blocks|key|6335489|text|Mathematica，52字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|6335490|RegionPlot[(x%5E2%2By%5E2-1)%5E3-x%5E2y%5E3<0,{x,-2,2},{y,-2,2}]|code-block|syntax|javascript|6335491|📷|atomic|offset|length|6335492|我们可以用ContourPlot再保存一个字节，但是后台会有一点问题：|unstyled|style|CODE|6335493|ContourPlot[(x%5E2%2By%5E2-1)%5E3-x%5E2y%5E3,{x,-2,2},{y,-2,2}]|6335494|6335495|对于另外一个字节，我们可以使用第四个公式和PolarPlot来代替：|6335496|PolarPlot[(x=Sin@t)Abs@Cos@t%5E.5/(x%2B7/5)-2x%2B2,{t,0,7}]|6335497|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://i.stack.imgur.com/wpOOf.png|imageAlt|1|https://i.stack.imgur.com/1rKZp.png|2|https://i.stack.imgur.com/Xm75X.png^0|0|0|0|1|0|0|5|B|0|0|0|1|1|0|L|9|0|0|0|1|2^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|1C|8|@]|9|@$J|1D|K|1E|1|1F]]|A|$]]|$1|L|3|M|5|N|7|1G|8|@$J|1H|K|1I|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1J|8|@]|9|@]|A|$E|F]]|$1|S|3|H|5|I|7|1K|8|@]|9|@$J|1L|K|1M|1|1N]]|A|$]]|$1|T|3|U|5|N|7|1O|8|@$J|1P|K|1Q|O|P]]|9|@]|A|$]]|$1|V|3|W|5|D|7|1R|8|@]|9|@]|A|$E|F]]|$1|X|3|H|5|I|7|1S|8|@]|9|@$J|1T|K|1U|1|1V]]|A|$]]]|Y|$Z|$5|10|11|12|A|$13|14|15|-4]]|16|$5|10|11|12|A|$13|17|15|-4]]|18|$5|10|11|12|A|$13|19|15|-4]]]]

<h2>Mathematica, 52 bytes</h2>

<pre><code>RegionPlot[(x^2+y^2-1)^3-x^2y^3&lt;0,{x,-2,2},{y,-2,2}]
</code></pre>

<p><a href="https://i.stack.imgur.com/wpOOf.png"><img src="https://i.stack.imgur.com/wpOOf.png" alt="enter image description here"></a></p>

<p>We could save one more byte with <code>ContourPlot</code> but then the background gets a bit trippy:</p>

<pre><code>ContourPlot[(x^2+y^2-1)^3-x^2y^3,{x,-2,2},{y,-2,2}]
</code></pre>

<p><a href="https://i.stack.imgur.com/1rKZp.png"><img src="https://i.stack.imgur.com/1rKZp.png" alt="enter image description here"></a></p>

<p>For one <em>additional</em> byte we could use the fourth formula and <code>PolarPlot</code> instead:</p>

<pre><code>PolarPlot[(x=Sin@t)Abs@Cos@t^.5/(x+7/5)-2x+2,{t,0,7}]
</code></pre>

<p><a href="https://i.stack.imgur.com/Xm75X.png"><img src="https://i.stack.imgur.com/Xm75X.png" alt="enter image description here"></a></p>


blocks|key|6335326|text|MATLAB+/+Octave，31字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|6335327|ezplot('(x%5E2%2By%5E2-1)%5E3-x%5E2*y%5E3')|code-block|syntax|javascript|6335328|📷|atomic|offset|length|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://i.stack.imgur.com/gDK58.png|imageAlt^0|0|0|0|1|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|V|8|@]|9|@$J|W|K|X|1|Y]]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R|S|-4]]]]

<h1>MATLAB / Octave, 31 bytes</h1>
<pre><code>ezplot('(x^2+y^2-1)^3-x^2*y^3')
</code></pre>
<p><a href="https://i.stack.imgur.com/gDK58.png" rel="noreferrer"><img src="https://i.stack.imgur.com/gDK58.png" alt="enter image description here" /></a></p>


