Minimize function using dynamic programming - algorithm

How to minimize function y12 + y22 + ... + yn2 with constraints y1*y2*...*yn = c; y1,y2,...,yn > 0 using dynamic programming? I have tried to solve this problem, but I have no idea how to create a recurrent function.


You need to think how to reduce the problem into "smaller problem"
D(i,c) = min { D(i-1, c/y) + y^2 | 1 <= y <= c }
In the above, you reduce the problem from y1,y2,....,yi to y1,...,y_{i-1}, and check all possible assignments for y_i - and chose the best out of them.
Base clauses will be:
D(0,0) = 0
D(i,0) = Infinity i>0
You can do a top-down or bottom-up DP solution with these recurrence formulas, assuming i,c are integers. (Might need to add stop clause of D(i,c) = Infinity if c is not natural


Based on your question and the clarification given, I think dynamic programming is not necessary. The minimal solution is to choose y1, ..., yn to be the prime factors of c (with repeats).
Example: Given c = 60, we let y1 = 2, y2 = 2, y3 = 3, y4 = 5.
Then the sum is y12 + y22 + y32 + y42 = 4 + 4 + 9 + 25 = 42.
If we took fewer factors, then the sum would be bigger:
y1 = 60, then sum = 602 = 3600.
y1 = 5, y2 = 6, then sum = 52 + 62 = 61.
y1 = 3, y2 = 4, y2 = 5, then sum = 32 + 42 + 45 = 50.
Informal justification - it is always beneficial to split the product:
Suppose c = ab, with a and b ≥ 2.
Then c2 = a2b2. (a2b2 > 2a2) and (a2b2 > 2b2) are both true.
Adding these inequalities we get 2a2b2 > 2a2 + 2b2.
Therefore c2 = a2b2 > a2 + b2.

Related

Finding min and max of a linear system

So I'm trying to make sense of a scenario in my class exercise which is to find the max and min value of a function. I have two vectors, w and v, of weights which are to sum to 1. The vectors are w = [0.6, 0.2, 0.2]^T v = [0.8, -0.2, 0.4]^T
These vectors form a linear combination of weights M = Aw + Bv, and A and B must sum to 1.
The function we are then optimizing is r = [0.1, 0.2, 0.1] • M
The constraints are as follows: 0 ≤ (0.6A + 0.8B) <= 1 , 0 ≤ (0.2A - 0.2B) <= 1 , 0 ≤ (0.2A + 0.4B) <= 1
The answer we should get are A = B = .5 for the minimum value of r which is 0.1. For the maximum we should get A = 2, B = -1 with r = 0.16. But the values I'm getting for the max are A = 3.5714286, B = -1.4285714, and for the min I'm getting A = B = 0.
Below is the code.
import pulp as p
from pulp import *
problem = LpProblem('Car Factory', LpMaximize)
A = LpVariable('Amound of w', cat=LpContinuous)
B = LpVariable('Amount of v', cat=LpContinuous)
#Objective Function
problem += (0.1)*(0.6*A + 0.8*B) + (0.2)*(0.2*A - 0.2*B) + (0.1)*(0.2*A + 0.4*B) , 'Objective Function'
#Constraints
problem += (0.6*A + 0.8*B) <= 1 , 'A'
problem += (0.6*A + 0.8*B) >= 0 , 'AL'
problem += (0.2*A - 0.2*B) <= 1, 'B'
problem += (0.2*A - 0.2*B) >= 0, 'BL'
problem += (0.2*A + 0.4*B) <= 1, 'C'
problem += (0.2*A + 0.4*B) >= 0, 'CL'
problem.solve()
print("Amount of w: ", A.varValue)
print("Amount of v: ", B.varValue)
print("total: ", value(problem.objective))
I'm sure it has to do with the set up which I'm just not seeing. And also is there a more efficient way to put this together?

I think you are missing a constraint, which would explain your deviation from the expected result. Where is your constraint that:
A + B == 1
Also, you are importing pulp twice, which may cause some confusion in the namespace of your code. Do one or the other, not both.
On expressing the problem more efficiently...? Nahh. You could treat your two column vectors as arrays of length 3 and do the math in your objective a bit differently, but it probably isn't worth it and your variables are just scalars, so I'd write it as you did. Now if the vectors were much larger, or if the variables were vectors, sure, I'd do something else.
pulp doesn't naturally handle vectors (like numpy arrays) to my knowledge. If you are going to be doing a lot of optimization in vector-matrix format and you are comfortable with the linear algebra, you might look at cvxpy which handles them naturally. If you're in a class that uses pulp, it's just fine to learn the basics.

