I am working on the QuickSort - Median Three Algorithm.
I have no problem with the first and last element sorting. But, when comes to the Median-three, I am slightly confused. I hope someone could help me on this.
Would be appreciate if someone could provide me some pseudocode?
My understanding is to get the middle index by doing this. (start + end) / 2 , then swap the middle pivot value to the first value, after all these done it should goes well with the normal quick sort ( partitioning and sorting).
Somehow, I couldn't get it works. Please help!
#Array Swap function
def swap(A,i,k):
temp=A[i]
A[i]=A[k]
A[k]=temp
# Get Middle pivot function
def middle(lista):
if len(lista) % 2 == 0:
result= len(lista) // 2 - 1
else:
result = len(lista) // 2
return result
def median(lista):
if len(lista) % 2 == 0:
return sorted(lista)[len(lista) // 2 - 1]
else:
return sorted(lista)[len(lista) // 2]
# Create partition function
def partition(A,start,end):
m = middle(A[start:end+1])
medianThree = [ A[start], A[m], A[end] ]
if A[start] == median(medianThree):
pivot_pos = start
elif A[m] == median(medianThree):
tempList = A[start:end+1]
pivot_pos = middle(A[start:end+1])
swap(A,start,pivot_pos+start)
elif A[end] == median(medianThree):
pivot_pos = end
#pivot = A[pivot_pos]
pivot = pivot_pos
# swap(A,start,end) // This line of code is to switch the first and last element pivot
swap(A,pivot,end)
p = A[pivot]
i = pivot + 1
for j in range(pivot+1,end+1):
if A[j] < p:
swap(A,i,j)
i+=1
swap(A,start,i-1)
return i-1
count = 0
#Quick sort algorithm
def quickSort(A,start,end):
global tot_comparisons
if start < end:
# This to create the partition based on the
pivot_pos = partition(A,start,end)
tot_comparisons += len(A[start:pivot_pos-1]) + len(A[pivot_pos+1:end])
# This to sort the the left partition
quickSort(A,start,pivot_pos -1)
#This to sort the right partition
quickSort(A,pivot_pos+1,end)
I am learning ruby and the way I am going about this is by learning and implementing sort algorithms. While working on selection sort, I tried to modify it as follows:
In every pass, instead of finding the smallest and moving it to the top or beginning of the array, find the smallest and the largest and move them to both ends
For every pass, increment the beginning and decrease the ending positions of the array that has to be looped through
While swapping, if the identified min and max are in positions that get swapped with each other, do the swap once (otherwise, two swaps will be done, 1 for the min and 1 for the max)
This doesn't seem to work in all cases. Am I missing something in the logic? If the logic is correct, I will revisit my implementation but for now I haven't been able to figure out what is wrong.
Please help.
Update: This is my code for the method doing this sort:
def mss(array)
start = 0;
stop = array.length - 1;
num_of_pass = 0
num_of_swap = 0
while (start <= stop) do
num_of_pass += 1
min_val = array[start]
max_val = array[stop]
min_pos = start
max_pos = stop
(start..stop).each do
|i|
if (min_val > array[i])
min_pos = i
min_val = array[i]
end
if (max_val < array[i])
max_pos = i
max_val = array[i]
end
end
if (min_pos > start)
array[start], array[min_pos] = array[min_pos], array[start]
num_of_swap += 1
end
if ((max_pos < stop) && (max_pos != start))
array[stop], array[max_pos] = array[max_pos], array[stop]
num_of_swap += 1
end
start += 1
stop -= 1
end
puts "length of array = #{array.length}"
puts "Number of passes = #{num_of_pass}"
puts "Number of swaps = #{num_of_swap}"
return array
end
The problem can be demonstrated with this input array
7 5 4 2 6
After searching the array the first time, we have
start = 0
stop = 4
min_pos = 3
min_val = 2
max_pos = 0 note: max_pos == start
max_val = 7
The first if statement will swap the 2 and 7, changing the array to
2 5 4 7 6
The second if statement does not move the 7 because max_pos == start. As a result, the 6 stays at the end of the array, which is not what you want.
I am taking a lesson on codecademy in which I am currently stuck do not know how to proceed - it is regarding return values.
The instructions are:
Write a method that takes an integer as an argument, and returns that integer times ten. Call times_ten in your code after you define it and print out its return value.
What is given in the script is:
def times_ten(integer)
# your code here
end
# call times_ten here
This is the example it gives but I am having a hard time understanding:
def first_squares(number_of_squares)
squares = []
idx = 0
while idx < number_of_squares
squares.push(idx * idx)
idx = idx + 1
end
return squares
end
puts("How many square numbers do you want?")
number_of_squares = gets.to_i
squares = first_squares(number_of_squares)
idx = 0
while idx < squares.length
puts(squares[idx])
idx = idx + 1
end
Thanks for your help
It should be:
def ten_times(n)
n*10 # you don't have to use 'return' explicitly
end
ten_times(n) -- but put in an actual integer instead of n (or maybe you have to puts or print it, depending on what they want)
Your example is not really related to your outcome.
The example script should be like this:
def ten_times(integer)
# integer * 10 #for implicit return
return integer * 10 #for explicit return
end
print ten_times(any number you want goes in here)
You can run the following code at www.rubyplus.biz:
Implicit return:
def times_ten(integer)
integer * 10
end
p times_ten(1)
Explicit return:
def times_ten(integer)
return integer * 10
end
p times_ten(2)
Below is my code to solve the following problem:
Write a function, nearest_larger(arr, i) which takes an array and an
index. The function should return another index, j: this should
satisfy:
(a) arr[i] < arr[j], AND
(b) there is no j2 closer to i than j where arr[i] < arr[j].
In case of ties choose the earliest (left-most)
of the two indices. If no number in arr is larger than arr[i],
return nil.
This was my attempt:
def nearest_larger(arr, i)
k = 1
loop do
jleft = i - k
jright = i + k
if (arr[i] < arr[jleft]) && (jleft >= 0)
return jleft
elsif (arr[i] < arr[jright]) && (jright < arr.length)
return jright
elsif (jleft < 0) && (jright >= arr.length)
return nil
end
k += 1
end
end
This is the actual correct answer
def nearest_larger(arr, idx)
diff = 1
loop do
left = idx - diff
right = idx + diff
if (left >= 0) && (arr[left] > arr[idx])
return left
elsif (right < arr.length) && (arr[right] > arr[idx])
return right
elsif (left < 0) && (right >= arr.length)
return nil
end
diff += 1
end
end
While my code works well for many of the values I tested when I use certain combinations like this:
x = [1,6,9,4,5]
puts nealest_larger(x, 4)
I get this error
calc.rb:8:in `<': comparison of Fixnum with nil failed (ArgumentError)
from calc.rb:8:in `block in nealest_larger'
from calc.rb:3:in `loop'
from calc.rb:3:in `nealest_larger'
from calc.rb:40:in `<main>'
Can someone tell me how my code differs from the actual answer, to me it seems like it should behave exactly the same but I must have missed some syntax or overlooked a piece of logic. I need another pair of eyes as I am unable to see the difference, thanks!
Your version uses array indices before you've checked that they're in-bounds, i.e., on the left-hand side of the &&. The working version checks for in-bounds first, then uses the index if it's legal. Because Ruby && short circuits, test && use approach avoids the problem you ran into with your implementation.
The 8th line of code in your solution compares the values in the array before first checking the bounds of the array. Notice the correct solution does those comparisons in the reverse order, and the && operator short circuits, avoiding the second (invalid) comparison.
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.