146 lines
8.6 KiB
Plaintext
146 lines
8.6 KiB
Plaintext
{
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"cell_type": "markdown",
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"# [Coin Sums](https://projecteuler.net/problem=31)\n",
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"\n",
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"There's a *very* cool way to solve this problem using [generating functions](https://en.wikipedia.org/wiki/Generating_function).\n",
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"\n",
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"## Generating functions\n",
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"If you're not familiar, a generating function is a way of expressing a sequence as coefficients of a power series. For example, the generating function for the Fibonacci sequence is\n",
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"$$0x^0 + 1x^1 + 1x^2 + 2x^3 + 3x^4 + 5x^5 + \\cdots$$\n",
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"\n",
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"The seemingly trivial series $(1, 1, 1, 1, \\ldots)$ has an equally simple generating function\n",
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"$$1 + x + x^2 + x^3 + x^4 + \\cdots$$\n",
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"which converges to $\\frac{1}{1-x}$ for $-1 < x < 1$ (note that we don't really care about the interval of convergence, because we're not going to be evaluating this function directly).\n",
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"\n",
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"If this is your first time seeing generating functions, it's probably not obvious how they can help us solve this problem. Let's look at some related, simpler problems to lead us towards understanding.\n",
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"\n",
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"## Trivial change-making\n",
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"First, consider an (almost trivial) question: how many ways can you make 0 cents, 1 cent, 2 cents, and so on using *only* 1 cent coins? It's not a trick question; there's one way for each, since you only have 1 cent coins to use! (1, 1, 1, 1, 1, 1, ...)\n",
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"\n",
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"Similarly, how many ways can you make 0 cents, 1 cent, 2 cents, etc. using only 2 cent coins? Answer: every even value has one way of making change, and every odd value has zero ways (1, 0, 1, 0, 1, 0, ...). Similarly, with only 5 cent coins, there is one way to make change for every multiple of 5, and zero ways for every other value (1, 0, 0, 0, 0, 1, ...).\n",
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"\n",
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"We can express the solutions to each of these questions as sequences, and therefore as generating functions:\n",
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"$$\\begin{align}\n",
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"1 + x + x^2 + x^3 + \\cdots = \\frac{1}{1-x}\\\\\n",
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"1 + x^2 + x^4 + x^6 + \\cdots = \\frac{1}{1-x^2}\\\\\n",
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"1 + x^5 + x^{10} + x^{15} + \\cdots = \\frac{1}{1-x^5}\\\\\n",
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"\\end{align}$$\n",
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"One way to interpret these functions that will be very useful later: for each term, the exponent represents the money amount you're trying to make change for, while the coefficient represents the number of ways to make change for that amount (in these functions, we only have 0 and 1 as coefficients, but sit tight...).\n",
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"\n",
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"## Not-so-trivial change-making\n",
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"Now we can jump to a more interesting question: if we have 1, 2, and 5 cent coins, how many ways can we make 7 cents? It's not too difficult to enumerate all six ways by hand:\n",
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"$$\\begin{align}\n",
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"5 + 2 \\\\\n",
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"5 + (2 \\times 1) \\\\\n",
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"(3 \\times 2) + 1 \\\\\n",
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"(2 \\times 2) + (3 \\times 1) \\\\\n",
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"2 + (5 \\times 1) \\\\\n",
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"(7 \\times 1)\n",
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"\\end{align}$$\n",
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"But we can also use the generating functions we came up with above to find the total. One way is to cut off each infinite series after the $x^7$ term, then multiply them all together."
