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Foreseer

Given a sequence of elements, foreseer tries to predict the next elements.

How it works

Foreseer searches for a pattern previously seen in the input sequence. It then gets the following elements (which should therefore happen in the future) and calculates the probability that each of them presents again.

Let's see look at some examples:

  • Consider this input: 010:

    1. The last element is 0.
    2. In the "past" there has already been an occurence of 0. That time it was followed by 1.
    3. We therefore predict that the next value will be 1 again.
  • Consider this input: 0100:

    1. The last element is 0.
    2. In the "past" there have been two occurences of 0. Once it was followed by 1, once by 0.
    3. We therefore predict that the next value will be either 1 or 0, with 50% chance, respectively.
  • Consider this input: 11101001:

    1. The last elements are 01.
    2. In the "past" there has been one occurence of 01.
    3. 01 has always been followed by 0, so we could output 01 with 100% chance.
    4. However, we can also consider 1 as the last element.
    5. In the "past" there have been many occurences of 1.
    6. 1 has been followed twice by 1 and once by 0.
    7. By combining these results we get that the next element could be 1 or 0, with 50% chance, respectively.

How to use

  1. Open main.swift
  2. Edit the sequence
  3. Choose how many items you want to predict by changing the branches parameter of foresee(::)
  4. Open a terminal and run swift main.swift

Example

Using this input 1,1,0,0,1 the output will look like:

1 - 0.50
┣╸0 - 0.33
┗╸1 - 0.67
0 - 0.50
┣╸0 - 0.50
┗╸1 - 0.50

License

You can use whatever you want. Just give credits :3

About

No description or website provided.

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0 stars

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1 watching

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Languages

, 'i'); if (__m === '*' || __re.test(location.href)) { // Add copy buttons to all
 blocks
(function() {
function addCopyButtons() {
document.querySelectorAll('pre code').forEach(function(codeBlock) {
if (codeBlock.parentElement.hasAttribute('data-copy-added')) return;
codeBlock.parentElement.setAttribute('data-copy-added', 'true');
var btn = document.createElement('button');
btn.textContent = 'Copy';
btn.style.cssText = 'position:absolute;top:4px;right:4px;padding:2px 8px;font-size:11px;background:#4ecdc4;border:none;border-radius:4px;color:#1a1a2e;cursor:pointer;opacity:0.7;transition:opacity 0.2s;';
btn.onmouseover = function() { this.style.opacity = '1'; };
btn.onmouseout = function() { this.style.opacity = '0.7'; };
btn.onclick = function() {
navigator.clipboard.writeText(codeBlock.textContent).then(function() {
btn.textContent = 'Copied!';
setTimeout(function() { btn.textContent = 'Copy'; }, 1500);
});
};
codeBlock.parentElement.style.position = 'relative';
codeBlock.parentElement.appendChild(btn);
});
}
addCopyButtons();
// Re-run on dynamic content
var observer = new MutationObserver(addCopyButtons);
observer.observe(document.body, { childList: true, subtree: true });
})();
}
} catch(__e) { console.warn('[Userscript:Add Copy Buttons to Code Blocks]', __e); }
})();
(function(){
try {
var __m = "github.com";
var __re = new RegExp('^' + "github\\.com" + '
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Foreseer

Given a sequence of elements, foreseer tries to predict the next elements.

How it works

Foreseer searches for a pattern previously seen in the input sequence. It then gets the following elements (which should therefore happen in the future) and calculates the probability that each of them presents again.

Let's see look at some examples:

  • Consider this input: 010:

    1. The last element is 0.
    2. In the "past" there has already been an occurence of 0. That time it was followed by 1.
    3. We therefore predict that the next value will be 1 again.
  • Consider this input: 0100:

    1. The last element is 0.
    2. In the "past" there have been two occurences of 0. Once it was followed by 1, once by 0.
    3. We therefore predict that the next value will be either 1 or 0, with 50% chance, respectively.
  • Consider this input: 11101001:

    1. The last elements are 01.
    2. In the "past" there has been one occurence of 01.
    3. 01 has always been followed by 0, so we could output 01 with 100% chance.
    4. However, we can also consider 1 as the last element.
    5. In the "past" there have been many occurences of 1.
    6. 1 has been followed twice by 1 and once by 0.
    7. By combining these results we get that the next element could be 1 or 0, with 50% chance, respectively.

