I have a variable and want to take a range of bits from that variable. I want the CLEANEST way to do this.
If x = 19767 and I want bit3 - bit8 (starting from the right):
100110100110111 is 19767 in binary.
I want the part in parenthesis 100110(100110)111 so the answer is 38.
What is the simplest/cleanest/most-elegant way to implement the following function with Ruby?
bit_range(orig_num, first_bit, last_bit)
PS. Bonus points for answers that are computationally less intensive.
19767.to_s(2)[-9..-4].to_i(2)
or
19767 >> 3 & 0x3f
Update:
Soup-to-nuts (why do people say that, anyway?) ...
class Fixnum
def bit_range low, high
len = high - low + 1
self >> low & ~(-1 >> len << len)
end
end
p 19767.bit_range(3, 8)
orig_num.to_s(2)[(-last_bit-1)..(-first_bit-1)].to_i(2)
Here is how this could be done using pure number operations:
class Fixnum
def slice(range_or_start, length = nil)
if length
start = range_or_start
else
range = range_or_start
start = range.begin
length = range.count
end
mask = 2 ** length - 1
self >> start & mask
end
end
def p n
puts "0b#{n.to_s(2)}"; n
end
p 0b100110100110111.slice(3..8) # 0b100110
p 0b100110100110111.slice(3, 6) # 0b100110
Just to show the speeds of the suggested answers:
require 'benchmark'
ORIG_NUMBER = 19767
def f(x,i,j)
b = x.to_s(2)
n = b.size
b[(n-j-1)...(n-i)].to_i(2)
end
class Fixnum
def bit_range low, high
len = high - low + 1
self >> low & ~(-1 >> len << len)
end
def slice(range_or_start, length = nil)
if length
start = range_or_start
else
range = range_or_start
start = range.begin
length = range.count
end
mask = 2 ** length - 1
self >> start & mask
end
end
def p n
puts "0b#{n.to_s(2)}"; n
end
n = 1_000_000
puts "Using #{ n } loops in Ruby #{ RUBY_VERSION }."
Benchmark.bm(21) do |b|
b.report('texasbruce') { n.times { ORIG_NUMBER.to_s(2)[(-8 - 1)..(-3 - 1)].to_i(2) } }
b.report('DigitalRoss string') { n.times { ORIG_NUMBER.to_s(2)[-9..-4].to_i(2) } }
b.report('DigitalRoss binary') { n.times { ORIG_NUMBER >> 3 & 0x3f } }
b.report('DigitalRoss bit_range') { n.times { 19767.bit_range(3, 8) } }
b.report('Philip') { n.times { f(ORIG_NUMBER, 3, 8) } }
b.report('Semyon Perepelitsa') { n.times { ORIG_NUMBER.slice(3..8) } }
end
And the output:
Using 1000000 loops in Ruby 1.9.3.
user system total real
texasbruce 1.240000 0.010000 1.250000 ( 1.243709)
DigitalRoss string 1.000000 0.000000 1.000000 ( 1.006843)
DigitalRoss binary 0.260000 0.000000 0.260000 ( 0.262319)
DigitalRoss bit_range 0.840000 0.000000 0.840000 ( 0.858603)
Philip 1.520000 0.000000 1.520000 ( 1.543751)
Semyon Perepelitsa 1.150000 0.010000 1.160000 ( 1.155422)
That's on my old MacBook Pro. Your mileage might vary.
Makes sense to define a function for that:
def f(x,i,j)
b = x.to_s(2)
n = b.size
b[(n-j-1)...(n-i)].to_i(2)
end
puts f(19767, 3, 8) # => 38
Expanding on the idea from DigitalRoss - instead of taking two arguments, you can pass a range:
class Fixnum
def bit_range range
len = range.last - range.first + 1
self >> range.first & ~(-1 >> len << len)
end
end
19767.bit_range 3..8
Related
I am trying to write a function scramble(str1, str2) that returns true if a portion of str1 characters can be rearranged to match str2, otherwise returns false. Only lower case letters (a-z) will be used. No punctuation or digits will be included. For example:
str1 = 'rkqodlw'; str2 = 'world' should return true.
str1 = 'cedewaraaossoqqyt'; str2 = 'codewars' should return true.
str1 = 'katas'; str2 = 'steak' should return false.
