# mentor question 3 is there a formula for this Classic List Threaded 6 messages Open this post in threaded view
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## mentor question 3 is there a formula for this

Hello,

I try now to solve this one :

## ntroduction

The diamond kata takes as its input a letter, and outputs it in a diamond shape. Given a letter, it prints a diamond starting with 'A', with the supplied letter at the widest point.

## Requirements

• The first row contains one 'A'.
• The last row contains one 'A'.
• All rows, except the first and last, have exactly two identical letters.
• All rows have as many trailing spaces as leading spaces. (This might be 0).
• The diamond is horizontally symmetric.
• The diamond is vertically symmetric.
• The diamond has a square shape (width equals height).
• The letters form a diamond shape.
• The top half has the letters in ascending order.
• The bottom half has the letters in descending order.
• The four corners (containing the spaces) are triangles.

## Examples

In the following examples, spaces are indicated by `·` characters.

Diamond for letter 'A':

``````A
``````

Diamond for letter 'C':

``````··A··
·B·B·
C···C
·B·B·
··A··

I noticed that if you take a quarter of it. you see this pattern

001
010
100

where a 0 is a space and a 1 is the character.

Is there a easy way to make some sort of formula so I can make a output of this ?

Roelof

``````
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## Re: mentor question 3 is there a formula for this

Formula? Well, it's pretty straight forward.

Let's start with the size of the diamond (it field, actually). In the example, there are two lines above and two lines below the "C" line.
C-A = 2. This number looks useful.
The dimension is 2+1+2 square.

How many spaces before a letter? C-letter i.e. C-A = 2 C-B = 1 and C-C = 0
The same number of spaces after the second/last occurrence of the letter on the line.
The number of spaces between the letters on a line is the size minus the number of spaces before and after plus the two copies of the current line's letter. For the "B" line, there is one space before and after, 2x"B", leaving 5 - 1 -1 - 2 = 1 space between. For the "A" line, the simple arithmetic would give you -1, which also confirms that just one "A" is appropriate with no space between first and last letter on the line. (For "A", it is first and last, once only.)

After that, you just need to iterate (\$A to: \$C), (\$(C-1) to: \$A by: -1) and do the arithmetic.

On Tue, Apr 28, 2020 at 1:23 AM Roelof Wobben via Pharo-users <[hidden email]> wrote:
Hello,

I try now to solve this one :

## ntroduction

The diamond kata takes as its input a letter, and outputs it in a diamond shape. Given a letter, it prints a diamond starting with 'A', with the supplied letter at the widest point.

## Requirements

• The first row contains one 'A'.
• The last row contains one 'A'.
• All rows, except the first and last, have exactly two identical letters.
• All rows have as many trailing spaces as leading spaces. (This might be 0).
• The diamond is horizontally symmetric.
• The diamond is vertically symmetric.
• The diamond has a square shape (width equals height).
• The letters form a diamond shape.
• The top half has the letters in ascending order.
• The bottom half has the letters in descending order.
• The four corners (containing the spaces) are triangles.

## Examples

In the following examples, spaces are indicated by `·` characters.

Diamond for letter 'A':

``````A
``````

Diamond for letter 'C':

``````··A··
·B·B·
C···C
·B·B·
··A··

I noticed that if you take a quarter of it. you see this pattern

001
010
100

where a 0 is a space and a 1 is the character.

Is there a easy way to make some sort of formula so I can make a output of this ?

Roelof

``````
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## Re: mentor question 3 is there a formula for this

Thanks,

And the 5 can also be calculated by   2 * (char - \$a)  + 1

Roelof

Op 28-4-2020 om 18:31 schreef Richard Sargent:
Formula? Well, it's pretty straight forward.

Let's start with the size of the diamond (it field, actually). In the example, there are two lines above and two lines below the "C" line.
C-A = 2. This number looks useful.
The dimension is 2+1+2 square.

How many spaces before a letter? C-letter i.e. C-A = 2 C-B = 1 and C-C = 0
The same number of spaces after the second/last occurrence of the letter on the line.
The number of spaces between the letters on a line is the size minus the number of spaces before and after plus the two copies of the current line's letter. For the "B" line, there is one space before and after, 2x"B", leaving 5 - 1 -1 - 2 = 1 space between. For the "A" line, the simple arithmetic would give you -1, which also confirms that just one "A" is appropriate with no space between first and last letter on the line. (For "A", it is first and last, once only.)

After that, you just need to iterate (\$A to: \$C), (\$(C-1) to: \$A by: -1) and do the arithmetic.

On Tue, Apr 28, 2020 at 1:23 AM Roelof Wobben via Pharo-users <[hidden email]> wrote:
Hello,

I try now to solve this one :

## ntroduction

The diamond kata takes as its input a letter, and outputs it in a diamond shape. Given a letter, it prints a diamond starting with 'A', with the supplied letter at the widest point.

## Requirements

• The first row contains one 'A'.
• The last row contains one 'A'.
• All rows, except the first and last, have exactly two identical letters.
• All rows have as many trailing spaces as leading spaces. (This might be 0).
• The diamond is horizontally symmetric.
• The diamond is vertically symmetric.
• The diamond has a square shape (width equals height).
• The letters form a diamond shape.
• The top half has the letters in ascending order.
• The bottom half has the letters in descending order.
• The four corners (containing the spaces) are triangles.

## Examples

In the following examples, spaces are indicated by `·` characters.

Diamond for letter 'A':

``````A
``````

Diamond for letter 'C':

``````··A··
·B·B·
C···C
·B·B·
··A··

I noticed that if you take a quarter of it. you see this pattern

001
010
100

where a 0 is a space and a 1 is the character.

Is there a easy way to make some sort of formula so I can make a output of this ?

Roelof

``````

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## Re: mentor question 3 is there a formula for this

In reply to this post by Richard Sargent
Building on Richard's advice, one way to break this down into a series of simpler tasks is
- first generate a square without any dots, i.e. ALL A's on the first line, ALL B's on second line, ALL C's on third line.
- second, work out when to output a dot rather than a letter.

cheers -ben

On Wed, 29 Apr 2020 at 00:32, Richard Sargent <[hidden email]> wrote:
Formula? Well, it's pretty straight forward.

Let's start with the size of the diamond (it field, actually). In the example, there are two lines above and two lines below the "C" line.
C-A = 2. This number looks useful.
The dimension is 2+1+2 square.

How many spaces before a letter? C-letter i.e. C-A = 2 C-B = 1 and C-C = 0
The same number of spaces after the second/last occurrence of the letter on the line.
The number of spaces between the letters on a line is the size minus the number of spaces before and after plus the two copies of the current line's letter. For the "B" line, there is one space before and after, 2x"B", leaving 5 - 1 -1 - 2 = 1 space between. For the "A" line, the simple arithmetic would give you -1, which also confirms that just one "A" is appropriate with no space between first and last letter on the line. (For "A", it is first and last, once only.)

After that, you just need to iterate (\$A to: \$C), (\$(C-1) to: \$A by: -1) and do the arithmetic.

On Tue, Apr 28, 2020 at 1:23 AM Roelof Wobben via Pharo-users <[hidden email]> wrote:
Hello,

I try now to solve this one :

## ntroduction

The diamond kata takes as its input a letter, and outputs it in a diamond shape. Given a letter, it prints a diamond starting with 'A', with the supplied letter at the widest point.

## Requirements

• The first row contains one 'A'.
• The last row contains one 'A'.
• All rows, except the first and last, have exactly two identical letters.
• All rows have as many trailing spaces as leading spaces. (This might be 0).
• The diamond is horizontally symmetric.
• The diamond is vertically symmetric.
• The diamond has a square shape (width equals height).
• The letters form a diamond shape.
• The top half has the letters in ascending order.
• The bottom half has the letters in descending order.
• The four corners (containing the spaces) are triangles.

## Examples

In the following examples, spaces are indicated by `·` characters.

Diamond for letter 'A':

``````A
``````

Diamond for letter 'C':

``````··A··
·B·B·
C···C
·B·B·
··A··

I noticed that if you take a quarter of it. you see this pattern

001
010
100

where a 0 is a space and a 1 is the character.

Is there a easy way to make some sort of formula so I can make a output of this ?

Roelof

``````
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## Re: mentor question 3 is there a formula for this

In reply to this post by Richard Sargent
Am I right here or terrible wrong ?

Roelof

Op 28-4-2020 om 20:53 schreef Roelof Wobben:
Thanks,

And the 5 can also be calculated by   2 * (char - \$a)  + 1

Roelof

Op 28-4-2020 om 18:31 schreef Richard Sargent:
Formula? Well, it's pretty straight forward.

Let's start with the size of the diamond (it field, actually). In the example, there are two lines above and two lines below the "C" line.
C-A = 2. This number looks useful.
The dimension is 2+1+2 square.