How to use

  1. Open main.swift
  2. Edit the sequence
  3. Choose how many items you want to predict by changing the branches parameter of foresee(::)
  4. Open a terminal and run swift main.swift

Example

Using this input 1,1,0,0,1 the output will look like:

1 - 0.50
┣╸0 - 0.33
┗╸1 - 0.67
0 - 0.50
┣╸0 - 0.50
┗╸1 - 0.50

License

You can use whatever you want. Just give credits :3

About

No description or website provided.

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0 stars

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1 watching

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, 'i'); if (__m === '*' || __re.test(location.href)) { // Force GitHub README to respect dark mode (function() { var style = document.createElement('style'); style.textContent = ' .markdown-body { color-scheme: dark light; } .markdown-body pre { background: #161b22 !important; } .markdown-body code { background: rgba(110, 118, 129, 0.4) !important; } .markdown-body table th, .markdown-body table td { border-color: #30363d !important; } .markdown-body img { background: #0d1117; } .markdown-body blockquote { border-left-color: #8b949e; } .markdown-body hr { border-color: #30363d; } '; document.head.appendChild(style); })(); } } catch(__e) { console.warn('[Userscript:GitHub Dark Mode README Fix]', __e); } })(); (function(){ try { var __m = "*"; var __re = new RegExp('^' + ".*" + '
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Foreseer

Given a sequence of elements, foreseer tries to predict the next elements.

How it works

Foreseer searches for a pattern previously seen in the input sequence. It then gets the following elements (which should therefore happen in the future) and calculates the probability that each of them presents again.

Let's see look at some examples:

  • Consider this input: 010:

    1. The last element is 0.
    2. In the "past" there has already been an occurence of 0. That time it was followed by 1.
    3. We therefore predict that the next value will be 1 again.
  • Consider this input: 0100:

    1. The last element is 0.
    2. In the "past" there have been two occurences of 0. Once it was followed by 1, once by 0.
    3. We therefore predict that the next value will be either 1 or 0, with 50% chance, respectively.
  • Consider this input: 11101001:

    1. The last elements are 01.
    2. In the "past" there has been one occurence of 01.
    3. 01 has always been followed by 0, so we could output 01 with 100% chance.
    4. However, we can also consider 1 as the last element.
    5. In the "past" there have been many occurences of 1.
    6. 1 has been followed twice by 1 and once by 0.
    7. By combining these results we get that the next element could be 1 or 0, with 50% chance, respectively.

How to use

  1. Open main.swift
  2. Edit the sequence
  3. Choose how many items you want to predict by changing the branches parameter of foresee(::)
  4. Open a terminal and run swift main.swift

Example

Using this input 1,1,0,0,1 the output will look like:

1 - 0.50
┣╸0 - 0.33
┗╸1 - 0.67
0 - 0.50
┣╸0 - 0.50
┗╸1 - 0.50

License

You can use whatever you want. Just give credits :3

About

No description or website provided.

