Bits and data.

The Scene at this step

Number

0fixed

Three gates / Half adder

Three gateshidden

Accepted within

± 0.01

A fixed value in the question

Wrong answers it recognises

• 3 ± 0.01

  Misconception: Confuses the largest value with how many values there are, forgetting that counting starts at 0.

  Feedback from the catalogue

  A computer counts from $0$⁠. Two bits have four patterns, 00, 01, 10 and 11, and they are the numbers $0$⁠ to $3$⁠: four numbers, the largest of which is $3$⁠. In general $n$⁠ bits give $2^n$⁠ values, the largest being $2^n - 1$⁠. So a byte, 8 bits, holds 256 values, from $0$⁠ to $255$⁠.

• 2 ± 0.01

  Misconception: Thinks each extra bit adds a fixed number of values instead of doubling them (2 bits give 2, 8 bits give twice what 4 give).

  Feedback from the catalogue

  Each extra bit doubles the number of values: every pattern of the earlier bits appears once with a 0 in the new bit and once with a 1. So $n$⁠ bits give $2 \times 2 \times \cdots \times 2 = 2^n$$2 \times 2 \times \cdots \times 2 =$$2^n$⁠ values: 4 bits give 16, and 8 bits give not 32 but $16 \times 16 = 256$⁠.

Hint

Worked solution

The Scene at this step

Number

0fixed

Three gates / Half adder

Three gateshidden

Choices

• Correct A = 1, B = 0

• Wrong A = 0, B = 1

  Misconception: Gives the rightmost bit the largest place value, reading the number back to front.

  Feedback from the catalogue

  As in decimal numbers, the rightmost digit is the ones, and the place values grow to its left. In base 2 the rightmost bit is worth $1$⁠, the next $2$⁠, then $4$⁠, $8$⁠ and so on. On the board’s switches A is the 2s place and B the 1s place, so the number $2$⁠ is A = 1, B = 0. Read backwards, $1101_2 = 13$⁠ turns into $1011_2 = 11$⁠.

• Wrong A = 1, B = 1, since two needs two switches

  Misconception: Takes the value of a binary number to be the number of 1s in it.

  Feedback from the catalogue

  Each 1 adds its place value, not $1$⁠. $1011_2$⁠ has three 1s, but they stand in the places of $8$⁠, $2$⁠ and $1$⁠, so the value is $8 + 2 + 1 = 11$⁠. And $10_2$⁠, with a single 1, is $2$⁠, because that 1 stands in the 2s place.

Once the prediction is locked in, the Scene changes

Number

2fixed

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Number

0

Three gates / Half adder

Three gateshidden

Read from the Scene at

Logic: input bit A, the 2s place of inputs/nodes/board/outputs/bitA

Read from the Scene at

Logic: input bit B, the 1s place of inputs/nodes/board/outputs/bitB

One setting that reaches it

Number

3

Hint

Worked solution

The Scene at this step

Number

3

Three gates / Half adder

Three gateshidden

Accepted within

± 0.01

A fixed value in the question

Wrong answers it recognises

• 1011 ± 0.01

  Misconception: Reads a binary numeral as if it were decimal: takes 10 in binary to be ten.

  Feedback from the catalogue

  The digits 0 and 1 are the same digits, but the place values differ. In base ten each place is worth 10 times the one to its right; in base 2 only twice: from right to left $1, 2, 4, 8, 16, \ldots$⁠ So $10_2$⁠ is $1 \times 2 + 0 = 2$⁠, not ten, and $1011_2$⁠ is $8 + 2 + 1 = 11$⁠. When in doubt, write the base as a subscript.

• 13 ± 0.01

  Misconception: Gives the rightmost bit the largest place value, reading the number back to front.

  Feedback from the catalogue

  As in decimal numbers, the rightmost digit is the ones, and the place values grow to its left. In base 2 the rightmost bit is worth $1$⁠, the next $2$⁠, then $4$⁠, $8$⁠ and so on. On the board’s switches A is the 2s place and B the 1s place, so the number $2$⁠ is A = 1, B = 0. Read backwards, $1101_2 = 13$⁠ turns into $1011_2 = 11$⁠.

• 3 ± 0.01

  Misconception: Takes the value of a binary number to be the number of 1s in it.

  Feedback from the catalogue

  Each 1 adds its place value, not $1$⁠. $1011_2$⁠ has three 1s, but they stand in the places of $8$⁠, $2$⁠ and $1$⁠, so the value is $8 + 2 + 1 = 11$⁠. And $10_2$⁠, with a single 1, is $2$⁠, because that 1 stands in the 2s place.