This is my code:
def scramble(s1, s2)
#sorts strings into arrays
first = s1.split("").sort
second = s2.split("").sort
correctLetters = 0
for i in 0...first.length
#check for occurrences of first letter
occurrencesFirst = first.count(s1[i])
for j in 0...second.length
#scan through second string
occurrencesSecond = second.count(s2[j])
#if letter to be tested is correct and occurrences of first less than occurrences of second
#meaning word cannot be formed
if (s2[j] == s1[i]) && occurrencesFirst < occurrencesSecond
return false
elsif s2[j] == s1[i]
correctLetters += 1
elsif first.count(s1[s2[j]]) == 0
return false
end
end
end
if correctLetters == 0
return false
end
return true
end
I need help optimising this code. Please give me suggestions.
Here is one efficient and Ruby-like way of doing that.
Code
def scramble(str1, str2)
h1 = char_counts(str1)
h2 = char_counts(str2)
h2.all? { |ch, nbr| nbr <= h1[ch] }
end
def char_counts(str)
str.each_char.with_object(Hash.new(0)) { |ch, h| h[ch] += 1 }
end
Examples
scramble('abecacdeba', 'abceae')
#=> true
scramble('abecacdeba', 'abweae')
#=> false
Explanation
The three steps are as follows.
str1 = 'abecacdeba'
str2 = 'abceae'
h1 = char_counts(str1)
#=> {"a"=>3, "b"=>2, "e"=>2, "c"=>2, "d"=>1}
h2 = char_counts(str2)
#=> {"a"=>2, "b"=>1, "c"=>1, "e"=>2}
h2.all? { |ch, nbr| nbr <= h1[ch] }
#=> true
The last statement is equivalent to
2 <= 3 && 1 <= 2 && 1 <= 2 && 2 <=2
The method char_counts constructs what is sometimes called a "counting hash". To understand how char_counts works, see Hash::new, especially the explanation of the effect of providing a default value as an argument of new. In brief, if a hash is defined h = Hash.new(0), then if h does not have a key k, h[k] returns the default value, here 0 (and the hash is not changed).
Suppose, for different data,
h1 = { "a"=>2 }
h2 = { "a"=>1, "b"=>2 }
Then we would find that 1 <= 2 #=> true but 2 <= 0 #=> false, so the method would return false. The second comparison is 2 <= h1["b"]. As h1 does not have a key "b", h1["b"] returns the default value, 0.
The method char_counts is effectively a short way of writing the method expressed as follows.
def char_counts(str)
h = {}
str.each_char do |ch|
h[ch] = 0 unless h.key?(ch) # instead of Hash.new(0)
h[ch] = h[c] + 1 # instead of h[c][ += 1
end
h # no need for this if use `each_with_object`
end
See Enumerable#each_with_object, String#each_char (preferable to String.chars, as the latter produces an unneeded temporary array whereas the former returns an enumerator) and Hash#key? (or Hash#has_key?, Hash#include? or Hash#member?).
An Alternative
def scramble(str1, str2)
str2.chars.difference(str1.chars).empty?
end
class Array
def difference(other)
h = other.each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
reject { |e| h[e] > 0 && h[e] -= 1 }
end
end
I have found the method Array#difference to be so useful I proposed it be added to the Ruby Core (here). The response has been, er, underwhelming.
One way:
def scramble(s1,s2)
s2.chars.uniq.all? { |c| s1.count(c) >= s2.count(c) }
end
Another way:
def scramble(s1,s2)
pool = s1.chars.group_by(&:itself)
s2.chars.all? { |c| pool[c]&.pop }
end
Yet another:
def scramble(s1,s2)
('a'..'z').all? { |c| s1.count(c) >= s2.count(c) }
end
Since this appears to be from codewars, I submitted my first two there. Both got accepted and the first one was a bit faster. Then I was shown solutions of others and saw someone using ('a'..'z') and it's fast, so I include that here.
