Password strength checker and generator

Check a password's entropy in bits and its average brute-force crack time at a chosen guess rate, or generate a password from a reproducible seed.

പുതുക്കിയത് പരിശോധിച്ച ഉദാഹരണങ്ങൾ: 6

Checked on this device only; nothing is stored.
= 10,000,000,000
പരീക്ഷിക്കുക
Entropy
bits
Entropy: 72.3 bits
ദശാംശ സ്ഥാനങ്ങൾ: 1; ഏറ്റവും അടുത്തതിലേക്ക്; തുല്യദൂരത്തിൽ ഇരട്ട അക്കത്തിലേക്ക്
നീളം
11characters
Character pool
95symbols
Average guesses to find it
2.844 × 10²¹
Average time to crack in seconds
284,400,000,000s
Average time to crack
9,010 years
This assumes every character was picked at random. Dictionary words, names, dates and keyboard patterns are guessed far sooner, so treat the result as an upper bound.
The password is checked on this device and never stored or sent. The share link does include it, so don't share the link for a real password.

This password has 11 characters from a pool of 95, about 72.3 bits of entropy. An attacker making 10,000,000,000 guesses per second needs 9,010 years on average to find it by brute force.

Average time to crack (log₁₀ seconds)

Online, rate-limited9.01 × 10¹² yearsOnline, no rate limit9.01 × 10¹⁰ yearsOffline, slow hash9.01 × 10⁹ yearsOffline, fast hash9,010 yearsOffline, large cluster90.1 years
Average time to crack by attacker speed വരികൾ: 5
AttackerGuesses per secondAverage time
Online, rate-limited109.01 × 10¹² years
Online, no rate limit1,0009.01 × 10¹⁰ years
Offline, slow hash10,0009.01 × 10⁹ years
Offline, fast hash10,000,000,0009,010 years
Offline, large cluster1,000,000,000,00090.1 years
കണക്കാക്കുന്ന വിധം S
  1. Character pool

    N=26 (lowercase)+26 (uppercase)+10 (digits)+33 (symbols and space)=95N = 26\ (\text{lowercase}) + 26\ (\text{uppercase}) + 10\ (\text{digits}) + 33\ (\text{symbols and space}) = 95
  2. Entropy

    H=Llog⁡2N=11×log⁡295=72.27 bitsH = L \log_2 N = 11 \times \log_2 95 = 72.27\ \text{bits}
  3. Average guesses

    2H−1=NL2=95112=2.844×10212^{H-1} = \frac{N^L}{2} = \frac{95^{11}}{2} = 2.844 \times 10^{21}

    On average an exhaustive search finds the password halfway through the space.

  4. Time at your attacker's speed

    t=2.844×102110,000,000,000 /s=284,400,000,000 st = \frac{2.844 \times 10^{21}}{10{,}000{,}000{,}000\ \text{/s}} = 284{,}400{,}000{,}000\ \text{s}

    About 9,010 years.

Password strength checker and generator സംബന്ധിച്ച്

Entropy counts how many guesses a randomly chosen password could take: H = L × log2 N bits, where L is the length and N the size of the character pool (26 lowercase, 26 uppercase, 10 digits and 33 symbols including space). A brute-force search finds the password after N^L ÷ 2 guesses on average, so the crack time is that number divided by the attacker's guesses per second.

The default, Tr0ub4dor&3 from the xkcd comic 936, draws on all 95 printable ASCII characters across 11 positions, which gives 72.3 bits; at 10 billion guesses per second an exhaustive search would average 9,010 years.

That figure is an upper bound, because it assumes every character was picked at random. Tr0ub4dor&3 is a dictionary word with common substitutions, which the comic puts at about 28 bits. Generated passwords come from xoshiro256** and a 64-bit seed, so they hold at most 64 bits whatever their length.

പരിഹരിച്ച ഉദാഹരണങ്ങൾ

Tr0ub4dor&3

I want to
Check a password
Password
Tr0ub4dor&3
Attacker's guesses per second
1e10
Character pool
95 symbols
നീളം
11 characters
Entropy
72.3 bits
Average time to crack in seconds
284,400,000,000 s

പരിശോധനയുടെ ഉറവിടം: Python 3.8: 11 * math.log2(95) = 72.2684…; Decimal(95)**11 / 2 / Decimal('1e10') = 284400046138.22998… s

password

I want to
Check a password
Password
password
Attacker's guesses per second
1e10
Character pool
26 symbols
Entropy
37.6 bits
Average guesses to find it
104,400,000,000
Average time to crack in seconds
10.44 s

പരിശോധനയുടെ ഉറവിടം: Python 3.8: 8 * math.log2(26) = 37.6035…; 26**8 // 2 = 104413532288; / 1e10 = 10.4414 s

Single character (edge)

I want to
Check a password
Password
a
Attacker's guesses per second
10
Character pool
26 symbols
നീളം
1 character
Entropy
4.7 bits
Average guesses to find it
13
Average time to crack in seconds
1.3 s

പരിശോധനയുടെ ഉറവിടം: Python 3.8: math.log2(26) = 4.7004; 26 / 2 = 13 guesses; 13 / 10 = 1.3 s

Seeded 16-character password

I want to
Generate a password
നീളം
16
Seed
2026
Attacker's guesses per second
1e10
Generated password
Ro_y2y-Z^~5cQiBL
Character pool
94 symbols
Entropy
104.9 bits

പരിശോധനയുടെ ഉറവിടം: Independent Python 3.8 port of splitmix64.c and xoshiro256starstar.c with the same rejection rule; 16 * math.log2(94) = 104.87

ചോദ്യങ്ങൾ

How long should a password be according to NIST?

