How can I get the number of iterations/steps that this method takes to find an answer?
def binary_search(array, n)
min = 0
max = (array.length) - 1
while min <= max
middle = (min + max) / 2
if array[middle] == n
return middle
elsif array[middle] > n
max = middle - 1
elsif array[middle] < n
min = middle + 1
end
end
"#{n} not found in this array"
end
One option to use instead of a counter is the .with_index keyword. To use this you'll need to use loop instead of while, but it should work the same. Here's a basic example with output.
arr = [1,2,3,4,5,6,7,8]
loop.with_index do |_, index| # The underscore is to ignore the first variable as it's not used
if (arr[index] % 2).zero?
puts "even: #{arr[index]}"
else
puts "odd: #{arr[index]}"
end
break if index.eql?(arr.length - 1)
end
=>
odd: 1
even: 2
odd: 3
even: 4
odd: 5
even: 6
odd: 7
even: 8
Just count the number of iterations.
Set a variable to 0 outside the loop
Add 1 to it inside the loop
When you return the index, return the count with it (return [middle, count]).
I assume the code to count numbers of interations required by binary_search is to be used for testing or optimization. If so, the method binary_search should be modified in such a way that to produce production code it is only necessary to remove (or comment out) lines of code, as opposed to modifying statements. Here is one way that might be done.
def binary_search(array, n)
# remove from production code lines marked -> #******
_bin_srch_iters = 0 #******
begin #******
min = 0
max = (array.length) - 1
loop do
_bin_srch_iters += 1 #******
middle = (min + max) / 2
break middle if array[middle] == n
break nil if min == max
if array[middle] > n
max = middle - 1
else # array[middle] < n
min = middle + 1
end
end
ensure #******
puts "binary_search reqd #{_bin_srch_iters} interations" #******
end #******
end
x = binary_search([1,3,6,7,9,11], 3)
# binary_search reqd 3 interations
#=> 1
binary_search([1,3,6,7,9,11], 5)
# binary_search reqd 3 interations
#=> nil
In IRB:
0.9999999999999998 == 1.0
# => false
0.99999999999999998 == 1.0
# => true
Also just showing the inspect values:
0.9999999999999998
# => 0.9999999999999998
0.99999999999999998
# => 1.0
In my case the value created by my program is unfortunately the first one, so I'm having trouble writing a test case which validates the value as equal to one.
I am deciding between the following options:
add round calls in the application code, however the application is already working, I just am unable to test
add round calls in the testing code
???
What would be your recommended approach? Is there a way I can configure my program to treat 0 with fifteen nines after the decimal as equal to 1.0? It feels a bit frustrating because sixteen nines after the decimal seems to be the cut off - I'm only one short.
Reading this article on how to compare small floating-point differences:
http://c-faq.com/fp/fpequal.html
I converted their proposed solutions to Ruby:
class Float
MY_EPSILON = 0.000000000000001
MY_TOLERANCE = 0.000000000000001
def eq_epsilon?(flt)
a = self
b = flt
(a - b).abs <= MY_EPSILON * a.abs
end
def self.reldif(a, b)
c = a.abs
d = b.abs
d = [c,d].max
d == 0.0 ? 0.0 : (a - b).abs / d
end
def eq_reldif?(flt)
Float.reldif(self, flt) <= MY_TOLERANCE
end
end
And thus we can run some test code:
f1 = 0.99999999999999998
f2 = 0.9999999999999998
f3 = 0.999999999999998
[f1, f2, f3].each { |f|
p f.eq_epsilon?(1.0)
p 1.0.eq_epsilon?(f)
}
puts "--------------"
[f1, f2, f3].each { |f|
p f.eq_reldif?(1.0)
p 1.0.eq_reldif?(f)
}
With output:
true
true
true
true
false
false
--------------
true
true
true
true
false
false
However more testing is probably needed to make sure it satisfies all your requirements.
Below I've posted the code to a non-working "divide and conquer" multiplication method in ruby(with debug prints). I cannot tell if its broken code, or a quirk in Ruby like how the L-shift(<<) operator doesn't push bits into the bit-bucket; this is unexpected compared to similar operations in C++.