blocks|key|906299|text|矩阵，36字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|906300|1e4t:w/EqXJX,JY,J%7CYl**JX,X%5EJ%7C.3%5E*&XG|code-block|syntax|javascript|906301|在MATL在线试一试|unstyled|offset|length|906302|📷|atomic|entityMap|0|LINK|mutability|MUTABLE|url|https://matl.io/?code=1e4t%253Aw%252FEqXJX%252CJY%252CJ%257CYl%252a%252aJX%252CX%255EJ%257C.3%255E%252a%2526XG&inputs=&version=19.7.4|1|IMAGE|IMMUTABLE|imageUrl|https://i.stack.imgur.com/gQJha.png|imageAlt^0|0|0|1|6|0|0|0|1|1^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|13|8|@]|9|@$J|14|K|15|1|16]]|A|$]]|$1|L|3|M|5|N|7|17|8|@]|9|@$J|18|K|19|1|1A]]|A|$]]]|O|$P|$5|Q|R|S|A|$T|U]]|V|$5|W|R|X|A|$Y|Z|10|-4]]]]

<h1>MATL, 36 bytes</h1>
<pre><code>1e4t:w/EqXJX,JY,J|Yl**JX,X^J|.3^*&amp;XG
</code></pre>
<p>Try it at <a href="https://matl.io/?code=1e4t%3Aw%2FEqXJX%2CJY%2CJ%7CYl%2a%2aJX%2CX%5EJ%7C.3%5E%2a%26XG&amp;inputs=&amp;version=19.7.4" rel="nofollow noreferrer">MATL Online</a></p>
<p><a href="https://i.stack.imgur.com/gQJha.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gQJha.png" alt="enter image description here" /></a></p>


blocks|key|906828|text|果冻，31+字节数|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|906829|*3×
²%2B’*3<çị⁾*+
25µNr÷21µ²ç@þUY|code-block|syntax|javascript|906830|在网上试试！|906831|或者使用35个字节使用♥，而不是*+fill:将中间行改为²%2B’*3<ç×9829Ọ»⁶。|unstyled|style|CODE|906832|++++++++++♥♥♥♥♥♥♥♥♥♥+++++++++++♥♥♥♥♥♥♥♥♥♥++++++++++
++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++++++
++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++++
+++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++++
+++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++++
++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++
+++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++
+++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++
++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++
++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++
++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++
++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++
++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++
++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++
++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++
++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++
++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++
++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++
++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++
++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++
+++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++
+++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++
+++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++
++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++
++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++
+++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++++
+++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++++
++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++++
++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++++
+++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++++++
++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++++++
++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++++++
+++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++++++++
++++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++++++++
+++++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++++++++++
++++++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++++++++++
+++++++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++++++++++++
++++++++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++++++++++++
+++++++++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++++++++++++++
+++++++++++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥+++++++++++++++++
++++++++++++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥++++++++++++++++++
+++++++++++++++++++♥♥♥♥♥♥♥♥♥♥♥♥♥+++++++++++++++++++
+++++++++++++++++++++♥♥♥♥♥♥♥♥♥+++++++++++++++++++++
++++++++++++++++++++++♥♥♥♥♥♥♥++++++++++++++++++++++
++++++++++++++++++++++++♥♥♥++++++++++++++++++++++++
+++++++++++++++++++++++++♥+++++++++++++++++++++++++|906833|怎么做？|906834|工具：|906835|📷|atomic|906836|一个51×51行的馈送分开的字符网格。|906837|*3×+-+Link+1:+y,+xSquared
*3++-+cube+y
++×+-+multiply+the+result+by+xSquared

²%2B’*3<çị⁾*++-+Link+2:+y,+xSquared
²+++++++++++-+square+y
+%2B++++++++++-+add+xSquared+to+the+result
++’+++++++++-+subtract+one
+++*3+++++++-+cube+the+result
+++++<++++++-+less+than?+(1+if+True,+0+if+False)
++++++ç+++++-+++++the+result+of+the+last+link+(1)+as+a+dyad:+f(y,+xSquared)
+++++++ị++++-+index+into+(1-based):
++++++++⁾*++-+the+string+"*+"+(there+is+a+trailing+space+on+this+line)

+++++++×9829Ọ»⁶+-+alternative+ending+of+link+2
+++++++×9829++++-+multiply+(vectorises)+by+9829
++++++++++++Ọ+++-+cast+to+character+(vectorises)+(9829+is+the+♥,+0+is+the+null-byte)
+++++++++++++»++-+maximum+(vectorises)+of+that+and:
++++++++++++++⁶+-+++++space+character+(ordinal+of+space+is+32,+greater+than+the+null-byte,+less+than+♥)

