The Lumber Room

"Consign them to dust and damp by way of preserving them"

Kalidasa’s Deepashikha

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(Another example of good vs bad translations from Sanskrit. Previously see here and here.)

One of Kālidāsa’s famous similes is in the following verse from the Raghuvaṃśa, in the context of describing the svayaṃvara of Indumatī. The various hopeful suitors of the princess, all kings from different regions, are lined up as she passes them one by one, her friend doing the introductions.

संचारिणी दीपशिखेव रात्रौ
यम् यम् व्यतीयाय पतिंवरा सा ।
नरेन्द्रमार्गाट्ट इव प्रपेदे
विवर्णभावम् स स भूमिपालः ॥ ६-६७


saṁcāriṇī dīpa-śikheva rātrau
yam yam vyatīyāya patiṁvarā sā |
narendra-mārga-aṭṭa iva prapede
vivarṇa-bhāvam sa sa bhūmipālaḥ || 6-67

Only today did I discover a decent translation into English. It’s by John Brough (1975/6):

As if a walking lamp-flame in the night
On the king’s highway, flanked with houses tall,
She moved, and lit each prince with hopeful light,
And, passing on, let each to darkness fall.

Every other translation I have seen really falls short. Witness the misunderstandings, and the killing of all feeling.

Here is Ryder (1904), who is usually good:

And every prince rejected while she sought
A husband, darkly frowned, as turrets, bright
One moment with the flame from torches caught,
Frown gloomily again and sink in night.

The idea is there, but requires too much effort to understand.

This is P. de Lacy Johnstone (1902):

Now as the Maid went by, each suitor-King,
Lit for a moment by her dazzling eyes,
Like wayside tower by passing lamp, sank back
In deepest gloom. …

Every king, whom Indumati passed by while choosing her husband, assumed a pale look as the houses on a high way are covered with darkness in the absence of lamps.

Whatsoever king the maiden intent on choosing her husband passed by, like the flame of a moving lamp at night, that same king turned pale, just as a mansion situate on the highway, is shrouded in darkness when left behind (by a moving light).

Desiraju Hanumanta Rao

67. pati.m varA sA= husband, selector, she – she who has come to select her husband, indumati; rAtrau sa.mcAriNI dIpa shikha iva= in night, moving, lamp’s, [glittering] flame, as with; ya.m ya.m= whom, whom; [bhUmi pAlam= king, whomever]; vyatIyAya= passed by; saH saH bhUmipAlaH= he, he, king – such and such a king; narendra mArga= on king’s, way; aTTa= a turret, or a balustrade; iva= like; vi+varNa bhAva.m= without, colour, aspect – they bore a colourless aspect; prapede= [that king] obtained – that king became colourless, he drew blank.
Princess indumati who came to choose her husband then moved like the glittering flame of a lamp on a king’s way, and whichever prince she left behind was suffused with pallor just like a turret or balustrade on the king’s way will be shrouded in darkness and becomes dim when left behind by the moving light on the king’s way. [6-67]

And this is representative of the average quality of Sanskrit-to-English translations, and how much beauty is lost.

Written by S

Mon, 2015-02-09 at 20:04:42

Posted in sanskrit

Tagged with

Some playing with Python

A long time ago, Diophantus (sort of) discussed integer solutions to the equation

$\displaystyle x^2 + y^2 = z^2$

(solutions to this equation are called Pythagorean triples).

Centuries later, in 1637, Fermat made a conjecture (now called Fermat’s Last Theorem, not because he uttered it in his dying breath, but because it was the last one to be proved — in ~1995) that

$\displaystyle x^n + y^n = z^n$

has no positive integer solutions for $n \ge 3$. In other words, his conjecture was that none of the following equations has a solution:

$\displaystyle x^3 + y^3 = z^3$

$\displaystyle x^4 + y^4 = z^4$

$\displaystyle x^5 + y^5 = z^5$

$\displaystyle x^6 + y^6 = z^6$

… and so on. An nth power cannot be partitioned into two nth powers.

