Daily Challenge #152 - Strongest Number in an Interval

Setup

The strength of an even number is determined by the amount of times we can successfully divide by 2 until we reach an odd number.

For example, if n = 12, then:
12 / 2 = 6
6 / 2 = 3
We have divided 12 in half twice and reached 3, so the strength of 12 is 2.

Given a closed interval [n, m], return the strongest even number in the interval. If multiple solutions exist, return the smallest of the possible solutions.

Examples

for [1, 2], return 2. (1 has a strenght of 0; 2 has a strength of 1)
for [5, 10], return 8.

Tests

[48, 56]
[129, 193]

Good luck~!

---

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Comments

[Vaibhav Yadav]

In Go.

func strongest(start, stop int) int {
    power := func(num int) int {
        numSlice := strings.Split(fmt.Sprintf("%b", num), "1")
        return len(numSlice[len(numSlice)-1])
    }

    hp := 0
    var result int

    for i := start; i <= stop; i++ {
        if hp < power(i) {
            hp = power(i)
            result = i
        }
    }

    return result
}

[Rashmin Dandekar]

Ruby

def get_strength(num)
  strength=0
  while !num.odd?
      num /= 2
      strength += 1
  end
  strength
end

def get_num_of_lowest_strength(nums)
  strength=Array.new
  nums.each do 
    |x| 
    strength.push [x, get_strength(x)]
  end  
  strength.sort_by{|k,v| [v,k]}.reverse
end

nums=[5..10,48..56,129..193]

nums.each do
  |num|
  x= get_num_of_lowest_strength(num)
  p x[0][0]
end

[Ryan Burmeister-Morrison]

Here's a Nim submission!

proc even(i: int): bool {.inline.} = i mod 2 == 0

proc numStrength(i: int): int =
  ## Calculate the strength of a number.
  ##
  ## A number's strength is determined by the number of times
  ## it can divided in half until it reaches an odd number.
  var i = i
  while i.even:
    i = i div 2
    inc(result)

proc strongestNum(n, m: int): int =
  ## Given the closed interval [n, m], calculate the strongest number.
  var topStrength = 0
  for i in n..m:
    let strength = numStrength(i)
    if strength > topStrength:
      topStrength = strength
      result = i

assert strongestNum(1, 2) == 2
assert strongestNum(5, 10) == 8
assert strongestNum(48, 56) == 48
assert strongestNum(129, 193 == 192

There's definitely room to improve this, but I'm just starting to look into Nim's standard library, and I don't know all that's available to me yet.

[Paul]

Did it long hand in js. Great puzzle!

const strength = num => {
  let count = 0;
  if (num === 0) {
    return 0;
  }
  while (num % 2 == 0) {
    num = num / 2;
    count++;
  }
  return count;
};

const strongestNumber = interval => {
  const [start, end] = interval;
  let strongest = start;

  for (let i = start; i <= end; i++) {
    if (strength(i) > strength(strongest)) {
      strongest = i;
    }
  }
  return strongest;
};

console.log(strongestNumber([1, 2]));
console.log(strongestNumber([5, 10]));
console.log(strongestNumber([48, 56]));
console.log(strongestNumber([129, 193]));

[Vitor Nogueira]

const calcMaxStrength = (numbers) => {
  const divider = 2;

  const isOdd = number => number % 2 === 0;

  const calcStrength = (dividend, strength = 0) => {
    if (isOdd(dividend)) {
      return calcStrength(dividend / divider, strength += 1);
    }

    return strength;
  };

  return Math.max(...numbers.map(number => calcStrength(number)));
};

console.log(calcMaxStrength([48, 56])); // 4
console.log(calcMaxStrength([129, 193])); // 0

[Amin]

Haskell

strongest :: (Int, Int) -> (Int, Int) -> (Int, Int)
strongest (strongestIndex, strongestValue) (index, value)
    | value > strongestValue = (index, value)
    | value == strongestValue && index < strongestIndex = (index, value)
    | otherwise = (strongestIndex, strongestValue)


strength :: Int -> Int
strength 0 = 0
strength number 
    | odd number = 0
    | otherwise = 1 + strength (div number 2)


strongestBetween :: Int -> Int -> Int
strongestBetween lowerbound upperbound =
    fst $ foldl1 strongest $ zip interval $ map strength interval

    where
        interval :: [Int]
        interval =
            [lowerbound..upperbound]


main :: IO ()
main = do
    print $ strongestBetween 1 2        -- 2
    print $ strongestBetween 5 10       -- 8
    print $ strongestBetween 48 56      -- 48
    print $ strongestBetween 129 193    -- 192

Try it.