Is it broken code (doesn't match the original algorithm) or unexpected behavior?
Pseudo code for original algorithm
def multiply(x,y,n, level)
#print "Level #{level}\n"
if n == 1
#print "\tx[#{x.to_s(2)}][#{y.to_s(2)}]\n\n"
return x*y
end
mask = 2**n - 2**(n/2)
xl = x >> (n / 2)
xr = x & ~mask
yl = y >> (n / 2)
yr = y & ~mask
print " #{n} | x = #{x.to_s(2)} = L[#{xl.to_s(2)}][#{xr.to_s(2)}]R \n"
print " #{n} | y = #{y.to_s(2)} = L[#{yl.to_s(2)}][#{yr.to_s(2)}]R \n"
#print "\t[#{xl.to_s(2)}][#{yr.to_s(2)}]\n"
#print "\t[#{xr.to_s(2)}][#{yr.to_s(2)}]\n"
#print "\t([#{xl.to_s(2)}]+[#{xr.to_s(2)}])([#{yl.to_s(2)}]+[#{yr.to_s(2)}])\n\n"
p1 = multiply( xl, yl, n/2, level+1)
p2 = multiply( xr, yr, n/2, level+1)
p3 = multiply( xl+xr, yl+yr, n/2, level+1)
return p1 * 2**n + (p3 - p1 - p2) * 2**(n/2) + p2
end
x = 21
y = 22
print "x = #{x} = #{x.to_s(2)}\n"
print "y = #{y} = #{y.to_s(2)}\n"
print "\nDC_multiply\t#{x}*#{y} = #{multiply(x,y,8, 1)} \nregular\t#{x}*#{y} = #{x*y}\n\n "
I am not familiar with the divide and conquer algorithm but i don't think it contains parts you can't do in Ruby.
Here is a quick attempt:
def multiplb(a,b)
#Break recursion when a or b has one digit
if a < 10 || b < 10
a * b
else
#Max number of digits of a and b
n = [a.to_s.length, b.to_s.length].max
# Steps to split numbers to high and low digits sub-numbers
# (1) to_s.split('') => Converting digits to string then arrays to ease splitting numbers digits
# (2) each_slice => Splitting both numbers to high(left) and low(right) digits groups
# (3) to_a , map, join, to_i => Simply returning digits to numbers
al, ar = a.to_s.split('').each_slice(n/2).to_a.map(&:join).map(&:to_i)
bl, br = b.to_s.split('').each_slice(n/2).to_a.map(&:join).map(&:to_i)
#Recursion
p1 = multiplb(al, bl)
p2 = multiplb(al + ar, bl + br)
p3 = multiplb(ar, br)
p1 * (10**n) + (p2 - p1 - p3) * (10**(n/2)) + p3
end
end
#Test
puts multiplb(1980, 2315)
# => 4583700 yeah that's correct :)
Here are some references to further explain part of the code:
Finding max of numbers => How do you find a min / max with Ruby?
Spliting an array to half => Splitting an array into equal parts in ruby
Turning a fixnum into array => Turning long fixed number to array Ruby
Hope it hepls !
I was playing around with Ruby and Latex to create a color coding set for a registor. I have the following block of code. When attempting to run this, band1 = 1e+02.
I tried band1 = (BigDecimal(i) * 100).to_f, thinking maybe there was some odd floating point issue. An integer multiplied by an integer should create an integer. I tried a variety of other things as well, but to no avail.
(1..9).each do |i| #Band 1
(0..9).each do |j| #Band 2
(0..11).each do |k| #Band 3
#Band 3 Start
#these are the colors of the resistor bands
b1 = $c_band12[i]
b2 = $c_band12[j]
b3 = $c_band3[k]
b4 = "Gold"
oms = ((i*100) + (j*10)) * $mult[k]
band1 = i*100
band2 = j
band3 = $mult[k]
end
end
end
Not sure what I'm missing. Should I be using each_with_index through these iterations? I tried this:
(1..9).each_with_index {|i, indexi| #Band 1
(0..9).each_with_index {|j, indexj| #Band 2
(0..11).each_with_index {|k, indexk| #Band 3
#Band 3 Start
#these are the colors of the resistor bands
b1 = $c_band12[i]
b2 = $c_band12[j]
b3 = $c_band3[k]
b4 = "Gold"
oms = ((i*100) + (j*10)) * $mult[k]
band1 = indexk * 100
and I got the same answer. I can't see why 1*100 should equate to such a large number.
edit: Additional info: If I have this: band1=i*10
then the calculation is correct. In fact, the calculation is correct up to 99.
