The middle-square method

Question

Introduction

The middle-square method is used for the generation of pseudorandom numbers. However, this is not a good method in practice, since its period is usually very short and has some severe weaknesses. How does this work? Let's take an example:

For the seed, we pick 123456:

Seed 123456

The seed squared (seed ×ばつ seed), is equal to:

Seed2 15241383936

We started with a 6-digit number. That means that the seed squared should deliver a 12-digit number. If this is not the case, leading zeroes are added to compensate:

Seed2 015241383936

We then take the middle part of the number, with the same size as the seed:

Seed2 015241383936
 ^^^^^^

This is then our new seed: 241383. We repeat the same process as shown above. We get the following:

0: 123456
 015241383936
 | |
1: 241383
 058265752689
 | |
2: 265752
 070624125504
 | |
3: 624125
 389532015625
 | |
4: 532015
 283039960225
 | |
5: 039960
 001596801600
 | |
6: 596801

And this keeps on in a while... Now we know what the middle-square method is, let's get to the challenge:

---

The Task

Every seed has a period. The period of a n-digit seed cannot be longer than 8ⁿ. For example, the seed 82. This would give the following sequence:

82 > 72 > 18 > 32 > 02 > 00 > 00 > 00 > 00 > 00
|____|____|____|____|____|____|____|____|____|___...
0 1 2 3 4 5 6 7 8 9

You can see that the period is equal to 5, before containing the same digit again. Your task is, when given a seed greater than 0 containing no leading zeroes, output the period of the seed. So, in this case, you need to output 5.

Another example is: 24, which gives the following:

24 > 57 > 24
|____|____|___...
0 1 2

As you can see, not all sequences end in 0. This cycle has a period of 1.

---

Test cases

Input > Output
24 > 1
82 > 5
123456 > 146
8989 > 68
789987 > 226

The pastebins with the sequences for 123456, 8989, 789987

This is code-golf, so the submission with the least amount of bytes wins!

You can assume that the input will never have an uneven number of digits.

• Dennis: Nit pick: That's not a period. Period implies that the sequence eventually goes back it its initial state. 24 is periodic (with period 2, I'd say), 82 is eventually periodic (with period 1).
• Luis Mendo: So "period" is the 0-index of the last state which is different from all previous states?
• Adnan: @LuisMendo Yes, that is correct. My mathematical knowledge isn't the best :p.
• ASCII-only: It'd be more like 'the number of iterations before it stabilizes'
• Adnan: @WashingtonGuedes See this pastebin. Does that make it more clear?

Answer 1

Pure bash, (削除) 162 131 116 113 (削除ここまで) 107

Saved 3 bytes by using $c...

Thanks @Dennis for help me to save 6 more bytes.

---- begin middleSquare ----

for((b=1ドル;i[c=10#$b]<2;)){ a=${#b}
printf -v b %0$[a*2]d $[c*c]
b=${b:a/2:a};((i[10#$b]++))
};echo ${#i[@]}

---- end middleSquare ----

for testCase in 24 82 123456 8989 789987 111111;do
 printf "%12s: " $testCase
 bash middleSquare $testCase
 done
 24: 2
 82: 5
 123456: 146
 8989: 68
 789987: 226
 111111: 374

Square formated, 131

---- begin middleSquare ----

for((b=1ドル;i[
10#$b]<2;1))
do a="${#b}" 
printf -v b\
 %0$[a*2]d \
$[10#$b**2];
b=${b:a/2:a}
((i[10#$b]++
));done;ech\
o ${#i[@]:0}

---- end middleSquare ----

for testCase in 24 82 123456 8989 789987 111111;do
 printf "%12s: %9d\n" $testCase $(
 bash middleSquare $testCase)
 done
 24: 2
 82: 5
 123456: 146
 8989: 68
 789987: 226
 111111: 374