25µNr÷21µ²ç@þUY+-+Main+link:+no+arguments
25++++++++++++++-+25
++µ+++++++++++++-+monadic+chain+separation+call+that+i
+++N++++++++++++-+negate+i
++++r+++++++++++-+inclusive+range(negated+i,+i)+->+[-25,+-24,+...,+24,+25]
+++++÷21++++++++-+divide+by+21+(vectorises)+->+[-1.1904761904761905,+-1.1428571428571428,+...,+1.1428571428571428,+1.1904761904761905]
++++++++µ+++++++-+monadic+chain+separation+call+that+r
+++++++++²++++++-+square+r+(vectorises)+(make+the+xSquared+values,+left+to+right)
+++++++++++++U++-+upend+r+(make+the+y+values,+top+to+bottom)
++++++++++++þ+++-+outer+product+using:
++++++++++ç@++++-+++++last+link+(2)+as+a+dyad+with+reversed+arguments:+f(y,+xSquared)
++++++++++++++Y+-+join+with+line+feeds
++++++++++++++++-+implicit+print|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/DennisMitchell/jelly|1|https://github.com/DennisMitchell/jelly/wiki/Code-page|2|https://tio.run/nexus/jelly#ATEAzv//KjPDlwrCsivigJkqMzzDp@G7i@KBviogCjI1wrVOcsO3MjHCtcKyw6dAw75VWf//|3|https://tio.run/nexus/jelly#ATcAyP//KjPDlwrCsivigJkqMzzDp8OXOTgyOeG7jMK74oG2CjI1wrVOcsO3MjHCtcKyw6dAw75VWf//|4|IMAGE|IMMUTABLE|imageUrl|https://i.stack.imgur.com/Kgehc.gif|imageAlt^0|0|2|0|6|3|1|0|0|0|6|2|0|B|1|G|1|T|F|4|5|3|0|0|0|0|0|1|4|0|0^^$0|@$1|2|3|4|5|6|7|1L|8|@]|9|@$A|1M|B|1N|1|1O]|$A|1P|B|1Q|1|1R]]|C|$]]|$1|D|3|E|5|F|7|1S|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1T|8|@]|9|@$A|1U|B|1V|1|1W]]|C|$]]|$1|K|3|L|5|M|7|1X|8|@$A|1Y|B|1Z|N|O]|$A|20|B|21|N|O]|$A|22|B|23|N|O]]|9|@$A|24|B|25|1|26]]|C|$]]|$1|P|3|Q|5|F|7|27|8|@]|9|@]|C|$G|H]]|$1|R|3|S|5|6|7|28|8|@]|9|@]|C|$]]|$1|T|3|U|5|M|7|29|8|@]|9|@]|C|$]]|$1|V|3|W|5|X|7|2A|8|@]|9|@$A|2B|B|2C|1|2D]]|C|$]]|$1|Y|3|Z|5|M|7|2E|8|@]|9|@]|C|$]]|$1|10|3|11|5|F|7|2F|8|@]|9|@]|C|$G|H]]]|12|$13|$5|14|15|16|C|$17|18]]|19|$5|14|15|16|C|$17|1A]]|1B|$5|14|15|16|C|$17|1C]]|1D|$5|14|15|16|C|$17|1E]]|1F|$5|1G|15|1H|C|$1I|1J|1K|-4]]]]

<h1><a href="https://github.com/DennisMitchell/jelly" rel="noreferrer">Jelly</a>, 31 <a href="https://github.com/DennisMitchell/jelly/wiki/Code-page" rel="noreferrer">bytes</a></h1>

<pre><code>*3×
²+’*3&lt;çị⁾* 
25µNr÷21µ²ç@þUY
</code></pre>

<p><strong><a href="https://tio.run/nexus/jelly#ATEAzv//KjPDlwrCsivigJkqMzzDp@G7i@KBviogCjI1wrVOcsO3MjHCtcKyw6dAw75VWf//" rel="noreferrer">Try it online!</a></strong></p>

<p>or <strong><a href="https://tio.run/nexus/jelly#ATcAyP//KjPDlwrCsivigJkqMzzDp8OXOTgyOeG7jMK74oG2CjI1wrVOcsO3MjHCtcKyw6dAw75VWf//" rel="noreferrer">35 bytes</a></strong> with a <code>♥</code>, rather than <code>*</code> fill: change middle line to <code>²+’*3&lt;ç×9829Ọ»⁶</code>.</p>