In your code, band1 has to be a Fixnum. Check with p band1.class. Not sure how you get "1e+02", maybe you print in some strange fashion, or you do band1 == 1e+02 which returns true in Ruby. You must use eql? to distinguish between 1 and 1.0:
1 == 1.0 # => true
1.eql?(1.0) # => false
Ruby may not be the optimal language for this but I'm sort of comfortable working with this in my terminal so that's what I'm going with.
I need to process the numbers from 1 to 666666 so I pin out all the numbers that contain 6 but doesn't contain 7, 8 or 9. The first number will be 6, the next 16, then 26 and so forth.
Then I needed it printed like this (6=6) (16=6) (26=6) and when I have ranges like 60 to 66 I need it printed like (60 THRU 66=6) (SPSS syntax).
I have this code and it works but it's neither beautiful nor very efficient so how could I optimize it?
(silly code may follow)
class Array
def to_ranges
array = self.compact.uniq.sort
ranges = []
if !array.empty?
# Initialize the left and right endpoints of the range
left, right = array.first, nil
array.each do |obj|
# If the right endpoint is set and obj is not equal to right's successor
# then we need to create a range.
if right && obj != right.succ
ranges << Range.new(left,right)
left = obj
end
right = obj
end
ranges << Range.new(left,right) unless left == right
end
ranges
end
end
write = ""
numbers = (1..666666).to_a
# split each number in an array containing it's ciphers
numbers = numbers.map { |i| i.to_s.split(//) }
# delete the arrays that doesn't contain 6 and the ones that contains 6 but also 8, 7 and 9
numbers = numbers.delete_if { |i| !i.include?('6') }
numbers = numbers.delete_if { |i| i.include?('7') }
numbers = numbers.delete_if { |i| i.include?('8') }
numbers = numbers.delete_if { |i| i.include?('9') }
# join the ciphers back into the original numbers
numbers = numbers.map { |i| i.join }
numbers = numbers.map { |i| i = Integer(i) }
# rangify consecutive numbers
numbers = numbers.to_ranges
# edit the ranges that go from 1..1 into just 1
numbers = numbers.map do |i|
if i.first == i.last
i = i.first
else
i = i
end
end
# string stuff
numbers = numbers.map { |i| i.to_s.gsub(".."," thru ") }
numbers = numbers.map { |i| "(" + i.to_s + "=6)"}
numbers.each { |i| write << " " + i }
File.open('numbers.txt','w') { |f| f.write(write) }
As I said it works for numbers even in the millions - but I'd like some advice on how to make prettier and more efficient.
I deleted my earlier attempt to parlez-vous-ruby? and made up for that. I know have an optimized version of x3ro's excellent example.
$,="\n"
puts ["(0=6)", "(6=6)", *(1.."66666".to_i(7)).collect {|i| i.to_s 7}.collect do |s|
s.include?('6')? "(#{s}0 THRU #{s}6=6)" : "(#{s}6=6)"
end ]
Compared to x3ro's version
... It is down to three lines
... 204.2 x faster (to 66666666)
... has byte-identical output
It uses all my ideas for optimization
gen numbers based on modulo 7 digits (so base-7 numbers)
generate the last digit 'smart': this is what compresses the ranges
So... what are the timings? This was testing with 8 digits (to 66666666, or 823544 lines of output):
$ time ./x3ro.rb > /dev/null
real 8m37.749s
user 8m36.700s
sys 0m0.976s
$ time ./my.rb > /dev/null
real 0m2.535s
user 0m2.460s
sys 0m0.072s
Even though the performance is actually good, it isn't even close to the C optimized version I posted before: I couldn't run my.rb to 6666666666 (6x10) because of OutOfMemory. When running to 9 digits, this is the comparative result:
sehe#meerkat:/tmp$ time ./my.rb > /dev/null
real 0m21.764s
user 0m21.289s
sys 0m0.476s
sehe#meerkat:/tmp$ time ./t2 > /dev/null
real 0m1.424s
user 0m1.408s
sys 0m0.012s
The C version is still some 15x faster... which is only fair considering that it runs on the bare metal.
