So I am testing my algorithm and trying to print all prime numbers within a range. I think the code is logical but it keeps printing the wrong output i.e the unfiltered list of numbers.
function sumPrimes(num) {
// Check all numbers for primality
let a = []
let b = []
for (let i = 2; i <= num; i ) {
a.push(i)
b.push(i)
}
//console.log(a)
return a.filter(function(item) {
for(let j = 0; j < b.length; j ) {
if(item !== b[j] && item % b[j] != 0) {
return true
}
}
return false;
})
}
sumPrimes(977);
CodePudding user response:
There are few issues. First j should start being equal to 2. Otherwise it will see every value has a factor of 1. We can also skip your check to make sure b[j] is not item if we don't go through the entire span of b and instead just up to item. The other main issue is that the return true and return false are backwards.
The next issues are that you don't really need some parts. For example b[j] will always be equal to j 2 so there is no need to build b.
function sumPrimes(num) {
// Check all numbers for primality
let a = []
for (let i = 2; i <= num; i ) {
a.push(i)
}
return a.filter(function(item) {
for(let j = 2; j < item; j ) {
if(item % j === 0) {
// j is a factor so we shouldn't keep it and instead skip it
return false;
}
}
// We know that all of the values up to item are not factors so we keep it
return true;
})
}
console.log(sumPrimes(977));
Plus we can skip adding every value to a if we apply the filtering as we add the values.
function getPrimesUpTo(num) {
let primes = [];
// Check each value
for (let i = 2; i <= num; i ) {
let hasFactors = false;
// Search for any factors before adding it to our list of primes
for (let possibleFactor = 2; possibleFactor < i; possibleFactor ) {
if (i % possibleFactor === 0) {
hasFactors = true;
break;
}
}
if (!hasFactors) {
primes.push(i);
}
}
return primes;
}
console.log(getPrimesUpTo(977));