<pre><code>          ♥♥♥♥♥♥♥♥♥♥           ♥♥♥♥♥♥♥♥♥♥          
        ♥♥♥♥♥♥♥♥♥♥♥♥♥♥       ♥♥♥♥♥♥♥♥♥♥♥♥♥♥        
      ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥   ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥      
     ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥ ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥     
     ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥ ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥     
    ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥    
   ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥   
   ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥   
  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥  
  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥  
  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥  
  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥  
  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥  
  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥  
  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥  
  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥  
  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥  
  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥  
  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥  
  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥  
   ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥   
   ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥   
   ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥   
    ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥    
    ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥    
     ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥     
     ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥     
      ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥      
      ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥      
       ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥       
        ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥        
        ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥        
         ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥         
          ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥          
           ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥           
            ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥            
             ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥             
              ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥              
               ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥               
                 ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥                 
                  ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥                  
                   ♥♥♥♥♥♥♥♥♥♥♥♥♥                   
                     ♥♥♥♥♥♥♥♥♥                     
                      ♥♥♥♥♥♥♥                      
                        ♥♥♥                        
                         ♥                         
</code></pre>

<h3>How?</h3>

<p>Implements: <a href="https://i.stack.imgur.com/Kgehc.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/Kgehc.gif" alt="equation"></a> over a 51 by 51 line feed separated grid of characters.</p>

<pre><code>*3× - Link 1: y, xSquared
*3  - cube y
  × - multiply the result by xSquared

²+’*3&lt;çị⁾*  - Link 2: y, xSquared
²           - square y
 +          - add xSquared to the result
  ’         - subtract one
   *3       - cube the result
     &lt;      - less than? (1 if True, 0 if False)
      ç     -     the result of the last link (1) as a dyad: f(y, xSquared)
       ị    - index into (1-based):
        ⁾*  - the string "* " (there is a trailing space on this line)

       ×9829Ọ»⁶ - alternative ending of link 2
       ×9829    - multiply (vectorises) by 9829
            Ọ   - cast to character (vectorises) (9829 is the ♥, 0 is the null-byte)
             »  - maximum (vectorises) of that and:
              ⁶ -     space character (ordinal of space is 32, greater than the null-byte, less than ♥)

25µNr÷21µ²ç@þUY - Main link: no arguments
25              - 25
  µ             - monadic chain separation call that i
   N            - negate i
    r           - inclusive range(negated i, i) -&gt; [-25, -24, ..., 24, 25]
     ÷21        - divide by 21 (vectorises) -&gt; [-1.1904761904761905, -1.1428571428571428, ..., 1.1428571428571428, 1.1904761904761905]
        µ       - monadic chain separation call that r
         ²      - square r (vectorises) (make the xSquared values, left to right)
             U  - upend r (make the y values, top to bottom)
            þ   - outer product using:
          ç@    -     last link (2) as a dyad with reversed arguments: f(y, xSquared)
              Y - join with line feeds
                - implicit print
</code></pre>


blocks|key|6335527|text|SageMath，66字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|6335528|polar_plot(sin(x)*abs(cos(x))%5E.5/(sin(x)%2B7/5)-2*sin(x)%2B2,(0,2*pi))|code-block|syntax|javascript|6335529|在网上试试|unstyled|offset|length|6335530|不幸的是，implicit_plot和parametric_plot的长名称(加上需要为前者定义隐式定义的x之外的变量)使得前三个选项太长，无法考虑。|style|CODE|6335531|感谢busukxuan和Andrea+Lazzarotto的字节|entityMap|0|LINK|mutability|MUTABLE|url|http://sagecell.sagemath.org/?z=eJwryM9JLIovyMkv0SjOzNOo0NRKTCrWSM4vBjI14_RM9aHC2ub6ppq6RlpQnpGOhoGOkVZBpqYmANUwE4E=&lang=sage^0|0|0|0|5|0|0|5|D|J|F|1H|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|10|8|@]|9|@$J|11|K|12|1|13]]|A|$]]|$1|L|3|M|5|I|7|14|8|@$J|15|K|16|N|O]|$J|17|K|18|N|O]|$J|19|K|1A|N|O]]|9|@]|A|$]]|$1|P|3|Q|5|I|7|1B|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