Hope you enjoyed it, and can I please have your votes if only for learning Ruby for the purpose :)
(Can you tell I'm proud? This is my first encounter with ruby; I started the ruby koans 2 hours ago...)
Edit by #johndouthat:
Very nice! The use of base7 is very clever and this a great job for your first ruby trial :)
Here's a slight modification of your snippet that will let you test 10+ digits without getting an OutOfMemory error:
puts ["(0=6)", "(6=6)"]
(1.."66666666".to_i(7)).each do |i|
s = i.to_s(7)
puts s.include?('6') ? "(#{s}0 THRU #{s}6=6)" : "(#{s}6=6)"
end
# before:
real 0m26.714s
user 0m23.368s
sys 0m2.865s
# after
real 0m15.894s
user 0m13.258s
sys 0m1.724s
Exploiting patterns in the numbers, you can short-circuit lots of the loops, like this:
If you define a prefix as the 100s place and everything before it,
and define the suffix as everything in the 10s and 1s place, then, looping
through each possible prefix:
If the prefix is blank (i.e. you're testing 0-99), then there are 13 possible matches
elsif the prefix contains a 7, 8, or 9, there are no possible matches.
elsif the prefix contains a 6, there are 49 possible matches (a 7x7 grid)
else, there are 13 possible matches. (see the image below)
(the code doesn't yet exclude numbers that aren't specifically in the range, but it's pretty close)
number_range = (1..666_666)
prefix_range = ((number_range.first / 100)..(number_range.last / 100))
for p in prefix_range
ps = p.to_s
# TODO: if p == prefix_range.last or p == prefix_range.first,
# TODO: test to see if number_range.include?("#{ps}6".to_i), etc...
if ps == '0'
puts "(6=6) (16=6) (26=6) (36=6) (46=6) (56=6) (60 thru 66) "
elsif ps =~ /7|8|9/
# there are no candidate suffixes if the prefix contains 7, 8, or 9.
elsif ps =~ /6/
# If the prefix contains a 6, then there are 49 candidate suffixes
for i in (0..6)
print "(#{ps}#{i}0 thru #{ps}#{i}6) "
end
puts
else
# If the prefix doesn't contain 6, 7, 8, or 9, then there are only 13 candidate suffixes.
puts "(#{ps}06=6) (#{ps}16=6) (#{ps}26=6) (#{ps}36=6) (#{ps}46=6) (#{ps}56=6) (#{ps}60 thru #{ps}66) "
end
end
Which prints out the following:
(6=6) (16=6) (26=6) (36=6) (46=6) (56=6) (60 thru 66)
(106=6) (116=6) (126=6) (136=6) (146=6) (156=6) (160 thru 166)
(206=6) (216=6) (226=6) (236=6) (246=6) (256=6) (260 thru 266)
(306=6) (316=6) (326=6) (336=6) (346=6) (356=6) (360 thru 366)
(406=6) (416=6) (426=6) (436=6) (446=6) (456=6) (460 thru 466)
(506=6) (516=6) (526=6) (536=6) (546=6) (556=6) (560 thru 566)
(600 thru 606) (610 thru 616) (620 thru 626) (630 thru 636) (640 thru 646) (650 thru 656) (660 thru 666)
(1006=6) (1016=6) (1026=6) (1036=6) (1046=6) (1056=6) (1060 thru 1066)
(1106=6) (1116=6) (1126=6) (1136=6) (1146=6) (1156=6) (1160 thru 1166)
(1206=6) (1216=6) (1226=6) (1236=6) (1246=6) (1256=6) (1260 thru 1266)
(1306=6) (1316=6) (1326=6) (1336=6) (1346=6) (1356=6) (1360 thru 1366)
(1406=6) (1416=6) (1426=6) (1436=6) (1446=6) (1456=6) (1460 thru 1466)
(1506=6) (1516=6) (1526=6) (1536=6) (1546=6) (1556=6) (1560 thru 1566)
(1600 thru 1606) (1610 thru 1616) (1620 thru 1626) (1630 thru 1636) (1640 thru 1646) (1650 thru 1656) (1660 thru 1666)
etc...
