Did 28 in ipython.
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@ -202,30 +202,82 @@ div#notebook {
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<p>I would try to create a function $f(n)$ which yields the sum of the outmost ring of a n by n spiral.</p>
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<p>When we know the corner of a certain spiral we can calculate it's total like $f_n = 4 c_n + 6 (n - 1)$. We then only have to update the corner value for each spiral.</p>
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<p>For example:</p>
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<p>$f(1) = 1$</p>
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<p>$f(3) = 3 + 5 + 7 + 9 = 24$</p>
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<p>$f(5) = 13 + 17 + 21 + 25 = 76$</p>
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<p>When we have this function we calculate the solution simply by</p>
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<pre><code>s = sum([f(n) for n in range(1, 1002, 2)])</code></pre>
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<p>For each outer ring there is an initial corner value c ($c_3 = 3, c_5 = 76$). Once we have this value we can caluclate f like $f(n) = c_{n} + (c_n + n - 1) + (c_n + 2(n-1)) + (c_n + 3(n-1)) = 4c_n + 6 (n-1)$</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">def</span> <span class="nf">f</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="n">total</span> <span class="o">=</span> <span class="mi">1</span>
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<span class="k">if</span> <span class="n">n</span> <span class="o">==</span> <span class="mi">1</span><span class="p">:</span>
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<span class="n">current_corner</span> <span class="o">=</span> <span class="mi">3</span>
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<span class="k">return</span> <span class="mi">1</span>
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<span class="k">return</span> <span class="mi">0</span>
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<span class="n">s</span> <span class="o">=</span> <span class="nb">sum</span><span class="p">([</span><span class="n">f</span><span class="p">(</span><span class="n">n</span><span class="p">)</span> <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="mi">1002</span><span class="p">,</span> <span class="mi">2</span><span class="p">)])</span>
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<span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span> <span class="mi">1002</span><span class="p">,</span> <span class="mi">2</span><span class="p">):</span>
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<span class="n">total</span> <span class="o">+=</span> <span class="mi">4</span> <span class="o">*</span> <span class="n">current_corner</span> <span class="o">+</span> <span class="mi">6</span> <span class="o">*</span> <span class="p">(</span><span class="n">n</span> <span class="o">-</span> <span class="mi">1</span><span class="p">)</span>
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<span class="n">current_corner</span> <span class="o">+=</span> <span class="mi">4</span> <span class="o">*</span> <span class="n">n</span> <span class="o">-</span> <span class="mi">2</span>
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<span class="n">s</span> <span class="o">=</span> <span class="n">total</span>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="k">assert</span><span class="p">(</span><span class="n">s</span> <span class="o">==</span> <span class="mi">669171001</span><span class="p">)</span>
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<span class="n">s</span>
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<pre>669171001</pre>
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<p>The only missing piece is how could we calculate the current corner value for a certain n. The series for this is as follows:</p>
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<p>$c = 1, 3, 13, 31, 57, 91, 133, 183, 241, \dots$</p>
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<p>With some experimenting it can be seen that</p>
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<p>$c_n = (n - 1)^2 - (n - 2) = n^2 - 2n + 1 - n + 2 = n^2 - 3n + 3$.</p>
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<p>Now, we can insert $c_n$ into $f_n$ which gives as the sum of corners for the nth spiral:</p>
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<p>$f_n = 4n^2 - 12n + 12 + 6n - 6 = 4n^2 - 6n + 6$</p>
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<div class=" highlight hl-ipython3"><pre><span></span><span class="n">s</span> <span class="o">=</span> <span class="mi">1</span> <span class="o">+</span> <span class="nb">sum</span><span class="p">([</span><span class="mi">4</span> <span class="o">*</span> <span class="n">n</span> <span class="o">*</span> <span class="n">n</span> <span class="o">-</span> <span class="mi">6</span> <span class="o">*</span> <span class="n">n</span> <span class="o">+</span> <span class="mi">6</span> <span class="k">for</span> <span class="n">n</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">3</span><span class="p">,</span> <span class="mi">1002</span><span class="p">,</span> <span class="mi">2</span><span class="p">)])</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">s</span> <span class="o">==</span> <span class="mi">669171001</span><span class="p">)</span>
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<span class="k">assert</span><span class="p">(</span><span class="n">s</span> <span class="o">==</span> <span class="mi">669171001</span><span class="p">)</span>
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<span class="n">s</span>
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<span class="n">s</span>
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@ -234,6 +286,22 @@ div#notebook {
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<pre>669171001</pre>
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@ -27,39 +27,87 @@
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"cell_type": "markdown",
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"cell_type": "markdown",
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"metadata": {},
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"metadata": {},
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"source": [
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"source": [
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"I would try to create a function $f(n)$ which yields the sum of the outmost ring of a n by n spiral.\n",