About a century later, Euler proved the $n = 3$ case of Fermat’s conjecture, but generalized it in a different direction: he conjectured in 1769 that an nth power cannot be partitioned into fewer than n nth powers, namely

$\displaystyle z^n = \sum_{i = 1}^k x_i^n$

has no solutions with $k < n$. So his conjecture was that (among others) none of the following equations has a solution:

$\displaystyle z^3 = a^3 + b^3$

$\displaystyle z^4 = a^4 + b^4 + c^4$

$\displaystyle z^5 = a^5 + b^5 + c^5 + d^5$

$\displaystyle z^6 = a^6 + b^6 + c^6 + d^6 + e^6$

… and so on.

This conjecture stood for about two centuries, until abruptly it was found to be false, by Lander and Parkin who in 1966 simply did a direct search on the fastest (super)computer at the time, and found this counterexample:

$\displaystyle 27^5 + 84^5 + 110^5 + 133^5 = 144^5$

(It is still one of only three examples known, according to Wikipedia.)

Now, how might you find this solution on a computer today?

In his wonderful (as always) post at bit-player, Brian Hayes showed the following code:

import itertools as it

def four_fifths(n):
'''Return smallest positive integers ((a,b,c,d),e) such that
a^5 + b^5 + c^5 + d^5 = e^5; if no such tuple exists
with e < n, return the string 'Failed'.'''
fifths = [x**5 for x in range(n)]
combos = it.combinations_with_replacement(range(1,n), 4)
while True:
try:
cc = combos.next()
cc_sum = sum([fifths[i] for i in cc])
if cc_sum in fifths:
return(cc, fifths.index(cc_sum))
except StopIteration:
return('Failed')


to which, if you add (say) print four_fifths(150) and run it, it returns the correct answer fairly quickly: in about 47 seconds on my laptop.

The if cc_sum in fifths: line inside the loop is an $O(n)$ cost each time it’s run, so with a simple improvement to the code (using a set instead) and rewriting it a bit, we can write the following full program:

import itertools

def find_counterexample(n):
fifth_powers = [x**5 for x in range(n)]
fifth_powers_set = set(fifth_powers)
for xs in itertools.combinations_with_replacement(range(1, n), 4):
xs_sum = sum([fifth_powers[i] for i in xs])
if xs_sum in fifth_powers_set:
return (xs, fifth_powers.index(xs_sum))
return 'Failed'

print find_counterexample(150)


which finishes in about 8.5 seconds.

Great!

But there’s something unsatisfying about this solution, which is that it assumes there’s a solution with all four numbers on the LHS less than 150. After all, changing the function invocation to find_counterexample(145) makes it run a second faster even, but how could we know to do without already knowing the solution? Besides, we don’t have a fixed 8- or 10-second budget; what we’d really like is a program that keeps searching till it finds a solution or we abort it (or it runs out of memory or something), with no other fixed termination condition.

The above program used the given “n” as an upper bound to generate the combinations of 4 numbers; is there a way to generate all combinations when we don’t know an upper bound on them?

Yes! One of the things I learned from Knuth volume 4 is that if you simply write down each combination in descending order and order them lexicographically, the combinations you get for each upper bound are a prefix of the list of the next bigger one, i.e., for any upper bound, all the combinations form a prefix of the same infinite list, which starts as follows (line breaks for clarity):

1111,
2111, 2211, 2221, 2222,
3111, 3211, 3221, 3222, 3311, 3321, 3322, 3331, 3332, 3333,
4111, ...
... 9541, 9542, 9543, 9544, 9551, ... 9555, 9611, ...


There doesn’t seem to be a library function in Python to generate these though, so we can write our own. If we stare at the above list, we can figure out how to generate the next combination from a given one:

1. Walk backwards from the end, till you reach the beginning or find an element that’s less than the previous one.
2. Increase that element, set all the following elements to 1s, and continue.

We could write, say, the following code for it:

def all_combinations(r):
xs = [1] * r
while True:
yield xs
for i in range(r - 1, 0, -1):
if xs[i] < xs[i - 1]:
break
else:
i = 0
xs[i] += 1
xs[i + 1:] = [1] * (r - i - 1)


(The else block on a for loop is an interesting Python feature: it is executed if the loop wasn’t terminated with break.) We could even hard-code the r=4 case, as we’ll see later below.

For testing whether a given number is a fifth power, we can no longer simply lookup in a fixed precomputed set. We can do a binary search instead:

def is_fifth_power(n):
assert n > 0
lo = 0
hi = n
# Invariant: lo^5 < n <= hi^5
while hi - lo > 1:
mid = lo + (hi - lo) / 2
if mid ** 5 < n:
lo = mid
else:
hi = mid
return hi ** 5 == n


but it turns out that this is slower than one based on looking up in a growing set (as below).

