What's the correct algorithmic expression?
I've got the following logic which I'm trying to find a formula for it.
Say we have the constants A, B, C.
I'm trying to find a formula to calculate C while following this logic.
If A = 1 then whatever is B, C = 0
If B = 1 then whatever is A, C = 0
If A = 2, B = 2 or 3, C = 1
If A = 2, B = 4 or 5, C = 2
If A = 2, B = 6 or 7, C = 3
... etc for A = 2
If A = 3, B <= 2, C = 0
If A = 3, B is between 3 to 5, C = 1
If A = 3, B is between 6 to 8, C = 2
If A = 3, B is between 9 to 11, C = 3
... etc for A = 3
If A = 4, B <= 3, C = 0
If A = 4, B is between 4 to 7, C = 1
If A = 4, B is between 8 to 11, C = 2
If A = 4, B is between 12 to 15, C = 3
... etc for A = 4
Etc for all multiples of A.
I thought about doing it like that, in pseudo algorithms :
IF A = 1 OR B = 1
C = 0
ELSE
IF B < A
C = 0
ELSE
C = INT(B/2) // INT function is used to get the integer of the result
But unfortunately this doesn't cover all cases, and I can't seem to figure out how can I cover all cases. Any help is appreciated !
algorithms computer-science
New contributor
add a comment |
I've got the following logic which I'm trying to find a formula for it.
Say we have the constants A, B, C.
I'm trying to find a formula to calculate C while following this logic.
If A = 1 then whatever is B, C = 0
If B = 1 then whatever is A, C = 0
If A = 2, B = 2 or 3, C = 1
If A = 2, B = 4 or 5, C = 2
If A = 2, B = 6 or 7, C = 3
... etc for A = 2
If A = 3, B <= 2, C = 0
If A = 3, B is between 3 to 5, C = 1
If A = 3, B is between 6 to 8, C = 2
If A = 3, B is between 9 to 11, C = 3
... etc for A = 3
If A = 4, B <= 3, C = 0
If A = 4, B is between 4 to 7, C = 1
If A = 4, B is between 8 to 11, C = 2
If A = 4, B is between 12 to 15, C = 3
... etc for A = 4
Etc for all multiples of A.
I thought about doing it like that, in pseudo algorithms :
IF A = 1 OR B = 1
C = 0
ELSE
IF B < A
C = 0
ELSE
C = INT(B/2) // INT function is used to get the integer of the result
But unfortunately this doesn't cover all cases, and I can't seem to figure out how can I cover all cases. Any help is appreciated !
algorithms computer-science
New contributor
add a comment |
I've got the following logic which I'm trying to find a formula for it.
Say we have the constants A, B, C.
I'm trying to find a formula to calculate C while following this logic.
If A = 1 then whatever is B, C = 0
If B = 1 then whatever is A, C = 0
If A = 2, B = 2 or 3, C = 1
If A = 2, B = 4 or 5, C = 2
If A = 2, B = 6 or 7, C = 3
... etc for A = 2
If A = 3, B <= 2, C = 0
If A = 3, B is between 3 to 5, C = 1
If A = 3, B is between 6 to 8, C = 2
If A = 3, B is between 9 to 11, C = 3
... etc for A = 3
If A = 4, B <= 3, C = 0
If A = 4, B is between 4 to 7, C = 1
If A = 4, B is between 8 to 11, C = 2
If A = 4, B is between 12 to 15, C = 3
... etc for A = 4
Etc for all multiples of A.
I thought about doing it like that, in pseudo algorithms :
IF A = 1 OR B = 1
C = 0
ELSE
IF B < A
C = 0
ELSE
C = INT(B/2) // INT function is used to get the integer of the result
But unfortunately this doesn't cover all cases, and I can't seem to figure out how can I cover all cases. Any help is appreciated !
algorithms computer-science
New contributor
I've got the following logic which I'm trying to find a formula for it.
Say we have the constants A, B, C.
I'm trying to find a formula to calculate C while following this logic.
If A = 1 then whatever is B, C = 0
If B = 1 then whatever is A, C = 0
If A = 2, B = 2 or 3, C = 1
If A = 2, B = 4 or 5, C = 2
If A = 2, B = 6 or 7, C = 3
... etc for A = 2
If A = 3, B <= 2, C = 0
If A = 3, B is between 3 to 5, C = 1
If A = 3, B is between 6 to 8, C = 2
If A = 3, B is between 9 to 11, C = 3
... etc for A = 3
If A = 4, B <= 3, C = 0
If A = 4, B is between 4 to 7, C = 1
If A = 4, B is between 8 to 11, C = 2
If A = 4, B is between 12 to 15, C = 3
... etc for A = 4
Etc for all multiples of A.
I thought about doing it like that, in pseudo algorithms :
IF A = 1 OR B = 1
C = 0
ELSE
IF B < A
C = 0
ELSE
C = INT(B/2) // INT function is used to get the integer of the result
But unfortunately this doesn't cover all cases, and I can't seem to figure out how can I cover all cases. Any help is appreciated !
algorithms computer-science
algorithms computer-science
New contributor
New contributor
edited yesterday
New contributor
asked yesterday
gumakettell
32
32
New contributor
New contributor
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
If $A=1$ then $C=0$ else $C=lfloor B/Arfloor$.
What's the?
– gumakettell
yesterday
It is the integer part - exactly what you use(d).
– metamorphy
yesterday
Thank you for your help!
– gumakettell
yesterday
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
gumakettell is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060570%2fwhats-the-correct-algorithmic-expression%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $A=1$ then $C=0$ else $C=lfloor B/Arfloor$.
What's the?
– gumakettell
yesterday
It is the integer part - exactly what you use(d).
– metamorphy
yesterday
Thank you for your help!
– gumakettell
yesterday
add a comment |
If $A=1$ then $C=0$ else $C=lfloor B/Arfloor$.
What's the?
– gumakettell
yesterday
It is the integer part - exactly what you use(d).
– metamorphy
yesterday
Thank you for your help!
– gumakettell
yesterday
add a comment |
If $A=1$ then $C=0$ else $C=lfloor B/Arfloor$.
If $A=1$ then $C=0$ else $C=lfloor B/Arfloor$.
answered yesterday
metamorphy
3,5721521
3,5721521
What's the?
– gumakettell
yesterday
It is the integer part - exactly what you use(d).
– metamorphy
yesterday
Thank you for your help!
– gumakettell
yesterday
add a comment |
What's the?
– gumakettell
yesterday
It is the integer part - exactly what you use(d).
– metamorphy
yesterday
Thank you for your help!
– gumakettell
yesterday
What's the
?– gumakettell
yesterday
What's the
?– gumakettell
yesterday
It is the integer part - exactly what you use(d).
– metamorphy
yesterday
It is the integer part - exactly what you use(d).
– metamorphy
yesterday
Thank you for your help!
– gumakettell
yesterday
Thank you for your help!
– gumakettell
yesterday
add a comment |
gumakettell is a new contributor. Be nice, and check out our Code of Conduct.
gumakettell is a new contributor. Be nice, and check out our Code of Conduct.
gumakettell is a new contributor. Be nice, and check out our Code of Conduct.
gumakettell is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060570%2fwhats-the-correct-algorithmic-expression%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown