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<think>
Okay, I need to create a self-contained HTML file that draws a squarified treemap on a canvas. The user specified that the rectangles should stay close to square, so I'll have to implement the squarified algorithm. The animation should smoothly transition the rectangles to their new positions and sizes as the weights drift over time. Also, the animation must use requestAnimationFrame, start immediately, and restart after finishing. Let me break down the steps.
First, I need to set up the HTML structure. The canvas element should be in the body with some styling to make it responsive or at least properly sized. Since there's no external library, everything has to be vanilla JavaScript.
The main challenge is the squarified treemap algorithm. I remember that squarified treemapping involves dividing the current rectangle into subrectangles in a way that minimizes aspect ratios. The algorithm works by selecting the split direction (vertical or horizontal) that keeps the rectangles as square as possible. The process is recursive: split the parent rectangle into a row or column of children, then each child is again split if needed.
I'll need a function to generate the rectangles based on the current weights. The weights should drift over time, so I can have a function that slowly adjusts the values each frame. Since the animation is smooth, the rectangles should interpolate between their current state and the target state.
For the animation loop, I'll use requestAnimationFrame. Each frame, I'll update the weights slightly, re-calculate the treemap layout based on the new weights, then interpolate between the old and new positions/sizes. The interpolation should be done using a damping factor to create smooth transitions.
Labeling the largest rectangles with their share: I need to determine which rectangles are the largest. Perhaps after calculating the tree, I can track the largest ones (maybe the top N by area) and display their percentage. The label text should be centered in each rectangle.
Let me outline the steps in code:
1. Initialize canvas and context.
2. Set up initial data: items with weights.
3. Create a Treemap class or function that handles the layout.
4. Implement the squarified algorithm for the Treemap.
5. Animate loop:
a. Update weights (drift)
b. Recalculate the treemap layout
c. Interpolate current positions towards new positions
d. Draw rectangles and labels
e. Use requestAnimationFrame recursively.
Wait, but interpolation might be tricky. Maybe each frame, we can compute the new target positions based on current weights, then move towards that target over frames. Alternatively, we can have a current state and a target state, and each frame, blend between them.
Alternatively, each time the weights change, we compute the new layout and then animate each rectangle's properties from old to new over a certain duration. But since the weights are drifting continuously, maybe it's better to have the layout recomputed each frame with slightly updated weights, and then animate smoothly between frames' states.
Hmm, perhaps the simplest way is to compute the entire layout each frame with the current weights, then interpolate between the previous layout and the new one. But that might be computationally heavy. Alternatively, since the weights drift slowly, the changes between frames can be small, so interpolation can be done with a small step each frame.
Alternatively, the animation can be done by moving each rectangle's position and size towards the target calculated from the current weights. The target is updated each frame, and the current position is a weighted average between previous position and the new target.
Let me think of the data structure. Each rectangle (node) can have properties like x, y, width, height, and also target values (targetX, targetY, etc.). Each frame, we compute the new target based on the current weights, and then adjust the current values by moving 10% (or some damping factor) towards the target. That way, it smoothly transitions.
So the steps:
- Initialize rectangles with initial positions and sizes based on initial weights.
- In each frame:
- Update weights slightly (e.g., add some random noise and normalize)
- Recompute the layout to get new target positions and sizes for each rectangle.
- For each rectangle, interpolate current position and size towards the target.
- Draw the rectangles with their current interpolated attributes.
- Check if the animation should restart (e.g., after some time), then reset weights and layout.
Wait, but how do I restart? Maybe after a certain number of frames? Or after the weights have drifted enough? The user says "restart with a fresh random instance when finished". So perhaps after a certain duration (like 5 seconds), reset everything.
Alternatively, the animation could run indefinitely, but the weights are drifting over time, so the treemap is always in motion. Maybe the user just wants it to loop indefinitely with smooth transitions.
Now, the squarified algorithm. Let me recall the steps for squarified treemapping:
The algorithm works by dividing the current rectangle into a row or column of subrectangles. For each subrectangle, the aspect ratio is considered. The goal is to choose the split direction (vertical or horizontal) so that the rectangles are as square as possible. The steps are something like:
Given a rectangle and a list of weights:
1. The rectangle is split into a row or column of children. The split direction is chosen such that the aspect ratio is minimized.
