Fibonacci & for loop: how are the commands executed step by step? - bash

#!/bin/bash
a=0
b=1
echo "give a number:"
read n
clear
echo "the fibonacci sequence until $n:"
for (( i=0; i<n; i++ ))
do
echo -n "$a "
c=$((a + b))
a=$b
b=$c
done
If I interpret it well, this code echoes a $a value after every i++ jumps, then switches the variables as you can see, then on the next i++ loop jump it happens again until "i" reaches "n".
Question: if we want in every loop jump the value of the new "c" why shall we echo $a? I see the connection that: a=$b, b=$c, and c=$((a + b)) but i don't get it why do we refer to $a when doing echo?
Is there a more elegant solution?

You mean, “never ever calculate anything needlessly, ever”? It is possible, of course, but it depends on how much ugliness in the control logic you are willing to tolerate. In the example below, fibonacci1 calculates at most one extra element of the series that may not get printed out and fibonacci2 never calculates any extra series elements and everything makes it to the standard output.
Is any of that “elegant”? Probably not. This is actually a common problem most people encounter when coding (in languages other than purely functional ones): Most high(er)-level languages (unlike e.g. assemblers) provide predefined control structures that work great in typical and obvious cases (e.g. one control variable and one operation per iteration) but may become “suboptimal” in more complex scenarios.
A notoriously common example is a variable that stores a value from the previous iteration. Let’s assume you assign it at the very end of the loop. That works fine, but… Could you avoid the very last assignment (because it is useless), instead of leaving it to the compiler’s wisdom? Yes, you could, but then (e.g.) for ((init; condition; step)); do ...; ((previous = current)); done becomes (e.g.) for ((init;;)); do ...; ((step)); ((condition)) || break; ((previous = current)); done.
On one hand, a tiny bit of something (such as thin air) may have been “saved”. On the other hand, the code became assembler-like and harder to write, read and maintain.
To find a balance there^^^ and {not,} optimize when it {doesn’t,does} matter is a lifelong struggle. It may be something like CDO, which is like OCD, but sorted correctly.
fibonacci1() {
local -ai fib=(0 1)
local -i i
for ((i = $1; i > 2; i -= 2)) {
printf '%d %d ' "${fib[#]}"
fib=($((fib[0] + fib[1])) $((fib[0] + 2 * fib[1])))
}
echo "${fib[#]::i}"
}
fibonacci2() {
trap 'trap - return; echo' return
local -i a=0 b=1 i="$1"
((i)) || return 0
printf '%d' "$a"
((--i)) || return 0
printf ' %d' "$b"
for ((;;)); do
((--i)) || return 0
printf ' %d' "$((a += b))"
((--i)) || return 0
printf ' %d' "$((b += a))"
done
}
for ((i = 0; i <= 30; ++i)); do
for fibonacci in fibonacci{1,2}; do
echo -n "${fibonacci}(${i}): "
"$fibonacci" "$i"
done
done

Related

Beginner Shell, can't find the issue (array sorting)