Hint

Worked solution

The Scene at this step

Number

3

Three gates / Half adder

Three gateshidden

Accepted within

± 0.01

A fixed value in the question

Wrong answers it recognises

• 16 ± 0.01

  Misconception: Confuses the largest value with how many values there are, forgetting that counting starts at 0.

  Feedback from the catalogue

  A computer counts from $0$⁠. Two bits have four patterns, 00, 01, 10 and 11, and they are the numbers $0$⁠ to $3$⁠: four numbers, the largest of which is $3$⁠. In general $n$⁠ bits give $2^n$⁠ values, the largest being $2^n - 1$⁠. So a byte, 8 bits, holds 256 values, from $0$⁠ to $255$⁠.

• 8 ± 0.01

  Misconception: Thinks each extra bit adds a fixed number of values instead of doubling them (2 bits give 2, 8 bits give twice what 4 give).

  Feedback from the catalogue

  Each extra bit doubles the number of values: every pattern of the earlier bits appears once with a 0 in the new bit and once with a 1. So $n$⁠ bits give $2 \times 2 \times \cdots \times 2 = 2^n$$2 \times 2 \times \cdots \times 2 =$$2^n$⁠ values: 4 bits give 16, and 8 bits give not 32 but $16 \times 16 = 256$⁠.

Hint

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

3 bitfixed

Picture / Red, green, blue

Picturefixed

Once the prediction is locked in, the Scene changes

Picture / Red, green, blue

Red, green, bluefixed

The Scene then shows: 8 · Levels each channel can take, 2^bits/nodes/picture/outputs/levels

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

4 bitfixed

Picture / Red, green, blue

Picturefixed

Choices

• Wrong 32, twice as many

  Misconception: Thinks each extra bit adds a fixed number of values instead of doubling them (2 bits give 2, 8 bits give twice what 4 give).

  Feedback from the catalogue

  Each extra bit doubles the number of values: every pattern of the earlier bits appears once with a 0 in the new bit and once with a 1. So $n$⁠ bits give $2 \times 2 \times \cdots \times 2 = 2^n$$2 \times 2 \times \cdots \times 2 =$$2^n$⁠ values: 4 bits give 16, and 8 bits give not 32 but $16 \times 16 = 256$⁠.

• Wrong 64

  Feedback

  $64 = 2^6$⁠: that is what six bits give. Each of the four extra bits doubles.

• Correct 256

Once the prediction is locked in, the Scene changes

Bits per channel

8 bitfixed

Picture / Red, green, blue

Red, green, bluefixed

The Scene then shows: 256 · Levels each channel can take, 2^bits/nodes/picture/outputs/levels

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

3 bit

Picture / Red, green, blue

Red, green, blue

Read from the Scene at

Levels each channel can take, 2^bits/nodes/picture/outputs/levels

One setting that reaches it

Pixels across

16 px

Bits per channel

6 bit

Hint

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bitfixed

Picture / Red, green, blue

Red, green, blue

Accepted within

± 0.01

Computed from the Scene at this step’s setting

The largest value a channel stores, 2^bits − 1 (255 for 8 bits)/nodes/picture/outputs/maxLevel

Wrong answers it recognises

• 256 ± 0.01

  Misconception: Confuses the largest value with how many values there are, forgetting that counting starts at 0.

  Feedback from the catalogue

  A computer counts from $0$⁠. Two bits have four patterns, 00, 01, 10 and 11, and they are the numbers $0$⁠ to $3$⁠: four numbers, the largest of which is $3$⁠. In general $n$⁠ bits give $2^n$⁠ values, the largest being $2^n - 1$⁠. So a byte, 8 bits, holds 256 values, from $0$⁠ to $255$⁠.

Hint

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bit

Picture / Red, green, blue

Red, green, blue

Accepted within

± 0.01

A fixed value in the question

Wrong answers it recognises

• 19 ± 0.01

  Misconception: Gives the rightmost bit the largest place value, reading the number back to front.

  Feedback from the catalogue

  As in decimal numbers, the rightmost digit is the ones, and the place values grow to its left. In base 2 the rightmost bit is worth $1$⁠, the next $2$⁠, then $4$⁠, $8$⁠ and so on. On the board’s switches A is the 2s place and B the 1s place, so the number $2$⁠ is A = 1, B = 0. Read backwards, $1101_2 = 13$⁠ turns into $1011_2 = 11$⁠.

• 3 ± 0.01

  Misconception: Takes the value of a binary number to be the number of 1s in it.