NIST SP 800-63B-4 (2025) requires at least 15 characters for a password used on its own and at least 8 when it is one factor of a multi-factor login, and services should accept passwords of at least 64 characters. It forbids composition rules such as forced symbols and periodic forced changes, and instead requires checking each new password against a blocklist of common and breached passwords.

How long would it take to crack my password?

It depends on the pool, the length and the attacker's speed. A random 8-letter lowercase password has 26^8 ≈ 2.1 × 10^11 combinations and falls in about 10 seconds on average at 10 billion guesses per second, a fast offline attack on a leaked hash. A random 16-character password from 94 printable symbols would take about 5.9 × 10^13 years at the same speed.

What is password entropy?

Password entropy is the base-2 logarithm of the number of equally likely passwords the choosing method could produce, in bits; each extra bit doubles the guesses needed. A random password of length L from N symbols has L × log2 N bits, so 8 lowercase letters give 37.6 bits. It rates how the password was chosen, not how it looks, which is why human-picked words score far lower.

Is a passphrase better than a complex password?

Usually, if the words are chosen at random. Four words drawn from the EFF's 7,776-word dice list give 4 × log2 7,776 ≈ 51.7 bits, and six give 77.5 bits, more than a random 11-character password from all 95 printable characters (72.3 bits), while being easier to type and remember. Words a person picks, or a known quote, are far weaker.

Is this password generator safe for real accounts?

Only with a secret, random seed. The generator is reproducible by design: xoshiro256** seeded with a 64-bit number, so anyone who knows the seed and settings gets the same password, and it never holds more than 64 bits of entropy. For everyday accounts, a password manager's generator is the better choice because it draws from the operating system's cryptographic random source.

“Password strength checker and generator” എത്രത്തോളം കൃത്യമാണ്?

കൃത്യത നിങ്ങളുടെ ഇൻപുട്ടുകളെയും രീതിയുടെ അനുമാനങ്ങളെയും ആശ്രയിച്ചിരിക്കുന്നു. ദശാംശ ഗണിതം 50 സാർഥക അക്കങ്ങൾ ഉപയോഗിക്കുന്നു. എന്നാൽ അനുമാനക്കണക്കുകൾ, സംഖ്യാത്മക രീതികൾ, ഉറവിട ഡാറ്റ എന്നിവയ്ക്ക് കൃത്യത കുറവാകാം; പ്രദർശിപ്പിക്കുന്ന മൂല്യം റൗണ്ട് ചെയ്യുന്നത് ഈ പരിമിതികൾ നീക്കില്ല. സ്വതന്ത്ര ഉറവിടങ്ങളിലെ പരിഹാരങ്ങളുമായി പരിശോധിച്ച ഉദാഹരണങ്ങൾ: 6. ഉദാഹരണത്തിന്, “Tr0ub4dor&3” എന്നത് Python 3.8: 11 * math.log2(95) = 72.2684…; Decimal(95)**11 / 2 / Decimal('1e10') = 284400046138.22998… s ഉപയോഗിച്ച് പരിശോധിക്കുന്നു.

ഈ രീതിയുടെ ഉറവിടം എന്താണ്?

NIST SP 800-63B, Digital Identity Guidelines: Authentication — memorized secrets and rate limiting; Blackman & Vigna, Scrambled linear pseudorandom number generators, ACM TOMS 47(4), 2021 — xoshiro256** reference code.

ഈ കാൽക്കുലേറ്ററിനെക്കുറിച്ച്

H=Llog⁡2Nt=2H−1R=NL2RR=guesses per second\begin{aligned} H &= L \log_2 N \\ t &= \frac{2^{H-1}}{R} = \frac{N^L}{2R} \\ R &= \text{guesses per second}\end{aligned}

സ്രോതസ്സുകൾ

  1. NIST SP 800-63B, Digital Identity Guidelines: Authentication — memorized secrets and rate limiting
  2. Blackman & Vigna, Scrambled linear pseudorandom number generators, ACM TOMS 47(4), 2021 — xoshiro256** reference code

ആധാരങ്ങളുമായി പരിശോധിച്ചത്

ഈ കാൽക്കുലേറ്ററിൽ സ്വതന്ത്ര ആധാരങ്ങളിൽ നിന്ന് ഉത്തരങ്ങൾ ലഭിച്ച 6 പരിഹരിച്ച ഉദാഹരണങ്ങളുണ്ട്. ഇവ പരിശോധനാസമുച്ചയത്തിൽ പ്രവർത്തിക്കുന്നു; നിങ്ങൾക്ക് ഇവിടെയും പ്രവർത്തിപ്പിക്കാം.

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