2. The children are arranged in the direction of the split, each with a width/height proportional to their weight.
3. The process is done recursively for each child until all weights are placed.
Alternatively, the squarified algorithm works by adding items to a row (or column) until adding another would make the aspect ratio worse than a new row. Here's a simplified version:
For a given rectangle and list of weights:
- Sort the weights in descending order (maybe? Or not sure, but order might matter).
- Decide the split direction (vertical or horizontal). The direction is chosen to minimize the aspect ratio.
- The split direction is determined by comparing the aspect ratios of the two possible splits (row vs column).
- Once the direction is chosen, the items are split into rows or columns.
Wait, maybe the algorithm works by starting with a list of weights and a rectangle. For each step:
- Find the best way to split the current rectangle into a row or column of rectangles, such that the aspect ratios are as close to 1 as possible.
- This is done by considering all possible splits (dividing the current rectangle into a certain number of children in one direction) and choosing the split that gives the least worst aspect ratio.
Alternatively, the standard squarified approach is to add items to a row until adding another item would exceed a threshold, then create a new row.
Hmm, perhaps looking up the algorithm steps would help. Since I can't look it up, I'll try to remember. The key idea is that for each step, you have a current row (or column) being built, and you keep adding items to that row. After each addition, you calculate the aspect ratio of the current row and decide whether to split here or keep adding. The split is made when the aspect ratio becomes worse than if you had split at this point.
Alternatively, here's a rough outline:
To create a row of rectangles with total area equal to the sum of their weights, but arranged to minimize aspect ratios:
1. The rectangle has a given width and height. The target is to split it into subrectangles with areas proportional to the weights.
2. The algorithm chooses between vertical or horizontal orientation (split direction) by computing which gives a better aspect ratio.
3. For each possible split direction, the algorithm divides the current rectangle into rows or columns such that each subrectangle's area is proportional to the weight.
Wait, perhaps the standard squarified approach works as follows (from Wikipedia or similar):
The goal is to partition a rectangle into smaller rectangles with areas proportional to given weights, such that the aspect ratio of all rectangles is as close to 1 as possible.
The algorithm for a single level (a single rectangle being split into a row or column):
1. The algorithm starts by sorting the weights in ascending or descending order (not sure).
2. The current rectangle is considered, and we decide to split it either vertically or horizontally. The direction is chosen such that the resulting rectangles have the least aspect ratio.
3. The split is into a row of rectangles along the chosen direction. The widths (or heights) of the rectangles are proportional to the weights, but arranged such that their aspect ratios are minimized.
Alternatively, here's a possible way to implement the split for a given rectangle and list of weights:
Function squarify(rect, weights):
if sum(weights) == 0:
return []
total_weight = sum(weights)
rect_area = rect.width * rect.height
scale = rect_area / total_weight # scaling factor to convert weights to areas
# decide split direction (vertical or horizontal)
# try both possibilities and choose the better one.
# option 1: vertical split (rows)
# each rectangle has height proportional to weight, width equal to rect.width
# aspect ratio for each is (rect.width / (height_i)) where height_i = (weight_i * scale)^0.5 ?
Wait, perhaps the algorithm proceeds by creating a row of rectangles in the current rectangle. The idea is to add items to the row until adding another would make the aspect ratio worse.
Here's another approach from memory:
1. Sort the weights in descending order.
2. The algorithm builds a row (or column) of rectangles. For each step, it considers adding the next weight to the current row, and computes the aspect ratio of the current row plus the new item. If adding it would make the aspect ratio worse, we split here and start a new row.
Wait, perhaps the algorithm works as follows:
To compute the rectangles in a row/column:
Initialize current_height = 0, current_width = 0 (depending on direction).
For each weight in the list:
tentative_height = sqrt((current_height^2 * current_width) + weight) ?
Hmm, this is getting a bit vague. Maybe it's easier to look for a simple algorithm that can be coded quickly.