Old but with fancy output, 162

---- begin middleSquare ----

for((b=1ドル;i[10#$b
]<2;1))do a=${#b}
printf -v b %0$[a
*2]d $[10#$b**2]
b=${b:a/2:a};((i[
10#$b]++));print\
f "%9d %s\n" ${#\
i[@]} $b;done;ec\
ho -- ${#i[@]} --

---- end middleSquare ----

bash middleSquare 24
 1 57
 2 24
 2 57
-- 2 --
for testCase in 24 82 123456 8989 789987 111111
 do while read f v f
 do r=$v;done < <(
 bash middleSquare $testCase)
 printf "%12s: %11d\n" $testCase $r
 done
 24: 2
 82: 5
 123456: 146
 8989: 68
 789987: 226
 111111: 374

Answer 2

Jelly, (削除) 26 (削除ここまで) (削除) 24 (削除ここまで) 18 bytes

3DL5*
2:1ドル⁄2¤%¢μÐL·L’

Try it online!

How it works

3DL5* Helper link. No arguments.
3 Yield the original input.
 D Convert from integer to base 10.
 L Get l, the length of the decimal representation.
 5* Compute 10 ** l.
2:1ドル⁄2¤%¢μÐL·L’ Main link. Input: n (integer)
2 Square n.
 1ドル⁄2¤ Call the helper link and take the square root of the result.
 : Integer division; divide the left result by the right one.
 ¢ Call the helper link.
 % Take the left result modulo the right one.
 μ Convert the previous chain into a link, and begin a new chain.
 ÐL· Repeat the previous chain until the results are no longer unique,
 updating n in each iteration. Collect the intermediate results.
 L Get the length of the list of results.
 ’ Decrement.

Answer 3

JavaScript (ES7), 82 bytes

f=(n,p={},m=-1,l=n.length)=>p[n]?m:f(`${n*n+100**l}`.substr(l/2+1,l,p[n]=1),p,++m)

Accepts input in the form of a string, e.g. "82", and returns an integer. Simple tail recursive technique to check each seed in turn against a hash of seeds that have already been seen. I add 100**l to the square to ensure a consistent length.

• Neil: @Downgoat Accepts input in the form of a string.
• Downgoat: oh yeah, I guess I can't read :|
• Neil: @WashingtonGuedes No, that doesn't work when the intermediate value begins with enough zeros. (This is why I "wasted" 7 bytes adding 100**l.)
• Neil: @WashingtonGuedes There are cases where it doesn't work, for instance try following the chain from 5288.

Answer 4

Python (削除) 3 (削除ここまで) 2, (削除) 139 (削除ここまで) (削除) 114 (削除ここまで) 97 bytes

Thanks to Seeq for golfing off 25 bytes and thanks to Dennis for golfing off 17 bytes! Code:

s=`input()`;u=[];l=len(s)/2
while not s in u:u+=[s];s=`int(s)**2`.zfill(l*4)[l:3*l]
print~-len(u)

Can definitely be golfed further. This was also the code used to make the test cases :P.

Answer 5

Pyth, 21 bytes

tl.us_<>_`^N2/lz2lzsz

Try it online: Demonstration or Test Suite

edit: Found the edge case 1000, which didn't worked with my previous code. Fixed it for 1 byte.

Explanation:

tl.us_<>_`^N2/lz2lzsz implicit: z = input string
 .u sz apply the following instructions to N, starting with N = int(z), 
 until it runs into a loop:
 ^N2 square it
 ` convert it to a string
 _ reverse order
 > /lz2 remove the first len(z)/2
 < lz remove everything but the first len(z) 
 _ reverse order
 s convert to int
 .u returns the list of all intermediate values
 l compute the length of this list
t minus 1

• Ven: any reason to use sz instead of Q?
• Jakube: @user1737909 If I use Q, I have to replace all the lz with l`Qs.
• Ven: mh, it seems surprising Pyth doesn't share input. I guess it's really meant to allow for a second stdin read..?
• Jakube: @user1737909 Yes. The only possibility to share input is with .z and .Q, though they read multiple lines of input and store them in lists. But I haven't actually seen somebody use this feature. It's only 1 byte to evaluate a string or to stringify a number.
• Ven: Okay, so you can read stdin at most 4 times in Pyth, Qz.Q.z?