  Feedback from the catalogue

  Each 1 adds its place value, not $1$⁠. $1011_2$⁠ has three 1s, but they stand in the places of $8$⁠, $2$⁠ and $1$⁠, so the value is $8 + 2 + 1 = 11$⁠. And $10_2$⁠, with a single 1, is $2$⁠, because that 1 stands in the 2s place.

• 11000000 ± 0.01

  Misconception: Reads a binary numeral as if it were decimal: takes 10 in binary to be ten.

  Feedback from the catalogue

  The digits 0 and 1 are the same digits, but the place values differ. In base ten each place is worth 10 times the one to its right; in base 2 only twice: from right to left $1, 2, 4, 8, 16, \ldots$⁠ So $10_2$⁠ is $1 \times 2 + 0 = 2$⁠, not ten, and $1011_2$⁠ is $8 + 2 + 1 = 11$⁠. When in doubt, write the base as a subscript.

Hint

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bit

Picture / Red, green, blue

Red, green, blue

Choices

Shown in a random order.

• Correct $0110\,0100_2$⁠

• Wrong $0010\,0110_2$⁠

  Misconception: Gives the rightmost bit the largest place value, reading the number back to front.

  Feedback from the catalogue

  As in decimal numbers, the rightmost digit is the ones, and the place values grow to its left. In base 2 the rightmost bit is worth $1$⁠, the next $2$⁠, then $4$⁠, $8$⁠ and so on. On the board’s switches A is the 2s place and B the 1s place, so the number $2$⁠ is A = 1, B = 0. Read backwards, $1101_2 = 13$⁠ turns into $1011_2 = 11$⁠.

• Wrong $0000\,0100_2$⁠, the digits 1, 0, 0 at the end

  Misconception: Reads a binary numeral as if it were decimal: takes 10 in binary to be ten.

  Feedback from the catalogue

  The digits 0 and 1 are the same digits, but the place values differ. In base ten each place is worth 10 times the one to its right; in base 2 only twice: from right to left $1, 2, 4, 8, 16, \ldots$⁠ So $10_2$⁠ is $1 \times 2 + 0 = 2$⁠, not ten, and $1011_2$⁠ is $8 + 2 + 1 = 11$⁠. When in doubt, write the base as a subscript.

• Wrong $0101\,0000_2$⁠

  Feedback

  $0101\,0000_2 = 64 + 16 = 80$$0101\,0000_2 =$$64 + 16 =$$80$⁠. Start from the largest place that fits: $64$⁠, leaving $36$⁠.

Hint

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bit

Picture / Red, green, blue

Picture

Accepted within

± 0.01

A fixed value in the question

Wrong answers it recognises

• 500 ± 0.01

  Misconception: Thinks each extra bit adds a fixed number of values instead of doubling them (2 bits give 2, 8 bits give twice what 4 give).

  Feedback from the catalogue

  Each extra bit doubles the number of values: every pattern of the earlier bits appears once with a 0 in the new bit and once with a 1. So $n$⁠ bits give $2 \times 2 \times \cdots \times 2 = 2^n$$2 \times 2 \times \cdots \times 2 =$$2^n$⁠ values: 4 bits give 16, and 8 bits give not 32 but $16 \times 16 = 256$⁠.

Hint

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

4 bitfixed

Picture / Red, green, blue

Picture

Accepted within

± 0.01

Computed from the Scene at this step’s setting

Levels each channel can take, 2^bits/nodes/picture/outputs/levels

Wrong answers it recognises

• 1365 ± 0.5

  Misconception: Adds or divides the channels' levels instead of multiplying: takes the colours of a pixel to be 3 × the levels.

  Feedback from the catalogue

  A pixel’s colour is a combination of three levels: every level of red can go with every level of green and every level of blue. So the number of colours is a product, $L \times L \times L = L^3$$L \times L \times L =$$L^3$⁠, not $3L$⁠. At 8 bits per channel: $256^3 = 16{,}777{,}216$$256^3 =$$16{,}777{,}216$⁠ colours, not 768. Going back from colours to levels takes a cube root, not a division by $3$⁠.

Hint

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bit

Picture / Red, green, blue

Red, green, blue

Read from the Scene at

Bits a pixel takes, 3 channels × bits/nodes/picture/outputs/bitsPerPixel

One setting that reaches it

Pixels across

16 px

Bits per channel

4 bit

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bit

Picture / Red, green, blue

Red, green, blue

Accepted within

± 0.01

A fixed value in the question

Wrong answers it recognises

• 1515 ± 0.01

  Misconception: Reads each hexadecimal digit as a decimal number and writes them side by side, or adds them, ignoring that the left digit is worth 16 times as much.