Number of N-digit numbers that are divisible by given two numbers

One of my friends got this question in google coding contest. Here goes the question.
Find the number of N-digit numbers that are divisible by both X and Y.
Since the answer can be very large, print the answer modulo 10^9 + 7.
Note: 0 is not considered single-digit number.
Input: N, X, Y.
Constraints:
1 <= N <= 10000
1 <= X,Y <= 20
Eg-1 :
N = 2, X = 5, Y = 7
output : 2 (35 and 70 are the required numbers)
Eg-2 :
N = 1, X = 2, Y = 3
output : 1 (6 is the required number)
If the constraints on N were smaller, then it would be easy (ans = 10^N / LCM(X,Y) - 10^(N-1) / LCM(X,Y)).
But N is upto 1000, hence I am unable to solve it.

This question looks like it was intended to be more difficult, but I would do it pretty much the way you said:
ans = floor((10N-1)/LCM(X,Y)) - floor((10N-1-1)/LCM(X,Y))
The trick is to calculate the terms quickly.
Let M = LCM(X,Y), and say we have:
10a = Mqa + ra, and
10b = Mqb + rb
The we can easily calculate:
10a+b = M(Mqaqb + raqb + rbqa + floor(rarb/M)) + (rarb%M)
With that formula, we can calculate the quotient and remainder for 10N/M in just 2 log N steps using exponentiation by squaring: https://en.wikipedia.org/wiki/Exponentiation_by_squaring

Following python works for this question ,
import math
MOD = 1000000007
def sub(x,y):
return (x-y+MOD)%MOD
def mul(x,y):
return (x*y)%MOD
def power(x,y):
res = 1
x%=MOD
while y!=0:
if y&1 :
res = mul(res,x)
y>>=1
x = mul(x,x)
return res
def mod_inv(n):
return power(n,MOD-2)
x,y = [int(i) for i in input().split()]
m = math.lcm(x,y)
n = int(input())
a = -1
b = -1
total = 1
for i in range(n-1):
total = (total * 10)%m
b = total % m
total = (total*10)%m
a = total % m
l = power(10 , n-1)
r = power(10 , n)
ans = sub( sub(r , l) , sub(a,b) )
ans = mul(ans , mod_inv(m))
print(ans)
Approach for this question is pretty straight forward,
let, m = lcm(x,y)
let,
10^n -1 = m*x + a
10^(n-1) -1 = m*y + b
now from above two equations it is clear that our answer is equal to
(x - y)%MOD .
so,
(x-y) = ((10^n - 10^(n-1)) - (a-b)) / m
also , a = (10^n)%m and b = (10^(n-1))%m
using simple modular arithmetic rules we can easily calculate a and b in O(n) time.
also for subtraction and division performed in the formula we can use modular subtraction and division respectively.
Note: (a/b)%MOD = ( a * (mod_inverse(b, MOD)%MOD )%MOD

Algorithm to solve a Hacker earth problem

I have been working on a Hackerearth Problem. Here is the problem statement:
We have three variables a, b and c. We need to convert a to b and following operations are allowed:
1. Can decrement by 1.
2. Can decrement by 2.
3. Can multiply by c.
Minimum steps required to convert a to b.
Here is the algorithm I came up with:
Increment count to 0.
Loop through till a === b:
1. Perform (x = a * c), (y = a - 1) and (z = a - 2).
2. Among x, y and z, choose the one whose absolute difference with b is the least.
3. Update the value of a to the value chosen among x, y and z.
4. Increment the count by 1.
I can get pass the basic test case but all my advance cases are failing. I guess my logic is correct but due to the complexity it seems to fail.
Can someone suggest a more optimized solution.
Edit 1
Sample Code
function findMinStep(arr) {
let a = parseInt(arr[0]);
let b = parseInt(arr[1]);
let c = parseInt(arr[2]);
let numOfSteps = 0;
while(a !== b) {
let multiply = Math.abs(b - (a * c));
let decrement = Math.abs(b - (a - 1));
let doubleDecrement = Math.abs(b - (a - 2));
let abs = Math.min(multiply, decrement, doubleDecrement);
if(abs === multiply) a = a * c;
else if(abs === decrement) a -= 1;
else a -= 2;
numOfSteps += 1;
}
return numOfSteps.toString()
}
Sample Input: a = 3, b = 10, c = 2
Explanation: Multiply 3 with 2 to get 6, subtract 1 from 6 to get 5, multiply 5 with 2 to get 10.
Reason for tagging both Python and JS: Comfortable with both but I am not looking for code, just an optimized algorithm and analytical thinking.
Edit 2:
function findMinStep(arr) {
let a = parseInt(arr[0]);
let b = parseInt(arr[1]);
let c = parseInt(arr[2]);
let depth = 0;
let queue = [a, 'flag'];
if(a === b ) return 0
if(a > b) {
let output = Math.floor((a - b) / 2);
if((a - b) % 2) return output + 1;
return output
}
while(true) {
let current = queue.shift();
if(current === 'flag') {
depth += 1;
queue.push('flag');
continue;
}
let multiple = current * c;
let decrement = current - 1;
let doubleDecrement = current -2;
if (multiple !== b) queue.push(multiple);
else return depth + 1
if (decrement !== b) queue.push(decrement);
else return depth + 1
if (doubleDecrement !== b) queue.push(doubleDecrement);
else return depth + 1
}
}
Still times out. Any more suggestions?
Link for the question for you reference.