Working on a little script which put random numbers in a 10 000 size array and then sort all this array with the method ask during the course.
I've done this code but it seem that it begin to sort (when I test I have some "a" that are printed but not as much as supposed to and I don't understand why)
I'm believing the problem come fromes my test on val array, and it's probably a beginner error but I don't really know how to find the problem on th web as I don't really now which line is the problem.
I don't necessary need an answer, just some clues to find it could be good :)
Here is my code: (new to stackoverflow so I don't know how to put a good code view directly, if anyone can show me)
for i in `seq 1 10000`;
do
val[${i}]=$RANDOM
done
echo `date +"%M.%S.%3N"`
FLAG=0
until [ $FLAG -eq 1 ]
do
FLAG=1
for j in `seq 1 9999`;
do
if [ ${val[${j}]} -gt ${val[${j+1}]} ]
then
TMP=${val[${j}]}
val[${j}]=${val[${j+1}]}
val[${j+1}]=$TMP
FLAG=0
echo a
fi
done
done
echo `date +"%M.%S.%3N"`
as asked I can't really have a useful output as I just want the date before and after the sort operation. But the sort is just supposed to put values from lower to higher by taking them two by two and invert them if necessary. Doing this until no numbers are inverted.
Edit: I tried with manual number:
10 3 6 9 1
when running it by putting echo ${val[*]} in the for loop it just print 4 times the same list in the same order, so I'm guessing it doesn't work at all... Is my use of "if" wrong ?
Edit 2: At the begining, I did it in C# and I wanted to do it in shell then, firstly because I wanted to practice shell and then because I wanted to compare efficiency and time needed for the same thing. here is the C# code, working.
Random random = new Random();
int[] _tab = new int[100000];
for (int i = 0; i < _tab.Length; i++)
{
_tab[i] = random.Next(1, _tab.Length);
}
bool perm;
int tmp;
DateTime dt = DateTime.Now;
do
{
perm = false;
for (int i = 0; i < (_tab.Length - 1); i++)
{
if (_tab[i] > _tab[i + 1])
{
tmp = _tab[i];
_tab[i] = _tab[i + 1];
_tab[i + 1] = tmp;
perm = true;
}
}
}
while (perm == true);
Console.WriteLine((DateTime.Now - dt).TotalMilliseconds);
Console.Read();
Thanks :)
If my understanding that you want to know why this script is not producing an "a" indicating the ordering of the array of the numbers initially produced in the "for" loop is correct, then here is a solution:
The syntax is incorrect for your variable expansion. The ${var} cannot have math operators inside the braces, because they have different meaning here. In a normal non-associative array Zsh handles subscripts with some basic math support, so you can use ${array[var+1]} instead of ${array[${var+1}]} as you previously did.
I suspect the reason this came about - complicated, error prone POSIX syntax - would have been avoided by using simplified Zsh syntax, but as stated in an earlier comment, it would not be portable to other shells.
Some shells support similar features: Bash supports most, but not bare subscripts ($array[var]). Strings may be ordered in Zsh in a similar manner, but the math-context brackets (( and )) would have to be replaced with normal test brackets [[ and ]] and the array $val might have to be defined with special typeset options to make the strings compare in the desired manner; that is, they might have to be padded and right or left aligned. For comparing enumeration types, like Jan - Feb, it gets a little more complicated with associative arrays and case-conversion.
Here is the script with the appropriate changes, then again in simplified Zsh:
#!/bin/sh
for i in `seq 1 10000`;
do
val[$((i))]=$RANDOM
done
echo `date +"%M.%S.%3N"`
FLAG=0
until [ $FLAG -eq 1 ]
do
FLAG=1
for j in `seq 1 9999`;
do
if [ ${val[$((j))]} -gt ${val[$((j+1))]} ]
then
TMP=${val[$((j))]}
val[$((j))]=${val[$((j+1))]}
val[$((j+1))]=$TMP
FLAG=0
echo a
fi
done
done
echo `date +"%M.%S.%3N"`
Zsh:
#!/bin/zsh
foreach i ( {1..10000} )
val[i]=$RANDOM
end
echo `date +"%M.%S.%3N"`
FLAG=0
until ((FLAG))
do
FLAG=1
foreach j ( {1..9999} )
if (( val[j] > val[j+1] ))
then
TMP=$val[j]
val[j]=$val[j+1]
val[j+1]=$TMP
FLAG=0
echo a
fi
end
done
echo `date +"%M.%S.%3N"`

How to round a large number in Shell command?