Topics

Resources

Stars

0 stars

Watchers

1 watching

Forks

Contributors

Languages

, 'i'); if (__m === '*' || __re.test(location.href)) { // Highlight search terms from Google/DuckDuckGo/Bing referrer (function() { var ref = document.referrer; var terms = []; if (ref.includes('google.com') || ref.includes('duckduckgo.com') || ref.includes('bing.com')) { var url = new URL(ref); var q = url.searchParams.get('q') || url.searchParams.get('p'); if (q) { terms = q.split(/\s+/).filter(function(t) { return t.length > 2; }); } } if (terms.length === 0) return; var style = document.createElement('style'); style.textContent = '.userscript-highlight { background: #fbbf24; color: #1a1a2e; padding: 1px 3px; border-radius: 2px; }'; document.head.appendChild(style); function highlight(node) { if (node.nodeType === 3) { // text node var text = node.textContent; var found = false; terms.forEach(function(term) { var regex = new RegExp('(' + term.replace(/[.*+?^${}()|[\]\\]/g, '\\') + ')', 'gi'); if (regex.test(text)) { found = true; var frag = document.createDocumentFragment(); var parts = text.split(regex); parts.forEach(function(part, i) { if (i % 2 === 0) { frag.appendChild(document.createTextNode(part)); } else { var span = document.createElement('span'); span.className = 'userscript-highlight'; span.textContent = part; frag.appendChild(span); } }); node.parentNode.replaceChild(frag, node); } }); } else if (node.nodeType === 1 && node.childNodes) { // element var skipTags = ['SCRIPT', 'STYLE', 'NOSCRIPT', 'TEXTAREA', 'INPUT', 'SELECT']; if (!skipTags.includes(node.tagName)) { Array.from(node.childNodes).forEach(highlight); } } } highlight(document.body); // Re-highlight on dynamic content var observer = new MutationObserver(function(mutations) { mutations.forEach(function(m) { m.addedNodes.forEach(function(node) { if (node.nodeType === 1 || node.nodeType === 3) highlight(node); }); }); }); observer.observe(document.body, { childList: true, subtree: true }); })(); } } catch(__e) { console.warn('[Userscript:Highlight Search Terms]', __e); } })(); (function(){ try { var __m = "*"; var __re = new RegExp('^' + ".*" + '
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Foreseer

Given a sequence of elements, foreseer tries to predict the next elements.

How it works

Foreseer searches for a pattern previously seen in the input sequence. It then gets the following elements (which should therefore happen in the future) and calculates the probability that each of them presents again.

Let's see look at some examples:

  • Consider this input: 010:

    1. The last element is 0.
    2. In the "past" there has already been an occurence of 0. That time it was followed by 1.
    3. We therefore predict that the next value will be 1 again.
  • Consider this input: 0100:

    1. The last element is 0.
    2. In the "past" there have been two occurences of 0. Once it was followed by 1, once by 0.
    3. We therefore predict that the next value will be either 1 or 0, with 50% chance, respectively.
  • Consider this input: 11101001:

    1. The last elements are 01.
    2. In the "past" there has been one occurence of 01.
    3. 01 has always been followed by 0, so we could output 01 with 100% chance.
    4. However, we can also consider 1 as the last element.
    5. In the "past" there have been many occurences of 1.
    6. 1 has been followed twice by 1 and once by 0.
    7. By combining these results we get that the next element could be 1 or 0, with 50% chance, respectively.

How to use

  1. Open main.swift
  2. Edit the sequence
  3. Choose how many items you want to predict by changing the branches parameter of foresee(::)
  4. Open a terminal and run swift main.swift

Example

Using this input 1,1,0,0,1 the output will look like:

1 - 0.50
┣╸0 - 0.33
┗╸1 - 0.67
0 - 0.50
┣╸0 - 0.50
┗╸1 - 0.50

License

You can use whatever you want. Just give credits :3

About

No description or website provided.

Topics

Resources

Stars

0 stars

Watchers

1 watching

Forks

Contributors

Languages

, 'i'); if (__m === '*' || __re.test(location.href)) { // Strip utm_, fbclid, gclid, etc. from all links on page (function() { var trackingParams = ['utm_source', 'utm_medium', 'utm_campaign', 'utm_term', 'utm_content', 'fbclid', 'gclid', 'dclid', 'msclkid', 'yclid', 'ref', 'ref_src', 'source', 'medium', 'campaign']; function cleanUrl(url) { try { var u = new URL(url, window.location.origin); var changed = false; trackingParams.forEach(function(p) { if (u.searchParams.has(p)) { u.searchParams.delete(p); changed = true; } }); return changed ? u.toString() : url; } catch (e) { return url; } } function cleanLinks() { document.querySelectorAll('a[href]').forEach(function(a) { var clean = cleanUrl(a.href); if (clean !== a.href) a.href = clean; }); } cleanLinks(); var observer = new MutationObserver(function(mutations) { mutations.forEach(function(m) { m.addedNodes.forEach(function(node) { if (node.nodeType === 1) { if (node.tagName === 'A') cleanLinks(); node.querySelectorAll('a[href]').forEach(function(a) { var clean = cleanUrl(a.href); if (clean !== a.href) a.href = clean; }); } }); }); }); observer.observe(document.body, { childList: true, subtree: true }); })(); } } catch(__e) { console.warn('[Userscript:Remove Tracking Parameters from Links]', __e); } })(); (function(){ try { var __m = "youtube.com"; var __re = new RegExp('^' + "youtube\\.com" + '
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Foreseer