BFS
A greedy approach won't work here.
However it is already on the right track. Consider the graph G, where each node represents a value and each edge represents one of the operations and connects two values that are related by that operation (e.g.: 4 and 3 are connected by "subtract 1"). Using this graph, we can easily perform a BFS-search to find the shortest path:
def a_to_b(a, b, c):
visited = set()
state = {a}
depth = 0
while b not in state:
visited |= state
state = {v - 1 for v in state if v - 1 not in visited} | \
{v - 2 for v in state if v - 2 not in visited} | \
{v * c for v in state if v * c not in visited}
depth += 1
return 1
This query systematically tests all possible combinations of operations until it reaches b by testing stepwise. I.e. generate all values that can be reached with a single operation from a, then test all values that can be reached with two operations, etc., until b is among the generated values.
In depth analysis
(Assuming c >= 0, but can be generalized)
So far for the standard-approach that works with little analysis. This approach has the advantage that it works for any problem of this kind and is easy to implement. However it isn't very efficient and will reach it's limits fairly fast, once the numbers grow. So instead I'll show a way to analyze the problem in depth and gain a (far) more performant solution:
In a first step this answer will analyze the problem:
We need operations -->op such that a -->op b and -->op is a sequence of
subtract 1
subtract 2
multiply by c
First of all, what happens if we first subtract and afterwards multiply?
(a - x) * c = a * c - x * c
Next what happens, if we first multiply and afterwards subtract?
a * c - x'
Positional systems
Well, there's no simplifying transformation for this. But we've got the basic pieces to analyze more complicated chains of operations. Let's see what happens when we chain subtractions and multiplications alternatingly:
(((a - x) * c - x') * c - x'') * c - x'''=
((a * c - x * c - x') * c - x'') * c - x''' =
(a * c^2 - x * c^2 - x' * c - x'') * c - x''' =
a * c^3 - x * c^3 - x' * c^2 - x'' * c - x'''
Looks familiar? We're one step away from defining the difference between a and b in a positional system base c:
a * c^3 - x * c^3 - x' * c^2 - x'' * c - x''' = b
x * c^3 + x' * c^2 + x'' * c + x''' = a * c^3 - b
Unfortunately the above is still not quite what we need. All we can tell is that the LHS of the equation will always be >=0. In general, we first need to derive the proper exponent n (3 in the above example), s.t. it is minimal, nonnegative and a * c^n - b >= 0. Solving this for the individual coefficients (x, x', ...), where all coefficients are non-negative is a fairly trivial task.
We can show two things from the above:
if a < b and a < 0, there is no solution
solving as above and transforming all coefficients into the appropriate operations leads to the optimal solution
Proof of optimality
The second statement above can be proven by induction over n.
n = 0: In this case a - b < c, so there is only one -->op
n + 1: let d = a * c^(n + 1) - b. Let d' = d - m * c^(n + 1), where m is chosen, such that d' is minimal and nonnegative. Per induction-hypothesis d' can be generated optimally via a positional system. Leaving a difference of exactly m * c^n. This difference can not be covered more efficiently via lower-order terms than by m / 2 subtractions.
Algorithm (The TLDR-part)
Consider a * c^n - b as a number base c and try to find it's digits. The final number should have n + 1 digits, where each digit represents a certain number of subtractions. Multiple subtractions are represented by a single digit by addition of the subtracted values. E.g. 5 means -2 -2 -1. Working from the most significant to the least significant digit, the algorithm operates as follows:
perform the subtractions as specified by the digit
if the current digit is was the last, terminate
multiply by c and repeat from 1. with the next digit
E.g.:
a = 3, b = 10, c = 2
choose n = 2
a * c^n - b = 3 * 4 - 10 = 2
2 in binary is 010
steps performed: 3 - 0 = 3, 3 * 2 = 6, 6 - 1 = 5, 5 * 2 = 10
or
a = 2, b = 25, c = 6
choose n = 2
a * c^n - b = 47
47 base 6 is 115
steps performed: 2 - 1 = 1, 1 * 6 = 6, 6 - 1 = 5, 5 * 6 = 30, 30 - 2 - 2 - 1 = 25
in python:
def a_to_b(a, b, c):
# calculate n
n = 0
pow_c = 1
while a * pow_c - b < 0:
n += 1
pow_c *= 1
# calculate coefficients
d = a * pow_c - b
coeff = []
for i in range(0, n + 1):
coeff.append(d // pow_c) # calculate x and append to terms
d %= pow_c # remainder after eliminating ith term
pow_c //= c
# sum up subtractions and multiplications as defined by the coefficients
return n + sum(c // 2 + c % 2 for c in coeff)

