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How to get digits from a BigInt in javascript?
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I am working on problem n°104 of project Euler Problem 104 and would like to do it in javascript.
In order to solve this problem I need to compute large values of the Fibonacci sequence, but the numbers produced by this sequence are too large to be handle by classic Number, so I'm using BigInt supported in the latest versions of javascript.
Once I've got a particular result stored in a BigInt, I need to check it's 10 first, and last digits.
To get the digits from a Number we usually do something like in the code below, but when the number becomes very large, things go wrong:
let number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = BigInt(1234567891111111111111111111111111111)
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // unpredictable resultIt seems like the "toString()" methods is only using the precision of the Number type (2^53 I believe), thus we are quickly losing precision on the last digits of the BigInt number. The problem is I can't find other methods to extract those digits.
Edit :
I need the precision to be perfect because basicaly what i'm doing for example is :
Compute Fibonacci(500) = 280571172992510140037611932413038677189525
Get the 10 last digits of this number : 8677189525 (this is where is lose the precision)
And then to solve my problem I need to check that those 10 last digits contains all the digits from 1 to 9
javascript bigint
asked Mar 9 at 0:44
S. SylvainS. Sylvain
164
add a comment |
I am working on problem n°104 of project Euler Problem 104 and would like to do it in javascript.
In order to solve this problem I need to compute large values of the Fibonacci sequence, but the numbers produced by this sequence are too large to be handle by classic Number, so I'm using BigInt supported in the latest versions of javascript.
Once I've got a particular result stored in a BigInt, I need to check it's 10 first, and last digits.
To get the digits from a Number we usually do something like in the code below, but when the number becomes very large, things go wrong:
let number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = BigInt(1234567891111111111111111111111111111)
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // unpredictable resultIt seems like the "toString()" methods is only using the precision of the Number type (2^53 I believe), thus we are quickly losing precision on the last digits of the BigInt number. The problem is I can't find other methods to extract those digits.
Edit :
I need the precision to be perfect because basicaly what i'm doing for example is :
Compute Fibonacci(500) = 280571172992510140037611932413038677189525
Get the 10 last digits of this number : 8677189525 (this is where is lose the precision)
And then to solve my problem I need to check that those 10 last digits contains all the digits from 1 to 9
javascript bigint
asked Mar 9 at 0:44
S. SylvainS. Sylvain
164
Exactly how precise do you want the numbers to be? Questions typically get better answers when you explicitly state what your desired results are. Also, +1
– Caleb Goodman
Mar 9 at 0:49
add a comment |
I am working on problem n°104 of project Euler Problem 104 and would like to do it in javascript.
In order to solve this problem I need to compute large values of the Fibonacci sequence, but the numbers produced by this sequence are too large to be handle by classic Number, so I'm using BigInt supported in the latest versions of javascript.
Once I've got a particular result stored in a BigInt, I need to check it's 10 first, and last digits.
To get the digits from a Number we usually do something like in the code below, but when the number becomes very large, things go wrong:
let number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = BigInt(1234567891111111111111111111111111111)
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // unpredictable resultIt seems like the "toString()" methods is only using the precision of the Number type (2^53 I believe), thus we are quickly losing precision on the last digits of the BigInt number. The problem is I can't find other methods to extract those digits.
Edit :
I need the precision to be perfect because basicaly what i'm doing for example is :
Compute Fibonacci(500) = 280571172992510140037611932413038677189525
Get the 10 last digits of this number : 8677189525 (this is where is lose the precision)
And then to solve my problem I need to check that those 10 last digits contains all the digits from 1 to 9
javascript bigint
asked Mar 9 at 0:44
S. SylvainS. Sylvain
164
I am working on problem n°104 of project Euler Problem 104 and would like to do it in javascript.
In order to solve this problem I need to compute large values of the Fibonacci sequence, but the numbers produced by this sequence are too large to be handle by classic Number, so I'm using BigInt supported in the latest versions of javascript.
Once I've got a particular result stored in a BigInt, I need to check it's 10 first, and last digits.
To get the digits from a Number we usually do something like in the code below, but when the number becomes very large, things go wrong:
let number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = BigInt(1234567891111111111111111111111111111)
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // unpredictable resultIt seems like the "toString()" methods is only using the precision of the Number type (2^53 I believe), thus we are quickly losing precision on the last digits of the BigInt number. The problem is I can't find other methods to extract those digits.