Answer 6

MATL, 33 (削除) 35 (削除ここまで) (削除) 40 (削除ここまで) bytes

`t0)2^10GVnXK2/^/k10K^\vtun@>]n2-

Try it online!

` % do...while
 t % duplicate. Take input implicitly on first iteration
 0) % pick last value of array
 2^ % square
 10 % push 10
 GVn % number of digits of input
 XK % copy that to clipboard K
 2/ % divide by 2
 ^ % power
 /k % divide and floor. This removes rightmost digits from the square value
 10K^ % 10 ^ number of digits of input
 \ % modulo. This takes the central part of the squared number
 v % concatenate this new number to array of previous numbers
 tun@> % does the number of unique values exceed the iteration index?
] % if so: next iteration. Else: exit loop
n2- % desired result is the amount of numbers minus 2. Implicitly display

Answer 7

Oracle SQL 11.2, 184 bytes

WITH v(v,p,n)AS(SELECT:1,'0',-1 FROM DUAL UNION ALL SELECT SUBSTR(LPAD(POWER(v,2),LENGTH(v)*2,0),LENGTH(v)/2+1,LENGTH(v)),v,n+1 FROM v)CYCLE v SET c TO 1 DEFAULT 0 SELECT MAX(n)FROM v;

Un-golfed

WITH v(v,p,n) AS
(
 SELECT :1,'0',-1 FROM DUAL
 UNION ALL
 SELECT SUBSTR(LPAD(POWER(v,2),LENGTH(v)*2,0), LENGTH(v)/2+1, LENGTH(v)),v,n+1 FROM v
)
CYCLE v SET c TO 1 DEFAULT 0
SELECT MAX(n) FROM v;

It uses the build in cycle detection to stop the recursivity.

Answer 8

Julia, 64 bytes

f(n,A=[],p=10^endof(dec(n)))=n∈A?-1:f(n^2÷√p%p,[A...n],p)+1

Try it with Coding Ground.

Answer 9

Mathematica, 80 bytes

(a=10^⌊Log10@#+1⌋;Length@NestWhileList[⌊#^2/a^.5⌋~Mod~a&,#,Unequal,All]-2)&

Answer 10

CJam, 37 bytes

q{__,W*:D;~_*sD2/<D>]___|=:A;~A}g],((

Ran into an annoying stack-order issue that I can't immediately see how to resolve. It's also incredibly slow.

How it works: Each iteration pushes the new value on top of the stack, then we wrap the stack into an array, and see if it's the same as its union with itself (to see if it has any duplicate elements). When it has duplicate elements, stop, and see how many elements are in the stack.

Answer 11

Python 2, 82 bytes

def f(n,A=[],l=0):l=l or len(`n`)/2;return-(n in A)or-~f(n*n/10**l%100**l,A+[n],l)

Try it on Ideone.

Answer 12

Python, 124 bytes

def f(s,p=-1,n=0,m=[]):
 x=len(str(s))*2
 while n not in m:m+=[s];y=str(s*s).zfill(x);n=int(y[x/4:x*3/4]);p+=1;s=n
 return p

Answer 13

VBSCRIPT, 131 bytes

s=inputbox(c):l=len(s):do:t=t&","&s:s=space(l*2-len(s*s))&s*s:s=mid(s,l/2+1,l):i=i+1:loop until instr(t,","&s)>0:msgbox i-1

Best I could do with vbscript, first time poster so go easy on me!

• GamrCorps: Welcome to Programming Puzzles and Code Golf Stack Exchange! Great first post! I edited the formatting of your post a little bit to make it more readable and to conform to our standards more. Happy golfing!