Algorithm to shift overlapping intervals until no overlap is left

I have a sorted list of overlapping intervals, intervals are never contained in each other, e.g.,
[(7, 11), (9, 14), (12, 17)]
The constraint for the output is to keep every element as close as possible to its
origin (the middle of the interval), preserve the order of the input, and remove all overlap. Only an
approximate solution is necessary. The expected result for the example input would be:
[(5,9), (9, 14), (14, 19)]
I'm only aware of solutions that go about this in some simulation
style: shift each element by some value in a free direction and
iterate until all overlap has been removed.
Is there an existing algorithm to solve this?

find the overall average:
in our example:
(7 + 11 + 9 + 14 + 12 + 17)/6 = 11.667
find the total length:
(11-7) + (14-9) + (17-12) = 4 + 5 + 5 = 14;
find the new min/max;
14/2 = 7
11.667 - 7 = 4.667
11.667 + 7 = 18.667
you can round 'em
4.667 ~ 5
18.667 ~ 19
start from the min, creating the sections by the intervals
(5, (11-7)+5) = (5,9)
(9, (14-9)+9) = (9,14)
(14, (17-12)+14) = (14,19)
NOTE:
this method will not keep the elements as equal as possible to the originals, but will keep them as close as possible to the original considering their relative values (preserving the center)
EDIT:
if you want to keep the averages of all intervals as close as possible to the original, you can implement a mathematical solution.
our problem's input is:
a1=(a1,1, a1,2) , ... , an=(an,1,an,2)
we will define:
ai1 = a1,2-a1,1 // define the intervals
b1 = (d, d+ai1)
bn = (d + sum(ai1..ain-1), d + sum(ai1..ain) )
bi1 = b1,2-b1,1 // define the intervals
we need to find a 'd' such as:
s = sum( abs((a1,1+a1,2)/2 - (b1,1+b1,2)/2) )
min(s) is what we want
in our example:
a1 = (7,11), ai1 = 4, Aavg1 = 9
a2 = (9,14), ai2 = 5, Aavg2 = 11.5
a3 = (12,7), ai3 = 5, Aavg3 = 14.5
b1 = (d, d+4) Bavg1 = d+2
b2 = (d+4, d+9) Bavg2 = d+6.5
b3 = (d+9, d+14) Bavg3 = d+11.5
s = abs(9-(d+2)) + abs(11.5-(d+6.5)) + abs(14.5-(d+11.5)) = abs(7-d) + abs(5-d) + abs(3-d)
now calculcate the derivative to find min/max OR iterate over d to get a result. in our case you will need to iterate from 3 to 7
that should do the trick

Given that the solution must be order-preserving, we can formulate this problem as a linear program. Let [ai, bi] be the ith interval. Let variables xi be the left shift of the ith interval and yi be the right shift of the ith interval.
minimize sumi (xi + yi)
subject to
(*) for all i: bi - xi + yi ≤ ai+1 - xi+1 + yi+1
for all i: xi, yi ≥ 0
Rewrite constraint (*) by introducing a variable zi.
for all i: xi - yi - xi+1 + yi+1 - zi = 0
for all i: zi ≥ bi - ai+1
Now the problem is reduced to computing a minimum-cost circulation, which can be done in poly-time. I have a feeling, however, that there's a more direct solution to this problem.
The graph looks something like
(*)
---- | ----
/ z| \
/ i| \
/ xi | xi+1 \
|/ <---- v <---- \|
... (*) ...
----> ---->
yi yi+1

The "guess the number" game for arbitrary rational numbers?

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 ;-) )

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