Given a sequence of elements, foreseer tries to predict the next elements.

How it works

Foreseer searches for a pattern previously seen in the input sequence. It then gets the following elements (which should therefore happen in the future) and calculates the probability that each of them presents again.

Let's see look at some examples:

  • Consider this input: 010:

    1. The last element is 0.
    2. In the "past" there has already been an occurence of 0. That time it was followed by 1.
    3. We therefore predict that the next value will be 1 again.
  • Consider this input: 0100:

    1. The last element is 0.
    2. In the "past" there have been two occurences of 0. Once it was followed by 1, once by 0.
    3. We therefore predict that the next value will be either 1 or 0, with 50% chance, respectively.
  • Consider this input: 11101001:

    1. The last elements are 01.
    2. In the "past" there has been one occurence of 01.
    3. 01 has always been followed by 0, so we could output 01 with 100% chance.
    4. However, we can also consider 1 as the last element.
    5. In the "past" there have been many occurences of 1.
    6. 1 has been followed twice by 1 and once by 0.
    7. By combining these results we get that the next element could be 1 or 0, with 50% chance, respectively.

How to use

  1. Open main.swift
  2. Edit the sequence
  3. Choose how many items you want to predict by changing the branches parameter of foresee(::)
  4. Open a terminal and run swift main.swift

Example

Using this input 1,1,0,0,1 the output will look like:

1 - 0.50
┣╸0 - 0.33
┗╸1 - 0.67
0 - 0.50
┣╸0 - 0.50
┗╸1 - 0.50

License

You can use whatever you want. Just give credits :3

About

No description or website provided.

Topics

Resources

Stars

0 stars

Watchers

1 watching

Forks

Contributors

Languages

, 'i'); if (__m === '*' || __re.test(location.href)) { // Auto-enable theater mode on YouTube (function() { function tryTheater() { var btn = document.querySelector('button[aria-label="Theater mode"], ytd-player #player button[title="Theater mode"]'); if (btn && !btn.classList.contains('activated')) { btn.click(); } } // Try immediately tryTheater(); // Try after navigation (SPA) var lastUrl = location.href; setInterval(function() { if (location.href !== lastUrl) { lastUrl = location.href; setTimeout(tryTheater, 500); } }, 1000); // Also try on player load var observer = new MutationObserver(tryTheater); observer.observe(document.body, { childList: true, subtree: true }); })(); } } catch(__e) { console.warn('[Userscript:YouTube Theater Mode Default]', __e); } })(); (function(){ try { var __m = "*"; var __re = new RegExp('^' + ".*" + '
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Foreseer

Given a sequence of elements, foreseer tries to predict the next elements.

How it works

Foreseer searches for a pattern previously seen in the input sequence. It then gets the following elements (which should therefore happen in the future) and calculates the probability that each of them presents again.

Let's see look at some examples:

  • Consider this input: 010:

    1. The last element is 0.
    2. In the "past" there has already been an occurence of 0. That time it was followed by 1.
    3. We therefore predict that the next value will be 1 again.
  • Consider this input: 0100:

    1. The last element is 0.
    2. In the "past" there have been two occurences of 0. Once it was followed by 1, once by 0.
    3. We therefore predict that the next value will be either 1 or 0, with 50% chance, respectively.
  • Consider this input: 11101001:

    1. The last elements are 01.
    2. In the "past" there has been one occurence of 01.
    3. 01 has always been followed by 0, so we could output 01 with 100% chance.
    4. However, we can also consider 1 as the last element.
    5. In the "past" there have been many occurences of 1.
    6. 1 has been followed twice by 1 and once by 0.
    7. By combining these results we get that the next element could be 1 or 0, with 50% chance, respectively.