Edit :
I need the precision to be perfect because basicaly what i'm doing for example is :
Compute Fibonacci(500) = 280571172992510140037611932413038677189525
Get the 10 last digits of this number : 8677189525 (this is where is lose the precision)
And then to solve my problem I need to check that those 10 last digits contains all the digits from 1 to 9
let number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = BigInt(1234567891111111111111111111111111111)
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // unpredictable resultlet number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = BigInt(1234567891111111111111111111111111111)
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // unpredictable resultjavascript bigint
javascript bigint
asked Mar 9 at 0:44
S. SylvainS. Sylvain
164
asked Mar 9 at 0:44
S. SylvainS. Sylvain
164
edited Mar 9 at 0:56
S. Sylvain
asked Mar 9 at 0:44
S. SylvainS. Sylvain
164
asked Mar 9 at 0:44
S. SylvainS. Sylvain
164
asked Mar 9 at 0:44
S. SylvainS. Sylvain
164
164
Exactly how precise do you want the numbers to be? Questions typically get better answers when you explicitly state what your desired results are. Also, +1
– Caleb Goodman
Mar 9 at 0:49
add a comment |
Exactly how precise do you want the numbers to be? Questions typically get better answers when you explicitly state what your desired results are. Also, +1
– Caleb Goodman
Mar 9 at 0:49
Exactly how precise do you want the numbers to be? Questions typically get better answers when you explicitly state what your desired results are. Also, +1
– Caleb Goodman
Mar 9 at 0:49
Exactly how precise do you want the numbers to be? Questions typically get better answers when you explicitly state what your desired results are. Also, +1
– Caleb Goodman
Mar 9 at 0:49
add a comment |
1 Answer
1
active
oldest
votes
For big numbers, I think you should add the n suffix:
let number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = 1234567891111111111111111111111111111n // <-- n suffix, literal syntax
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // result
let bigNumber2 = BigInt('1234567891111111111111111111111111111') // <-- also works as a string, in case you can't use the literal for some reason
console.log(bigNumber2.toString())
console.log(bigNumber2.toString()[30]) // resultanswered Mar 9 at 0:50
Austin GrecoAustin Greco
21.8k44249
Thank you, it's quite tricky that we can't use the BigInt() initializer without using n or literal but I guess it makes sense
– S. Sylvain
Mar 9 at 1:02
add a comment |
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1 Answer
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For big numbers, I think you should add the n suffix:
let number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = 1234567891111111111111111111111111111n // <-- n suffix, literal syntax
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // result
let bigNumber2 = BigInt('1234567891111111111111111111111111111') // <-- also works as a string, in case you can't use the literal for some reason
console.log(bigNumber2.toString())
console.log(bigNumber2.toString()[30]) // resultanswered Mar 9 at 0:50
Austin GrecoAustin Greco
21.8k44249
Thank you, it's quite tricky that we can't use the BigInt() initializer without using n or literal but I guess it makes sense
– S. Sylvain
Mar 9 at 1:02
add a comment |
For big numbers, I think you should add the n suffix:
let number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = 1234567891111111111111111111111111111n // <-- n suffix, literal syntax
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // result
let bigNumber2 = BigInt('1234567891111111111111111111111111111') // <-- also works as a string, in case you can't use the literal for some reason
console.log(bigNumber2.toString())
console.log(bigNumber2.toString()[30]) // resultanswered Mar 9 at 0:50
Austin GrecoAustin Greco
21.8k44249
Thank you, it's quite tricky that we can't use the BigInt() initializer without using n or literal but I guess it makes sense
– S. Sylvain
Mar 9 at 1:02
add a comment |
For big numbers, I think you should add the n suffix:
let number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = 1234567891111111111111111111111111111n // <-- n suffix, literal syntax
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // result
let bigNumber2 = BigInt('1234567891111111111111111111111111111') // <-- also works as a string, in case you can't use the literal for some reason
console.log(bigNumber2.toString())
console.log(bigNumber2.toString()[30]) // resultanswered Mar 9 at 0:50
Austin GrecoAustin Greco
21.8k44249
For big numbers, I think you should add the n suffix:
let number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = 1234567891111111111111111111111111111n // <-- n suffix, literal syntax
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // result
let bigNumber2 = BigInt('1234567891111111111111111111111111111') // <-- also works as a string, in case you can't use the literal for some reason
console.log(bigNumber2.toString())
console.log(bigNumber2.toString()[30]) // resultlet number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = 1234567891111111111111111111111111111n // <-- n suffix, literal syntax
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // result
let bigNumber2 = BigInt('1234567891111111111111111111111111111') // <-- also works as a string, in case you can't use the literal for some reason
console.log(bigNumber2.toString())
console.log(bigNumber2.toString()[30]) // resultlet number = BigInt(123456789)
console.log(number.toString())
console.log(number.toString()[3]) // Result is fine
let bigNumber = 1234567891111111111111111111111111111n // <-- n suffix, literal syntax
console.log(bigNumber.toString())
console.log(bigNumber.toString()[30]) // result
let bigNumber2 = BigInt('1234567891111111111111111111111111111') // <-- also works as a string, in case you can't use the literal for some reason
console.log(bigNumber2.toString())
console.log(bigNumber2.toString()[30]) // resultanswered Mar 9 at 0:50
Austin GrecoAustin Greco
21.8k44249
answered Mar 9 at 0:50
Austin GrecoAustin Greco
21.8k44249
answered Mar 9 at 0:50
Austin GrecoAustin Greco
21.8k44249
answered Mar 9 at 0:50
Austin GrecoAustin Greco
21.8k44249
21.8k44249
Thank you, it's quite tricky that we can't use the BigInt() initializer without using n or literal but I guess it makes sense
– S. Sylvain
Mar 9 at 1:02
add a comment |
Thank you, it's quite tricky that we can't use the BigInt() initializer without using n or literal but I guess it makes sense
– S. Sylvain
Mar 9 at 1:02
Thank you, it's quite tricky that we can't use the BigInt() initializer without using n or literal but I guess it makes sense
– S. Sylvain
Mar 9 at 1:02
Thank you, it's quite tricky that we can't use the BigInt() initializer without using n or literal but I guess it makes sense
– S. Sylvain
Mar 9 at 1:02
add a comment |
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Exactly how precise do you want the numbers to be? Questions typically get better answers when you explicitly state what your desired results are. Also, +1
– Caleb Goodman
Mar 9 at 0:49