Programming Praxis

12 Responses to “Perfect Power Sequence”

1. Zack said

   February 16, 2018 at 12:13 PM

   Here is my implementation in Julia.

   function IsPerfectPower(x::Int64)
   y = ceil(Int64, sqrt(x))

   for z = 2:y
       if mod(x ^ (1 / z), 1) == 0; return true; end
   end
   
   return false
   

   end

   function genseq(n::Int64)
   y = zeros(Int64, n)
   y[1] = 1

   for x = 2:n
       if IsPerfectPower(x); y[x] = x; end
   end
   
   return y[y .> 0]
   

   end

   Performance benchmark: on an Intel i7 laptop running on Win7, it generates the sequence of the PP numbers that are less or equal to 10000 in about 0.1 sec, using approx. 85 MB of RAM.
2. Zack said

   February 16, 2018 at 12:14 PM

   Typo in my previous commend: 85 Kilobytes, not MB… Sorry about that!
3. matthew said

   February 16, 2018 at 1:51 PM

   import Data.List.Ordered
   f n = n*n : merge powers (f (n+1)) where powers = n*n*n : map (n*) powers
   main = print (take 20 (nub (1 : (f 2))))
   -- [1,4,8,9,16,25,27,32,36,49,64,81,100,121,125,128,144,169,196,216]
   
4. Paul said

   February 17, 2018 at 6:11 PM

   Two solutions in Python. I expected the first to be the fastest: merging already sorted sequences. But it turned out to be otherwise. The second is a little faster.

   from ma.mymath import isqrt  # integer square root
   from itertools import takewhile, count
   from functools import partial
   from operator import ge
   
   def genpows(n):
       """generate powers n^2, n^3, ...."""
       x = n
       while True:
           x *= n
           yield x
   
   def perpow(limit):
       """generate all powers (up to limit), merge them and handle duplicates"""
       maxn, last = isqrt(limit), 1
       its = [genpows(i) for i in range(2, maxn+1)]
       yield 1
       for i in takewhile(partial(ge, limit), merge(*its)):
           if i != last:
               yield i
               last = i
   
   def perpow2(limit):
       def nums(i):
           return takewhile(partial(ge, limit), 
               map(partial(pow, i), count(2)))
       powers = set((n for i in range(2, isqrt(limit)+1) for n in nums(i)))
       return [1] + sorted(powers)
   
5. Scott said

   February 17, 2018 at 11:30 PM

   Here is some Java. It generates up to 10000 in around 60ms on a 3.3GHz i5 iMac:

   import java.util.PriorityQueue;
   
   public class PerfectPowerSequenceGenerator {
     private static class NumberSequence {
       public long number;
       public long nextPerfectPowerValue;
       
       public NumberSequence(long number) {
         this.number = number;
         this.nextPerfectPowerValue = number * number;
       }
     }
   
     private PriorityQueue<NumberSequence> sequences = new PriorityQueue<>((a, b) -> {
       if (b.nextPerfectPowerValue > a.nextPerfectPowerValue) {
         return -1;
       } else if (b.nextPerfectPowerValue < a.nextPerfectPowerValue) {
         return 1;
       }
       return 0;
     });
   
     private long nextSequence;    // next sequence to add to the heap
     private long lastResult = 0;  // last result we returned, to avoid dupes
     
     public PerfectPowerSequenceGenerator() {
       nextSequence = 2;
     }
   
     public long next() {
       long result = 1;
       if (lastResult != 0 && !sequences.isEmpty()) {
         NumberSequence seq;
         do {
           seq = sequences.poll();
           result = seq.nextPerfectPowerValue;
           seq.nextPerfectPowerValue *= seq.number;
           sequences.add(seq);
         } while (result == lastResult);
       }
           
       if (nextSequence * nextSequence > result) {
         sequences.add(new NumberSequence(nextSequence++));
       }
       
       lastResult = result;
     
       return result;
     }
     
     public static void main(String[] args) {
       long startTime = System.nanoTime();
       PerfectPowerSequenceGenerator gen = new PerfectPowerSequenceGenerator();
       while (true) {
         long pp = gen.next();
         System.out.println(pp);
         if (pp >= 10000) {
           break;
         }
       }
       System.out.println("Ran in " + (System.nanoTime() - startTime) / 1000000 + "ms");
     }
   }
   