Putting everything together, we can write the following (very C-like) code:

largest_known_fifth_power = (0, 0)
known_fifth_powers = set()
def is_fifth_power(n):
global largest_known_fifth_power
while n > largest_known_fifth_power[0]:
m = largest_known_fifth_power[1] + 1
m5 = m ** 5
largest_known_fifth_power = (m5, m)
return n in known_fifth_powers

def fournums_with_replacement():
(x0, x1, x2, x3) = (1, 1, 1, 1)
while True:
yield (x0, x1, x2, x3)
if x3 < x2:
x3 += 1
continue
x3 = 1
if x2 < x1:
x2 += 1
continue
x2 = 1
if x1 < x0:
x1 += 1
continue
x1 = 1
x0 += 1
continue

if __name__ == '__main__':
tried = 0
for get in fournums_with_replacement():
tried += 1
if (tried % 1000000 == 0):
print tried, 'Trying:', get
rhs = get[0]**5 + get[1]**5 + get[2]**5 + get[3]**5
if is_fifth_power(rhs):
print 'Found:', get, rhs
break


which is both longer and slower (takes about 20 seconds) than the original program, but at least we have the satisfaction that it doesn’t depend on any externally known upper bound.

I originally started writing this post because I wanted to describe some experiments I did with profiling, but it’s late and I’m sleepy so I’ll just mention it.

python -m cProfile euler_conjecture.py


will print relevant output in the terminal:

         26916504 function calls in 26.991 seconds

Ordered by: standard name

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
1   18.555   18.555   26.991   26.991 euler_conjecture.py:1()
13458164    4.145    0.000    4.145    0.000 euler_conjecture.py:12(fournums_with_replacement)
13458163    4.292    0.000    4.292    0.000 euler_conjecture.py:3(is_fifth_power)
175    0.000    0.000    0.000    0.000 {method 'add' of 'set' objects}
1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}


Another way to view the same thing is to write the profile output to a file and read it with cprofilev:

python -m cProfile -o euler_profile.out euler_conjecture.py
cprofilev euler_profile.out


and visit http://localhost:4000 to view it.

Of course, simply translating this code to C++ makes it run much faster:

#include <array>
#include <iostream>
#include <map>
#include <utility>

typedef long long Int;
constexpr Int fifth_power(Int x) { return x * x * x * x * x; }

std::map<Int, int> known_fifth_powers = {{0, 0}};
bool is_fifth_power(Int n) {
while (n > known_fifth_powers.rbegin()->first) {
int m = known_fifth_powers.rbegin()->second  + 1;
known_fifth_powers[fifth_power(m)] = m;
}
return known_fifth_powers.count(n);
}

std::array<Int, 4> four_nums() {
static std::array<Int, 4> x = {1, 1, 1, 0};
int i = 3;
while (i > 0 && x[i] == x[i - 1]) --i;
x[i] += 1;
while (++i < 4) x[i] = 1;
return x;
}

std::ostream& operator<<(std::ostream& os, std::array<Int, 4> x) {
os << "(" << x[0] << ", " << x[1] << ", " << x[2] << ", " << x[3] << ")";
return os;
}

int main() {
while (true) {
std::array<Int, 4> get = four_nums();
Int rhs = fifth_power(get[0]) + fifth_power(get[1]) + fifth_power(get[2]) + fifth_power(get[3]);
if (is_fifth_power(rhs)) {
std::cout << "Found: " << get << " " << known_fifth_powers[rhs] << std::endl;
break;
}
}
}


and


clang++ -std=c++11 euler_conjecture.cc && time ./a.out


runs in 2.43s, or 0.36s if compiled with -O2.

But I don’t have a satisfactory answer to how to make our Python program which takes 20 seconds as fast as the 8.5-second known-upper-bound version.