<h2>SageMath, 66 bytes</h2>
<pre class="lang-py prettyprint-override"><code>polar_plot(sin(x)*abs(cos(x))^.5/(sin(x)+7/5)-2*sin(x)+2,(0,2*pi))
</code></pre>
<p><a href="http://sagecell.sagemath.org/?z=eJwryM9JLIovyMkv0SjOzNOo0NRKTCrWSM4vBjI14_RM9aHC2ub6ppq6RlpQnpGOhoGOkVZBpqYmANUwE4E=&amp;lang=sage" rel="noreferrer">Try it online</a></p>
<p>Unfortunately, the long names of <code>implicit_plot</code> and <code>parametric_plot</code> (plus the need to define a variable other than the implicitly-defined <code>x</code> for the former) make the first three options much too long to be considered.</p>
<p>Thanks to busukxuan and Andrea Lazzarotto for a byte</p>


blocks|key|6336115|text|Ruby，80字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|6336116|给露比一些爱。|unstyled|6336117|-2.step(2,0.2){%7Cy%7C-2.step(2,0.1){%7Cx%7CA2gt;<<'.+O'[(y*y-1%2Bx*=x)**3%2Bx*y**3<=>0]};puts}|code-block|syntax|javascript|6336118|输出：|6336119|+++++++++OOOOO+++++OOOOO+++++++++++++
++++++.OOOOOOOOO.OOOOOOOOO.++++++++++
+++++OOOOOOOOOOOOOOOOOOOOOOO+++++++++
+++++OOOOOOOOOOOOOOOOOOOOOOO+++++++++
+++++OOOOOOOOOOOOOOOOOOOOOOO+++++++++
++++++OOOOOOOOOOOOOOOOOOOOO++++++++++
++++++.OOOOOOOOOOOOOOOOOOO.++++++++++
++++++++OOOOOOOOOOOOOOOOO++++++++++++
+++++++++OOOOOOOOOOOOOOO+++++++++++++
+++++++++++OOOOOOOOOOO+++++++++++++++
++++++++++++++OOOOO++++++++++++++++++
++++++++++++++++.++++++++++++++++++++|6336120|上面和下面空空如也|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|S|8|@]|9|@]|A|$H|I]]|$1|J|3|K|5|6|7|T|8|@]|9|@]|A|$]]|$1|L|3|M|5|G|7|U|8|@]|9|@]|A|$H|I]]|$1|N|3|O|5|D|7|V|8|@]|9|@]|A|$]]]|P|$]]

<h2>Ruby, 80 bytes</h2>

<p>Show Ruby some love.</p>

<pre><code>-2.step(2,0.2){|y|-2.step(2,0.1){|x|$&gt;&lt;&lt;'. O'[(y*y-1+x*=x)**3+x*y**3&lt;=&gt;0]};puts}
</code></pre>

<h3>Output:</h3>

<pre><code>         OOOOO     OOOOO             
      .OOOOOOOOO.OOOOOOOOO.          
     OOOOOOOOOOOOOOOOOOOOOOO         
     OOOOOOOOOOOOOOOOOOOOOOO         
     OOOOOOOOOOOOOOOOOOOOOOO         
      OOOOOOOOOOOOOOOOOOOOO          
      .OOOOOOOOOOOOOOOOOOO.          
        OOOOOOOOOOOOOOOOO            
         OOOOOOOOOOOOOOO             
           OOOOOOOOOOO               
              OOOOO                  
                .                    
</code></pre>

<p>with some empty lines above and below</p>


blocks|key|906581|text|英国广播公司|type|unstyled|depth|inlineStyleRanges|entityRanges|data|906582|+BASIC，80|header-two|906583|明天就会打完高尔夫球。|906584|http://www.bbcbasic.co.uk/bbcwin/bbcwin.html下载解释器|offset|length|906585|ORIGIN99,99F.u=-56TO56x=u/50b=x%5E2%5E.33/2r=(b%5E2-x%5E2%2B1)%5E.5*50L.u,b*50-r,u,r%2Bb*50N.|code-block|syntax|javascript|906586|用第一个方程画一个填充的心，我们重新排列，取一个实数的立方根，一个实数和两个复数，但我们只对实根感兴趣。然后我们重新排列，得到y中的二次型|style|CODE|906587|x%5E2%2By%5E2-1=(x%5E0.66)y

y%5E2-(x%5E0.66)y%2Bx%5E2-1=0|906588|使用ay%5E2%2Bby%2Bc=0类型表达式的标准二次公式，并利用a=1这一事实，我们得到|906589|y+=+-b/2+%2B/-+sqrt((b/2)%5E2-c)