In Mac terminal, I would like to round a large number.
For example,
At 10^13th place:
1234567812345678 --> 1230000000000000
Or at 10^12th place:
1234567812345678 --> 1235000000000000
So I would like to specify the place, and then get the rounded number.
How do I do this?
You can use arithmetic expansion:
$ val=1234567812345678
$ echo $(( ${val: -13:1} < 5 ? val - val % 10**13 : val - val % 10**13 + 10**13 ))
1230000000000000
$ echo $(( ${val: -12:1} < 5 ? val - val % 10**12 : val - val % 10**12 + 10**12 ))
1235000000000000
This checks if the most significant removed digit is 5 or greater, and if it is, the last significant unremoved digit is increased by one; then we subtract the division remainder from the (potentially modified) initial value.
If you don't want to have to write it this way, you can wrap it in a little function:
round () {
echo $(( ${1: -$2:1} < 5 ? $1 - $1 % 10**$2 : $1 - $1 % 10**$2 + 10**$2 ))
}
which can then be used like this:
$ round "$val" 13
1230000000000000
$ round "$val" 12
1235000000000000
Notice that quoting $val isn't strictly necessary here, it's just a good habit.
If the one-liner is too cryptic, this is a more readable version of the same:
round () {
local rounded=$(( $1 - $1 % 10**$2 )) # Truncate
# Check if most significant removed digit is >= 5
if (( ${1: -$2:1} >= 5 )); then
(( rounded += 10**$2 ))
fi
echo $rounded
}
Apart from arithmetic expansion, this also uses parameter expansion to get a substring: ${1: -$2:1} stands for "take $1, count $2 from the back, take one character". There has to be a space before -$2 (or is has to be in parentheses) because otherwise it would be interpreted as a different expansion, checking if $1 is unset or null, which we don't want.
awk's [s]printf function can do rounding for you, within the limits of double-precision floating-point arithmetic:
$ for p in 13 12; do
awk -v p="$p" '{ n = sprintf("%.0f", $0 / 10^p); print n * 10^p }' <<<1234567812345678
done
1230000000000000
1235000000000000
For a pure bash implementation, see Benjamin W.'s helpful answer.
Actually, if you want to round to n significant digits you might be best served by mixing up traditional math and strings.
Serious debugging is left to the student, but this is what I quickly came up with for bash shell and hope MAC is close enough:
function rounder
{
local value=$1;
local digits=${2:-3};
local zeros="$( eval "printf '0%.0s' {1..$digits}" )"; #proper zeros
# a bit of shell magic that repats the '0' $digits times.
if (( value > 1$zeros )); then
# large enough to require rounding
local length=${#value};
local digits_1=$(( $digits + 1 )); #digits + 1
local tval="${value:0:$digits_1}"; #leading digits, plus one
tval=$(( $tval + 5 )); #half-add
local tlength=${#tval}; #check if carried a digit
local zerox="";
if (( tlength > length )); then
zerox="0";
fi
value="${tval:0:$digits}${zeros:0:$((length-$digits))}$zerox";
fi
echo "$value";
}
See how this can be done much shorter, but that's another exercise for the student.
Avoiding floating point math due to the inherit problems within.
All sorts of special cases, like negative numbers, are not covered.

Efficient way to bit mask every number of a file list using bitwise operators

I have a file which contains a list of numbers defined as follow :
var1=0x00000001
var2=0x00000002
var3=0x00000008
var4=0x00000020
var5=0x00000040
var6=0x00000080
var7=0x00000100
var8=0x00000200
var9=0x00000400
var10=0x00000800
var11=0x000001000
var12=0x000002000
var13=0x000004000
var14=0x000008000
var15=0x00010000
var16=0x00020000
var17=0x00040000
var18=0x10000000
var19=0x20000000
var20=0x40000000
var21=0x80000000
I want to write something like this:
decValue=2147483650
printf -v hexValue "%x" "${decValue}"
echo $hexValue
IFS="="
while read name ID x
do
test $((${hexValue} & ${ID})) = 0 && continue
array+=("${name}")
done < "$FILE_NAME"
It returns :
80000002
var2 var9 var11 var12 var14 var17
But, in this specific case I just would like to return :
var21 var2
Other example, if decValue=12288 I would like to return var11 and var12.
Bitwise operators is a good tool to solve this issue ?
Use
printf -v hexValue "%#x" "${decValue}"
(or use ${decimalValue} in the test inside the loop)
As it is now, $hexValue ends up being 80000002 (as your own echo statement shows), and this is later interpreted as a decimal number when you want it to be interpreted as a hexadeximal one.
Passing %#x as format specifier to printf will make $hexValue have the value 0x80000002.
You'll also have to take another good look at the var table; there are a number of gaps in it. 0x4 is missing between var2 and var3, 0x10 is missing between var3 and var4, and between var17 and var18, the whole block from 0x80000 to 0x8000000 is gone as well. You're not going to get the results you expect for variables that have any of these bits set.
It might also be worth a thought to generate the bitmasks on the fly rather than holding them precalcuated in a file. One possible approach for that is
for((i = 0; (1 << i) <= $hexValue; ++i))
do
test $(($hexValue & (1 << i))) = 0 && continue
# Note: this will remember (zero-based) bit numbers rather than variable
# names because there are no named variables any longer
array+=($i)
done
In this, the bitshift expression 1 << i gives the number 2i, or put another way: 1 << i has the same value as var$((i + 1)) would in a repaired table.
Or use perl as:
perl -F= -slnE 'say $_ if( hex($F[1]) & $num )' -- -num=12288 < file
prints:
var11=0x000001000
var12=0x000002000
or
perl -F= -slnE 'say $F[0] if( hex($F[1]) & $num )' -- -num=12288 < file
prints
var11
var12

Shell Scripts: Nested for Loop iterating only once

Im trying to automate a download of multiple files from a server. There is a nested loop structure to pass through directories. The files in one directory are numbered and I'm just iterating through them using a counter. The indication that a directory is devoid of all required files is when a 404 is encountered. However, the break statement in the if statement seems to break out of both loops during this case. I tried giving break 1 to indicate that only one loop is to be stopped, but to no avail.
page=$1
chapStart=$2
chapEnd=$3
k=$2
for i in {$2..$3}
do
j=1
while true
do
if ((j < 10))
then
pg=0"$j"
else
pg=$j
fi
res=$(wget --timeout=10 http://www.somesite.com/content/content/"$page"/"$k"/0"$pg".file)
if(($? != 0))
then
break 1
fi
let ++j
done
let ++k
done
I need the break to exit just the while loop when encountered.
Edit
As per chepner's correct answer, the problem was in the for loop structure rather than the break statement, which functions rather beautifully. Updated, working code is:
page=$1
chapStart=$2
chapEnd=$3
k=$2
for((i=$2;i<=$3;i++));
do
j=1
while true
do
if ((j < 10))
then
pg=0"$j"
else
pg=$j
fi
res=$(wget --timeout=10 http://www.somesite.com/content/content/"$page"/"$k"/0"$pg".file)
if(($? != 0))
then
break 1
fi
let ++j
done
let ++k
done
You cannot use parameter expansion inside brace expansions. Your outer loop only has one iteration, with i set to the literal string {$2..$3}. You should use one of
for ((i=$2; i<=$3; i++)); do
or
for i in $(seq $2 $3); do
instead.

How can I shift digits in bash?