Edit [2015-05-08]: I wrote some benchmarking code to compare all the different “combination” functions.

import itertools

# Copied from the Python documentation
def itertools_equivalent(iterable, r):
pool = tuple(iterable)
n = len(pool)
if not n and r:
return
indices = [0] * r
yield tuple(pool[i] for i in indices)
while True:
for i in reversed(range(r)):
if indices[i] != n - 1:
break
else:
return
indices[i:] = [indices[i] + 1] * (r - i)
yield tuple(pool[i] for i in indices)

# Above function, specialized to first argument being range(1, n)
def itertools_equivalent_specialized(n, r):
indices = [1] * r
yield indices
while True:
for i in reversed(range(r)):
if indices[i] != n - 1:
break
else:
return
indices[i:] = [indices[i] + 1] * (r - i)
yield indices

# Function to generate all combinations of 4 elements
def all_combinations_pythonic(r):
xs = [1] * r
while True:
yield xs
for i in range(r - 1, 0, -1):
if xs[i] < xs[i - 1]:
break
else:
i = 0
xs[i] += 1
xs[i + 1:] = [1] * (r - i - 1)

# Above function, written in a more explicit C-like way
def all_combinations_clike(r):
xs = [1] * r
while True:
yield xs
i = r - 1
while i > 0 and xs[i] == xs[i - 1]:
i -= 1
xs[i] += 1
while i < r - 1:
i += 1
xs[i] = 1

# Above two functions, specialized to r = 4, using tuple over list.
def fournums():
(x0, x1, x2, x3) = (1, 1, 1, 1)
while True:
yield (x0, x1, x2, x3)
if x3 < x2:
x3 += 1
continue
x3 = 1
if x2 < x1:
x2 += 1
continue
x2 = 1
if x1 < x0:
x1 += 1
continue
x1 = 1
x0 += 1
continue

# Benchmarks for all functions defined above (and the library function)
def benchmark_itertools(n):
for xs in itertools.combinations_with_replacement(range(1, n), 4):
if xs[0] >= n:
break
def benchmark_itertools_try(n):
combinations = itertools.combinations_with_replacement(range(1, n), 4)
while True:
try:
xs = combinations.next()
if xs[0] >= n:
break
except StopIteration:
return
def benchmark_itertools_equivalent(n):
for xs in itertools_equivalent(range(1, n), 4):
if xs[0] >= n:
break
def benchmark_itertools_equivalent_specialized(n):
for xs in itertools_equivalent_specialized(n, 4):
if xs[0] >= n:
break
def benchmark_all_combinations_pythonic(n):
for xs in all_combinations_pythonic(4):
if xs[0] >= n:
break
def benchmark_all_combinations_clike(n):
for xs in all_combinations_clike(4):
if xs[0] >= n:
break
def benchmark_fournums(n):
for xs in fournums():
if xs[0] >= n:
break

if __name__ == '__main__':
benchmark_itertools(150)
benchmark_itertools_try(150)
benchmark_itertools_equivalent(150)
benchmark_itertools_equivalent_specialized(150)
benchmark_all_combinations_pythonic(150)
benchmark_all_combinations_clike(150)
benchmark_fournums(150)


As you can see, I chose inside the benchmarking function the same statement that would cause all_combinations to terminate, and have no effect for the other combination functions.
When run with

python -m cProfile benchmark_combinations.py

the results include:

    2.817 benchmark_combinations.py:80(benchmark_itertools)
8.583 benchmark_combinations.py:84(benchmark_itertools_try)
126.980 benchmark_combinations.py:93(benchmark_itertools_equivalent)
46.635 benchmark_combinations.py:97(benchmark_itertools_equivalent_specialized)
44.032 benchmark_combinations.py:101(benchmark_all_combinations_pythonic)
18.049 benchmark_combinations.py:105(benchmark_all_combinations_clike)
10.923 benchmark_combinations.py:109(benchmark_fournums)


Lessons:

• Calling itertools.combinations_with_replacement is by far the fastest, taking about 2.7 seconds. It turns out that it’s written in C, so this would be hard to beat. (Still, writing it in a try block is seriously bad.)
• The “equivalent” Python code from the itertools documentation (benchmark_itertools_combinations_with_replacment) is about 50x slower.
• Gets slightly better when specialized to numbers.
• Simply generating all combinations without an upper bound is actually faster.
• It can be made even faster by writing it in a more C-like way.
• The tuples version with the loop unrolled manually is rather fast when seen in this light, less than 4x slower than the library version.

Written by S

Sun, 2015-02-08 at 00:03:38