diff --git a/content/Chapter_01/03_Collisions_in_Hashing.ipynb b/content/Chapter_01/03_Collisions_in_Hashing.ipynb index 8f21a1bf..6c6ad9f1 100644 --- a/content/Chapter_01/03_Collisions_in_Hashing.ipynb +++ b/content/Chapter_01/03_Collisions_in_Hashing.ipynb @@ -33,67 +33,14 @@ "What if the hash values were just assigned at random, without taking into account which of them have already been assigned? If there are a large number of distinct values and a relatively small number of individuals, then it seems reasonable to think that the chance of a collision will be small. For example, if there are 1,000 available hash values and only 5 individuals, it doesn't seem likely that you'll get a collision if you just pick a random sequence of 5 values for the 5 individuals." ] }, - { - "cell_type": "code", - "execution_count": null, - "metadata": { - "editable": true, - "slideshow": { - "slide_type": "" - }, - "tags": [ - "remove-cell" - ] - }, - "outputs": [ - { - "data": { - "image/jpeg": 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- "text/html": [ - "\n", - " \n", - " " - ], - "text/plain": [ - "" - ] - }, - "metadata": { - "scrapbook": { - "mime_prefix": "", - "name": "vid_prob_model" - } - }, - "output_type": "display_data" - } - ], - "source": [ - "#| label: vid-prob-model\n", - "#| echo: false\n", - "from IPython.display import YouTubeVideo\n", - "YouTubeVideo(\"uFLsmVFcHXw\")" - ] - }, { "cell_type": "markdown", - "metadata": { - "editable": true, - "slideshow": { - "slide_type": "" - }, - "tags": [] - }, + "metadata": {}, "source": [ - "```{dropdown} 🎥 See More\n", - "![](#vid-prob-model)\n", - "```" + ":::{iframe} https://www.youtube.com/embed/uFLsmVFcHXw\n", + ":width: 100%\n", + ":title: Probability Model\n", + ":::" ] }, { @@ -126,66 +73,14 @@ "If you look back to Part (i) in the example about random number generators in the previous section, you will find that it is the same as our current question, in the case where $N = 10$ and $n=2$. We can just follow the same process to get our solution here." ] }, - { - "cell_type": "code", - "execution_count": null, - "metadata": { - "editable": true, - "slideshow": { - "slide_type": "" - }, - "tags": [ - "remove-cell" - ] - }, - "outputs": [ - { - "data": { - "image/jpeg": 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- "text/html": [ - "\n", - " \n", - " " - ], - "text/plain": [ - "" - ] - }, - "metadata": { - "scrapbook": { - "mime_prefix": "", - "name": "vid_collision_prob" - } - }, - "output_type": "display_data" - } - ], - "source": [ - "#| label: vid-collision-prob\n", - "#| echo: false\n", - "YouTubeVideo('cD26s3DW-J8')" - ] - }, { "cell_type": "markdown", - "metadata": { - "editable": true, - "slideshow": { - "slide_type": "" - }, - "tags": [] - }, + "metadata": {}, "source": [ - "```{dropdown} 🎥 See More\n", - "![](#vid-collision-prob)\n", - "```" + ":::{iframe} https://www.youtube.com/embed/cD26s3DW-J8\n", + ":width: 100%\n", + ":title: Collision Probability\n", + ":::" ] }, { diff --git a/content/Chapter_01/04_Birthday_Problem.ipynb b/content/Chapter_01/04_Birthday_Problem.ipynb index c1148d3c..5684ab69 100644 --- a/content/Chapter_01/04_Birthday_Problem.ipynb +++ b/content/Chapter_01/04_Birthday_Problem.ipynb @@ -58,61 +58,14 @@ "The problem is commonly solved under the assumptions that each year consists of 365 days and that each person is equally likely to be born on any of the 365 days regardless of the birthdays of others. " ] }, - { - "cell_type": "code", - "execution_count": null, - "metadata": { - "editable": true, - "slideshow": { - "slide_type": "" - }, - "tags": [ - "remove-cell" - ] - }, - "outputs": [ - { - "data": { - "image/jpeg": 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- "text/html": [ - "\n", - " \n", - " " - ], - "text/plain": [ - "" - ] - }, - "metadata": { - "scrapbook": { - "mime_prefix": "", - "name": "vid141" - } - }, - "output_type": "display_data" - } - ], - "source": [ - "#| label: vid-bday-assum\n", - "#| echo: false\n", - "from IPython.display import YouTubeVideo\n", - "YouTubeVideo('A88MJdZLe3A')" - ] - }, { "cell_type": "markdown", "metadata": {}, "source": [ - "```{dropdown} 🎥 See More\n", - "![](#vid-bday-assum)\n", - "```" + ":::{iframe} https://www.youtube.com/embed/A88MJdZLe3A\n", + ":width: 100%\n", + ":title: Birthday Assumptions\n", + ":::" ] }, { @@ -156,51 +109,14 @@ "$$" ] }, - { - "cell_type": "code", - "execution_count": null, - "metadata": { - "tags": [ - "remove-cell" - ] - }, - "outputs": [ - { - "data": { - "text/html": [ - "\n", - " \n", - " " - ], - "text/plain": [ - "" - ] - }, - "execution_count": 3, - "metadata": {}, - "output_type": "execute_result" - } - ], - "source": [ - "#| label: vid-note-code\n", - "#| echo: false\n", - "from IPython.display import YouTubeVideo\n", - "YouTubeVideo(\"ci7bPVOpfsk\")" - ] - }, { "cell_type": "markdown", "metadata": {}, "source": [ - "```{dropdown} 🎥 See More\n", - "![](#vid-note-code)\n", - "```" + ":::{iframe} https://www.youtube.com/embed/ci7bPVOpfsk\n", + ":width: 100%\n", + ":title: Notes on the Code\n", + ":::" ] }, { diff --git a/content/Chapter_01/05_An_Exponential_Approximation.ipynb b/content/Chapter_01/05_An_Exponential_Approximation.ipynb index 05c0b25c..081d11f7 100644 --- a/content/Chapter_01/05_An_Exponential_Approximation.ipynb +++ b/content/Chapter_01/05_An_Exponential_Approximation.ipynb @@ -62,51 +62,14 @@ "The main steps in the approximation will be used repeatedly in this course, so we will set them out in some detail here." ] }, - { - "cell_type": "code", - "execution_count": null, - "metadata": { - "tags": [ - "remove-cell" - ] - }, - "outputs": [ - { - "data": { - "text/html": [ - "\n", - " \n", - " " - ], - "text/plain": [ - "" - ] - }, - "execution_count": 1, - "metadata": {}, - "output_type": "execute_result" - } - ], - "source": [ - "#| label: vid-approx-collide\n", - "#| echo: false\n", - "from IPython.display import YouTubeVideo\n", - "YouTubeVideo(\"rUhunl890KM\")" - ] - }, { "cell_type": "markdown", "metadata": {}, "source": [ - "```{dropdown} 🎥 See More\n", - "![](#vid-approx-collide)\n", - "```" + ":::{iframe} https://www.youtube.com/embed/rUhunl890KM\n", + ":width: 100%\n", + ":title: Approximate Chance on Collision\n", + ":::" ] }, {