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"When we know the corner of a certain spiral we can calculate it's total like $f_n = 4 c_n + 6 (n - 1)$. We then only have to update the corner value for each spiral. "
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"\n",
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"For example:\n",
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"\n",
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"$f(1) = 1$\n",
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"\n",
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"$f(3) = 3 + 5 + 7 + 9 = 24$\n",
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"\n",
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"$f(5) = 13 + 17 + 21 + 25 = 76$\n",
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"\n",
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"When we have this function we calculate the solution simply by\n",
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"\n",
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"~~~\n",
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"s = sum([f(n) for n in range(1, 1002, 2)])\n",
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"~~~\n",
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"\n",
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"For each outer ring there is an initial corner value c ($c_3 = 3, c_5 = 76$). Once we have this value we can caluclate f like $f(n) = c_{n} + (c_n + n - 1) + (c_n + 2(n-1)) + (c_n + 3(n-1)) = 4c_n + 6 (n-1)$"
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]
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]
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{
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{
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"cell_type": "code",
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"cell_type": "code",
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"execution_count": null,
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"execution_count": 1,
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"collapsed": true
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"outputs": [],
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"outputs": [],
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"source": [
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"source": [
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"def f(n):\n",
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"total = 1\n",
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" if n == 1:\n",
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"current_corner = 3\n",
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" return 1\n",
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" return 0\n",
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"\n",
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"\n",
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"s = sum([f(n) for n in range(1, 1002, 2)])\n",
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"for n in range(3, 1002, 2):\n",
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" total += 4 * current_corner + 6 * (n - 1)\n",
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" current_corner += 4 * n - 2\n",
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"\n",
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"s = total"
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]
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"669171001"
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],
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"source": [
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"assert(s == 669171001)\n",
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"s"
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"The only missing piece is how could we calculate the current corner value for a certain n. The series for this is as follows:\n",
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"\n",
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"$c = 1, 3, 13, 31, 57, 91, 133, 183, 241, \\dots$\n",
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"\n",
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"With some experimenting it can be seen that\n",
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"\n",
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"$c_n = (n - 1)^2 - (n - 2) = n^2 - 2n + 1 - n + 2 = n^2 - 3n + 3$.\n",
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"\n",
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"Now, we can insert $c_n$ into $f_n$ which gives as the sum of corners for the nth spiral:\n",
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"\n",
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"$f_n = 4n^2 - 12n + 12 + 6n - 6 = 4n^2 - 6n + 6$"
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]
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},
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{
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{
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"669171001"
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},
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"execution_count": 3,
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"metadata": {},
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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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"s = 1 + sum([4 * n * n - 6 * n + 6 for n in range(3, 1002, 2)])\n",
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"assert(s == 669171001)\n",
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"assert(s == 669171001)\n",
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"s"
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"s"
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@ -82,7 +130,7 @@
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"name": "python",
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"name": "python",
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"nbconvert_exporter": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"pygments_lexer": "ipython3",
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"version": "3.6.3"
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"version": "3.5.4"
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},
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},
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"tags": [
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"tags": [
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"spiral",
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"spiral",
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Reference in New Issue
Block a user