6. matthew said

   February 18, 2018 at 12:06 AM

   The queue solution works nicely in Python:

   import heapq
   def gen(i0,j0,f):
       global q
       q = [(f(i0,j0),i0,j0)]
       last = None
       while True:
           (n,i,j) = heapq.heappop(q)
           heapq.heappush(q,(f(i+1,j),i+1,j))
           if i == i0: heapq.heappush(q,(f(i,j+1),i,j+1))
           if n != last: yield n
           last = n
   
   g = gen(2,2,lambda i,j: i**j)
   s = [1]+[next(g) for _ in range(19)]
   print(s); print(q)
   # [1,4,8,9,16,25,27,32,36,49,64,81,100,121,125,128,144,169,196,216]
   # [(225,15,2),(243,3,5),(256,2,8),(256,4,4),(729,3,6),(2187,3,7),(343,7,3)]
   

   It’s quite important how the queue is extended – doing

           if j == j0: heapq.heappush(q,(f(i+1,j),i+1,j))
           heapq.heappush(q,(f(i,j+1),i,j+1))
   

   still works, but the queue gets much longer.

   Incidentally, the code on the solution page seems to be garbled – I think WordPress have messed something up with angle brackets in code.
7. matthew said

   February 18, 2018 at 12:36 AM

   By the way, this is https://oeis.org/A001597
8. Paul said

   February 18, 2018 at 2:15 PM

   @matthew: I tried the priority method too and it worked nicely, but a little slow in Python. The method below takes about 0.1 sec for all powers below 10^9.

   from ma.mymath import isqrt  # integer square root
   from itertools import takewhile, count
   from functools import partial
   from operator import ge
   
   _small_pp = [1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100]
   
   def pp2(limit):
       def powers(i):
           return takewhile(partial(ge, limit), 
               map(partial(pow, i), count(2)))
       nmax = isqrt(limit)
       # remove the perfect powers from the range of bases to loop over
       if nmax > 120:  # 120 is the first pp after 100 (_small_pp)
           excl = pp2(nmax)
       else:
           excl = _small_pp
       bases = set(range(2, nmax+1)) - set(excl)
       allpowers = (n for base in bases for n in powers(base))
       return [1] + sorted(allpowers)
   
9. matthew said

   February 18, 2018 at 3:08 PM

   @Paul: did you do a speed comparison with my priority queue function above?
10. Paul said

    February 18, 2018 at 4:25 PM

    @matthew: I see now that your method works very fast (91 msec for yours and 121 ms for pp2 for all powers up to 10^9). I must have messed up something with my implementation.
11. matthew said

    February 18, 2018 at 4:29 PM

    @Paul: that’s about what I made it – also the queue uses less memory, I’m running the program just printing out all the numbers, up to 25528390307779969 after about 30 mins.

    For the queue, as mentioned above, it’s significant how the queue is extended.
12. Daniel said

    February 19, 2018 at 5:49 AM

    Here’s an implementation in C that uses the approach from the solution.

    #include <stdbool.h>
    #include <stdio.h>
    #include <stdlib.h>
    
    typedef unsigned long u_long;
    
    typedef struct {
      u_long base;
      u_long exponent;
      u_long value;
    } item_t;
    
    typedef struct {
      item_t* array;
      size_t count;
      size_t capacity;
    } heap_t;
    
    typedef struct {
      heap_t* heap_p;
      size_t n_generated;
    } generator_t;
    
    void init_heap(heap_t** heap_pp) {
      *heap_pp = malloc(sizeof(heap_t));
      heap_t* heap_p = *heap_pp;
      heap_p->array = malloc(sizeof(item_t));
      heap_p->count = 0;
      heap_p->capacity = 1;
    }
    