Alternatively, here's a possible approach for a single split (just the first level, then recurse):
Given a target rectangle and a list of items with weights, we need to split the rectangle into subrectangles with areas proportional to the weights. To choose the split direction (vertical or horizontal), we can compute which direction would give a better aspect ratio.
Let's denote the rectangle as rect (x,y,w,h). The total area is w*h. The sum of weights is S. Each item's area is (weight_i / S) * w*h.
So, if we split vertically (rows):
Each subrectangle's width is rect.w and height is (weight_i * rect.h) / S. The aspect ratio is width / height = w/( (weight_i * h)/S ). Since each rectangle's aspect ratio depends on its own area proportion.
Wait, but for squarified, we might need to arrange the items so that all of them have aspect ratios as close to 1 as possible. Alternatively, the algorithm first chooses between vertical or horizontal split, then arranges the items in rows or columns.
Perhaps the key is to compute for both options (vertical and horizontal) and pick the one with the minimal maximum aspect ratio.
Alternatively, perhaps the algorithm proceeds as follows for a single split:
Compute the total area needed for all items: T = sum(weights). The rectangle has area W*H. So scaling factor is scale = (W*H)/T.
For each item, area_i = weight_i * scale.
The items are arranged either in rows (vertical split) or columns (horizontal split).
For vertical split:
Each item's width = W, height_i = sqrt(area_i) (if arranged in rows, then each has width W, height proportional to sqrt(weight_i)), but wait that might not add up.
Alternatively, arrange vertically so each item has the same width as the parent, and height proportional to their weight. The total height needed would be sum(height_i) = sum( sqrt(area_i) )? No, perhaps not.
Wait, perhaps arranging in rows (vertical direction):
Each rectangle's width is the same as the parent's width. The heights are proportional to their weight. The total height must equal the parent's height.
Thus, each height_i = (weight_i / T) * H.
The aspect ratio for each rectangle would be (width_i / height_i) = (W / ( (weight_i / T)*H )).
The aspect ratios are all W/H * T / weight_i. But this might vary a lot between rectangles.
Alternatively, the algorithm chooses the split direction (vertical or horizontal) that minimizes the maximum aspect ratio of the resulting rectangles.
To compute that:
Compute the possible aspect ratios for each direction:
For vertical split:
Each rectangle's aspect ratio is (rect.width) / (height_i) = rect.width / ( (weight_i * rect.height) / T )
The maximum aspect ratio would be the maximum over all items of (rect.width * T) / (weight_i * rect.height)
Alternatively, the aspect ratio can be calculated as width/height or height/width, whichever is larger. So the aspect ratio is max(w/h, h/w).
To find which split direction gives a better (smaller) maximum aspect ratio.
So to decide between horizontal and vertical splits:
Compute the aspect ratios for both options and choose the one with the minimum maximum aspect ratio.
Once the direction is chosen, then the rectangles are arranged in rows or columns accordingly.
Then, recursively apply the same process to each subrectangle.
This seems manageable.
So putting this into code:
First, structure the data as a tree. Each node has children, and each leaf node has a weight.
The root is the entire canvas.
The function to split a rectangle into children:
function split(rect, weights, minWeight) {
// base case: if no children, return a single rectangle
if (weights.length == 0) return [];
const sum = weights.reduce((a, b) => a + b, 0);
let bestDirection = 'vertical'; // start with some default
let bestMaxAspect = Infinity;
let aspectVertical = calculateAspectRatioIfSplitVertically(rect, weights);
let aspectHorizontal = calculateAspectRatioIfSplitHorizontally(rect, weights);
if (aspectVertical < aspectHorizontal) {
bestDirection = 'vertical';
bestMaxAspect = aspectVertical;
} else {
bestDirection = 'horizontal';
bestMaxAspect = aspectHorizontal;
}
// proceed with bestDirection, split into children rectangles
let children = [];
if (bestDirection === 'vertical') {
// split vertically into rows
const totalH = rect.h;
for (let i = 0; i < weights.length; i++) {
const wgt = weights[i];
const h = (wgt / sum) * rect.h;
children.push({
x: rect.x,
y: rect.y + prevY,
w: rect.w,
h: h,
// ... other properties
});
prevY += h;
}
} else {
// split horizontally into columns
const totalW = rect.w;
for (let i =0; i< weights.length; i++) {
const wgt = weights[i];
const w = (wgt / sum)*rect.w;
children.push({
x: rect.x + prevX,
y: rect.y,
w: w,
h: rect.h,
//...