Note I don't speak ruby, but I intend to dohave done a ruby version later just for speed comparison :)
If you just iterate all numbers from 0 to 117648 (ruby <<< 'print "666666".to_i(7)') and print them in base-7 notation, you'll at least have discarded any numbers containing 7,8,9. This includes the optimization suggestion by MrE, apart from lifting the problem to simple int arithmetic instead of char-sequence manipulations.
All that remains, is to check for the presence of at least one 6. This would make the algorithm skip at most 6 items in a row, so I deem it less unimportant (the average number of skippable items on the total range is 40%).
Simple benchmark to 6666666666
(Note that this means outputting 222,009,073 (222M) lines of 6-y numbers)
Staying close to this idea, I wrote this quite highly optimized C code (I don't speak ruby) to demonstrate the idea. I ran it to 282475248 (congruent to 6666666666 (mod 7)) so it was more of a benchmark to measure: 0m26.5s
#include <stdio.h>
static char buf[11];
char* const bufend = buf+10;
char* genbase7(int n)
{
char* it = bufend; int has6 = 0;
do
{
has6 |= 6 == (*--it = n%7);
n/=7;
} while(n);
return has6? it : 0;
}
void asciify(char* rawdigits)
{
do { *rawdigits += '0'; }
while (++rawdigits != bufend);
}
int main()
{
*bufend = 0; // init
long i;
for (i=6; i<=282475248; i++)
{
char* b7 = genbase7(i);
if (b7)
{
asciify(b7);
puts(b7);
}
}
}
I also benchmarked another approach, which unsurprisingly ran in less than half the time because
this version directly manipulates the results in ascii string form, ready for display
this version shortcuts the has6 flag for deeper recursion levels
this version also optimizes the 'twiddling' of the last digit when it is required to be '6'
the code is simply shorter...
Running time: 0m12.8s
#include <stdio.h>
#include <string.h>
inline void recursive_permute2(char* const b, char* const m, char* const e, int has6)
{
if (m<e)
for (*m = '0'; *m<'7'; (*m)++)
recursive_permute2(b, m+1, e, has6 || (*m=='6'));
else
if (has6)
for (*e = '0'; *e<'7'; (*e)++)
puts(b);
else /* optimize for last digit must be 6 */
puts((*e='6', b));
}
inline void recursive_permute(char* const b, char* const e)
{
recursive_permute2(b, b, e-1, 0);
}
int main()
{
char buf[] = "0000000000";
recursive_permute(buf, buf+sizeof(buf)/sizeof(*buf)-1);
}
Benchmarks measured with:
gcc -O4 t6.c -o t6
time ./t6 > /dev/null
$range_start = -1
$range_end = -1
$f = File.open('numbers.txt','w')
def output_number(i)
if $range_end == i-1
$range_end = i
elsif $range_start < $range_end
$f.puts "(#{$range_start} thru #{$range_end})"
$range_start = $range_end = i
else
$f.puts "(#{$range_start}=6)" if $range_start > 0 # no range, print out previous number
$range_start = $range_end = i
end
end
'1'.upto('666') do |n|
next unless n =~ /6/ # keep only numbers that contain 6
next if n =~ /[789]/ # remove nubmers that contain 7, 8 or 9
output_number n.to_i
end
if $range_start < $range_end
$f.puts "(#{$range_start} thru #{$range_end})"
end
$f.close
puts "Ruby is beautiful :)"
I came up with this piece of code, which I tried to keep more or less in FP-styling. Probably not much more efficient (as it has been said, with basic number logic you will be able to increase performance, for example by skipping from 19xx to 2000 directly, but that I will leave up to you :)
def check(n)
n = n.to_s
n.include?('6') and
not n.include?('7') and
not n.include?('8') and
not n.include?('9')
end
def spss(ranges)
ranges.each do |range|
if range.first === range.last
puts "(" + range.first.to_s + "=6)"
else
puts "(" + range.first.to_s + " THRU " + range.last.to_s + "=6)"
