Colour by another list

Challenge 11 of 16

The question can be about one list while the shape is drawn from another, as long as both are read at the same item number. The width decides the colour and the height decides the picture.

Set the width a bar has to beat.

Your code JavaScript

Set the width a bar has to beat.

Console print() output

0 / 30 frames · 400×400

Target

Yours

Yours over Target — ghosted shapes are in the target but missing from yours

Results

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Penalties: none this run

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Lesson Choosing from a List

Every loop in this course has done the same thing to every item. That is what a loop is for, and it is also its limit: there has been no way to say that one of them is different.

An if inside the body is the way.

for (var i = 0; i < tall.length; i = i + 1) {
if (tall[i] > 180) {
shapeColour(RED);
} else {
shapeColour(GREY);
}
rectangle(20 + i * 64, 320 - tall[i], 48, tall[i]);
}

The question is asked afresh on every turn, about that turn's item. Six bars, six separate answers, and the tall ones come out red.

Two kinds of question are worth telling apart, because they pick out different things.

A question about the item — tall[i] > 180 — is about the value, so it picks out shapes by what they are. Move the values around in the list and the same bars stay red; they have just moved.

A question about the counter — i == 3, or i < sizes.length / 2 — is about the place, so it picks out shapes by where they are in the row. Move the values around and the same positions stay lit whatever is now in them.

An if with no else simply skips the turn. Nothing is drawn, no gap is left, the loop moves on. That is how you draw only some of a list.

A turn can look at more than its own item. The square brackets take a sum, so tall[i - 1] is the one before — which is how you ask whether the line is going up or down. Start that loop at 1, because item 0 has nothing behind it to compare with.

And a variable made before the loop lives through all of it. Set it to 0, add one inside the if, and by the end it is a count of how many passed. Or start it at 0 and let every item that beats it become the new value, and by the end it is the biggest. Neither of those is anything a single item knows — they are facts about the whole list, worked out one turn at a time.