I am currently reading Michael Soltys' Analysis of Algorithms (2nd Edition), and Problem 1.13 of the subsection titled Invariance reads:

Let $n$ be an odd number, and suppose that we have the set $\{1,2,\dots,2n\}$. We pick any two numbers $a$, $b$ in the set, delete them from the set, and replace them with $|{a-b}|$. Continue repeating this until just one number remains in the set; show that this remaining number must be odd.

However, I picked $n=3$ and performed the following.

  • I start with $\{1,2,3,4,5,6\}$.
  • I pick $1$ and $2$; I end up with $(\{1,2,3,4,5,6\}-\{1,2\})\cup\{|{1-2}|\}=\{1,3,4,5,6\}$.
  • I pick $1$ and $6$; I end up with $\{3,4,5\}$.
  • I pick $3$ and $5$; I end up with $\{2,4\}$.
  • And finally, I pick $2$ and $4$; I end up with $\{2\}$.

Clearly, $2$ is not an odd number.

Is there something I misunderstood in my attempt?


1 Answer 1


I think they don't mean to consider a "set" of numbers (where $\{3, 4, 5, 5\} = \{3, 4, 5\}$) but rather a list or multiset of numbers.

In that case the result is true: initially the sum of the entries of the list is $n(2n+1)$ which is odd, and at each step the parity of the sum of the entries is preserved. When you choose $a, b$, the sum changes from its old value $s$ to $s-a-b+ |b-a|$. If $a, b$ are even then $-a-b+|b-a|$ is clearly even, if they are both odd then $-a-b$ is even as is $|b-a|$, and if one is even and one odd then $-a-b$ is odd and $|b-a|$ is odd so $-a-b+|b-a|$ is odd+odd=even.

  • 1
    $\begingroup$ Thank you for your clarification. That makes sense. I personally used a different invariant, reading: "After $n>0$ operations, the number of odd numbers in the set/list is odd." Indeed, the invariant holds when $n=0$. After one operation, there are three cases: (1) Both $a,b$ are even. In that case, the list would 'lose' one even number and no odd number. (2) Both $a, b$ are odd. In that case, the list would 'lose' two odd numbers and gain an even one. However, subtracting $2$ from an even natural number keeps it odd. (3) $a,b$ are of different parity. Then, we only 'lose' an even number. $\endgroup$ Mar 13 at 18:23
  • $\begingroup$ Two corrections to my comment above. (1) I meant, $n\geq 0$, not $n>0$. (2) In the second case, I meant, "subtracting $2$ from an odd natural number keeps it odd." $\endgroup$ Mar 13 at 20:51

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