How many spaces before a letter? C-letter i.e. C-A = 2 C-B = 1 and C-C = 0
The same number of spaces after the second/last occurrence of the letter on the line.
The number of spaces between the letters on a line is the size minus the number of spaces before and after plus the two copies of the current line's letter. For the "B" line, there is one space before and after, 2x"B", leaving 5 - 1 -1 - 2 = 1 space between. For the "A" line, the simple arithmetic would give you -1, which also confirms that just one "A" is appropriate with no space between first and last letter on the line. (For "A", it is first and last, once only.)

After that, you just need to iterate (\$A to: \$C), (\$(C-1) to: \$A by: -1) and do the arithmetic.

On Tue, Apr 28, 2020 at 1:23 AM Roelof Wobben via Pharo-users <[hidden email]> wrote:
Hello,

I try now to solve this one :

## ntroduction

The diamond kata takes as its input a letter, and outputs it in a diamond shape. Given a letter, it prints a diamond starting with 'A', with the supplied letter at the widest point.

## Requirements

• The first row contains one 'A'.
• The last row contains one 'A'.
• All rows, except the first and last, have exactly two identical letters.
• All rows have as many trailing spaces as leading spaces. (This might be 0).
• The diamond is horizontally symmetric.
• The diamond is vertically symmetric.
• The diamond has a square shape (width equals height).
• The letters form a diamond shape.
• The top half has the letters in ascending order.
• The bottom half has the letters in descending order.
• The four corners (containing the spaces) are triangles.

## Examples

In the following examples, spaces are indicated by `·` characters.

Diamond for letter 'A':

``````A
``````

Diamond for letter 'C':

``````··A··
·B·B·
C···C
·B·B·
··A··

I noticed that if you take a quarter of it. you see this pattern

001
010
100

where a 0 is a space and a 1 is the character.

Is there a easy way to make some sort of formula so I can make a output of this ?

Roelof

``````

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|

## Re: mentor question 3 is there a formula for this

In reply to this post by Richard Sargent
On Wed, Apr 29, 2020 at 11:30 AM Roelof Wobben <[hidden email]> wrote:
Am I right here or terrible wrong ?

If you are referring to calculating the size of the square, you are correct.

Roelof

Op 28-4-2020 om 20:53 schreef Roelof Wobben:
Thanks,

And the 5 can also be calculated by   2 * (char - \$a)  + 1

Roelof

Op 28-4-2020 om 18:31 schreef Richard Sargent:
Formula? Well, it's pretty straight forward.

Let's start with the size of the diamond (it field, actually). In the example, there are two lines above and two lines below the "C" line.
C-A = 2. This number looks useful.
The dimension is 2+1+2 square.
Yes. That is what I wrote.

How many spaces before a letter? C-letter i.e. C-A = 2 C-B = 1 and C-C = 0
The same number of spaces after the second/last occurrence of the letter on the line.
The number of spaces between the letters on a line is the size minus the number of spaces before and after plus the two copies of the current line's letter. For the "B" line, there is one space before and after, 2x"B", leaving 5 - 1 -1 - 2 = 1 space between. For the "A" line, the simple arithmetic would give you -1, which also confirms that just one "A" is appropriate with no space between first and last letter on the line. (For "A", it is first and last, once only.)

After that, you just need to iterate (\$A to: \$C), (\$(C-1) to: \$A by: -1) and do the arithmetic.

On Tue, Apr 28, 2020 at 1:23 AM Roelof Wobben via Pharo-users <[hidden email]> wrote:
Hello,

I try now to solve this one :

## ntroduction

The diamond kata takes as its input a letter, and outputs it in a diamond shape. Given a letter, it prints a diamond starting with 'A', with the supplied letter at the widest point.

## Requirements

• The first row contains one 'A'.
• The last row contains one 'A'.
• All rows, except the first and last, have exactly two identical letters.
• All rows have as many trailing spaces as leading spaces. (This might be 0).
• The diamond is horizontally symmetric.
• The diamond is vertically symmetric.
• The diamond has a square shape (width equals height).
• The letters form a diamond shape.
• The top half has the letters in ascending order.
• The bottom half has the letters in descending order.
• The four corners (containing the spaces) are triangles.

## Examples

In the following examples, spaces are indicated by `·` characters.

Diamond for letter 'A':

``````A
``````

Diamond for letter 'C':

``````··A··
·B·B·
C···C
·B·B·
··A··

I noticed that if you take a quarter of it. you see this pattern

001
010
100

where a 0 is a space and a 1 is the character.

Is there a easy way to make some sort of formula so I can make a output of this ?

Roelof

``````