y+=+-b/2+%2B/-+sqrt((b/2)%5E2-x%5E2%2B1)|906590|其中-b/2+=+(x%5E0.66)/2+(由程序中的b表示)。|906591|我们只需从左到右扫描x的值，在每一对有效值之间画一条垂直线，以便y建立一个实体形状。|906592|ORIGIN99,99F.u=-56TO56x=u/50b=x%5E2%5E.33/2r=(b%5E2-x%5E2%2B1)%5E.5*50L.u,b*50-r,u,r%2Bb*50N.|906593|ungolfed|906594|ORIGIN99,99
FORu=-56TO56
++x=u/50
++b=x%5E2%5E.33/2
++r=(b%5E2-x%5E2%2B1)%5E.5*50
++LINEu,b*50%2Br,u,b*50-r
NEXT|entityMap|0|LINK|mutability|MUTABLE|url|http://www.bbcbasic.co.uk/bbcwin/bbcwin.html^0|0|0|0|0|18|0|0|0|1R|1|0|0|2|B|T|1|0|0|2|H|Q|1|0|A|1|W|1|0|0|0^^$0|@$1|2|3|4|5|6|7|1G|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1H|8|@]|9|@]|A|$]]|$1|E|3|F|5|6|7|1I|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|1J|8|@]|9|@$I|1K|J|1L|1|1M]]|A|$]]|$1|K|3|L|5|M|7|1N|8|@]|9|@]|A|$N|O]]|$1|P|3|Q|5|6|7|1O|8|@$I|1P|J|1Q|R|S]]|9|@]|A|$]]|$1|T|3|U|5|M|7|1R|8|@]|9|@]|A|$N|O]]|$1|V|3|W|5|6|7|1S|8|@$I|1T|J|1U|R|S]|$I|1V|J|1W|R|S]]|9|@]|A|$]]|$1|X|3|Y|5|M|7|1X|8|@]|9|@]|A|$N|O]]|$1|Z|3|10|5|6|7|1Y|8|@$I|1Z|J|20|R|S]|$I|21|J|22|R|S]]|9|@]|A|$]]|$1|11|3|12|5|6|7|23|8|@$I|24|J|25|R|S]|$I|26|J|27|R|S]]|9|@]|A|$]]|$1|13|3|14|5|M|7|28|8|@]|9|@]|A|$N|O]]|$1|15|3|16|5|D|7|29|8|@]|9|@]|A|$]]|$1|17|3|18|5|M|7|2A|8|@]|9|@]|A|$N|O]]]|19|$1A|$5|1B|1C|1D|A|$1E|1F]]]]

<h1>BBC BASIC, 80</h1>
<p>Will finish golfing tomorrow.</p>
<p>Download interpreter at <a href="http://www.bbcbasic.co.uk/bbcwin/bbcwin.html" rel="nofollow noreferrer">http://www.bbcbasic.co.uk/bbcwin/bbcwin.html</a></p>
<pre><code>ORIGIN99,99F.u=-56TO56x=u/50b=x^2^.33/2r=(b^2-x^2+1)^.5*50L.u,b*50-r,u,r+b*50N.
</code></pre>
<p>Draws a filled heart using the first equation, which we rearrange and take the cube root of (there are 3 cube roots of a real number, one real and 2 complex but we only interested in the real root here.) We then rearrange again to get a quadratic in <code>y</code></p>
<pre><code>x^2+y^2-1=(x^0.66)y

y^2-(x^0.66)y+x^2-1=0
</code></pre>
<p>Using the standard quadratic formula for expressions of the type <code>ay^2+by+c=0</code>, and taking advantage of the fact that <code>a</code>=1 we get</p>
<pre><code>y = -b/2 +/- sqrt((b/2)^2-c)

y = -b/2 +/- sqrt((b/2)^2-x^2+1)
</code></pre>
<p>where <code>-b/2 = (x^0.66)/2</code> (represented by <code>b</code> in the program.)</p>
<p>We simply scan through the values of <code>x</code> from left to right, drawing a vertical line between each pair of valid values for <code>y</code> to build up a solid shape.</p>
<pre><code>ORIGIN99,99F.u=-56TO56x=u/50b=x^2^.33/2r=(b^2-x^2+1)^.5*50L.u,b*50-r,u,r+b*50N.
</code></pre>
<p><strong>ungolfed</strong></p>
<pre><code>ORIGIN99,99
FORu=-56TO56
  x=u/50
  b=x^2^.33/2
  r=(b^2-x^2+1)^.5*50
  LINEu,b*50+r,u,b*50-r
NEXT
</code></pre>