[Cent]

Not the most efficient answer, but it is!

const findStrongestNumber = () => {
 let interval = [
   parseInt(document.querySelector("#num1").value),
   parseInt(document.querySelector("#num2").value)
 ];

  let numStrength = [0, 0];

  for(let i = 0; i < interval.length; i++) {
    let num = interval[i];

    while ((num % 2) !== 1) {
      num/=2;
      numStrength[i]++;
    }
  }

  let answer = 0;
  if(numStrength[0] > numStrength[1]) {
    answer = interval[0];
  } else if (numStrength[1] > numStrength[0]) {
    answer = interval[1]
  } else {
    if(interval[0] < interval[1]) {
      answer = interval[0];
    } else {
      answer = interval[1];
    }
  }

  document.querySelector("#answer").innerHTML = answer;
}

CodePen

[PranshuKhandal]

Hello guys, this is my first time here, and this is my recursive approach:

function strongest(from, to, times = 0) {
    if (from > to) return false;
    if (to - from < 1) return from * (2 ** times);
    if (to - from < 2) return (from + (from % 2)) * (2 ** times);
    return strongest((from + (from % 2)) / 2, (to - (to % 2)) / 2, times + 1);
}

EDIT without times:

function strongest(from, to) {
    if (from > to) return false;
    if (to - from < 1) return from;
    if (to - from < 2) return from + (from % 2);
    return 2 * strongest((from + (from % 2)) / 2, (to - (to % 2)) / 2);
}

[Brett Martin]

Inefficient Python w/ tail recursion:

def strongness_tail(n, a):
    return a if n % 2 == 1 else strongness_tail(n/2, 1 + a)

def strongness(n):
    return strongness_tail(n, 0)

def strongest_even(n, m):
    strongest = [-1, -1]
    for i in range(n, m+1):
        strength = strongness(i)
        if strength > strongest[0]:
            strongest = [strength, i]

    return strongest[1]

[Idan Arye]

The optimization of using bit operations to find the "strength" of a number is a low hanging fruit, but the relationship between the number's strength and its binary representation can be taken much farther than that - to the point of solving this in logarithmic(!) time:

fn strongest_number(min: usize, max: usize) -> usize {
    assert!(min <= max);
    if min == max {
        // Otherwise the second `while` loop will not end
        return min;
    }
    let mut prefix_mask = !0usize;
    while 0 != (prefix_mask & max) {
        prefix_mask <<= 1;
    }
    while (prefix_mask & max) == (prefix_mask & min) {
        prefix_mask |= prefix_mask >> 1;
    }
    if (prefix_mask & min) == min {
        min
    } else {
        prefix_mask & max
    }
}

fn main() {
    assert_eq!(strongest_number(1, 2), 2);
    assert_eq!(strongest_number(5, 10), 8);
    assert_eq!(strongest_number(48, 56), 48);
    assert_eq!(strongest_number(129, 193), 192);
}

Explanation:

You can compare two numbers, padded to the same length, by finding their longest common prefix and then looking at the highest digit that's different. This work in any base, but farther ahead we'll use some properties unique to binary so let's stick with that.

For example, if we take 129 and 193 and convert them to binary:

129: 0000000010000001
193: 0000000011000001

We can see that both start with 000000001, and then 129 has 0000001 while 193 has 1000001. So the highest nonshared digit is 0 for 129 and 1 for 193 - which is why 193 is higher (yup - that's the direction the causation works here. Though this is math, so one can twist it to be the other way around...)

Lemma: Any number between the lower and higher bounds will share that same prefix with them.
Proof: If it doesn't, then at least one digit is different. Take the most significant digit where the number is different than the shared prefix. If it's 0 for the number and 1 for the prefix - this means the number is lower than the lower bound. If it's 1 for the number and 0 for the prefix - this means the number is higher than the higher bound.

So, the strongest number, being in the range, must also have that prefix. The strongest number with that prefix is just the prefix with only zeroes after it (remember - the number of digits is fixed!). That's also the lowest number in that range, so if it's equal to the lower bound - that's the strongest number in the range. Otherwise - the next strongest number is the prefix, a single one after it, and after that only zeroes. This number is guaranteed to not be higher than the higher bound, because they share the original common prefix plus a single one after it - and that number is the lowest in that prefix, since after that it's all zeroes.

[La blatte]

That is just so clean, well thought and well explained! Congrats!