  Feedback from the catalogue

  In base 16 each digit is worth $0$⁠ to $15$⁠ (A to F are $10$⁠ to $15$⁠), and the left digit is worth 16 times the right one. So $\mathrm{FF}_{16} = 15 \times 16 + 15 = 255$$\mathrm{FF}_{16} =$$15 \times 16 + 15 =$$255$⁠: not “15 15” and not $15 + 15$⁠. Likewise $\mathrm{B3}_{16} = 11 \times 16 + 3 = 179$$\mathrm{B3}_{16} =$$11 \times 16 + 3 =$$179$⁠. Two hexadecimal digits are exactly one byte, $0$⁠ to $255$⁠.

• 30 ± 0.01

  Misconception: Reads each hexadecimal digit as a decimal number and writes them side by side, or adds them, ignoring that the left digit is worth 16 times as much.

  Feedback from the catalogue

  In base 16 each digit is worth $0$⁠ to $15$⁠ (A to F are $10$⁠ to $15$⁠), and the left digit is worth 16 times the right one. So $\mathrm{FF}_{16} = 15 \times 16 + 15 = 255$$\mathrm{FF}_{16} =$$15 \times 16 + 15 =$$255$⁠: not “15 15” and not $15 + 15$⁠. Likewise $\mathrm{B3}_{16} = 11 \times 16 + 3 = 179$$\mathrm{B3}_{16} =$$11 \times 16 + 3 =$$179$⁠. Two hexadecimal digits are exactly one byte, $0$⁠ to $255$⁠.

Hint

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bit

Picture / Red, green, blue

Red, green, blue

Choices

Shown in a random order.

• Correct Orange: full red, half green, no blue

• Wrong Azure: full blue, half green, no red

  Misconception: Reads a #RRGGBB colour code in the wrong order, as blue, green, red.

  Feedback from the catalogue

  In a web colour code the order is always red, green, blue: the first two digits after the # are red, the middle two green, and the last two blue. #FF8000 is full red, green 128 and no blue: orange. Read backwards it would be a light blue, $(0, 128, 255)$⁠.

• Wrong Grey, since the values average out in the middle

  Feedback

  Grey is three equal channels, like #808080. Here the channels are very different.

• Wrong Green

  Feedback

  80 is the green, in the middle, but the first FF is full red.

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bit

Picture / Red, green, blue

Red, green, blue

Accepted within

± 0.01

A fixed value in the question

Wrong answers it recognises

• 113 ± 0.01

  Misconception: Reads each hexadecimal digit as a decimal number and writes them side by side, or adds them, ignoring that the left digit is worth 16 times as much.

  Feedback from the catalogue

  In base 16 each digit is worth $0$⁠ to $15$⁠ (A to F are $10$⁠ to $15$⁠), and the left digit is worth 16 times the right one. So $\mathrm{FF}_{16} = 15 \times 16 + 15 = 255$$\mathrm{FF}_{16} =$$15 \times 16 + 15 =$$255$⁠: not “15 15” and not $15 + 15$⁠. Likewise $\mathrm{B3}_{16} = 11 \times 16 + 3 = 179$$\mathrm{B3}_{16} =$$11 \times 16 + 3 =$$179$⁠. Two hexadecimal digits are exactly one byte, $0$⁠ to $255$⁠.

• 14 ± 0.01

  Misconception: Reads each hexadecimal digit as a decimal number and writes them side by side, or adds them, ignoring that the left digit is worth 16 times as much.

  Feedback from the catalogue

  In base 16 each digit is worth $0$⁠ to $15$⁠ (A to F are $10$⁠ to $15$⁠), and the left digit is worth 16 times the right one. So $\mathrm{FF}_{16} = 15 \times 16 + 15 = 255$$\mathrm{FF}_{16} =$$15 \times 16 + 15 =$$255$⁠: not “15 15” and not $15 + 15$⁠. Likewise $\mathrm{B3}_{16} = 11 \times 16 + 3 = 179$$\mathrm{B3}_{16} =$$11 \times 16 + 3 =$$179$⁠. Two hexadecimal digits are exactly one byte, $0$⁠ to $255$⁠.

Hint

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bit

Picture / Red, green, blue

Red, green, blue

Choices

Shown in a random order.

• Correct Yellow

• Wrong Brown, as when you mix red and green paint

  Misconception: Mixes coloured light as if it were paint: expects red and green light to make brown.

  Feedback from the catalogue

  Paint absorbs light, so mixing paints darkens. A screen gives out light: when red and green are lit together the eye receives both and sees yellow, brighter than either. All three at full give white, and all three off give black. This is additive mixing, and it is how a pixel of three small lights shows any colour.

