When solving the problem of coin exchange by greedy algorithm, why will we will always have the correct result if the coin values are $1, a, a^2, \cdots, a^n$, where $a\ge 2$ and $n\gt 0$?

For example, if $a = 3$, $n =3$, we get the following coin values: 1, 3, 9, 27. When total exchange is 16, answer is 4 coins, as returned by greedy algorithm, $16=9+3+3+1$.

  • $\begingroup$ Welcome to Computer Science! I am very interested in how you know "we will always have the correct result" in these cases. Could you please provide a url or reference in the question? $\endgroup$
    – John L.
    Nov 3 '18 at 5:55
  • $\begingroup$ You can use LaTeX to typeset mathematics. I edited to show you how; we also have a brief tutorial. $\endgroup$
    – John L.
    Nov 3 '18 at 10:58
  • 2
    $\begingroup$ I just updated the question on behalf of the asker. It should be clear now. $\endgroup$
    – John L.
    Nov 3 '18 at 17:36
  • $\begingroup$ Thanks, @Apass.Jack -- that's much clearer. $\endgroup$ Nov 3 '18 at 18:46
  • $\begingroup$ @Apass.Jack, thanks for the corrections! The fact is that there is no basis for my words. I said this without any arguments, I just checked some numbers, it was so. $\endgroup$
    – BoberMod
    Nov 3 '18 at 22:49

For the coin values: 1, 3, 9, 27, suppose $16 = a_0 + 3a_1 + 9a_2 + 27a_3$ is the coin exchange with minimal number of coins. Note that $16\equiv a_0 \equiv 1(\operatorname{mod}3)$. If $a_0 \ge 3$, then we can subtract 3 from $a_0$ and add 1 to $a_1$ to obtain coin exchange with less number of coins. So, $a_0$ must be the remainder of 16 divided by 3. Hopefully, you can see where I am going.

All numbers or variables below are nonnegative integers. $(d_nd_{n-1}\cdots d_0)_a$ means the integer with those digits in base $a$, i.e., $(d_nd_{n-1}\cdots d_0)_a=\sum_{i=0}^{n}d_ia^i$ where $d_i\lt a $ for $i\le n$.

Here is the general theorem.

(Greedy algorithm works for power coins) Let $a\ge2$ and $n\ge0$. If $e_na^n+(e_{n-1}e_{n-2}\cdots e_0)_a=\sum_{i=0}^nc_ia^i$, then $ \sum_{i=0}^ne_i\le\sum_{i=0}^nc_i$.

Proof by mathematical induction on $n\ge0$.

The base case, when $n=0$ is trivial.

Suppose it is true for some $n\ge0$. Let us check the case of $n+1$. Suppose $m=e_{n+1}a^{n+1}+(e_ne_{n-1}\cdots e_0)_a=\sum_{i=0}^{n+1}c_ia^i$. Note that $e_0$ is the remainder of $m$ divided by $a$.

If $c_{0}\ge a$, we can subtract $a$ from $c_0$ and add $1$ to $c_1$ to reduce sum $\sum_{i=0}^{n+1}c_i$. Without loss of generality we can assume $c_{0}\lt a$, i.e., $c_0$ is also the remainder of $m$ divided by $a$.

Now let $m'=(m-e_0)/a=(m-c_0)/a$. Then $m'=e_{n+1}a^{n}+(e_{n}e_{n-1}\cdots e_1)_a=\sum_{i=0}^{n}c_{i+1}a^i$. By induction hypothesis, we know $\sum_{i=0}^ne_{i+1}\le\sum_{i=0}^nc_{i+1}$. Adding $e_0=c_0$ to both sides, we obtained the wanted inequality for the case of $n+1$. Q.E.D.

What is left to be explained is that $\sum_{i=0}^ne_i$ does come from the greedy algorithm. That easy task will be left for interested readers to verify.

  • $\begingroup$ This should always work if any but the smallest coin have a value that is a multiple of the previous coin. $\endgroup$
    – gnasher729
    Nov 3 '18 at 16:02
  • $\begingroup$ @gnasher729 Sharp observation! The proof also works after slight modification. $\endgroup$
    – John L.
    Nov 3 '18 at 16:47

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