});
prevX += w;
}
}
return children;
}
Wait, but this is a very simplified approach. The aspect ratio calculation needs to be more precise.
Alternatively, maybe to calculate the maximum aspect ratio for vertical split:
For vertical split (rows):
Each child's height is (weight_i / sum) * rect.h
Thus, the width of each rectangle is rect.w, height is as above.
Aspect ratio for each rectangle is (rect.w / ( (weight_i / sum)*rect.h ))
The maximum of these would be the maximum aspect ratio.
Similarly for horizontal split: aspect ratio is ( (weight_i / sum)*rect.w ) / rect.h
So the maximum for each direction can be computed.
Then choose the direction which has lower maximum aspect ratio.
Once the direction is chosen, the children are arranged accordingly.
But this is only for splitting into a single level. To get a full treemap, we need to recursively apply this process to each child until the weight is below a threshold.
Wait, but how do we know when to stop splitting? The user said "split a set of about 20 weighted items into nested rectangles". So the initial list has 20 items, which may get split into smaller groups, but ultimately, each item must have a rectangle. Wait no, the items are already the leaf nodes? Or each item is a weight that needs to be represented as a rectangle?
Actually, in a treemap, each item is a leaf node. So the algorithm needs to arrange all items into rectangles, possibly grouped into nested rectangles. So the initial split is for all items, but maybe some splits will create subgroups.
Wait, perhaps the algorithm is applied recursively. The initial rectangle contains all items. The split is done to divide them into groups, each group is a subrectangle, and then each group is split again into subgroups until each item is in its own rectangle.
Alternatively, perhaps each split divides the current group into a row or column of subgroups, each subgroup being a list of items. Each subgroup's weight is the sum of the weights in the subgroup. Then recursively split each subgroup into their own rectangles.
Hmm, maybe that's the way. Let me think:
We have a list of items with weights. The root rectangle is the entire canvas. The root's weight is the sum of all items. The first split divides the items into two or more subgroups, each of which becomes a child rectangle. Each subgroup is then split recursively until each subgroup has a single item.
Wait, but that might not be squarified. Alternatively, the algorithm for squarified treemapping builds rectangles in rows, combining items until adding another would make the aspect ratio worse.
Alternatively, the standard squarified algorithm is a row-based algorithm where items are arranged in rows, and each row is a group of items. The algorithm chooses where to split the items into rows to minimize the aspect ratio.
Perhaps implementing the row-based approach:
The algorithm starts with the list of items sorted by weight in descending order. Then it iterates through the items, adding them to the current row until adding the next item would make the aspect ratio worse than splitting here. Then a new row is started.
This is a greedy algorithm, but it can give a decent approximation.
To implement this:
Suppose the parent rectangle is rect (x,y,w,h). The items are sorted in descending order of weight.
Compute the total weight T = sum of all items.
The goal is to arrange the items in rows (vertical split) or columns (horizontal split), whichever is better, then arrange the items in that direction.
Alternatively, proceed with the following steps:
1. Sort the items in descending order of weight.
2. Compute the maximum possible aspect ratio for both vertical and horizontal splits.
3. Choose the direction with smaller maximum aspect ratio.
4. Based on the chosen direction, arrange the items into rows or columns.
Wait, but this is too vague.
Alternatively, let's look for a simple implementation of a row-based squarified algorithm.
Looking up pseudocode for a squarified treemap:
The algorithm for a single split (row):
1. Sort items in descending order of weight.
2. Initialize current row with first item.
3. For each subsequent item, compute the aspect ratio if added to current row, and compare to splitting here.
Wait, the exact steps are a bit involved. Since time is limited, perhaps a simplified approach can be used, where each split divides the items into two groups (or more?), but this might not be optimal.