    void clear_heap(heap_t** heap_pp) {
      free((*heap_pp)->array);
      free(*heap_pp);
      *heap_pp = NULL;
    }
    
    void swap(item_t* a, item_t* b) {
      item_t tmp = *a;
      *a = *b;
      *b = tmp;
    }
    
    void heap_insert(heap_t* heap_p, item_t item) {
      size_t new_count = heap_p->count + 1;
      if (new_count > heap_p->capacity) {
        size_t new_capacity = 2 * (heap_p->capacity + 1) - 1;
        heap_p->array = realloc(heap_p->array, new_capacity * sizeof(item_t));
        heap_p->capacity = new_capacity;
      }
      item_t* array = heap_p->array;
      // reheapify upward
      size_t idx = heap_p->count;
      array[idx] = item;
      while (1) {
        if (idx == 0) break;
        size_t parent_idx = (idx-1)/2;
        u_long value = item.value;
        u_long parent_value = array[parent_idx].value;
        if (value > parent_value) break;
        swap(&array[idx], &array[parent_idx]);
        idx = parent_idx;
      }
      heap_p->count = new_count;
    }
    
    item_t heap_peak(heap_t* heap_p) {
      return (heap_p->array)[0];
    }
    
    // Popping from an empty heap is not supported
    item_t heap_pop(heap_t* heap_p) {
      item_t popped_item = heap_peak(heap_p);
      size_t new_count = heap_p->count - 1;
      if (new_count > 0) {
        item_t* array = heap_p->array;
        array[0] = array[new_count];
        size_t idx = 0;
        while (1) {
          u_long value = array[idx].value;
          size_t left_idx = 2 * idx + 1;
          size_t right_idx = left_idx + 1;
          if (left_idx > new_count - 1) break;
          u_long min_child_value = array[left_idx].value;
          size_t min_child_idx = left_idx;
          if (right_idx <= new_count - 1) {
            u_long right_child_value = array[right_idx].value;
            if (right_child_value < min_child_value) {
              min_child_value = right_child_value;
              min_child_idx = right_idx;
            }
          }
          if (value <= min_child_value) break;
          swap(&array[idx], &array[min_child_idx]);
          idx = min_child_idx;
        }
      }
      heap_p->count = new_count;
      return popped_item;
    }
    
    void init_generator(generator_t** generator_pp) {
      *generator_pp = malloc(sizeof(generator_t));
      generator_t* generator_p = *generator_pp;
      init_heap(&(generator_p->heap_p));
      item_t item = {.base = 2, .exponent = 2, .value = 4};
      heap_insert(generator_p->heap_p, item);
      generator_p->n_generated = 0;
    }
    
    void clear_generator(generator_t** generator_pp) {
      clear_heap(&(*generator_pp)->heap_p);
      free(*generator_pp);
      *generator_pp = NULL;
    }
    
    u_long next(generator_t* generator) {
      ++(generator->n_generated);
      if (generator->n_generated == 1) return 1;
      heap_t* heap_p = generator->heap_p;
      u_long value = heap_peak(heap_p).value;
      while (heap_peak(heap_p).value == value) {
        item_t item = heap_pop(heap_p);
        if (item.exponent == 2) {
          item_t new_item;
          new_item.base = item.base + 1;
          new_item.exponent = 2;
          new_item.value = new_item.base * new_item.base;
          heap_insert(heap_p, new_item);
        }
        ++(item.exponent);
        item.value *= item.base;
        heap_insert(heap_p, item);
      }
      return value;
    }
    
    int main(void) {
      generator_t* generator;
      init_generator(&generator);
      u_long limit = 1764;
      while (1) {
        u_long value = next(generator);
        printf("%lu", value);
        if (value >= limit) break;
        printf(",");
      }
      clear_generator(&generator);
      return 0;
    }
    

    Output:

    1,4,8,9,16,25,27,32,36,49,64,81,100,121,125,128,144,169,196,216,225,243,256,289,324,343,361,400,441,484,512,529,576,625,676,729,784,841,900,961,1000,1024,1089,1156,1225,1296,1331,1369,1444,1521,1600,1681,1728,1764