The codewars "performance tests" aren't shown explicitly but they're all up to about 45000 letters long. So I benchmarked these solutions as well as Cary's (yours was too slow to be included) on shuffles of the alphabet repeated to be about that long (and doing it 100 times):
user system total real
Stefan 1 0.812000 0.000000 0.812000 ( 0.811765)
Stefan 2 2.141000 0.000000 2.141000 ( 2.127585)
Other 0.125000 0.000000 0.125000 ( 0.122248)
Cary 1 2.562000 0.000000 2.562000 ( 2.575366)
Cary 2 3.094000 0.000000 3.094000 ( 3.106834)
Moral of the story? String#count is fast here. Like, ridiculously fast. Almost unbelievably fast (I actually had to run extra tests to believe it). It counts through about 1.9 billion letters per second (100 times 26 letters times 2 strings of ~45000 letters, all in 0.12 seconds). Note that the difference to my own first solution is just that I do s2.chars.uniq, and that increases the time from 0.12 seconds to 0.81 seconds. Meaning this double pass through one string takes about six times as long as the 52 passes for counting. The counting is about 150 times faster. I did expect it to be very fast, because it presumably just searches a byte in an array of bytes using C code (edit: looks like it does), but this speed still surprised me.
Code:
require 'benchmark'
def scramble_stefan1(s1,s2)
s2.chars.uniq.all? { |c| s1.count(c) >= s2.count(c) }
end
def scramble_stefan2(s1,s2)
pool = s1.chars.group_by(&:itself)
s2.chars.all? { |c| pool[c]&.pop }
end
def scramble_other(s1,s2)
('a'..'z').all? { |c| s1.count(c) >= s2.count(c) }
end
def scramble_cary1(str1, str2)
h1 = char_counts(str1)
h2 = char_counts(str2)
h2.all? { |ch, nbr| nbr <= h1[ch] }
end
def char_counts(str)
str.each_char.with_object(Hash.new(0)) { |ch, h| h[ch] += 1 }
end
def scramble_cary2(str1, str2)
str2.chars.difference(str1.chars).empty?
end
class Array
def difference(other)
h = other.each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
reject { |e| h[e] > 0 && h[e] -= 1 }
end
end
Benchmark.bmbm do |x|
n = 100
s1 = (('a'..'z').to_a * (45000 / 26)).shuffle.join
s2 = s1.chars.shuffle.join
x.report('Stefan 1') { n.times { scramble_stefan1(s1, s2) } }
x.report('Stefan 2') { n.times { scramble_stefan2(s1, s2) } }
x.report('Other') { n.times { scramble_other(s1, s2) } }
x.report('Cary 1') { n.times { scramble_cary1(s1, s2) } }
x.report('Cary 2') { n.times { scramble_cary2(s1, s2) } }
end
Given two numbers, say (14, 18), the problem is to find the sum of all the numbers in this range, 14, 15, 16, 17, 18 recursively. Now, I have done this using loops but I have trouble doing this recursively.
Here is my recursive solution:
def sum_cumulative_recursive(a,b)
total = 0
#base case is a == b, the stopping condition
if a - b == 0
puts "sum is: "
return total + a
end
if b - a == 0
puts "sum is: "
return total + b
end
#case 1: a > b, start from b, and increment recursively
if a > b
until b > a
puts "case 1"
total = b + sum_cumulative_recursive(a, b+1)
return total
end
end
#case 2: a < b, start from a, and increment recursively
if a < b
until a > b
puts "case 2"
total = a + sum_cumulative_recursive(a+1, b)
return total
end
end
end
Here are some sample test cases:
puts first.sum_cumulative_recursive(4, 2)
puts first.sum_cumulative_recursive(14, 18)
puts first.sum_cumulative_recursive(-2,-2)
My solution works for cases where a > b, and a < b, but it doesn't work for a == b.
How can I fix this code so that it works?
Thank you for your time.
def sum_cumulative_recursive(a,b)
return a if a == b
a, b = [a,b].sort
a + sum_cumulative_recursive(a + 1, b)
end
EDIT
Here is the most efficient solution I could see from some informal benchmarks:
def sum_cumulative_recursive(a,b)
return a if a == b
a, b = b, a if a > b
a + sum_cumulative_recursive(a + 1, b)
end
Using:
Benchmark.measure { sum_cumulative_recursive(14,139) }
Benchmark for my initial response: 0.005733
Benchmark for #Ajedi32's response: 0.000371
Benchmark for my new response: 0.000115
I was also surprised to see that in some cases, the recursive solution approaches or exceeds the efficiency of the more natural inject solution:
Benchmark.measure { 10.times { (1000..5000).inject(:+) } }
# => 0.010000 0.000000 0.010000 ( 0.027827)
Benchmark.measure { 10.times { sum_cumulative_recursive(1000,5000) } }
# => 0.010000 0.010000 0.020000 ( 0.019441)
Though you run into stack level too deep errors if you take it too far...
I'd do it like this:
def sum_cumulative_recursive(a, b)
a, b = a.to_i, b.to_i # Only works with ints
return sum_cumulative_recursive(b, a) if a > b
return a if a == b
return a + sum_cumulative_recursive(a+1, b)
end
Here's one way of doing it. I assume this is just an exercise, as the sum of the elements of a range r is of course just (r.first+r.last)*(f.last-r.first+1)/2.
def sum_range(range)
return nil if range.last < range.first
case range.size
when 1 then range.first
when 2 then range.first + range.last
else
range.first + range.last + sum_range(range.first+1..range.last-1)
end
end
sum_range(14..18) #=> 80
sum_range(14..14) #=> 14
sum_range(14..140) #=> 9779
sum_range(14..139) #=> 9639
Another solution would be to have a front-end invocation that fixes out-of-order arguments, then a private recursive back-end which does the actual work. I find this is useful to avoid repeated checks of arguments once you've established they're clean.
def sum_cumulative_recursive(a, b)
a, b = b, a if b < a
_worker_bee_(a, b)
end
private
def _worker_bee_(a, b)
a < b ? (a + _worker_bee_(a+1,b-1) + b) : a == b ? a : 0
end
This variant would cut the stack requirement in half by summing from both ends.
If you don't like that approach and/or you really want to trim the stack size:
def sum_cumulative_recursive(a, b)
if a < b
mid = (a + b) / 2
sum_cumulative_recursive(a, mid) + sum_cumulative_recursive(mid+1, b)
elsif a == b
a
else
sum_cumulative_recursive(b, a)
end
end
This should keep the stack size to O(log |b-a|).
While trying some memoization techniques I stumbled on this benchmark results which is against my expectations. Seems I am making some silly mistake, does someone see what I am doing wrong here (the benchmark gives similar results for memoized and non memoized code)?
require 'benchmark'
# -----------------------------------------
class FactorialClass
##sequence = [1]
def self.of( n )
##sequence[n] || n * of( n - 1 )
end
end
# -----------------------------------------
def factorial_with_memoization
sequence = [1]
lambda = Proc.new do |n|
sequence[n] || n * lambda.call( n - 1 )
end
end
f = factorial_with_memoization()
# -----------------------------------------
def factorial n
n == 0 ? 1 : n * factorial( n - 1 )
end
# -----------------------------------------
count = 1_000
n = 1000
without_memoization = Benchmark.measure do
count.times do
factorial(n)
end
end
with_memoization_lambda = Benchmark.measure do
count.times do
f.call(n)
end
end
with_memoization_class = Benchmark.measure do
count.times do
FactorialClass.of(n)
end
end
puts "Without memoization : #{ without_memoization }"
puts "With memoization using lambda : #{ with_memoization_lambda }"
puts "With memoization using class : #{ with_memoization_class }"
** The results are: **
Without memoization : 1.210000 0.100000 1.310000 ( 1.309675)
With memoization using lambda : 1.750000 0.100000 1.850000 ( 1.858737)
With memoization using class : 1.270000 0.090000 1.360000 ( 1.358375)
You never assign any memorized value to the cache. As #xlembouras said, you didn't memorize anything.