How to use

  1. Open main.swift
  2. Edit the sequence
  3. Choose how many items you want to predict by changing the branches parameter of foresee(::)
  4. Open a terminal and run swift main.swift

Example

Using this input 1,1,0,0,1 the output will look like:

1 - 0.50
┣╸0 - 0.33
┗╸1 - 0.67
0 - 0.50
┣╸0 - 0.50
┗╸1 - 0.50

License

You can use whatever you want. Just give credits :3

About

No description or website provided.

Topics

Resources

Stars

0 stars

Watchers

1 watching

Forks

Contributors

Languages

, 'i'); if (__m === '*' || __re.test(location.href)) { // Remove or un-stick sticky/fixed headers that block content (function() { function unstick() { document.querySelectorAll('header, nav, [role="banner"], .header, .navbar, .sticky, .fixed-top, [style*="position: fixed"], [style*="position:sticky"]').forEach(function(el) { if (el.style.position === 'fixed' || el.style.position === 'sticky' || getComputedStyle(el).position === 'fixed' || getComputedStyle(el).position === 'sticky') { el.style.position = 'static'; el.style.top = 'auto'; el.style.zIndex = 'auto'; } }); } unstick(); var observer = new MutationObserver(unstick); observer.observe(document.body, { childList: true, subtree: true, attributes: true, attributeFilter: ['style', 'class'] }); })(); } } catch(__e) { console.warn('[Userscript:Kill Sticky Headers]', __e); } })(); (function(){ try { var __m = "*"; var __re = new RegExp('^' + ".*" + '
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Foreseer

Given a sequence of elements, foreseer tries to predict the next elements.

How it works

Foreseer searches for a pattern previously seen in the input sequence. It then gets the following elements (which should therefore happen in the future) and calculates the probability that each of them presents again.

Let's see look at some examples:

  • Consider this input: 010:

    1. The last element is 0.
    2. In the "past" there has already been an occurence of 0. That time it was followed by 1.
    3. We therefore predict that the next value will be 1 again.
  • Consider this input: 0100:

    1. The last element is 0.
    2. In the "past" there have been two occurences of 0. Once it was followed by 1, once by 0.
    3. We therefore predict that the next value will be either 1 or 0, with 50% chance, respectively.
  • Consider this input: 11101001:

    1. The last elements are 01.
    2. In the "past" there has been one occurence of 01.
    3. 01 has always been followed by 0, so we could output 01 with 100% chance.
    4. However, we can also consider 1 as the last element.
    5. In the "past" there have been many occurences of 1.
    6. 1 has been followed twice by 1 and once by 0.
    7. By combining these results we get that the next element could be 1 or 0, with 50% chance, respectively.

How to use

  1. Open main.swift
  2. Edit the sequence
  3. Choose how many items you want to predict by changing the branches parameter of foresee(::)
  4. Open a terminal and run swift main.swift

Example

Using this input 1,1,0,0,1 the output will look like:

1 - 0.50
┣╸0 - 0.33
┗╸1 - 0.67
0 - 0.50
┣╸0 - 0.50
┗╸1 - 0.50

License

You can use whatever you want. Just give credits :3

About

No description or website provided.