end
end
end
range = (1..666666)
range = range.select { |n| check(n) }
range = range.inject([0..0]) do |ranges, n|
temp = ranges.last
if temp.last + 1 === n
ranges.pop
ranges.push(temp.first..n)
else
ranges.push(n..n)
end
end
spss(range)
My first answer was trying to be too clever. Here is a much simpler version
class MutablePrintingCandidateRange < Struct.new(:first, :last)
def to_s
if self.first == nil and self.last == nil
''
elsif self.first == self.last
"(#{self.first}=6)"
else
"(#{self.first} thru #{self.last})"
end
end
def <<(x)
if self.first == nil and self.last == nil
self.first = self.last = x
elsif self.last == x - 1
self.last = x
else
puts(self) # print the candidates
self.first = self.last = x # reset the range
end
end
end
and how to use it:
numer_range = (1..666_666)
current_range = MutablePrintingCandidateRange.new
for i in numer_range
candidate = i.to_s
if candidate =~ /6/ and candidate !~ /7|8|9/
# number contains a 6, but not a 7, 8, or 9
current_range << i
end
end
puts current_range
Basic observation: If the current number is (say) 1900 you know that you can safely skip up to at least 2000...
(I didn't bother updating my C solution for formatting. Instead I went with x3ro's excellent ruby version and optimized that)
Undeleted:
I still am not sure whether the changed range-notation behaviour isn't actually what the OP wants: This version changes the behaviour of breaking up ranges that are actually contiguous modulo 6; I wouldn't be surprised the OP actually expected
.
....
(555536=6)
(555546=6)
(555556 THRU 666666=6)
instead of
....
(666640 THRU 666646=6)
(666650 THRU 666656=6)
(666660 THRU 666666=6)
I'll let the OP decide, and here is the modified version, which runs in 18% of the time as x3ro's version (3.2s instead of 17.0s when generating up to 6666666 (7x6)).
def check(n)
n.to_s(7).include?('6')
end
def spss(ranges)
ranges.each do |range|
if range.first === range.last
puts "(" + range.first.to_s(7) + "=6)"
else
puts "(" + range.first.to_s(7) + " THRU " + range.last.to_s(7) + "=6)"
end
end
end
range = (1..117648)
range = range.select { |n| check(n) }
range = range.inject([0..0]) do |ranges, n|
temp = ranges.last
if temp.last + 1 === n
ranges.pop
ranges.push(temp.first..n)
else
ranges.push(n..n)
end
end
spss(range)
My answer below is not complete, but just to show a path (I might come back and continue the answer):
There are only two cases:
1) All the digits besides the lowest one is either absent or not 6
6, 16, ...
2) At least one digit besides the lowest one includes 6
60--66, 160--166, 600--606, ...
Cases in (1) do not include any continuous numbers because they all have 6 in the lowest digit, and are different from one another. Cases in (2) all appear as continuous ranges where the lowest digit continues from 0 to 6. Any single continuation in (2) is not continuous with another one in (2) or with anything from (1) because a number one less than xxxxx0 will be xxxxy9, and a number one more than xxxxxx6 will be xxxxxx7, and hence be excluded.
Therefore, the question reduces to the following:
3)
Get all strings between "" to "66666" that do not include "6"
For each of them ("xxx"), output the string "(xxx6=6)"
4)
Get all strings between "" to "66666" that include at least one "6"
For each of them ("xxx"), output the string "(xxx0 THRU xxx6=6)"
The killer here is
numbers = (1..666666).to_a
Range supports iterations so you would be better off by going over the whole range and accumulating numbers that include your segments in blocks. When one block is finished and supplanted by another you could write it out.