I have a homework assignment that is asking to shift a decimal number by a specified amount of digits. More clearly this bash script will take two input arguments, the first is the number(maximum 9 digits) that the shift will be performed on and the second is the number(-9 to 9) of digits to shift. Another requirement is that when a digit is shifted off the end, it should be attached to the other end of the number. One headache of a requirement is that we cannot use control statements of any kind: no loops, no if, and switch cases.
Example: 12345 3 should come out to 345000012 and 12345 -3 should be 12345000
I know that if I mod 12345 by 10^3 I get 345 and then if I divide 12345 by 10^3 I get 12 and then I can just concatenate those two variables together to get 34512. I am not quite sure if that is exactly correct but that is the closest I can get as of now. As far as the -3 shift, I know that 10^-3 is .001 and would work however when I try using 10^-3 in bash I get an error.
I am just lost at this point, any tips would be greatly appreciated.
EDIT: After several hours of bashing (pun intended) my head against this problem, I finally came up with a script that for the most part works. I would post the code right now but I fear another student hopelessly lost might stumble upon it. I will check back and post what I came up with in a week or two. I was able to do it with mods and division. Thank you all for the responses, it really helped me to open up and think about the problem from different angles.
Here's a hint:
echo ${string:0:3}
echo ${#string}
Edit (2011-02-11):
Here's my solution. I added some additional parameters with defaults.
rotate-string ()
{
local s=${1:-1} p=${2:--1} w=${3:-8} c=${4:-0} r l
printf -vr '%0*d' $w 0 # save $w zeros in $r
r=${r//0/$c}$s # change the zeros to the character in $c, append the string
r=${r: -w} # save the last $w characters of $r
l=${r: -p%w} # get the last part of $r ($p mod %w characters)
echo "$l${r::w-${#l}}" # output the Last part on the Left and the Right part which starts at the beginning and goes for ($w minus the_length_of_the_Left_part) characters
}
usage: rotate-string string positions-to-rotate width fill-character
example: rotate-string abc -4 9 =
result: ==abc====
Arguments can be omitted starting from the end and these defaults will be used:
fill-character: "0"
width: 8
positions-to-rotate: -1
string: "1"
More examples:
$ rotate-string
00000010
$ rotate-string 123 4
01230000
Fun stuff:
$ for i in {126..6}; do printf '%s\r' "$(rotate-string Dennis $i 20 .)"; sleep .05; done; printf '\n'
$ while true; do for i in {10..1} {1..10}; do printf '%s\r' "$(rotate-string : $i 10 .)"; sleep .1; done; done
$ while true; do for i in {40..2} {2..40}; do printf '%s\r' "$(rotate-string '/\' $i 40 '_')"; sleep .02; done; done
$ d=0; while true; do for i in {1..10} {10..1}; do printf '%s\r' "$(rotate-string $d $i 10 '_')"; sleep .02; done; ((d=++d%10)); done
$ d=0; while true; do for i in {1..10}; do printf '%s\r' "$(rotate-string $d $i 10 '_')"; sleep .2; ((d=++d%10)); done; done
$ shape='▁▂▃▄▅▆▇█▇▆▅▄▃▂▁'; while true; do for ((i=1; i<=COLUMNS; i++)); do printf '%s\r' "$(rotate-string "$shape" $i $COLUMNS ' ')"; done; done
In the absence of control structures, you need to use recursion, with index values as "choice selections", which is how functional programming often works.
#!/bin/sh
#
# cshift NUMBER N
cshift() {
let num=10#$1
num=`printf '%09d' $num`
lshift="${num:1:8}${num:0:1}"
rshift="${num:8:1}${num:0:8}"
next=( "cshift $lshift $(($2 + 1))" "echo $num" "cshift $rshift $(( $2 - 1 ))" )
x=$(( $2 == 0 ? 1 : $2 < 0 ? 0 : 2 ))
eval "${next[x]}"
}
cshift $1 $2
and, the testing:
$ for ((i=-9;i<=9;i++)); do cshift 12345 $i ; done
000012345
500001234
450000123
345000012
234500001
123450000
012345000
001234500
000123450
000012345
500001234
450000123
345000012
234500001
123450000
012345000
001234500
000123450
000012345
You can also do some math on the indexes and avoid the recursion, but I don't mind making the computer work harder so I don't have to. It's easy to think of how to do the shift by one in either direction, and then I use an evaluated choice that is selected by the signum of the shift value, outputting a value and stopping when the shift value is zero.

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