N-bit Variation on Subset-Sum

Question

For another challenge I am writing, I need to verify that test cases are solveable with bounded integers. Specifically, I need to verify the following, for a non-empty array of integers A and an integer bit width n:

1. All integers a in A satisfy -2**(n-1) <= a < 2**(n-1) (representable with n-bit two's complement integers).
2. The length of A is less than 2**n.
3. The sum of A satisfies -2**(n-1) <= sum(A) < 2**(n-1).
4. All combinations of elements in A satisfy all of the above conditions.

Naturally, I've decided to outsource this problem to you!

Given an array of integers A and a positive integer bit width n, verify that A satisfies the conditions above.

Test Cases

[0, 0, 0], 2: True
[0, 0, 0, 0], 2: False (violates #2)
[1, 2, 3, 4, 5], 8: True
[1, 2, 3, 4, 5], 2: False (violates all conditions)
[1, 2, 3, 4, 5], 5: True
[-3, 4, 1], 4: True
[10, 0, -10], 4: False (violates #1 and #4)
[27, -59, 20, 6, 10, 53, -21, 16], 8: False (violates #4)
[-34, 56, 41, -4, -14, -54, 30, 38], 16: True
[-38, -1, -11, 127, -35, -47, 28, 89, -8, -12, 77, 55, 75, 75, -80, -22], 7: False (violates #4)
[-123, -85, 6, 121, -5, 12, 52, 31, 64, 0, 6, 101, 128, -72, -123, 12], 12: True

Reference Implementation (Python 3)

#!/usr/bin/env python3
from itertools import combinations
from ast import literal_eval


def check_sum(L, n):
  return -2**(n-1) <= sum(L) < 2**(n-1)


def check_len(L, n):
  return len(L) < 2**n


def check_elems(L, n):
  return all(-2**(n-1) <= a < 2**(n-1) for a in L)


A = literal_eval(input())
n = int(input())
OUTPUT_STR = "{}, {}: {}".format(A, n, "{}")

if not (check_elems(A, n) and check_len(A, n) and check_sum(A, n)):
  print(OUTPUT_STR.format(False))
  exit()

for k in range(1, len(A)):
  for b in combinations(A, k):
    if not check_sum(b, n):
      print(OUTPUT_STR.format(False))
      exit()

print(OUTPUT_STR.format(True))

Try it online!

Answer 1

Wolfram Language (Mathematica), 40 bytes

Max[x=2Tr/@Subsets@#,-x-1,Tr[1^#]]<2^#2&

Try it online!

Condition 1 is implied by checking condition 3 for all subsets, including the one-element ones. So we take the max of

• twice the sum of each subset,
• one less than twice the negative of the sum of each subset, and
• the length of the whole set

and check if that's less than 2^#2 (where #2 is the bit-width input).

At the cost of only 6 more bytes, we can replace Subsets@# with GatherBy[#,Arg], which is much more efficient because it computes only the two worst-case subsets: the subset of all nonnegative values, and the subset of all negative values. (This works because Arg has a value of 0 on the former and π on the latter.)

Answer 2

Jelly, 19 bytes

ŒPS€;⁸L¤ḟ⁹’2*$ŒRṖ¤Ṇ

Try it online!

It suffices to check that mapped sum of powerset + length of set is in the required range.

Answer 3

05AB1E, 13 12 11 bytes

Saved 1 byte thanks to Mr. Xcoder

æO·D±¹gMIo‹

Try it online!

Explanation

æ             # powerset of first input
 O            # sum each subset
  ·           # multiply each element by 2
   D          # duplicate
    ±         # bitwise negation of each element in the copy
     ¹g       # push length of first input
       M      # get the maximum value on the stack
        Io    # push 2**<second input>
          ‹   # compare

Answer 4

JavaScript (ES6), 75 63 58 bytes

a=>n=>!a.some(e=>(a.length|2*(e<0?l-=e:u+=e))>>n,u=0,l=-1)

The sum of any subset of a lies between the sums of the negative and nonnegative elements, so checking the two sums suffices for everything except case 2. Edit: Saved 12 17 bytes thanks to @Arnauld.

Answer 5

Jelly, 21 20 bytes

»0,«0$S€~2¦Ḥ;LṀ<2*Ɠ¤

Try it online!

Linear time complexity solution. Turns out that I over-estimated the time complexity

in The Nineteenth Byte, 2017-12-11 13-15-03Z, by user202729

@NewSandboxedPosts "Real" subset sum problem are much harder. This one can be done in linearithmic time...

because now I realize sorting the array is completely unnecessary.