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]
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"cell_type": "code",
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"execution_count": 1,
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"id": "74c5fe00",
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"metadata": {},
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"data": {
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"text/plain": [
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"x^18 + x^17 + 2*x^16 + 2*x^15 + 3*x^14 + 4*x^13 + 5*x^12 + 6*x^11 + 5*x^10 + 6*x^9 + 5*x^8 + 6*x^7 + 5*x^6 + 4*x^5 + 3*x^4 + 2*x^3 + 2*x^2 + x + 1"
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]
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},
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"execution_count": 1,
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"output_type": "execute_result"
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}
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],
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"source": [
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"G(x) = (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7)*(1 + x^2 + x^4 + x^6)*(1 + x^5)\n",
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"G(x).expand()"
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]
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},
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{
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"cell_type": "markdown",
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"id": "03d52995",
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"metadata": {},
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"source": [
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"Our answer is then simply the coefficient of the $x^7$ term (sometimes denoted $[x^7]G(x)$), which is 6, as expected.\n",
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"\n",
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"But how in the world does *that* work? Consider just part of the above calculation.\n",
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"$$(1 + x^2 + x^4 + x^6)(1 + x^5) = 1 + x^5 + x^2 + x^7 + x^4 + x^9 + x^6 + x^{11}$$\n",
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"The terms of the resulting product are written in the same order as you would probably get when distributing by hand. I'm going to try to give an interpretation for the first few resulting terms (remember what we said before about the meanings of the exponents and coefficients):\n",
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"* $1 \\times 1 = 1$: There's **one way** to make change for **0 cents** with only 2 cent coins, and **one way** to make change for **0 cents** with only 5 cent coins. Therefore, there's **one way** to make change for **0 cents** using both 2 cent and 5 cent coins.\n",
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"* $1 \\times x^5 = x^5$: There's **one way** to make change for **0 cents** with only 2 cent coins, and **one way** to make change for **5 cents** with only 5 cent coins. Therefore, there's **one way** to make change for **5 cents** using both 2 cent and 5 cent coins.\n",
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"* $x^2 \\times 1 = x^2$: There's **one way** to make change for **2 cents** with only 2 cent coins, and **one way** to make change for **0 cents** with only 5 cent coins. Therefore, there's **one way** to make change for **2 cents** using both 2 cent and 5 cent coins.\n",
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"* $x^2 \\times x^5 = x^7$: There's **one way** to make change for **2 cents** with only 2 cent coins, and **one way** to make change for **5 cents** with only 5 cent coins. Therefore, there's **one way** to make change for **7 cents** using both 2 cent and 5 cent coins.\n",
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"\n",
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"Now let's try multiplying this result by the rest of the function:\n",
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"$$(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7)(1 + x^2 + x^4 + x^5 + x^6 + x^7 + x^9 + x^{11})$$\n",
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"Here, I'll do interpretations of a select few terms:\n",
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"* $1 \\times x^7 = x^7$: There's **one way** to make change for **0 cents** with only 1 cent coins, and **one way** to make change for **7 cents** with 2 and 5 cent coins. Therefore, there's **one way** to make change for **7 cents** using 1, 2, and 5 cent coins.\n",
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"* $x \\times x^6 = x^7$: There's **one way** to make change for **1 cent** with only 1 cent coins, and **one way** to make change for **6 cents** with 2 and 5 cent coins. Therefore, there's **one way** to make change for **7 cents** using 1, 2, and 5 cent coins.\n",
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"* $x^3 \\times x^4 = x^7$: There's **one way** to make change for **3 cents** with only 1 cent coins, and **one way** to make change for **4 cents** with 2 and 5 cent coins. Therefore, there's **one way** to make change for **7 cents** using 1, 2, and 5 cent coins.\n",
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"\n",
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"Notice that each of these select terms represent a different way to make change for 7 cents. Therefore, together they represent 3 ways to make change for 7 cents with 1, 2, and 5 cent coins. There are 3 more products that result in $x^7$ ($x^2 \\times x^5$, $x^5 \\times x^2$, and $x^7 \\times 1$), and when we are done distributing, we combine like terms, which results in $6x^7$. Once again, the exponent represents the amount we are making change for, and the coefficient represents the number of ways to make change for that amount with the denominations we're using.\n",
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"\n",
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"## The original problem\n",
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"Just like the simpler problems above, each $n$th term in this generating function gives the number of ways to make change for $n$ pence using 1, 2, 5, 10, 20, 50, 100, and 200 pence coins.\n",
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"$$G(x) = \\frac{1}{(1-x)(1-x^2)(1-x^5)(1-x^{10})(1-x^{20})(1-x^{50})(1-x^{100})(1-x^{200})}$$\n",
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"\n",
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"We could proceed like before and cut off each generating function after the $x^{200}$ term, then multiply and find $[x^{200}]G(x)$. However, another possibility is to use the function as written, and instead calculate its 200th [Taylor polynomial](https://en.wikipedia.org/wiki/Taylor_series)."
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]
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"text/plain": [
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"73682"
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]
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"output_type": "execute_result"
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}
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],
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"source": [
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"f(x) = 1/prod(1 - x^k for k in (1, 2, 5, 10, 20, 50, 100, 200))\n",
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"f(x).taylor(x, 0, 200).coefficient(x^200)"
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]
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}
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