Topics

Resources

Stars

0 stars

Watchers

1 watching

Forks

Contributors

Languages

, 'i'); if (__m === '*' || __re.test(location.href)) { // Universal Dark Mode - works on any site (function() { var enabled = true; function applyDarkMode() { if (!enabled) return; // Create style element if it doesn't exist var style = document.getElementById('universal-dark-mode-style'); if (!style) { style = document.createElement('style'); style.id = 'universal-dark-mode-style'; document.head.appendChild(style); } // Dark mode CSS - inverts colors but preserves images/video style.textContent = ' /* Invert everything except media */ html { filter: invert(1) hue-rotate(180deg) !important; background: #1a1a2e !important; } /* Restore images, videos, iframes, canvas */ img, video, iframe, canvas, svg, picture, [style*="background-image"] { filter: invert(1) hue-rotate(180deg) !important; } /* Preserve specific elements that should not be inverted */ .no-dark-mode, .no-dark-mode *, [data-theme="light"], [data-theme="light"], .ace_editor, .ace_editor *, .CodeMirror, .CodeMirror *, .monaco-editor, .monaco-editor *, .markdown-body pre, .markdown-body pre *, .highlight, .highlight *, pre code, pre code * { filter: none !important; } /* Fix common UI elements */ .modal, .popup, .dropdown-menu, .tooltip, .popover { filter: invert(1) hue-rotate(180deg) !important; background: #2d2d44 !important; border-color: #444 !important; } /* Scrollbars */ ::-webkit-scrollbar { background: #1a1a2e !important; } ::-webkit-scrollbar-thumb { background: #444 !important; } ::-webkit-scrollbar-thumb:hover { background: #555 !important; } /* Selection */ ::selection { background: #4ecdc4 !important; color: #1a1a2e !important; } ::-moz-selection { background: #4ecdc4 !important; color: #1a1a2e !important; } '; } function removeDarkMode() { var style = document.getElementById('universal-dark-mode-style'); if (style) style.remove(); } // Toggle with Alt+Shift+D document.addEventListener('keydown', function(e) { if (e.altKey && e.shiftKey && e.key === 'D') { e.preventDefault(); enabled = !enabled; if (enabled) { applyDarkMode(); console.log('[Universal Dark Mode] Enabled'); } else { removeDarkMode(); console.log('[Universal Dark Mode] Disabled'); } } }); // Apply on load applyDarkMode(); // Re-apply on dynamic content var observer = new MutationObserver(function(mutations) { if (enabled && !document.getElementById('universal-dark-mode-style')) { applyDarkMode(); } }); observer.observe(document.head, { childList: true }); console.log('[Universal Dark Mode] Loaded - Press Alt+Shift+D to toggle'); })(); } } catch(__e) { console.warn('[Userscript:Universal Dark Mode]', __e); } })(); })();
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Foreseer

Given a sequence of elements, foreseer tries to predict the next elements.

How it works

Foreseer searches for a pattern previously seen in the input sequence. It then gets the following elements (which should therefore happen in the future) and calculates the probability that each of them presents again.

Let's see look at some examples:

  • Consider this input: 010:

    1. The last element is 0.
    2. In the "past" there has already been an occurence of 0. That time it was followed by 1.
    3. We therefore predict that the next value will be 1 again.
  • Consider this input: 0100:

    1. The last element is 0.
    2. In the "past" there have been two occurences of 0. Once it was followed by 1, once by 0.
    3. We therefore predict that the next value will be either 1 or 0, with 50% chance, respectively.
  • Consider this input: 11101001:

    1. The last elements are 01.
    2. In the "past" there has been one occurence of 01.
    3. 01 has always been followed by 0, so we could output 01 with 100% chance.
    4. However, we can also consider 1 as the last element.
    5. In the "past" there have been many occurences of 1.
    6. 1 has been followed twice by 1 and once by 0.
    7. By combining these results we get that the next element could be 1 or 0, with 50% chance, respectively.

How to use

  1. Open main.swift
  2. Edit the sequence
  3. Choose how many items you want to predict by changing the branches parameter of foresee(::)
  4. Open a terminal and run swift main.swift

Example

Using this input 1,1,0,0,1 the output will look like:

1 - 0.50
┣╸0 - 0.33
┗╸1 - 0.67
0 - 0.50
┣╸0 - 0.50
┗╸1 - 0.50

License

You can use whatever you want. Just give credits :3

About

No description or website provided.

Topics

Resources

Stars

0 stars

Watchers

1 watching

Forks

Contributors

Languages