continued_fraction((low + high)/2, bound)
answer = ask(mid)
if -1 == answer:
low = mid
elif 1 == answer:
high = mid
else:
print "Thanks for playing!"
print "I needed %d guesses and %d operations" % (guesses, operations[0])
It appears slightly more efficient in guesses than the previous solution, and does a lot fewer operations. For 101/1024 it required 19 guesses and 251 operations. For .98765 it needed 27 guesses and 623 operations. For 0.0123456789 it required 66 guesses and 889 operations. And for giggles and grins, for 0.0123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789 (that's 10 copies of the previous one) it required 665 guesses and 23289 operations.
You can sort rational numbers in a given interval by for example the pair (denominator, numerator). Then to play the game you can
Find the interval [0, N] using the doubling-step approach
Given an interval [a, b] shoot for the rational with smallest denominator in the interval that is the closest to the center of the interval
this is however probably still O(log(num/den) + den) (not sure and it's too early in the morning here to make me think clearly ;-) )