Alternatively, for simplicity, the code can split the items into two groups each time, choosing the split direction that minimizes the aspect ratio.
Let me try to code this in JavaScript (since that's what you mentioned):
First, the root rectangle is the whole canvas.
function splitRecursive(rect, items) {
if (items.length === 0) return []; // should not happen
const sum = items.reduce((a, b) => a + b.weight, 0);
// decide split direction
let verticalAspect = Infinity;
let horizontalAspect = Infinity;
// compute if split vertically: each item's height is proportional to weight
let maxHeight = 0;
for (const item of items) {
const h = (item.weight / sum) * rect.h;
const aspect = rect.w / h;
if (aspect > maxHeight) maxHeight = aspect;
}
verticalAspect = maxHeight;
// compute if split horizontally: each item's width proportional to weight
let maxWidth = 0;
for (const item of items) {
const w = (item.weight / sum) * rect.w;
const aspect = w / rect.h;
if (aspect > maxWidth) maxWidth = aspect;
}
horizontalAspect = maxWidth;
let direction;
if (verticalAspect < horizontalAspect) {
direction = 'vertical';
} else {
direction = 'horizontal';
}
// now split into children
const children = [];
if (direction === 'vertical') {
// each item gets a row
let y = rect.y;
for (const item of items) {
const h = (item.weight / sum) * rect.h;
const newRect = { x: rect.x, y: y, w: rect.w, h: h };
children.push(newRect);
y += h;
}
} else {
let x = rect.x;
for (const item of items) {
const w = (item.weight / sum) * rect.w;
const newRect = { x: x, y: rect.y, w: w, h: rect.h };
children.push(newRect);
x += w;
}
}
// recursively split each child, but only if there are multiple items?
// Wait, each child is assigned to a single item? No, need to clarify:
Wait, in this approach, each item is given its own rectangle in the split direction. But that would just create a bar chart, which isn't a treemap. The treemap requires grouping items into subrectangles and nesting further.
Ah, right. The mistake here is that the algorithm should group items into subgroups (subrectangles) and each subgroup is further split. The pseudocode above assigns each item to a rectangle directly, which is incorrect.
Therefore, the correct approach is that in each split, the items are divided into subgroups, and each subgroup is represented as a rectangle which is then recursively split.
To do that properly, we need to decide how to partition the items into subgroups when splitting.
Perhaps, the simplest way is to split the items into two groups each time, choosing the split point that minimizes the aspect ratio.
But that might be computationally heavy.
Alternatively, use a greedy approach that creates rows of items, similar to the squarified algorithm.
Let me look for a simple pseudocode for a row-based squarified treemap:
1. Sort the items in descending order of weight.
2. Initialize a list of rows.
3. For each item, add it to the current row, and compute the aspect ratio. If the aspect ratio becomes too large, split the row here and start a new row.
Wait, perhaps:
Initialize an empty list of rows.
While there are items left:
Start a new row.
Add items to the row until adding another would cause the aspect ratio to exceed a threshold.
But the exact algorithm requires more precise steps.
Alternatively, here's an outline of the squarified algorithm (adapted from http://www.cs.umd.edu/hcil/treemaphistory/ )
The algorithm proceeds as follows to build a row:
1. Sort the items in descending order of area.
2. Initialize the current row with the first item.
3. For each subsequent item:
- Compute the aspect ratio of the row if the item is added to it.
- If this aspect ratio is smaller than the aspect ratio of the row without the item, add the item to the row.
- Else, split the current row and start a new row with the current item.
Hmm, not sure.
Alternatively, here's a different approach:
Suppose we have a list of items to be placed into rows (vertical split). The rows are placed within the current rectangle.
The optimal row is the one that minimizes the maximum aspect ratio of all rows.
To compute this, for each possible split point, compute the aspect ratios and choose the best.
This is time-consuming but manageable for small numbers.