---

Explanation:

»0,«0$S€~2¦Ḥ;LṀ<2*Ɠ¤    Main link. Example list: [-1, 0, 1]
»0                      Maximize with 0. Get [0, 0, 1]
  ,                     Pair with
   «0$                    minimize with 0. Get [-1, 0, 0]
      S€                Sum €ach. Get [1, -1]
        ~               Inverse
          ¦               at element
         2                2. (list[2] = ~list[2]) Get [-1, 2]
           Ḥ            Unhalve (double, ×2). Get [-2, 4]
            ;           Concatenate with
             L            Length (3). Get [-2, 4, 3]
              Ṁ         Maximum of the list (4).
               <   ¤    Still less than
                2         two
                 *        raise to the power of
                  Ɠ       eval(input())

Answer 6

JavaScript (ES6), 114 bytes

Takes input in currying syntax (A)(n). Returns a boolean.

A=>n=>!(A.reduce((a,x)=>[...a,...a.map(y=>[x,...y])],[[]]).some(a=>(s=eval(a.join`+`),s<0?~s:s)>>n-1)|A.length>>n)

Test cases

let f =

A=>n=>!(A.reduce((a,x)=>[...a,...a.map(y=>[x,...y])],[[]]).some(a=>(s=eval(a.join`+`),s<0?~s:s)>>n-1)|A.length>>n)

console.log('[Truthy]')
console.log(f([0, 0, 0])(2))
console.log(f([1, 2, 3, 4, 5])(8))
console.log(f([1, 2, 3, 4, 5])(5))
console.log(f([-3, 4, 1])(4))
console.log(f([-34, 56, 41, -4, -14, -54, 30, 38])(16))
console.log(f([-123, -85, 6, 121, -5, 12, 52, 31, 64, 0, 6, 101, 128, -72, -123, 12])(12))

console.log('[Falsy]')
console.log(f([0, 0, 0, 0])(2)) // violates #2
console.log(f([1, 2, 3, 4, 5])(2)) // violates all conditions
console.log(f([10, 0, -10])(4)) // violates #1 and #4
console.log(f([27, -59, 20, 6, 10, 53, -21, 16])(8)) // violates #4
console.log(f([-38, -1, -11, 127, -35, -47, 28, 89, -8, -12, 77, 55, 75, 75, -80, -22])(7)) // violates #4

Answer 7

Clojure, 121 117 bytes

#(let[l(int(Math/pow 2(dec %2)))](every?(set(range(- l)l))(cons(count %)(for[i(vals(group-by pos? %))](apply + i)))))

Well that was a bit dumb, splitting into positive and negative values is a lot better than sorting. Original, but surprisingly not much longer:

#(let[l(int(Math/pow 2(dec %2)))S(sort %)R reductions](every?(set(range(- l)l))(concat[(count S)](R + S)(R +(into()S)))))

This works by checking prefix sums of the sequence in ascending and descending order, I think it isn't necessary to generate all combinations of elements in A.

(into () S) is in effect same as (reverse S), as lists grow from the head. I couldn't figure out a way to use cons instead of concat when there are two lists to cons to. :/

Answer 8

Jelly, 15 bytes

ŒPS€Ḥ;~$;LṀl2<Ɠ

Try it online!

Explanation

ŒPS€Ḥ;~$;LṀl2<Ɠ ~ Monadic full program.

ŒP              ~ Powerset.
  S€            ~ The sum of each subset.
    Ḥ           ~ Double (element-wise).
     ;~$        ~ Append the list of their bitwise complements.
        ;L      ~ Append the length of the first input.
          Ṁ     ~ And get the maximum.
           l2   ~ Base-2 logarithm.
             <Ɠ ~ Is smaller than the second input (from stdin)?

Saved 1 byte thanks to caird coinheringaahing (reading the second input from STDIN instead of CLA).

Answer 9

Pyth, 20 bytes

>EleS++KyMsMyQmt_dKl

Try it here!

Answer 10

Husk, 14 bytes

≥Lḋ▲ṁ§eLöa→DΣṖ

Going with brute force by looping over all sublists, since splitting into positive and negative parts takes more bytes. Try it online!

Explanation

≥Lḋ▲ṁ§eLöa→DΣṖ  Implicit inputs, say A=[1,2,3,4,5] and n=5
             Ṗ  Powerset of A: [[],[1],[2],[1,2],..,[1,2,3,4,5]]
    ṁ           Map and concatenate:
                  Argument: a sublist, say S=[1,3,4]
            Σ     Sum: 8
           D      Double: 16
          →       Increment: 17
        öa        Absolute value: 17
     §eL          Pair with length of S: [3,17]
                Result is [0,1,1,3,1,5,2,7,..,5,31]
   ▲            Maximum: 31
  ḋ             Convert to binary: [1,1,1,1,1]
 L              Length: 5
≥               Is it at most n: 1

Answer 11

Python 2, 72 71 70 bytes

def f(x,n):t=2*sum(k*(k<0)for k in x);max(2*sum(x)-t,~t,len(x))>>n>0>_

Output is via presence/absence of an error.

Try it online!