blocks|key|906996|text|英国广播公司|type|unstyled|depth|inlineStyleRanges|entityRanges|data|906997|+BASIC+(BBC微音波)(126个字节)|header-two|906998|这是水平河街+回答的一个变体，但对于最初的BBC微缩版的BBC来说。|offset|length|906999|您可以将以下任何一项复制并粘贴到JSBeeb在https://bbc.godbolt.org/上的新粘贴功能中(一种JavaScript支持的BCMicro仿真器)，粘贴之后，在“屏幕”内按Enter以显示新的提示，然后键入RUN，然后按Enter以运行任一版本。|907000|打高尔夫的。(126个字节)|907001|1MO.1:V.29,64;64;:MOVE-57,20:GC.0,1:F.U=-56TO56:X=U/50:IFX<>0TH.B=X%5E2%5E.33/2
2R=(B%5E2-X%5E2%2B1)%5E.5*50:DRAWU,B*50%2BR:DRAW+U,B*50-R:N.|code-block|syntax|javascript|907002|不打高尔夫球。(176字节)|907003|1+MODE+1
2+VDU+29,64;64;
3+MOVE+-57,20
4+GCOL+0,1
5+FOR+U+=+-56+TO+56
6+X+=+U/50
7+IF+X+<>+0+THEN+B+=+X%5E2%5E.33/2
8+R+=+(B%5E2-X%5E2%2B1)%5E.5*50
9+DRAW+U,B*50%2BR
10+DRAW+U,B*50-R
11+NEXT|907004|第7行需要对X进行检查，以防止当X在循环过程中变为0时出现“日志范围”错误。除了这个，差不多一样。在真正的机器上绘制要慢得多，JSBeeb很好地模拟了这一点。|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/users/15599/level-river-st|1|https://codegolf.stackexchange.com/a/110000|2|https://bbc.godbolt.org/^0|0|0|2|4|0|7|2|1|0|N|O|2|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|19|8|@]|9|@]|A|$]]|$1|E|3|F|5|6|7|1A|8|@]|9|@$G|1B|H|1C|1|1D]|$G|1E|H|1F|1|1G]]|A|$]]|$1|I|3|J|5|6|7|1H|8|@]|9|@$G|1I|H|1J|1|1K]]|A|$]]|$1|K|3|L|5|6|7|1L|8|@]|9|@]|A|$]]|$1|M|3|N|5|O|7|1M|8|@]|9|@]|A|$P|Q]]|$1|R|3|S|5|6|7|1N|8|@]|9|@]|A|$]]|$1|T|3|U|5|O|7|1O|8|@]|9|@]|A|$P|Q]]|$1|V|3|W|5|6|7|1P|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]|14|$5|Z|10|11|A|$12|15]]|16|$5|Z|10|11|A|$12|17]]]]

<h1>BBC BASIC (BBC Micro) (126 bytes)</h1>

<p>This is a variation of <a href="https://codegolf.stackexchange.com/users/15599/level-river-st">Level River Street's</a> <a href="https://codegolf.stackexchange.com/a/110000">answer</a>, but for the original BBC Micro's iteration of BBC BASIC.</p>

<p>You can copy and paste either of the following into JSBeeb's new paste functionality at <a href="https://bbc.godbolt.org/" rel="nofollow noreferrer">https://bbc.godbolt.org/</a> (A JavaScript powered BBC Micro emulator), once it has pasted, press Enter inside the 'screen' to show a new prompt, then type RUN, then press Enter to run either version.</p>

<p>golfed. (126 bytes)</p>

<pre><code>1MO.1:V.29,64;64;:MOVE-57,20:GC.0,1:F.U=-56TO56:X=U/50:IFX&lt;&gt;0TH.B=X^2^.33/2
2R=(B^2-X^2+1)^.5*50:DRAWU,B*50+R:DRAW U,B*50-R:N.
</code></pre>

<p>ungolfed. (176 bytes)</p>

<pre><code>1 MODE 1
2 VDU 29,64;64;
3 MOVE -57,20
4 GCOL 0,1
5 FOR U = -56 TO 56
6 X = U/50
7 IF X &lt;&gt; 0 THEN B = X^2^.33/2
8 R = (B^2-X^2+1)^.5*50
9 DRAW U,B*50+R
10 DRAW U,B*50-R
11 NEXT
</code></pre>