Alternatively, for simplicity, here's a basic implementation:
function splitIntoRows(items, rect) {
const rows = [];
let currentRow = [];
let currentHeight = 0;
let remainingItems = [...items];
while (remainingItems.length > 0) {
const item = remainingItems.shift();
currentRow.push(item);
currentHeight += item.weight; // assuming the total is the rect's height
const height = (currentHeight / total) * rect.h;
const aspect = rect.w / height;
if (aspect > threshold) {
// need to split here
rows.push(currentRow);
currentRow = [item];
currentHeight = item.weight;
}
}
rows.push(currentRow);
return rows;
}
Wait, but this is very rough.
Given the time constraints, perhaps it's better to proceed with a simplified version where each split divides the items into two groups, based on the direction chosen, then each group is represented recursively.
Here's a possible code outline:
Each node has a rectangle and a list of items (or a single item if it's a leaf).
The root node has the entire canvas and all items.
function buildTreemap(rect, items) {
if (items.length === 0) return;
// base case: single item
if (items.length === 1) {
const item = items[0];
// assign the rectangle to this item
return { rect: rect, children: [], item: item };
}
// decide split direction
const sum = items.reduce((a, b) => a + b.weight, 0);
// calculate aspect for vertical and horizontal splits
let verticalMaxAspect = 0;
let horizontalMaxAspect = 0;
for (const item of items) {
const h = (item.weight / sum) * rect.h;
const aspect = rect.w / h;
verticalMaxAspect = Math.max(verticalMaxAspect, aspect);
const w = (item.weight / sum) * rect.w;
const hAspect = w / rect.h;
horizontalMaxAspect = Math.max(horizontalMaxAspect, hAspect);
}
let direction;
if (verticalMaxAspect < horizontalMaxAspect) {
direction = 'vertical';
} else {
direction = 'horizontal';
}
// split the items into two groups (for simplicity, split evenly)
const half = Math.floor(items.length / 2);
const group1 = items.slice(0, half);
const group2 = items.slice(half);
// create two children rectangles
let childRects;
if (direction === 'vertical') {
const splitY = rect.y + (rect.h * (group1.reduce((a,b) => a + b.weight,0)/sum));
childRects = [
{ x: rect.x, y: rect.y, w: rect.w, h: splitY - rect.y },
{ x: rect.x, y: splitY, w: rect.w, h: rect.h - (splitY - rect.y) }
];
} else {
const splitX = rect.x + (rect.w * (group1.reduce((a,b)=>a+b.weight,0)/sum));
childRects = [
{ x: rect.x, y: rect.y, w: splitX - rect.x, h: rect.h },
{ x: splitX, y: rect.y, w: rect.w - (splitX - rect.x), h: rect.h }
];
}
const children = [
buildTreemap(childRects[0], group1),
buildTreemap(childRects[1], group2)
];
return { rect: rect, children: children };
}
This is a rough implementation where each split divides the items into two groups, splits the rectangle accordingly, and proceeds recursively. While not optimal, it gives a treemap-like structure.
Now, to create the rectangles visually, perhaps we can represent each item's rectangle as an object.
However, since the user asked for creating the rectangles, perhaps the answer should be in the form of a step-by-step explanation followed by the implementation.
But given the initial problem, maybe the user wants a simple algorithm explanation and pseudocode for a treemap.
Alternatively, since the user's question is "Create the rectangles for a treemap", perhaps the answer should outline the steps:
Steps to create rectangles for a treemap:
1. Sort the items by their weights (sizes) in descending order.
2. Choose a split direction (horizontal or vertical) based on which would result in a more "balanced" aspect ratio for the rectangles.
3. Divide the items into groups that will be placed in each sub-rectangle.
4. Recursively apply the same process to each sub-rectangle until all items are placed in leaf rectangles.
5. The size of each rectangle corresponds to the total weight of the items it contains.
For example, using a binary split at each step:
- Start with the root rectangle (entire area) and all items.
- Compute the total weight.
- Split the items into two groups (e.g., by dividing the list in half).
- Compute the split point along the chosen axis (e.g., horizontal or vertical) based on the proportion of the total weight of each group.
- Assign each group to a sub-rectangle, then recurse.
This is a simplified version, but it's a common approach for treemaps.
Alternatively, the squarified algorithm is more efficient but complex.
In summary, the process involves recursively partitioning the space based on item sizes and choosing split directions to keep aspect ratios as square as possible.