class FactorialClass
##sequence = [1]
def self.of( n )
# ##sequence[n] get evaluated to nil unless n is 0, always!
##sequence[n] || n * of( n - 1 )
end
end
You need to manually assign memorized value to the cache, after you finish the computation.
class FactorialClass
##sequence = [1]
def self.of( n )
##sequence[n] = (##sequence[n] || n * of( n - 1 ))
end
end
However, does memorization really works for your factorial computation? No.
f(n) = n * f(n-1) = n * ((n-1) * f(n-2)) = ... = n * ((n-1) * (... * (3 * f(2))))
All the recursion step calculates the factorial of a new value (from 2 to n), which hasn't been calculated before. The memorization won't get hit at any step.
I have a hash table:
hash = Hash.new(0)
hash[:key] = hash[:key] + 1 # Line 1
hash[:key] += 1 # Line 2
Line 1 and Line 2 do the same thing. Looks like line 1 needs to query hash by key two times while line 2 only once. Is that true? Or they are actually same?
I created a ruby script to benchmark it
require 'benchmark'
def my_case1()
#hash[:key] = #hash[:key] + 1
end
def my_case2()
#hash[:key] += 1
end
n = 10000000
Benchmark.bm do |test|
test.report("case 1") {
#hash = Hash.new(1)
#hash[:key] = 0
n.times do; my_case1(); end
}
test.report("case 2") {
#hash = Hash.new(1)
#hash[:key] = 0
n.times do; my_case2(); end
}
end
Here is the result
user system total real
case 1 3.620000 0.080000 3.700000 ( 4.253319)
case 2 3.560000 0.080000 3.640000 ( 4.178699)
It looks hash[:key] += 1 is slightly better.
#sza beat me to it :)
Here is my example irb session:
> require 'benchmark'
=> true
> n = 10000000
=> 10000000
> Benchmark.bm do |x|
> hash = Hash.new(0)
> x.report("Case 1:") { n.times do; hash[:key] = hash[:key] + 1; end }
> hash = Hash.new(0)
> x.report("Case 2:") { n.times do; hash[:key] += 1; end }
> end
user system total real
Case 1: 1.070000 0.000000 1.070000 ( 1.071366)
Case 2: 1.040000 0.000000 1.040000 ( 1.043644)
The Ruby Language Specification spells out the algorithm for evaluating abbreviated indexing assignment expressions quite clearly. It is something like this:
primary_expression[indexing_argument_list] ω= expression
# ω can be any operator, in this example, it is +
is (roughly) evaluated like
o = primary_expression
*l = indexing_argument_list
v = o.[](*l)
w = expression
l << (v ω w)
o.[]=(*l)
In particular, you can see that both the getter and the setter are called exactly once.
You can also see that by looking at the informal desugaring:
hash[:key] += 1
# is syntactic sugar for
hash[:key] = hash[:key] + 1
# which is syntactic sugar for
hash.[]=(:key, hash.[](:key).+(1))
Again, you see that both the setter and the getter are called exactly once.
second one is the customary way of doing it. It is more efficient.
What is the best way to generate a long string quickly in ruby? This works, but is very slow:
str = ""
length = 100000
(1..length).each {|i| str += "0"}
I've also noticed that creating a string of a decent length and then appending that to an existing string up to the desired length works much faster:
str = ""
incrementor = ""
length = 100000
(1..1000).each {|i| incrementor += "0"}
(1..100).each {|i| str += incrementor}
Any other suggestions?
str = "0" * 999999
Another relatively quick option is
str = '%0999999d' % 0
Though benchmarking
require 'benchmark'
Benchmark.bm(9) do |x|
x.report('format :') { '%099999999d' % 0 }
x.report('multiply:') { '0' * 99999999 }
end
Shows that multiplication is still faster
user system total real
format : 0.300000 0.080000 0.380000 ( 0.405345)
multiply: 0.080000 0.080000 0.160000 ( 0.172504)