• Wrong White

  Feedback

  White needs all three lights at full; here the blue is off.

Worked solution

The Scene at this step

Pixels across

32 pxfixed

Bits per channel

8 bitfixed

Picture / Red, green, blue

Picture

Choices

• Wrong 2 times

  Misconception: Thinks doubling the pixels across doubles the picture's pixels and its size, forgetting that the height doubles too.

  Feedback from the catalogue

  The picture keeps its shape, 4 : 3: when you double the pixels across, the pixels down double as well. The number of pixels is width times height, so it grows $2 \times 2 = 4$⁠ times, and so does the file. It is the same sum as for an area: a square with twice the side is four times as big.

• Correct 4 times

• Wrong 8 times

  Feedback

  The width and the height double, but the depth, the bits per channel, stays the same: $2 \times 2 = 4$⁠.

Once the prediction is locked in, the Scene changes

Pixels across

64 pxfixed

The Scene then shows: 9216 · Uncompressed size in bytes: pixels × 3 channels × bits ÷ 8/nodes/picture/outputs/bytes

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Worked solution

The Scene at this step

Pixels across

64 pxfixed

Bits per channel

8 bitfixed

Picture / Red, green, blue

Picture

Accepted within

± 0.01

Computed from the Scene at this step’s setting

Uncompressed size in bytes: pixels × 3 channels × bits ÷ 8/nodes/picture/outputs/bytes

Wrong answers it recognises

• 3072 ± 0.01

  Misconception: Stores one number per pixel, forgetting its three channels.

  Feedback from the catalogue

  A colour pixel is three numbers: red, green and blue. In true colour each of them is a byte, so a pixel takes three bytes, 24 bits. The uncompressed size of a picture is the number of pixels times $3$⁠ times the bits per channel, divided by $8$⁠.

• 73730 ± 0.01

  Misconception: Mixes up bits and bytes: gives a size in bits as bytes, off by a factor of 8.

  Feedback from the catalogue

  A byte is eight bits. File sizes are measured in bytes (B, kB, MB), and network speeds usually in bits per second (Mbit/s). To go from bits to bytes, divide by $8$⁠; from bytes to bits, multiply by $8$⁠. An answer eight times too big is nearly always a count of bits.

Hint

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bit

Picture / Red, green, blue

Picture

Read from the Scene at

Uncompressed size in bytes: pixels × 3 channels × bits ÷ 8/nodes/picture/outputs/bytes

Read from the Scene at

Bits a pixel takes, 3 channels × bits/nodes/picture/outputs/bitsPerPixel

One setting that reaches it

Pixels across

32 px

Bits per channel

2 bit

Hint

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bit

Picture / Red, green, blue

Picture

Accepted within

± 0.01

A fixed value in the question

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bit

Picture / Red, green, blue

Picture

Choices

Shown in a random order.

• Correct 5

• Wrong 40

  Misconception: Mixes up bits and bytes: gives a size in bits as bytes, off by a factor of 8.

  Feedback from the catalogue

  A byte is eight bits. File sizes are measured in bytes (B, kB, MB), and network speeds usually in bits per second (Mbit/s). To go from bits to bytes, divide by $8$⁠; from bytes to bits, multiply by $8$⁠. An answer eight times too big is nearly always a count of bits.

• Wrong 1, one word

  Feedback

  The computer does not know what a word is; it stores each character separately, the second L too.

Worked solution

The Scene at this step

Pixels across

64 px

Bits per channel

8 bit

Picture / Red, green, blue

Picture

Choices

Shown in a random order.

• Correct 8

• Wrong 4, a byte per letter

  Misconception: Assumes every character takes one byte, as in ASCII, including Hebrew letters.

  Feedback from the catalogue

  ASCII has only 128 characters, Latin letters, digits and signs, and each fits in a byte. To write every language, Unicode gives a number to every character in the world, more than a hundred thousand, and that no longer fits in one byte. In the UTF-8 encoding a Latin character is still one byte, a Hebrew letter is two bytes, and an emoji is four.

• Wrong 64

  Misconception: Mixes up bits and bytes: gives a size in bits as bytes, off by a factor of 8.

  Feedback from the catalogue

  A byte is eight bits. File sizes are measured in bytes (B, kB, MB), and network speeds usually in bits per second (Mbit/s). To go from bits to bytes, divide by $8$⁠; from bytes to bits, multiply by $8$⁠. An answer eight times too big is nearly always a count of bits.

Worked solution