<p>Line 7 requires a check for X, to prevent a "Log range" error when X becomes 0 half way through the loop. Other than this, it's pretty much the same. It's a lot slower to draw on a real machine, which JSBeeb emulates pretty well.</p>


blocks|key|907030|text|Google工作表(条件格式)，73|type|header-two|depth|inlineStyleRanges|entityRanges|data|907031|在范围A1:A1+100上设置条件格式：|unstyled|907032|=(2500/9-(Column()-25)%5E2-(ROW()-25)%5E2)%5E3>(Column()-25)%5E2*(ROW()-25)%5E3/.06|code-block|syntax|javascript|907033|您必须调整您的列/行，以便使100x100单元格适合您的屏幕。(10个像素是一个很好的值。)改变心脏样细胞的颜色。|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|O|8|@]|9|@]|A|$H|I]]|$1|J|3|K|5|D|7|P|8|@]|9|@]|A|$]]]|L|$]]

<h1>Google Sheets (Conditional Formatting), 73</h1>
<p>Set conditional formatting on range A1:CV100:</p>
<pre><code>=(2500/9-(Column()-25)^2-(ROW()-25)^2)^3&gt;(Column()-25)^2*(ROW()-25)^3/.06
</code></pre>
<p>You'll have to adjust your columns/rows so that 100x100 cells fit on your screen. (10 pixels is a good value.) Colors the cells in the cardioid.</p>


<p>Your challenge is to draw/plot one of the hearts below. The size must be at least 50x50 pixels (vector graphics is OK). You may choose which of the hearts you want to draw/plot. Axes, grid lines etc. are accepted. There must be <strong>at least</strong> 100 distinct coordinates/points on the graph. The heart can be filled with color if you want to.</p>

<p><a href="https://i.stack.imgur.com/Vnin0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vnin0.jpg" alt="enter image description here"></a></p>

<p>The equations are either:</p>

<p><a href="https://i.stack.imgur.com/Kgehc.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Kgehc.gif" alt="enter image description here"></a></p>

<p><strong>or</strong></p>

<p><a href="https://i.stack.imgur.com/DCI6L.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DCI6L.gif" alt="enter image description here"></a></p>

<p><a href="https://latex.codecogs.com/gif.latex?y=%7Bcos(t)%7D%5E%7B0.5%7D%5Ccdot|t|%5E%7B0.3%7D" rel="nofollow noreferrer"><img src="https://latex.codecogs.com/gif.latex?y=%7Bcos(t)%7D%5E%7B0.5%7D%5Ccdot|t|%5E%7B0.3%7D" alt="enter image description here"></a></p>

<p><code>t</code> is in the range <code>[-1, 1]</code>.</p>

<p><strong>or</strong></p>

<p><a href="https://i.stack.imgur.com/fFH5v.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fFH5v.gif" alt="enter image description here"></a></p>

<p><strong>or</strong></p>

<p><a href="https://latex.codecogs.com/gif.latex?r=%5Cfrac%7Bsin(t)%5Csqrt%7B|cos(t)|%7D%7D%7Bsin(t)+%5Cfrac%7B7%7D%7B5%7D%7D-2sin(t)+2" rel="nofollow noreferrer"><img src="https://latex.codecogs.com/gif.latex?r=%5Cfrac%7Bsin(t)%5Csqrt%7B|cos(t)|%7D%7D%7Bsin(t)+%5Cfrac%7B7%7D%7B5%7D%7D-2sin(t)+2" alt="enter image description here"></a></p>

<p>This is <a href="/questions/tagged/code-golf" class="post-tag" title="show questions tagged &#39;code-golf&#39;" rel="tag">code-golf</a> so the shortest code (in each language) in bytes wins. Please provide the output (or link to an online interpreter).</p>


Draw/plot a heart graph

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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你的挑战是画出下面的一颗心。大小必须至少为50x50像素(矢量图形是可以的)。你可以从你想画的那颗心中选择一个。轴，网格线等被接受。图上必须至少有100个不同的坐标/点。如果你想的话，心脏可以充满颜色。📷方程式是：📷或📷📷t在[-1, 1]范围内。或📷或📷这是密码-高尔夫，因此以字节表示的最短代码(在每种语言中)获胜。请提供输出(或链接到在线解释器)。

问绘制/绘制心电图
EN

或

或

或

回答 9

Code Golf用户

Mathematica，52字节

Code Golf用户

MATLAB / Octave，31字节

Code Golf用户

矩阵，36字节

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问绘制/绘制心电图EN

或

或

或

回答 9

Code Golf用户

Mathematica，52字节

Code Golf用户

MATLAB / Octave，31字节

Code Golf用户

矩阵，36字节

社区

活动

圈层

关于

腾讯云开发者

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问绘制/绘制心电图
EN